Motion in General

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

PROPOSITION 98. THEOREM.

802. There are three principal forces which can be put in place, and into which other forces must be resolved, in order that a body can move on a curve that does not exist in a plane; these individual forces are normal to each other; and of these one is the tangential force, and the remaining two are normal to that force, of which one lies in the given plane, and the direction of the other is normal to this plane, and nothing remains of the original forces to change the actions of these forces.

DEMONSTRATION.

With the plane APQ assumed fixed [in space] (Fig. 75) and relative to that axis AP, an element Mm is described by a body. From the points M and m the perpendiculars MQ and mq are sent to the fixed plane and from the points Q and q, perpendiculars QP and qp are sent to the axis AP. Now, if the body is not acted on by any force, then it progresses along the line Mm produced with the speed that is has in Mm ; therefore in an equal small interval of time, equal to the time in which it traversed Mm, it arrives at n, with the element mn described equal to and in the same direction as that put for the element Mm. Whereby by also sending a perpendicular nr from n to the plane APQ, then the elements Qq and qr are also equal to each other and placed in the same direction ; because of this, the perpendicular rπ sent from r to the axis AP cuts off the element pπ = Pp. Let the speed in which the first element Mm is described, correspond to the height v, and in the first place the tangential force is considered, which has a direction along mn and the whole force is taken up with changing the speed. This tangential force T is put in place with the existing force of gravity equal to 1, and we have [p. 340] dv = T .Mm , and the element mn is completed with a speed corresponding to the height v + dv. Following this, in the plane Mr there is considered a force having a direction ms normal to the direction Mm of the body. This therefore has the effect that the body can deflect from mn and progress along the element mv placed in the same plane as Mr. Let this normal force be equal to N, and since the radius of osculation of the elements Mm and mν , with the perpendicular νε sent along fromν to mn is equal to mνεν [In triangle 2 mεν , we can set ds = Rdθ , where ds = mν and dθ = εν / mν , giving the required result for the radius R], henceEULER’S MECHANICA VOL. 1. Chapter Five (part e). Translated and annotated by Ian Bruce. 2 v .νε = N mν 2 page 489 (561). Truly νε is the sine of the angle nmν . On account of which, mν 2 vsin .nmν = N .mv = N .Mm , and thus sin .nmv = N .Mm . 2v [Note that Fig. 75 is drawn in perspective, so that the elements Mm = mn = mν , and the force normal to mn acts in this manner as previously shown for centripetal forces in a plane.] The third force is normal to each of those set up on mn and ms, thus in order that its direction is along the normal mt to the plane Mr. This force neither impedes the actions of the preceding forces, nor is it allowed itself to be impeded by their actions. Therefore the whole effect of this force is to draw the body away from the plane Mr; the body is drawn from that plane from ν to μ , thus so that the plane νmμ is normal to the plane Mr, and the angleνmμ is the result of this force. Therefore from the same argument that we put in place for the preceding force, by evaluating the force in the same way, if this force is M, then it is given by :EULER’S MECHANICA VOL. 1. Chapter Five (part e). Translated and annotated by Ian Bruce. sin .νmμ = M .Mm . 2v page 490 Therefore these three forces likewise have the effect that the body, after it has described the element Mm moves to the element mμ , with an increase in speed clearly corresponding to the height v + T .Mm. Moreover, any other forces acting on the body can also be resolved in this way into the forces, the directions of which lie along the directions mn, ms, mt. As we have determined the effect of these forces on the body, so [p. 341] likewise the effect of any forces can also become known. Q.E.D. Corollary 1. 803. By taking νμ in the plane nr π , and by sending a perpendicular μρ from μ to the plane APQ, μρ is parallel to rn itself. Therefore the three coordinates for the points M, m and μ are AP, PQ, QM; Ap, pq, qm, and Aπ , πρ , ρμ . Corollary 2. 804. Whereby if from μ the perpendicular μη is sent to mv, it is normal to the plane Mr; and likewise in a similar manner ρθ , which is perpendicular to qr, is normal to the same plane. On account of which, as ρ and μ on the line ρμ put parallel to this plane, then we have ρθ = μη and θη = ρμ . Corollary 3. 805. If the normal qT is drawn to Qq in the fixed plane APQ, this line qT is normal to the plane Mr. Therefore since mt is also normal to the same plane, then mt is parallel to qT; and between these the distance is the height mq.

Corollary 4.

806. The three coordinates are called AP = x , PQ = y and QM = z . And we have : Pp = pπ = dx , pq = y + dy , qm = z + dz and πρ = y + 2dy + ddy and ρμ = z + 2dz + ddz = θη. But [from a binomial expansion on extracting the equivalent of qr] and hence: Again we have [p. 342]:EULER’S MECHANICA VOL. 1. Chapter Five (part e). Translated and annotated by Ian Bruce. πr = y + 2dy and rn = z + 2dz . page 491 Then and [Thus, the length mμ is compounded from the lengths pπ or dx in the x-direction, πρ − pq = y + 2dy + ddy - y-dy = dy + ddy in the y-direction, and ρμ − qm = z + 2dz + ddz − z − dz = dz + ddz , in the z-direction; from which the length corresponding to mμ is extracted by a binomial expansion.] Corollary 5.
807. Since mq , θη and rν are parallel to each other, in the same plane and terminated by the lines qr and mν , then [The gradient of the line mη is the same as the gradient of the line mν in the plane mνrq , as are qθ and qr] For it is the case that : and Whereby and hence Thus, it is found that :EULER’S MECHANICA VOL. 1. Chapter Five (part e). Translated and annotated by Ian Bruce. page 492 [We apply the sine rule to the triangle nmν : it is easily found that the sine of the angle mnν in this triangle can be found from the right-angled triangle with hypotenuse Mn in the plane Mr with base MM’ parallel to Qr, to be given by sin .mnν = dx 2 + dy 2 dx 2 + dy 2 + dz 2 , while the length mν is equal to mn = dx 2 + dy 2 + dz 2 , where sums of powers of higher orders are ignored; from these on applying the sine rule, the result quoted emerges. The diagram here shows the coordinates of some of the points, and may be of some assistance to you if you want to establish the result for yourself.] Corollary 6.
808. Since rρ = −ddy and Qq : Pp = rρ : ρθ , it follows that On this account we have : [In triangle ημm , which projects normal to the plane Mr, we have from above, On using the sine rule, and noting that both the large angles are essentially right, we have, sin .νmμ / νμ = 1 / mμ , sin .νmμ = νμ / mμ , which gives the result quoted on neglecting the higher order terms in the denominator.]EULER’S MECHANICA VOL. 1. Chapter Five (part e). Translated and annotated by Ian Bruce. page 493 Corollary 7.
809. Therefore from the three given forces T, N and M that the body is acted upon, there arises the three following equations : and from which the speed of the body at individual points as well as the curve itself becomes known. [p. 343] Corollary 8.
810. The two latter equations joined together with v eliminated give this equation : For which, the nature of the surface in which the curve described by the body lies, can be assumed to be expressed by the equation,. Scholium.
811. Therefore from this proposition we can deduce the first rules, from which the motion of a body acted on can be deduced, in order that it does not move in the same plane. Indeed we have shown that all the forces can be resolved in terms of three, the effects of which we have determined ; and thus, whatever forces are proposed to be acting, whatever motion they produce on the body can become known. Indeed it is apparent, if the [second normal] force M is not present, then the motion of the body is entirely in its own plane, which does not concern us here. But if the tangential force T vanishes with the forces M and N left, then the body describes a non planar orbit, but yet is still carried around uniformly. From which the position of the orbit is generally known, and it is necessary to find the intersection of this inclined plane, in which the elements Mm and mμ are present, with the plane APQ. [The task of the next proposition.]EULER’S MECHANICA VOL. 1. Chapter Five (part e). Translated and annotated by Ian Bruce. page 494 PROPOSITION 99. PROBLEM.
812. To determine the inclination of the plane, in which the elements Mm and mμ (Fig.

76. described by a body have been placed, relative to the fixed plane APQ and to find the line of intersection of the two planes. [p. 344] SOLUTION. In the plane, of which we seek the inclination, the three points M, m and μ are given; therefore in this plane a certain line is placed of which the points pass through both planes. [This is the element mM ; the plane APQ can be considered as horizontal while the eye looks down and sees the plane containing QM, the z coordinate, rising vertically, as in the previous proposition. The point μ arises as previously due to the actions of the forces M and N, and the lines mM and mμ define a plane, which is tangential to the surface on which the body moves.] Whereby if the line mM is produced, then it meets the line qQ [both in the plane APQ] produced in S, while the point S then lies in the plane Mmμ as well as in the fixed plane APQ; therefore the line formed by the intersection of the two planes passes through S. Therefore with the lengths remaining as before: AP = x, PQ = y and QM = z and with the elements of the abscissa Pp and pπ equal to each other, then qm − QM : Qq = QM : QS and hence : For the position of the line QS is known from the angle PQS, the sine of which is equal toEULER’S MECHANICA VOL. 1. Chapter Five (part e). Translated and annotated by Ian Bruce. page 495 In addition, the point n is also situated in the plane Mmμ ; on account of this the line passing through n and μ or drawn parallel to this passes through M extant in the same plane. Moreover this line crosses the plane APQ in the point R of the line QP produced, and QR is known from this ratio rn − ρμ : rρ = QM : QR ; hence we have zddy zddy QR = ddz and thus PR = ddz − y = zddy − yddz . ddz [For rn − rμ = nμ = −ddz ; rρ = −ddy , QM = z, giving QR as shown, etc.] For Qq : Pp = QS : PT with ST drawn perpendicular to AP. From which there arises PT = zdx . dz Furthermore, Pp : ( pq − PQ ) = PT : ( PQ + ST ) and thus zdy PQ + ST = dz and ST = zdy − y. dz The line RS produced cuts the axis AP in O and hence PR − ST : PT = PR : PO , from which it is found that: and [p. 345] Again from these, the tangent of the angle POR is equal to [Note that the numerator of PO can be factored to give dx( zddy − yddz ) , which cancels with the numerator of PR above to give the simpler expression shown]; thus the position of the intersection RO of the plane Mmμ with the fixed plane APQ is known.[On the annotated diagram, the distance AO along the fixed axis AP is known, and the angle θ to AP has been found.] Moreover, the mutual inclination of these planes is found by sending the perpendicular QV from Q to RS ; for then the tangent of the angle MQ of inclination is equal to QV . Indeed [as the right triangles RVQ and RPO are similar] this becomes : and thus from which the angle of the inclination between the two planes Mmμ and APQ is determined. Q.E.I.EULER’S MECHANICA VOL. 1. Chapter Five (part e). Translated and annotated by Ian Bruce. page 496 Corollary 1.

813. If the angle POR is always the same, then and the tangent of the angle is equal to α. This equation on integration gives αdx + dy + βdz = 0 and αx + y + βz = f . [Note that β is defined in the next cor.] From which equation it is known that the whole orbit described by the body lies in the same plane in this case. Corollary 2.
814. If indeed it is the case that αx + y + β z = f , then αdx + dy + βdz = 0 and ddy + βddz = 0. Hence we have : ddy [on substituting β = − ddz into the equation for PO above, and simplifying,] and because − dy − βdz = αdx , then and likewise AO is constant. Corollary 3. [p. 346]
815. Again with the angle POR remaining constant or αx + y + β z = f , then the tangent of the angle of inclination of the planes Mmμ and APQ is equal to : Whereby this angle itself is constant. ddy [This is readily seen on substituting β = − ddz into and using the result of Cor. 2.]EULER’S MECHANICA VOL. 1. Chapter Five (part e). Translated and annotated by Ian Bruce. page 497 Corollary 4.
816. For neither can the point of intersection O be put as invariable, unless likewise the orbit described by the body becomes a plane. For let AO = f and putting x − f = t [= OP] and dx = dt , then: [from the expression set equal to t, we have] and hence This is multiplied by t, from which it is found that : Which equation can be integrated, in account of dt being constant; for it becomes This divided by tt and integrated gives : y αz = t + β or y = az + βx − βf . t Which is seen to be a plane surface. [The first result follows by setting tddy d ( tdy − ydt ) d ( tdz − zdt ) = tdy − ydt and likewise tdztddz = tdz − zdt and noting that ddt = 0; while the tdy − ydt − zdt second case follows directly from tdy − ydt α ( tdz − zdt ) = on integrating by parts, where t2 t2 integrals cancel.] Corollary 5.
817. But if the tangent of the angle of inclination of the planes Mmμ and APQ is constant, an equation of the kind αx + y + βz = f is not produced. And elsewhere it has been shown that the orbit described by the body is then not by necessity a plane. Corollary 6.
818. Whereby, with the curve described by the body not being planar, then neither the point O nor the angle POR can be take as invariable. Moreover if these are variables, then no more can the angle between the planes Mmμ and APQ be considered as constant. [p. 347]EULER’S MECHANICA VOL. 1. Chapter Five (part e). Translated and annotated by Ian Bruce. page 498 Corollary 7.
819. The line of intersection RO, which is called the line of the nodes in astronomy, if it does not have a constant position, turns about the point S. For the element mMS has been put in the plane of the elements Mm from the preceding argument. Whereby the intersection of RO and the preceding line cross over each other at S. Corollary 8.
820. Therefore this point S is at that place, where [ AP – TP = ] − zdx and ST = AT = xdzdz zdy − ydz . dz From which the position of the point S is known. Corollary 9.
821. If the point S is put invariable, then xdz − zdx = adz and zdy − ydz = bdz , hence it is found that x − a = α z and y − b = β z . Therefore in this case the orbit described by the body is not only planar, but also it is a straight line, since the projection of Qqρ gives a straight line in the x-y plane, and since y − b = βz , Mmμ is also a straight line in the y-z plane. Scholion.
822. Now from the principles established, which are concerned with the non planar motion of a body on a surface, the description itself can be divided into two parts as before with regard to motion in a plane. In the first of these we instruct how to find the curve described by a body from given forces, and in the second truly it is shown, if the curve is given which the body describes, what kind of forces are required to do this. Here only the curve itself needs to be given, or likewise also the speed of the body at individual points. [p. 348]EULER’S MECHANICA VOL. 1. Chapter Five (part e). Translated and annotated by Ian Bruce. page 499EULER’S MECHANICA VOL. 1. Chapter Five (part e). Translated and annotated by Ian Bruce. page 500 PROPOSITION 100. PROBLEM.
823. If a body is acted on by three forces, of which the directions Mf, Mg and MQ (Fig.

77. are parallel to the three coordinate axes AP, PQ and QM, to determine the motion of the body and the orbit in which it moves. SOLUTION. Since Mf and Mg are parallel to AP and PQ, the plane fMg is parallel to the plane APQ. In this plane, Mi is drawn parallel to the element Qq; and this line Mi is also placed in the [vertical] plane Mq. A perpendicular Qd is sent from Q to the element mM produced; and from f and g the perpendiculars fi and gk are sent to Mi. Then from i and k the perpendiculars ib and kc fall on Md. Moreover fi and gk are perpendicular to the plane Mq, since the plane fMg is normal to the plane Mq. Now with AP = x, PQ = y and QM = z remaining as before, let the force pulling the body along Mf be equal to P, the force which pulls along the body along Mg is equal to Q, and the force pulling along MQ is equal to R. Therefore these forces, in order that their effect can become known, must be resolved into forces acting along the tangent Mm, placed along the normal to Mm in the plane Mq, and normal to the plane Mq. Since the angle Mfi =Qqp it is the case that [p. 349] and Pdy ( dx 2 + dy 2 ) expresses the force arising from P acting along if , and if P alone acts, then by (802) : Following this, the force pulling along Mi is equal to : Pdx . 2 2 ( dx + dy ) Again this force can be resolved into a force pulling along bi equal to : and a force pulling along Mb equal to : Therefore from P we have the contributions :EULER’S MECHANICA VOL. 1. Chapter Five (part e). Translated and annotated by Ian Bruce. page 501 In a like manner the force Q, the direction of which is Mg, can be resolved into the force acting along kg equal to : Qdx ( dx 2 + dy 2 ) and the force along Mk equal to : Qdy ( dx 2 + dy 2 ) . This is finally resolved into the force along ck equal to : and the force along Mc equal to : Whereby, if this force alone acts, we have : Finally the force R, having the direction MQ, is resolved into the force along Md equal to : and the force along dQ equal to : Therefore from the force R there becomes : Therefore from these three forces P, Q, and R acting at the same time, the tangential force arising from all : the normal force in the plane Mq which is in place : and the second normal force :EULER’S MECHANICA VOL. 1. Chapter Five (part e). Translated and annotated by Ian Bruce. page 502 With these values T, N and M substituted in place in the equations (809), the three following equations are produced : and Which equations determine the equations of motion of the body. Q.E.I. [p. 350]EULER’S MECHANICA VOL. 1. Chapter Five (part e). page 503 Translated and annotated by Ian Bruce. Corollary 1.

824. These two last equations give that ratio: From which it is found : Corollary 2.
825. Therefore the plane Mmμ (Fig. 76), in which the elements Mm and mμ are present, can be defined in this manner. Let −Qdz + Rdy and the tangent of the angle POR = Pdz − Rdx and the tangent of the angle, which the plane Mmμ makes with the plane APQ, Corollary 3.
826. If the force P vanishes, there is found : from the two equations found above. And from the third there is produced : the integral of which is Corollary 4.
827. Therefore from this hypothesis the time, or ∫ ( dx 2 + dy 2 + dz 2 ) v , From which it is understood that the motion of the body progressing along parallel to the axis AP is uniform. [p. 351]EULER’S MECHANICA VOL. 1. Chapter Five (part e). Translated and annotated by Ian Bruce. page 504 Corollary 5.
828. Again from the same hypothesis, we have : since ddy : ddz = Q : R (824). From which equations the curve itself described by the body can be determined. Scholium.
829. From the resolution of forces into three forces that we have considered in this proposition, it is easily seen how all forces which can be devised can generally be reduced. Whereby, with these forces given, it is not difficult to find the curve described by the body, also for whatever cases are proposed this theorem has the maximum usefulness. PROPOSITIO 101. PROBLEM.
830. If a body is always acted on by a force towards the axis AP along the perpendicular MP dropped to the body (Fig. 78), then it is required to find the motion of the body. SOLUTION. With the coordinates drawn as before AP = x, PQ = y and QM = z then MP = ( y 2 + z 2 ) . Let the force acting along MP be equal to V and that is resolved into two components acting along MQ and Mg, where Mg is parallel and equal to PQ [p. 352]. Therefore the force acting along MQ is equal to Vz . With these compared with the 2 2 ( y +z ) preceding proposition, we have P = 0, Q = Vy ( y +z ) 2 2 and R = Vz . ( y2 + z2 ) Whereby we have : ddy : ddz = Q : R = y : z (828) and yddz = zddy or yddz − zddy = 0, the integral of this equation is : ydz − zdy = bdx. [on integrating by parts, and recalling that ddx = 0] Again we have Vy ( y +z ) 2 2 =− 2addy , from (828). dx 2EULER’S MECHANICA VOL. 1. Chapter Five (part e). Translated and annotated by Ian Bruce. page 505 Which equations can be solved together to determine the curve described by the body. Moreover the speed of the body is given by the equation : vdx 2 = a( dx 2 + dy 2 + dz 2 ) , from (826). or the speed itself is equal to a( dx 2 + dy 2 + dz 2 ) . dx Q.E.I. Corollary 1.
831. Putting dx = pdy, on account of constant dx : With these substituted these equations are obtained in order that the curve described can become known : Corollary 2.
832. If again on putting z = qy, these equations are transformed into : Which also contain three variables. Scholium 1.
833. It is appropriate to use examples in order that this can be explained most clearly. On account of which we produce some, in which the force V is made to depend on the distance MP, and that we put proportional to some power of the distance MP, for which it is possible to compare this motion with the motion in a plane [p. 353] arising from a centripetal force in proportion to a certain power of the distance. There is a great similarity between these cases [and those previously considered], as if the centre of force in the plane has its place taken by the axis of the force to which the body is attracted. And if the body is thus initially projected, so that it does not progress along the axis AP, then the motion of the body is in the plane PQM, and the body is always attracted to the point P, the centre of force.EULER’S MECHANICA VOL. 1. Chapter Five (part e). Translated and annotated by Ian Bruce. page 506 Example 1.
834. Let the force V be directly proportional to the distance MP and put V = ( y2 + z2 ) . f Hence we have : y 2adp = 3 (831) and on integrating : f p dy Since moreover we have, by (832) : The integral of which is : Whereby we have : which equation expresses the projection of the described curve on the y, z plane normal to the axis AP, which is therefore seen to be an ellipse, the centre of which lies on the axis AP. Then since dx = pdy, it is the case that dx = dy af ( 2cf − y 2 ) , which equation expresses the projection of the curve sought in the x, y plane APQ. And thus this is the curve of the sine of Leibniz, with the abscissa x taken as the arc, of which the sine is the applied line y. Example 2.
835. If the force V varies inversely as the square of the distance MP or V= ff ff = since z = qy. [p. 354] y2 + z2 yy( 1+ q 2 ) On account of which we have : ff y ( 1+ q ) 2 2 3 2 = 2adp . p 3dy 2 y dq Moreover since dy = bp (832), then we have f 2 dq ( 1+ q ) 2 the integral of which is : 3 2 = 2 abdp ; p2EULER’S MECHANICA VOL. 1. Chapter Five (part e). Translated and annotated by Ian Bruce. y 2 dq with the value bdy put in place of p. Hence this becomes : page 507 and on integrating : Whereby, when q = zy , there is produced : Which is the equation of the projected curve described in plane normal to the axis AP ; that can therefore be recalled as the section of a cone, of which either focus is in the position of the axis AP. Scholium 2.
836. From these it is understood that the projections of the curves described in the plane normal to the axis AP are congruent with the curves that bodies describe in that plane with the same force acting. Moreover, neither is this wonderful ; for the motion, that we consider in place of this, can be reduced to the motion in a plane made normal to the axis AP by a body always attracted to the axis, provided the plane is considered to move with a uniform speed along the axis AP. And for this progressive motion, since it is made uniformly in direction, it is unable to disturb the motion of the body in the plane. Now from this arrangement it can be deduced from (827) where, if the force P vanishes, the motion of the body [p. 355] has been shown to progress along the axis uniformly. On account of which just as the force P becomes zero, then also the motion sought can be reduced to the motion in the plane. Clearly this can only happen, if a backwards motion of such a size can be imparted to the plane normal to the axis AP, as that found for the progressive motion along AP (827). Scholium 3.
837. Yet meanwhile this fact does merit the most attention, that we have found with so much ease in the examples proposed, the equations between the orthogonal coordinates for the projections of the curve on the plane normal to the axis AP, and thus for the curves described by the body if the progressive motion should disappear. For in the first part of this chapter, in which we considered the motion generated in the plane by a centripetal force, the labour involved much more work and the comparison of the arcs of circles, in order that we could arrive at the ordinary equations for the curves described. Therefore with the greater generality, which most often results in increased difficulty in finding the solutions to questions, it is not an impediment in this case, but rather the solutions can be easier found in particular cases that had been felt more difficult to solve.EULER’S MECHANICA VOL. 1. Chapter Five (part e). Translated and annotated by Ian Bruce. page 508 Corollary 3.
838. In the case of this proposition, the plane of the elements Mm and mμ (Fig. 76) is easily found. For since ddy : ddz = y : z then PO = 0 and AO = x , and O falls on P. [p. 356] Again the tangent of the angle PQR = ydz − zdy b = z since ydz − zdy = bdx . Therefore zdx the cotangent of the angle POR varies as QM. And then the tangent of the angle of inclination of the plane Mmμ to the plane APQ is equal to ( z 2 +b 2 ) . y Scholium 4.
839. Since the case in which the force P vanishes can be reduced to motion in a plane, so also the case in which either Q or R disappears can be reduced to motion in a plane. For if the axis is taken on a line normal to AP in the plane APQ, then the force Q lies in a direction parallel to the axis and the remaining forces P and R now act as Q and R before. But if the axis is taken normal to AP and to the plane APQ, then the force R takes the place of the force P parallel to the axis. Clearly as the coordinates x, y and z are able to be commuted with respect to each other, so also similarly the order of the forces P, Q and R can be declared. PROPOSITION 102. PROBLEM.
840. If a body is acted on at individual points M (Fig. 79) by two forces, the first in the direction MA, and the other, the direction of which is MQ along the normal sent from M to the plane APQ, then it is required to determine the motion of the body M and its orbit. SOLUTION. With MP drawn, which is the normal to AP, the force MA is resolved into forces acting along Mf parallel to AP and the other along MP. Truly the force along MP can be resolved into forces acting along [p. 357] MQ and Mg. Therefore with AP = x, PQ = y and QM = z put as before, and the force pulling along MA equal to V and the force along MQ equal to W and with the resolution of the forces put in place and with a comparison made with Prop. 100 (823) it is found that andEULER’S MECHANICA VOL. 1. Chapter Five (part e). Translated and annotated by Ian Bruce. And from the equations of the same proposition there are produced : From these it is found : and hence on integration : or This value substituted in place of v gives these equations : From which the described curve can be determined. Q.E.I. Corollary 1.
841. The time in which the body comes as far as M, is equal to Moreover since then that time is given by : which becomes known from the quadrature of the given curve in the plane APQ. Corollary 2.
842. If we put y = px and z = qx, the following equations are produced [p. 358] page 509EULER’S MECHANICA VOL. 1. Chapter Five (part e). Translated and annotated by Ian Bruce. page 510 Which are helpful in making the curve known. Example.
843. If the force V is proportional to the distance MA, and the force W is proportional to the perpendicular MQ, on putting and thus Whereby from the equations of the preceding corollary, these equations become : The integral of this equation is x 2 = C − 2a 3 fdx 2 , x 4 dp 2 [Euler has not indicated how he solved this equation. However, if we set dx then dp = dx and ddp = ddx − dx dt = − dx dt ; on substituting these into the above t = dp 2 2 t t t 6 3 2 t 3 equation, we find that : x dx = 4a ft dx − 2a fxtdt . This can be rearranged to give the integrals : ∫ ∫x ∫x 2 3 tdt ; and which gives the above result on integration, and xdx = 4a3 f t dx 5 − 2a f 4 treating t and x as independent variables,] from which there arises [on setting C = c2, ] and the integral of this : for the equation of the projected curve described in the plane APQ, which therefore is an ellipse, the centre of which has been put at A. From the value of dp found, again from which with the values of dp and ddp substituted into the other equation, there arises : On putting q = e ∫ rdx , there is produced :EULER’S MECHANICA VOL. 1. Chapter Five (part e). On setting r = x2 page 511 Translated and annotated by Ian Bruce. u , there comes into being : ( c2 − x2 ) Which on putting t = ( c2 − x2 ) or x = c , the equation is transformed into : x ( 1+tt ) of which we will show the integral later. In order to know the plane, in which the elements Mm and mμ (Fig. 76) are situated, on account of : [p. 359] with h in place of f + g it is found that : and the tangent of the angle of the plane Mmμ with the plane APQ is equal to xxdp , is given by : a a Then the time, which the body takes to arrive at M, since it is equal to ∫ Thus this time is proportional to the angle, of which the sine is the abscissa x, with the total sine taken equal to c, or of which the sine is cx , if the total sine is taken as equal to
844. From which it is evident that the motion of the body projected on the plane APQ makes equal angles around A in equal times, and the time of one revolution is proportional to f . Corollary 3.
845. For the projection of the curve described in the plane APQ makes a circle, if n = 0 and − a 2af = cc ; the centre of which is at A and the radius = c. Therefore we have y = ( c 2 − x 2 ) . And z is given from this equation: gddz gxdz − hzdx = cc − xx , which extends to dx the case of the preceding example equally widely, even if only this particular case is considered.EULER’S MECHANICA VOL. 1. Chapter Five (part e). page 512 Translated and annotated by Ian Bruce. Corollary 4. gddz
846. In order to discover the value of z from the equation dx = gxdz − hzdx , I put cc − xx z = e ∫ rdx . From which being done, this differential equation of the first degree is produced: Put r = u and there is produced : ( c − x2 ) 2 [p. 360] With gh or f +g = m 2 this equation arises : g in which the indeterminates can now be separated from each other in turn. Corollary 5.
847. Truly, and Therefore with the constant added and with the given numbers, we have then: and hence [These integrals are easily shown to be true, on differentiation.] ∫
848. Since truly lz = rdx and r = Corollary 6. u , then we have ( c − x2 ) 2EULER’S MECHANICA VOL. 1. Chapter Five (part e). On putting ∫ Translated and annotated by Ian Bruce. dx = s , then there is ( c2 − x2 ) and 2 ms or on putting e −1 = t ,so that ds = dt2mt−1 , then we have and hence it becomes : Corollary 7. [p. 361]
849. Now from the value of z we have and And again putting y = ( c 2 − x 2 ) there will be Corollary 8.
850. Hence from these there is found : and tang. PQR = page 513EULER’S MECHANICA VOL. 1. Chapter Five (part e). page 514 Translated and annotated by Ian Bruce. In a similar manner, from these the angle of inclination between the plane is found in which the body is moving and the plane APQ. [p. 362] Scholium.
851. The application to finding the value of z and the inclination of the orbit is very difficult on account of the imaginary quantities occurring in turn. On this account we are unwilling to tarry longer with the intersections of the curve described by the body with the plane APQ, to be determined. Moreover since this is the great question of the moment in astronomy in finding the motion of the nodes, the following proposition has been designated to this business, in which, we investigate when the body by its own motion shall arrive in the plane APQ. For the body completes part of its motion above the plane, and part below; and whether it is above or below, the body is always drawn from the other towards this plane by a force W in the direct ratio of the distance from this plane. Truly the point in the plane APQ, through which the body passes from the upper part to the lower part, is called the descending node, and the point in which it reverts to the upper, is called the ascending node. PROPOSITION 103. PROBLEM.
852. If the body is always attracted partially to some fixed point A (Fig. 79) in the ratio of the distances from the same, and partially normally to the plane APQ in the ratio of the given distances from this plane as well; it is required to determine the nodes or the points in which the body arrives at this plane, and besides also the points, [p. 363] at which the body is at a maximum distance from the plane. SOLUTION. With the three coordinates x, y and z remaining as before, and the force, by which the body is drawn towards A , is equal to ( x2 + y2 + z2 ) ; and the force, by which the body is f drawn towards the plane APQ , is equal to gz and on putting f +g = m 2 it is evident that g the body is incident in the plane APQ, when z = 0. But z = 0, as often as (847). This comes about whenever u =∝ , by (846). Therefore since, by (845) with centre A the circle BQC is described (Fig. 80) with radius AB = c and the body is moving along the region BQC and in place of this equation, this equivalent equation is taken [multiplying by c changes the angles into arcs]:EULER’S MECHANICA VOL. 1. Chapter Five (part e). Translated and annotated by Ian Bruce. page 515 The integral of which is in which expresses the arc, of which the tangent [SC] is cu , and m expresses the arc BQ, the sine of which is AP = x. [Thus, the integrals are taken to represent the arcs of which the angles arctan and arcsin of add up to one right angle.] Let C be the arc BQC or the quadrant of the circle, and the tangent CS = cu , to which there m corresponds the arc CR. Hence the above equation is changed into this: 1 CQR = CQ and CR = m.CQ. From this, it is evident that u =∝ , [p. 364] if the angle m CAR is right, or equal to 3, or 5, or 7, etc. right angles. And thus as a consequence, the motion of the body has an infinite value of u, if the arc CR is made successively into the degrees of the following sequence of degrees, with the values [corresponding to up or down in the diagram in turn] : 90, − 90, − 270, − 450 , − 630 etc. Moreover then the arc CQ contains the degrees : 90 , − 90 , − 270 , − 450 , − 630 , etc. and thus the arc BQ : m m m m m 90 − 90 , 90 + 90 , 90 + 270 , 90 + 450 , 90 + 630 , etc. m m m m m Whereby if the body were perchance at a node, then it would arrive at the other nodes successively by the absolute angular motion about A, at angles of 180 , 360 , 540 , 720 , etc. degrees. m m m m Therefore two adjacent nodes are separated by an angle of 180 degrees. And the m ascending or descending nodes are distant from the following node of the same kind by an angle of 360 degrees, i. e, by the angle m
853. g ( g+ f ) degrees. Thus demonstrating the first part of the proposition. It follows that the body is at the greatest distance from the plane APQ, when dz = 0; and that comes about, wheneverEULER’S MECHANICA VOL. 1. Chapter Five (part e). Translated and annotated by Ian Bruce. page 516 (848). Moreover now it is the case that u = 0. [from (846)]. Therefore with the previous construction kept in place that avoids putting u = 0, then u is zero as often as CR = 0 or – 180 or – 360, etc. degrees. Then moreover the arc CQ contains the degrees : 0, − 180 , − 360 , − 540 etc. m m m and thus the arc BQ has the degrees : 90, 90 + 180 , 90 + 360 , 90 + 540 , etc. m m m Therefore the maximum distance of the body from the plane APQ is equally distant from the nearest node on each sides. Thus demonstating the second part of the proposition. [p. 365]

Corollary 1.

852. Therefore the nodes finally return to the same point, if m is a rational number or f +g f +g is a square number. But if g is not a square number, then the body is never g incident at the same point in the plane APQ , in which before it was finally incident. Corollary 2.
853. If g is a positive number or the body is always attracted to the plane by a positive force, then m is a number greater than one. Therefore the interval between the two nodes is less than an angle of 180 degrees. Whereby the nodes are progressing backwards, and the distance of a node from the position of the previous node is distant by the angle 180( m −1 ) degrees. m Corollary 3.
854. If the body is always repelled from the plane APQ, then g becomes a negative number m = g− f . Whereby if g > f , then m is a real number, but less than unity ; g therefore then the nodes as a consequence are progressing more quickly by the amount which f is less distant than g. And if f = g, then the body escapes from the node and never returns to the surface APQ . But if f > g, then the body always departs form this plane. [p. 366] Corollary 4.
855. Since the body is at a maximum distance from the plane APQ, when it is true that the maximum distance itself can be had, if this is substituted in the value of z found (847). Moreover this maximum distance is equal to 2k , which is therefore the same −1 everywhere.EULER’S MECHANICA VOL. 1. Chapter Five (part e). Translated and annotated by Ian Bruce. page 517 Corollary 5.
856. If this circle BQC (Fig. 80) were the projection of the orbit described by the body in the plane APQ, then the tangent of the angle of inclination of the plane of the orbit to the 2 plane APQ is equal to 2m k in the places where the body is at a maximum distance from c −1 2 this plane, or equal to 2mc k with k put in place of k , since the constant k must avoid −1 having imaginary values. [An imaginary value of k has to be assumed initially.] Corollary 6.
857. With the same hypothesis kept for the places, where the body is incident in the plane APQ, the tangent of the angle of incidence is equal to dz ( c 2 − k 2 ) with the equality cdx Thus this tangent becomes equal to and with the value of z put in place from (847) this is equal to 2mkc −1 , [p. 367] or with a suitable value put in place of k this tangent becomes equal to 2mk . c

Corollary 7.

858. Therefore the tangent of the angle of inclination of the orbit described by a body to the plane APQ, if the body is at a maximum distance from the plane APQ, is to the same tangent, if the body is incident in this plane, as m to 1. Therefore the ratio can be made : as the separation of the two nodes is to 180 degrees, thus the inclination of the orbit, if the body is at a maximum distance from the nodes, to the inclination of the orbit described by the body, if the body changes from one node to the other.

Scholium.

859. Indeed this proposition is seen to have some use in astronomy, because there that force, by which the body is attracted to the fixed point A, we can make proportional to the distance, truly for celestial bodies that have in place a force that is inversely proportional to the square of the distance. Yet this is extremely useful, if the orbits of bodies do not depart much from circles; for there is no interest in orbits that depart from circles, in whatever way the centripetal force depends on the distances. On account of which, when the orbits of planets do not depart greatly with circular orbits, this proposition [p. 368] is able to be adapted successfully to this motion. And it then becomes most useful, as the curve corresponding to f can be found, which can itself be to the distance from the centre as the force of gravity to the centripetal force. The other force, which draws the body to the given plane, is effective as some proportional of the distance; yet if that does not happen, then the letter g must be considered as a variable, from which truly the approximate motions of the nodes can be gathered from all the values of g, by choosing as it were the mean value. In lunar motion the motion of the nodes merits special attention, clearly which will happen to be nearly our preceding determination. Moreover it is observed from the opposition of the preceding node that the nodes differ by nearly 43’, thus in order that From which the lunar force is known to be always pulling from behind towards the plane of the ecliptic.