# The Mathematical Principles of Natural Philosophy (1846)/BookII-VI

SECTION VI.
Of the motion and resistance of funependulous bodies.

PROPOSITION XXIV. THEOREM XIX.
The quantities of matter in funependulous bodies, whose centres of oscillation are equally distant from the centre of suspension, are in a ratio compounded of the ratio of the weights and the duplicate ratio of the times of the oscillations in vacuo.

For the velocity which a given force can generate in a given matter in a given time is as the force and the time directly, and the matter inversely. The greater the force or the time is, or the less the matter, the greater velocity will be generated. This is manifest from the second Law of Motion. Now if pendulums are of the same length, the motive forces in places equally distant from the perpendicular are as the weights: and therefore if two bodies by oscillating describe equal arcs, and those arcs are divided into equal parts; since the times in which the bodies describe each of the correspondent parts of the arcs are as the times of the whole oscillations, the velocities in the correspondent parts of the oscillations will be to each other as the motive forces and the whole times of the oscillations directly, and the quantities of matter reciprocally: and therefore the quantities of matter are as the forces and the times of the oscillations directly and the velocities reciprocally. But the velocities reciprocally are as the times, and therefore the times directly and the velocities reciprocally are as the squares of the times; and therefore the quantities of matter are as the motive forces and the squares of the times, that is, as the weights and the squares of the times.   Q.E.D.

Cor. 1. Therefore if the times are equal, the quantities of matter in each of the bodies are as the weights.

Cor. 2. If the weights are equal, the quantities of matter will be as the squares of the times.

Cor. 3. If the quantities of matter are equal, the weights will be reciprocally as the squares of the times.

Cor. 4. Whence since the squares of the times, caeteris paribus, are as the lengths of the pendulums, therefore if both the times and quantities of matter are equal, the weights will be as the lengths of the pendulums.

Cor. 5. And universally, the quantity of matter in the pendulous body is as the weight and the square of the time directly, and the length of the pendulum inversely.

Cor. 6. But in a non-resisting medium, the quantity of matter in the pendulous body is as the comparative weight and the square of the time directly, and the length of the pendulum inversely. For the comparative weight is the motive force of the body in any heavy medium, as was shewn above; and therefore does the same thing in such a non-resisting medium as the absolute weight does in a vacuum.

Cor. 7. And hence appears a method both of comparing bodies one among another, as to the quantity of matter in each; and of comparing the weights of the same body in different places, to know the variation of its gravity. And by experiments made with the greatest accuracy, I have always found the quantity of matter in bodies to be proportional to their weight.

PROPOSITION XXV. THEOREM XX.
Funependulous bodies that are, in any medium, resisted in the ratio of the moments of time, and funependulous bodies that move in a non-resisting medium of the same specific gravity, perform their oscillations in a cycloid in the same time, and describe proportional parts of arcs together.

Let AB be an arc of a cycloid, which a body D, by vibrating in a non-resisting medium, shall describe in any time. Bisect that arc in C, so that C may be the lowest point thereof; and the accelerative force with which the body is urged in any place D, or d or E, will be as the length of the arc CD, or Cd, or CE. Let that force be expressed by that same arc; and since the resistance is as the moment of the time, and therefore given, let it be expressed by the given part CO of the cycloidal arc, and take the arc Od in the same ratio to the arc CD that the arc OB has to the arc CB: and the force with which the body in d is urged in a resisting medium, being the excess of the force Cd above the resistance CO, will be expressed by the arc Od, and will therefore be to the force with which the body D is urged in a non-resisting medium in the place D, as the arc Od to the arc CD; and therefore also in the place B, as the arc OB to the arc CB. Therefore if two bodies D, d go from the place Bc and are urged by these forces; since the forces at the beginning are as the arc CB and OB, the first velocities and arcs first described will be in the same ratio. Let those arcs be BD and Bd, and the remaining arcs CD, Od, will be in the same ratio. Therefore the forces, being proportional to those arcs CD, Od, will remain in the same ratio as at the beginning, and therefore the bodies will continue describing together arcs in the same ratio. Therefore the forces and velocities and the remaining arcs CD, Od, will be always as the whole arcs CB, OB, and therefore those remaining arcs will be described together. Therefore the two bodies D and d will arrive together at the places C and O; that which moves in the non-resisting medium, at the place C, and the other, in the resisting medium, at the place O. Now since the velocities in C and O are as the arcs CB, OB, the arcs which the bodies describe when they go farther will be in the same ratio. Let those arcs be CE and Oe. The force with which the body D in a non-resisting medium is retarded in E is as CE, and the force with which the body d in the resisting medium is retarded in e, is as the sum of the force Ce and the resistance CO, that is, as Oe; and therefore the forces with which the bodies are retarded are as the arcs CB, OB, proportional to the arcs CE, Oe; and therefore the velocities, retarded in that given ratio, remain in the same given ratio. Therefore the velocities and the arcs described with those velocities are always to each other in that given ratio of the arcs CB and OB; and therefore if the entire arcs AB, aB are taken in the same ratio, the bodies D and d will describe those arcs together, and in the places A and a will lose all their motion together. Therefore the whole oscillations are isochronal, or are performed in equal times; and any parts of the arcs, as BD, Bd, or BE, Be, that are described together, are proportional to the whole arcs BA, Ba.   Q.E.D.

Cor. Therefore the swiftest motion in a resisting medium does not fall upon the lowest point C, but is found in that point O, in which the whole arc described Ba is bisected. And the body, proceeding from thence to a, is retarded at the same rate with which it was accelerated before in its descent from B to O.

PROPOSITION XXVI. THEOREM XXI.
Funependulous bodies, that are resisted in the ratio of the velocity, have their oscillations in a cycloid isochronal.

For if two bodies, equally distant from their centres of suspension, describe, in oscillating, unequal arcs, and the velocities in the correspondent parts of the arcs be to each other as the whole arcs; the resistances, proportional to the velocities, will be also to each other as the same arcs. Therefore if these resistances be subducted from or added to the motive forces arising from gravity which are as the same arcs, the differences or sums will be to each other in the same ratio of the arcs; and since the increments and decrements of the velocities are as these differences or sums, the velocities will be always as the whole arcs; therefore if the velocities are in any one case as the whole arcs, they will remain always in the same ratio. But at the beginning of the motion, when the bodies begin to descend and describe those arcs, the forces, which at that time are proportional to the arcs, will generate velocities proportional to the arcs. Therefore the velocities will be always as the whole arcs to be described, and therefore those arcs will be described in the same time.   Q.E.D.

PROPOSITION XXVII. THEOREM XXII.
If funependulous bodies are resisted in the duplicate ratio of their velocities, the differences between the times of the oscillations in a resisting medium, and the times of the oscillations in a non-resisting medium of the same, specific gravity, will be proportional to the arcs described in oscillating nearly.

For let equal pendulums in a resisting medium describe the unequal arcs A, B; and the resistance of the body in the arc A will be to the resistance of the body in the correspondent part of the arc B in the duplicate ratio of the velocities, that is, as AA to BB nearly. If the resistance in the arc B were to the resistance in the arc A as AB to AA, the times in the arcs A and B would be equal (by the last Prop.) Therefore the resistance AA in the arc A, or AB in the arc B, causes the excess of the time in the arc A above the time in a non-resisting medium; and the resistance BB causes the excess of the time in the arc B above the time in a non-resisting medium. But those excesses are as the efficient forces AB and BB nearly, that is, as the arcs A and B.   Q.E.D.

Cor. 1. Hence from the times of the oscillations in unequal arcs in a resisting medium, may be known the times of the oscillations in a non-resisting medium of the same specific gravity. For the difference of the times will be to the excess of the time in the lesser arc above the time in a non-resisting medium as the difference of the arcs to the lesser arc.

Cor. 2. The shorter oscillations are more isochronal, and very short ones are performed nearly in the same times as in a non-resisting medium. But the times of those which are performed in greater arcs are a little greater, because the resistance in the descent of the body, by which the time is prolonged, is greater, in proportion to the length described in the descent than the resistance in the subsequent ascent, by which the time is contracted. But the time of the oscillations, both short and long, seems to be prolonged in some measure by the motion of the medium. For retarded bodies are resisted somewhat less in proportion to the velocity, and accelerated bodies somewhat more than those that proceed uniformly forwards; because the medium, by the motion it has received from the bodies, going forwards the same way with them, is more agitated in the former case, and less in the latter; and so conspires more or less with the bodies moved. Therefore it resists the pendulums in their descent more, and in their ascent less, than in proportion to the velocity; and these two causes concurring prolong the time.

PROPOSITION XXVIII. THEOREM XXIII.
If a funependulous body, oscillating in a cycloid, be resisted in the ratio of the moments of the time, its resistance will be to the force of gravity as the excess of the arc described in the whole descent above the arc described in the subsequent ascent to twice the length of the pendulum.

Let BC represent the arc described in the descent, Ca the arc described in the ascent, and Aa the difference of the arcs: and things remaining as they were constructed and demonstrated in Prop. XXV, the force with which the oscillating body is urged in any place D will be to the force of resistance as the arc CD to the arc CO, which is half of that difference Aa. Therefore the force with which the oscillating body is urged at the beginning or the highest point of the cycloid, that is, the force of gravity, will be to the resistance as the arc of the cycloid, between that highest point and lowest point C, is to the arc CO; that is (doubling those arcs), as the whole cycloidal arc, or twice the length of the pendulum, to the arc Aa.   Q.E.D.

PROPOSITION XXIX. PROBLEM VI.
Supposing that a body oscillating in a cycloid is resisted in a duplicate ratio of the velocity: to find the resistance in each place.

Let Ba be an arc described in one entire oscillation, C the lowest point

of the cycloid, and CZ half the whole cycloidal arc, equal to the length of the pendulum; and let it be required to find the resistance of the body in any place D. Cut the indefinite right line OQ in the points O, S, P, Q, so that (erecting the perpendiculars OK, ST, PI, QE, and with the centre O, and the aysmptotes OK, OQ, describing the hyperbola TIGE cutting the perpendiculars ST, PI, QE in T, I, and E, and through the point I drawing KF, parallel to the asymptote OQ, meeting the asymptote OK in K, and the perpendiculars ST and QE in L and F) the hyperbolic area PIEQ may be to the hyperbolic area PITS as the arc BC, described in the descent of the body, to the arc Ca described in the ascent; and that the area IEF may be to the area ILT as OQ to OS. Then with the perpendicular MN cut off the hyperbolic area PINM, and let that area be to the hyperbolic area PIEQ as the arc CZ to the arc BC described in the descent. And if the perpendicular RG cut off the hyperbolic area PIGR, which shall be to the area PIEQ as any arc CD to the arc BC described in the whole descent, the resistance in any place D will be to the force of gravity as the area $\scriptstyle \frac{OR}{OQ}$ IEF - IGH to the area PINM.

For since the forces arising from gravity with which the body is urged in the places Z, B, D, a, are as the arcs CZ, CB, CD, Ca and those arcs are as the areas PINM, PIEQ, PIGR, PITS; let those areas be the exponents both of the arcs and of the forces respectively. Let Dd be a very small space described by the body in its descent: and let it be expressed by the very small area RGgr comprehended between the parallels RG, rg; and produce rg to h, so that GHhg and RGgr may be the contemporaneous decrements of the areas IGH, PIGR. And the increment GHhg - $\scriptstyle \frac{Rr}{OQ}$ IEF, or Rr $\scriptstyle \times$ HG - $\scriptstyle \frac{Rr}{OQ}$ IEF, of the area $\scriptstyle \frac{OR}{OQ}$ IEF - IGH will be to the decrement RGgr, or Rr $\scriptstyle \times$ RG, of the area PIGR, as HG - $\scriptstyle \frac{IEF}{OQ}$ to RG; and therefore as OR $\scriptstyle \times$ HG - $\scriptstyle \frac{OR}{OQ}$ IEF to OR $\scriptstyle \times$ GR or OP $\scriptstyle \times$ PI, that is (because of the equal quantities OR $\scriptstyle \times$ HG, OR $\scriptstyle \times$ HR - OR $\scriptstyle \times$ GR, ORHK - OPIK, PIHR and PIGR + IGH), as PIGR + IGH - $\scriptstyle \frac{OR}{OQ}$ IEF to OPIK. Therefore if the area $\scriptstyle \frac{OR}{OQ}$ IEF - IGH be called Y, and RGgr the decrement of the area PIGR be given, the increment of the area Y will be as PIGR - Y.

Then if V represent the force arising from the gravity, proportional to the arc CD to be described, by which the body is acted upon in D, and R be put for the resistance, V - R will be the whole force with which the body is urged in D. Therefore the increment of the velocity is as V - R and the particle of time in which it is generated conjunctly. But the velocity itself is as the contemporaneous increment of the space described directly and the same particle of time inversely. Therefore, since the resistance is, by the supposition, as the square of the velocity, the increment of the resistance will (by Lem. II) be as the velocity and the increment of the velocity conjunctly, that is, as the moment of the space and V - R conjunctly; and, therefore, if the moment of the space be given, as V - R; that is, if for the force V we put its exponent PIGR, and the resistance R be expressed by any other area Z, as PIGR - Z.

Therefore the area PIGR uniformly decreasing by the subduction of given moments, the area Y increases in proportion of PIGR - Y, and the area Z in proportion of PIGR - Z. And therefore if the areas Y and Z begin together, and at the beginning are equal, these, by the addition of equal moments, will continue to be equal and in like manner decreasing by equal moments, will vanish together. And, vice versa, if they together begin and vanish, they will have equal moments and be always equal; and that, because if the resistance Z be augmented, the velocity together with the arc Ca, described in the ascent of the body, will be diminished; and the point in which all the motion together with the resistance ceases coming nearer to the point C, the resistance vanishes sooner than the area Y. And the contrary will happen when the resistance is diminished.

Now the area Z begins and ends where the resistance is nothing, that is, at the beginning of the motion where the arc CD is equal to the arc CB,

and the right line RG falls upon the right line QE; and at the end of the motion where the arc CD is equal to the arc Ca, and RG falls upon the right line ST. And the area Y or $\scriptstyle \frac{OR}{OQ}$ IEF - IGH begins and ends also where the resistance is nothing, and therefore where $\scriptstyle \frac{OR}{OQ}$ IEF and IGH are equal; that is (by the construction), where the right line RG falls successively upon the right lines QE and ST. Therefore those areas begin and vanish together, and are therefore always equal. Therefore the area $\scriptstyle \frac{OR}{OQ}$ IEF - IGH is equal to the area Z, by which the resistance is expressed, and therefore is to the area PINM, by which the gravity is expressed, as the resistance to the gravity.   Q.E.D.

Cor. 1. Therefore the resistance in the lowest place C is to the force of gravity as the area $\scriptstyle \frac{OP}{OQ}$ IEF to the area PINM.

Cor. 2. But it becomes greatest where the area PIHR is to the area IEF as OR to OQ. For in that case its moment (that is, PIGR - Y) becomes nothing.

Cor. 3. Hence also may be known the velocity in each place, as being in the subduplicate ratio of the resistance, and at the beginning of the motion equal to the velocity of the body oscillating in the same cycloid without any resistance.

However, by reason of the difficulty of the calculation by which the resistance and the velocity are found by this Proposition, we have thought fit to subjoin the Proposition following.

PROPOSITION XXX. THEOREM XXIV.
If a right line aB be equal to the arc of a cycloid which an oscillating body describes, and at each of its points D the perpendiculars DK be erected, which shall be to the length of the pendulum as the resistance of the body in the corresponding points of the arc to the force of gravity; I say, that the difference between the arc described in the whole descent and the arc described in the whole subsequent ascent drawn into half the sum of the same arcs will be equal to the area BKa which all those perpendiculars take up.

Let the arc of the cycloid, described in one entire oscillation, be expressed by the right line aB, equal to it, and the arc which would have been described in vacuo by the length AB. Bisect AB in C, and the point C will represent B the lowest point of the cycloid, and CD will be as the force arising from gravity, with which the body in D is urged in the direction of the tangent of the cycloid, and will have the same ratio to the length of the pendulum as the force in D has to the force of gravity. Let that force, therefore, be expressed by that length CD, and the force of gravity by the length of the pendulum; and if in DE you take DK in the same ratio to the length of the pendulum as the resistance has to the gravity, DK will be the exponent of the resistance. From the centre C with the interval CA or CB describe a semi-circle BEeA. Let the body describe, in the least time, the space Dd; and, erecting the perpendiculars DE, de, meeting the circumference in E and e, they will be as the velocities which the body descending in vacuo from the point B would acquire in the places D and d. This appears by Prop LII, Book I. Let therefore, these velocities be expressed by those perpendiculars DE, de; and let DF be the velocity which it acquires in D by falling from B in the resisting medium. And if from the centre C with the interval CF we describe the circle FfM meeting the right lines de and AB in f and M, then M will be the place to which it would thenceforward, without farther resistance, ascend, and df the velocity it would acquire in d. Whence, also, if Fg represent the moment of the velocity which the body D, in describing the least space Dd, loses by the resistance of the medium; and CN be taken equal to Cg; then will N be the place to which the body, if it met no farther resistance, would thenceforward ascend, and MN will be the decrement of the ascent arising from the loss of that velocity. Draw Fm perpendicular to df, and the decrement Fg of the velocity DF generated by the resistance DK will be to the increment fm of the same velocity, generated by the force CD, as the generating force DK to the generating force CD. But because of the similar triangles Fmf, Fhg, FDC, fm is to Fm or Dd as CD to DF; and, ex aequo, Fg to Dd as DK to DF. Also Fh is to Fg as DF to CF; and, ex aequo perturbatè, Fh or MN to Dd as DK to CF or CM; and therefore the sum of all the MN $\scriptstyle \times$ CM will be equal to the sum of all the Dd $\scriptstyle \times$ DK. At the moveable point M suppose always a rectangular ordinate erected equal to the indeterminate CM, which by a continual motion is drawn into the whole length Aa; and the trapezium described by that motion, or its equal, the rectangle Aa $\scriptstyle \times$ ½aB, will be equal to the sum of all the MN $\scriptstyle \times$ CM, and therefore to the sum of all the Dd $\scriptstyle \times$ DK, that is, to the area BKVTa.   Q.E.D.

Cor. Hence from the law of resistance, and the difference Aa of the arcs Ca, CB, may be collected the proportion of the resistance to the gravity nearly.

For if the resistance DK be uniform, the figure BKTa will be a rectangle under Ba and DK; and thence the rectangle under ½Ba and Aa will be equal to the rectangle under Ba and DK, and DK will be equal to ½Aa. Wherefore since DK is the exponent of the resistance, and the length of the pendulum the exponent of the gravity, the resistance will be to the gravity as ½Aa to the length of the pendulum; altogether as in Prop. XXVIII is demonstrated.

If the resistance be as the velocity, the figure BKTa will be nearly an ellipsis. For if a body, in a non-resisting medium, by one entire oscillation, should describe the length BA, the velocity in any place D would be as the ordinate DE of the circle described on the diameter AB. Therefore since Ba in the resisting medium, and BA in the non-resisting one, are described nearly in the same times; and therefore the velocities in each of the points of Ba are to the velocities in the correspondent points of the length BA nearly as Ba is to BA, the velocity in the point D in the resisting medium will be as the ordinate of the circle or ellipsis described upon the diameter Ba; and therefore the figure BKVTa will be nearly an ellipsis. Since the resistance is supposed proportional to the velocity, let OV be the exponent of the resistance in the middle point O; and an ellipsis BRVSa described with the centre O, and the semi-axes OB, OV, will be nearly equal to the figure BKVTa, and to its equal the rectangle Aa $\scriptstyle \times$ BO. Therefore Aa $\scriptstyle \times$ BO is to OV $\scriptstyle \times$ BO as the area of this ellipsis to OV $\scriptstyle \times$ BO; that is, Aa is to OV as the area of the semi-circle to the square of the radius, or as 11 to 7 nearly; and, therefore, $\scriptstyle \frac{7}{11}$Aa is to the length of the pendulum as the resistance of the oscillating body in O to its gravity.

Now if the resistance DK be in the duplicate ratio of the velocity, the figure BKVTa will be almost a parabola having V for its vertex and OV for its axis, and therefore will be nearly equal to the rectangle under Ba and OV. Therefore the rectangle under ½Ba and Aa is equal to the rectangle ⅔Ba $\scriptstyle \times$ OV, and therefore OV is equal to ¾Aa; and therefore the resistance in O made to the oscillating body is to its gravity as ¾Aa to the length of the pendulum.

And I take these conclusions to be accurate enough for practical uses. For since an ellipsis or parabola BRVSa falls in with the figure BKVTa in the middle point V, that figure, if greater towards the part BRV or VSa than the other, is less towards the contrary part, and is therefore nearly equal to it.

PROPOSITION XXXI. THEOREM XXV.
If the resistance made to an oscillating body in each of the proportional parts of the arcs described be augmented or diminished in a given ratio, the difference between the arc described in the descent and the arc described in the subsequent ascent will be augmented or diminished in the same ratio.

For that difference arises from the retardation of the pendulum by the resistance of the medium, and therefore is as the whole retardation and the retarding resistance proportional thereto. In the foregoing Proposition the rectangle under the right line ½aB and the difference Aa of the arcs CB, Ca, was equal to the area BKTa. And that area, if the length aB remains, is augmented or diminished in the ratio of the ordinates DK; that is, in the ratio of the resistance and is therefore as the length aB and the resistance conjunctly. And therefore the rectangle under Aa and ½aB is as aB and the resistance conjunctly, and therefore Aa is as the resistance.   Q.E.D.

Cor. 1. Hence if the resistance be as the velocity, the difference of the arcs in the same medium will be as the whole arc described: and the contrary.

Cor. 2. If the resistance be in the duplicate ratio of the velocity, that difference will be in the duplicate ratio of the whole arc: and the contrary.

Cor. 3. And universally, if the resistance be in the triplicate or any other ratio of the velocity, the difference will be in the same ratio of the whole arc: and the contrary.

Cor. 4. If the resistance be partly in the simple ratio of the velocity, and partly in the duplicate ratio of the same, the difference will be partly in the ratio of the whole arc, and partly in the duplicate ratio of it: and the contrary. So that the law and ratio of the resistance will be the same for the velocity as the law and ratio of that difference for the length of the arc.

Cor. 5. And therefore if a pendulum describe successively unequal arcs, and we can find the ratio of the increment or decrement of this difference for the length of the arc described, there will be had also the ratio of the increment or decrement of the resistance for a greater or less velocity.

GENERAL SCHOLIUM.

From these propositions we may find the resistance of mediums by pendulums oscillating therein. I found the resistance of the air by the following experiments. I suspended a wooden globe or ball weighing 57$\scriptstyle \frac{7}{22}$ ounces troy, its diameter 6$\scriptstyle \frac{7}{8}$ London inches, by a fine thread on a firm hook, so that the distance between the hook and the centre of oscillation of the globe was 10½ feet. I marked on the thread a point 10 feet and 1 inch distant from the centre of suspension; and even with that point I placed a ruler divided into inches, by the help whereof I observed the lengths of the arcs described by the pendulum. Then I numbered the oscillations in which the globe would lose $\scriptstyle \frac{1}{8}$ part of its motion. If the pendulum was drawn aside from the perpendicular to the distance of 2 inches, and thence let go, so that in its whole descent it described an arc of 2 inches, and in the first whole oscillation, compounded of the descent and subsequent ascent, an arc of almost 4 inches, the same in 164 oscillations lost $\scriptstyle \frac{1}{8}$ part of its motion, so as in its last ascent to describe an arc of 1¾ inches. If in the first descent it described an arc of 4 inches, it lost $\scriptstyle \frac{1}{8}$ part of its motion in 121 oscillations, so as in its last ascent to describe an arc of 3½ inches. If in the first descent it described an arc of 8, 16, 32, or 64 inches, it lost $\scriptstyle \frac{1}{8}$ part of its motion in 69, 35½, 18½, 9⅔ oscillations, respectively. Therefore the difference between the arcs described in the first descent and the last ascent was in the 1st, 2d, 3d, 4th, 5th, 6th cases, ¼, ½, 1, 2, 4, 8 inches respectively. Divide those differences by the number of oscillations in each case, and in one mean oscillation, wherein an arc of 3¾, 7½, 15, 30, 60, 120 inches was described, the difference of the arcs described in the descent and subsequent ascent will be $\scriptstyle \frac{1}{656}$, $\scriptstyle \frac{1}{242}$, $\scriptstyle \frac{1}{69}$, $\scriptstyle \frac{4}{71}$, $\scriptstyle \frac{8}{37}$, $\scriptstyle \frac{24}{29}$ parts of an inch, respectively. But these differences in the greater oscillations are in the duplicate ratio of the arcs described nearly, but in lesser oscillations something greater than in that ratio; and therefore (by Cor. 2, Prop. XXXI of this Book) the resistance of the globe, when it moves very swift, is in the duplicate ratio of the velocity, nearly; and when it moves slowly, somewhat greater than in that ratio.

Now let V represent the greatest velocity in any oscillation, and let A, B, and C be given quantities, and let us suppose the difference of the arcs to be AV + $\scriptstyle BV^{\frac{3}{2}}$ + CV² . Since the greatest velocities are in the cycloid as ½ the arcs described in oscillating, and in the circle as ½ the chords of those arcs; and therefore in equal arcs are greater in the cycloid than in the circle in the ratio of ½ the arcs to their chords; but the times in the circle are greater than in the cycloid, in a reciprocal ratio of the velocity; it is plain that the differences of the arcs (which are as the resistance and the square of the time conjunctly) are nearly the same in both curves: for in the cycloid those differences must be on the one hand augmented, with the resistance, in about the duplicate ratio of the arc to the chord, because of the velocity augmented in the simple ratio of the same; and on the other hand diminished, with the square of the time, in the same duplicate ratio. Therefore to reduce these observations to the cycloid, we must take the same differences of the arcs as were observed in the circle, and suppose the greatest velocities analogous to the half, or the whole arcs, that is, to the numbers ½, 1, 2, 4, 8, 16. Therefore in the 2d, 4th, and 6th cases, put 1, 4, and 16 for V; and the difference of the arcs in the 2d case will become $\scriptstyle \frac{\frac{1}{2}}{121}$ = A + B + C; in the 4th case $\scriptstyle \frac{2}{35\frac{1}{2}}$ = 4A + 8B + 16C; in the 6th $\scriptstyle \frac{8}{9\frac{2}{3}}$ = 16A + 64B + 256C. These equations reduced give A = 0,0000916, B = 0,0010847, and C = 0,0029558. Therefore the difference of the arcs is as 0,0000916V + $\scriptstyle 0,0010847V^{\frac{3}{2}}$ + 0,0029558V²: and therefore since (by Cor. Prop. XXX, applied to this case) the resistance of the globe in the middle of the arc described in oscillating, where the velocity is V, is to its weight as $\scriptstyle \frac{7}{11}$AV + $\scriptstyle \frac{7}{10}BV^{\frac{3}{2}}$ + ¾CV² to the length of the pendulum, if for A, B, and C you put the numbers found, the resistance of the globe will be to its weight as 0,0000583V + $\scriptstyle 0,0007593V^{\frac{3}{2}}$ + 0,0022169V² to the length of the pendulum between the centre of suspension and the ruler, that is, to 121 inches. Therefore since V in the second case represents 1, in the 4th case 4, and in the 6th case 16, the resistance will be to the weight of the globe, in the 2d case, as 0,0030345 to 121; in the 4th, as 0,041748 to 121; in the 6th, as 0,61705 to 121.

The arc, which the point marked in the thread described in the 6th case, was of 120 - $\scriptstyle \frac{8}{9\frac{2}{3}}$, or 119$\scriptstyle \frac{5}{29}$ inches. And therefore since the radius was 121 inches, and the length of the pendulum between the point of suspension and the centre of the globe was 126 inches, the arc which the centre of the globe described was 124$\scriptstyle \frac{3}{31}$ inches. Because the greatest velocity of the oscillating body, by reason of the resistance of the air, does not fall on the lowest point of the arc described, but near the middle place of the whole arc, this velocity will be nearly the same as if the globe in its whole descent in a non-resisting medium should describe 62$\scriptstyle \frac{3}{62}$ inches, the half of that arc, and that in a cycloid, to which we have above reduced the motion of the pendulum; and therefore that velocity will be equal to that which the globe would acquire by falling perpendicularly from a height equal to the versed sine of that arc. But that versed sine in the cycloid is to that arc 62$\scriptstyle \frac{3}{62}$ as the same arc to twice the length of the pendulum 252, and therefore equal to 15,278 inches. Therefore the velocity of the pendulum is the same which a body would acquire by falling, and in its fall describing a space of 15,278 inches. Therefore with such a velocity the globe meets with a resistance which is to its weight as 0,61705 to 121, or (if we take that part only of the resistance which is in the duplicate ratio of the velocity) as 0,56752 to 121.

I found, by an hydrostatical experiment, that the weight of this wooden globe was to the weight of a globe of water of the same magnitude as 55 to 97: and therefore since 121 is to 213,4 in the same ratio, the resistance made to this globe of water, moving forwards with the above-mentioned velocity, will be to its weight as 0,56752 to 213,4, that is, as 1 to 376$\scriptstyle \frac{1}{50}$. Whence since the weight of a globe of water, in the time in which the globe with a velocity uniformly continued describes a length of 30,556 inches, will generate all that velocity in the falling globe, it is manifest that the force of resistance uniformly continued in the same time will take away a velocity, which will be less than the other in the ratio of 1 to 376$\scriptstyle \frac{1}{50}$, that is, the $\scriptstyle \frac{1}{376\frac{1}{50}}$ part of the whole velocity. And therefore in the time that the globe, with the same velocity uniformly continued, would describe the length of its semi-diameter, or 3$\scriptstyle \frac{7}{16}$ inches, it would lose the $\scriptstyle \frac{1}{3342}$ part of its motion.

I also counted the oscillations in which the pendulum lost ¼ part of its motion. In the following table the upper numbers denote the length of the arc described in the first descent, expressed in inches and parts of an inch; the middle numbers denote the length of the arc described in the last ascent; and in the lowest place are the numbers of the oscillations. I give an account of this experiment, as being more accurate than that in which only $\scriptstyle \frac{1}{8}$ part of the motion was lost. I leave the calculation to such as are disposed to make it.

 First descent 2 4 8 16 32 64 Last ascent 1½ 3 6 12 24 48 Numb. of oscill. 374 272 162½ 83⅓ 41⅔ 22⅔

I afterward suspended a leaden globe of 2 inches in diameter, weighing 26¼ ounces troy by the same thread, so that between the centre of the globe and the point of suspension there was an interval of 10½ feet, and I counted the oscillations in which a given part of the motion was lost. The first of the following tables exhibits the number of oscillations in which $\scriptstyle \frac{1}{8}$ part of the whole motion was lost; the second the number of oscillations in which there was lost part of the same.

 First descent 1 2 4 8 16 32 64 Last ascent $\scriptstyle \frac{7}{8}$ $\scriptstyle \frac{7}{4}$ 3½ 7 14 28 56 Numb, of oscill. 226 228 193 140 90½ 53 30 First descent 1 2 4 8 16 32 64 Last ascent ¾ 1½ 3 6 12 24 48 Numb. of oscill. 510 518 420 318 204 121 70

Selecting in the first table the 3d, 5th, and 7th observations, and expressing the greatest velocities in these observations particularly by the numbers 1, 4, 16 respectively, and generally by the quantity V as above, there will come out in the 3d observation $\scriptstyle \frac{\frac{1}{2}}{193}$ = A + B + C, in the 5th observation $\scriptstyle \frac{2}{90\frac{1}{2}}$ = 4A + 8B + 16C, in the 7th observation $\scriptstyle \frac{8}{30}$ = 16A + 64B + 256C. These equations reduced give A = 0,001414, B = 0,000297, C = 0,000879. And thence the resistance of the globe moving with the velocity V will be to its weight 26¼ ounces in the same ratio as 0,0009V + $\scriptstyle 0,000208V^{\frac{3}{2}}$ + 0,000659V² to 121 inches, the length of the pendulum. And if we regard that part only of the resistance which is in the duplicate ratio of the velocity, it will be to the weight of the globe as 0,000659V² to 121 inches. But this part of the resistance in the first experiment was to the weight of the wooden globe of 57$\scriptstyle \frac{7}{22}$ ounces as 0,002217V² to 121; and thence the resistance of the wooden globe is to the resistance of the leaden one (their velocities being equal) as 57$\scriptstyle \frac{7}{22}$ into 0,002217 to 26¼ into 0,000659, that is, as 7⅓ to 1. The diameters of the two globes were 6$\scriptstyle \frac{7}{8}$ and 2 inches, and the squares of these are to each other as 47¼ and 4, or 11$\scriptstyle \frac{13}{16}$ and 1, nearly. Therefore the resistances of these equally swift globes were in less than a duplicate ratio of the diameters. But we have not yet considered the resistance of the thread, which was certainly very considerable, and ought to be subducted from the resistance of the pendulums here found. I could not determine this accurately, but I found it greater than a third part of the whole resistance of the lesser pendulum; and thence I gathered that the resistances of the globes, when the resistance of the thread is subducted, are nearly in the duplicate ratio of their diameters. For the ratio of 7⅓ - ⅓ to 1 - ⅓, or 10½ to 1 is not very different from the duplicate ratio of the diameters 11$\scriptstyle \frac{13}{16}$ to 1.

Since the resistance of the thread is of less moment in greater globes, I tried the experiment also with a globe whose diameter was 18¾ inches. The length of the pendulum between the point of suspension and the centre of oscillation was 122½ inches, and between the point of suspension and the knot in the thread 109½ inches. The arc described by the knot at the first descent of the pendulum was 32 inches. The arc described by the same knot in the last ascent after five oscillations was 28 inches. The sum of the arcs, or the whole arc described in one mean oscillation, was 60 inches. The difference of the arcs 4 inches. The $\scriptstyle \frac{1}{10}$ part of this, or the difference between the descent and ascent in one mean oscillation, is $\scriptstyle \frac{2}{5}$ of an inch. Then as the radius 109½ to the radius 122½, so is the whole arc of 60 inches described by the knot in one mean oscillation to the whole arc of 67$\scriptstyle \frac{1}{8}$ inches described by the centre of the globe in one mean oscillation; and so is the difference $\scriptstyle \frac{3}{5}$ to a new difference 0,4475. If the length of the arc described were to remain, and the length of the pendulum should be augmented in the ratio of 126 to 122½, the time of the oscillation would be augmented, and the velocity of the pendulum would be diminished in the subduplicate of that ratio; so that the difference 0,4475 of the arcs described in the descent and subsequent ascent would remain. Then if the arc described be augmented in the ratio of 124$\scriptstyle \frac{3}{31}$ to 67$\scriptstyle \frac{1}{8}$, that difference 0.4475 would be augmented in the duplicate of that ratio, and so would become 1,5295. These things would be so upon the supposition that the resistance of the pendulum were in the duplicate ratio of the velocity. Therefore if the pendulum describe the whole arc of 124$\scriptstyle \frac{3}{31}$ inches, and its length between the point of suspension and the centre of oscillation be 126 inches, the difference of the arcs described in the descent and subsequent ascent would be 1,5295 inches. And this difference multiplied into the weight of the pendulous globe, which was 208 ounces, produces 318,136. Again; in the pendulum above-mentioned, made of a wooden globe, when its centre of oscillation, being 126 inches from the point of suspension, described the whole arc of 124$\scriptstyle \frac{3}{31}$inches, the difference of the arcs described in the descent and ascent was $\scriptstyle \frac{126}{121}$ into $\scriptstyle \frac{8}{9\frac{2}{3}}$. This multiplied into the weight of the globe, which was 57$\scriptstyle \frac{7}{22}$ ounces, produces 49,396. But I multiply these differences into the weights of the globes, in order to find their resistances. For the differences arise from the resistances, and are as the resistances directly and the weights inversely. Therefore the resistances are as the numbers 318,136 and 49,396. But that part of the resistance of the lesser globe, which is in the duplicate ratio of the velocity, was to the whole resistance as 0,56752 tor 0,61675, that is, as 45,453 to 49,396; whereas that part of the resistance of the greater globe is almost equal to its whole resistance; and so those parts are nearly as 318,136 and 45,453, that is, as 7 and 1. But the diameters of the globes are 18¾ and 6$\scriptstyle \frac{7}{8}$; and their squares 351$\scriptstyle \frac{9}{16}$ and 47$\scriptstyle \frac{17}{64}$ are as 7,438 and 1, that is, as the resistances of the globes 7 and 1, nearly. The difference of these ratios is scarce greater than may arise from the resistance of the thread. Therefore those parts of the resistances which are, when the globes are equal, as the squares of the velocities, are also, when the velocities are equal, as the squares of the diameters of the globes.

But the greatest of the globes I used in these experiments was not perfectly spherical, and therefore in this calculation I have, for brevity's sake, neglected some little niceties; being not very solicitous for an accurate calculus in an experiment that was not very accurate. So that I could wish that these experiments were tried again with other globes, of a larger size, more in number, and more accurately formed; since the demonstration of a vacuum depends thereon. If the globes be taken in a geometrical proportion, as suppose whose diameters are 4, 8, 16, 32 inches; one may collect from the progression observed in the experiments what would happen if the globes were still larger.

In order to compare the resistances of different fluids with each other, I made the following trials. I procured a wooden vessel 4 feet long, 1 foot broad, and 1 foot high. This vessel, being uncovered, I filled with spring water, and, having immersed pendulums therein, I made them oscillate in the water. And I found that a leaden globe weighing 166$\scriptstyle \frac{1}{6}$ ounces, and in diameter 3$\scriptstyle \frac{5}{8}$ inches, moved therein as it is set down in the following table; the length of the pendulum from the point of suspension to a certain point marked in the thread being 126 inches, and to the centre of oscillation 134$\scriptstyle \frac{3}{8}$ inches.

 The arc described in the first descent, by a point marked in the thread was inches. 64 . 32 . 16 . 8 . 4 . 2 . 1 . ½ . ¼ The arc described in the last ascent was inches. 48 . 24 . 12 . 6 . 3 . 1½ . ¾ . $\scriptstyle \frac{3}{8}$ . $\scriptstyle \frac{3}{16}$ The difference of the arcs, proportional to the motion lost, was inches. 16 . 8 . 4 . 2 . 1 . ½ . ¼ . $\scriptstyle \frac{1}{8}$ . $\scriptstyle \frac{1}{16}$ The number of the oscillations in water. $\scriptstyle \frac{29}{60}$ . 1$\scriptstyle \frac{1}{5}$ . 3 . 7 . 11¼ . 12⅔ . 13⅓ The number of the oscillations in air. 85½ . 287 . 535

In the experiments of the 4th column there were equal motions lost in 535 oscillations made in the air, and 1$\scriptstyle \frac{1}{5}$ in water. The oscillations in the air were indeed a little swifter than those in the water. But if the oscillations in the water were accelerated in such a ratio that the motions of the pendulums might be equally swift in both mediums, there would be still the same number 1$\scriptstyle \frac{1}{5}$ of oscillations in the water, and by these the same quantity of motion would be lost as before; because the resistance it increased, and the square of the time diminished in the same duplicate ratio. The pendulums, therefore, being of equal velocities, there were equal motions lost in 535 oscillations in the air, and 1$\scriptstyle \frac{1}{5}$ in the water; and therefore the resistance of the pendulum in the water is to its resistance in the air as 535 to 1$\scriptstyle \frac{1}{5}$. This is the proportion of the whole resistances in the case of the 4th column.

Now let AV + CV² represent the difference of the arcs described in the descent and subsequent ascent by the globe moving in air with the greatest velocity V; and since the greatest velocity is in the case of the 4th column to the greatest velocity in the case of the 1st column as 1 to 8; and that difference of the arcs in the case of the 4th column to the difference in the case of the 1st column as $\scriptstyle \frac{2}{535}$ to $\scriptstyle \frac{16}{85\frac{1}{2}}$, or as 85½ to 4280; put in these cases 1 and 8 for the velocities, and 85½ and 4280 for the differences of the arcs, and A + C will be = 85½, and 8A + 64C = 4280 or A + 8C = 535; and then by reducing these equations, there will come out 7C = 449½ and C = 64$\scriptstyle \frac{3}{14}$ and A = 21$\scriptstyle \frac{2}{7}$; and therefore the resistance, which is as $\scriptstyle \frac{7}{11}$AV + ¾CV², will become as 13$\scriptstyle \frac{6}{11}$V + 48$\scriptstyle \frac{9}{56}$V². Therefore in the case of the 4th column, where the velocity was 1, the whole resistance is to its part proportional to the square of the velocity as 13$\scriptstyle \frac{6}{11}$ + 48$\scriptstyle \frac{9}{56}$ or 61$\scriptstyle \frac{12}{17}$ to 48$\scriptstyle \frac{9}{56}$; and therefore the resistance of the pendulum in water is to that part of the resistance in air, which is proportional to the square of the velocity, and which in swift motions is the only part that deserves consideration, as 61$\scriptstyle \frac{12}{17}$ to 48$\scriptstyle \frac{9}{56}$ and 535 to 1$\scriptstyle \frac{1}{5}$ conjunctly, that is, as 571 to 1. If the whole thread of the pendulum oscillating in the water had been immersed, its resistance would have been still greater; so that the resistance of the pendulum oscillating in the water, that is, that part which is proportional to the square of the velocity, and which only needs to be considered in swift bodies, is to the resistance of the same whole pendulum, oscillating in air with the same velocity, as about 850 to 1, that is as, the density of water to the density of air, nearly.

In this calculation we ought also to have taken in that part of the resistance of the pendulum in the water which was as the square of the velocity; but I found (which will perhaps seem strange) that the resistance in the water was augmented in more than a duplicate ratio of the velocity. In searching after the cause, I thought upon this, that the vessel was too narrow for the magnitude of the pendulous globe, and by its narrowness obstructed the motion of the water as it yielded to the oscillating globe. For when I immersed a pendulous globe, whose diameter was one inch only, the resistance was augmented nearly in a duplicate ratio of the velocity, I tried this by making a pendulum of two globes, of which the lesser and lower oscillated in the water, and the greater and higher was fastened to the thread just above the water, and, by oscillating in the air, assisted the motion of the pendulum, and continued it longer. The experiments made by this contrivance proved according to the following table.

 Arc descr. in first descent 16 . 8 . 4 . 2 . 1 . ½ . ¼ Arc descr. in last ascent 12 . 6 . 3 . 1½ . ¾ . $\scriptstyle \frac{3}{8}$ . $\scriptstyle \frac{3}{16}$ Diff. of arcs, proport. to motion lost 4 . 2 . 1 . ½ . ¼ . $\scriptstyle \frac{1}{8}$ . $\scriptstyle \frac{1}{16}$ Number of oscillations 3$\scriptstyle \frac{3}{8}$ . 6½ . 12$\scriptstyle \frac{1}{12}$ . 21$\scriptstyle \frac{1}{5}$ . 34 . 53 . 62$\scriptstyle \frac{1}{5}$

Lastly, since it is the opinion of some that there is a certain aethereal medium extremely rare and subtile, which freely pervades the pores of all bodies; and from such a medium, so pervading the pores of bodies, some resistance must needs arise; in order to try whether the resistance, which we experience in bodies in motion, be made upon their outward superficies only, or whether their internal parts meet with any considerable resistance upon their superficies, I thought of the following experiment. I suspended a round deal box by a thread 11 feet long, on a steel hook, by means of a ring of the same metal, so as to make a pendulum of the aforesaid length. The hook had a sharp hollow edge on its upper part, so that the upper arc of the ring pressing on the edge might move the more freely; and the thread was fastened to the lower arc of the ring. The pendulum being thus prepared, I drew it aside from the perpendicular to the distance of about 6 feet, and that in a plane perpendicular to the edge of the hook, lest the ring, while the pendulum oscillated, should slide to and fro on the edge of the hook: for the point of suspension, in which the ring touches the hook, ought to remain immovable. I therefore accurately noted the place to which the pendulum was brought, and letting it go, I marked three other places, to which it returned at the end of the 1st, 2d, and 3d oscillation. This I often repeated, that I might find those places as accurately as possible. Then I filled the box with lead and other heavy metals that were near at hand. But, first, I weighed the box when empty, and that part of the thread that went round it, and half the remaining part, extended between the hook and the suspended box; for the thread so extended always acts upon the pendulum, when drawn aside from the perpendicular, with half its weight. To this weight I added the weight of the air contained in the box. And this whole weight was about $\scriptstyle \frac{1}{78}$ of the weight of the box when filled with the metals. Then because the box when full of the metals, by extending the thread with its weight, increased the length of the pendulum, I shortened the thread so as to make the length of the pendulum, when oscillating, the same as before. Then drawing aside the pendulum to the place first marked, and letting it go, I reckoned about 77 oscillations before the box returned to the second mark, and as many afterwards before it came to the third mark, and as many after that before it came to the fourth mark. From whence I conclude that the whole resistance of the box, when full, had not a greater proportion to the resistance of the box, when empty, than 78 to 77. For if their resistances were equal, the box, when full, by reason of its vis insita, which was 78 times greater than the vis insita of the same when empty, ought to have continued its oscillating motion so much the longer, and therefore to have returned to those marks at the end of 78 oscillations. But it returned to them at the end of 77 oscillations.
Let, therefore, A represent the resistance of the box upon its external superficies, and B the resistance of the empty box on its internal superficies; and if the resistances to the internal parts of bodies equally swift be as the matter, or the number of particles that are resisted, then 78B will be the resistance made to the internal parts of the box, when full; and therefore the whole resistance A + B of the empty box will be to the whole resistance A + 78B of the full box as 77 to 78, and, by division, A + B to 77B as 77 to 1; and thence A + B to B as 77 $\scriptstyle \times$ 77 to 1, and, by division again, A to B as 5928 to 1. Therefore the resistance of the empty box in its internal parts will be above 5000 times less than the resistance on its external superficies. This reasoning depends upon the supposition that the greater resistance of the full box arises not from any other latent cause, but only from the action of some subtile fluid upon the included metal.