The Mathematical Principles of Natural Philosophy

Definitions

Definition 1

The quantity of matter is its measure, arising from its density and its volume multiplied together.

Air of twice the density, for example, in twice the space, is four times the quantity, and in three times the space, six times the quantity. In symbols, with $\rho$ for the density and $V$ for the volume, $m\propto \rho V$. The same is to be understood of snow, and of fine dust or powders, made denser by compression or by melting, and of all bodies made denser to different degrees by any causes whatever. I take no account here of a medium, if there is any such thing, that passes freely between the parts of bodies. It is this quantity that I mean from here on, everywhere, by the word body or mass. It is known from the weight of each body, for it is proportional to the weight, as I have found by very accurate experiments with pendulums, which I will report later.¹⁷

Definition 2

The quantity of motion is its measure, arising from the velocity and the quantity of matter multiplied together.

The motion of the whole is the sum of the motions of all its parts. In a body twice as large, then, moving at the same speed, the motion is twice as great, and at twice the speed it is four times as great: the quantity of motion is proportional to the mass times the velocity, $mv$.¹⁸

Definition 3

The vis insita, or inherent force of matter, is a power of resisting, by which every body, as far as it is able, keeps on in its present state, whether that is a state of rest or of moving uniformly forward in a straight line.

This force is always proportional to the body whose force it is, and it differs in nothing from the inactivity of the mass except in the way we think of it. Because matter is inactive, a body is not moved out of its state of rest or motion without difficulty. For this reason the vis insita may be given the very fitting name vis inertiae, the force of inactivity, or inertia. But a body exerts this force only when another force, impressed on it, tries to change its condition, and its exercise of the force can be regarded both as resistance and as impulse. It is resistance in so far as the body, to keep its present state, withstands the force impressed on it; it is impulse in so far as the body, by not readily giving way to the impressed force of another, tries to change the state of that other. Resistance is usually ascribed to bodies at rest, and impulse to bodies in motion. But motion and rest, as they are commonly understood, are distinguished only relative to each other, and the bodies commonly taken to be at rest are not always truly at rest.¹⁹

Definition 4

An impressed force is an action exerted on a body to change its state, either of rest or of moving uniformly forward in a straight line.

This force consists in the action alone, and it does not remain in the body once the action is over, for a body keeps each new state it acquires by its force of inertia alone. Impressed forces have different origins: they come from a blow, from pressure, or from centripetal force.

Definition 5

A centripetal force is one by which bodies are drawn or pushed, or in any way tend, toward a point as to a center.

Gravity is a force of this kind, by which bodies tend toward the center of the Earth. So is magnetism, by which iron tends toward a lodestone, and so is that force, whatever it may be, by which the planets are continually drawn aside from the straight-line motions they would otherwise follow and made to go round in curved orbits. A stone whirled in a sling tries to fly away from the hand that turns it, and by that effort it stretches the sling, all the harder the faster it is whirled; and the moment it is let go, it flies off. The force that opposes this effort, by which the sling continually pulls the stone back toward the hand and keeps it in its orbit, I call the centripetal force, because it is directed toward the hand as the center of the orbit. The same is true of all bodies that go round in any orbits. They all try to move away from the centers of their orbits, and if an opposing force did not restrain them and hold them in their orbits, they would fly off in straight lines with uniform motion. That opposing force is the one I therefore call centripetal.²⁰

A projectile, if there were no force of gravity, would not curve toward the Earth but would go off from it in a straight line, and with uniform motion if the resistance of the air were taken away. It is by its gravity that it is continually drawn aside from its straight course and made to bend toward the Earth, more or less according to the strength of its gravity and the speed of its motion. The less its gravity in proportion to its mass, or the greater the speed with which it is launched, the less it will bend away from a straight course and the farther it will go. Suppose a lead ball, fired by gunpowder from the top of a mountain at a given speed, in a direction parallel to the horizon, is carried along a curved line for two miles before it falls to the ground. With the resistance of the air taken away, the same ball fired at twice or ten times the speed would fly twice or ten times as far. By increasing the speed we can increase at will the distance to which it is thrown, and lessen the curvature of the line it traces, until at last it might come down 10, 30 or 90 degrees of the way around the Earth, or even go all the way round the Earth before it falls; or, finally, it might never fall to the Earth at all, but go on into the spaces of the heavens and continue its motion indefinitely.²¹

In the same way that a projectile can be made by the force of gravity to go round in an orbit and circle the whole Earth, the Moon too may be continually drawn aside toward the Earth, out of the straight path it would follow by its inherent force, and made to go round in the orbit it now travels. The force that does this may be gravity, if the Moon has gravity, or any other force that pushes it toward the Earth; and without some such force the Moon could not be kept in its orbit. If this force were too small, it would not turn the Moon enough from a straight course; if it were too great, it would turn it too much, and draw the Moon down out of its orbit toward the Earth. The force must be of exactly the right size. It is the business of mathematicians to find the force that will keep a body exactly in a given orbit at a given speed, and vice versa, to determine the curved path into which a body launched from a given place at a given speed will be turned aside from its natural straight path by a given force.²²

The quantity of any centripetal force may be considered as of three kinds: absolute, accelerative and motive.

Definition 6

The absolute quantity of a centripetal force is its measure, proportional to the strength of the cause that spreads it from the center through the spaces around.

The magnetic force, for example, is greater in one lodestone and less in another, according to their size and strength.

Definition 7

The accelerative quantity of a centripetal force is its measure, proportional to the velocity it produces in a given time.

The force of the same lodestone, for example, is greater at a smaller distance and less at a greater one. The force of gravity, too, is greater in valleys and less on the tops of very high mountains, and less still (as will be shown later) at greater distances from the body of the Earth. But at equal distances it is the same everywhere, because (once the resistance of the air is taken away or allowed for) it accelerates all falling bodies equally, whether heavy or light, large or small.

Definition 8

The motive quantity of a centripetal force is its measure, proportional to the momentum it produces in a given time.

This kind of quantity is the centripetal tendency, or inclination, of the whole body toward the center: its weight, as I might call it. It is always known by the force, equal and opposite to it, that is just enough to stop the body from falling. Weight in this sense is greater in a larger body and less in a smaller one, and in the same body it is greater near the Earth and less at greater distances from it.

For brevity, these quantities of forces may be called simply motive, accelerative and absolute forces; and, to tell them apart, they may be referred to the bodies that tend toward the center, to the places of those bodies, and to the center of force toward which they tend. That is, I refer the motive force to the body, as the striving and tendency of the whole toward a center, made up of the tendencies of all its parts together. I refer the accelerative force to the place of the body, as a kind of power spread out from the center to all the places around it, to move the bodies that are in them. And I refer the absolute force to the center, as having some cause without which the motive forces would not be spread through the spaces around it, whether that cause is some central body (such as the lodestone at the center of the magnetic force, or the Earth at the center of the force of gravity) or something else that has not yet come to light. For here I mean only to give a mathematical idea of these forces, without considering their physical causes and seats.

The accelerative force therefore stands in the same relation to the motive force as speed does to momentum. For momentum arises from the speed multiplied by the mass, and the motive force arises from the accelerative force multiplied by the same mass, since the sum of the actions of the accelerative force on the separate particles of the body is the motive force of the whole. Hence near the surface of the Earth, where the accelerative gravity, the force that produces gravity in all bodies, is the same, the motive gravity, or weight, is proportional to the body. But if we were to go up into higher regions, where the accelerative gravity is less, the weight would be less as well, and it would always be proportional to the body multiplied by the accelerative gravity: if $g$ stands for the accelerative gravity, the weight is proportional to $mg$. So in regions where the accelerative gravity is reduced to one half, a body a half or a third as large will weigh a quarter or a sixth as much.²³

In the same sense I call attractions and impulses accelerative and motive. I use the words attraction, impulse, or tendency of any kind toward a center interchangeably, one for another, treating these forces not physically but mathematically. The reader should not imagine, then, that by these words I anywhere mean to define the kind or manner of any action, or its cause or physical reason, or that I attribute forces, in a true and physical sense, to centers (which are only mathematical points), whenever I happen to speak of centers as attracting, or as having attractive powers.

Scholium

So far I have given definitions of the less familiar words, and explained the sense in which I want them understood in what follows. I do not define time, space, place and motion, since they are well known to everyone. I must point out only that ordinary people conceive of these quantities solely through their relation to things the senses can perceive. Certain prejudices arise from this, and to remove them it will be useful to divide each of these quantities into absolute and relative, true and apparent, mathematical and common.²⁴

1. Absolute, true, and mathematical time, of itself and by its own nature, flows evenly, without regard to anything external, and by another name is called duration. Relative, apparent, and common time is some perceptible and external measure of duration by means of motion, whether accurate or uneven, which is commonly used in place of true time: an hour, a day, a month, a year.

2. Absolute space, by its own nature, without regard to anything external, always remains the same and immovable. Relative space is some movable dimension or measure of the absolute spaces, which our senses fix by its position relative to bodies, and which is commonly taken for immovable space. Such is the extent of a space below ground, in the air, or in the heavens, fixed by its position relative to the Earth. Absolute and relative space are the same in shape and size, but they do not always remain one and the same space. If the Earth moves, for instance, a space of our air that stays always the same relative to the Earth will at one time be one part of the absolute space through which the air passes, and at another time another part of it; understood absolutely, it is continually changing.

3. Place is a part of space that a body takes up, and it is absolute or relative as the space it occupies is absolute or relative. I say a part of space: not the body's position, nor its outer surface. For the places of equal solids are always equal, but their surfaces, because their shapes differ, are often unequal. Positions, properly speaking, have no quantity, and they are not so much places themselves as properties of places. The motion of the whole is the same as the sum of the motions of the parts; that is, the movement of the whole out of its place is the same as the sum of the movements of the parts out of their places. The place of the whole is therefore the same as the sum of the places of the parts, and for that reason it is internal, and inside the whole body.

4. Absolute motion is the movement of a body from one absolute place to another, and relative motion the movement from one relative place to another. In a ship under sail, for example, the relative place of a body is the part of the ship that the body occupies, or the part of the hollow of the hull that it fills, which therefore moves along with the ship; and relative rest is the body's staying in the same part of the ship, or of its hollow. But real, absolute rest is the body's staying in the same part of that immovable space in which the ship itself, its hollow and everything it contains are moving. Therefore, if the Earth is really at rest, the body that is at rest relative to the ship will really and absolutely move with the same speed that the ship has on the Earth. But if the Earth also moves, the body's true and absolute motion will arise partly from the true motion of the Earth in immovable space, and partly from the relative motion of the ship on the Earth. And if the body also moves relative to the ship, its true motion will arise partly from the true motion of the Earth in immovable space, and partly from the relative motions both of the ship on the Earth and of the body in the ship; and from these relative motions will arise the relative motion of the body on the Earth. Suppose, for example, that the part of the Earth where the ship is moves truly toward the east with a speed of 10,010 units, while the ship itself, with a fresh wind and full sails, is carried toward the west with a speed of 10 of those units, and a sailor walks in the ship toward the east with 1 unit of speed. Then the sailor moves truly and absolutely in immovable space toward the east with a speed of 10,001 units, and relative to the Earth toward the west with a speed of 9 units.

In astronomy, absolute time is distinguished from relative time by the equation, or correction, of common time. The natural days are in fact unequal, though they are commonly treated as equal and used as a measure of time, and astronomers correct this inequality so as to work out the motions of the heavens more accurately. It may be that there is no such thing as an even motion by which time can be measured exactly. All motions may speed up or slow down, but the true, even flow of absolute time is subject to no change. The duration, or continuance, of the existence of things is the same whether motions are fast or slow or there are none at all. It ought therefore to be distinguished from what are only perceptible measures of it, from which we infer it by means of the astronomical equation. That this equation is needed for fixing the times of an event in the heavens is shown by experiments with the pendulum clock as well as by the eclipses of the satellites of Jupiter.²⁵

Just as the order of the parts of time cannot change, so neither can the order of the parts of space. Suppose those parts were moved out of their places: they would then be moved, if the expression may be allowed, out of themselves. For times and spaces are, as it were, the places of themselves as well as of all other things. All things are placed in time as to order of succession, and in space as to order of position. It is by their essence or nature that they are places, and it is absurd that the primary places of things should be movable. These, then, are the absolute places, and movements out of these places are the only absolute motions.²⁶

But because the parts of space cannot be seen or told apart from one another by our senses, we use perceptible measures of them in their place. We define all places from the positions and distances of things from some body that we regard as immovable, and then we reckon all motions with respect to those places, thinking of bodies as carried from some of them to others. So instead of absolute places and motions we use relative ones, and in ordinary affairs this causes no inconvenience. But in natural philosophy, the study of nature, we ought to set our senses aside and consider things in themselves, as distinct from what are only perceptible measures of them. For it may be that there is no body really at rest to which the places and motions of others could be referred.

We can, however, tell rest and motion, absolute and relative, apart from each other by their properties, causes and effects. It is a property of rest that bodies really at rest are at rest relative to one another. It is possible that in the remote regions of the fixed stars, or perhaps far beyond them, there is some body absolutely at rest; but it is impossible to know, from the positions of bodies relative to one another in our regions, whether any of them keeps the same position relative to that remote body. It follows that absolute rest cannot be determined from the positions of bodies in our regions.

It is a property of motion that parts which keep given positions relative to their wholes share in the motions of those wholes. For all the parts of a revolving body tend to move away from the axis of motion, and the impetus of a body moving forward arises from the combined impetus of all its parts. It follows that if surrounding bodies move, the bodies that are at rest relative to them inside them share in their motion. For this reason the true and absolute motion of a body cannot be determined from its movement away from bodies that only seem to be at rest, since the outside bodies must not only appear to be at rest but really be at rest. Otherwise all the bodies inside them, besides their movement away from the bodies around them, also share in the true motions of those surrounding bodies; and even if that movement away did not take place, they would not really be at rest but would only seem to be. The surrounding bodies stand in the same relation to the bodies they surround as the outer part of a whole does to the inner part, or as the shell does to the kernel; and if the shell moves, the kernel also moves, as part of the whole, without any movement away from the shell.

A closely related property is that when a place moves, whatever is placed in it moves along with it; and so a body that moves away from a place that is itself in motion also shares in the motion of its place. For this reason all motions away from places that are in motion are only parts of whole and absolute motions. Every whole motion is made up of the motion of the body out of its first place, and the motion of this place out of its own place, and so on, until we come to some immovable place, as in the example of the sailor above. Whole and absolute motions, therefore, can be determined only by immovable places, and that is why I referred absolute motions to immovable places above, and relative motions to movable places. Now the only immovable places are those that, from infinity to infinity, all keep the same given positions relative to one another, and so must always remain unmoved; and these make up what I call immovable space.

The causes by which true and relative motions are distinguished from each other are the forces impressed on bodies to produce motion. True motion is neither produced nor changed except by some force impressed on the moving body itself, but relative motion can be produced or changed without any force being impressed on that body. It is enough to impress some force only on the other bodies with which it is compared, so that, as they give way, the relation in which its relative rest or motion consisted is changed. Again, true motion is always changed in some way by any force impressed on the moving body, but relative motion is not necessarily changed by such forces. For if the same forces are also impressed on the other bodies with which the comparison is made, so that their relative position is kept, then the condition in which the relative motion consists will be kept as well. Any relative motion, then, can be changed while the true motion stays unchanged, and a relative motion can be kept while the true motion undergoes some change. For these reasons true motion in no way consists in such relations.

The effects that distinguish absolute from relative motion are the forces of moving away from the axis of circular motion. In circular motion that is purely relative there are no such forces, but in true and absolute circular motion they are greater or smaller in proportion to the quantity of the motion. Suppose a bucket, hung by a long cord, is turned around so many times that the cord is tightly twisted, and is then filled with water and held still together with the water. Then, by the sudden action of another force, it is set whirling the opposite way, and while the cord untwists, the bucket keeps up this motion for some time. The surface of the water will at first be flat, as it was before the bucket began to move. But the bucket, gradually passing its motion on to the water, will make it begin noticeably to revolve, to draw back little by little from the middle and climb the sides of the bucket, forming itself into a hollow, concave shape (as I have found by experiment); and the faster the motion becomes, the higher the water will rise, until at last, revolving in the same time as the bucket, it comes to rest relative to it. This rise of the water shows its tendency to move away from the axis of its motion; and the true and absolute circular motion of the water, which here is exactly opposite to its relative motion, makes itself known and can be measured by this tendency. At first, when the relative motion of the water in the bucket was greatest, it produced no tendency to move away from the axis. The water showed no tendency toward the rim and no rise toward the sides of the bucket, but kept a flat surface, and so its true circular motion had not yet begun. Afterward, when the relative motion of the water had decreased, its rise toward the sides of the bucket proved its tendency to move away from the axis; and this tendency showed that the real circular motion of the water was continually increasing, until it reached its greatest quantity when the water was at rest relative to the bucket. This tendency, therefore, does not depend on any movement of the water relative to the bodies around it, and true circular motion cannot be defined by such movements. Any one revolving body has only one real circular motion, corresponding to only one power of tending to move away from its axis of motion, which is its proper and exact effect. But the relative motions of one and the same body are countless, according to the various relations it has to outside bodies, and like other relations they have no real effect at all, except so far as they may share in that one true motion. This means that in the system of those who suppose that our heavens, revolving below the sphere of the fixed stars, carry the planets along with them, the parts of those heavens, and the planets, which are indeed at rest relative to their heavens, nevertheless really move. For they change their positions relative to one another, which never happens to bodies truly at rest; and being carried along with their heavens, they share in the heavens' motions and, as parts of revolving wholes, tend to move away from the axis of those motions.²⁷

Relative quantities, then, are not the quantities themselves whose names they bear, but the perceptible measures of them, accurate or inaccurate, that are commonly used in place of the quantities measured. And if the meaning of words is to be settled by their use, then by the names time, space, place and motion their measures should properly be understood, and the expression will be unusual, and purely mathematical, if the measured quantities themselves are meant. For this reason those who interpret these words in the Holy Scriptures as meaning the measured quantities do violence to the Scriptures. And those who confuse real quantities with their relations and common measures corrupt the purity of mathematical and philosophical truths no less.

It is indeed a matter of great difficulty to discover the true motions of particular bodies and actually tell them apart from the apparent ones, because the parts of that immovable space in which those motions take place can in no way be observed by our senses. Yet the case is not altogether hopeless. We have some evidence to guide us, partly from the apparent motions, which are the differences of the true motions, and partly from the forces, which are the causes and effects of the true motions. For instance, if two globes, kept at a given distance from each other by a cord connecting them, were revolving about their common center of gravity, we could discover from the tension of the cord the tendency of the globes to move away from the axis of their motion, and from that we could calculate the quantity of their circular motion. Then, if equal forces were impressed at the same moment on alternate faces of the globes, to increase or decrease their circular motion, we could infer the increase or decrease of their motion from the increase or decrease of the cord's tension. From this we would find on which faces the forces ought to be impressed to increase the motion of the globes most; that is, we could discover their rear faces, the ones that follow behind in the circular motion. Once we knew the faces that follow, and so the opposite faces that lead, we would also know the direction of their motion. In this way we could find both the quantity and the direction of this circular motion even in an immense vacuum, where there was nothing external or perceptible with which the globes could be compared. Now suppose that some remote bodies were placed in that space which always kept a given position relative to one another, as the fixed stars do in our regions. We could not then tell, from the relative movement of the globes among those bodies, whether the motion belonged to the globes or to the bodies. But if we observed the cord and found that its tension was the very tension that the motion of the globes required, we could conclude that the motion was in the globes and that the bodies were at rest; and then, finally, from the movement of the globes among the bodies, we would find the direction of their motion. But how we are to infer the true motions from their causes, effects and apparent differences, and, vice versa, how from the motions, whether true or apparent, we may come to know their causes and effects, will be explained more fully in the treatise that follows. It was for this purpose that I wrote it.

Axioms, or the Laws of Motion

Law 1

Every body continues in its state of rest, or of uniform motion in a straight line, unless it is compelled to change that state by forces impressed upon it.

Thrown objects keep moving, except so far as the air slows them and gravity pulls them down. A spinning top, whose parts are constantly drawn away from straight-line motion by the cohesion that holds them together, does not stop spinning except as the air slows it. The planets and comets, which are far larger and meet less resistance in the open spaces of the heavens, keep both their forward motion and their circling motion for much longer.

Law 2

The change of motion is proportional to the motive force impressed, and takes place along the straight line in which that force is impressed.

If a force produces some motion, double the force produces double the motion and triple the force triple the motion, whether it acts all at once or a little at a time. The new motion always points the same way as the force that produced it, so if the body was already moving, the new motion is added to the old when the two point the same way and subtracted when they point opposite ways. When they are at an angle, the two combine into a new motion whose direction depends on both.²⁸

Law 3

To every action there is always an equal and opposite reaction: or, the mutual actions of two bodies upon each other are always equal, and directed to contrary parts.

Whatever pulls or presses on something is pulled or pressed by it just as much. If you press a stone with your finger, the stone presses your finger. If a horse pulls a stone tied to a rope, the horse is pulled back toward the stone just as hard, if I may put it that way, because the stretched rope, in trying to go slack, pulls the horse toward the stone as much as it pulls the stone toward the horse, and holds back the one as much as it moves the other forward. If one body strikes another and changes its motion, its own motion changes by the same amount in the opposite direction, since the pressure between them is mutual and equal. The changes are equal in the motions of the bodies, that is, in their momenta, and not in their speeds, provided nothing else interferes. Because the momenta change equally, the changes in speed, which go in opposite directions, are inversely proportional to the bodies: $\frac{\Delta v_A}{\Delta v_B}=\frac{m_B}{m_A}$. The law holds for attractions too, as the next Scholium will prove.

Corollary 1

A body acted on by two forces together moves along the diagonal of a parallelogram in the same time in which it would move along the sides under the two forces acting separately. (Plate I, Fig. 1)

Suppose that in a given time the force M, acting alone on the body at A, would carry it at a steady speed from A to B, and that the force N, acting alone at the same place, would carry it from A to C. Complete the parallelogram ABDC. The two forces acting together then carry the body along the diagonal from A to D in the same time. The force N acts along the line AC, which is parallel to BD, and so it does not change at all the speed toward BD that the force M produced (by Law 2). The body therefore reaches the line BD in the same time whether N acts or not, and at the end of that time it is somewhere on BD. By the same argument, at the end of the same time it is somewhere on the line CD, and so it is at D, where the two lines meet. It travels from A to D in a straight line (by Law 1).

Corollary 2

From this we can see how any one direct force AD is made up of any two slanting forces AB and BD, and, conversely, how any one direct force AD can be split into any two slanting forces AB and BD. Mechanics confirms this combining and splitting of forces abundantly. (Plate I, Fig. 2)

Take, for example, a wheel with center O, from whose unequal radii OM and ON the weights A and P hang by the cords MA and NP, and suppose we want to know how strongly each weight works to turn the wheel. Through O draw the straight line KOL, meeting the cords at right angles at K and L. With center O and radius OL, the greater of the two distances, draw a circle meeting the cord MA at D; draw OD, and make AC parallel to it and DC perpendicular to it. It makes no difference whether the points K, L and D of the cords are fixed to the wheel or not, so the weights have the same effect whether they hang from K and L or from D and L. Let the line AD represent the whole force of the weight A, and split it into the forces AC and CD. The force AC pulls the radius OD straight out from the center and does nothing to turn the wheel. The force DC pulls on the radius DO at right angles, and so it has the same effect as if it pulled at right angles on the equal radius OL; that is, the same effect as the weight P, if the ratio of P to A is the ratio of the force DC to the force DA. Because the triangles ADC and DOK are similar, that ratio is the ratio of OK to OD, or to OL. The weights A and P therefore balance when they are inversely proportional to the radii OK and OL, which lie on one straight line: $A\times OK=P\times OL$. This is the familiar rule of the balance, the lever and the wheel. If either weight is heavier than the rule allows, it turns the wheel with correspondingly greater force.

Now suppose the weight p, equal to the weight P, hangs partly from the cord Np and partly rests on the slanting plane pG. Draw pH perpendicular to the horizon and NH perpendicular to the plane pG. If pH represents the downward force of the weight p, that force can be split into the forces pN and HN. Imagine a second plane, pQ, perpendicular to the cord pN and meeting the plane pG along a level line. If the weight p were held up by the two planes pQ and pG alone, it would press on each at right angles, on pQ with the force pN and on pG with the force HN. Take the plane pQ away, and let the weight stretch the cord instead: the cord, which now holds the weight up in place of the plane, is stretched by the same force pN that pressed on the plane. The tension in this slanting cord pN is therefore in the same ratio to the tension in the vertical cord PN as pN is to pH. Consequently, if the ratio of the weight p to the weight A equals the inverse ratio of the shortest distances of the cords PN and AM from the center of the wheel, multiplied by the ratio of pH to pN, that is, if $\frac{p}{A}=\frac{OK}{OL}\times \frac{pH}{pN}$, the two weights have equal power to turn the wheel and hold each other up, as anyone can confirm by experiment.

The weight p, pressing on those two slanting planes, can also be regarded as a wedge driven between the two inner faces of a body it is splitting, and the forces of the wedge and the mallet follow. The weight is driven along pH toward both planes, by its own weight or by the blow of a mallet; the force with which it presses on the plane pQ is in the same ratio to that driving force as pN is to pH, and in the same ratio to the force with which it presses on the plane pG as pN is to NH. The force of the screw comes from a similar splitting of forces, since a screw is simply a wedge driven by a lever. The use of this corollary therefore extends very widely, and the extent of its use confirms its truth. The whole of mechanics, as different authors have demonstrated it in different ways, rests on what has just been said: the forces of machines built from wheels, pulleys, levers, cords and weights, lifting straight up or at a slant, and of the other simple machines, are easily derived from it, and so is the force with which the tendons move the bones of animals.²⁹

Corollary 3

The quantity of motion, found by adding the motions that point one way and subtracting those that point the opposite way, is not changed when bodies act on one another.

By Law 3 an action and the reaction against it are equal, and so by Law 2 they make equal changes in the motions, in opposite directions. If the bodies are moving the same way, whatever motion the one in front gains, the one behind loses, and the sum stays as it was. If they meet head-on, each loses the same amount, and the difference between the two opposite motions stays as it was. This is the principle now called the conservation of momentum.

For example, let a spherical body A have three times the mass of a spherical body B, and let A move with 2 units of velocity while B follows it along the same straight line with 10. The momentum of A is then in the same ratio to the momentum of B as 6 is to 10, so take their momenta to be 6 and 10 units, 16 in all. If, when they collide, A gains 3, 4 or 5 units of momentum, B loses as many, and after the rebound A goes on with 9, 10 or 11 units and B with 7, 6 or 5, the total remaining 16. If A gains 9, 10, 11 or 12 units, and so goes on with 15, 16, 17 or 18, then B, losing what A gains, either goes on with 1 unit, having lost 9; or stops, having lost all 10 units of its forward momentum; or goes back with 1 unit, having lost all its momentum and, if I may put it so, one unit more; or goes back with 2 units, because 12 units of forward momentum have been taken from it. In each case the sums of the momenta in the same direction, $15+1$ or $16+0$, and the differences of the momenta in opposite directions, $17-1$ and $18-2$, come to 16 units, as before the collision.

Once the momenta after the collision are known, so are the speeds, since the speed after is in the same ratio to the speed before as the momentum after is to the momentum before. In the last case the momentum of A was 6 units before the collision and 18 after, and its speed was 2 units before; its speed after is therefore 6 units, the ratio of 6 to 18 units of momentum being the same as the ratio of 2 to 6 units of speed.

If the bodies are not spheres, or if they move along different straight lines and strike each other at a slant, their motions after the collision are found by first determining the plane that touches both bodies at the point of contact. The motion of each body is then split into two parts, one perpendicular to that plane and the other parallel to it (by Corollary 2). Since the bodies act on each other only along a line perpendicular to the plane, the parallel motions are unchanged by the collision, while the perpendicular motions receive equal changes in opposite directions, in such a way that the sum of the motions in the same direction and the difference of those in opposite directions stay as they were. Collisions of this kind sometimes also set bodies spinning about their own centers. I do not consider such cases in what follows; it would take too long to demonstrate everything that belongs to the subject.

Corollary 4

The common center of gravity of two or more bodies does not change its state of motion or rest when the bodies act on one another. Hence the common center of gravity of all bodies that act on one another is either at rest or moving uniformly in a straight line, as long as nothing outside acts on them or obstructs them.

If two points move uniformly in straight lines, and the distance between them is divided in a fixed ratio, the dividing point is either at rest or moves uniformly in a straight line. This is proved later, in Lemma 23 and its corollary, for points moving in one plane, and a similar argument proves it for points that are not. If any number of bodies move uniformly in straight lines, the common center of gravity of any two of them divides the line joining their centers in a fixed ratio, and so it is either at rest or moves uniformly in a straight line. The common center of those two and of a third body divides the distance between the center of the pair and the center of the third in a fixed ratio, and so it, too, is either at rest or moves uniformly in a straight line; the same holds for the common center of these three and a fourth body, and so on without limit. In a system of bodies that do not act on one another at all, and on which nothing acts from outside, each body moves uniformly in a straight line, and the common center of gravity of all of them is therefore either at rest or moves uniformly forward in a straight line.³⁰

Now consider two bodies that do act on each other. The distances of their centers from their common center of gravity are inversely proportional to the bodies, and so the momenta with which they move toward or away from that center are equal. Their changes of motion are equal and opposite, so their action on each other neither speeds up nor slows down their common center, nor changes its state of motion or rest at all.

In a system of several bodies, when any two of them act on each other, their common center of gravity does not change its state, and still less does the common center of gravity of the other bodies, which the action does not involve. The common center of gravity of all the bodies divides the line between these two centers into parts inversely proportional to the total masses of the two groups, so that while the two centers keep their state of motion or rest, the common center of all keeps its state too. The action of any two bodies on each other therefore never changes the state of the common center of all; and since every action in such a system either takes place between two bodies or is made up of actions between pairs, no action within the system ever changes that state.

When the bodies do not act on one another, that center is either at rest or moves uniformly forward in some straight line. It therefore continues in that state, in spite of all the actions of the bodies on one another, unless a force acting on the whole system from outside compels it to change. The same law thus holds for a system of many bodies as for a single body, as regards continuing at rest or in motion, and the forward motion of a single body or of a whole system is always to be reckoned from the motion of its center of gravity.

Corollary 5

Bodies enclosed in a given space move in the same way among themselves whether that space is at rest or moving uniformly forward in a straight line without turning.

In both cases the differences of the motions that point the same way, and the sums of those that point opposite ways, are the same at the start, by assumption; and the collisions and impulses between the bodies arise from these sums and differences. The effects of the collisions are therefore the same in both cases (by Law 2), and the motions of the bodies among themselves remain the same in the one case as in the other. A ship gives a clear proof of this: everything on board moves in the same way whether the ship is at rest or moving uniformly forward in a straight line.³¹

Corollary 6

If bodies moving in any way among themselves are pushed along parallel lines by equal accelerative forces, they all go on moving among themselves just as they would if those forces were not acting.

These forces act equally on all the bodies, in proportion to their masses, and along parallel lines, so they change the velocities of all the bodies equally (by Law 2), and so they never alter the positions or motions of the bodies relative to one another.³²

Scholium

In this Scholium Newton tests the laws against experience: Galileo's results on falling bodies and projectiles, the collision experiments made with pendulums, attraction between a lodestone and iron, and the simple machines.

So far I have set out principles that mathematicians accept and that a great many experiments confirm. From the first two laws and the first two corollaries, Galileo found that the distance a falling body covers is proportional to the square of the time it has been falling, $s\propto t^2$, and that a thrown body moves along a parabola; experience agrees with both, except so far as the air slows these motions a little. A falling body's gravity acts as a steady force, adding equal amounts of speed in equal intervals of time, so that over the whole fall the speed gained is proportional to the time. The distance covered in a given time is proportional to the speed and the time multiplied together, and so it is proportional to the square of the time. When a body is thrown straight up, its steady gravity takes away speed in proportion to the time; the time it takes to reach its highest point is proportional to the speed it starts with, and the height it reaches, being proportional to that speed multiplied by that time, is proportional to the square of the starting speed.

When a body is thrown in any direction, the motion from the throw combines with the motion from its gravity. Suppose the body A (Plate I, Fig. 3), moving by its throw alone, would travel along the straight line AB in a given time, and by its fall alone would drop through the height AC in the same time. Complete the parallelogram ABDC. With both motions combined, the body is at D at the end of that time, and the curve AED that it traces is a parabola, touched by the straight line AB at A, whose drop BD below that line is proportional to the square of AB: $BD\propto A{B}^2$.

The results proved about the times in which pendulums swing rest on the same laws and corollaries, and pendulum clocks confirm them every day. From the same laws, together with the third, Sir Christopher Wren, Dr. Wallis and Mr. Huygens, the greatest geometers of our time, each worked out independently the rules for the collision and rebound of hard bodies, and they reported their discoveries to the Royal Society at about the same time, in exact agreement with one another. Dr. Wallis was the first to publish, Sir Christopher Wren next, and Mr. Huygens last. It was Sir Christopher Wren, however, who demonstrated the rules before the Royal Society with an experiment using pendulums, which Mr. Mariotte soon afterward saw fit to explain in a book devoted entirely to the subject.³³

To make this experiment agree exactly with the theory, we must allow both for the resistance of the air and for the elastic force of the colliding bodies. Let the balls A and B hang from the points C and D by the parallel strings AC and BD, of equal length (Plate I, Fig. 4). With C and D as centers and the strings as radii, draw the semicircles EAF and GBH, which the radii CA and DB cut in half.

To measure the loss caused by the air, take ball B away, pull ball A back to any point R on the arc EAF, and let it go; suppose that after one swing it returns only as far as V. Then RV is the loss caused by the air. Mark off ST, a fourth of RV, in the middle of it, so that RS equals TV and RS is in the same ratio to ST as 3 is to 2; ST then represents, very nearly, the loss during the swing down from S to A.

Now put ball B back in its place and let ball A fall from S. At the bottom, where the collision takes place, its speed will be very nearly what it would have been had it fallen from T in a vacuum, and so the chord of the arc TA can represent that speed; for it is well known to geometers that a pendulum's speed at its lowest point is proportional to the chord of the arc through which it has swung down.³⁴

Suppose that after the collision ball A rises to s and ball B to k. Take ball B away and find the point v from which ball A, let go, would return after one swing to a point r such that st is a fourth of rv, placed in its middle so that rs equals tv. The chord of the arc tA then represents the speed of ball A at the bottom just after the collision, since t is where it would have risen if there were no air. In the same way the point k, to which ball B rises, is corrected to the point l to which it would have risen in a vacuum, and every part of the experiment can be tested as though we really were in a vacuum.

Multiplying the body A by the chord of the arc TA, which represents its speed, gives (if I may call it a product) its momentum at the bottom just before the collision, and multiplying it by the chord of tA gives its momentum just after. In the same way, the body B multiplied by the chord of the arc Bl gives its momentum just after the collision. When the two balls are let go together from different places, the momentum of each is found in the same way, before and after the collision, and the momenta can then be compared and the effects of the collision worked out.

I tested the matter in this way with pendulums ten feet long, using unequal balls as well as equal ones, and making them meet after falling through large distances, of 8, 12 or 16 feet. Whenever the balls met head-on, I always found, to within 3 inches, that their motions changed by equal amounts in opposite directions, and so that action and reaction were always equal. When ball A struck ball B, which was at rest, with 9 units of momentum, and lost 7 and went on with 2, ball B was carried off with those 7 units. When the balls met moving in opposite directions, A with 12 units of momentum and B with 6, and A bounced back with 2, then B bounced back with 8, so that each lost 14 units: taking 12 units from A's momentum leaves nothing, and taking 2 more gives 2 units in the opposite direction, while taking 14 units from B's 6 gives 8 units in the opposite direction. When both balls were moving the same way, A, the faster, with 14 units and B, the slower, with 5, and after the collision A went on with 5, then B went on with 14, 9 units having passed from A to B. The other cases went the same way. In every collision the quantity of motion, that is, the sum of the momenta pointing the same way or the difference of those pointing opposite ways, was unchanged.

An error of an inch or two in the measurements is easily accounted for by the difficulty of doing everything exactly. It was not easy to release the two pendulums so precisely together that the balls struck each other at the lowest point, AB, nor to mark the points s and k to which they rose after the collision, and some error may also have come from uneven density inside the balls themselves, or from other irregularities in their texture.

It might be objected that the rule this experiment was meant to prove assumes the bodies to be perfectly hard, or at least perfectly elastic, and that no such bodies exist in nature. I should therefore add that the experiments just reported do not depend on hardness at all, and succeed as well with soft bodies as with hard ones. To test the rule on bodies that are not perfectly hard, we need only reduce the rebound in proportion to their elasticity. On the theory of Wren and Huygens, perfectly hard bodies separate with the same speed at which they meet, and this can be asserted with even more confidence of perfectly elastic bodies. In bodies that are not perfectly elastic, the speed of separation must be reduced along with the elastic force, because that force, unless the parts of the bodies are bruised by the collision or stretched as they are under a hammer, is as far as I can tell fixed and definite, and makes the bodies separate with a relative speed that is a fixed fraction of the relative speed at which they met.

I tested this with balls of wool, wound tight and strongly compressed. I first let the pendulums go and measured the rebound, which gave the amount of the balls' elastic force, and from that I calculated the rebounds that ought to occur in other collisions. The experiments made afterward agreed with the calculation: the balls always separated with a relative speed about 5/9 of the relative speed at which they had met. Balls of steel rebounded with almost the same speed, balls of cork with a little less, and for balls of glass the ratio was about 15 to 16. The third law, then, as far as it concerns impacts and rebounds, is confirmed by a theory that agrees exactly with experience.³⁵

For attractions, I prove the point briefly as follows. Place an obstacle between two bodies A and B that attract each other, so as to keep them apart. If A were drawn toward B more strongly than B toward A, the obstacle would be pressed harder by A than by B and would not remain in equilibrium; the stronger pressure would prevail, and the two bodies and the obstacle together would move off toward the side where B lies, and in empty space would go on forever, faster and faster. That is absurd, and contrary to the first law, by which the system must remain at rest or move uniformly in a straight line. The two bodies must therefore press equally on the obstacle, and so attract each other equally. I have tried the experiment with a lodestone and a piece of iron: placed in separate small vessels and floated near each other in still water, neither pushes the other along, but, being equally attracted, each holds against the other's pressure until at last they come to rest in equilibrium.

The gravitation between the Earth and its parts is mutual in the same way. Let the Earth FI (Plate I, Fig. 5) be cut by any plane EG into the two parts EGF and EGI; their weights toward each other are equal. To see this, cut the larger part EGI by a second plane HK, parallel to EG, into the pieces EGKH and HKI, choosing HK so that HKI is equal to the part EFG cut off first. The middle piece EGKH evidently has no tendency of its own toward either side, but hangs in equilibrium between the two, while the outer piece HKI bears with its whole weight on the middle piece and presses it toward the other outer part EGF. The force with which EGI, the pieces HKI and EGKH together, tends toward EGF is therefore equal to the weight of HKI, which is the weight of EGF, and so the two parts EGI and EGF weigh equally toward each other, as was to be proved. If these weights were unequal, the whole Earth, floating in the unresisting ether (the thin matter then supposed to fill space), would yield to the greater weight and move off away from it forever.

Just as, in collisions, bodies balance when their speeds are inversely proportional to their inherent forces, that is, to their masses, so in mechanical instruments two agents balance, and each holds back the opposing pressure of the other, when their speeds, measured along the directions of their forces, are inversely proportional to the forces.

Two weights therefore have equal power to move the arms of a balance when they are inversely proportional to the speeds at which they rise and fall as the balance swings. If they rise and fall vertically, this means they are inversely proportional to the distances of their points of suspension from the pivot; if slanting planes or other obstacles turn them aside so that they rise and fall at a slant, they balance when they are inversely proportional to the vertical heights through which they rise and fall, because gravity acts vertically downward.

In a pulley, or a set of pulleys, a hand pulling the rope straight holds up the weight, whether the weight rises vertically or at a slant, when the ratio of the hand's force to the weight equals the ratio of the speed of the weight's vertical rise to the speed of the hand.

In clocks and similar instruments built from a train of wheels, the forces that drive the wheels and the forces that resist them hold each other in balance when they are inversely proportional to the speeds of the parts of the wheels on which they act.

The force with which a screw presses on a body has the same ratio to the force of the hand turning its handle as the speed of the handle, at the place where the hand pushes it, has to the forward speed of the screw toward the body.

The forces with which a wedge presses apart the two parts of the wood it splits have the same ratio to the force of the mallet on the wedge as the speed at which the wedge advances, in the direction in which the mallet drives it, has to the speed at which the parts of the wood give way, along lines perpendicular to the sides of the wedge. The same account holds for every machine.

The power and use of machines consist only in this, that by reducing the speed we can increase the force, and the reverse. Every suitable machine thus solves the problem to move a given weight with a given power, or to overcome any other given resistance with a given force. If a machine is designed so that the speeds of the agent and of the resistance are inversely proportional to their forces, the agent will just hold the resistance in balance, and with a greater difference of speeds it will overcome it. If the difference of speeds is great enough to overcome all the resistance that usually arises from the friction of bodies sliding past one another, from the cohesion of bodies that are to be separated, and from the weights of bodies that are to be raised, the force left over will produce an acceleration proportional to it, both in the parts of the machine and in the resisting body. But mechanics is not my subject here. I wished only to show by these examples how widely the third law of motion applies and how certain it is. If the action of the agent is measured by its force and speed together, and the reaction of the obstacle by the speeds of its several parts together with the forces of resistance arising from their friction, cohesion, weight and acceleration, then in every kind of machine the action and the reaction will be found equal; and as far as the action is transmitted through the intervening parts and finally reaches the resisting body, its direction will always be opposite to that of the reaction.³⁶

Section 1: The Method of First and Last Ratios

These eleven lemmas set out the method of limits on which the proofs of Book One depend: how a curve may be treated as the end of a sequence of straight-sided figures, and how quantities may be compared at the moment they vanish. Newton explains in the Scholium at the end of the Section why he chose this method over the older ones.

Lemma 1

Quantities, and ratios of quantities, that during any finite time move steadily toward equality, and before the end of that time come nearer to each other than by any given difference, become equal in the end.

If you deny it, suppose that in the end they are unequal, and let D be their final difference. Then they cannot come nearer to equality than by that given difference D, which is against the supposition.³⁷

Lemma 2

Take any figure AacE, bounded by the straight lines Aa and AE and by the curve acE, and inscribe in it any number of parallelograms Ab, Bc, Cd, and so on, standing on equal bases AB, BC, CD, and so on, with their sides Bb, Cc, Dd, and so on, parallel to the side Aa of the figure. Complete the parallelograms aKbl, bLcm, cMdn, and so on. Then let the width of these parallelograms be made smaller, and their number larger, without limit. I say that the limiting ratios that the inscribed figure AKbLcMdD, the circumscribed figure AalbmcndoE, and the curved figure AabcdE have to one another are ratios of equality. (Plate I, Fig. 6)

The difference between the inscribed and the circumscribed figures is the sum of the parallelograms Kl, Lm, Mn, Do. Since all their bases are equal, that sum is the rectangle whose base is one of those bases, Kb, and whose height is the sum of their heights, Aa; that is, it is the rectangle ABla. But because its width AB is supposed to shrink without limit, this rectangle becomes smaller than any given area. Therefore the inscribed and circumscribed figures become equal to each other in the end (by Lemma 1), and the curved figure, which lies between them, becomes equal to either, all the more. Q.E.D.³⁸

Lemma 3

The same limiting ratios are still ratios of equality when the widths AB, BC, DC, and so on, of the parallelograms are unequal, provided they all shrink without limit.

Take AF equal to the greatest width, and complete the parallelogram FAaf. This parallelogram will be greater than the difference between the inscribed and circumscribed figures. But because its width AF shrinks without limit, it will become smaller than any given rectangle. Q.E.D.

Corollary 1. Hence the final sum of those vanishing parallelograms coincides with the curved figure in every part.³⁹

Corollary 2. Still more surely, the straight-sided figure formed by the chords of the vanishing arcs ab, bc, cd, and so on, coincides in the end with the curved figure.

Corollary 3. The same is true of the circumscribed straight-sided figure formed by the tangents of the same arcs.

Corollary 4. These final figures, therefore, so far as their boundaries acE are concerned, are not straight-sided figures at all, but the curved limits of straight-sided figures.

Lemma 4

Suppose that in two figures AacE and PprT you inscribe, as before, two rows of parallelograms, the same number in each row, and that when their widths shrink without limit, the limiting ratio of each parallelogram in one figure to the corresponding parallelogram in the other is one and the same. I say that the two figures AacE and PprT are in that same ratio to each other. (Plate I, Fig. 7)

Each parallelogram in the one figure is in the same ratio to the corresponding parallelogram in the other, and so, by adding the terms of these equal ratios together, the sum of all the parallelograms in the one is in that same ratio to the sum of all the parallelograms in the other. The one figure is in that ratio to the other as well, because each figure is in a ratio of equality to its own sum of parallelograms (by Lemma 3). Q.E.D.

Corollary. Hence if two quantities of any kind are divided in any way into the same number of parts, and those parts, as their number grows and their size shrinks without limit, are in a given ratio to each other, the first to the first, the second to the second, and so on in order, then the whole quantities are in that same given ratio to each other. For if, in the figures of this lemma, the ratios of the parallelograms to one another are taken equal to the ratios of the parts, the sum of the parts will always be proportional to the sum of the parallelograms. Therefore as the number of parallelograms and parts grows and their size shrinks without limit, those sums will be in the limiting ratio of a parallelogram in the one figure to the corresponding parallelogram in the other; that is (by the supposition), in the limiting ratio of any part of the one quantity to the corresponding part of the other.

Lemma 5

In similar figures, corresponding sides of every sort, curved or straight, are proportional, and the areas are proportional to the squares of corresponding sides.

Lemma 6

Suppose an arc ACB, fixed in position, is spanned by its chord AB, and at some point A, where the curvature is continuous on both sides, is touched by a straight line AD extended both ways. If the points A and B then approach each other and meet, I say that the angle BAD between the chord and the tangent will shrink without limit and in the end vanish. (Plate II, Fig. 1)

If that angle does not vanish, the arc ACB will make with the tangent AD an angle equal to a straight-sided angle, and so the curvature at the point A will not be continuous, which is against the supposition.

Lemma 7

With the same suppositions, I say that the limiting ratio of the arc, the chord and the tangent, any one to any other, is a ratio of equality. (Plate II, Fig. 1)

As the point B approaches the point A, always think of AB and AD as extended to the distant points b and d, and draw bd parallel to the secant BD; and let the arc Acb always be similar to the arc ACB. Then, when the points A and B come together, the angle dAb will vanish (by the preceding lemma). The straight lines Ab and Ad, which always keep a finite length, and the arc Acb between them, will therefore coincide and become equal to one another. The straight lines AB and AD and the arc ACB between them are always proportional to these, so they will vanish and, in the end, be in a ratio of equality. Q.E.D.

Corollary 1. It follows that if through B (Plate II, Fig. 2) we draw BF parallel to the tangent, always cutting at F some straight line AF through A, this line BF will in the end be in a ratio of equality with the vanishing arc ACB. For if the parallelogram AFBD is completed, BF is always equal to AD.

Corollary 2. And if more straight lines, such as BE, BD, AF and AG, are drawn through B and A, cutting the tangent AD and its parallel BF, then the limiting ratio of all the abscissas AD, AE, BF and BG, and of the chord and the arc AB, any one to any other, will be a ratio of equality.⁴⁰

Corollary 3. In all our reasoning about limiting ratios, therefore, we may freely use any one of these lines in place of any other.

Lemma 8

Suppose the straight lines AR and BR, together with the arc ACB, the chord AB and the tangent AD, make three triangles RAB, RACB and RAD, and the points A and B approach each other and meet. I say that in the end these vanishing triangles are similar in shape, and their limiting ratio is a ratio of equality. (Plate II, Fig. 1)

As the point B approaches the point A, always think of AB, AD and AR as extended to the distant points b, d and r, with rbd drawn parallel to RD, and let the arc Acb always be similar to the arc ACB. Then, when the points A and B come together, the angle bAd will vanish. Therefore the three triangles rAb, rAcb and rAd, which always keep a finite size, will coincide, and for that reason become both similar and equal. The triangles RAB, RACB and RAD are always similar to these and proportional to them, and so in the end they too become both similar and equal to one another. Q.E.D.

Corollary. In all our reasoning about limiting ratios, then, we may use any one of these triangles in place of any other, as we please.

Lemma 9

Suppose a straight line AE and a curve ABC, both fixed in position, cut each other at a given angle at A, and that BD and CE are drawn as ordinates to that straight line at another given angle, meeting the curve at B and C. If the points B and C then approach the point A together and meet there, I say that in the end the areas of the triangles ABD and ACE are in the same ratio as the squares of their sides. (Plate II, Fig. 3)

As the points B and C approach the point A, always suppose AD extended to the distant points d and e, so that Ad and Ae are proportional to AD and AE; and suppose the ordinates db and ec drawn parallel to the ordinates DB and EC, meeting AB and AC extended at b and c. Let the curve Abc be similar to the curve ABC, and draw the straight line Ag touching both curves at A and cutting the ordinates DB, EC, db and ec at F, G, f and g. Then, keeping the length Ae the same, let the points B and C come together at the point A. As the angle cAg vanishes, the curved areas Abd and Ace will coincide with the straight-sided areas Afd and Age, and so they will be in the same ratio as the squares of the sides Ad and Ae (by Lemma 5). But the areas ABD and ACE are always proportional to these areas, just as the sides AD and AE are proportional to these sides. In the end, therefore, the areas ABD and ACE are in the same ratio as the squares of the sides AD and AE. Q.E.D.⁴¹

Lemma 10

The distances that a body travels under any finite force, whether that force is fixed and unchanging or is continually increasing or continually decreasing, are, at the very beginning of the motion, in the same ratio as the squares of the times.

Let the times be represented by the lines AD and AE, and the speeds gained in those times by the ordinates DB and EC. The distances traveled at these speeds will be proportional to the areas ABD and ACE that those ordinates sweep out; that is, at the very beginning of the motion, they will be in the same ratio as the squares of the times AD and AE (by Lemma 9). Q.E.D.

Corollary 1. From this one can easily infer that when bodies travel similar parts of similar figures in proportional times, their errors, that is, their deviations, are very nearly proportional to the squares of the times in which they arise. This holds provided the deviations are produced by equal forces applied in the same way to the bodies, and are measured by the distances of the bodies from the places in the similar figures that they would have reached in those proportional times without the action of those forces.

Corollary 2. But the deviations produced by proportional forces, applied in the same way to the bodies at similar parts of the similar figures, are proportional to the forces and the squares of the times multiplied together.

Corollary 3. The same is to be understood of any distances whatever traveled by bodies under different forces. At the very beginning of the motion, all of them are proportional to the forces multiplied by the squares of the times, $s\propto F{t}^2$.

Corollary 4. The forces are therefore proportional to the distances traveled at the very beginning of the motion divided by the squares of the times, $F\propto s/t^2$.

Corollary 5. And the squares of the times are proportional to the distances traveled divided by the forces, $t^2\propto s/F$.

For a constant force, Lemma 10 is Galileo's rule that a falling body's distance is proportional to the square of the time; the lemma extends it to any finite force, at the start of the motion.

Scholium

When quantities of different kinds, not fixed in size, are compared, and one is said to be proportional to another, or inversely proportional to it, the meaning is that the first increases or decreases in the same ratio as the second, or as its reciprocal. And if one is said to be proportional to two or more others, or inversely proportional to them, or to some in the one way and some in the other, the meaning is that the first increases or decreases in the ratio made by multiplying together the ratios in which the others, or their reciprocals, increase or decrease. For example, if A is said to be proportional to B and to C, and inversely proportional to D, the meaning is that A increases or decreases in the same ratio as $B\times C\times \frac{1}{D}$; that is, A and $\frac{BC}{D}$ are in a given ratio to each other. In the modern sign, this is written $A\propto \frac{BC}{D}$.

Lemma 11

In every curve that has a finite curvature at the point of contact, the vanishing subtense of the angle of contact is, in the end, proportional to the square of the subtense of the arc that ends at the same point. (Plate II, Fig. 4)

Case 1. Let AB be the arc, AD its tangent, BD the subtense of the angle of contact, perpendicular to the tangent, and AB the subtense of the arc. Draw BG perpendicular to the subtense AB, and AG perpendicular to the tangent AD, meeting at G. Then let the points D, B and G move to the points d, b and g, and suppose J to be the final intersection of the lines BG and AG when the points D and B have reached A. Clearly the distance GJ can be made less than any assignable length. Now the circles through the points A, B, G and through A, b, g have AG and Ag as diameters, since the angles at B and b are right angles, and from the nature of these circles $A{B}^2=AG\times BD$, and $A{b}^2=Ag\times bd$. Therefore the ratio of $A{B}^2$ to $A{b}^2$ is the ratio of AG to Ag multiplied by the ratio of BD to bd. But because GJ can be taken shorter than any assignable length, the ratio of AG to Ag can be made to differ from a ratio of equality by less than any assignable difference. The ratio of $A{B}^2$ to $A{b}^2$ can therefore be made to differ from the ratio of BD to bd by less than any assignable difference. Therefore the limiting ratio of $A{B}^2$ to $A{b}^2$ is the same as the limiting ratio of BD to bd (by Lemma 1). Q.E.D.⁴²

Case 2. Now let BD be inclined to AD at any given angle. The limiting ratio of BD to bd will always be the same as before, and so the same as the ratio of $A{B}^2$ to $A{b}^2$. Q.E.D.

Case 3. And suppose the angle D is not given, but the straight line BD passes through a given point, or is fixed by any other condition whatever. Even so, the angles D and d, being fixed by the same rule, will always draw nearer to equality, and come closer to each other than by any assigned difference, and so in the end they will be equal (by Lemma 1). The ratio of the lines BD and bd is therefore the same as before.

Corollary 1. Therefore, since the tangents AD and Ad, the arcs AB and Ab, and their sines BC and bc become equal in the end to the chords AB and Ab, their squares will in the end be proportional to the subtenses BD and bd.

Corollary 2. Their squares are also, in the end, proportional to the versed sines of the arcs, where these versed sines bisect the chords and pass through a given point. For those versed sines are proportional to the subtenses BD and bd.

Corollary 3. Therefore the versed sine is proportional to the square of the time in which a body travels the arc at a given speed.

Corollary 4. The straight-sided triangles ADB and Adb are in the end proportional to the cubes of the sides AD and Ad, and to the 3/2 power of the sides DB and db, since they are proportional to AD times DB and to Ad times db: $ADB\propto A{D}^3\propto D{B}^{3/2}$. In the same way the triangles ABC and Abc are in the end proportional to the cubes of the sides BC and bc. What I call the 3/2 power is the square root of the cube, which is the same as the first power multiplied by the square root.

Corollary 5. Because DB and db are in the end parallel and proportional to the squares of the lines AD and Ad, the final curved areas ADB and Adb will be two thirds of the straight-sided triangles ADB and Adb (by the nature of the parabola, which is the curve whose distance from a tangent grows as the square of the distance along it), and the segments AB and Ab, between the arc and the chord, will be one third of the same triangles. Those areas and those segments will therefore be proportional to the cubes both of the tangents AD and Ad and of the chords and arcs AB and Ab.

Proposition 6, by which Book One measures a centripetal force from the versed sine of a very small arc and the time taken to travel it, is proved from Corollaries 2 and 3 of this lemma, and Newton adds there that Corollary 4 of Lemma 10 proves it as well.

Scholium

Throughout, however, we have supposed that the angle of contact is neither infinitely greater nor infinitely smaller than the angles of contact that circles make with their tangents; that is, that the curvature at the point A is neither infinitely small nor infinitely great, or that the distance AJ has a finite size. For DB may be taken proportional to AD${}^3$, and in that case no circle can be drawn through the point A between the tangent AD and the curve AB, so the angle of contact will be infinitely smaller than those of circles. By the same reasoning, if DB is made proportional in turn to AD${}^4$, AD${}^5$, AD${}^6$, AD${}^7$, and so on, we shall have a series of angles of contact going on without limit, each infinitely smaller than the one before it. And if DB is made proportional in turn to AD${}^2$, AD${}^{3/2}$, AD${}^{4/3}$, AD${}^{5/4}$, AD${}^{6/5}$, AD${}^{7/6}$, and so on, we shall have another infinite series of angles of contact, the first of the same kind as those of circles, the second infinitely greater, and each one after it infinitely greater than the one before. Between any two of these angles, moreover, another series of intermediate angles of contact can be placed, going on without limit in both directions, each infinitely greater or infinitely smaller than the one before it. For example, between the terms AD${}^2$ and AD${}^3$ the series AD${}^{13/6}$, AD${}^{11/5}$, AD${}^{9/4}$, AD${}^{7/3}$, AD${}^{5/2}$, AD${}^{8/3}$, AD${}^{11/4}$, AD${}^{14/5}$, AD${}^{17/6}$, and so on, could be placed. And again, between any two angles of this series a new series of intermediate angles can be placed, differing from one another by infinite steps. Nature knows no limit here.

What has been proved about curved lines and the surfaces they enclose can easily be applied to the curved surfaces and volumes of solids. I have set out these lemmas first to avoid the tedium of working through long and involved proofs by contradiction, in the manner of the ancient geometers. Proofs are made shorter by the method of indivisibles. But because the assumption of indivisibles seems rather harsh, and that method is therefore thought less geometrical, I chose instead to base the proofs of the propositions that follow on the first and last sums and ratios of nascent and vanishing quantities, that is, on the limits of those sums and ratios, and so to set out first, as briefly as I could, the proofs of those limits. This achieves the same thing as the method of indivisibles, and now that its principles have been proved, we may use them more safely. Therefore if from now on I happen to treat quantities as made up of particles, or use small curved lines in place of straight ones, I do not mean indivisibles, but vanishing divisible quantities; not the sums and ratios of definite parts, but always the limits of sums and ratios. The force of such proofs always depends on the method set out in the preceding lemmas.⁴³

It may be objected that there is no last ratio of vanishing quantities, because before the quantities have vanished, their ratio is not the last one, and after they have vanished, there is no ratio at all. But by the same argument it could be claimed that a body arriving at a certain place and stopping there has no last speed, because before it reaches the place its speed is not the last one, and when it has arrived it has none. The answer is easy. By the last speed is meant the speed with which the body is moving neither before it arrives at its last place and the motion stops, nor after, but at the very instant it arrives; that is, the speed with which the body reaches its last place and with which the motion stops. In the same way, the last ratio of vanishing quantities means the ratio of the quantities, not before they vanish, and not afterward, but the ratio with which they vanish. Likewise the first ratio of nascent quantities is the ratio with which they begin to be, and the first or last sum is the sum with which they begin or cease to be, or to increase or decrease. There is a limit that the speed at the end of the motion can reach but not exceed, and this is the last speed. There is a like limit for all quantities and proportions that begin and cease to be. Since such limits are certain and definite, finding them is a strictly geometrical problem, and whatever is geometrical we may properly use in finding and proving anything else that is also geometrical.

It may also be objected that if the last ratios of vanishing quantities are given, their last sizes will be given too, and so every quantity will be made up of indivisibles, which is contrary to what Euclid proved about incommensurables in the tenth book of his Elements. But this objection rests on a false assumption. The last ratios with which quantities vanish are not really ratios of last quantities, but limits toward which the ratios of quantities that decrease without limit always converge. The ratios come nearer to these limits than by any given difference, but never pass them, nor actually reach them, until the quantities have shrunk without limit. This is clearer in the case of quantities that grow without limit. If two quantities whose difference is given grow without limit, their last ratio will be given, namely a ratio of equality; but it does not follow that the last or greatest quantities themselves, whose ratio that is, are given. Therefore, if in what follows, to make things easier to understand, I happen to speak of quantities as least, or vanishing, or last, you are not to think of quantities of any definite size, but of quantities thought of as always decreasing without end.⁴⁴

Section 2: Finding Centripetal Forces

Proposition 1, Theorem 1

When a body orbits a fixed center of force, the radius drawn from the body to the center sweeps out areas that lie in one fixed plane, and these areas are proportional to the times in which they are swept out. (Plate II, Fig. 5)

This is the law of areas, which Kepler had established for the orbit of Mars from Tycho Brahe's observations. Newton derives it from the first law and the first corollary, together with the limiting argument of Lemma 3, by replacing the continuous pull of the center with a series of separate pushes.

Divide the time into equal parts, and suppose that in the first part the body, moving by its inherent force, travels along the straight line AB. In the second part, if nothing interfered, it would go straight on to c, along a line Bc equal to AB (by Law 1), and the radii AS, BS and cS would sweep out the two equal triangles ASB and BSc.

Now suppose that when the body arrives at B, a centripetal force acts on it all at once with a single strong push, turning it aside from the line Bc so that it goes on along the straight line BC instead. Draw cC parallel to BS, meeting BC at C. At the end of the second part of the time the body will be at C, still in the plane of the triangle ASB (by Corollary 1 of the Laws).

Join SC. Because SB and Cc are parallel, the triangles SBC and SBc, which stand on the same base SB, are equal; and since SBc is equal to SAB, so is SBC. The same argument holds at each later corner. If the centripetal force acts in turn at C, D, E and so on, making the body travel along the straight lines CD, DE, EF in successive equal intervals, all these lines lie in one plane, and each triangle equals the one before it: SCD equals SBC, SDE equals SCD, and SEF equals SDE. In equal times, then, equal areas are swept out in one fixed plane, and by adding them up, any two totals such as SADS and SAFS are in the same ratio as the times taken to sweep them out.

Finally, let the number of triangles increase and their width decrease without limit. In the limit the outline ADF becomes a curve (by Lemma 3, Corollary 4), and the centripetal force, which continually draws the body back from the tangent of that curve, acts without interruption. The areas SADS and SAFS were proportional to the times at every stage, and so they remain proportional to the times in this limiting case. Q.E.D.

Nothing in the argument depends on how strong the pushes are, or on how their strength varies from one corner to the next; it uses only the fact that every push is aimed at S. The law of areas therefore holds for any central force whatever, and Proposition 2 proves the converse.⁴⁵

Corollary 1. For a body attracted toward a fixed center, in space with no resistance, the speed is inversely proportional to the perpendicular dropped from the center onto the tangent of the orbit. The speeds at A, B, C, D and E are proportional to the bases AB, BC, CD, DE and EF of equal triangles, and since a triangle's area is half its base times its height, equal triangles have bases inversely proportional to their heights, which here are the perpendiculars. If the perpendicular is called p, then $v\propto 1/p$.⁴⁶

Corollary 2. Take the chords AB and BC of two arcs that the body travels one after the other, in equal times, in space with no resistance, and complete the parallelogram ABCV. As the arcs shrink without limit, the diagonal BV approaches a final position, and in that position, extended both ways, it passes through the center of force.

Corollary 3. Take the chords AB, BC and DE, EF of arcs traveled in equal times, in space with no resistance, and complete the parallelograms ABCV and DEFZ. As the arcs shrink without limit, the force at B and the force at E come to be in the same ratio as the diagonals BV and EZ. The body's motions along BC and EF are compounded of the motions Bc and BV, and Ef and EZ (by Corollary 1 of the Laws). In the proof of the proposition, BV and EZ are equal to Cc and Ff, which were produced by the pushes of the centripetal force at B and E, and so they are proportional to those pushes.

Corollary 4. In space with no resistance, the forces that draw bodies away from straight-line motion and bend them into curved orbits are proportional to the versed sines of arcs traveled in equal times. The versed sine meant here is the short line from the middle of the arc to the chord across it; in the limit, as the arcs shrink, it points toward the center of force and bisects the chord. The corollary holds because these versed sines are half the diagonals of Corollary 3.

Corollary 5. The ratio of one of these forces to the force of gravity is therefore the same as the ratio of its versed sine to the vertical versed sine of the parabolic arc that a projectile traces in the same time.

Corollary 6. All of this still holds when the planes in which the bodies move, together with the centers of force in those planes, are not at rest but move uniformly forward in straight lines (by Corollary 5 of the Laws).

Proposition 2, Theorem 2

Suppose a body moves along a curve that lies in a plane, and a radius drawn from it to some point sweeps out areas about that point proportional to the times, the point being either fixed or moving uniformly forward in a straight line. Then the body is driven by a centripetal force directed toward that point.

Case 1. A body moving along a curve is turned aside from a straight course by some force acting on it (by Law 1). That force makes the body sweep out, in equal times, the equal and very small triangles SAB, SBC, SCD and so on about the fixed point S. Since the triangles SBc and SBC are equal and stand on the same base SB, the line cC is parallel to SB (by Euclid's Elements, Book 1, Proposition 40), and so at B the force acts along a line parallel to cC, that is, along BS (by Law 2). In the same way, at C it acts along a line parallel to dD, that is, along CS, and so on. The force therefore always points toward the fixed point S. Q.E.D.

Case 2. It makes no difference whether the surface on which the body moves is at rest, or moves uniformly forward in a straight line, carrying the body, its path and the point S along with it (by Corollary 5 of the Laws).

Corollary 1. In space or in media with no resistance, if the areas are not proportional to the times, the forces do not point at the place where the radii meet. They lean forward, toward the direction of motion, if the rate at which areas are swept out is increasing, and backward if it is decreasing.

Corollary 2. Even in a resisting medium, if the rate at which areas are swept out is increasing, the forces lean away from the place where the radii meet, toward the direction of motion.

Scholium

A body may be driven by a centripetal force made up of several forces, and in that case the proposition means that their combined force points toward S. A force that always acts at right angles to the surface of the motion would pull the body out of its plane, but it would neither enlarge nor reduce the area swept out, and so it can be left out when the forces are combined.

Proposition 3, Theorem 3

Proposition 3 extends the result to a center that is itself moving, as the Earth is while the Moon circles it.

Suppose a radius drawn from a body to the center of a second body, which may be moving in any way at all, sweeps out areas about that center proportional to the times. Then the first body is driven by a combination of two forces: the centripetal force toward the second body, and the whole accelerative force that drives the second body.

Call the first body L and the second T. Suppose both bodies are acted on, along parallel lines, by a new force equal and opposite to the force that drives T. The first body L will then go on sweeping out the same areas about T as before (by Corollary 6 of the Laws), while the force on T is cancelled, so that T, left to itself, either stays at rest or moves uniformly forward in a straight line (by Law 1). The first body, driven now by the difference of the forces, that is, by the force left over, still sweeps out areas about T proportional to the times, and so that left-over force is directed toward T as its center (by Proposition 2). Q.E.D.

Corollary 1. Suppose, then, that the body L, by a radius drawn to T, sweeps out areas proportional to the times. From the whole force driving L, whether that force is single or made up of several as in Corollary 2 of the Laws, subtract the whole accelerative force driving T, using the same corollary. What remains of the force on L is directed toward T as its center.

Corollary 2. If these areas are nearly proportional to the times, the remaining force is directed nearly toward T.

Corollary 3. Conversely, if the remaining force is directed nearly toward T, the areas are nearly proportional to the times.

Corollary 4. Suppose the body L, by a radius drawn to T, sweeps out areas that are far from proportional to the times, while T is at rest or moves uniformly in a straight line. Then either no centripetal force at all is directed toward T, or that force is mixed with very strong forces of other kinds, and the total of all of them, if there are many, is directed toward some other center, fixed or moving. The same holds when T moves in any way at all, provided we take the centripetal force that remains after subtracting the whole force acting on T.

Scholium

Because the even sweeping out of areas shows that there is a center toward which the force that most affects the body is directed, the force that draws it away from straight-line motion and holds it in its orbit, why should we not use the even sweeping out of areas, in what follows, as the sign of a center about which all circular motion in free space takes place?

Proposition 4, Theorem 4

Proposition 4 gives the force needed to hold a body on a circle, a result Huygens had published without proof in 1673. Its sixth corollary is the first appearance of the inverse-square law in Book One.

Bodies that move uniformly around different circles are held there by centripetal forces that point toward the centers of the circles. Each force is proportional to the square of the arc the body travels in a given time, divided by the radius of its circle.

The forces point toward the centers of the circles (by Proposition 2 and Proposition 1, Corollary 2). They are proportional to the versed sines of the very small arcs traveled in equal times (by Proposition 1, Corollary 4), and in the limit each of these versed sines is the square of its arc divided by the diameter of the circle (by Lemma 7). The arcs traveled in very small equal times are in the same ratio as the arcs traveled in any equal times, and the diameters are in the same ratio as the radii. Each force is therefore proportional to the square of the arc traveled in a given time, divided by the radius: $F\propto \frac{{\left(\mathrm{\text{arc}}\right)}^2}{r}$. Q.E.D.

Corollary 1. Since the arcs are proportional to the speeds of the bodies, the centripetal force is proportional to the square of the speed divided by the radius: $F\propto \frac{v^2}{r}$. For a stone whirled on a sling, doubling the speed on the same length of cord means the cord must pull four times as hard.

Corollary 2. The periodic time is proportional to the radius divided by the speed, $T\propto r/v$, so the centripetal force is proportional to the radius divided by the square of the periodic time: $F\propto \frac{r}{T^2}$.

Corollary 3. If the periodic times are equal, the speeds are proportional to the radii, and then the centripetal forces are proportional to the radii as well, $F\propto r$. The reverse also holds.

Corollary 4. If the periodic times and the speeds are both proportional to the square root of the radii, the centripetal forces are all equal; and the reverse also holds.

Corollary 5. If the periodic times are proportional to the radii, the speeds are equal, and the centripetal forces are inversely proportional to the radii, $F\propto 1/r$. The reverse also holds.

Corollary 6. If the periodic times are proportional to the 3/2 power of the radii, then the speeds are inversely proportional to the square root of the radii, and the centripetal forces are inversely proportional to the square of the radii; and the reverse also holds. In symbols, if $T^2\propto r^3$, then $F\propto 1/r^2$. The hypothesis is Kepler's third law, which the planets obey, so for planets in nearly circular orbits this corollary gives an attraction toward the Sun inversely proportional to the square of the distance. Proposition 15 shows that the same 3/2-power rule holds for elliptical orbits under an inverse-square force, and Book Three applies the result to the planets and their moons.⁴⁷

Corollary 7. In general, if the periodic time is proportional to some power $R^n$ of the radius R, then the speed is inversely proportional to the power $R^{n-1}$ of the radius, and the centripetal force is inversely proportional to the power $R^{2n-1}$ of the radius; and the reverse also holds.

Corollary 8. The same results hold for the times, speeds and forces of bodies that trace similar parts of any similar figures, provided the centers occupy similar positions in those figures. This is seen by applying the preceding proof to those figures, which is easily done: put the even sweeping out of areas in place of uniform motion, and the distances of the bodies from the centers in place of the radii.

Corollary 9. It also follows from the same proof that if a body moves uniformly around a circle under a given centripetal force, the arc it travels in any time is the mean proportional between the diameter of the circle and the distance the same body would fall from rest, under the same force, in the same time. In other words, the diameter divided by the arc equals the arc divided by the distance fallen: ${\left(\mathrm{\text{arc}}\right)}^2=2r\times s$.

Scholium

The case of the sixth corollary holds for the heavenly bodies, as Sir Christopher Wren, Dr. Hooke and Dr. Halley have each observed, and so in what follows I will deal at length with a centripetal force that decreases as the square of the distance from the center. The proposition and its corollaries also allow us to compare a centripetal force with any other known force, such as gravity. If a body orbits in a circle concentric with the Earth, held there by its gravity, that gravity is its centripetal force; and from the fall of heavy bodies we can work out both the time of one full orbit and the arc traveled in any given time (by Corollary 9). Mr. Huygens, in his excellent book On the Pendulum Clock (De horologio oscillatorio), used propositions of this kind to compare the force of gravity with the centrifugal forces of revolving bodies.^48,49

The proposition can also be proved in another way. Inscribe in any circle a polygon with any number of sides, and let a body move at a given speed along the sides, bouncing off the circle at each corner. At each bounce it strikes the circle with a force proportional to its speed, so the total force in a given time is proportional to the speed multiplied by the number of bounces. If the shape of the polygon is fixed, this is proportional to the length traveled in the given time multiplied by that length divided by the radius, that is, to the square of the length divided by the radius. When the sides shrink without limit and the polygon merges with the circle, the total force becomes proportional to the square of the arc traveled in a given time, divided by the radius. This is the centrifugal force with which the body presses on the circle, and the force with which the circle continually pushes the body back toward the center is equal and opposite to it.

Proposition 5, Problem 1

Given a figure that a body traces under forces directed toward some common center, and given the body's speed at any places on it, to find that center. (Plate III, Fig. 1)

Let the three straight lines PT, TQV and VR touch the traced figure at three points P, Q and R, and meet at T and V. On the tangents erect the perpendiculars PA, QB and RC, inversely proportional to the body's speeds at P, Q and R, the points from which they are raised. That is, make the ratio of PA to QB the same as the ratio of the speed at Q to the speed at P, and the ratio of QB to RC the same as the ratio of the speed at R to the speed at Q. Through the ends A, B and C of the perpendiculars draw AD, DBE and EC at right angles to them, meeting at D and E. The straight lines TD and VE, extended, will meet at S, the center required.

For the perpendiculars dropped from the center S onto the tangents PT and QT are inversely proportional to the body's speeds at P and Q (by Proposition 1, Corollary 1). By the construction, then, they are proportional to the perpendiculars AP and BQ, that is, to the perpendiculars dropped from the point D onto the same tangents, since AD and DB run parallel to the tangents at the distances AP and BQ. From this it follows easily that the points S, D and T lie on one straight line: every point whose distances from the two tangents through T are in a fixed ratio lies on one straight line through T. By the same argument the points S, E and V also lie on one straight line, and so the center S is at the point where the straight lines TD and VE meet. Q.E.D.

Proposition 6, Theorem 5

Proposition 6 turns the versed sine of Proposition 1, Corollary 4 into a measure of the force at a single point of an orbit. Its fifth corollary is the method by which Propositions 7 to 10 below find their forces, and Section 3 uses it again for the conic sections.

In space with no resistance, suppose a body orbits a fixed center, and in the shortest time traces an arc that is just coming into being; and suppose the versed sine of that arc is drawn so that it bisects the chord, and is extended to pass through the center of force. Then the centripetal force at the middle of the arc is proportional to the versed sine divided by the square of the time.

For the versed sine traced in a given time is proportional to the force (by Proposition 1, Corollary 4). If the time is increased in any ratio, the arc increases in the same ratio, and the versed sine increases as the square of that ratio (by Lemma 11, Corollaries 2 and 3); so the versed sine is proportional to the force multiplied by the square of the time. Divide both sides by the square of the time, and the force is proportional to the versed sine divided by the square of the time: $F\propto \frac{\mathrm{\text{versed sine}}}{t^2}$. Q.E.D.

The same result can also be proved easily from Lemma 10, Corollary 4.

Corollary 1. Suppose a body P, orbiting the center S (Plate III, Fig. 2), traces a curved line APQ, which the straight line ZPR touches at some point P. From any other point Q of the curve draw QR parallel to the distance SP, meeting the tangent at R, and draw QT perpendicular to the distance SP. Then the centripetal force is inversely proportional to the product $\frac{S{P}^2\times Q{T}^2}{QR}$, provided this quantity is given the value it ends up with as the points P and Q come together. For QR is equal to the versed sine of twice the arc QP, an arc whose middle is P; and twice the triangle SQP, or $SP\times QT$, is proportional to the time in which that double arc is traced, and so it can stand for the time.

Corollary 2. By the same reasoning, the centripetal force is inversely proportional to the product $\frac{S{Y}^2\times Q{P}^2}{QR}$, if SY is the perpendicular from the center of force onto PR, the tangent of the orbit. For the products $SY\times QP$ and $SP\times QT$ are equal in the limit, as Q comes to P.⁵⁰

Corollary 3. Suppose the orbit is a circle, or else touches or cuts a circle "concentrically," that is, meets it at the smallest possible angle of contact or of section, so that the two have the same curvature and the same radius of curvature at the point P. Let PV be a chord of this circle, drawn from the body through the center of force. Then the centripetal force is inversely proportional to the product $S{Y}^2\times PV$. For PV is equal to $\frac{Q{P}^2}{QR}$, in the limit as Q comes to P.⁵¹

Corollary 4. On the same assumptions, the centripetal force is proportional to the square of the speed divided by that chord, $F\propto v^2/PV$. For the speed is inversely proportional to the perpendicular from S onto the tangent (by Proposition 1, Corollary 1).

Corollary 5. Hence if any curved figure APQ is given, and in it a point S toward which a centripetal force is always directed, we can find the law of centripetal force by which a body P will be continually drawn back from a straight course, held on the perimeter of that figure, and made to trace it in a continual revolution. That is, we have to calculate either the product $\frac{S{P}^2\times Q{T}^2}{QR}$ or the product $S{Y}^2\times PV$, to which this force is inversely proportional. We will give examples of this in the problems that follow.

Proposition 7, Problem 2

If a body moves on the circumference of a circle, to find the law of a centripetal force directed toward any given point. (Plate III, Fig. 3)

Let VQPA be the circumference of the circle, S the given point toward which the force tends as its center, P the body moving on the circumference, Q the next place it will move to, and PRZ the tangent of the circle at its first place. Through the point S draw the chord PV and the diameter VA of the circle, and join AP. Draw QT perpendicular to SP, and extend it to meet the tangent PR at Z. Lastly, through the point Q draw LR parallel to SP, meeting the circle at L and the tangent PZ at R. Because the triangles ZQR, ZTP and VPA are similar, the ratio of $R{P}^2$, which equals the product QR times RL, to $Q{T}^2$ is the same as the ratio of $A{V}^2$ to $P{V}^2$. Therefore $\frac{QRL\times P{V}^2}{A{V}^2}$ is equal to $Q{T}^2$. Multiply these equal quantities by $\frac{S{P}^2}{QR}$, and, as the points P and Q come together, write PV for RL; then $\frac{S{P}^2\times P{V}^3}{A{V}^2}=\frac{S{P}^2\times Q{T}^2}{QR}$. Therefore the centripetal force is inversely proportional to $\frac{S{P}^2\times P{V}^3}{A{V}^2}$ (by Proposition 6, Corollaries 1 and 5), that is, since $A{V}^2$ is given, inversely proportional to the square of the distance or height SP multiplied by the cube of the chord PV: $F\propto \frac{1}{S{P}^2\times P{V}^3}$. Q.E.I.

The same, another way.

Drop the perpendicular SY onto the tangent PR, extended. Because the triangles SYP and VPA are similar, the ratio of AV to PV is the same as the ratio of SP to SY, and so $\frac{SP\times PV}{AV}=SY$, and $\frac{S{P}^2\times P{V}^3}{A{V}^2}=S{Y}^2\times PV$. Hence the centripetal force is inversely proportional to $\frac{S{P}^2\times P{V}^3}{A{V}^2}$ (by Proposition 6, Corollaries 3 and 5), that is, since AV is given, inversely proportional to $S{P}^2\times P{V}^3$. Q.E.I.

Corollary 1. It follows that if the given point S, toward which the centripetal force always tends, is placed on the circumference of the circle, say at V, the centripetal force is inversely proportional to the fifth power of the height SP, $F\propto 1/S{P}^5$.

Corollary 2. Compare the force by which the body P orbits the center of force S on the circle APTV (Plate III, Fig. 4) with the force by which the same body could orbit any other center of force R on the same circle, in the same periodic time. The ratio of the first force to the second is the ratio of $R{P}^2\times SP$ to the cube of the straight line SG, drawn from the first center S parallel to the body's distance PR from the second center R, and meeting the tangent PG of the orbit at G. For by the construction of this proposition, the ratio of the first force to the second is the ratio of $R{P}^2\times P{T}^3$ to $S{P}^2\times P{V}^3$, which is the ratio of $SP\times R{P}^2$ to $\frac{S{P}^3\times P{V}^3}{P{T}^3}$, or, because the triangles PSG and TPV are similar, to $S{G}^3$.

Corollary 3. Take the force by which the body P orbits the center of force S in any orbit at all, and the force by which the same body could orbit any other center of force R in the same orbit and in the same periodic time. The ratio of the first to the second is the ratio of the product $SP\times R{P}^2$, the body's distance from the first center S times the square of its distance from the second center R, to the cube of the straight line SG, drawn from S parallel to RP and meeting the tangent PG of the orbit at G. For the force in this orbit at any point P is the same as in a circle of the same curvature.

Proposition 8, Problem 3

If a body moves on the semicircle PQA, to find the law of the centripetal force directed toward a point S so far away that all the lines PS, RS drawn to it can be taken as parallel. (Plate III, Fig. 5)

From C, the center of the semicircle, draw the radius CA, cutting the parallels at right angles at M and N, and join CP. Because the triangles CPM, PZT and RZQ are similar, the ratio of $C{P}^2$ to $P{M}^2$ is the same as the ratio of $P{R}^2$ to $Q{T}^2$. From the nature of the circle, $P{R}^2$ is equal to the product $QR\times \overline{RN+QN}$, or, as the points P and Q come together, to the product $QR\times 2PM$. Therefore the ratio of $C{P}^2$ to $P{M}^2$ is the same as the ratio of $QR\times 2PM$ to $Q{T}^2$. It follows that $\frac{Q{T}^2}{QR}=\frac{2P{M}^3}{C{P}^2}$, and $\frac{Q{T}^2\times S{P}^2}{QR}=\frac{2P{M}^3\times S{P}^2}{C{P}^2}$. Therefore the centripetal force is inversely proportional to $\frac{2P{M}^3\times S{P}^2}{C{P}^2}$ (by Proposition 6, Corollaries 1 and 5), that is, leaving out the given ratio $\frac{2S{P}^2}{C{P}^2}$, inversely proportional to $P{M}^3$: $F\propto 1/P{M}^3$. Q.E.I.

The same result also follows easily from the preceding proposition.

Scholium

By similar reasoning, a body will move on an ellipse, or even on a hyperbola or a parabola, under a centripetal force that is inversely proportional to the cube of the ordinate and directed toward an infinitely distant center of force.

Proposition 9, Problem 4

If a body orbits on a spiral PQS that cuts all the radii SP, SQ and so on at a given angle, to find the law of the centripetal force directed toward the center of that spiral. (Plate III, Fig. 6)

Suppose the indefinitely small angle PSQ is given. Then, since all the angles are given, the figure SPRQT is fixed in shape: only its size can change. The ratio $\frac{QT}{QR}$ is therefore also given, and $\frac{Q{T}^2}{QR}$ is proportional to QT, that is, since the figure is fixed in shape, proportional to SP. Now suppose the angle PSQ is changed in any way. The straight line QR, which subtends the angle of contact QPR, will change in proportion to the square of PR or QT (by Lemma 11). Therefore the ratio $\frac{Q{T}^2}{QR}$ stays the same as before, that is, proportional to SP. Then $\frac{Q{T}^2\times S{P}^2}{QR}$ is proportional to $S{P}^3$, and so the centripetal force is inversely proportional to the cube of the distance SP (by Proposition 6, Corollaries 1 and 5): $F\propto 1/S{P}^3$. Q.E.I.⁵²

The same, another way.

The perpendicular SY dropped onto the tangent, and the chord PV of the circle that cuts the spiral concentrically, are each in a given ratio to the height SP. It follows that $S{P}^3$ is proportional to $S{Y}^2\times PV$, which is inversely proportional to the centripetal force (by Proposition 6, Corollaries 3 and 5).

Lemma 12

All parallelograms drawn around any pair of conjugate diameters of a given ellipse or hyperbola are equal to one another.

This is proved by the writers on the conic sections.⁵³

Proposition 10, Problem 5

Proposition 10 takes the ellipse with the force directed toward its center. Proposition 11, at the start of the next Section, takes the force toward a focus, which is the case of the planets.

If a body orbits on an ellipse, to find the law of the centripetal force directed toward the center of the ellipse. (Plate IV, Fig. 1)

Suppose CA and CB are the semi-axes of the ellipse; GP and DK are conjugate diameters; PF and QT are perpendiculars to those diameters; Qv is an ordinate to the diameter GP; and the parallelogram QvPR is completed. Then, by the properties of the conic sections, the ratio of the product Pv times vG to $Q{v}^2$ is the ratio of $P{C}^2$ to $C{D}^2$; and because the triangles QvT and PCF are similar, the ratio of $Q{v}^2$ to $Q{T}^2$ is the ratio of $P{C}^2$ to $P{F}^2$. Combining the two, the ratio of Pv times vG to $Q{T}^2$ is the product of the ratio of $P{C}^2$ to $C{D}^2$ and the ratio of $P{C}^2$ to $P{F}^2$. That is, the ratio of vG to $\frac{Q{T}^2}{Pv}$ is the ratio of $P{C}^2$ to $\frac{C{D}^2\times P{F}^2}{P{C}^2}$. Now put QR in place of Pv; put $BC\times CA$ in place of $CD\times PF$ (by Lemma 12); and, as the points P and Q come together, put 2 PC in place of vG. Multiplying the outer terms of the proportion together and the inner terms together, we get $\frac{Q{T}^2\times P{C}^2}{QR}$ equal to $\frac{2B{C}^2\times C{A}^2}{PC}$. The centripetal force is therefore inversely proportional to $\frac{2B{C}^2\times C{A}^2}{PC}$ (by Proposition 6, Corollary 5); that is, since $2B{C}^2\times C{A}^2$ is given, inversely proportional to $\frac{1}{PC}$; that is, proportional to the distance PC itself, $F\propto PC$. Q.E.I.

The same, another way.

On the straight line PG, on the other side of the point T, take the point u so that Tu is equal to Tv. Then take uV such that the ratio of uV to vG is the same as the ratio of $D{C}^2$ to $P{C}^2$. Since the ratio of $Q{v}^2$ to Pv times vG is the ratio of $D{C}^2$ to $P{C}^2$ (by the conic sections), we have $Q{v}^2=Pv\times uV$. Add the product uP times Pv to both sides, and the square of the chord of the arc PQ will equal the product VP times Pv. It follows that a circle that touches the conic section at P and passes through the point Q will also pass through the point V. Now let the points P and Q come together. The ratio of uV to vG, which is the same as the ratio of $D{C}^2$ to $P{C}^2$, becomes the ratio of PV to PG, or of PV to 2 PC, and so PV is equal to $\frac{2D{C}^2}{PC}$. Therefore the force by which the body P orbits on the ellipse is inversely proportional to $\frac{2D{C}^2}{PC}\times P{F}^2$ (by Proposition 6, Corollary 3); that is, since $2D{C}^2\times P{F}^2$ is given, proportional to PC. Q.E.I.

Corollary 1. Therefore the force is proportional to the distance of the body from the center of the ellipse. And conversely, if the force is proportional to the distance, the body will move on an ellipse whose center is at the center of force, or possibly on a circle, which is what the ellipse becomes when its axes are equal.

Corollary 2. And the periodic times of the revolutions in all ellipses whatever about the same center will be equal. For in similar ellipses those times are equal (by Proposition 4, Corollaries 3 and 8). In ellipses that share the same major axis, the times are proportional to the whole area of each ellipse divided by the part of the area swept out in a given time; that is, to the minor axis divided by the speed of the body at its principal vertex; that is, to that minor axis divided by the ordinate at the same point of the common axis. And since the minor axes and those ordinates are in the same ratio, the quotients are equal, and so are the times. Under a force proportional to distance, the time of one revolution does not depend on the size or the shape of the orbit.

Scholium

If the ellipse has its center moved off to an infinite distance, it becomes a parabola, and the body will move on this parabola; the force, now directed toward a center infinitely far away, will become uniform. This is Galileo's theorem, that a projectile under a uniform gravity moves on a parabola. And if the parabolic section of the cone becomes a hyperbola, by a change in the slope of the plane that cuts the cone, the body will move on the perimeter of this hyperbola, with its centripetal force turned into a centrifugal force. In the circle or the ellipse, suppose the forces are directed toward the center of the figure, placed on the abscissa. If the ordinates are increased or reduced in any given ratio, or even if the angle at which they slope to the abscissa is changed, those forces are always increased or reduced in the ratio of the distances from the center, provided the periodic times stay the same. In the same way, in any figure at all, if the ordinates are increased or reduced in any given ratio, or their slope is changed in any way, while the periodic time stays the same, the forces directed toward any center placed on the abscissa are increased or reduced, at the different ordinates, in the ratio of the distances from the center.

Section 3: The Motion of Bodies in Eccentric Conic Sections

Proposition 11, Problem 6

A body moves in an ellipse: it is required to find the law of the centripetal force directed toward a focus of the ellipse. (Plate IV, Fig. 2)

Kepler had found that Mars moves in an ellipse with the Sun at one focus. Proposition 11 asks what force, directed toward that focus, will hold a body on such an orbit, and answers it with the measure of force from Proposition 6 and the properties of the ellipse; Propositions 12 and 13 do the same for the hyperbola and the parabola.^54,55

Let S be the focus of the ellipse. Draw SP, cutting the diameter DK (the diameter parallel to the tangent at P) in E, and cutting the ordinate Qv in x; and complete the parallelogram QxPR. It is evident that EP is equal to the greater semi-axis AC. For draw HI from the other focus H parallel to EC. Because CS and CH are equal, ES and EI are equal too, and so EP is half the sum of PS and PI. Because HI and PR are parallel and the angles IPR and HPZ are equal, PI is equal to PH; so EP is half the sum of PS and PH, and these two together are equal to the whole axis $2AC$.

Draw QT perpendicular to SP, and let L stand for the principal latus rectum of the ellipse, that is, for $\frac{2B{C}^2}{AC}$. Then the ratio of $L\times QR$ to $L\times Pv$ is the same as the ratio of QR to Pv, which is the ratio of PE, or AC, to PC. The ratio of $L\times Pv$ to the rectangle GvP is the ratio of L to Gv. The ratio of GvP to $Q{v}^2$ is the ratio of $P{C}^2$ to $C{D}^2$. When the points Q and P come together, the ratio of $Q{v}^2$ to $Q{x}^2$ becomes a ratio of equality (by Lemma 7, Corollary 2). And the ratio of $Q{x}^2$, or $Q{v}^2$, to $Q{T}^2$ is the ratio of $E{P}^2$ to $P{F}^2$, that is, of $C{A}^2$ to $P{F}^2$, which is the ratio of $C{D}^2$ to $C{B}^2$ (by Lemma 12).⁵⁶

Compounding all these ratios, by multiplying their first terms together and their second terms together, we find that the ratio of $L\times QR$ to $Q{T}^2$ is the same as the ratio of $AC\times L\times P{C}^2\times C{D}^2$, or $2C{B}^2\times P{C}^2\times C{D}^2$, to $PC\times Gv\times C{D}^2\times C{B}^2$, and so the same as the ratio of $2PC$ to Gv. But when Q and P come together, $2PC$ and Gv are equal. Therefore the quantities $L\times QR$ and $Q{T}^2$, which are proportional to these, are also equal. Multiply both of these equals by $\frac{S{P}^2}{QR}$, and $L\times S{P}^2$ becomes equal to $\frac{S{P}^2\times Q{T}^2}{QR}$. Therefore the centripetal force is inversely proportional to $L\times S{P}^2$ (by Proposition 6, Corollaries 1 and 5); and since L is fixed for a given ellipse, the force is inversely proportional to the square of the distance SP, $F\propto \frac{1}{S{P}^2}$. Q.E.I.

The same, another way.

The force directed toward the center of the ellipse, by which the body P can travel around that ellipse, is proportional to the distance CP of the body from the center C (by Proposition 10, Corollary 1). Draw CE parallel to the tangent PR of the ellipse, meeting PS in E. Then the force by which the same body P can travel around the same ellipse about any other point S of the ellipse is proportional to $\frac{P{E}^3}{S{P}^2}$ (by Proposition 7, Corollary 3). If the point S is the focus of the ellipse, PE is given, and so the force is inversely proportional to $S{P}^2$. Q.E.I.

We could have reduced this problem to the parabola and the hyperbola as briefly as we did the fifth problem (Proposition 10). But because of the importance of the problem and its use in what follows, I shall confirm the other cases by proofs of their own.

Proposition 11 obtains the inverse square from the shape of a single orbit, the force being directed to the focus, whereas Proposition 4, Corollary 6 obtained it, for circles, from the relation between the periods and distances of different orbits; Kepler's first law and his third thus each lead to the same law of force, $F\propto 1/r^2$.

Proposition 12, Problem 7

Suppose a body moves in a hyperbola: it is required to find the law of the centripetal force directed toward the focus of that figure. (Plate V, Fig. 1)

Let CA and CB be the semi-axes of the hyperbola; PG and KD another pair of conjugate diameters, KD being parallel to the tangent at P; PF a perpendicular to the diameter KD; and Qv an ordinate to the diameter GP. Draw SP, cutting the diameter DK in E and the ordinate Qv in x, and complete the parallelogram QRPx. It is evident that EP is equal to the semi-transverse axis AC, half the axis that runs through the foci. For draw HI from the other focus H of the hyperbola parallel to EC. Because CS and CH are equal, ES and EI are equal too, and so EP is half the difference of PS and PI. Because IH and PR are parallel and the angles IPR and HPZ are equal, PI is equal to PH; so EP is half the difference of PS and PH, and that difference is equal to the whole axis $2AC$.

Draw QT perpendicular to SP, and let L stand for the principal latus rectum of the hyperbola, that is, for $\frac{2B{C}^2}{AC}$. Then the ratio of $L\times QR$ to $L\times Pv$ is the same as the ratio of QR to Pv, or of Px to Pv; and because the triangles Pxv and PEC are similar, this is the ratio of PE to PC, or of AC to PC.⁵⁷

Next, the ratio of $L\times Pv$ to $Gv\times Pv$ is the ratio of L to Gv. By the properties of the conic sections, the ratio of the rectangle GvP to $Q{v}^2$ is the ratio of $P{C}^2$ to $C{D}^2$. When the points Q and P come together, the ratio of $Q{v}^2$ to $Q{x}^2$ becomes a ratio of equality (by Lemma 7, Corollary 2). And the ratio of $Q{x}^2$, or $Q{v}^2$, to $Q{T}^2$ is the ratio of $E{P}^2$ to $P{F}^2$, that is, of $C{A}^2$ to $P{F}^2$, which is the ratio of $C{D}^2$ to $C{B}^2$ (by Lemma 12). Compounding all these ratios, we find that the ratio of $L\times QR$ to $Q{T}^2$ is the same as the ratio of $AC\times L\times P{C}^2\times C{D}^2$, or $2C{B}^2\times P{C}^2\times C{D}^2$, to $PC\times Gv\times C{D}^2\times C{B}^2$, and so the same as the ratio of $2PC$ to Gv. But when P and Q come together, $2PC$ and Gv are equal. Therefore the quantities $L\times QR$ and $Q{T}^2$, which are proportional to these, are also equal. Multiply both of these equals by $\frac{S{P}^2}{QR}$, and we have $L\times S{P}^2$ equal to $\frac{S{P}^2\times Q{T}^2}{QR}$. Therefore the centripetal force is inversely proportional to $L\times S{P}^2$ (by Proposition 6, Corollaries 1 and 5), that is, inversely proportional to the square of the distance SP. Q.E.I.

The same, another way.

Find the force directed toward the center C of the hyperbola. It will be proportional to the distance CP. From this, the force directed toward the focus S will be proportional to $\frac{P{E}^3}{S{P}^2}$ (by Proposition 7, Corollary 3); and because PE is given, it is inversely proportional to $S{P}^2$. Q.E.I.

In the same way it can be shown that if the centripetal force is changed into a centrifugal force, one that pushes the body away from the focus, the body will move in the conjugate hyperbola.

Lemma 13

The latus rectum of a parabola belonging to any vertex is four times the distance of that vertex from the focus of the figure.

This is proved by the writers on the conic sections.

Lemma 14

The perpendicular dropped from the focus of a parabola onto its tangent is the mean proportional between the distance of the focus from the point of contact and its distance from the principal vertex of the figure. (Plate V, Fig. 2)

Let AP be the parabola, S its focus, A its principal vertex, P the point of contact, PO an ordinate to the principal diameter, PM the tangent meeting the principal diameter in M, and SN the perpendicular from the focus onto the tangent. Join AN. Because MS and SP are equal, MN and NP are equal, and MA and AO are equal, the straight lines AN and OP are parallel. The triangle SAN therefore has a right angle at A and is similar to the equal triangles SNM and SNP. Therefore the ratio of PS to SN is the same as the ratio of SN to SA, $S{N}^2=PS\times SA$. Q.E.D.

Corollary 1. The ratio of $P{S}^2$ to $S{N}^2$ is the same as the ratio of PS to SA.

Corollary 2. And because SA is given, $S{N}^2$ is proportional to PS.

Corollary 3. And any tangent PM meets the straight line SN, drawn from the focus perpendicular to that tangent, at a point of the straight line AN, which touches the parabola at the principal vertex.

Proposition 13, Problem 8

If a body moves along the perimeter of a parabola: it is required to find the law of the centripetal force directed toward the focus of that figure. (Plate V, Fig. 3)

Keep the construction of the preceding lemma, and let P be the body on the perimeter of the parabola. From the place Q that it reaches next, draw QR parallel to SP and QT perpendicular to it, and also Qv parallel to the tangent, meeting the diameter PG in v and the distance SP in x. Because the triangles Pxv and SPM are similar, and the sides SP and SM of the second are equal, the sides Px (or QR) and Pv of the first are also equal. But by the conic sections the square of the ordinate Qv is equal to the rectangle formed by the latus rectum and the segment Pv of the diameter, that is, to the rectangle $4PS\times Pv$ (by Lemma 13), or $4PS\times QR$. And when the points P and Q come together, the ratio of Qv to Qx becomes a ratio of equality (by Lemma 7, Corollary 2). In this case, therefore, $Q{x}^2$ becomes equal to the rectangle $4PS\times QR$.

But because the triangles QxT and SPN are similar, the ratio of $Q{x}^2$ to $Q{T}^2$ is the ratio of $P{S}^2$ to $S{N}^2$, which is the ratio of PS to SA (by Lemma 14, Corollary 1), and so the ratio of $4PS\times QR$ to $4SA\times QR$. Therefore $Q{T}^2$ and $4SA\times QR$ are equal (by Euclid's Elements, Book 5, Proposition 9). Multiply these equals by $\frac{S{P}^2}{QR}$, and $\frac{S{P}^2\times Q{T}^2}{QR}$ becomes equal to $S{P}^2\times 4SA$. Therefore the centripetal force is inversely proportional to $S{P}^2\times 4SA$ (by Proposition 6, Corollaries 1 and 5); and because $4SA$ is given, it is inversely proportional to the square of the distance SP. Q.E.I.

Corollary 1. From the last three propositions it follows that if any body P leaves the place P with any speed, in the direction of any straight line PR, and is at the same time driven by a centripetal force inversely proportional to the square of the distance of its places from the center, the body will move in one of the conic sections, with its focus at the center of force; and the reverse also holds. For when the focus, the point of contact and the position of the tangent are given, a conic section can be drawn that has a given curvature at that point. But the curvature is fixed by the given centripetal force and the given speed of the body; and two different orbits that touch each other cannot be traced by the same centripetal force with the same speed.

Corollary 2. Suppose the speed with which the body leaves its place P is such that in some very small moment of time it would carry the body along the short line PR, and the centripetal force is such that in the same time it would move the body through the distance QR. Then the body will move in one of the conic sections, whose principal latus rectum is the limiting value of $\frac{Q{T}^2}{QR}$ as the short lines PR and QR are made smaller without limit. In these corollaries I count the circle as an ellipse, and I except the case in which the body falls straight down to the center.

Corollary 1 states the converse of the three propositions: under an inverse-square force every orbit is a conic section with its focus at the center of force, and Proposition 17 constructs the orbit from the starting place, direction and speed.

Proposition 14, Theorem 6

If several bodies revolve about one common center, and the centripetal force is inversely proportional to the square of the distance of their places from the center, then the principal latera recta of their orbits are proportional to the squares of the areas that the bodies, by radii drawn to the center, sweep out in the same time. (Plate VI, Fig. 1)

The latus rectum L is equal to the limiting value of $\frac{Q{T}^2}{QR}$ when the points P and Q come together (by Proposition 13, Corollary 2). But in a given time the short line QR is proportional to the centripetal force that produces it, that is, by hypothesis, inversely proportional to $S{P}^2$. Therefore $\frac{Q{T}^2}{QR}$ is proportional to $Q{T}^2\times S{P}^2$; that is, the latus rectum L is proportional to the square of the area $QT\times SP$. Q.E.D.

Corollary. Hence the whole area of the ellipse, and the product of the axes, which is proportional to it, is proportional to the square root of the latus rectum multiplied by the periodic time. For the whole area is proportional to the area $QT\times SP$ swept out in a given time, multiplied by the periodic time.

Proposition 15, Theorem 7

Under the same assumptions, the periodic times in ellipses are proportional to the 3/2 power of their greater axes.

This is the elliptical case of the 3/2-power rule of Proposition 4, Corollary 6, and the form of Kepler's third law that Book Three applies to the planets.

For the lesser axis is the mean proportional between the greater axis and the latus rectum, and therefore the product of the axes is proportional to the square root of the latus rectum multiplied by the 3/2 power of the greater axis. But this product is also proportional to the square root of the latus rectum multiplied by the periodic time (by Proposition 14, Corollary). Take away from both sides the factor of the square root of the latus rectum, and what remains is the 3/2 power of the greater axis, proportional to the periodic time: with a the semi-major axis, $T\propto a^{3/2}$, or $T^2\propto a^3$. Q.E.D.

Corollary. Therefore the periodic times in ellipses are the same as in circles whose diameters are equal to the greater axes of the ellipses.

Proposition 16, Theorem 8

Under the same assumptions, let straight lines be drawn touching the orbits at the bodies, and let perpendiculars be dropped onto these tangents from the common focus. Then the speeds of the bodies are inversely proportional to the perpendiculars and proportional to the square roots of the principal latera recta. (Plate VI, Fig. 2)

From the focus S draw SY perpendicular to the tangent PR. The speed of the body P is then inversely proportional to the square root of the quantity $\frac{S{Y}^2}{L}$. For that speed is proportional to the very small arc PQ traveled in a given moment of time, that is, to the tangent PR (by Lemma 7). Since the ratio of PR to QT is the same as the ratio of SP to SY, the speed is proportional to $\frac{SP\times QT}{SY}$, which is $SP\times QT$ divided by SY. But $SP\times QT$ is proportional to the area swept out in the given time, that is, to the square root of the latus rectum (by Proposition 14). Hence $v\propto \frac{\sqrt{L}}{SY}$. Q.E.D.

Corollary 1. The principal latera recta are proportional to the squares of the perpendiculars multiplied by the squares of the speeds.

Corollary 2. The speeds of bodies at their greatest and least distances from the common focus are inversely proportional to those distances and proportional to the square roots of the principal latera recta. For at those places the perpendiculars are the distances themselves.

Corollary 3. And therefore the speed in a conic section at its greatest or least distance from the focus, divided by the speed in a circle at the same distance from the center, equals the square root of the principal latus rectum divided by the square root of twice that distance.

Corollary 4. The speeds of bodies revolving in ellipses, at their mean distances from the common focus, are the same as those of bodies revolving in circles at the same distances; that is, they are inversely proportional to the square roots of the distances (by Proposition 4, Corollary 6).

For the perpendiculars are now the lesser semi-axes, and these are proportional to the mean proportionals between the distances and the latera recta, $\sqrt{a\times L}$. The speed is proportional to the square root of the latus rectum divided by the perpendicular (by this proposition), and dividing $\sqrt{L}$ by $\sqrt{a\times L}$ leaves the speeds inversely proportional to the square roots of the distances, $v\propto 1/\sqrt{a}$.

Corollary 5. In the same figure, or even in different figures whose principal latera recta are equal, the speed of a body is inversely proportional to the perpendicular dropped from the focus onto the tangent.

Corollary 6. In a parabola the speed is inversely proportional to the square root of the distance of the body from the focus of the figure; in the ellipse it varies more than this rule gives, and in the hyperbola less. For the perpendicular dropped from the focus onto the tangent of a parabola is proportional to the square root of the distance (by Lemma 14, Corollary 2). In the hyperbola the perpendicular varies less than this, and in the ellipse more.

Corollary 7. In a parabola the speed of a body at any distance from the focus is $\sqrt{2}$ times the speed of a body revolving in a circle at the same distance from the center; in the ellipse the ratio is less than this, and in the hyperbola greater. For the speed at the vertex of a parabola is in this ratio (by Corollary 2 of this proposition), and the same ratio holds at all distances (by Corollary 6 of this proposition and by Proposition 4). Hence also, in a parabola the speed everywhere equals the speed of a body revolving in a circle at half the distance; in the ellipse it is less, and in the hyperbola greater.

The parabolic speed of Corollary 7, the circular speed multiplied by $\sqrt{2}$, is what is now called the escape speed: a body moving faster than this at a given distance follows a hyperbola and does not return.

Corollary 8. Compare the speed of a body revolving in any conic section with the speed of a body revolving in a circle at a distance of half the principal latus rectum of the section: the ratio of the first to the second is the ratio of that distance to the perpendicular dropped from the focus onto the tangent of the section. This is clear from Corollary 5.

Corollary 9. The speeds of bodies revolving in two circles are inversely proportional to the square roots of their distances (by Proposition 4, Corollary 6). Combining this with Corollary 8, the speed of a body revolving in a conic section, divided by the speed of a body revolving in a circle at the same distance, equals the mean proportional between that common distance and half the principal latus rectum of the section, divided by the perpendicular dropped from the common focus onto the tangent of the section.

Proposition 17, Problem 9

Suppose the centripetal force is inversely proportional to the squares of the distances of places from the center, and the absolute quantity of that force is known. It is required to determine the line that a body will trace when it is let go from a given place, with a given speed, in the direction of a given straight line.

Proposition 17 runs the problem the other way: the force is given, and the orbit is to be found from the starting place, direction and speed.

Let the centripetal force directed toward the point S be such as will make the body p revolve in some given orbit pq (Plate VI, Fig. 3), and suppose that the speed of this body at the place p is known. Now suppose the body P is let go from the place P with a given speed in the direction of the line PR, and is at once turned aside from that straight line into the conic section PQ by the centripetal force. The straight line PR will therefore touch the conic section at P. Suppose likewise that the straight line pr touches the orbit pq at p. If perpendiculars are dropped from S onto these tangents, the ratio of the principal latus rectum of the conic section to that of the orbit is the ratio of the squares of the perpendiculars multiplied by the ratio of the squares of the speeds (by Proposition 16, Corollary 1), and so this latus rectum is given. Let this latus rectum be L. The focus S of the conic section is also given. Let the angle RPH be the supplement of the angle RPS, so that the two together make two right angles; then the line PH, on which the other focus H lies, is given in position.

Drop SK perpendicular to PH, and erect the conjugate semi-axis BC. In what follows KPH stands for the product of KP and PH, SPH for the product of SP and PH, and $\overline{SP+P{H}^2}$ for the square of $SP+PH$, that is, ${(SP+PH)}^2$. This done, we have $S{P}^2-2KPH+P{H}^2=S{H}^2$ = $4C{H}^2$ = $4B{H}^2-4B{C}^2$ = $\overline{SP+P{H}^2}-L\times \overline{SP+PH}$ = $S{P}^2+2SPH+P{H}^2-L\times \overline{SP+PH}$. Add $2KPH-S{P}^2-P{H}^2+L\times \overline{SP+PH}$ to both sides, and we have $L\times \overline{SP+PH}$ = $2SPH+2KPH$; in other words, the ratio of $SP+PH$ to PH is the same as the ratio of $2SP+2KP$ to L. From this, PH is given both in length and in position.

That is, if the speed of the body at P is such that the latus rectum L is less than $2SP+2KP$, PH will lie on the same side of the tangent PR as the line SP, and so the figure will be an ellipse, which is also given, since its foci S and H and its principal axis $SP+PH$ are given. But if the speed of the body is so great that the latus rectum L becomes equal to $2PS+2KP$, the length PH will be infinite, and so the figure will be a parabola, with its axis SH parallel to the line PK, and it is given by this. And if the body leaves its place P with a still greater speed, the length PH must be taken on the other side of the tangent; then the tangent passes between the foci, and the figure will be a hyperbola, with its principal axis equal to the difference of the lines SP and PH, and so it is given. For if the body, in these cases, revolves in a conic section found in this way, it has been proved in Propositions 11, 12 and 13 that the centripetal force will be inversely proportional to the square of the distance of the body from the center of force S. We have therefore correctly determined the line PQ that a body let go from a given place P, with a given speed, in the direction of the straight line PR given in position, would trace under such a force. Q.E.F.

Corollary 1. Hence in every conic section, when the principal vertex D, the latus rectum L and the focus S are given, the other focus H is given, by making the ratio of DH to DS equal to the ratio of the latus rectum to the difference between the latus rectum and $4DS$. For when P is at the vertex D, the proportion of the proposition (the ratio of $SP+PH$ to PH equals the ratio of $2PS+2KP$ to L) becomes: the ratio of $DS+DH$ to DH equals the ratio of $4DS$ to L. Subtracting the second term from the first on each side, the ratio of DS to DH equals the ratio of $4DS-L$ to L.

Corollary 2. If the speed of a body at the principal vertex D is given, the orbit can therefore easily be found, namely, by making the ratio of its latus rectum to twice the distance DS equal to the square of the ratio of this given speed to the speed of a body revolving in a circle at the distance DS (by Proposition 16, Corollary 3), and then making the ratio of DH to DS equal to the ratio of the latus rectum to the difference between the latus rectum and $4DS$.

Corollary 3. Hence also, if a body moves in any conic section and is forced out of its orbit by any push, you can find the orbit in which it will continue afterward. For by compounding the body's own motion with the motion that the push alone would produce, you will have the motion with which the body leaves a given place of the push, in the direction of a straight line given in position.

Corollary 4. And if the body is continually disturbed by the action of some outside force, we can know its course approximately, by collecting the changes that this force makes at certain points, and estimating the continual changes it will undergo at the places in between from the pattern that appears in the progress of the series.

Scholium

Suppose a body P (Plate VI, Fig. 4), driven by a centripetal force directed toward any given point R, moves along the perimeter of any given conic section whose center is C, and the law of the centripetal force is required. Draw CG parallel to the radius RP, meeting the tangent PG of the orbit in G. The force required will be proportional to $\frac{C{G}^3}{R{P}^2}$ (by Proposition 10, Corollary 1 and its Scholium, and Proposition 7, Corollary 3).

Section 4: Finding Elliptical, Parabolic and Hyperbolic Orbits from a Given Focus

Section 3 found the force that holds a body in a conic orbit about a focus. This section and the next turn to pure geometry and the reverse task: drawing the conic itself from data an astronomer could have, here a focus, such as the Sun, together with points the orbit passes through or straight lines it touches.

Lemma 15

Suppose that from the two foci S and H of any ellipse or hyperbola we draw the straight lines SV and HV to any third point V, so that HV is equal to the principal axis of the figure (the axis on which the foci lie), and that SV is bisected at T by the perpendicular TR dropped onto it. Then that perpendicular TR touches the conic section somewhere; and conversely, if it does touch it, HV is equal to the principal axis of the figure. (Plate VII, Fig. 1)

Let the perpendicular TR cut the straight line HV, extended if necessary, at R, and join SR. Because TS and TV are equal, the straight lines SR and VR are equal, and so are the angles TRS and TRV. The point R therefore lies on the conic section, since its distances SR and HR from the two foci add up to HV, the principal axis (in the hyperbola, their difference does); and the perpendicular TR touches the conic section there. The reverse also holds. Q.E.D.

Proposition 18, Problem 10

Given a focus and the principal axes, to trace elliptical and hyperbolic trajectories that pass through given points and touch straight lines given in position. (Plate VII, Fig. 2)

Let S be the common focus of the figures, AB the length of the principal axis of any one trajectory, P a point through which the trajectory must pass, and TR a straight line it must touch. With center P and radius $AB-SP$ if the orbit is an ellipse, or $AB+SP$ if it is a hyperbola, draw the circle HG. Drop the perpendicular ST onto the tangent TR and extend it to V, so that TV is equal to ST; then with center V and radius AB draw the circle FH. In this way, whether we are given two points P and p, or two tangents TR and tr, or a point P and a tangent TR, we draw two circles. Let H be the point where they cross, and with the foci S and H and the given axis, trace the trajectory. I say that this trajectory is the one required. Because $PH+SP$ in the ellipse, and $PH-SP$ in the hyperbola, is equal to the axis, the trajectory traced passes through the point P; and it touches the straight line TR (by Lemma 15). By the same argument it passes through the two points P and p, or touches the two straight lines TR and tr. Q.E.F.

Proposition 19, Problem 11

About a given focus, to trace a parabolic trajectory that passes through given points and touches straight lines given in position. (Plate VII, Fig. 3)

Let S be the focus, P a point, and TR a tangent of the trajectory to be traced. With center P and radius PS, draw the circle FG. From the focus drop ST perpendicular onto the tangent, and extend it to V, so that TV is equal to ST. In the same way, draw a second circle fg if a second point p is given, or find a second point v if a second tangent tr is given. Then draw the straight line IF so that it touches the two circles FG and fg if two points P and p are given; or passes through the two points V and v if two tangents TR and tr are given; or touches the circle FG and passes through the point V if the point P and the tangent TR are given. Drop the perpendicular SI onto FI and bisect it at K, and trace a parabola with the axis SK and the principal vertex K. I say that this parabola is the one required. Because SK is equal to IK, and SP to FP, it passes through the point P, every point of a parabola being as far from its focus as from the line IF. And because ST is equal to TV and STR is a right angle, it touches the straight line TR (by Lemma 14, Corollary 3). Q.E.F.⁵⁸

Proposition 20, Problem 12

About a given focus, to trace any trajectory fixed in shape that passes through given points and touches straight lines given in position.

Case 1. About the focus S (Plate VII, Fig. 4) we are to trace a trajectory ABC that passes through two points B and C. Because the trajectory is fixed in shape, the ratio of its principal axis to the distance between its foci is given. Take KB in that ratio to BS, and LC in the same ratio to CS. With centers B and C and radii BK and CL, draw two circles, and onto the straight line KL that touches them at K and L, drop the perpendicular SG. Cut SG at A and a so that the ratio of GA to AS, and of Ga to aS, equals that of KB to BS; and with the axis Aa and the vertices A and a, trace a trajectory. I say that this trajectory is the one required. Let H be the other focus of the figure traced. Since the ratio of GA to AS is the same as that of Ga to aS, subtracting the terms gives $Ga-GA$, or Aa, in the same ratio to $aS-AS$, or SH. This is the ratio that the principal axis of the figure required has to the distance between its foci, and so the figure traced has the same shape as the figure required. And since KB has the same ratio to BS as LC has to CS, this figure passes through the points B and C, as is clear from the properties of the conic sections.

Case 2. About the focus S (Plate VII, Fig. 5) we are to trace a trajectory that touches two straight lines TR and tr somewhere. From the focus drop the perpendiculars ST and St onto those tangents, and extend them to V and v, so that TV and tv are equal to TS and tS. Bisect Vv at O, and erect the perpendicular OH, of unlimited length. Cut the straight line VS, extended without limit, at K and k, so that the ratio of VK to KS, and of Vk to kS, equals the ratio of the principal axis of the trajectory required to the distance between its foci. On the diameter Kk draw a circle cutting OH at H, and with the foci S and H and a principal axis equal to VH, trace a trajectory. I say that this trajectory is the one required. Bisect Kk at X, and join HX, HS, HV and Hv. The ratio of VK to KS is the same as that of Vk to kS; adding the terms, it is the same as that of $VK+Vk$ to $KS+kS$, and subtracting them, as that of $Vk-VK$ to $kS-KS$. That is, it equals the ratio of twice VX to twice KX, and also that of twice KX to twice SX; and therefore, since HX is equal to KX, the ratio of VX to HX equals that of HX to SX, and the triangles VXH and HXS are similar. Hence the ratio of VH to SH equals that of VX to XH, and therefore that of VK to KS. Thus VH, the principal axis of the trajectory traced, has the same ratio to SH, the distance between its foci, as the principal axis of the trajectory required has to the distance between its foci, and the two have the same shape. And since VH and vH are each equal to the principal axis, and VS and vS are bisected at right angles by the straight lines TR and tr, it is clear that those lines touch the trajectory traced (by Lemma 15). Q.E.F.

Case 3. About the focus S (Plate VII, Fig. 6) we are to trace a trajectory that touches a straight line TR at a given point R. Drop the perpendicular ST onto the straight line TR, and extend it to V, so that TV is equal to ST. Join VR, and cut the straight line VS, extended without limit, at K and k, so that the ratio of VK to SK, and of Vk to Sk, equals the ratio of the principal axis of the ellipse required to the distance between its foci. Draw a circle on the diameter Kk, cutting the straight line VR, extended, at H; then with the foci S and H and a principal axis equal to VH, trace a trajectory. I say that this trajectory is the one required. The ratio of VH to SH equals that of VK to SK, as was shown in Case 2, and so equals the ratio of the principal axis of the trajectory required to the distance between its foci; the trajectory traced therefore has the same shape as the one required. And it is certain from the properties of the conic sections that the straight line TR, which bisects the angle VRS, touches the trajectory at the point R. Q.E.F.

Case 4. About the focus S (Plate VII, Fig. 7) we are to trace a trajectory APB that touches a straight line TR, passes through any given point P off the tangent, and is similar to the figure apb, drawn with the principal axis ab and the foci s and h. Drop the perpendicular ST onto the tangent TR, and extend it to V, so that TV is equal to ST. Make the angles hsq and shq equal to the angles VSP and SVP; then, with center q and a radius whose ratio to ab equals that of SP to VS, draw a circle cutting the figure apb at p. Join sp, and draw SH so that the ratio of SH to sh equals that of SP to sp, the angle PSH equals the angle psh, and the angle VSH equals the angle psq. Then with the foci S and H and a principal axis AB equal to the distance VH, trace a conic section. I say that this conic section is the one required. Draw sv so that the ratio of sv to sp equals that of sh to sq, the angle vsp equals the angle hsq, and the angle vsh equals the angle psq. The triangles svh and spq are then similar, and so the ratio of vh to pq equals that of sh to sq; and this, because the triangles VSP and hsq are similar, equals the ratio of VS to SP, or of ab to pq. Therefore vh and ab are equal. Next, because the triangles VSH and vsh are similar, the ratio of VH to SH equals that of vh to sh. That is, the axis of the conic section just traced has the same ratio to the distance between its foci as the axis ab has to the distance between the foci s and h, and so the figure just traced is similar to the figure apb. Because the triangle PSH is similar to the triangle psh, this figure passes through the point P; and because VH is equal to its axis, and VS is bisected at right angles by the straight line TR, the figure touches the straight line TR. Q.E.F.

Lemma 16

From three given points, to draw to a fourth point, which is not given, three straight lines whose differences are either given or zero.

Case 1. Let the given points be A, B and C (Plate VIII, Fig. 1), and let Z be the fourth point, which we are to find. Because the difference of the lines AZ and BZ is given, the point Z lies on a hyperbola whose foci are A and B and whose principal axis is that given difference. Let MN be that axis. Take PM so that the ratio of PM to MA equals that of MN to AB, erect PR perpendicular to AB, and drop ZR perpendicular to PR. Then, from the nature of the hyperbola (PR being its directrix on the side of A), the ratio of ZR to AZ equals that of MN to AB. By the same argument, Z lies on a second hyperbola, whose foci are A and C and whose principal axis is the difference between AZ and CZ; and a line QS perpendicular to AC can be drawn such that, if a perpendicular ZS is dropped onto it from any point Z of this hyperbola, the ratio of ZS to AZ equals the ratio of the difference between AZ and CZ to AC. The ratios of ZR and of ZS to AZ are therefore given, and so is the ratio of ZR to ZS. If, then, the straight lines RP and SQ meet at T, and TZ and TA are drawn, the figure TRZS is fixed in shape, and the straight line TZ, somewhere on which the point Z lies, is given in position. The straight line TA and the angle ATZ are given as well; and because the ratios of AZ and of TZ to ZS are given, their ratio to each other is given too. From this the triangle ATZ, whose vertex is the point Z, is given. Q.E.I.

Case 2. If two of the three lines, for example AZ and BZ, are equal, draw the straight line TZ so that it bisects the straight line AB; then find the triangle ATZ as above. Q.E.I.

Case 3. If all three are equal, the point Z lies at the center of a circle that passes through the points A, B and C. Q.E.I.

This lemma, which is a problem in the form of a lemma, is also solved in Apollonius's book On Tangencies, as restored by Viète.⁵⁹

Proposition 21, Problem 13

About a given focus, to trace a trajectory that passes through given points and touches straight lines given in position.

Let the focus S (Plate VIII, Fig. 2), the point P and the tangent TR be given, and suppose that the other focus H is to be found. Drop the perpendicular ST onto the tangent and extend it to Y, so that TY is equal to ST; then YH is equal to the principal axis. Join SP and HP; then SP is the difference between HP and the principal axis. In this way, if more tangents TR or more points P are given, we always obtain as many lines YH or PH, drawn from those points Y or P to the focus H, which either are equal to the axis or differ from it by given lengths SP; and which are therefore either equal to one another or have given differences. From this the other focus H is given (by Lemma 16). And once the foci and the length of the axis are known (the axis being either YH or, if the trajectory is an ellipse, $PH+SP$, or $PH-SP$ if it is a hyperbola), the trajectory is given. Q.E.I.⁶⁰

Scholium

When the trajectory is a hyperbola, I do not include its conjugate hyperbola under the name of this trajectory. A body going on in a continuous motion can never pass out of one hyperbola into its conjugate.⁶¹

The case in which three points are given is more easily solved as follows. Let B, C and D (Plate VIII, Fig. 3) be the given points. Join BC and CD and extend them to E and F, so that the ratio of EB to EC equals that of SB to SC, and the ratio of FC to FD equals that of SC to SD. Draw EF and extend it, and drop onto it the perpendiculars SG and BH; and on GS, extended without limit, take A and a so that the ratio of GA to AS, and of Ga to aS, equals that of HB to BS. Then A is the vertex and Aa the principal axis of the trajectory. The trajectory is an ellipse, a parabola or a hyperbola depending on whether GA is greater than, equal to, or less than AS. In the first case the point a falls on the same side of the line GF as the point A; in the second it goes off to an infinite distance; in the third it falls on the other side of the line GF. For if the perpendiculars CI and DK are dropped onto GF, the ratio of IC to HB equals that of EC to EB, that is, of SC to SB; and, exchanging the middle terms, the ratio of IC to SC equals that of HB to SB, or of GA to SA. By the same argument, KD has the same ratio to SD. The points B, C and D therefore lie on a conic section traced about the focus S, in such a way that the straight lines drawn from the focus S to all the points of the section, and the perpendiculars dropped from the same points onto the straight line GF, are in that given ratio.

That excellent geometer Monsieur De la Hire has solved this problem in much the same way in his Conics, Book 8, Proposition 25.⁶²

Section 5: Finding Orbits When Neither Focus Is Given

This Section constructs conic sections from given points and given tangents when neither focus is known. Lemmas 17 to 21 are the pure geometry of conics that the constructions need, and Proposition 22 draws the conic through five given points.

Lemma 17

Take any quadrilateral ABDC inscribed in a given conic section, and extend its four sides AB, CD, AC, DB as far as needed. From any point P of the conic, draw four straight lines PQ, PR, PS, PT, one to each side, each at a given angle to its side. Then the product $PQ\times PR$ of the two lines drawn to the opposite sides AB, CD is in a given ratio to the product $PS\times PT$ of the two lines drawn to the other two opposite sides AC, BD.

Case 1. Suppose first that the lines drawn to one pair of opposite sides are parallel to one of the other sides: PQ and PR parallel to the side AC (Plate VIII, Fig. 4), and PS and PT parallel to the side AB. Suppose also that one pair of opposite sides, AC and BD, are parallel to each other. The straight line that bisects these parallel sides is then one of the diameters of the conic section, since a diameter bisects every chord parallel to a given one, and it also bisects RQ. Let O be the midpoint of RQ. Then PO is an ordinate to that diameter. Extend PO to K so that OK equals PO, and OK is the ordinate on the other side of the diameter, so that K also lies on the conic. Since the points A, B, P and K lie on the conic section, and PK crosses AB at a given angle, the product PQ times QK, $PQ\times QK$, is in a given ratio to the product AQ times QB, $AQ\times QB$ (by Apollonius's Conics, Book 3, Propositions 17, 19, 21 and 23). But QK and PR are equal, because QK is the difference of OK and OQ, and PR is the difference of OP and OR, and the lines in each pair are equal. The product PQ times QK is then equal to $PQ\times PR$, and therefore $PQ\times PR$ is in a given ratio to the product AQ times QB, that is, to $PS\times PT$, since PS equals AQ and PT equals QB. Q.E.D.⁶³

Case 2. Suppose next that the opposite sides AC and BD of the quadrilateral are not parallel (Plate VIII, Fig. 5). Draw Bd parallel to AC, meeting the straight line ST at t and the conic section at d. Join Cd, cutting PQ at r, and draw DM parallel to PQ, cutting Cd at M and AB at N. Because the triangles BTt and DBN are similar, the ratio of Bt, which equals PQ, to Tt is the same as the ratio of DN to NB. In the same way, the ratio of Rr to AQ, which equals PS, is the same as the ratio of DM to AN. Multiply these two proportions together, first terms by first terms and second terms by second terms: the ratio of $PQ\times Rr$ to $PS\times Tt$ equals the ratio of the product ND times DM to the product AN times NB, $\frac{PQ\times Rr}{PS\times Tt}=\frac{ND\times DM}{AN\times NB}$. The quadrilateral ABdC has its sides AC and Bd parallel, so the ratio of $PQ\times Pr$ to $PS\times Pt$ is that same ratio (by Case 1). Subtracting, since PR is the difference of Pr and Rr, and PT the difference of Pt and Tt, the ratio of $PQ\times PR$ to $PS\times PT$ is the same as well, and so it is given. Q.E.D.

Case 3. Suppose, last, that the four lines PQ, PR, PS, PT are not parallel to the sides AC and AB, but slope to them at any angles (Plate VIII, Fig. 6). In their place draw Pq and Pr parallel to AC, and Ps and Pt parallel to AB. Because the angles of the triangles PQq, PRr, PSs and PTt are given, the ratios of PQ to Pq, PR to Pr, PS to Ps and PT to Pt are also given. Therefore the ratio of $PQ\times PR$ to $Pq\times Pr$ is given, and so is the ratio of $PS\times PT$ to $Ps\times Pt$. But by what has already been proved, the ratio of $Pq\times Pr$ to $Ps\times Pt$ is given, and therefore so is the ratio of $PQ\times PR$ to $PS\times PT$. Q.E.D.

Lemma 18

With the same things supposed, if the product $PQ\times PR$ of the lines drawn to two opposite sides of the quadrilateral is in a given ratio to the product $PS\times PT$ of the lines drawn to the other two sides, then the point P from which the lines are drawn lies on a conic section drawn through the four corners of the quadrilateral. (Plate VIII, Fig. 7)

Imagine a conic section drawn through the points A, B, C, D and through one of the infinitely many points P, say p. I say that the point P always lies on this conic. Suppose it does not. Join AP, and let it cut the conic at some point other than P, say b. From the points p and b draw the straight lines pq, pr, ps, pt and bk, bn, bf, bd to the sides of the quadrilateral at the given angles. Then the ratio of $bk\times bn$ to $bf\times bd$ equals the ratio of $pq\times pr$ to $ps\times pt$ (by Lemma 17), and this in turn equals the ratio of $PQ\times PR$ to $PS\times PT$, by the supposition. Because the quadrilaterals bkAf and PQAS are similar, the ratio of bk to bf equals the ratio of PQ to PS. Dividing the terms of the first proportion by the corresponding terms of this one, the ratio of bn to bd equals the ratio of PR to PT. The quadrilaterals Dnbd and DRPT, whose angles are equal, are therefore similar, and since they share the corner D, their diagonals Db and DP lie along one line. Hence b lies where the straight lines AP and DP meet, and therefore coincides with the point P. Wherever the point P is taken, then, it lies on the conic section assigned. Q.E.D.

Corollary. Suppose three straight lines PQ, PR, PS are drawn from a common point P to three other straight lines AB, CD, AC fixed in position, one to each, at three given angles, and suppose that the product $PQ\times PR$ of two of the lines drawn is in a given ratio to the square $P{S}^2$ of the third. Then the point P lies on a conic section that touches the lines AB and CD at A and C; and the reverse also holds. For let the three straight lines AB, CD, AC keep their positions, and let the line BD move up to the line AC until it coincides with it; let the line PT likewise come to coincide with the line PS. The product $PS\times PT$ then becomes $P{S}^2$, and the straight lines AB and CD, which used to cut the curve at the points A and B, and C and D, can no longer cut it at those points, which have now merged, but only touch it there.

Scholium

In this lemma the term "conic section" is to be taken in a broad sense, including the straight-line section through the vertex of the cone as well as the circular section parallel to its base. If the point p happens to lie on a straight line joining the points A and D, or C and B, the conic section turns into two straight lines: one is the line on which the point p falls, and the other is the line joining the other two of the four points. If the two opposite angles of the quadrilateral together make two right angles, and the four lines PQ, PR, PS, PT are drawn to its sides at right angles, or at any other equal angles, and the product $PQ\times PR$ of two of the lines drawn, PQ and PR, equals the product $PS\times PT$ of the other two, PS and PT, then the conic section becomes a circle. The same happens if the four lines are drawn at any angles, and the ratio of the product $PQ\times PR$ of one pair to the product $PS\times PT$ of the other pair equals the ratio of the product of the sines of the angles S and T, at which the last two lines PS and PT are drawn, to the product of the sines of the angles Q and R, at which the first two, PQ and PR, are drawn: $\frac{PQ\times PR}{PS\times PT}=\frac{sinS\times sinT}{sinQ\times sinR}$. In every other case the locus of the point P is one of the three figures commonly called the conic sections. In place of the quadrilateral ABCD we may also use a four-sided figure in which two opposite sides cross each other like diagonals. And one or two of the four points A, B, C, D may be supposed to move off to an infinite distance, so that the sides of the figure that meet at those points become parallel. In that case the conic section passes through the other points and runs off in the direction of the parallels without limit.

Lemma 19

To find a point P from which four straight lines PQ, PR, PS, PT, drawn to four other straight lines AB, CD, AC, BD fixed in position, one to each, at given angles, make the product $PQ\times PR$ of any two of the lines drawn stand in a given ratio to the product $PS\times PT$ of the other two. (Plate VIII, Fig. 8)

Suppose the lines AB and CD, to which the two straight lines PQ and PR forming one of the products are drawn, meet the other two lines fixed in position at the points A, B, C, D. From one of these, say A, draw any straight line AH on which you want to find the point P. Let it cut the opposite lines BD and CD at H and I. Because all the angles of the figure are given, the ratios of PQ to PA and of PA to PS are given, and therefore so is the ratio of PQ to PS. Dividing the given ratio of $PQ\times PR$ to $PS\times PT$ by this ratio, the ratio of PR to PT is given. Multiplying by the given ratios of PI to PR and of PT to PH, the ratio of PI to PH is given, and therefore the point P is given. Q.E.I.

Corollary 1. A tangent can also be drawn in this way at any point D of the locus of all the points P. When the points P and D come together, that is, when AH is drawn through the point D, the chord PD becomes a tangent. In that case the limiting ratio of the vanishing lines IP and PH is found as above. Draw CF parallel to AD, meeting BD at F, and divide it at E in that same limiting ratio; then DE is the tangent, because CF and the vanishing line IH are parallel, and are divided in the same ratio at E and P.

Corollary 2. The locus of all the points P can also be determined in this way. Through any one of the points A, B, C, D, say A (Plate IX, Fig. 1), draw AE touching the locus, and through any other point, say B, draw BF parallel to the tangent and meeting the locus at F; the point F is found by this lemma. Bisect BF at G, and draw the straight line AG, extended without limit. This is the position of the diameter to which BG and FG are ordinates. Let AG meet the locus again at H. Then AH is its diameter, the transverse diameter (latus transversum), and the ratio of the latus rectum belonging to it to AH is the same as the ratio of $B{G}^2$ to $AG\times GH$. If AG never meets the locus again, so that the line AH is infinite, the locus is a parabola, and its latus rectum belonging to the diameter AG is $\frac{B{G}^2}{AG}$. But if AG does meet the locus again somewhere, the locus is a hyperbola when the points A and H lie on the same side of the point G, and an ellipse when the point G falls between the points A and H; unless the angle AGB is a right angle and at the same time $B{G}^2$ equals the product AG times GH, in which case the locus is a circle.

In this corollary, then, we have given a solution of the famous problem of the ancients concerning four lines, which Euclid began and Apollonius carried on; and it is not an analytical calculation but a geometrical construction, of the kind the ancients required.⁶⁴

Lemma 20

Let ASPQ be any parallelogram whose two opposite corners A and P lie on a conic section, and let the sides AQ and AS of one of these angles, extended without limit, meet the same conic section at B and C. From the meeting points B and C draw two straight lines BD and CD to any fifth point D of the conic section, meeting the other two sides PS and PQ of the parallelogram, extended without limit, at T and R. Then the parts PR and PT cut off from those sides are always in a given ratio to each other. And the reverse: if the parts cut off are in a given ratio to each other, the locus of the point D is a conic section passing through the four points A, B, C, P. (Plate IX, Fig. 2)

Case 1. Join BP and CP, and from the point D draw two straight lines DG and DE. The first, DG, is parallel to AB and meets PB, PQ and CA at H, I and G; the other, DE, is parallel to AC and meets PC, PS and AB at F, K and E. The ratio of the product $DE\times DF$ to the product $DG\times DH$ is then given (by Lemma 17). But the ratio of PQ to DE, which equals IQ, is the same as the ratio of PB to HB, and so the same as the ratio of PT to DH; exchanging the middle terms, the ratio of PQ to PT equals the ratio of DE to DH. Likewise the ratio of PR to DF is the same as the ratio of RC to DC, and so the same as the ratio of IG, which equals PS, to DG; exchanging the middle terms, the ratio of PR to PS equals the ratio of DF to DG. Multiplying these ratios together, the ratio of the product $PQ\times PR$ to the product $PS\times PT$ equals the ratio of the product $DE\times DF$ to the product $DG\times DH$, $\frac{PQ\times PR}{PS\times PT}=\frac{DE\times DF}{DG\times DH}$, and so it is given. But PQ and PS are given, and therefore the ratio of PR to PT is given. Q.E.D.

Case 2. If instead PR and PT are supposed to be in a given ratio to each other, then running the same reasoning backward, the ratio of the product $DE\times DF$ to the product $DG\times DH$ is given, and so the point D lies on a conic section passing through the points A, B, C, P, as its locus (by Lemma 18). Q.E.D.

Corollary 1. Hence if we draw BC, cutting PQ at r, and on PT take Pt so that the ratio of Pt to Pr equals the ratio of PT to PR, then Bt touches the conic section at the point B. For suppose the point D runs into the point B, so that the chord BD vanishes and BT becomes a tangent; then CD and BT coincide with CB and Bt.

Corollary 2. Conversely, if Bt is a tangent, and the lines BD and CD meet at any point D of a conic section, the ratio of PR to PT equals the ratio of Pr to Pt. And the reverse also holds: if the ratio of PR to PT equals the ratio of Pr to Pt, then BD and CD meet at some point D of a conic section.

Corollary 3. One conic section cannot cut another conic section in more than four points. For suppose it could, and let two conic sections pass through the five points A, B, C, P, O. Let the straight line BD cut them at the points D and d, and let the straight line Cd cut the straight line PQ at q. Then the ratio of PR to PT equals the ratio of Pq to PT, since both conics pass through O and so have the same given ratio. Hence PR and Pq are equal, which contradicts the supposition that the two conics are different.

Lemma 21

Let two straight lines BM and CM, extended without limit, turn about the fixed points B and C as poles, and let their meeting point M trace a third straight line MN fixed in position. Let two other straight lines BD and CD, extended without limit, be drawn through the same points B and C, making given angles MBD and MCD with the first two. I say that the meeting point D of these two straight lines BD and CD traces a conic section passing through the points B and C. And the reverse: if the meeting point D of the straight lines BD and CD traces a conic section passing through the given points B, C, A, and the angle DBM is always equal to the given angle ABC, and the angle DCM always equal to the given angle ACB, then the point M lies on a straight line fixed in position, as its locus. (Plate IX, Fig. 3)

On the straight line MN let a point N be given, and when the moving point M falls on the fixed point N, let the moving point D fall on a fixed point P. Join CN, BN, CP and BP, and from the point P draw the straight lines PT and PR, meeting BD and CD at T and R, so that the angle BPT equals the given angle BNM and the angle CPR equals the given angle CNM. Since the angles MBD and NBP are equal by supposition, as are the angles MCD and NCP, take away the angles NBD and NCD that they have in common, and what remains is equal: the angles NBM and PBT are equal, and so are NCM and PCR. The triangles NBM and PBT are therefore similar, and so are the triangles NCM and PCR. The ratio of PT to NM therefore equals the ratio of PB to NB, and the ratio of PR to NM equals the ratio of PC to NC. But the points B, C, N, P are fixed. PT and PR each have a given ratio to NM, and hence a given ratio to each other. Therefore the point D, where the moving straight lines BT and CR continually meet, lies on a conic section passing through the points B, C, P (by Lemma 20). Q.E.D.

Conversely, suppose the moving point D lies on a conic section passing through the given points B, C, A (Plate IX, Fig. 4), and the angle DBM is always equal to the given angle ABC, and the angle DCM always equal to the given angle ACB; and suppose that when the point D falls in turn on any two fixed points p and P of the conic section, the moving point M falls in turn on two fixed points n and N. Through these points n and N draw the straight line nN: this line nN is the locus of the moving point M, wherever it goes. For suppose, if possible, that the point M lies on some curved line. Then the point D lies on a conic section passing through the five points B, C, A, p, P, while the point M always lies on a curved line. But by what was proved above, the point D also lies on a conic section passing through the same five points B, C, A, p, P when the point M always lies on a straight line. Two conic sections would then both pass through the same five given points, which contradicts Lemma 20, Corollary 3. It is therefore absurd to suppose that the point M lies on a curved line. Q.E.D.

Proposition 22, Problem 14

To draw a trajectory that passes through five given points. (Plate IX, Fig. 5)

Let the five given points be A, B, C, P, D. From any one of them, say A, draw the straight lines AB and AC to any other two, say B and C, which may be called the poles, and through the fourth point P draw the lines TPS and PRQ parallel to them. Then from the two poles B and C draw through the fifth point D two lines BDT and CRD, extended without limit, meeting the lines just drawn, TPS and PRQ (the first with the first and the second with the second), at T and R. Now draw the straight line tr parallel to TR, cutting off from the straight lines PT and PR any segments Pt and Pr proportional to PT and PR. If the straight lines Bt and Cr are drawn through their ends t and r and the poles B and C, meeting at d, that point d lies on the required trajectory. For the point d lies on a conic section passing through the four points A, B, C, P (by Lemma 20), and as the lines Rr and Tt vanish, the point d comes to coincide with the point D. Therefore the conic section passes through the five points A, B, C, P, D. Q.E.D.

The same, another way. (Plate IX, Fig. 6)

Join any three of the given points, say A, B, C. Let the angles ABC and ACB, keeping their given sizes, turn about two of these points, B and C, as poles. Lay the legs BA and CA first on the point D, then on the point P, and mark the points M and N where the other legs BL and CL cross each other in the two cases. Draw the straight line MN, extended without limit, and let the moving angles turn about their poles B and C in such a way that the crossing point of the legs BL and CL, or BM and CM, which we now call m, always falls on that line MN. Then the crossing point of the legs BA and CA, or BD and CD, which we now call d, traces the required trajectory PADdB. For the point d lies on a conic section passing through the points B and C (by Lemma 21), and when the point m comes to coincide with the points L, M, N, the point d comes, by the construction, to coincide with the points A, D, P. A conic section is therefore traced that passes through the five points A, B, C, P, D. Q.E.F.

Corollary 1. Hence a straight line can easily be drawn that touches the trajectory at any given point B. Let the point d come to coincide with the point B, and the straight line Bd becomes the required tangent.

Corollary 2. The centers, diameters and latera recta of the trajectories can also be found, as in Lemma 19, Corollary 2.

Scholium

The first of these constructions (Plate IX, Fig. 5) becomes somewhat simpler if we join BP, and on that line, extended if need be, take Bp so that the ratio of Bp to BP equals the ratio of PR to PT; then through p draw the straight line pe, extended without limit, parallel to SPT, and on it always take pe equal to Pr; and draw the straight lines Be and Cr to meet at d. For since the ratios of Pr to Pt, PR to PT, pB to PB and pe to Pt are all the same, pe and Pr are always equal. This is the quickest way of finding the points of the trajectory, unless you would rather draw the curve mechanically, as in the second construction.

Proposition 23, Problem 15

To trace a trajectory that passes through four given points and touches a straight line given in position. (Plate X, Fig. 1)

Case 1. Suppose that HB is the given tangent, B the point where it touches, and C, D and P the three other given points. Join BC; draw PS parallel to BH and PQ parallel to BC, and complete the parallelogram BSPQ. Draw BD cutting SP at T, and CD cutting PQ at R. Finally, draw any line tr parallel to TR, cutting off from PQ and PS the segments Pr and Pt, proportional to PR and PT respectively. If Cr and Bt are then drawn, the point d where they meet always lies on the trajectory to be traced (by Lemma 20).

The same, another way. (Plate X, Fig. 2)

Let the angle CBH, of a given size, turn about the pole B, and let the radius DC, a straight line extended both ways, turn about the pole C. Mark the points M and N where the leg BC of the angle cuts that radius at the moments when the other leg, BH, meets the same radius at the points P and D. Then draw the line MN, of unlimited length, and let the radius CP or CD and the leg BC of the angle always meet on this line. The point where the other leg, BH, meets the radius will then trace the trajectory required.

Suppose that in the constructions of the preceding problem, Proposition 22, the point A moves until it coincides with the point B. Then the lines CA and CB coincide, and the line AB, in its final position, becomes the tangent BH; so the constructions given there become the same as the constructions given here. The point where the leg BH meets the radius therefore traces a conic section that passes through the points C, D and P and touches the line BH at the point B. Q.E.F.

Case 2. Suppose the four given points B, C, D and P (Plate X, Fig. 3) all lie off the tangent HI. Join them in pairs by the lines BD and CP, which meet at G and cut the tangent at H and I. Cut the tangent at A so that the ratio of HA to IA is the same as the ratio of two products: the mean proportional between CG and GP times the mean proportional between BH and HD, to the mean proportional between GD and GB times the mean proportional between PI and IC. Then A is the point where the trajectory touches the tangent; in symbols, $\frac{HA}{IA}=\frac{\sqrt{CG\times GP}\times \sqrt{BH\times HD}}{\sqrt{GD\times GB}\times \sqrt{PI\times IC}}$. For suppose HX, drawn parallel to the straight line PI, cuts the trajectory at any points X and Y. By the properties of the conic sections, the point A must be so placed that the ratio of $H{A}^2$ to $A{I}^2$ is compounded of two ratios: the ratio of the product $XH\times HY$ to the product $BH\times HD$, which equals the ratio of $CG\times GP$ to $DG\times GB$; and the ratio of $BH\times HD$ to $PI\times IC$. Once the point of contact A is found, the trajectory is traced as in the first case. Q.E.F. The point A may be taken either between the points H and I or outside them, and so two different trajectories can be traced.

Proposition 24, Problem 16

To trace a trajectory that passes through three given points and touches two straight lines given in position. (Plate X, Fig. 4)

Suppose that HI and KL are the given tangents, and B, C and D the given points. Through any two of those points, say B and D, draw the straight line BD, of unlimited length, meeting the tangents at the points H and K. In the same way, through any other two of the points, say C and D, draw the straight line CD, of unlimited length, meeting the tangents at the points I and L. Cut the lines so drawn at R and S, so that the ratio of HR to KR is the same as the ratio of the mean proportional between BH and HD to the mean proportional between BK and KD; and so that the ratio of IS to LS is the same as the ratio of the mean proportional between CI and ID to the mean proportional between CL and LD. In symbols, the rule is $\frac{HR}{KR}=\frac{\sqrt{BH\times HD}}{\sqrt{BK\times KD}}$ and $\frac{IS}{LS}=\frac{\sqrt{CI\times ID}}{\sqrt{CL\times LD}}$. You may cut them as you please, either between the points K and H, and I and L, or outside them. Then draw RS cutting the tangents at A and P; A and P are the points of contact.

For suppose instead that the points of contact A and P lie anywhere else on the tangents. Through one of the points H, I, K and L that lies on either tangent, say I on the tangent HI, draw a straight line IT parallel to the other tangent KL, meeting the curve at X and Y, and on that line take IZ equal to the mean proportional between IX and IY. By the properties of the conic sections, the ratio of the product $IX\times IY$, or $I{Z}^2$, to $L{P}^2$ is the same as the ratio of the product $CI\times ID$ to the product $CL\times LD$; by the construction, this is the ratio of $S{I}^2$ to $S{L}^2$. It follows that the ratio of IZ to LP is the same as that of SI to SL, and therefore the points S, P and Z lie on one straight line. Next, since the tangents meet at G, the ratio of the product $IX\times IY$, or $I{Z}^2$, to $I{A}^2$ is the same as the ratio of $G{P}^2$ to $G{A}^2$, again by the properties of the conic sections. Consequently the ratio of IZ to IA is the same as that of GP to GA. Hence the points P, Z and A lie on one straight line, and therefore the points S, P and A lie on one straight line. The same argument proves that the points R, P and A lie on one straight line. The points of contact A and P therefore lie on the straight line RS. Once these points are found, the trajectory can be traced as in the first case of the preceding problem. Q.E.F.

In this proposition, and in Case 2 of the one before it, the constructions are the same whether or not the straight line XY cuts the trajectory at X and Y; they do not depend on that cutting. But once the constructions have been proved for the case in which that line does cut the trajectory, they are also known for the case in which it does not, and so, for brevity's sake, I give no further proof of them.

Lemma 22

To transform figures into other figures of the same kind. (Plate X, Fig. 5)

Suppose that some figure HGI is to be transformed. Draw, as you please, two parallel lines AO and BL, cutting at A and B any third line AB given in position. From any point G of the figure, draw a straight line GD parallel to OA until it meets the straight line AB. Then from some given point O on the line OA, draw the straight line OD to the point D, meeting BL at d. From that meeting point raise the straight line dg, making some given angle with the straight line BL, and having the same ratio to Od as DG has to OD. Then g is the point of the new figure hgi that corresponds to the point G. In the same way, each point of the first figure gives a corresponding point of the new figure. If we therefore imagine the point G carried by a continuous motion through all the points of the first figure, the point g will likewise be carried by a continuous motion through all the points of the new figure, and will trace it. To tell them apart, let us call DG the first ordinate, dg the new ordinate, AD the first abscissa and ad the new abscissa; O the pole, OD the cutting radius, OA the first ordinate radius, and Oa, which completes the parallelogram OABa, the new ordinate radius.

I say, then, that if the point G lies on a straight line given in position, the point g also lies on a straight line given in position. If the point G lies on a conic section, the point g also lies on a conic section; and here I count the circle as one of the conic sections. Further, if the point G lies on a line of the third analytical order, the point g also lies on a line of the third order, and so on for curves of higher orders. The two lines on which the points G and g lie are always of the same analytical order. For the ratio of ad to OA is the same as the ratios of Od to OD, of dg to DG, and of AB to AD; and therefore AD is equal to $\frac{OA\times AB}{ad}$ and DG is equal to $\frac{OA\times dg}{ad}$. Now suppose the point G lies on a straight line. Then in any equation that expresses the relation between the abscissa AD and the ordinate DG, those variable lines AD and DG rise no higher than the first power. If in this equation we write $\frac{OA\times AB}{ad}$ in place of AD, and $\frac{OA\times dg}{ad}$ in place of DG, a new equation results in which the new abscissa ad and the new ordinate dg also rise only to the first power, and which must therefore represent a straight line. But if AD and DG, or either of them, had risen to the second power in the first equation, ad and dg would likewise have risen to the second power in the second equation; and so on for the third or higher powers. The variable lines ad and dg in the second equation, and AD and DG in the first, always rise to the same power, and therefore the lines on which the points G and g lie are of the same analytical order.⁶⁵

I say further that if any straight line touches the curve in the first figure, the same straight line, carried over into the new figure in the same way as the curve, will touch the curve in the new figure, and the reverse. For suppose any two points of the curve in the first figure approach each other until they coincide. The same points, carried over, will approach each other until they coincide in the new figure, and therefore the straight lines that join them will become, together, tangents to the curves in both figures. I could have proved these claims in a more geometrical form, but I aim to be brief.

Hence, if one figure made of straight lines is to be transformed into another, we need only carry over the points where the straight lines of the first figure cross, and through the carried-over points draw straight lines in the new figure. But if a curved figure is to be transformed, we must carry over the points, the tangents and the other straight lines by which the curve is defined. This lemma is useful in solving the harder problems, because it allows us to transform the given figures, if they are complicated, into simpler ones. In this way any straight lines that converge to a point are transformed into parallels, by taking as the first ordinate radius any straight line that passes through the point where the converging lines meet. By this choice that meeting point is sent off to an infinite distance, and parallel lines are lines that tend to a point infinitely far away. Once the problem is solved in the new figure, transforming the new figure back into the first by the reverse operations gives the solution required.⁶⁶

This lemma is also useful in solving solid problems. Whenever two conic sections occur by whose intersection a problem can be solved, either of them, if it is a hyperbola or a parabola, can be transformed into an ellipse, and this ellipse can then easily be changed into a circle. In the same way, in constructing plane problems, a straight line and a conic section can be transformed into a straight line and a circle.⁶⁷

Proposition 25, Problem 17

To trace a trajectory that passes through two given points and touches three straight lines given in position. (Plate X, Fig. 6)

Through the point where any two of the tangents meet, and the point where the third tangent meets the straight line through the two given points, draw a straight line of unlimited length. Taking this line as the first ordinate radius, transform the figure into a new figure by the preceding lemma. In the new figure those two tangents become parallel to each other, and the third tangent becomes parallel to the straight line through the two given points. Suppose hi and kl are those two parallel tangents, ik the third tangent, and hl a straight line parallel to it through the points a and b through which the conic section must pass in this new figure. Complete the parallelogram hikl, and cut the straight lines hi, ik and kl at c, d and e so that the ratio of hc to the square root of the product ahb, that is, of $ah\times hb$, of ic to id, and of ke to kd, is in each case the ratio of the sum of the straight lines hi and kl to the sum of three lines: the straight line ik, and the square roots of the products ahb and alb. Then c, d and e are the points of contact, and in symbols the rule is $\frac{hc}{\sqrt{ah\times hb}}=\frac{ic}{id}=\frac{ke}{kd}=\frac{hi+kl}{ik+\sqrt{ah\times hb}+\sqrt{al\times lb}}$.

For by the properties of the conic sections, the ratios of $h{c}^2$ to the product ahb, of $i{c}^2$ to $i{d}^2$, of $k{e}^2$ to $k{d}^2$, and of $e{l}^2$ to the product alb, are all the same. Therefore the ratios of hc to the square root of ahb, of ic to id, of ke to kd, and of el to the square root of alb, are all the square root of that ratio. Adding the terms, they are all the same as the given ratio of the sum of all the first terms, which is $hi+kl$, to the sum of all the second terms, which is $\sqrt{ahb}+ik+\sqrt{alb}$. From that given ratio, then, we have the points of contact c, d and e in the new figure. Carry these points back into the first figure by the reverse operations of the last lemma, and the trajectory is traced there by Proposition 22. Q.E.F. According as the points a and b fall between the points h and l or outside them, the points c, d and e must be taken either between the points h, i, k and l or outside them. If one of the points a and b falls between the points h and l, and the other outside them, the problem is impossible.

Proposition 26, Problem 18

To trace a trajectory that passes through a given point and touches four straight lines given in position. (Plate XI, Fig. 1)

From the point where any two of the tangents cross to the point where the other two cross, draw a straight line of unlimited length. Taking this line as the first ordinate radius, transform the figure into a new figure (by Lemma 22); the two pairs of tangents, each pair of which met on the first ordinate radius, now become parallel. Let hi and kl, and ik and hl, be those pairs of parallels, completing the parallelogram hikl. Let p be the point in this new figure that corresponds to the given point in the first figure. Through O, the center of the figure, draw pq, with Oq equal to Op; then q is the other point through which the conic section must pass in this new figure, since a conic that touches all four sides of a parallelogram has its center at the center of the parallelogram. Carry this point back into the first figure by the reverse operation of Lemma 22, and there we have the two points through which the trajectory is to be traced. Through those points the trajectory can be traced by Proposition 25. Q.E.F.

Lemma 23

If two straight lines AC and BD, given in position and ending at the given points A and B, are in a given ratio to each other, and the straight line CD, which joins their other ends C and D, which are not fixed, is cut at K in a given ratio, then I say that the point K lies on a straight line given in position. (Plate XI, Fig. 2)

Let the straight lines AC and BD meet at E. On BE take BG so that the ratio of BG to AE is the ratio of BD to AC, and let FD always be equal to the given line EG. Then by the construction the ratio of EC to GD, that is, to EF, is the same as that of AC to BD, and so it is a given ratio; and therefore the triangle EFC is fixed in shape. Cut CF at L so that the ratio of CL to CF is that of CK to CD. Because that ratio is given, the triangle EFL is fixed in shape, and therefore the point L lies on the straight line EL, which is given in position. Join LK; the triangles CLK and CFD are similar. Because FD is a given line, and LK has a given ratio to FD, LK is also given. Take EH equal to it, and ELKH is always a parallelogram. The point K therefore always lies on the side HK of that parallelogram, and HK is given in position. Q.E.D.

Corollary. Because the figure EFLC is fixed in shape, the three straight lines EF, EL and EC, that is, GD, HK and EC, are in given ratios to each other.

Lemma 24

If three straight lines touch any conic section, two of them parallel and given in position, then I say that the semi-diameter of the section that is parallel to those two is a mean proportional between the segments of those two cut off between the points of contact and the third tangent. (Plate XI, Fig. 3)

Let AF and GB be the two parallels, touching the conic section ADB at A and B; let EF be the third straight line, touching the conic section at I and meeting the first two tangents at F and G; and let CD be the semi-diameter of the figure parallel to those tangents. I say that AF, CD and BG are in continued proportion: the ratio of AF to CD is the same as that of CD to BG, so that $C{D}^2=AF\times BG$.

For suppose the conjugate diameters AB and DM meet the tangent FG at E and H and cut each other at C, and complete the parallelogram IKCL. From the nature of the conic sections, the ratio of EC to CA is the same as that of CA to CL, that is, $C{A}^2=EC\times CL$. Taking differences of the terms, the ratio of $EC-CA$ to $CA-CL$, or of EA to AL, is the same ratio again, so that the ratio of EA to AL equals that of EC to CA. Adding the terms, the ratio of EA to $EA+AL$, or EL, is the same as that of EC to $EC+CA$, or EB. Therefore, because the triangles EAF, ELI, ECH and EBG are similar, the ratio of AF to LI is the same as that of CH to BG. Likewise, from the nature of the conic sections, the ratio of LI, or CK, to CD is the same as that of CD to CH, that is, $C{D}^2=CK\times CH$. And therefore, by perturbed proportion (ex æquo perturbate), the ratio of AF to CD is the same as that of CD to BG. Q.E.D.^68,69

Corollary 1. Hence, if two tangents FG and PQ meet the two parallel tangents AF and BG at F and G, and at P and Q, and cut each other at O, then by perturbed proportion the ratio of AF to BQ is the same as that of AP to BG; taking differences, it is the same as that of FP to GQ; and therefore it is the same as that of FO to OG.

Corollary 2. It follows also that the two straight lines PG and FQ, drawn through the points P and G, and F and Q, meet on the straight line ACB, which passes through the center of the figure and the points of contact A and B.

Lemma 25

If the four sides of a parallelogram, extended without limit, touch any conic section, and are cut by a fifth tangent, then I say the following. Take, on any two adjacent sides, the segments that end at opposite corners of the parallelogram. The ratio of either segment to the side from which it is cut off is the same as the ratio of the part of the other adjacent side lying between its point of contact and the third side to the other segment. (Plate XI, Fig. 4)

Let the four sides ML, IK, KL and MI of the parallelogram MLIK touch the conic section at A, B, C and D, and let the fifth tangent FQ cut those sides at F, Q, H and E. Take the segments ME and KQ of the sides MI and KI, or the segments KH and MF of the sides KL and ML. I say that the ratio of ME to MI is the same as that of BK to KQ, and the ratio of KH to KL the same as that of AM to MF. For the ratio of ME to EI is the same as that of AM, or its equal BK, to BQ (by Lemma 24, Corollary 1); and so, adding the terms, the ratio of ME to MI is the same as that of BK to KQ. Q.E.D. Likewise the ratio of KH to HL is the same as that of BK, or its equal AM, to AF; and so, taking differences, the ratio of KH to KL is the same as that of AM to MF. Q.E.D.⁷⁰

Corollary 1. If a parallelogram IKLM drawn around a given conic section is given, the product $KQ\times ME$ is given, and so is the product $KH\times MF$, which is equal to it. For because the triangles KQH and MFE are similar, those products are equal.

Corollary 2. And if a sixth tangent eq is drawn, meeting the tangents KI and MI at q and e, the product $KQ\times ME$ is equal to the product $Kq\times Me$. The ratio of KQ to Me is then the same as that of Kq to ME, and so, taking differences, the same as that of Qq to Ee.

Corollary 3. It follows also that if Eq and eQ are joined and bisected, and a straight line is drawn through the two points of bisection, this straight line passes through the center of the conic section. For since the ratio of Qq to Ee is the same as that of KQ to Me, the same straight line passes through the middle of all the lines Eq, eQ and MK (by Lemma 23), and the middle point of the straight line MK is the center of the section.⁷¹

Proposition 27, Problem 19

To trace a trajectory that touches five straight lines given in position. (Plate XI, Fig. 5)

Proposition 22 traced a conic through five given points; this one does the same for five given tangents, and the corollaries of Lemmas 24 and 25 supply the construction.

Suppose ABG, BCF, GCD, FDE and EA are the tangents given in position. Bisect at M and N the diagonals AF and BE of the quadrilateral ABFE, which is bounded by any four of them; then the straight line MN, drawn through the points of bisection, passes through the center of the trajectory (by Lemma 25, Corollary 3). Again, bisect at P and Q the diagonals (if I may call them that) BD and GF of the quadrilateral BGDF, bounded by any other four of the tangents; the straight line PQ, drawn through these points of bisection, also passes through the center of the trajectory. The center is therefore given, at the point where the two bisecting lines meet. Suppose it to be O. Parallel to any tangent BC, draw KL at such a distance that the center O lies midway between the two parallels; this KL will touch the trajectory to be traced. Let it cut any other two tangents, GCD and FDE, at L and K. Through the points C and K, and F and L, where the tangents that are not parallel, CL and FK, meet the parallel tangents CF and KL, draw CK and FL, meeting at R. The straight line OR, drawn and extended, then cuts the parallel tangents CF and KL at their points of contact (by Lemma 24, Corollary 2). By the same method the other points of contact can be found, and then the trajectory can be traced (by Proposition 22). Q.E.F.⁷²

Scholium

The propositions so far include the problems in which either the center or the asymptotes of the trajectories are given. For when points, tangents and the center are given, as many more points and as many more tangents are given at equal distances on the other side of the center. And an asymptote should be counted as a tangent, with its infinitely distant end (if we may say so) as its point of contact. Imagine the point of contact of any tangent moved away without limit, and the tangent will turn into an asymptote; the constructions of the problems above will then turn into the constructions of the problems in which the asymptote is given.

Once the trajectory is traced, we can find its axes and foci in the following way. In the construction and figure of Lemma 21 (Plate XII, Fig. 1), let the legs BP and CP of the moving angles PBN and PCN, whose meeting point traced the trajectory, be made parallel to each other; and keeping them parallel, let them turn about their poles B and C in that figure. Meanwhile let the other legs of those angles, CN and BN, trace by their meeting point K or k the circle BKGC. Let O be the center of this circle. From this center drop the perpendicular OH onto the ruler MN, the fixed straight line on which the legs CN and BN met while the trajectory was being traced, and let OH meet the circle at K and L. When those other legs CK and BK meet at the point K nearest the ruler, the first legs CP and BP will be parallel to the greater axis and perpendicular to the lesser; and the reverse happens if those legs meet at the farthest point L. Hence if the center of the trajectory is given, the axes are given; and once they are given, the foci are easily found.

Moreover, the squares of the axes are in the same ratio as KH and LH, and from this it is easy to trace a trajectory fixed in shape through four given points. For if two of the given points are made the poles C and B, the third gives the moving angles PCK and PBK; and once those are given, the circle BGKC can be drawn. Then, because the trajectory is fixed in shape, the ratio of OH to OK is given, and therefore OH itself is given. About the center O, with radius OH, draw another circle. The straight line that touches this circle and passes through the meeting point of the legs CK and BK, when the first legs CP and BP meet at the fourth given point, will be the ruler MN, by means of which the trajectory can be traced. It follows, conversely, that a quadrilateral fixed in shape can be inscribed in a given conic section, apart from a few cases that are impossible.

There are also other lemmas by whose help trajectories fixed in shape can be traced through given points and touching given lines. One of them is this: if a straight line is drawn through any point given in position, cutting a given conic section in two points, and the distance between the two intersections is bisected, the point of bisection will lie on another conic section of the same kind as the first, with its axes parallel to the axes of the first. But I hasten on to things of greater use.

Lemma 26

To place the three corners of a triangle, fixed in shape and size, on as many straight lines given in position, provided they are not all parallel to one another, so that each corner touches its own line. (Plate XII, Fig. 2)

Three straight lines AB, AC and BC, of unlimited length, are given in position, and the triangle DEF is to be placed so that its corner D touches the line AB, its corner E the line AC, and its corner F the line BC. On DE, DF and EF draw three segments of circles, DRE, DGF and EMF, each of which holds an angle equal, respectively, to the angles BAC, ABC and ACB; that is, any angle drawn in the segment and standing on its chord equals the given angle. These segments are to be drawn on such sides of the lines DE, DF and EF that the letters DRED run round in the same order as the letters BACB, the letters DGFD in the same order as ABCA, and the letters EMFE in the same order as ACBA. Then complete the segments into whole circles. Let the first two circles cut each other at G, and let P and Q be their centers. Join GP and PQ, and take Ga so that the ratio of Ga to AB is the same as that of GP to PQ. About the center G, with radius Ga, draw a circle cutting the first circle DGE at a. Join aD, cutting the second circle DFG at b, and aE, cutting the third circle EMF at c. Complete the figure ABCdef similar and equal to the figure abcDEF. I say that this does what was required.

For draw Fc meeting aD at n, and join aG, bG, QG, QD and PD. By the construction, the angle EaD is equal to the angle CAB, and the angle acF is equal to the angle ACB; so the triangle anc has the same angles as the triangle ABC. The angle anc, or FnD, is therefore equal to the angle ABC, and so to the angle FbD; and therefore the point n falls on the point b. Moreover, the angle GPQ, which is half the angle GPD at the center, is equal to the angle GaD at the circumference. And the angle GQP, which is half the angle GQD at the center, is equal to the supplement of the angle GbD at the circumference (what that angle lacks of two right angles), and therefore equal to the angle Gab. For this reason the triangles GPQ and Gab are similar, and the ratio of Ga to ab is the same as that of GP to PQ, which by the construction is the ratio of Ga to AB. Therefore ab and AB are equal, and consequently the triangles abc and ABC, which we have just proved similar, are also equal. And therefore, since the corners D, E and F of the triangle DEF touch the sides ab, ac and bc of the triangle abc respectively, the figure ABCdef can be completed similar and equal to the figure abcDEF, and completing it solves the problem. Q.E.F.

Corollary. Hence a straight line can be drawn whose parts, of given lengths, are cut off between three straight lines given in position. Suppose the triangle DEF turns into a straight line, its point D coming down onto the side EF and its sides DE and DF being laid in one straight line. Its given part DE is to be fitted between the straight lines AB and AC, given in position, and its given part DF between the straight lines AB and BC, given in position. Applying the construction above to this case solves the problem.

Proposition 28, Problem 20

To trace a trajectory fixed in shape and size, given parts of which are cut off between three straight lines given in position. (Plate XII, Fig. 3)

Suppose a trajectory is to be traced that is similar and equal to the curve DEF, and that is cut by three straight lines AB, AC and BC, given in position, into parts similar and equal to the given parts DE and EF of this curve.

Draw the straight lines DE, EF and DF, and place the corners D, E and F of this triangle DEF so that they touch those straight lines given in position (by Lemma 26). Then around the triangle trace the trajectory, similar and equal to the curve DEF. Q.E.F.

Lemma 27

To construct a quadrilateral fixed in shape whose corners are so placed on four straight lines given in position, which are neither all parallel to one another nor all meeting at one point, that each corner touches its own line. (Plate XIII, Fig. 1)

Let the four straight lines ABC, AD, BD and CE be given in position, the first cutting the second at A, the third at B and the fourth at C. Suppose a quadrilateral fghi is to be constructed similar to the quadrilateral FGHI, whose corner f, equal to the given angle F, touches the straight line ABC, and whose other corners g, h and i, equal to the other given angles G, H and I, touch the other lines AD, BD and CE respectively. Join FH, and on FG, FH and FI draw as many segments of circles, FSG, FTH and FVI: the first, FSG, holding an angle equal to the angle BAD; the second, FTH, an angle equal to the angle CBD; and the third, FVI, an angle equal to the angle ACE. The segments are to be drawn on such sides of the lines FG, FH and FI that the letters FSGF run round in the same order as the letters BADB, the letters FTHF in the same order as CBDC, and the letters FVIF in the same order as ACEA. Complete the segments into whole circles, and let P be the center of the first circle FSG, and Q the center of the second, FTH. Join PQ and extend it both ways, and on it take QR so that the ratio of QR to PQ is the same as that of BC to AB. QR is taken on the side of the point Q that puts the letters P, Q, R in the same order as the letters A, B, C. About the center R, with radius RF, draw a fourth circle FNc, cutting the third circle FVI at c. Join Fc, cutting the first circle at a and the second at b. Draw aG, bH and cI, and let the figure ABCfghi be made similar to the figure abcFGHI; then the quadrilateral fghi is the one required.

For let the first two circles, FSG and FTH, cut each other at K. Join PK, QK, RK, aK, bK and cK, and extend QP to L. The angles FaK, FbK and FcK at the circumferences are half the angles FPK, FQK and FRK at the centers, and so they are equal to the angles LPK, LQK and LRK, which are the halves of those angles. The figure PQRK therefore has the same angles as the figure abcK and is similar to it, and consequently the ratio of ab to bc is the same as that of PQ to QR, that is, of AB to BC. But by the construction the angles fAg, fBh and fCi are equal to the angles FaG, FbH and FcI. The figure ABCfghi can therefore be completed similar to the figure abcFGHI. When that is done, a quadrilateral fghi has been constructed similar to the quadrilateral FGHI, which with its corners f, g, h and i touches the straight lines ABC, AD, BD and CE. Q.E.F.

Corollary. It follows that a straight line can be drawn whose parts, cut off in a given order between four straight lines given in position, are in a given proportion to one another. Let the angles FGH and GHI be opened out until the straight lines FG, GH and HI lie in one straight line. Constructing the problem for this case draws a straight line fghi whose parts fg, gh and hi, cut off between the four straight lines given in position, AB and AD, AD and BD, and BD and CE, are proportional to the lines FG, GH and HI, and keep the same order among themselves. But the same thing can be done more quickly in the following way.

Extend AB to K (Plate XIII, Fig. 2) and BD to L, so that the ratio of BK to AB is the same as that of HI to GH, and the ratio of DL to BD the same as that of GI to FG; and join KL, meeting the straight line CE at i. Extend iL to M, so that the ratio of LM to iL is the same as that of GH to HI. Then draw MQ parallel to LB, meeting the straight line AD at g, and join gi, cutting AB and BD at f and h. I say that this does what was required.

For let Mg cut the straight line AB at Q, and AD cut the straight line KL at S; and draw AP parallel to BD, meeting iL at P. Then the ratios of gM to Lh, of gi to hi, of Mi to Li, of GI to HI, of AK to BK, and of AP to BL are all the same. Cut DL at R so that the ratio of DL to RL is that same ratio. Because the ratios of gS to gM, of AS to AP, and of DS to DL are equal, it follows by equality of ratios (ex æquo) that the ratio of gS to Lh is the same as that of AS to BL, and of DS to RL. Taking differences of these terms, the ratio of $BL-RL$ to $Lh-BL$ is the same as that of $AS-DS$ to $gS-AS$. That is, the ratio of BR to Bh is the same as that of AD to Ag, and therefore the same as that of BD to gQ. And, exchanging the middle terms, the ratio of BR to BD is the same as that of Bh to gQ, or of fh to fg. But by the construction the line BL was cut at D and R in the same ratio as the line FI at G and H, and therefore the ratio of BR to BD is the same as that of FH to FG. Therefore the ratio of fh to fg is the same as that of FH to FG. Since, then, the ratio of gi to hi is likewise that of Mi to Li, that is, of GI to HI, it is clear that the lines FI and fi are cut in the same proportions, at G and H and at g and h. Q.E.F.

In the construction of this corollary, after the line LK is drawn cutting CE at i, we may instead extend iE to V, so that the ratio of EV to Ei is the same as that of FH to HI, and then draw Vf parallel to BD. It comes to the same thing if, about the center i with radius IH, we draw a circle cutting BD at X, extend iX to Y so that iY is equal to IF, and then draw Yf parallel to BD.

Sir Christopher Wren and Dr. Wallis long ago gave other solutions of this problem.

Proposition 29, Problem 21

To trace a trajectory fixed in shape that is cut by four straight lines given in position into parts given in order, in shape, and in proportion.

Suppose a trajectory is to be traced that is similar to the curve FGHI (Plate XIII, Fig. 3), and whose parts, similar and proportional to the parts FG, GH and HI of that curve, are cut off between the straight lines AB and AD, AD and BD, and BD and CE, given in position: the first part between the first pair of those lines, the second between the second, and the third between the third. Draw the straight lines FG, GH, HI and FI, and construct a quadrilateral fghi similar to the quadrilateral FGHI, whose corners f, g, h and i touch the straight lines given in position, AB, AD, BD and CE, each in its order (by Lemma 27). Then trace a trajectory around this quadrilateral, and that trajectory will be similar to the curve FGHI.

Scholium

This problem can also be constructed in the following way. Joining FG, GH, HI and FI (Plate XIII, Fig. 4), extend GF to V, join FH and IG, and make the angles CAK and DAL equal to the angles FGH and VFH. Let AK and AL meet the straight line BD at K and L, and from these points draw KM and LN. Let KM make the angle AKM equal to the angle GHI, and let the ratio of KM to AK be the same as that of HI to GH. Let LN make the angle ALN equal to the angle FHI, and let the ratio of LN to AL be the same as that of HI to FH. AK, KM, AL and LN are drawn on such sides of the lines AD, AK and AL that the letters CAKMC, ALKA and DALND run round in the same order as the letters FGHIF. Draw MN, meeting the straight line CE at i. Make the angle iEP equal to the angle IGF, and let the ratio of PE to Ei be the same as that of FG to GI. Through P draw PQf, making with the straight line ADE an angle PQE equal to the angle FIG, and meeting the straight line AB at f; and join fi. PE and PQ are drawn on such sides of the lines CE and PE that the letters PEiP and PEQP run round in the same order as the letters FGHIF. Then, if a quadrilateral fghi similar to the quadrilateral FGHI is constructed on the line fi, with its letters in the same order, and a trajectory fixed in shape is drawn around it, the problem is solved.

So much for finding the orbits. It remains to determine the motions of bodies in the orbits so found.

Section 6: Finding Motions in Given Orbits

Proposition 30, Problem 22

To find, at any given time, the place of a body moving along a given parabolic path.

Let S be the focus of the parabola and A its principal vertex (Plate XIV, Fig. 1), and suppose that $4AS\times M$ is equal to the parabolic area APS that is to be cut off: the area that the radius SP has swept out since the body left the vertex, or that it will sweep out before the body arrives there. The size of that area is known from the time, to which it is proportional. Bisect AS in G, and erect the perpendicular GH equal to $3M$. A circle drawn about the center H, with radius HS, will then cut the parabola in P, the place required.

For drop PO perpendicular to the axis, and draw PH. Since HP equals the radius HS, and GS equals AG, the square of HP is $A{G}^2+G{H}^2$; and since HP is also the slanting side of a right triangle whose other sides are $AO-AG$ and $PO-GH$, we have $A{G}^2+G{H}^2$ $(=H{P}^2=\overline{AO-AG}{|}^2+\overline{PO-GH}{|}^2)$ $=A{O}^2+P{O}^2-2GAO-2GH+PO+A{G}^2+G{H}^2$, where 2GAO stands for twice the product of GA and AO. Taking $A{G}^2+G{H}^2$ from both sides leaves $2GH\times PO$$(=A{O}^2+P{O}^2-2GAO)$$=A{O}^2+\frac{3}{4}P{O}^2$, the last step because in the parabola $P{O}^2=4AS\times AO$, so that twice GA times AO, which is AS times AO, equals a quarter of the square of PO. In place of $A{O}^2$ write $AO\times \frac{P{O}^2}{4AS}$; then divide every term by 3 PO and multiply by 2 AS, and we have $\frac{4}{3}GH\times AS$ $(=\frac{1}{6}AO\times PO+\frac{1}{2}AS\times PO=$$\frac{AO+3AS}{6}\times PO=$$\frac{4AO-3SO}{6}\times PO=$ the area $\overline{APO-SPO}|=$ the area APS. That is, it equals the parabolic area APO, which is two thirds of AO times PO, less the triangle SPO, which is half of SO times PO; and that difference is the area APS. But GH was taken equal to $3M$, and so $\frac{4}{3}GH\times AS$ is $4AS\times M$. The area APS that is cut off is therefore equal to the area $4AS\times M$ that was to be cut off. Q.E.D.⁷³