The Rectilinear Motion Of A Free Point Acted On By Absolute Forces

PROPOSITION 24. THEOREM.

189. When the directions of the motion and of the force are along the same straight line, the motion will be rectilinear.

DEMONSTRATION.

Every body by its inertial force tries to continue its motion in the direction that it always performs, unless it is impeded (65). Truly there are two effects of a force on the motion of a body that we can show : the one by which the direction is unchanged, and the other by which the speed is unchanged. But the direction remains unchanged if the direction of the force lies in the direction that the point moves (128). Therefore in this case the point goes on traveling in a straight line. Q. E. D.

Corollary 1.

190. We will only consider the case in this chapter in which the motion of the point and the direction of the force acting are placed on the same straight line. Corollary 2.

191. Moreover we see that this consistency can come about in two ways, as clearly either both [the inertial force and the applied force] are acting in the same direction, [p. 77] or in opposite directions. In the one case the speed of the point is increased, and in the other it is decreased 128).

Scholium.

192. There are two things that have to be consider in this rectilinear motion, the first of which is the force upon which some point mass is acted, and the other is indeed the speed that the point has at any place. To these we may add also the time in which some interval of distance is traversed. Indeed these three variables are thus comparable, since with one given the remaining two can always be determined. In the first place therefore we will consider some force as given : then truly on that account we will find either the corresponding speed or time from the given force.

PROPOSITION 25. PROBLEM.

193. The point A is resting on the line (Fig. 22) AP, and is to be pulled forwards by a uniform force, or which acts with the same strength everywhere, and the speed of the point is to be determined at any position P.

SOLUTION.

The mass or the force of inertia of the point is set out by the letter A, and the [external] force by the letter g, which is constant since it is the same amount everywhere. Let the distance AP = x, and the required speed at P is put equal to c. An element of distance Pp is taken, which is equal to dx; and the increment of the speed dc is acquired by the point on completing the element Pp, acted on by the force g [p. 78]. With these put in place, it follows that A ngdx cdc = (157), since the force constantly pulls downwards, and on this account we put the motion to be accelerating.

From this equation, if it is integrated, there arises .Const 2 += A ngx cc , which constant must be determined from the condition that the speed of the point vanishes at A. Therefore from this it follows that c = 0, and x = 0, gives the constant as zero. On account of which we have A ngx A ngx ccc 22 or == . Q. E. I. Corollary I. 194. Therefore the point A always falls along the straight line AP, and the speed at any point is as the square root of the distance now traversed. EULER’S MECHANICA VOL. 1. Chapter Three (part a). Translated and annotated by Ian Bruce. page 94 Corollary 2. 195. From these also, the descents of many points acted on by uniform or constant forces can be compared; for the speeds are in the ratio directly of the square root of the forces and the distances traversed, and inversely as the square root of the masses. Scholium 1. 196. This case agrees above all with the fall of bodies on the earth : for the force of gravity, which in turn takes the place of the force, is uniform for not too great distances from the surface of the earth. In so much as the weight of any body from the highest mountain to the deepest valley is found to be the same ; moreover from the weight the force of gravity is found. Therefore in the free fall of weights the speeds are as the square roots of the distances traversed. This is the proposal of Galileo himself, that he discovered first from experiment and then by reasoning [p. 79]. Moreover the descent should be made in a space from which the air has been evacuated, since the air resists the motion and this effect is avoided. Scholium 2. 197. In an empty space, which can be effected with the help of pneumatic pumps, it has been shown by many experiments that any bodies fall equally. From which it follows, if there should be no air, all bodies that fall from equal heights gain equal speeds. On account of this, if g designates the force of gravity, by which any body A is moved, then A g is always the same constant. Hence the force of gravity is proportional to the quantity of matter in the body on which it acts. But that force is none other than the weight of the body ; whereby the weight in the Princip. Phil. confirms this too, and that besides is in agreement with pendulum experiments. Corollary 3. 198. Therefore any body on the surface of the earth fallen from a given height will acquire a step of the speed. Therefore with the height known, from which the body descended, [the speed] acquired from the descent will be known likewise. Scholium 3. 199. Therefore in order that we can measure these speeds, these heights are to put in place from which a weight falling to the surface of the earth acquires an equal speed [p. 80]. Indeed this height cannot be substituted in place of the speed itself, since the speeds are in the square root ratio to the height. Yet truly it will be possible for the height to denote the square of the speed. DEFINITION 15. 200. Hereafter we will call the height corresponding to the speed that height, from which a weight falling to the surface of the earth, acquires that same speed. EULER’S MECHANICA VOL. 1. Chapter Three (part a). Translated and annotated by Ian Bruce. page 95 Corollary 1. 201. This height must therefore be as the square of the speed, to which it refers. With the speed c arising and with the due height v, v shall be as c2. [There is thus to be a proportionality between c2 and v. People had a lot of trouble with non-standardised units in these days, and preferred to use proportionalities instead, whenever they could. Here of course an independent variable v is introduced for the uniform height that can be integrated, without worrying about the more troublesome speed c. ] Scholium 1. 202. This far we have expressed the speed on a straight line, which can be traversed by the speed in a given time. But in the following it will be more convenient to introduce the corresponding height in place of this [Euler calls this the height owing or due, as one would do in a financial transaction; there is hence the understanding that something is conserved or changed from one form into another, in this analysis of a falling body; Euler uses the distance fallen as a means of representing the speed.] On account of this we put v = cc and vc = . We will therefore have in the preceding problem this equation : A ngx v 2 = . Corollary 2. 203. Therefore in what follows, in place of the speed c it will be permitted to put v or the square root of height that corresponds to the speed. [p. 81; Note the use of the symbols : v here refers to verticalis for vertical (height), c refers to celeritas for speed, s refers to spatium or interval of distance; some of these symbols have endured to the present time, as for example c for the speed of light.] Corollary 3. 204. If the force g denotes that of gravity itself, then x will be the height corresponding to the speed c, and thus v = x. For indeed A ngx v 2 = , from which it therefore follows that 2g An = . From this we have gained a convenience, as we have determined the value of the letter n, which must maintain the same value in all cases (155). Scholium 2. 205. Since g signifies the force of gravity [i. e. the weight of the body; do not confuse g with our symbol for the acceleration of gravity, which it does not represent], then A g is a constant quantity (197). Therefore we can put this as 1, as that is allowed, since the force to the [mass] does not [yet] have a defined ratio. And hence it easily shows the ratio of A g in all cases, or the value of the applied force to the [mass of the] body. Certainly the EULER’S MECHANICA VOL. 1. Chapter Three (part a). Translated and annotated by Ian Bruce. page 96 ratio A g to 1 or g : A as the force g, acting on the body A, is to the weight [this should be mass], that the same body may have in our part of the world. Therefore the letter A will no longer denote the quantity of matter, but the weight A of the body itself, [since A.1 is the weight] if it should be placed on the surface of the earth. In this way we will compare all forces with weights, since that will add a great deal of light to the measurement of forces. Corollary 4. 206. Since g An 2 = , g denotes the force of gravity [i. e. the weight of the body and not the acceleration] and if 1=g A , then 2 1=n . That value will always be retained, if the speeds are to be expressed in terms of the appropriate square roots of the heights. And thus in our situation, this becomes A gx A gdx vdv == and . [p. 82. Thus, Euler chooses as his working dimensions not distance and time but acceleration and time. The time is measured in seconds, and with the acceleration of gravity A/g = 1, then 2 1=n .] Corollary 5. 207. On this account in the general law A npds cdc = (157), if the height v of the corresponding speed is c, then 2 dvcdc = , and thus on account of 2 1=n this law is obtained A pds dv = , where p is to A as the force p is to the weight of the body A. [For Euler has defined c 2 = v and hence dv = 2cdc = A pds A npds = 2 when 2 1=n .] Corollary 6. 208. In a like manner, indeed the equations set out in (161) and (163) : ,and 2dsAcnprdxnpdyAcdc == by substituting v in place of c 2 and 2 1 in place of n, are transformed into ,2and AvdsprdxpdyAdv == where p to A has the ratio in the manner given. Corollary 7. 209. And in (165) it is found that 2 p Avr = or pr = 2Av. Likewise in (165) there is obtained ,pdsAdv = and in the case of (167) there is obtained .pdsAdv −= And in this manner we have reduced the previous variable quantities n and c to fixed values. EULER’S MECHANICA VOL. 1. Chapter Three (part a). Translated and annotated by Ian Bruce. page 97 PROPOSITION 26. THEOREM. 210. The heights, by which equal small bodies fall to acquire equal speeds, vary inversely as the forces, under the hypothesis of different uniform [gravitational] forces. DEMONSTRATION. Let the mass or weight of some small body on the surface of the earth be A, the force [on some other celestial body] some constant value g and the corresponding height [p. 83] for this speed to be acquired be v. Truly the height shall be x, by which the small body A falling under the action of the force g shall acquire a speed [c2 = ] 2 A ngx v = (202). But 2 1=n (206), hence A gx v = or Av = gx. [On the Earth, g/A = 1 and v = x.] Whereby, since the speeds have been produced by different forces and the bodies have been made equal, then the quantity Av is constant, and thus also gx. On account of this, x will vary inversely as g, i. e. the height, by which the body A acquires the speed v by the action of the force g, varies inversely as the force g. Q. E. D. [No doubt by now the reader has thrown up his or her hands in horror and said : what about the conservation of mechanical energy ? The factor n = ½ has arisen from the requirement that the acceleration of gravity is one. The problem for us lies in the lack of understanding that there was at the time about kinetic and potential energy and the conservation of the sum of these for a falling body. Euler has been very clever and set up his differential equations so that they can be scaled, and when it comes down to doing a numerical example, as with the flight of the cannonball in Ch. 4, he gets the correct answers. Thus, the answers right themselves when known experimental values are put in place for the time of fall of a body. It has been convenient to put the acceleration of gravity arbitrarily as 1, from which by (209) n = ½ (for Euler does his best to get rid of constants that always appear, a tradition that has been followed by theoreticians ever since!); but if the time is measured in seconds and the distance in scruples or thousandth parts of Rhenish feet, then the acceleration of gravity is not 1, but something around 32 ft/s2 or 32000 scruples/s2 . If fact, from the equation gHV 22 = , for the speed V of a body dropped from rest from a height H is given by HHHgV 250625002 =×≈= , the scaling factor used by Euler later. Hence when experimental results are imposed, the correct value for the acceleration of gravity results, and all is well.] Corollary 1. 211. Newton has shown that the impressed forces on the same body put in place on the surface, and acted upon towards the centre of, the Sun, Jupiter, Saturn, or the Earth, are as 10000, 835, 525, and 400. Therefore the heights from which the body acquires equal speeds in falling on the surface of the Sun, Jupiter, Saturn, and the Earth, are between themselves as .and,, 410 1 525 1 835 1 10000 1 EULER’S MECHANICA VOL. 1. Chapter Three (part a). Translated and annotated by Ian Bruce. page 98 Corollary 2. 212. Moreover Newton understood likewise that all bodies fall at equal rates on these surfaces, just as on the surface of the earth. Therefore there is no need to add the condition that the bodies are equal, for from the heights which are in the ratio to each other ,and,, 410 1 525 1 835 1 10000 1 on the surfaces of the Sun, Jupiter, Saturn, and the Earth, any bodies dropped will acquire the same increase in their speed. [p. 84] Scholium 1. 213. It is understood from these that there is a two-fold effect on the body by any force : on the one hand, by which a certain force or effort is impressed on the body, and on the other, by how the body may be moved by that force. The one that is to be considered mainly in statics is the weight and how it should be measured, that the body has in an equal attempt to fall downwards, and it may be called the absolute strength of the force. In turn, truly the effect should be measured by the acceleration or the change in the speed, that is impressed on the body in a given time : this is proportional to that force divided by the mass of the body (154). This effect is called by Newton the accelerating force, and therefore the strength of the accelerating force is proportional to absolute force applied to the mass of the body, or the weight applied. On account of which, since A pds dv = (207) and A p denotes the strength of the acceleration, then dv is equal to the product of the acceleration and the element of distance travelled. Thus the absolute force of gravity is proportional to the mass of the bodies on which it acts ; for the effect of these downwards is the cause or the weight that we have shown to be in proportion to the mass. Moreover the accelerating force of gravity is equal on all bodies, since they all fall equally and they gain equal speeds in equal intervals of time. Corollary 3. 214. Hence the sizes of the accelerations are to each other as the absolute forces, if the bodies are of equal mass. Whereby since the strength of the acceleration due to gravity is taken as 1, as we have put in place before (205), [p. 85] then the strength of the accelerating due to gravity on the surface of the sun is equal to 24.290; the strength of the acceleration on the surface of Jupiter is 2.036; the strength of the acceleration on the surface of Saturn is 1.280. And the strength of gravity has been taken on the moon by Newton as 3 1 . Corollary 4. 215. Whereby if from Proposition 25 the fall of the body to the surface of the Earth is to be accounted for, then 1=A g , as we have done in (205). But truly the fall of bodies on the surface of the sun requires 290.24=A g ; or on the surface of Jupiter, 036.2=A g ; on the surface of Saturn 280.1=A g ; and for the fall of bodies on the surface of the moon, it will be 3 1=A g . EULER’S MECHANICA VOL. 1. Chapter Three (part a). Translated and annotated by Ian Bruce. page 99 Scholium 2. 216. Here we assume with Newton that all celestial bodies are similar to our Earth and bodies placed on the surface of these have a force pulling then to their centre, which is always similar to the force of terrestrial on bodies. Therefore, from Newton’s exposition, it is apparent that a body, the weight of which here is 1 pound [lb.], will weigh on the surface of the Sun 24.290 lb ;on the surface of Jupiter, it will be 036.2 lb; on the surface of Saturn 280.1 lb ; and on the surface of the moon one third of a pound. [p. 86] Scholium 3. 217. Moreover in order that the nature of the gravitational forces can be more easily compared for celestial bodies, the individual equal elements of the bodies are understood to be equally affected by gravity. From which it follows, since now it agrees with experiment, that the forces of gravity which act on any bodies, are themselves proportional to the masses or quantities of matter. Truly it has been shown previously, that if the forces are in proportion to the masses of the bodies, then their effect on moving bodies is the equal (136). On which account it is shown from these that all bodies on the surface of the earth should descent equally, and likewise for all celestial bodies. PROPOSITION 27. PROBLEM. 218. For the point A (Fig. 22) is to be moved forwards through the distance AP by a uniform force, and it is required to find the time in which the distance AP is completed. SOLUTION. As before, let the force acting be g, the distance AP = x and the height corresponding to the speed at P should be v; on this account .,2 1 A gx vn == Therefore the speed itself at P is equal to .A gx v = The time therefore be found, in which the element Pp = dx is traversed, varies as gx Adx . Let the time in which the distance AP is completed be t and put , gx Amdx dt = it is necessary to perform a single experiment to determine the value of the letter [p. 87] m, so that the time in the given measurement can be found, reckoned in seconds. Indeed from that equation the time t will be produced by integration and we have ,2 g Axmt = to which there is no need to add on a constant quantity, since for the position x = 0 the time t vanishes, as it should. Therefore on determining m by experiment we have the time to fall given by g Axmt 2= sec. Moreover, so that the measurement of the time by this means shall be absolute, it is necessary that x likewise is shown to be measured in agreement with this constant [i. e. in some standard unit of length]: therefore we will always determine the interval x in EULER’S MECHANICA VOL. 1. Chapter Three (part a). Translated and annotated by Ian Bruce. page 100 scruples, i. e. thousandth parts of Rhenish feet ; for indeed the fraction g A is expressed in absolute numbers, in order that it will not be necessary to have a certain measurement for that. Therefore with the letter m defined, that we will make soon, the complete solution to the problem will be had. Q. E. I. Corollary 1. 219. If g designates the force of gravity, then 1=g A (205) ; on account of this the time in which the body will fall to the earth from the height of x scruples of Rhenish feet, will be xm2 seconds. Corollary 2. 220. Moreover from experiments it has been ascertained that in a time of one second a body falls through a height of 15625 scruples of Rhenish feet [4.904 metres, which is in good agreement with the modern accepted value, although we do not know where the measurements were made for the full significant figures]. When on account of this, if the height is put as x = 15625, a time of one second will be produced : t = 1. But since xmt 2= , then 1562521 m= , i. e. = 250m. Hence the letter m is found to be 250 1=m . [p. 88; when we compare this equation with the present value of the acceleration of gravity of around 9.8 m/s2 in which case 2 2 1 gtx = , then in Euler’s units of scruples, the acceleration of gravity g is given by 15625 × 2 = 31,250 scruples/s2 , corresponding to 9.81 m/s2 ; we have discussed this in a slightly different way previously at the end of (210).] Corollary 3. 221. Since indeed the letter m retains the same value in all cases, in the case of the problem, the time will be g Axt 125 1= sec. Therefore with the distance x expressed in scruples of Rhenish feet, the time as the number of seconds is given by g Ax 125 1 for this space to be traversed [by falling from rest]. Corollary 4. 222. And in all straight forward cases this value of m found can be applied. Indeed let the element of distance to be described be ds, with the height v corresponding to the speed in which this is traversed, the element of time is .and ∫== v ds v mds mtdt From this equation, if v and s are expressed in scruples of Rhenish feet and we put 250 1=m , the time will be produced in seconds : ∫= v dst 250 1 sec. EULER’S MECHANICA VOL. 1. Chapter Three (part a). Translated and annotated by Ian Bruce. page 101 Scholium 1. 223. Therefore from this, since we specify the speeds by the square roots of the corresponding heights, which we take again for convenience in the following, as we always find the measure of the absolute time. For truly the experiment we have used from which the height is found that a weight falls in a time of one second, as Huygens found by experiments with pendulums to be 15 Paris feet, 1 digit, 18 12 lines, i. e. in decimal fractions 15.0796 Parisian feet. Moreover the ratio of Rhenish feet to Parisian feet we use is 1000 to 1035, from which the height to fall in one second comes to be 15.625 Rhenish feet or 15,625 [p. 89] scruples of the same feet; and this is the measure we prefer to have rather than the Parisian one, since this number is a square and in this way we avoid frequent root extractions. Besides, the number 250 that is found is easily remembered, by which ∫ v ds (with s and v expressed in scruples of Rhenish feet) must be divided in order to find the time in seconds. Corollary 5. 224. Since A g denotes the strength of the acceleration of the force (213), the times will be, in which any distances are traversed under uniform forces, in the square root ratio composed from the distances and inversely as the strengths of the accelerations. [i. e. 2 ats ∝ ] Corollary 6. 225. With the speed c put in place, that the point A acquires with a force g acting over the height x; c varies as A gx (193). Hence ct will be as x, since t is as g Ax . Consequently the distances travelled through are in the ratio composed from the times in which they are described, and the speed that they gain in the descent, for whatever the forces acting in a uniform manner. Corollary 7. 226. And the distances that are traversed in equal intervals of time are as the strengths of the accelerations of the forces acting. [p. 90] Corollary 8. 227. Therefore the distances through which bodies fall in equal intervals of time on the surfaces of the Sun, Jupiter, Saturn, the Moon, and the Earth, are amongst themselves as 24390, 2036, 1280, 333, 1000. (214). EULER’S MECHANICA VOL. 1. Chapter Three (part a). Translated and annotated by Ian Bruce. page 102 Corollary 9. 228. For the hypothesis of the same acceleration under the same force, the times in which any distances are traversed , are as the speeds acquired, and so for the times as with the speeds, are in the ratio of the square roots of the distances described. Scholium 2. 229. Here we have always put the descending bodies to start the descent from rest or their initial speed to be zero. In the following indeed we will investigate these motions, which arise when the bodies are themselves in motion at the start and now they have some speed. Moreover with these both the times and the distances ought to be understood, which in the beginning have their own point where the speed vanishes, and all the equations found are thus comparable, as with the vanishing of c or v likewise x and t vanish. PROPOSITION 28. THEOREM. 230. For a body falling through the distance AP (Fig. 22), as we have put in place as hitherto, the [final] speed at P is of such a size, that if it progressed uniformly at this speed for the same time in which the body had fallen through AP, then it would be able to complete a distance twice as great as AP. [p. 91] DEMONSTRATION. With everything kept, that we put in place in the preceding proposition, with the body A, the force g, with the distance described x, with the speed acquired at P v and with the time of the descent t, then g Axt 125 1= (221) and A gx v = (206). Then on account of this we have x v A g = , and thus . 125 2 125 v x v xt == But this expression also gives the time, in which the distance 2x will be traversed with a uniform speed of v , since v x2 is divided by 250, and then we find the number to be expressed in seconds (220). Consequently the distance 2x is travelled in the same time with the speed v , in which the distance x is fallen under a uniform acceleration. Q. E. D. [Thus the final speed is twice the average speed for the motion of a body released from rest under gravity.] Corollary 1. 231. Therefore a body acted on by a uniform force falling in a time t through a distance x will acquire as great a speed, as that by which a body can progress uniformly through the same distance in half the time t. EULER’S MECHANICA VOL. 1. Chapter Three (part a). Translated and annotated by Ian Bruce. page 103 Corollary 2. 232. Since on the surface of the earth bodies are falling in a time of one second a distance of 15625 scruples of Rhenish feet, their final speed acquired in the fall will be as great as that with a uniform motion over a distance of 31250 in one second, or 15625 scruples traversed in a time of half a second. [p. 92] Corollary 3. 233. When the speeds are expressed, as we have established, from the square roots of the heights through which they have fallen, the speed will be as 15625 or as great as 125, in which time a body in one second can complete a distance of 31250 scruples. [For which the constant of proportionality can now be evaluated : thus a distance of 15625 corresponds to a speed of 125 in the square root proportionality, for which the same distance of 15625 scruples corresponds to an actual speed of 31250 scruples/sec. = 250 × 125; or the true speed = 250 × v for the height v fallen. Thus, the constant of proportionality is 250.] Corollary 4. 234. It is therefore easy to assign the distance that will be traversed in a time of one second with the speed expressed by v . Indeed it happens that the distances described in the same time are in the same ratio as the speeds, thus 125 is to v thus as 31250 scruples is to 250 v . For which by a factor 250 v expresses the distance in scruples traveled in one second, if indeed for the height v it may be shown by such a proportionality that the distance can be completed with a speed v in a second, and which is clearly is equal to the motion. Example 1. 235. The fall of a body from a height of 1000 ped., will be v = 1000000 in scruples, whereby for this descent it will acquire as much speed as in one second it travels a distance 250000 scruples, i. e. it will be able to complete 250 feet in one second. Corollary 5. 236. And reciprocally, if the speed is expressed by the distance that is traversed in one second, as we did in the beginning, then this can hence be reduced to our method of taking square roots for the corresponding heights. Indeed if that distance is a scrup., and the corresponding height for this speed is v scrup., then it follows that 250 v = a and 62500 2 av = scrup. [p. 93] EULER’S MECHANICA VOL. 1. Chapter Three (part a). Translated and annotated by Ian Bruce. page 104 Example 2. 237. Let the body have such a speed that in a second it is able to traverse a distance of 1000 feet or 1000000 scrup., then the height that corresponds to this speed will be 62500 0001000000000= scrup. or 1600 feet. Scholium. 238. Therefore the way in which each speed is to be expressed in turn is clear, and how it may be required that the one can be reduced to the other. For initially we were expressing the speeds by the distances travelled in a second, or in some other interval of time. Afterwards, truly it was seen that the speeds were shown to be in agreement with the corresponding heights. Now truly we show how each way can be adapted to measure the speed. PROPOSITION 29. PROBLEM. 239. The body now initially at B has a given speed along the line BP, and with a uniform force present acting along the line BP (Fig. 23); the speed of this body is required at any point P of the line BP. SOLUTION. As before let the force of gravity be g and the body A. Indeed the speed that it has in the beginning at B is due to the height c. We call BP = x, and the speed at P, that we are looking for, shall correspond to the height v. As before (207) on account of the constant force g acting on the element Pp = dx , it follows that A gdx dv = . [For in (207) Euler has defined c2 = v and hence dv = 2cdc = A pds A npds = 2 .] By integrating [p. 94] we therefore have .Constv A gx += , which constant quantity is to be determined from this, since by taking x = 0 then v = c (by hyp.); and therefore the constant is equal to c. Consequently we have : A gx cv += and speed itself : .A gx cv += Q. E. I. Corollary 1. 240. If we put 1=A g , as in the case of ordinary gravity, then v = c + x. Therefore the height corresponding to the speed at P is the sum of the initial height at B and the distance traversed. EULER’S MECHANICA VOL. 1. Chapter Three (part a). Translated and annotated by Ian Bruce. page 105 Scholium. 241. The solution of the problem comes to mind in another way; for we put the motion along BP with the initial speed c in B to be considered as part of the motion from rest along the line AP, as we have put the term, in which the body as it came from A to B has the proposed speed c . Therefore if this distance AB = k, then A gk c = (206) and A gx A xkg cv +== + )( , as now it has been found. Moreover the distance AB is g Ac . PROPOSITION 30. PROBLEM. 242. With everything put in place as in the preceding proposition, to determine the time in which the distance BP (Fig. 23) is run through. SOLUTION. Let the time of for the distance BP = t, then )( A gx c mdxdt + = (218) ; )( A gx c+ is the speed that the body has in traversing the element Pp , as we have found in the preceding proposition. On integrating, this therefore becomes .Const)(2 ++= A gx g mA ct This constant amount is to be added from this can be defined, since for the position x = 0 , it must make t = 0. Therefore the constant produced will be : ..Const 2 cg mA−= Consequently we have .)( 22 cct g mA A gx g mA −+= With c and x expressed in scruples of Rhenish feet put in place, 250 1=m (220) , and there comes about the time expressed in seconds. Q. E. I. Corollary. 243. Since in our terrestrial regions 1=A g , the time in which the distance BP is completed with the speed given at the start at B by c , is equal to ccv 125 1 250 1 )( −+ seconds, if c and x are expressed in scruples of Rhenish feet. Scholium. 244. In a similar way we can affirm another solution for this problem, which is added to the previous as a scholium. For on putting the line AB = k, from which the body falling from A to B gains the speed corresponding to the height c, the time will be, in which this EULER’S MECHANICA VOL. 1. Chapter Three (part a). Translated and annotated by Ian Bruce. page 106 distance is completed, equal to g Ak 125 1 and the time, in which the distance AP is run through, will be g Ak g xkA 125 1)( 125 1 −

• . Therefore g Ack = (241). Consequently this time sought for the distance BP becomes cc A gx g A 125 1 125 )( −+ , as we found before, if 250 1 is put in place of m. And these are the solutions of the problems concerned with the descent bodies according to the hypothesis of uniform forces that had to be set out. Therefore I can go on to rectilinear ascent, in which the direction of the speed is in the opposite direction to the force, that also is now uniform or that I make constant. PROPOSITION 31. PROBLEM.

245. The body has a given speed in the upwards direction and with a uniform force pulling in downwards ; the speed of the body is required at any point in the interval BA that it has travelled through while rising. SOLUTION. It is evident that in this case that the progression of the motion along the straight line to be one of retardation (191), since the direction of its motion is acting against the direction of the force. And thus the speed of the body at B is due to the altitude c, and the body is put in place arriving at P. The altitude to which the speed at this place is owed, is called v, and the distance BP now traversed is called x. Take Pp = dx, and the altitude to which the speed is owed [i. e. corresponding] at p will be v

• dv. Moreover since the force, that I put equal to g, is contrary in direction to the motion, the total speed is diminished. On account of this , it is necessary to put dv equal to A gdx− , with A denoting the mass of the body. And thus A gdx dv =− , and on integration, it becomes A gx vC =− . In order that the constant C can be found, put x = 0, in which case v is changed into c; and hence it becomes C = c. From which the equation itself is produced : A gx A gx cvvc −==− or , which determines the speed of the body to be described in any point of the ascent. Q. E. I. Corollary 1.

246. The speed therefore vanishes when A gx c = , i. e. when it arrives at the height g Acx = . Let BA be equal to that height A gc , hence A gBA c . = , from which it is understood that BA is the height itself, from which the body A acted on by the force g by falling acquired the speed c (206). Therefore the body with that speed, that it gained by falling through the EULER’S MECHANICA VOL. 1. Chapter Three (part a). Translated and annotated by Ian Bruce. page 107 given height, progresses upwards again to reach the same height that it had at the start of its motion when it was sent off. Corollary 2.
247. Besides the body ascending through the distance BA has the same speeds at the individual points, that it has at the same points if it is descending from A. For with AP = y the speed at P descending, coming from A, is equal to A gy ; but the speed at the same place P in the ascent from B leaves )( A gx c − . Moreover, since g AcBAyx ==+ , it is apparent that these expressions of the speed are equal , surely .A gx A gy c −= Corollary 3.
248. Therefore the motion of the ascending body agrees with the motion of the descending body, and the speeds of each at the same points are equal, i. e. which are placed at the same distance from the upper point at which the speed is zero. Corollary 4.
249. From these it is likewise evident that the time of ascent through the distance BA is equal to the time of descent through the same distance. Whereby by calling BA = a, the time of descent is equal to g Aa 125 1 sec. (221), and that must be equal to the time of ascent through BA; or to the place with the value g Ac , the time of the whole ascent will be cg A 125 . Corollary 5.
250. In a similar manner the time of the ascent through any part BP can be defined; for keeping AP = y the time for either the ascent or the descent through AP g Ay 125 1= , that taken from the time of the whole ascent , which is equal to c125 1 , there is left the time to pass through the part BP. This distance is given by ,xy g Ac −= whereby the time to ascend through BP is given by .)(125125 A gx g A g A cc −− Scholium 1.
251. Evidently in the first place the equation of the ascent has been shown from the action of the force. Since indeed in the ascent the force takes away as much from the speed, as it adds on during the descent, it is evident that there is a complete equality between each motion, and one can be turned into the other without any distinction, unless the succession of the time, by knowing which the one case is only turned into the other. EULER’S MECHANICA VOL. 1. Chapter Three (part a). Translated and annotated by Ian Bruce. page 108 Indeed the reasoning is also the same of all the motion produced by the absolute forces: for the body can be turned around in its path with the same speeds, if indeed in the return motion the impressed forces are the same, as those in the departing motion, but to be clearly in the opposite direction. Thus the planets would go round the sun in contrary orbits, which now move in ellipses, if in the first place they had been sent off in a motion contrary to these. For through the same element of distance, the effect of the force in changing the speed is always the same, and for that reason it shall be negative, when the body returns. Moreover the effect, which is devoted to changing the direction of motion of the body, in each case remains the same, which is, that the body in coming and going retraces the same path. But this will become more apparent below, where motion of this kind will be established. But truly, if it is decreased by resistance, this similitude between the ascending and the descending motion vanishes : for in each case the resistance to the motion is made less, and neither is the effect of this on either of the other in the opposite direction, as usually comes about for absolute forces acting. Scholium 2.
252. Enough therefore with rectilinear motion, which arises from uniform forces, and it is time to move on to different kinds of forces, which in different places exert other forces on bodies, and that needs to be examined, for whatever the motions of the bodies, in as much as they shall be on a straight line, they will vary from the above. Indeed all the forces that we observe in the world suffer from this kind of difference, that no force can be assigned to a body put in any place that will equally affect the body. Thus the planets, where they are closer to the sun, [p. 100] are attracted to the sun by a stronger force; and also where a body more removed from the surface of the earth, the weight or the exertion downwards shall be less on that body. It almost comes about in the same manner, where we observe the magnet to attract an iron filing more at shorter distances, and the attraction to be weaker at greater distances. Therefore we will elicit the laws, for whatever ratio of the forces with distance they hold, according to which the motion of the body acted upon can be changed. And in the first place we are indeed to contemplate forces that vary according to the distance of the body from a fixed point raised to some power. Definition 16.
253. That fixed point is called the centre of attraction, to which bodies are attracted by a force, which depends on the distance from this point, or which is as some fraction of this distance. Corollary 1.
254. Therefore given the distance from this centre of force in which the body placed is drawn to the centre by as large a force, as if it should be the force of gravity acting on this, and placed on the surface of the earth. Corollary 2.
255. Therefore with this distance and the law of the attraction known, clearly given by a function of the distance, to which the attraction is proportional, the ratio of this force is EULER’S MECHANICA VOL. 1. Chapter Three (part a). Translated and annotated by Ian Bruce. page 109 known, for any position of the body trying to fall towards the centre of force, to the force of gravity on the same body acted on if it should be on the surface of the earth. [p. 101] Corollary 3.
256. Thus in this manner, any variable forces are allowed to be compared with the force of gravity, since the effect of this force on the body is known, and thus also the effect of any force on the body can be determined. Scholium 1.
257. I put this attraction of the centres of the forces similar to the force of gravity, thus in order that likewise the forces of different bodies placed in the same position are as the masses themselves, and thus the magnitudes of the accelerations are all the same (212). Therefore in handling these problems, it is not necessary to call upon the mass of the body in the computation of the motion, but only the magnitude of the acceleration, to which the force of attraction to the centre divided by the mass is in proportion. Moreover it can be compared with the acceleration of the force of gravity, that we put equal to one, and we will compare all the accelerations due to the magnitudes of these forces with this acceleration of unity, clearly homogeneous quantities. Corollary 4.
258. Thus when we talk about the forces as being as the distances from the centre of force or proportional to a certain function of these, it is not only the forces that the bodies have to the centre, but also the accelerations associated with these force, i. e. it is the ratio of the force to the mass, that should be understood. Corollary 5.
259. Therefore since the direction of the force which presses upon the body, always pulls towards the centre of force, [p. 102] it is evident, if the body is either at rest or it has motion, the direction of which passes through the centre of motion, then the body must be moving on this straight line perpetually crossing through the centre of force (189). Definition 17.
260. The force, which presses upon bodes to the centre of this kind of force, is called the centripetal force. And that, if it is negative, in order that the body is repelled from the centre, is called the centrifugal force. Corollary 1.
261. Since this will be the question about the motion, the centripetal force will be for us the magnitude of the acceleration, or the force pulling the body towards the centre divided by the mass of the body. EULER’S MECHANICA VOL. 1. Chapter Three (part a). Translated and annotated by Ian Bruce. page 110 Corollary 2.
262. Therefore the effort or the striving that the body has towards the centre of force, is expressed by the strength of the centripetal force [i. e. the acceleration] multiplied by the mass of the body. On account of which it will be to the weight of the same body, if it were put in place on the surface of the earth, as the strength of the centripetal force or the strength of the acceleration to unity. (257). Scholion.
263. Newton, who mainly talks about the centripetal force [in the Definitions in the Principia], has paid attention to three ways in which the effects of the same can be measured. In the first case, it provides a measure of the absolute quantity of the effectiveness of the centre of the force, without regard to the [mass of the] body being attracted ; thus he asserts, in the case of the larger loadstone the greater the absolute quantity of the centripetal force present, and in the case of the lesser lodestone, the corresponding centripetal force is smaller too [p. 103; See Def. VI. Principia ; though the mass of the orbiting body would not cancel, if such a motion were possible, as in e/m experiments with a uniform magnetic field.] And in a similar manner, following this theorem, the absolute quantity is greater on the sun than it is on the earth. Moreover this comparison is to be understood from the similarity of the centres of force, i. e. according to the same function of the distances, and with attractions; indeed the comparison does not have differences of this kind at the same place. Hence this absolute quantity of force is to be measured from a known effort, which the body has exerted on it at a given distance from the given centre of force. Moreover in place of this consideration, I put in place the distance into which the body can be put with a force equal to its weight pushing towards the centre (254). In the second place the strength of the centripetal force has the magnitude of an acceleration, which is perceived by the senses at the object itself, where the centripetal force itself is acting (261); indeed it is measured by the ratio of the effort applied to the mass. In the third case, the strength of the centripetal force leads us to the magnitude of the motive force, which is specified by the force the bodies experience on approaching the centre of force; the motive force is the quantity of motion, and that is usually measured by the product of the speed by the mass, and which is produced in a given time proportional to this effort itself. And for this force p, and the mass A, the increment in the speed in a given element of time varies as A p (154), that multiplied by the mass A gives the increment in the quantity of motion, that is proportional to the force p. [According to Cohen, on p. 406 of his translation of the Principia, which is very good on this point, Newton’s summary of the three effects of centripetal force are : absolute, accelerative, and motive. One may take the first to mean that the centripetal force lies in the category of absolute forces; the second that it involves an acceleration; and the third that the momentum of the body is related to it from the impulse divided by time relation, as Euler sketches above. One may raise the odd eyebrow at Def. 8, which tells us that it is the opposing centrifugal force which prevents the body from falling…… Well!! It may also be appropriate to note what the modern usage is by contrast, at least according to this translator’s understanding : Newton’s Second Law, which he never actually enunciated EULER’S MECHANICA VOL. 1. Chapter Three (part a). Translated and annotated by Ian Bruce. page 111 himself in the form ‘F = ma ’ is a cause – effect relation. The cause can be some phenomenologically determined relation such as the law of universal gravitation, Coulomb’s law of electrostatics, etc: this takes care of the first property. The effect is what the mass does in response. If the force is always applied at right angles to the direction of motion, then we get the centripetal acceleration, which is a purely kinematic quantity: which is the second property. The third property is an amalgamation of these two, whereby the rate of change of momentum can be used as a measure of the central force, as for example for a moon rotating around a planet, etc. This is all viewed in the realm of classical physics. Newton was to correct his unfortunate assertion made above, when he discussed cannon balls falling towards the centre of the earth when in orbit.] PROPOSITION 32. PROBLEM.
264. Let C be the centre of the forces (Fig. 25) that attract bodies in some ratio of the distance, and by this force a body at rest at A is drawn forwards; the speed of this body is then sought at any point in the interval AC. [p. 104] SOLUTION. Let AC = a, AP = x; and the speed that the body has at P is that corresponding to having fallen from the height v. The attraction shall be given as the ratio of the distance raised to some power n, and f is taken as the distance from C, at which the force on the body towards C is equal to the weight of the body, if it should be placed on the surface of the earth. Therefore the strength of the acceleration, by which the body at P is pulled towards C, will be as the strength of gravity, that I put equal to 1, as CPn, i. e. as nn fxa to)( − ; on account of which the acceleration is expressed by )( n n f xa− . Therefore by taking Pp = dx then )( n n f dxxa dv − = . For dv is equal to dx multiplied by the strength of the acceleration (213). This integrated equation produces )1( )( 1 n n fn xa Cv + − + −= . For the constant C to be defines put x = 0, in which case by hypothesis it must become v = 0; therefore )1( 1 n n fn aC +

= . It is therefore found that )1( )( 11 n nn fn xaa v + −− ++ = . Or by putting a – x = CP = y it becomes )1( 11 n nn fn ya v + − ++ = . From which equation the speed of the body at any point of the interval AC is known. Q. E. I. [Thus, the force on the body at position a – x is given by some function n xak )( − , and the acceleration is mxak n /)( − , where m is the mass of the accelerated body and k is a constant of proportionality. However, when the body is at f, it is considered to have the force equal to its weight acting on it, which is just m, as g is taken as equal to 1; hence EULER’S MECHANICA VOL. 1. Chapter Three (part a). Translated and annotated by Ian Bruce. page 112 1.mkf n = , or n fmk /= , giving the force as nn fxam /)( − , and the acceleration as nn fxa /)( − . In addition, for motion under gravity 2 2 Vvg = , where V is the final speed at the point and g is taken as 1, then VdV = gdv and also adxVdV = for the motion under the new force; hence adx = dv, where a is the acceleration under the new force, and the point masses under gravity and under the new force have the same speed and increment in the speed; as these are not in proportion in general, each point must correspond to a different release point. This gives the ratio of the accelerations under gravity and the force, which is a function of the distance. From which it follows that )( n n f dxxa dv − = , which can then be integrated as above. We would now proceed in a slightly different manner , and set the acceleration ;// dxVdVdtdV = in which case nn fxadxVdV /)(/ −= and )1( )(2 2 1 1 n n fn xa CvV + − + −== , etc. We may note also that this first integral is in fact just the conservation of energy, as the sum of the kinetic and potential energy is related to the constant C. This relation obviously breaks down when infinite quantities are involved.] Corollary 1. 265. If n + 1 is a positive number, yn + 1 vanishes when y = 0. Therefore in this case [p. 105] the altitude corresponding to the speed, that the body has on arriving at C, will correspond to )1( 1 n n fn a + + . But if n + 1 is a negative number, yn + 1 will become infinitely large when y is made zero : hence in this case the body on arriving at C will have an infinitely large speed. Corollary 2. 266. But if n + 1 = 0 or n = -1, the value found from the equation itself may not be known on account of the numerator and the denominator vanishing. Because of this, it will be necessary to repeat the differential equation. Moreover, it follows that xa fdx dv − = , the integral of which is )( xaflCv −−= . And it must be that C = fla, on account of which )]log([ y a y a xa a fflflv === − . Which is the true value of v, when n has the value - 1, i. e. when the centripetal force varies inversely with the distance from the centre of the force. Corollary 3. 267. Therefore in this case, n = – 1, when the body arrives at the centre C, its speed is infinitely great; for it shall be that v = fl ∞ . This infinite step is to be deplored, and if a nearby value is taken, it is finite ; however if n + 1 should exceed zero a little, then the speed at C suddenly becomes finite. EULER’S MECHANICA VOL. 1. Chapter Three (part a). Translated and annotated by Ian Bruce. page 113 Corollary 4. 268. Moreover since n + 1 should be a positive number, since then the height corresponding to the speed at C is )1( 1 n n fn a + + , then the speeds of many bodies falling [p. 106] towards the centre C, and which they have at C, are as 2 1+n a , i. e. as the 2 1+n power of the distances from which they have began the fall. Scholium 1. 269. Moreover, after the body arrives at C from A in C, where then it shall keep on moving forwards, it is not so easy to be defined. Indeed it is observed, that if y is made negative in the expression found, the height corresponding to the speed at Q should be emerging ; which if it is positive, then the body again returns to Q ; but truly if it is negative, from the evidence, this body never reaches beyond C into the region CQ. In truth this way of continuing the motion is not always possible to be adhered to; often indeed the hypothesis itself, by which the attractive force is placed before and beyond C towards the centre is opposite. In as much as the body proving to be at P, since it is being pulled down, when it arrives at Q, it is pushed up by an equal force, if CQ = CP. This force, on account of the nature of the force which is acting on the body at Q, is negative with the former ratio and thus is to be expressed by a negative quantity. Therefore the force at P expressed by n n n n f y f xa or )( − must be the negative of itself, when – y is put in place of y, and that never happens, unless n is either an odd number or a fraction, of which the numerator and denominator are uneven. Therefore for these cases the value of v is produces, when the body arrives at Q ; always in the remaining cases, since in calculating the force acting on the body at Q when indeed with the value not in agreement, the quantity elicited for the letter v is not the true value [p. 107]. If indeed n is an even number, the attracting force at Q by making y negative is equal to the force at P and clearly falls in the same place. From which it shall be, that as the body crosses the centre C on the line CQ must continue to fall to infinity, that calculation also makes clear. Because when it disagrees with the hypothesis, it is seen that in these cases the motion of the body, after it has arrived at C, cannot be defined by the formula defined. Moreover it is seen to be more absurd, when 2 1=n or another fraction of this kind, which changes yn into an imaginary quantity with –y put in place of y; because that may indicate that that not only is the body not attracted to C, but the force of attraction also becomes imaginary, which is indeed not possible to understand. Corollary 5. 270. Therefore if n is an odd number, the value of v itself, which is )1( 11 n nn fn ya + − ++ , does not change with – y put in place of + y, since the even number n + 1 of y avoids the [sign EULER’S MECHANICA VOL. 1. Chapter Three (part a). Translated and annotated by Ian Bruce. page 114 change in the] exponent. From which it is apparent that the speed of the body at Q is equal to that which it had before at P, if indeed CQ = CP. Therefore the motion is equal in the manner in which the body recedes in the direction CQ, by which before it approached along AC; and it shall reach as far as B, thus in order that CB = AC, where it loses all its speed. And thus it reverts again in the same way to C, and then it arrives at A again. Which reciprocal motion, unless decreased by friction, will be carried out indefinitely.[p. 108] Corollary 6. 271. Nevertheless the case when n = – 1, since – 1 is an odd number is to be undertaken. For with y made negative y aflv −= , which is an imaginary amount. From which it is seen that the body never goes beyond C. Hence another judgement is seen to be brought down, when n is a negative number, even if it is odd. For a similar example of this kind occurs beyond, if n = – 3 (355). Scholium 2. 272. Indeed this is seen to be less in agreement with the truth; for the reason is hardly apparent why the body with its infinite speed that it acquires at C should be about to progress into CB rather than another region, especially when the direction of this infinite speed should follow into this region. But whatever it shall be, here the calculation rather than our judgement being trusted and established, the jump if it is made from the infinite to the finite, is not thoroughly understood. Moreover, this opinion is further confirmed by a similar example for which a full explanation is given below, (665), if n = – 2; for in this case the speed of the body arriving at C is also infinite and directed along CB ; by no less truth, in this case the body does not progress beyond C, but suddenly reverts from C equally and approaches towards A. From which it is understood, that as often as an infinite speed should arise at C, judgement about the further motion of the body should be suspended. So for the time being only this shall be done, until we come to considering motion along curves [p. 109] ; and with these indeed which are rectilinear, and [the resolution of this problem is] clearly connected to these(762). For neither then is the calculation which is put in place subject to this inconvenience, as it is in disagreement with the hypothesis ; but whatever is put equal to the centripetal force is not in opposition to the calculation. Scholion 3. 273. But always, when the speed at C is not infinitely great, because that happens, when the size of the number n + 1 is positive, the whole motion of the body is known by our judgement, even if the calculation is insufficient. For if the speed at C is finite and has the direction along CB, that by necessity is should have, then it may not be possible to happen, as no motion can be continued along CB. But in a like manner the motion may be continuing to recede from C, when before it was approaching along AC, and at some point Q it has the same speed that before it had at the point P placed at an equal distance from C, thus as can be understood from § 251. Therefore the motion occurs perpetually between A to B and back again, and in returning the body completes the motion. EULER’S MECHANICA VOL. 1. Chapter Three (part a). Translated and annotated by Ian Bruce. page 115 PROPOSITION 33. PROBLEM. 274. With the attraction from the centre C (Fig. 26) to be in some ratio of a multiple of the distances, the body at D now has a given speed ; the point A is required on the line CD produced, from which the descent of the body towards C begins, so that it has acquired this speed when it arrives at D. [p. 110] SOLUTION. With n denoting as above the exponent of the ratio of multiplication, in what shall be the centripetal force, and f the distance at which the centripetal force is equal to the force of gravity; let CD = b, the speed at D corresponding to the height h, and the distance CA sought which is put equal to q. Since here therefore q denotes that same distance as a [i. e. CA] in the above proposition, and b likewise that of y [i. e. CP], and h likewise here represents v, this equation is formed: n nn fn bq h )1( 11 + − ++ = . [c. f. )1( 11 n nn fn ya v + − ++ = above.] From which 1 1 111 ))1((and)1( +++=++= +++ nnnnnn hfnbqhfnbq . Moreover in the particular case, when n = -1, there is obtained beqflh f h b q == henceand , where e is the number, the logarithm of which is unity. Q. E. I. Corollary 1. 275. If the centripetal force varies directly as the distance, then n = 1 and )2( 2 fhbq += . Which is always a finite quantity, but only if b, f and h are such. Likewise it arises, provided n + 1 is a positive number. And also in the case n = - 1 the distance q is never infinite. Corollary 2. 276. But if n + 1 is a negative number, for example – m, since n = - m – 1, then m hmbf f mm m bq −+ + = 1 1 . , for which the height is infinite when h is given by m m mb f h 1+ = , and if h is a quantity greater than this, [p. 111] q is negative, or rather infinitely greater, or even imaginary. From which it is understood from these cases that only by falling from infinity is the body able to acquire as much speed at D. EULER’S MECHANICA VOL. 1. Chapter Three (part a). Translated and annotated by Ian Bruce. page 116 Corollary 3. 277. By keeping n + 1 equal to the negative number – m and the distance h for the point A at infinity will be m m mb f h 1+ = . And the distance from the centre C at which the body, falling from the infinite distance, will have the speed h , which is equal to m mh f f . Corollary 4. 278. If the centripetal force is inversely proportional to the square of the distance, then m = 1. On account of which bhf bf q − = 2 2 . When, therefore b f h 2 = , the distance AC, i. e. q, shall be infinitely great. Corollary 5. 279. If this problem is joined with the preceding one, the motion of the body can be easily determined, since it begins to descent to C from B with the speed h . From the preceding indeed the descent of the body is observed to be made from A, since it begins to descend in the part from D with the speed h , CP is called equal to y, and the speed, that the body has at P, corresponding to the height v, is )1( 11 n nn fn yq v + − ++ = (264). [p. 112] But )1(11 nnn hfnbq ++= ++ . Hence it becomes . )1()1( )1( 1111 hv n nn n nnn fn yb fn yhfnb +== + − + −++ ++++ Corollary 6. 280. From this expression for v, when the descent begins from D with the speed corresponding to the height h, it does not differ from that which is produced, if the descent were made from rest, except that this is quantity is always a distance greater than h. Scholium. 281. Since the time delay, in which the distance AC or any part of this is completed for any hypothetical centripetal force (Fig. 25), that is easily known from the known speeds. Generally the time for any letter n cannot be shown in a finite number of terms, clearly the time to traverse AP is found to equal ∫ ++ −− + ))(( )1( 11 nn n xaa fndx , which quantity generally neither can be integrated nor reduced to the quadrature of any known curve. But yet in various cases of n itself, it can be expressed neatly enough , on account of which from the general case set out, particular special cases will be examined in the following propositions.

PROPOSITION 34. PROBLEM.

282. If the centripetal force is in proportion to the distance from the centre C (Fig. 27) and the body falls from A as far as C, [p. 113] it is required to determine the time in which the body completes any part of this distance. SOLUTION. With AC = a, and the distance from the centre C, in which the centripetal force is equal to the force of gravity , equal to f, some part of the distance CP = y and the speed at P corresponding to the height v. Therefore the time, in which the distance CP is completed, is equal to ∫ v dy ; with the fraction 250 1 ignored, since this can be used for the known time in seconds and can be added as desired. Truly from Prop. 32 (264) by making n = 1 .hence, 2 )( 2 2222 f ya f ya vv −− == From which the time to travel through PC ∫∫ − = − = )( 2 )( 2 2222 ya ady a f ya fdy . Upon AC the quadrant of a circle is constructed AME, and to this the lines CE and PM as axis. From which is made, as agreed, the arc EM = ∫ − )( 22 ya ady . On account of which the time to traverse PC becomes = a fEM 2. . The time therefore of the total descent through AC will be a fAME 2. . Hence the time of descent through AP = a fAM 2. . From these therefore the time of descent through any distance travelled through can become known, and that in seconds, if these expressions are divided by 250 and the length f is shown in thousandth parts of Rhenish feet. Q. E. I. Corollary 1.
283. Let 1 : π denote the ratio of the diameter to the circumference , then it becomes 2AME : a = π : 1 and 2 π=a AME . Hence on account of this, the time of descent through AC is equal to f22 π . [p. 114] Since that does not depend on the height dropped or travelled through a, but whatever amount this shall be, it keeps the same value. Therefore all bodies, which are released towards this centre, reach that in equal amounts of time.

Scholium.

284. This equality of the time follows from the expression for the time f ya 2 )( 22 − , in which a and y are required to have one dimension. Indeed the amount comes about, the times, in which any distances a are travelled through, must be equal to each other. (46). Corollary 2.
285. If besides there should be another of centre of force of this kind, but with a different effectiveness provided, thus in order that the distance at which the centripetal force is equal to the force of gravity is F, the times of the descents shall be to each other as Ff ad . But the effectiveness of each are themselves in this case in the inverse ratio of the distances f : F; indeed these are as the forces, which these forces exercise at equal distances. Wherefore the times of descent to the different centres of force are in the inverse ratio of the square roots of their effectiveness. Which ratio indeed holds in all similar centres of force in place, if the distances traversed are equal to each other, as will be taught in what follows.