PROPOSITION 95. Problem.

856. If a body moving on a surface is acted on by any forces, to define the normal forces, evidently the normal force pressing [on the surface], the deflecting force [in the tangent plane normal to the curve], and the force along the tangent, with all arising from resolution.

Solution.

Whatever the forces acting should be, these can be reduced to three forces, the directions of which are along the three coordinates x, y, z [at M]. Now let the force at M (Fig. 92), drawing the body parallel to the abscissa PA, be equal to E, the force drawing the body parallel to the direction QP to be equal to F, and the force drawing the body along MQ to be equal to G. Hence the forces are each to be resolved in turn into three parts, clearly a normal pressing force, a normal force of deflection [in the plane of the tangent], and a force along the tangent [to the curve at M]. Moreover since these three directions are normal to each other, from each of these forces E, F et G, the normal and tangential forces themselves can be produced, if these are taken by the cosine of the angle that the directions of these forces make with these [other directions]. We may begin with the force along the tangent, the direction of which is MT, for which the equation arises [from Prop. 93] : and

Hence the cosine of the angle QMT, that the direction the force G makes with the force along the tangent, is equal to : [p. 478] If the force G is multiplied by this, then the tangential force due to G produced, equal to

Now the cosine of the angle that MT makes with the direction of the force F, which is parallel to QP , is equal to Hence the tangential force from F that has arisen is equal to Again the cosine of the angle that the direction of the force E makes with MT, is equal toEULER’S MECHANICA VOL. 2. Chapter 4b. page 735 Translated and annotated by Ian Bruce. and thus the tangential force arising from E is equal to [Thus, the direction ratios of Mm are ( dx , dy , dz ), and the components E cos α , F cos β ,G cos γ lies along this element, where cos α = dx , etc] dx 2 + dy 2 + dz 2 Now the normal pressing force is considered (Fig. 91), and the direction of this is along MN, with the equations or From which it follows that Therefore the cosine of the angle that the direction of the force G makes with MN is MQ

MN 1 (1+ P 2 + Q 2 ) and thus the normal force arising from G is equal to G . 2 2 (1+ P + Q ) Again the cosine of the angle that the direction of the force F, which is parallel to QP, makes with MN, is equal to Hence the normal pressing force that has come from the force F is : − FQ (1+ P 2 + Q 2 ) . And in a like manner the normal pressing force arising from the force E (Fig. 94) is equal to : − EP . (1+ P 2 + Q 2 )EULER’S MECHANICA VOL. 2. Chapter 4b. Translated and annotated by Ian Bruce. page 736 And finally, since the force of deflection is in the direction MG : as then the cosine of the angle that MG makes with the direction of the force G , is equal to [p. 479] [MQ/MG in Fig. 94]: whereby the deflecting force arising from the force G is equal to : Again the cosine of the angle that MG makes with the direction of the force F, is equal to: [as PG + EG is the projection of the horizontal component QG on the y-axis] on account of which the deflecting force arising is equal to : And then the cosine of the angle that the direction of the force E makes with MG is equal to : [-PE is the projection of QG on the x-axis; note that all the positive forces act along the negative directions of their axis. ] Hence the deflecting force arising from the force E is equal to :

Moreover since previously we have called the tangential force T, the normal pressing force M, and the deflecting force N, we have reduced the three proposed forces E, F, and G to these; namely and and Q.E.I.

Corollary 1.

857. Therefore if the body is acted on by three forces E, F and G, on putting v for the height corresponding to the speed at M, then (849), if in place of T there is put the tangential force arising from the resolution of the forces E, F and G. [As this is the only force that does work, as we now understand the physics.] Corollary 2. [p. 480]

858. If in addition, the body is moving in a resistive medium and the resistance at M is equal to R, then (850) Corollary 3.

859. If, in the equation (851) found, in which the effect of the deflecting force N has been determined, in place of N the deflecting force arising from the resolution of the forces E, F and G is substituted, then there is produced :

Corollary 4.

860. Therefore if the two equations are solved together with the elimination of v, the equation is produced, which joined with the local equation for the surface dz = Pdx + Qdy , determines the path described on the surface by the body. Corollary 5.

861. Moreover the force by which the surface is pressed along its normal, both by the normal pressing force M as well as the centrifugal force that has arisen, is equal to: (845), on substituting the value found in place of M.

Corollary 6.

862. Now from the equation in corollary 3, it has been found that in which with the value substituted there, the total pressing force is equal to [p. 481] : Scholium.

863. Since the three forces E, F, G have directions normal to each other in turn, the equivalent force from these is equal to ( E 2 + F 2 + G 2 ) . Now we have found the three forces M, N and T to be equivalent to these three forces, the directions of which also are normal to each other in turn ; whereby from these three, the equivalent force is equal to ( M 2 + N 2 + T 2 ) . On account of which, if in place of M, N and T the values found from E, F and G are substituted, there must be produced ( E 2 + F 2 + G 2 ) ; which is the thing to have checked in setting up a calculation. Moreover this serves the part of a criterion, however involved the calculation, by completely resolving the question that the solution has been rightly or wrongly set up. Now from this criterion these formulae are found to be correctly established.

PROPOSITION 96. Problem.

864. According to the hypothesis of gravity g acting downwards uniformly, to determine the line that a body projected on some surface in vacuo describes. Solution. [p. 482] Let APQ (Fig. 92) be a horizontal plane and the point M is given both on the surface, and on the line described by the body. Hence MQ is vertical and therefore in the direction of the force of gravity g. On putting AP = x, PQ = y and QM = z and with the equation expressing the nature of the surface dz = Pdx + Qdy let the speed at M, in which the element Mm is traversed, correspond to the height v. Therefore as this problem is a case of the preceding, for it becomes G = g, E = 0 and F = 0, these two equations are obtained : dv = − gdz (857) and (859). Again let a height equal to b correspond to the speed that the body is to have, if it arrives in the horizontal plane APQ; then v = b − gz . Now through the other equation :

Hence it becomes : Which equation with the help of the equation dz = Pdx + Qdy is changed into this : which integrated gives : Therefore in which a special case has to be investigated, or can be integrated. If that happens, then v is obtained by differentials of the first degree, and since it is v = b − gz , an equation of the differential of the first degree arises expressing the nature of the described curve. Now the pressing force on the surface along the normal is equal to :

which with the differentials of the second order removed, (861) changes into this : Q.E.I.

Corollary 1. [p. 483]

865. Hence the speed of any motion the body according to the hypothesis of uniform gravity g directed downwards can be known on some surface can become known from the height only, as if everywhere the body is moving in the same plane. Corollary 2.

866. If the minute time, in which the element Mm is completed, is put as dt, then Hence by the equation found, we have:

Corollary 3.

867. From these equations uncovered, it is found that : and

Hence

Now the radius of osculation of this described curve is equal to : Scholium.

868. We will explain more broadly in the following problems the use of these formulas in particular cases, in which a certain kind of surfaces is considered, to which we add examples of individual surfaces. [p. 484]

PROPOSITION 97. Problem.

869. According to the hypothesis of uniform gravity g acting downwards, to determine the motion of a body on the surface of cylinder of any kind, the axis of which is vertical. Solution. Because the axis of the cylinder is put as vertical, all the sections are equal to each other; therefore let ABQC (Fig. 95) be the base of the cylinder, on the surface of which the body is moving. Putting AP = x, PQ = y and let z be the height of the body at the point Q on the surface of the cylinder. Hence the nature of this surface of the cylinder is expressed by this equation: 0 = Pdx + Qdy or Qdy = − Pdx . Moreover this equation arises from the general equation dz = Pdx + Qdy , if P and Q are made infinitely large quantities, or with the coefficient of dz vanishing, if we may assume this. On account of which in the equations found before, P and Q must be considered as infinitely large quantities, even if they are quantities of finite magnitude [i. e. relative to each other; recall that P = ∂∂xz and Q = ∂∂yz , here with a negative sign.] Moreover P and Q are functions of x et y only, and nor is z is present in these. Therefore from these there is obtained from the deductions in the calculation : v = b − gz andEULER’S MECHANICA VOL. 2. Chapter 4b. Translated and annotated by Ian Bruce. page 742 on account of the ratio P : Q = dy : −dx. But the logarithm of the second equation is : [p. 485] Thus the equation becomes : Hence there arises : or and the integral of this is [corrected in the O. O edition]: where ∫ (dx + dy ) denotes the arc of the base BQC traversed in the horizontal motion. 2 2 If the increment of the time, in which the element Mm is completed, is put as dt, then [From the equation vc = dx 2 + dy 2 + dz 2 it follows that the constant α = dx 2 + dy 2 c . Note by Paul St. in the O. O.] and again, Whereby the times are in proportion to the corresponding arcs in the base Moreover the equation gives the equation for the curve described on the surface of the cylinder, if this surface is considered to be set out in a plane; for then ∫ (dx + dy ) denotes the abscissa on the 2 2 horizontal axis and z the applied vertical line. Now we have the projection of the curve described in the vertical plane in which we have AC cutting the horizontal plane, if withEULER’S MECHANICA VOL. 2. Chapter 4b. Translated and annotated by Ian Bruce. page 743 the help of the equation y is eliminated, in order that an equation is produced between x then and z, which are the coordinates of this projection. Clearly since dy = − Pdx Q and thus where this value of x must be substituted into P and Q in place of y. Now the pressing force that the surface sustains, arises only from the centrifugal force on account of the force g being placed in the surface itself, and is equal to : by which force the body is trying to recede from the axis of the cylinder, if this expression is positive. Q.E.I. [p. 486] [There was at the time, and even now, unfortunately, the notion held by some people that bodies try to flee from the centre when involved in circular motion, and that seems to have originated with Huygens in his Horologium. The reader may wish to compare Euler’s early solutions with more modern solution to these and further problems, such as presented by Whittaker in his Analytical Dynamics, in Ch. 4, The Soluble Problems of Particle Dynamics. ] Corollary 1.
870. Therefore the curve, that the body describes on the surface of the cylinder if the surface is set out as a plane, changes into a parabola, clearly that trajectory that a body describes in the vertical plane. Corollary 2.
871. If the motion of the body on the surface of the cylinder is considered to be composed from a vertical motion, as it progresses either up or down, and from the horizontal, then the horizontal motion is uniform, since the times t are proportional to ∫ (dx + dy ) , i. e. 2 2 to the arcs traversed in the horizontal motion. Now the vertical motion is either uniformly accelerated or decelerated. Corollary 3.
872. Hence if the horizontal motion vanishes, then the body either ascends or descends along a straight line, and generally if the body can freely ascend or descend. And this case is produced if c = 0, when ∫ (dx + dy ) vanishes. 2 2EULER’S MECHANICA VOL. 2. Chapter 4b. Translated and annotated by Ian Bruce. page 744 Example. [p. 487]
873. Let the base of the cylinder be a circle, the quadrant of which is BQC and the radius AB = a; then the equation is x 2 + y 2 = a 2 and xdx + ydy = 0. Hence the equations become P = x and Q = y = (a 2 − x 2 ) . Now the projection of the line described by this on the surface of the cylinder in the vertical plane erected from AC, is expressed by the equation : Therefore if we set c = 0, then dx = 0 and x is constant; whereby in this case the projection is a straight line. Now the curve, which is the projection for whatever value of c, is constructed with the help of the rectification of the circle. Moreover the pressing force, that the surface of the cylinder sustains, is equal to on account of the equation Whereby the pressing force is constant everywhere and proportional to c. Corollary 4.
874. On account of the same equation the pressing force generally that any cylinder sustains, Scholium. [p. 488]
875. But not only can the motion of a body on the surface of an erect cylinder be easily determined with the help of resolution, but also, if the axis of the cylinder should be horizontal, with the same ease the motion of the body becomes known. And indeed if the body does not have a horizontal motion along the cylinder, then the body perpetually remains on the same section of the cylinder as that is moving on a given line. But if now the motion is agreed to be horizontal, this remains the same always and does not disturb the other motion; and from these motions taken together the motion of the body truly becomes known easily.EULER’S MECHANICA VOL. 2. Chapter 4b. Translated and annotated by Ian Bruce. page 745 PROPOSITION 98. Problem.
876. If a body is moving on the surface of a solid of revolution, of which the axis is the vertical line AL (Fig. 96), in vacuo with gravity g acting uniformly, to determine the motion of the body on a surface of this kind. Solution. The solid of revolution is generated by the rotation of the curve AM about the vertical axis AL; all the sections of this are horizontal circles, of which the radii are applied lines of the curve AM. Therefore the equation expressing the nature of this surface is dz = xdx + ydy , Z with Z some function z; indeed it is given by : ∫ 2 Zdz = x 2 + y 2 = LM 2 . Therefore if the equation is given between AL = z and LM = 2 Zdz for the curve AM , then Z is also given. [p. 489] With these put in ∫ y place, therefore P = Zx and Q = Z ; from which values if substituted the two following equations are obtained, from which the described curve as well as the motion on that surface become known : v = b − gz and Whereby this becomes : Putting x 2 + y 2 = u 2 ; the function u is a certain function of z, clearly ∫ u 2 = 2 Zdz , and the above equation changes into this : Now the projection in the horizontal plane is obtained through the equation between x ∫ and y, if from the equation x 2 + y 2 = 2 Zdz the value of z is substituted in terms of xEULER’S MECHANICA VOL. 2. Chapter 4b. Translated and annotated by Ian Bruce. and y ; and the arc of this projection is page 746 ∫ (dx + dy ) . Moreover in the vertical plane the 2 2 projection is obtained by eliminating y; with which done the equation is produced : which equation, if it is divided by u, allows the construction. Now the pressing force, that the surface sustains towards the axis, is equal to : Q.E.I. Corollary1.
877. If the increment of the time in which the element Mm is completed, is put as dt, then and the integral of this is : [From equation (876) v = c3( dx 2 + dy 2 + dz 2 ) it follows that the constant α = c ( xdy − ydx ) 2 c . Note in the O.O.] Therefore the time becomes as : ∫ with ydx denoting the area of the projection in the horizontal plane. Corollary 2. [p. 490]
878. If the body is considered to be moving in the projection in the horizontal plane, then the speed of this at Q corresponds to the height from which motion in the projection the motion on the surface itself can be found.EULER’S MECHANICA VOL. 2. Chapter 4b. page 747 Translated and annotated by Ian Bruce. Corollary 3.
879. Therefore let BQC (Fig. 97) be the projection of the curve in the horizontal plane, in which the body is moving, thus so that the motion of this corresponds to the motion of the body on the surface itself; the time in xy ∫ which the arc BQ is completed, is as 2 − ydx ,or with the negative as ∫ xy ydx − 2 , i. e. as the area BAQ drawn by the radius AQ. Corollary 4.
880. Moreover the element of the area BAQ is ydx− xdy . Therefore with the tangent QT 2 drawn, and on sending the perpendicular AT to that from A, then 3 3 Whereby the height corresponding to the speed at Q is equal to c 2 = c 2 on putting AT p AT = p. Corollary 5.
881. Therefore the body moving in the projection likewise is moving on that surface, and if it is moving freely attracted by some centripetal force to the centre A (Book I. (587)). Corollary 6. [p. 491]
882. The point B corresponds to the start of the motion made on the surface, and since the direction for the first motion on the surface is given, the perpendicular to the tangent at B is given. Therefore let AB = f and the perpendicular to the tangent is equal to h ; then the 3 height corresponding to the speed at B is equal to c 2 . h Corollary 7.
883. Therefore the centripetal force tending towards A, which acts, as the body in the projection BQC is moving freely, is equal to 2c 3 dp with u put in place fore p 3 du ( x2 + y2 ) . Now the equation between u and z expresses the nature of the curve, and the rotation of this curve thus has given rise to the proposed surface, and is therefore given.EULER’S MECHANICA VOL. 2. Chapter 4b. page 748 Translated and annotated by Ian Bruce. Corollary 8.
884. Again it is the case that These values, if they are substituted in the equation found, give or Corollary 9.
885. Therefore this quantity, with the differential of this divided by du, gives the centripetal force required at A, as the body in the projection BQC is moving freely, in a motion corresponding to the motion of the body on the surface. Corollary 10.
886. If c = 0, also making p = 0. Whereby in this case the projection is in the horizontal plane with a straight line passing through A, upon which the body thus approaching A is thus attracted, as it is the centripetal force Corollary 11. [p. 492]
887. If the direction of the body is first horizontal, the tangent at B is normal to AB and thus h = f. Now in this case the speed on the surface is equal to the speed in the 3 projection ; whereby if i is the value of z, and if u = f, then b − gi = c 2 . f

Corollary 12.

888. If besides the centripetal force at B is equal to the centrifugal force, then the curve BQC is a circle and therefore so also the curve described on the surface, and both the motion on the surface as well as that in the projection are equal. Let it be the case, when u = f and z = i, that dz = mdu ; as p = f and u = f and z = i in the case in which a circle is described, then 2c3 = gmf 3 . [See (892)] Corollary 13.

889. If there is put in place π : 1 as the ratio of the periphery to the diameter, then the perimeter of our circle is equal to 2πf , which divided by the speed gives the time of one period in the circle, that hence becomes equal to c 3 , i. e. f2 2π 2 f gm gmf , 2 . Hence the f pendulum completing whole isochronous oscillations with this period is equal to m in the same hypothesis of gravity. Corollary 14. [p. 493]

890. Therefore on the surface which is generated by the rotation of the curve AM (Fig. 96) about the vertical axis AL, a projected body can describe a circle of radius LM in the same time that a pendulum of length equal to the subnormal LS can complete a whole oscillation.

Corollary 15.

891. Therefore if the curve AM is a parabola, all the periods along horizontal circles in the parabolic conoid are completed in equal times; and pendulums in the same times by performing whole oscillations with a length equal to half the parameter.

Scholium 1.

892. Whatever curve AM is assumed, if the centripetal force motion in the projection towards A is defined from a given formula and that is put equal to the centrifugal force 3 and joined together with the equation b − gi = c 2 , there is produced 2c3 = gmf 3 ; for in f this case as the projection is a circle as is the curve described on the surface. Now so that this can be more apparent, there is put dz = qdu, and the (884) centripetal force comes about equal toEULER’S MECHANICA VOL. 2. Chapter 4b. Translated and annotated by Ian Bruce. page 750 Now putting u = f, z = i and bf 2 = gf 2i + c3 and q = m; the centripetal force is equal to : 3 which put equal to the centrifugal force 2c2 , gives 2c3 = gmf 3 . f Example. [p. 494]

893. Let the surface be a circular cone or AM (Fig. 96 and Fig. 97) a right line inclined at some angle to the axis AL; whereby then z = mu and dz = mdu. Therefore the centripetal force tending towards A which is produced, in order that the body projected in BQC is moving freely, is equal to Therefore this force is composed from a constant force and from a force that varies inversely as the cube of the distance from the centre A. If c = 0, then the projection is a right line passing through A and the centripetal force is equal to gm , or is 1+ m 2 constant. Therefore the body approaches A with a uniformly accelerated motion; now the motion on the surface of the cone agrees with the descent or the ascent on a right line inclined equally and the acceleration is the same. But if the projection is curvilinear and the tangent at B is normal to AB, then 3 i = mf and b − gmf = c 2 on putting AB = f. Therefore in this case we have f b= c + gmf f2 3 3 , and if the body is revolving in a horizontal circle, it becomes above: 2c3 = gmf 3 . Hence it comes about that the speed of the body corresponds to the height c 3 = gmf . And the periods completed in the same times in this circle, in which 2 f2 f pendulums with lengths m complete whole oscillations. But if it is not the case that

3 3 c gmf + 2c3 = gmf 3 , now still b = , the curve BQC has a certain tangent at B normal to f2 AB, but this projection is not a circle. But if 2c3 is almost equal to gmf 3 , then the curve does not depart much from a circle ; but it has various apses, at which the tangent is perpendicular to the radius. Now the positions of these apses has been determined by proposition 91 of the preceding book. For since the centripetal force is if this is compared with that centripetal force P3 , [p. 495] on account of y = u then y and thus on putting u = f , the line of the apses also is distant from the preceding apse by the angle But since 2c 3 is approximately equal to gmf 3 , the angle intercepted between the two 2 2 apses is equal to 180 m 3+1 degrees. Hence as m 3+1 is always greater than 13 , then the angle intercepted is greater than1030 55’. Scholium 2. 894. I do not include here the example of the surface of the sphere, but I resolve the motion on that boundary in the following proposition, since this matter is worthy of being handled most carefully. For if a pendulum is not set in motion following a vertical plane, then the body moves on the surface of a sphere and the body describes a circle or another not less elegant curves, as becomes known from any experiment set up. Indeed the case, in which a pendulum completes a circle, has now been published in the Acta Erud. Lips. A. 1715, [p.242] by the most celebrated Johann Bernoulli under the title De centro turbinationis inventa nova. [Opera omnia, Tom. 2, p. 187]. But if the curve is not circular, no one as far as I know, has consider either the motion of this pendulum nor determined it. [Isaac Newton had considered an analogous problem in the Principia, London, 1687, Book. I sect. X Prop. LV and LVI, in connection with the apses of the moon, where gravity follows the inverse square law and therefore is not constant; see Cohen’s translation and commentary.]EULER’S MECHANICA VOL. 2. Chapter 4b. Translated and annotated by Ian Bruce. page 752 Definition 6. [p. 496] 895. The rotational [or whirling] motion of a pendulum is the name given to the motion imparted to a pendulum which is not in a vertical plane. Therefore in this case the pendulum is not moving in the same vertical plane, but describes some curve on the surface of the sphere of which the radius is the length of the pendulum in place. [This is now called the spherical pendulum.]

PROPOSITION 99. Problem.

896. Pendulums are set in motion in rotational motion ; to determine the motion and the curved line described on the surface of a sphere. Solution. Since the moving body is bound by the pendulum, it moves on the surface of a sphere of which the radius is the length of the pendulum. Let this length or the radius of the sphere be equal to a ; from the nature of the circle AM (Fig. 96 and Fig. 97) it follows that z = a − (a 2 − u 2 ) . Hence we have and thus From these is found the centripetal force attracting towards A, which put in place, in order that the body can move freely in the projection BQC , is equal to And this equation is obtained for the curve BQC : which is sufficient for the construction of the projection BQC. Let the tangent at B be perpendicular to the radius AB, which must always happen somewhere, unless the projection is a straight line, since the centripetal force decreases with decreasing u. Putting AB = f; then i = a − (a 2 − f 2 ) and [p. 497] If besides it should be thatEULER’S MECHANICA VOL. 2. Chapter 4b. Translated and annotated by Ian Bruce. page 753 then the body is rotating in a circle, the radius of which is f, with a speed corresponding to the height Now the length of the pendulum completing whole oscillations in the same time is equal to (a 2 − f 2 ) . But if it is not the case that 2c3 = gf 4 (a 2 − f 2 ) , and now the difference is very small, then the curve BQC does not depart much from the circle. In order that the positions of cusps of this curve can be found, just as in Proposition 91 of the preceding book : Hence this becomes : and Since the curve is nearly a circle, put u = f and with which done the is produced : Hence it follows that the interval between the apses is the angle : Clearly the position for an angle of such a size, at which the pendulum is at a maximum distance from the axis, is placed at intervals with the position at which the pendulum is closest to the axis. Q.E.I.

Corollary 1.

897. Therefore in order that the pendulum AB = a (Fig. 98) describes the circle BCDE in g .BO 2 a rotational motion, it is necessary that the speed corresponds to the height 2.AO .

Corollary 2.

898. Now the length of the pendulum, that completes the smallest whole oscillations in the same time, in which the period is completed in the circle BDCE, is equal to AO.

Corollary 3. [p. 498]

899. Therefore the times, in which different circles are traversed by the rotating pendulum AB, are in the square root ratio of the heights AO.

Corollary 4. 900. Therefore since the pendulum of length a makes a maximum horizontal circle, the radius of which is a, an infinitely great speed is required and each period is completed in an infinitely short time.

Corollary 5.

901. If the radius of the circle BO should be very small with respect to the pendulum AB = a, the periods of the rotational motion are in agreement with the whole oscillations of the same pendulum. Corollary 6.
902. If the curve described by the pendulum should not be a circle, but a close figure and BO is very small, then the angle between two apses is 900 or a right angle.

Corollary 7.

903. Now in this case the curve described by the body is an ellipse having the centre at A. Which can be gathered from the centripetal force, which then becomes proportional to the distances.EULER’S MECHANICA VOL. 2. Chapter 4b. Translated and annotated by Ian Bruce. page 755 Corollary 8. [p. 499]
904. Moreover since the larger the radius BO becomes, from that also the greater becomes the angle intercepted between the two apses. And on making BO = BA, the angle here is 1800. Corollary 9.
905. If the angle BAO is 30 degrees, then BO = 12 BA or f = 12 a . Therefore the angle intercepted between the two apses is 360 degrees or 990 50’. Hence this figure 13 abcdefghik, etc is of the projection in the horizontal plane (Fig. 99), in which the highest apses are at a, c, e, g, i and the deepest at b, d, f, h, k. Corollary 10.
906. Therefore in this figure the line of the apses is moving in succession; for each period progresses by almost 390 in succession.EULER’S MECHANICA VOL. 2. Chapter 4b. Translated and annotated by Ian Bruce. page 756 Corollary 11. [p. 500]
907. But if that angle BAO should be less than 30 degrees, then the progression of the apses is less also. For whatever the angle BAO which is known at some point, I resolve the fraction in a series, which is the following : Hence in one period the line of the apses is moved forwards by the angle 135 f 2 degrees a2 as an approximation, if f is very small. Corollary 12.
908. From these it is apparent that the movement of the line of the apses in individual periods is almost in the square ratio of the sines of the angle BAO.