THE CURVILINEAR MOTION OF FREE POINTS ACTED ON BY ABSOLUTE FORCES OF ANY KIND

DEFINITION 21.

543. A body describes the curved line AMB (Fig. 47) when acted upon by a force. The tangential force on the body is the component of the force along the direction of the tangent TMt to the curve at the point M.

Corollary 1.

544. Therefore the tangential force exerts no other effect on the body while the element Mm is traversed, except that the motion of the body is either accelerated or retarded, since clearly the body is pulled either following the direction of MT or Mt.

Corollary 2.

545. Therefore when a body is moving along a curved line, the tangential force continually changes its direction and exerts its influence at another place.

Scholium.

546. Indeed a tangential force is hardly ever able to arise of its own accord in the nature of things; truly nothing is known of this wider use. For a force acting has a direction that can always be resolved into two parts, one of which is placed along direction of the tangent. [p. 226] DEFINITION 22.
547. The normal force is the force acting on the body describing the line AMB (Fig. 47), the direction of which MR is normal to the element of the curve Mm or the tangent MT .EULER’S MECHANICA VOL. 1. Chapter Five (part a). Translated and annotated by Ian Bruce. page 304 Corollary 1.
548. Therefore the normal force can neither increase nor decrease the speed of the body, since its direction MR is always at right angles to the direction of motion. (164). Corollary 2.
549. The effect of this force is agreed upon, as we show in what follows, as only the direction of the body can be changed and affected, because by itself the body progresses along a straight line, and the action of the normal force makes it move along a curve (165). Scholium 1.
550. If a body moves in the same plane, and also the directions of the forces acting on it are in put in the same plane, then the individual forces can be resolved into two parts, one of which is the normal, and the other the tangent, as is apparent from the principles of statics. Whereby when we have determined the effect of the tangential and normal forces on the body, then likewise also, [p. 227] the effect of any oblique force is also known. Moreover we call all forces acting on the body oblique, which are neither along the normal nor the tangent. Scholion 2.
551. Hence the first division of this chapter arises. For in the first part we consider these motions that have their paths in the same plane, and likewise all the forces are agreed to be acting in the same plane. Following this, we are to consider motions that follow paths that do not lie in the same plane; for which it is understood that it is not sufficient to resolve the individual forces into two parts, for these are required to be resolved into three parts, on account of the three dimensions in which the body is moving.

page 305 Translated and annotated by Ian Bruce. PROPOSITION 70. PROBLEM. 552. If a body, as it traverses the element Mm (Fig. 47) in a plane, is acted on by two forces, the one normal and the other tangential, to determine the effect of each in altering the motion of the body. SOLUTION. Let the speed of the body describing the element Mm correspond to the height v, the force pulling along the normal MR is equal to N, and the tangential force pulling along MT is equal to T, the element Mm = ds and the radius of osculation at M = r. [Which we now usually call the radius of curvature.] To determine the effect of the force N, since the direction of this is along the normal at Mm, we use the formula (165), which is npr = Ac 2 . [p. 228] p This becomes pr = 2 Av (209). But here N has been put in place of A for us ; for we understand by N not only the impression of the force N on the body, but the strength of the acceleration or the absolute force divided by the mass of the body, [which is A] (213). p Whereby here in place of A must be substituted N, with which done we have : Nr = 2v . Q. E.D. for the first part. [Thus in modern terms, we have the centripetal acceleration at the point M , ac = V 2 / r , where V is the tangential speed at this point.] Then to determine the effect of the tangential force T, I use this rule : Acdc = npds (166), or in place of this, it is modified to give : Adv = pds (209). And on p account of the reasons offered, here in place of A I substitute T, and it becomes dv = Tds . Q. E.D. for the second part. [We may wish to consider Tds as the increase in the kinetic energy of the body supplied by the force T acting along the tangent on a unit mass through the increment ds, while dv is the corresponding change in a hypothetical gravitational potential energy of a unit mass under unit gravitational acceleration. One needs to look at Euler’s work in Ch. 2 to see how relations equivalent to these can be found without recourse to work–energy relations. ]EULER’S MECHANICA VOL. 1. Chapter Five (part a). Translated and annotated by Ian Bruce. page 306 Corollary 1. 553. Because the acceleration of gravity is put equal to 1 [here called force, but in the sense force per unit mass], the normal acceleration is to the acceleration of gravity as the height corresponding to the speed to half the radius of osculation [curvature] ; which ratio follows from the equation Nr = 2v . [In modern terms, the centripetal acceleration ac = V 2 / r = 2 gh / r = 2.1.v / r = N . ] Corollary 2. 554. Therefore with the given acceleration to the normal, and the curve that the body describes, the speed of the body is known at once, for the speed is expressed by v , and v= Nr . 2 Corollary 3. 555. Truly the increment of the height corresponding to the speed is always equal to the product of the tangential acceleration and the element of distance traversed by the body [p. 229]. Or the element of v is to the element of distance described ds as the tangential acceleration is to the acceleration of gravity. PROPOSITION 71. PROBLEM. 556. If the body, while it traverses the element Mm (Fig. 47), is acted on by some oblique force in the direction MC, then it is required to determine the effect of this force in changing the motion of the body. SOLUTION. Let this oblique force be in the ratio to the force of gravity, if on the surface of the earth, as P to 1, the element Mm = ds and the speed at M corresponds to the height v. Since the obliqueness of the force MC has been given, the angle CMT is given, and on account of this the triangles CMT and Mmt, which come from the perpendiculars dropped from C to MT and from m to MC, are as shown. We can therefore put Mr = dy and mr = dx, then ds 2 = dx 2 + dy 2 , and the ratio between both ds, dx, and dy is given. They are now brought together with these equations (161, 163) that have been presented before (208). For these that we wish to find here areEULER’S MECHANICA VOL. 1. Chapter Five (part a). Translated and annotated by Ian Bruce. page 307 completely similar to these previous ones,; only in this respect do they differ, as for us here the ratio is P:1, while there it was p : A. On this account, we have : dv = Pdy , and with the radius of osculation MR at M put equal to r this equation is found : Pr dx = 2vds p (208), I write P in place of A . Of these equations : dv = Pdy defines the increment of the speed, as the body travels through the element vds shows the lines of the curvatures Mm. [p. 230] Truly the other, Pr dx = 2vds or r = 2Pdx at M described by the body. Hence the whole effect of the oblique force on the motion of the body can become known. Q. E. I. [The component of P along the curve at M is Pcosθ = Pdy / ds : hence at .ds = dv , where at is the tangential acceleration, as in the previous chapters ; the component of P normal to the curve, or the normal acceleration an , is given by an = P sinθ = Pdx / ds = V 2 / r = 2v / r , where P is taken as the acceleration of a body of unit mass along MC. ] Corollary 1. 557. If the oblique force is put in place at an obtuse angle with the element Mm (Fig. 48), everything remains as before, except that the line element mr =dy must be taken ( 1 − λ2 ) = μ . Hence dv = − Pdy , and the other equation Pr dx = 2dvds remains as before. Corollary 2. 558. Therefore if the direction of the force MC falls between the normal MR and the element Mm as in Fig. 47, then the motion of the body is one of acceleration. But if CM falls outside each, as in Fig. 48, the motion is one of retardation. Corollary 3. 559. If the direction of the force MC falls on the tangent MT, then the angle mMr, Mr is made equal to Mm or dx = 0 and dy = ds. Therefore we have dv = Pds and r =∝ , which indicates that the direction of the body is not affected by this tangential force. Corollary 4. 560. If the direction of the force MC (Fig. 48) on the other part Mt, in which case Mr = dx = 0 and dy = ds , giving dv = − Pds and r =∝ as before. Therefore only theEULER’S MECHANICA VOL. 1. Chapter Five (part a). Translated and annotated by Ian Bruce. page 308 tangential speed of the body is affected, and the direction of the motion is clearly not changed.(544). [p. 231] Corollary 5. 561. The direction of the force MC falls on the normal MR, where the effect of the normal force is known. Hence in this case dy = 0 and dx = ds . And consequently dv = 0 and Pr = 2v is produced. Hence the normal force does not affect the speed, but only the direction of the motion. (548). Scholium 1. 562. Therefore both the effect of the normal force and of the tangential force on the motion of the body is known. On which account, when all the forces, as many as act on the body, are put in a plane in the same way in two’s, they can be resolved with one part normal and the other tangential, and the effect of any forces on the motion of the body can become known. Scholium 2. 563. Hence it will be most convenient for motion in the same plane to be subdivided, as in the first place the directions of all the forces acting are parallel to each other, as in our region, the directions of the weights are observed to be parallel to each other. Then we will consider the case, in which the directions of all the forces converge to a single point, to which also the body is always attracted, and which is the case of the centripetal force, such as the singular discoveries made by Newton Part I of the Princ. Phil. Nat. Truly in the third place we introduce and investigate any forces acting on a body [p. 232], which the motion shall always be about to follow. Moreover we will turn from these individual cases, as at first we propose direct questions, then indeed also inverse questions, as far as we have done at this stage, that we may be able to resolve. Always finally, as much as is permitted, we will progress from the more simple to the more complicated and difficult cases. [Euler now solves simple projectile motion using his general equations.]EULER’S MECHANICA VOL. 1. Chapter Five (part a). Translated and annotated by Ian Bruce. page 309 PROPOSITION 72. PROBLEM. 564. If there is a constant force the direction of which is normal everywhere to the right line AB (Fig. 49) and if a body is projected from A with a given speed along the direction AH, then it is required to find the curve AMDB described by the body, and the motion of the body on this curve. SOLUTION. The force acting is called g, and the speed with which the body is projected from A corresponds to the height c and the cosine of the angle HAB = λ, with the whole sine taken equal to 1. Now with the body travelling through the element Mm = ds , the speed at M corresponds to the height v, and PM = y and likewise the radius of osculation at M = r. On account of which Mr = dx and mr = dy , and it is apparent that this case can be referred to Fig. 48, where the motion of the body is slowing down. [Note however that the elemental triangle in Fig. 48 has been rotated counterclockwise about M, resulting in x and y being interchanged in the diagram, but not in the equations.] Therefore we have : dv = − gdy and grdx = 2vds (557). From the first of these equations there arises : v = C − gy , and C must be found from this, that with y = 0, it must become v = c. Therefore we have : [p. 233] v = c − gy . [Note also that v and c are actually multiplied by 1, which is the assumed acceleration of gravity in Euler’s units on earth, in order that the equation has the correct dimensions of L2/T2. Thus, g also is one on earth, but it has been made more general in the equation. It is probably a good idea to refresh our memory of the derivation of the radius of curvature for a function y(x) at some regular point. In the extra sketch, the component of the force or acceleration along the normal to the curve shown is g cosθ = gdx / ds , while the component along the tangent is g sinθ = gdy / ds . Hence the equation grdx = 2vds can be reduced to the familiar form V 2 / r = gdx / ds . The radius of curvature is givenEULER’S MECHANICA VOL. 1. Chapter Five (part a). page 310 Translated and annotated by Ian Bruce. 2 2 2 by r = ds / dθ = ( ds / dx )( dx / dθ ; now the element of arc ds = dx + dy , and dy / dx = tanθ , leading to d 2 y / dx 2 = sec 2 θ .dθ / dx and hence dθ / dx = ( d 2 y / dx 2 ) /( 1 + tan 2 θ ) = ddy / ds 2 . From which it follows that r = ( ds / dx )( ds 2 / ddy ) = ds 3 / dxddy . The negative sign is applied for a convex curve as we have here. ] Now from the other equation, this value found is put in place of v and the equation is produced : grdx = c − gy . 2ds 3 ds with constant dx put in place [i. e. x is an independent variable], with Truly, r = − dxddy − gds 2 which substituted, the equation becomes : 2ddy = c − gy or 2cddy = 2 gyddy − gdx 2 − gdy 2 with dx 2 + dy 2 in place of ds 2 . The integral of this equation can be found: dx = ds C . c − gy [This can be shown be assuming ( ds / dx )2 = p( y ) , differentiating w.r.t. x, and − gds 2 substituting in the original equation 2ddy = c − gy . ] = λ c −cgy and Moreover with y = 0 we designate C = λ2c . We therefore have : dx ds hence λdy c = dx . ( c( 1− λ2 )− gy ) The integral of which is : C − 2λ ( c( 1 − λ2 ) − gy ) = gx = 2λ c( 1 − λ2 ) − 2λ ( c( 1 − λ2 ) − gy ) , c with the constant C determined, so that y vanishes by putting x = 0. For the sake of brevity, the sine of the angle HAB or ( 1 − λ2 ) = μ , then gx = 2λμc − 2λ ( μ 2c 2 − gcy ) and hence, μx y= λ −gx 2 . 4λ2 c μgxg 2 x2 4λ2 c Hence also we have : v=c− λ + andEULER’S MECHANICA VOL. 1. Chapter Five (part a). Translated and annotated by Ian Bruce. 2 μgx g 2 x2 g 2 x2 μ 2 μgx ds 2 = dx 2 ( 1 + 2 − 3 + 4 2 ) = dx2 ( 1 − λc + 4 2 ) . λ λ c 4λ c λ 4λ c page 311 2 2 Consequently the equation arises : dsv = dx2 and ∫ λc ds = x v λ c is equal to the time, in which the arc AM is travelled through. Q.E.I. Corollary 1. 565. Therefore the body falls back on the horizontal line AB at the point B with 4λμc AB = g . Indeed the time, in which the body is turning above AB and the curve ADB is completed, is given by 4μ c . g Corollary 2. 566. Moreover 2λμ denotes the sine of the angle, which is twice the sine of the angle 2χ c HAB. Whereby, if the sine of this angle is called χ, [p. 234] then AB = g . From which it is apparent that the distance AB to be a maximum, if χ = 1 and thus the angle HAB is half a right angle, if indeed the body is projected with the same speed c . Corollary 3. 567. It is also understood that the horizontal motion of the body is uniform. For the times, in which any arc is described, are proportional to the corresponding abscissa on the line AB. Corollary 4. 568. If the body is always projected with the same speed c , but at different angles with AB, the times in which the motion above AB is completed, are to each other as the sines of the angles HAB (565). Corollary 5. 569. The maximum height DE, to which the body can reach, is the vertical line through 2λμc the point E, taken from AE = g . From which it is apparent that AE is half of AB. μ 2c Truly the maximum height itself DE is equal to g , which hence is in proportion to the square of the sine of the angle HAB.EULER’S MECHANICA VOL. 1. Chapter Five (part a). page 312 Translated and annotated by Ian Bruce. Corollary 6. gx 2 μx 570. From the equation y = λ − 2 it is seen that the curve ADB is a parabola, the axis 4λ c 2 2 of which is the line DE and the parameter is equal to AE = 4λ c . Therefore the DE g parameter is proportional to the square of the cosine of the angle HAB. [p. 235] Corollary 8. 571. Hence the vertex of this parabola is the point D, and the distance DF of the focus F from the vertex D is equal to λgc . Whereby if the line MF is drawn, this will be 2 MF = DQ + DF from the nature of the parabola. [For MF 2 = ( y − a )2 + y 2 = ( y + a )2 . ] Corollary 8. c − gy 572. Moreover also MF = DE − MP + DF [= gc − ( y − a ) + a] = gc − y = g . Since truly v = c − gy is the height corresponding to the speed at M = g .MF . Therefore it is also apparent that AF = gc , [as c = g .AF ]. Corollary 9. 573. Therefore it is evident that the body describing this parabola has the same speed at some point M, as the same body acted on by the same uniform force falling straight down can acquire from that height, which is equal to the distance of the point M from the focus of the parabola. Corollary 10. AE = 2λμ = χ . From which it is evident that 574. The cosine of the angle FAE is equal to AF the angle FAE is equal to the complement of twice the angle HAE to the right angle or rather the excess of twice the angle over the right angle. [ cos FAE = 2λμ = sin 2 HAB ; FAE = 90 - 2HAB or 2HAB - 90] Corollary 11. 575. Since the angle AFD is the supplement of the angle AFE, then the angle AFD is twice the angle HAB. [p. 236] [AFD = 180 - AFE = 90 + FAE = 2HAB. ] Scholium 1. 576. Therefore the construction of the parabola is easily deduced from these, that describes the projection of the body with a given speed in a given direction. For with AG drawn normal to AB, and the angle GAF taken equal to twice the angle HAE, and in this direction AF is taken equal to the height, from which the same body falling in a straight line acquires the speed equal to this, with which it is projected from A, from which the focus point F of the parabola sought is found (573, 575). Again the normal DE through FEULER’S MECHANICA VOL. 1. Chapter Five (part a). Translated and annotated by Ian Bruce. page 313 drawn to the line AB is the axis of this parabola, and the vertex D is found by taking DF = AF 2− FE . Therefore when the parameter is equal to 4DF, the description of the parabola is brought into view. Scholium 2. 577. If g = 1, then we have the case of terrestrial gravity. Concerning which, if there should be no air present, which might impede the motion of the body on account of resistance, all projected bodies move in parabolas. Galileo first elicited this truth, and after him all the writers of mechanics have demonstrated this fact. And indeed it can be arrived at in a much shorter way without differential equations and differentiation, but we have preferred to use a general method, which can be widely extended, rather than apply ourselves too much with the particular solutions, which can only be adapted to that one case. [p. 237] PROPOSITION 73. PROBLEM. 578. If the body at A (Fig. 50) with a given speed and projected in a given direction is always attracted to the line AB in some ratio of some function of the distances from this line, to determine the curve ADB, that the body thus acted upon describes and the whole motion on this curve. SOLUTION. Let the speed at A = c and the sine of the angle HAB = μ , with the whole sine taken as 1, and the cosine equal to λ. The speed of the body at M corresponds to the height v and the element Mm = ds , and likewise the abscissa AP = x, with the applied line PM = y and the radius of curvature MR = r. Moreover while the body is traversing the element Mm, it is meanwhile pulled by a given force , which shall be equal to P, for a certain function of y, in the direction MP. With these put in place, dv = − Pdy , et Pr dx = 2dvds (557). Moreover, ∫ Pdy = Y and Y is such a function of y, which is made equal to 0 with y put equal to 0. Hence, on this account, v = c − Y . Then from the first equation, the value of dv , is substituted into the other equation Pr dx = 2dvds , and there is obtained : P, i. e. −dy − dv = 2dyds . v rdx dyds It is true, and with ds made constant, that with r = ddx ,EULER’S MECHANICA VOL. 1. Chapter Five (part a). page 314 Translated and annotated by Ian Bruce. [The reason for introducing the new curvature condition is of course to allow the ddx and by integrating, variables to be separated.] and hence −vdv = 2dx 2 l C − l v = l dx 2 . ds The constant C is determined from this , that at the point in which v = c , put dx =λ. ds According to which, l C = l c + 2lλ again with the numbers taken λ2cds 2 = vdx 2 . [p. 238] From this equation the element of the time is at once found : ds = dx . v λ c Moreover the equation of the curve AMDB is obtained from the two equations found : v = c − Y et λ2cds 2 = vdx 2 ; by eliminating v there arises : λ2cds 2 = cdx 2 − Ydx2 = λ2cdx 2 + λ2cdy 2 . Bus since 1 − λ2 = μ 2 , it gives dx = λdy c ( μ 2 c −Y ) , in which equation, since the variables x and y can be separated from each other, the curve ADB can be constructed. Q.E.I. Corollary 1. 579. Therefore the times in which any arcs AM are described, are in the ratio of the corresponding abscissae AP. Indeed the time to traverse AM is equal to x . λ c [Since ds = dx .] v λ c Scholium 1. 580. The motion of the body on the curve AM can be considered to be resolved into two other motions, of which the one becomes lines parallel to the line AB, and the other follows the perpendiculars to this line AB. In that motion the body progresses following the line AB, and indeed it either rises or falls with respect to the line AB. Now indeed it is evident that the progression of the horizontal motion is not to be changed by a force, the direction of which is always perpendicular to AB, and for this reason this motion must always be regular and made with the same speed, which arises from the resolution of the initial motion. Since the direction of the initial motion is along the line AH, the speed of this c to the progressive speed along AB is as the whole sine 1 to the cosine of the angle HAB, which is λ. [p. 239] Therefore the speed of progression along AB is λ c , from which the time, in which the horizontal motion is completed through the distance AP = x , comes to equal x , as we have found. λ cEULER’S MECHANICA VOL. 1. Chapter Five (part a). Translated and annotated by Ian Bruce. page 315 Corollary 2. 581. If the body with the speed c rises from A perpendicularly along AC and we take AL = PM = y , then the speed at L is equal to the speed at M, clearly v . For dv = − Pdy and v = c − Y , as we have found for the point M. Corollary 3. 582. If AC is the total height, to which the body projected up from A with the speed c is able to reach, then the height CL is that by which the body acquires the same speed by falling, that it had at M. Corollary 4. 583. The maximum height DE is found by making dy = 0 , in which case Y = μ 2c , from which equation the value of y extracted must give the height DE, and the speed at D is of such as size as that acquired by the body falling from the height CI. [We can perhaps also see this from energy conservation: The initial kinetic energy of the body at A is partitioned into a constant amount for the horizontal motion Eh, and an amount that is all transformed into potential energy at D, Ev. Now, Eh + Ev is constant, and is equal to the potential energy at C if the body is projected straight up, in which case the amount of kinetic energy gained in falling is the same as Eh , as the potential energy is the same at D and I in each case.] Corollary 5. 584. Moreover, we have also found that v = λ cds2 . Whereby, when at the point D , 2 2 dx dx = ds , and the speed at D = λ c , which is equal to the horizontal speed that the body has progressing along AB. [p. 240] Scholium 2. 585. Indeed the size of AB is not apparent from the equation, since it cannot be integrated. Yet it is evident that the parts AE and EB are required to be equal, and the branch DB is similar and equal to the arc DA. For after the body arrives at D, it is accelerated again in a similar way, in which it was retarded before along AD; because the force is the same in these same distances from AB and in this way the motion is again perfectly restored. Scholium 3. 586. Hence also the inverse problem can be resolved easily, where for a given curve ADB and with the speed of the body at A, the law of the force is sought which must be put in place, in order that the body moves on that curve. For from the equation :EULER’S MECHANICA VOL. 1. Chapter Five (part a). page 316 Translated and annotated by Ian Bruce. dx = λ cdy − 2λ cdyddy λdy c it is found that Y = μ 2c − and dY = Pdy = with dx 2 2 dx dx 2 ( μ c −Y ) 2 2 2 ds with dx placed constant, then P = 2λ cds , which is equal to constant. Since r = − dxddy 3 2 2 3 rdx the force acting on the body at M, following the direction MP. PROPOSITION 74. PROBLEM. 587. A body is projected at A and everywhere it is pulled (Fig. 51) towards the centre of force C by some centripetal force, and it is required to determine the nature of the curve AM, in which the body is moving, and the motion of the body on this curve. [p. 241] SOLUTION. The centripetal force acting on the body at M towards C is put equal to P, and the speed of the body at M shall correspond to the height v. Then the distance CM is called y, the element of length Mm is equal to ds and the perpendicular CT to the tangent of the curve MT is called p, also the element Mr is dx and the radius of curvature MR is called r. Therefore because of the similar triangles Mmr and CMT ds ( y 2 − p 2 ) = ydy and dx ( y 2 − p 2 ) = pdy [where MT = ( y 2 − p 2 ) ]. And the ydy radius of curvature is found r = dp . [For on drawing the other tangent at m, and considering the equality of the small angle dp ydy ds .MT = between these tangents and the angle MCm, MT = ds or r = dp . See sketch r dp added to (601)] With these put in place we have : dv = − Pdy and Pr dx = 2vds (557). With these −2 dsdy equations collated with the elimination of P , we have : rdvdx = −2vdsdy or dv = rdx . v ydy y ds , there is produced : Substituting dp in place of r and p in place of dx dv = −2dp . v p Which on integrating gives : v = C2 . p This constant C is defined from the given initial speed at A, which corresponds to the height c, and with the direction of projection, that we define thus, in order that with the distance CA = a , and the perpendicular CD to the tangent at AD is equal to h. On account of which C = ch2 andEULER’S MECHANICA VOL. 1. Chapter Five (part a). page 317 Translated and annotated by Ian Bruce. 2 v = ch2 . p Hence we have ds = v pds , which is equal to the element of the time required to traverse h c the element Mm. Consequently as pds = 2MCm the time [i. e. twice the area of the elemental triangle MCm, leading to Kepler’s law for equal areas described in equal times 2 in the inverse square law instance; we note also that v = ch2 is the formula for the p conservation of angular momentum of the unit point mass], in which the arc AM is traversed, = 2.ACM . h c 2 The curve itself can be determined by substituting ch2 in place of v in the equation p Pr dx = 2vds , and with this done the equation is produced : Pp 2rdx = 2ch2ds . This ydy y ds , is equation, with the values dp and p substituted in the equation, in place of r and dx changed into this : Pdy = 2ch 2 dp . p3 Which equation, whatever the function of y P shall be, can be constructed on account of the separate variables. Q.E.I. [p. 242] Corollary 1. 588. Because the time, in which the arc AM is traversed, is 2.ACM , the times, in which h c any arcs can be described, are as the areas taken described by the arc and with the line drawn from the centre C. Corollary 2. 2 589. Then since v = ch2 , it follows that p v = h p c . Therefore the speed of the body at any location of the curve traversed varies inversely as the perpendicular from the centre C sent to the tangent at that point. Scholium 1. 590. This description of the equality of the areas has been established in Newton’s first proposition, from which nearly everything is deduced. Moreover these two properties are the most general and they only require that the direction of the centripetal force is always directed towards the centre of the circle. For whatever the size of the centripetal force, either by a function of the lengths CM, or otherwise expressed, yet each is prevailing equally. For as in these calculations that we have come across, from the calculation the centripetal force P might be removed, but yet in these only the direction is abandoned in the consideration.EULER’S MECHANICA VOL. 1. Chapter Five (part a). Translated and annotated by Ian Bruce. page 318 Corollary 3. 591. It is necessary that the centripetal force itself should be known for the given curve described by the body, [p. 243] and from that the equation of the curve can be found. For Pdy = 2ch 2 dp expresses the nature of the curve, if P is a given quantity. p3 Corollary 4. ydy 592. Since dp = r , also we can put P = P= 2ch 2 y . This is De Moivre’s Theorem, p 3r 2ch 2 dp ,truly that Keill first contended to have come across. p 3 dy [Abraham de Moivre (1667–1754) : Some simple properties of the conic sections deduced from the nature of the foci; with general theorems of centripetal force, by means of which the law of the centripetal force tending to the foci of the sections, the velocities of bodies revolving in them, and the description of the orbits, may be easily determined. Philosophical Transactions 1717, p. 622; cf. Miscellanea analytica de seriebus et quadraturis, London 1730, p.233. John Keill (1671–1721) : Of the laws of centripetal force, Philosophical Transactions 1708, p. 174; " The learned Mr. Halley having showed me a theorem, by which the law of centripetal force can be exhibited in finite quantities, which was communicated to him by Mr. De Moivre, who said that Mr. Is. Newton had before discovered a similar theorem, and as the demonstration of the theorem is very easy, I wish to communicate it to the public, with some other thoughts in the same subject." Noted by Paul Stackel in the 1912 edition of Tom I, which is available from the Gallica website, which is used in this translation. You will have to use JSTOR perhaps to access the papers quoted, to which I do not have acces any more.] Scholium 2. 593. These equations are useful in two ways : For in the first place from the given law of the centripetal force, the nature of the curve can be determined that a body projected describes. Then also in turn with the help of these equations, if the curve is given, that the body describes around the centre of force C, it is possible to define the centripetal force bringing about the motion at any place, in order that the body is free to move on this curve. Corollary 5. 594. Since also dv = − Pdy , it is evident, if P is a function of y, the speed everywhere depends on the distance of the body from the centre and at the same distances the body must have the same speed.EULER’S MECHANICA VOL. 1. Chapter Five (part a). page 319 Translated and annotated by Ian Bruce. Corollary 6. 595. Therefore just as P is a function of y, so also the curve described can thus be compared, as with equal distances from the centre the tangents dropped are always equal to each other, since the speeds vary inversely as the lengths of these perpendiculars (589). [p. 244 : Later of course considered to be an aspect of the conservation of the angular momentum of the rotating body.] Scholium 3. 596. And always, if P is a function of y, a line EC can be assigned, along which a body acted on by the centripetal force in falling to the individual points N has the same speed, that it has on the curve AM at points M with the points equally distant from the centre. For take CN = CM = y and Nn = mr = dy , if the speed at N is equal to the speed at M, clearly corresponding to the height v, it is also the case, as the body traverses through the element Nn, that dv = − Pdy . From which it is understood that the speed at n is always equal to the speed at m. And thus the body has a speed at the individual points of the line EC, as great as it has on the curve AM at the same distances from C. If therefore the initial motion starts from E at which the speed is zero, the speed of the body at the individual points on the curve AM is known from the given line EC. Therefore we will use this line EC in later results in defining the speeds of the motion of the body and we will call that the distance defining the speeds. Corollary 7. 597. Therefore with the speed given at A, clearly corresponding to the height c, from this the whole distance EC can be found. For only the distant point E from C needs to be taken, as the body falling from E is acted upon by this centripetal force and acquires a speed equal to c at A. [p. 245] Corollary 8. 598. If the angle MCm = dw, then dw = dx

y p= to y 2 dw ( y 2 dw 2 + dy 2 ) pdy y ( y2 − p2 ) . Hence there arises . Truly the time element, in which the angle MCm is resolved, is equal y 2 dw pds pydy

. Therefore this element of time is equal to . h c h c h c( y 2 − p 2 ) Corollary 9. 599. If we wish to measure the angular speed, or that in which MCm is traversed, by dividing this angle by the time, the angular speed at M is produced equal to h 2c . y Therefore the speed of the angle [our angular velocity] varies inversely as the square of the distance of the body from the centre C.EULER’S MECHANICA VOL. 1. Chapter Five (part a). page 320 Translated and annotated by Ian Bruce. Scholium 4. 600. Moreover since the curve that a body describes acted upon by a given centripetal force, cannot be known otherwise than from the equation between the distance of the body from the centre and the perpendicular dropped on the tangent from the centre, generally to be judged with difficulty, such is the curve found, when we are accustomed to explain the nature of the curves by equations between orthogonal coordinates. Yet nevertheless equations of this kind between the distance from the centre and the perpendicular to the tangent, provided they are either of algebraic or differential forms, in which the variables can be separated from each other, are sufficient for the curves sought to be constructed. But where is possible to investigate thoroughly the nature and order of these curves, [p. 246] we will give here the method in which these equations between the distance and the perpendicular can be reduced to ordinary equations between the coordinates. PROPOSITION 75. PROBLEM. 601. The nature of the curve AM (Fig. 52), that a body acted upon by some centripetal force is to be described, in order the equation between the orthogonal coordinates CP and PM referred to fixed axes AC can be defined. SOLUTION. As before with the initial speed at A = c , AC = a and with the perpendicular from C dropped to the tangent at A equal to h and in addition CM = y, CT = p and the centripetal force at M is equal to P, which is a certain function of y, then CP is called equal to x and PM = z, then CM = ( x 2 + z 2 ) = y or z = ( y 2 − x 2 ) ; for we keep y in place of z in the calculation, since in this manner the calculation is easier and shorter, and after handling the whole operation, with z in readiness to be introduced in place of y. With these in place : Mm = ( dx 2 + dz 2 ) = ( y 2 dy 2 − 2 yxdydx + y 2 dx 2 ) ( y −x ) 2 2 and Mr = − ydx + xdy . ( y2 − x2 ) [This use of this auxillary coordinate y, which is just the radial distance of the body from the centre of force, eases the calculation, where dz2 is found from z 2 = y 2 − x 2 and mr = dy, giving Mr 2 = Mm2 − dy 2 , leading to Mr.] Therefore we have [see Prop. 74 and following sketch for the similar triangles]:EULER’S MECHANICA VOL. 1. Chapter Five (part a). page 321 Translated and annotated by Ian Bruce. − ydx + xdy . Mr = p = [cos(θ ) =] Mm 2 2 y ( y dy − 2 yxdydx + y 2 dx 2 ) and consequently p= yxdy − y 2 dx ( y 2 dy 2 − 2 yxdydx + y 2 dx 2 ) . But for the curve AM , we have elicited the equation previously (591) : Pdy = 2ch 2 dp . p3 Moreover we set Y for the integral of Pdy thus assumed, [Y is related to the work done by the centripetal force as the body moves along y, and hence is the potential energy function for unit mass and gravity;] so that it vanishes with y = a. With this 2 done, then Y = C − ch2 , where the p constant C on account of p = h, if y = a, must be equal to c. Hence there comes about : Y = cp 2 − ch 2 , [p. 247] from which p2 there is produced : p= h c . ( c −Y ) [We note that h c is the initial angular momentum of unit mass at the point in the orbit where the speed and the distance from C are perpendicular; while Y is the height corresponding to the change in the potenial energy of unit mass in a uniform field of g = 1, and hence c − Y is the speed elsewhere corresponding to the perpendicular distance p, giving the same angular momentum. ] Which quantity, when Y is a function of y, is also a certain function of y. Hence for this reason, we have this equation : [ p =] We put x = uy and h c ( c −Y ) h c

( c −Y ) yxdy − y 2 dx ( y 2 dy 2 − 2 yxdydx + y 2 dx 2 ) = Q for the sake of brevity, then Q = . − y 2 dx ( dy 2 −u 2 dy 2 + y 2 du 2 ) , from which there arises : Qdy hdy c du

= 2 2 2 ( 1−u ) y ( y −Q ) y ( cy 2 − ch 2 − y 2Y ) with h c restored in place of Q. Therefore which equation, when the variables u y ( c −Y ) and y in that equation are separated [note that u is the cosine of the angle θ’ between MC and PC; thus y and θ’ are the polar coordinates of M as noted in the next cor.], can always be constructed. Q.E.I.EULER’S MECHANICA VOL. 1. Chapter Five (part a). page 322 Translated and annotated by Ian Bruce. Corollary 1. 602. Since u = xy , u expresses the cosine of the angle MCA. And for this reason, this final equation is separated into the distance of the body from the centre and the cosine of the angle ACM. Indeed from this equation there emerges immediately the equation between x and z. [Thus, in this remarkable manner, the equation has been separated into angular and radial components for the general case. ] Corollary 2. 603. Truly the equation cannot be algebraic, unless a circle commensurable with the arc hdy c ∫ y ( cy −ch − y Y ) denotes the arc of 2 2 2 ∫ ( 1du−u ) . [Note that on putting u = cosθ’ , this 2 integrand becomes –dθ’.] Corollary 3. 604. Therefore whenever hdy c y ( cy 2 − ch 2 − y 2Y ) can be reduced to a form of the kind λdZ and λ is a rational number, so an algebraic equation for the curve sought can ( A2 − Z 2 ) be shown. [p. 248] Scholium 1. 605. But if du is equal to a quantity of the kind ( 1−u 2 ) λdZ ( A2 − Z 2 ) , with the integration carried put with imaginary logarithms, this equation is obtained : λ λ ⎛ ( A 2 −C 2 ) −C −1 ⎞ ⎛ ( A 2 − Z 2 ) + Z −1 ⎞ ⎜ ⎟ ⎜ ⎟ .

( 1−u 2 ) −u −1 ⎜⎝ ( A 2 −C 2 ) + C −1 ⎟⎠ ⎜⎝ ( A 2 − Z 2 ) − Z −1 ⎟⎠ [Note that if u = cos θ , where we have now taken the angle MCA = θ, then ( 1−u 2 ) + u −1 ( 1−u 2 ) + u −1 ( 1−u ) −u −1 2 θ + i cos θ = −ei .− 2θ ; similarly, if Z = A cos θ , then = sin sin θ −i cos θ λ ( ) ⎛ ( A2 − Z 2 ) + Z −1 ⎞ A sin θ + Ai cos θ λ = ( −1 )λ ei .− 2λθ , and similarly, if C = A cos α , ⎟ ⎜⎜

⎟ 2 2 A sin θ − Ai cos θ ⎝ ( A − Z ) − Z −1 ⎠ λ ⎛ ( A2 −C 2 ) −C −1 ⎞ ⎟⎟ = ( −1 )λ ei .2λα . Hence, the result of the integration can be then ⎜⎜ 2 2 1 − + − ( A C ) C ⎠ ⎝ written in the convenient form : ei .θ = i( Aei .θ )λ .( Aei .−α )λ / A2λ ; hence, u = cos θ gives the expression for u subsequently quoted below.EULER’S MECHANICA VOL. 1. Chapter Five (part a). page 323 Translated and annotated by Ian Bruce. Note also that the angle θ in the original integral can be written as θ + i cos θ .] 2iθ = iθ + iθ = log eiθ − log e −iθ = log sin sin θ −i cos θ Truly here C is a constant quantity to be determined from this equation, since if CM ( y ) = CA( a ) , then likewise too, x = a or u = 1. Moreover from the above equation, this equation can be constructed : ( ( A −C ) −C −1) ( ( A − Z ) + Z −1) −( ( A −C ) +C −1) ( ( A − Z ) − Z −1) . u= 2 λ 2 2 λ 2 2 λ 2 2 λ 2 2 A2 λ −1 Which, whenever λ is a rational number, it is free to be returned from the affects of the imaginary − 1 and transformed into an algebraic equation of a certain order. [Various values of λ corrrespond to well-known curves to be investigated; e. g. λ = 12 ,1 and correspond to ellipses with different centripetal forces.] Corollary 4. 606. Since x = uy , this equation is put in place : ( ( A −C ) −C −1) ( ( A − Z ) + Z −1) y −( ( A −C ) +C −1) ( ( A − Z ) − Z −1) y , x= 2 λ 2 2 λ 2 2 2 λ 2 2 λ 2 A 2 λ −1 which, since Z is a function of y and y = ( x 2 + z 2 ) ,can easily be changed into an equation between x and z. Scholium 2. 607. The above equation can also be transformed into this : 1 1 ⎡⎛ λ λ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎤ 2 2 2 2 2 2 1 Z= ⎢⎜ ( 1 − u ) + u − 1 ⎟ ⎜ ( A − C ) + C − 1 ⎟ − ⎜ ( 1 − u ) − u − 1 ⎟ ⎜ ( A − C ) − C − 1 ⎟⎥ , 2 −1 ⎢⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎥⎦ ⎣ [p. 249] which is more convenient, if λ1 is a positive whole number. [Thus, from ei .θ = i( Aei .θ )λ .( Aei .−α )λ / A2λ , we have eiθ = ei( θ +π / 2 ) / λ .eiα ; from which Z = A cos θ can be found.] Scholion 3. 608. But if λ is a negative number, equal to = − μ , we have : ( ( A −C ) +C −1) ( ( A − Z ) − Z −1) −( ( A −C ) −C −1) ( ( A − Z ) + Z −1) = u 2 μ 2 2 2 μ 2 2 μ 2 2 μ 2 A 2 μ −1 From which it is apparent, if λ is negative, the value of u can only be negative, indeed that which is understood from the differential equation. Truly in a similar manner, ⎡ Z = 1 ⎢⎛⎜ 2 −1 ⎢⎝ ⎣ 1 1 ⎤ μ μ ⎞ ⎛ ⎞ ⎛ ⎞ 2 2 2 ( 1 − u ) − u − 1 ⎟ ⎜ ( A − C ) + C − 1 ⎟ − ⎜ ( 1 − u ) + u − 1 ⎟ ⎛⎜ ( A2 − C 2 ) − C − 1 ⎞⎟⎥ . ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎥⎦ 2EULER’S MECHANICA VOL. 1. Chapter Five (part a). Translated and annotated by Ian Bruce. page 324 Corollary 5. 609. If λ = 1 , then u = Z ( A −C ) −C ( A 2 − Z 2 ) Zy ( A2 −C 2 ) −Cy ( A 2 − Z 2 ) and x = . 2 A A2 2 2 If λ = −1 , also u or x has to be taken negative. Corollary 6. 610. If λ = 2 , then u= 2 Z ( A2 − 2C 2 ) ( A 2 − Z 2 ) − 2C ( A2 − 2 Z 2 ) ( A2 −C 2 ) . A4 But if λ = 12 , then Z = C − 2Cu 2 + 2u ( A2 − C 2 )( 1 − u 2 ). [p. 250] PROPOSITION 76. THEOREM. 611. If the centripetal force is as some function of the distance from the centre C (Fig. 53), and a body at A is projected following the direction normal to AC with a speed, of which the corresponding height has a ratio to half AC as the centripetal force at A is to the force of gravity 1, then this body moves uniformly on the circumference of the circle AMBA, the centre of which is C. DEMONSTRATION. Indeed the body moves in this circle ; since the distance of this from the centre of the circle does not change ever from the centripetal force acting towards C, also for any function of the distance if it is in proportion to the centripetal force. And since it is acting towards the centre C, the direction of the force acting is always normal to any small portion of the curve, along which the body is moving. On this account it is always acted on by a normal force, and nowhere by one along the tangent, and hence, and for this reason this speed always remains the same (561), and thus the body describes the periphery with a uniform motion. Then since the centripetal or normal force is always the same, put this equal to g and likewise the constant speed corresponds to the height c and the radius AC = a, which quantity a is everywhere shown to be the radius of curvature. Therefore with these in place ag = 2c (561)[ dv = 0 and Pr = 2v ]. From which this ratio arises : As the height to the speed of the body, with which it is initially projected from A, there corresponds c to half the distance AC, 12 a , thus as the centripetal force g to the force of gravity 1. Q.E.D. [p. 251]EULER’S MECHANICA VOL. 1. Chapter Five (part a). Translated and annotated by Ian Bruce. page 325 Corollary 1. 612. Therefore when the body has once described an arc of the circle, the centre of which is itself the centre of force C, then it rotates for ever on the periphery of that circle. If indeed the centripetal force only depends on the distances from the centre, thus as with equal distances, the centripetal force is always equal. Corollary 2. 613. With the ratio of the diameter to the periphery put as 1 : π, the periphery of the circle in which the body is moving is equal to 2πa. Then since the speed, with which the body is moving, is equal to c= ag , the time of one period along the whole periphery 2 is equal to 2π 2a . g Corollary 3. 614. If therefore many bodies are moving in different circles, the time of the revolutions are in the square root ratio composed as directly from the radii of the circles and inversely as the centripetal forces. Corollary 4. 615. If the body is projected from A perpendicularly to AC, but with a speed either greater or less than ag , then the body describes the arc of a circle, the radius of which 2 is either greater or less than AC. [p. 252] Corollary 5. 616. Therefore in this case, when the body begins to move along the arc of the circle, the centre of which is not at C, suddenly it approaches more towards, or recedes from, the centre of the force. And this sudden motion arises from the action of another centripetal force, unless perhaps the centripetal force is the same everywhere. Scholium. 617. Moreover the body is projected with some speed from A (Fig. 54), but the direction of this is a perpendicular crossing the line AC through the centre of force C, and the curve BMAND has this property, in order that the potions AMB and AND of this, placed on this side and the other side of AC are similar and equal to each other, and AC is the axis and diameter of this curve. For since, as we have now agreed, the centripetal force is proportional to a certain function of the distances from the centre, the body either above or below AC is acted on by an equal force at the same distance from C, and for this reason this body must move towards A in the same way along DNA, as it moves away from A along AMB, and also it has the same speed at the homologous points M and N.EULER’S MECHANICA VOL. 1. Chapter Five (part a). Translated and annotated by Ian Bruce. page 326 Corollary 6. 618. Therefore every line drawn from the centre C, which is normal to the curve, is likewise a diameter of the curve ; thus in order that the parts of the curve on placed on this or that side of this line are equal and similar to each other.[p. 253] PROPOSITION 77. THEOREM. 619. If more bodies are moving around the centre of force C (Fig. 55) and describe the similar curves AM and am about C, the speeds at the similar points M and m are in the square root ratio composed from the ratios of the homologous sides and of the centripetal forces at the homologous points M and m. DEMONSTRATION. Because AC : aC = MC : mC, also the radius of osculation at M is in the same ratio to the radius of osculation at m, as also is the perpendicular CT to Ct. From proposition 74 (587) it is truly apparent that Pr dx = 2vds , [i. e. the centripetal force equation normal to the curve with radius of curvature r] or, with y ds , put in place of dx p Ppr = 2vy . Hence therefore the speed is given by : v= Ppr . 2y From which this ratio is produced : the speed at M is to the speed at m in the square root ratio composed from the direct ratios of the centripetal forces at M and m and the perpendiculars CT to cT, and in the inverse ratio of the distances MC to mC. Since truly CT is to cT as MC is to mC and the radius of osculation at M is to the radius of osculation at m as MC is to mC, then the speed at M to the speed at m is in the square root ratio composed from the ratios of the centripetal forces at M and m and of the homologous sides MC to mC. Q.E.D. Corollary 1. 620. If therefore we call AC = A, aC = a, CD = H and Cd= h, and likewise the speed at A= C and at a = c , the angular speed at M = H C2 [p. 254] and the angular speed at MC m = h C2 (599). But since H : h = MC : mC = A : a, the angular speeds at M and m are as mC C to ac , i. e. in a constant ratio. AEULER’S MECHANICA VOL. 1. Chapter Five (part a). Translated and annotated by Ian Bruce. page 327 Corollary 2. 621. Therefore the times, in which the equal angles ACM and aCm or the homologous distances AM and am are completed, are reciprocally as the angular speeds at M and m, i. e. directly as the homologous sides and reciprocally as the speeds at the homologous points. Scholium 1. 622. Indeed the speeds at the homologous points maintain the same ratio everywhere. For the speed at M = HCTC and the speed at m = hCtc (589). On account of which, since H : h = CT : Ct, the speed at M to the speed at m is in the ratio C to A to the speed at a. c , i. e. as the speed at Corollary 3. 623. Moreover from this proposition it is evident that the speed at A, C , to the speed at a, c , are in the square root ratio composed from the ratios of the centripetal forces at A and a and of the homologous sides A and a. Whereby if the said centripetal force at A = G and the centripetal force at a = g, then we have the ratio : C : c = AG : ag . [p. 255] Corollary 4. 624. Consequently the time to pass through AM is to the time to pass through am as A : G a , i. e. in ratio composed with the square root from the direct proportions of the g homologous sides and inversely as the centripetal forces at the points A and a. Therefore this ratio is constant, and the whole times of the revolutions must keep the same values between each other. Scholium 2. 625. Also in this case, in which several similar figures are described around the centre C, the centripetal forces in the homologous points must maintain the same ratio. For since the force P = 2ch 2 y (592), the centripetal force at M to the centripetal force at m is p 3r directly as the square of the speed at A to the square of the speed at a and inversely as AC to aC, which is a constant ratio. On account of which, when more similar figures are to be described around the centre C, it is required that a centripetal force of this kind of the distances can be expressed by a function, which presents the centripetal forces in the same ratio at the homologous positions. Clearly with the centripetal force P put at the distance y and Q at the distance my, P to Q must have a constant ratio, in which y is not present. For unless this is done, then it cannot be made to happen, that more similar figures can be described around the centre C.EULER’S MECHANICA VOL. 1. Chapter Five (part a). Translated and annotated by Ian Bruce. page 328 Corollary 5. 626. Moreover it is not possible to obtain this result, unless P is a certain power of y, such as yn . [p. 256] For in this case, fn Q= mn y n , and the ratio P : Q is 1 : mn, which is n f constant. Corollary 6. 627. Therefore unless the centripetal force is in proportion to a certain power of the distance from the centre C, then indeed it is not possible to happen, that more similar figures can be described around the centre C. And only with these cases do they have the properties in place, that we have elicited from this proposition. Corollary 7. n n yn 628. Moreover, if P = n , then G = A n and g = a n . Therefore the speeds at the f f f n +1 n+1 homologous places maintain the ratio A 2 to a 2 . Corollary 8. 629. And the times, in which similar arcs AM and am are completed, are in the ratio 1− n 1− n A 2 to a 2 or in the ratio of a multiple of the homologous sides, the exponent of which is 1− n . 2 Scholium 3. 630. In the preceding and in this proposition all the theorems are in place, which Huygens put as an appendix concerning centrifugal forces to his tract de Horologio. [p. 257] And that part of the appendix that I have placed here, is the part usually deduced from these propositions.

PROPOSITION 78. PROBLEM.

631. If the centre of force C attracts directly in the ratio of the distances (Fig. 56) and a body is projected from A following the direction normal to a given radius AC, to determine the curve AMDBH that the body describes, and the speed of the body at particular points. SOLUTION. The distance CA is put equal to a, and the perpendicular dropped from C on the direction of the motion is equal to a. Truly the speed at A corresponds to the height c. On arriving at M, the distance CM = y; and the perpendicular CT from C sent to the tangent at M is put equal to p, and the speed at M corresponds to the height v. Again the distance at which the centripetal force is equal to the force of gravity is f; and y the centripetal force at M = f , with the force y of gravity equal to 1. When these are compared with Prop.75 (601), P = f and y 2 −a2 Y = 2f . On account of which we have : p = a 2cf a + 2cf − y 2 2 2 For truly v = a 2c (587), from which it follows that v = p element of the curve is equal to ydy y2 − p2 a 2 + 2cf − y 2 . Moreover, as the 2f , the element of time is equal to : ydy 2 f a y + 2cfy 2 − y 4 − 2a 2 cf 2 . 2 . For the perpendicular MP dropped from M at CA is called CP = x, and x = uy. [p. 258] With these in place, ady 2cf du = 2 2 ( 1−u ) y 2cfy − 2a 2 cf − y 4 + a 2 y 2 (601). Put y = 1 2 ( q + 2 cf 2+ a ) , and there comes about 4 a cf du = ( 1−u 2 ) From which, collated with the formula − 12 dq (( a 2 − 2 cf 4 a 2cf λdZ ( A2 − Z 2 ) )2 − q 2 . (604) there is produced : λ = 12 , Z = −q and A = a 2 − 2cf . 4a 2 cf Moreover from this equation there arises ; Z = −q = C − 2Cu 2 + 2u ( A2 − C 2 )( 1 − u 2 ) = a 2 + 2cf − 12 . y 4a 2 cf (610). The constant amount C can be determined from this equation, since by making u = 1 it comes about that y = a. Hence therefore, C = 2cf − a 2 , thus 4a 2 cf ( A2 − C 2 ) = 0 . With these in place we have : 1 − 1 = ( a − 2cf )u 2cf y2 4 a 2 cf 2 2 = ( a 2 − 2cf ) x 2 4 a 2 cfy 2 on account of u = xy . Therefore this equation results : a 2 y 2 − a 2 2cf = ( a 2 − 2cf )x 2 . The applied line MP is put equal to z, and y 2 = x 2 + z 2 . Hence the following equation is produced between the orthogonal coordinates for the curve sought: a 2 z 2 + 2cfx 2 = 2a 2cf . This is the equation for an ellipse, the centre of which is situated at C, and AB = 2a is the length of one axis, and indeed the length of the other DH = 2 2cf . Q.E.I. Corollary 1.
632. The height v corresponding to the speed at M is equal to a 2 + 2cf − y 2 . Moreover since 2f a 2 + 2cf = AC 2 + CD 2 = AD 2 , then v = AD 2−fCM . 2 2 Corollary 2.
633. In a similar manner the perpendicular CT dropped on the tangent MT is given by : AC .CD p= and the tangent itself : [p. 259] 2 2 ( AD −CM ) MT = if indeed AC > CD. ( AD 2 .CM 2 −CM 4 − AC 2 .CD 2 ) ( AD 2 −CM 2 ) = ( AD 2 −CD 2 ).CP .PM AC .CD ( AD 2 −CM 2 )EULER’S MECHANICA VOL. 1. Chapter Five (part a). page 331 Translated and annotated by Ian Bruce. Corollary 3.
634. In this case, when AC > CD, AB is the transverse axis of the ellipse, and on which the foci F and G are placed. But CF = CH = ( AC 2 − CD 2 ) , and hence MT = CF 2 .CP .PM . AC .CD ( AD 2 −CM 2 )

Corollary 4.

635. Therefore the sine of the angle TMC, which the direction of the body at M makes with the radius MC , is equal to : AC .CD CM ( AD 2 −CM 2 ) and the cosine is equal to: CF 2 .CP .PM . CM . AC .CD ( AD 2 −CM 2 )

Scholion 1.

636. The distance determining the speeds CE (596, 597) is always equal to the sub- tangent AD. For by putting CE = k the body falls from E towards C drawn by the centripetal force; the body must have, when it arrives at A, a speed corresponding to the height c. On account of which, k = ( a 2 + 2cf ) (275). Moreover since AC = a and CD = 2cf , then CE = ( AC 2 + CD 2 ) = AD . Corollary 5.
637. The height corresponding to the speed, which the body moving along the line EC can acquire, when it comes to C , is given by : k 2 = a + 2cf = AD 2 . 2f 2f 2f 2 Corollary 6.
638. The time, in which the arc AM is absolved, is equal to 2.ACM (588) as h = a in this a c case. [p. 260] On this account the time of the whole revolution around the perimeter of the ellipse ADBHA is equal to
639. Area of Ellipse . With the ratio of the diameter to the a c periphery put in place 1 : π , the elliptic distance is equal to πa 2cf . Consequently the time of one revolution is 2π 2 f .

Corollary 7.

639. Therefore if more bodies are rotating in ellipses around the same centre of force attracting in the direct ratio of the distances, the times for the whole revolutions are equal to each other.

Scholium 2.

640. When the initial direction of the body at the point A is not put normal to the radius AC, the calculation does not present an ellipse for the curve described by the body, but another curve of the fourth order, which yet cannot be considered to be satisfactory. The reason for this disagreement between the calculation and the truth depends on this, that the expression of the sine of the angle, which is put in place by the curve with the radius, taken in y and u always avoids being equal to 1, with y put equal to a and u = 1, even if the following hypothesis must produce another quantity. For the sine of the angle, which the curve makes with the radius, is given by : ydu dy −u dy 2 + y 2 du 2 2 2 , which expression with u made equal to 1 evidently departs from unity, when yet it should give ah . On account of this it is evident, unless h is put equal to a, a calculation set up in this way can never agree with the truth, if indeed the equation between u and y must be found. Therefore this rule must always be stretched, [p. 261] as often as the curve to be investigated follows the teaching of Prop.75 (601). Moreover the method handled in this proposition has only been found to be convenient for algebraic curves ; and indeed it is required to use another method, if the curves are transcendental. But all algebraic curves enjoy the use of this property, so that in these from any point it is possible to draw a perpendicular. [In the Opera Omnia there is a note added by Paul Stackel : ‘This proposition seized upon by Euler without demonstration is easily shown to be false by considering Neil’s parabola ’ . However , this curve is not a closed orbit, as it is a 3 function of the form y = ax 2 , and it appears that Euler is concerned here with closed or negative energy orbits, for which there must be turning points in a physical sense, at which the radius is perpendicualar to the curve, so that the body can return to its original position. There are of course open or positive energy hyperbolic and parabolic orbits where this condition is satisfied, but which do not have a finite period. Comets, for example, may have bound orbits and return periodically, or be free on hyperbolic orbits and never return. Other motions such as the two dimensional mass on a massless spring are performing bound orbits. I have not had time to check the results in this last proposition; if someone feels they would like to do so, I would be glad to hear from them regarding their conclusions. ] Concerning which, as often as the body revolves on an algebraic curve around the centre of curvature, always one of more points can be assigned, at which the radius is perpendicular to the curve. Therefore the body should be put to begin moving at points of this kind, and the calculation is always in agreement with the truth. Moreover from such a solution the speed of the body is easily found at any other place, and hence by the inverse method from the given speed of the body at a point, at which the radius is not normal toEULER’S MECHANICA VOL. 1. Chapter Five (part a). Translated and annotated by Ian Bruce. page 333 the curve, the speed may be found at a place, where the radius falls normal to the curve. Moreover how this should be effected, is evident from the following proposition. Truly from the properties of algebraic curves recalled above, where from any given point a perpendicular can be dropped on these, but not all transcendental curves agree with this. For in the logarithmic spiral no radius can be drawn from the centre normal to the curve, but always with at some constant angle that has been found. When finally the ratio can be realised, whereby ydu dy −u dy 2 + y 2 du 2 2 2 u is always made equal to 1, can be changed into one, since still any other angle too, besides a right angle, can be produced from that formula, u departing from one should be considered 1, [p. 262] and the element du is to vanish before dy, unless the tangent at A is normal to the radius AC. Hence on this account by putting u = 1 the element dy 2 ( 1 − u 2 ) is not to be neglected with the ratio y 2du 2 , since each vanishes, as ydu in the numerator. By which it happens, that the sine of any arbitrary angle from that formula by making u = 1 can be expressed. But since this warning in the calculation cannot be observed, except the case h = a the calculation cannot show the true curve anywhere.