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Chapter 6a

The Curvilinear Motion Of A Free Point In A Resistive Medium

by Euler
March 20, 2025 31 minutes  • 6523 words

PROPOSITION 104. THEOREM.

  1. If a body is moving in a medium with resistance acted on by some number of absolute forces, the resistive force does not disturb the action of the other absolute forces in any way, except that the tangential force arising from that is diminished.

DEMONSTRATION.

From the last chapter it has been explained well enough that all absolute forces can be resolved into two forces, the tangential and the normal, if the motion is to be in the same plane. But if the body does not move in the same plane, then three equivalent forces can be assigned in place of any number of forces acting, of which one is the tangential and two are normal. But the force that the resistance exerts on the body, is always put to agree with the direction of the body (117). On account of which the resistive force has to be referred to the tangential force that it diminishes, [p. 370] since the motion of the body is slowed down, and indeed it does not in short affect the normal forces. Therefore it is evident that the resistance has no effect on the absolute forces, except in as much as the tangential force arising from these is diminished by the resistance. Q.E.D.

Corollary 1.

  1. Therefore the whole effect of the resistance is consistent with changing the speed of the body and leaves the direction unchanged, except in as much as the action of the normal force varies with the variation of the speed.

Corollary 2.

  1. Therefore except besides by the aid of [normal] absolute forces, it is not possible for the body to move along a curve, but always to progress along a straight line, while it loses its motion.

Scholium 1.

  1. Therefore in this chapter, in which we treat curvilinear motion, it is necessary that we consider absolute forces likewise and these are of such a kind that they can be resolved to give a normal force, and which are different from what we discussed in Chapter IV. On this account the first force we consider pulls towards a point at an infinite or the direction of this force is always kept parallel to itself. From there we progress to centripetal forces and to other forces set out in whatever manner. And hence also we submit to our analysis motions not contained in the same plane, [p. 371] of such kinds as arise from motion in a resistive medium.

Corollary 3.

  1. If the tangential force is T, and either the one normal is N , or the two normals are N and M , and the force of resistance is R, then the rules containing the effect of these forces that we gave in the previous chapter are also to be applied here, except that we put T – R in place of T for these.

Scholium 2.

  1. As the force of the resistance is made to depend on the speed of the body, it is necessary to be explained by a rule for the resistance, and the explanation has been widely set out in Chapter IV. Truly in this chapter a wide range of resistances are uncovered to be treated, which did not find a place in that previous chapter. Besides, this treatment thus has been subdivided, in order that at first we can determine the curve described and the motion of the body, from the given absolute forces and the resistance. Then if the curve and the absolute force is given, from these we deduce the resistance. Following this, in the third place, from the given curve and the resistance the absolute force in a given direction is to be investigated. Then from the given curve, with both the speed of the body at individual points and the resistive force, the absolute force and its direction can be found. But in the first part of this chapter [p. 372] the division of motion in coplanar and non-coplanar parts is agreed upon.

PROPOSITION 105. PROBLEM.

  1. If a body is moving in a medium with some resistance and is acted upon by some absolute forces, yet thus, so that the motion is completed in the same plane, to define the rules that the body observes in its motion.

SOLUTION.

The body describes the curve AMB on account of the forces acting (Fig.81); let the speed of this at the point M correspond to the height v and the element of the curve Mm = ds. Again the normal force is put equal to N, and hence the direction of this force is along the normal MN to the curve, truly the tangential force arising from the same absolute forces is equal to T, the direction of which is MT, the tangent of the curve at M. And the force of the resistance at M is equal to R. With these put in place, the motion of the body can be defined from the normal force N and from the force along the tangent T – R (864). Now let the radius of osculation at M be equal to r and soEULER’S MECHANICA VOL. 1. Chapter Six (part a). Translated and annotated by Ian Bruce. N = 2rv and dv = ( T − R )ds page 547 (552). From these two equations if v is eliminated, the equation is produced expressing the nature of the curve, and likewise the speed of the body at individual points can be observed from the equation N = 2rv . Q.E.I. [p. 373] Corollary 1. 867. Therefore there arises v = Nr . Hence there is found : dv = Ndr +2 rdN . Which value, if 2 in place of v, and placed in the equation dv = ( T − R )ds in place of dv and in R is put Nr 2 the equation for the curve described by the body is produced. Corollary 2. 868. If in R v should have a single dimension, since that comes about if the resistance is proportional to the square of the speed, then the equation dv = ( T − R )ds is able to be separated and each force can be determined from that. And this equation solved with v = Nr gives a simpler equation for the curve described. 2

Scholium.

  1. Besides this case, in which v has a single dimension in R, many others are given, for which the equation dv = ( T − R )ds can be integrated; but there is no need to explain these, as v is eliminated in any case. Truly we have noted the case for this particular idea, since in fact it pertains to the resistance of fluids and which therefore we will examine with more care before the others.

PROPOSITION 106. PROBLEM.

  1. The force acts normally everywhere to the given line in the position AP (Fig.82) and the body moves in a medium with some resistance [p. 374] ; it is required to determine the curve AMB in which the body moves, and the motion of the body.

SOLUTION.

Let the force which acts on the body at M be equal to P, the direction of which is therefore MP. The speed of the body at M corresponds to the height v and the force of the resistance there is equal to R. The element Mm is taken, and with mp drawn then AP = x, PM = y and Mm = ds. Then we have Pp = Mr = dx and mr = dy. Again the tangent MT is drawn, and on that the perpendicular PT from P is drawn. With these done, the force P is resolved Pdy .PT = Pdx and the tangential P .PT = into the normal PPM components. ds PM ds Moreover since this tangential force retards the motion of the body, the negative value of this must be taken. Therefore with the radius of osculation at M put equal to r then we = 2rv and dv = − Pdy − Rds (866). From which equations the curve itself as have Pdx ds well as the motion of the curve can be found. Q.E.I.

Corollary 1.

ds . On this 871. With dx placed constant, the radius of osculation is given by r = − dxddy 3 account we have the following equation : P=− 2vddy . ds 2 Which value of P substituted in the other equation gives the equation : dv = 2vdyddy − Rds or, since dyddy = dsdds , this equation : dv = 2vdds − Rds . ds ds 2 [The result dyddy = dsdds can be shown in a straight–forward manner on differentiating dx , as ddx = 0.] dy = ds cos θ and dx = ds sinθ , and noting that dθ = dds ds dy Whereby the equation is put in place, for whatever the force P might have been, only the direction of this is along MP. Corollary 2. [p. 375] 872. If the law of the resistance is some ratio of the multiple of the speed, and the m exponent of the resistance is some quantity of the variable q, thus in order that R = v m , q then we have this equation : m dv = 2vdds − v mds , ds q of which the integral is : [This can be shown in an inductive manner by differentiation of the result.] Corollary 3. dds − ds , the integral of 873. If according to the same hypothesis, m = 1, then dv = 2ds v q which is : If in addition the resistance is constant or q = c, thenEULER’S MECHANICA VOL. 1. Chapter Six (part a). page 549 Translated and annotated by Ian Bruce. Therefore in this case the time in which the arc AM is completed is equal to :

PROPOSITION 107. PROBLEM.

  1. If both the force and the resisting medium is constant and the direction of the force is along the direction MP (Fig.82) normal to the given line AP, and the medium has a resistance in the ratio of the square of the speed, to determine the motion of the projected body.

SOLUTION.

The force pulling the body towards AP is always equal to g, and the exponent of the resistance is equal to c, and the rest of the denominated terms remain as before. Therefore we have : P = g and R = vc . Hence there arises the following equations : with dx taken as constant, and s From solving these equations together we now find: e c v = ads2 (873). 2 dx Whereby, when it becomes : the equation is produced on eliminating v : which contains the nature of the curve described. Since dsdds = dyddy, it can also become : On putting dx = pds, then dpds dds = − p and dy = ds ( 1 − pp ) . With these put in place there arises this equation :EULER’S MECHANICA VOL. 1. Chapter Six (part a). page 550 Translated and annotated by Ian Bruce. which suffices to construct the described curve. Indeed the equation of this integral is : Truly with dx restored in place of p it is found that: ds Which is a differential equation of the first order, and which cannot be reduced to any simpler form. Q.E.I. Corollary 1. 875. A differential equation of the third order can be immediately found for the described gds 2 curve. For on account of v = − 2ddy then we have dv = − gdy + gds 2 d 3 y , from the values 2ddy 2 there is produced the of which substituted into the equation [p. 377] dv = − gdy − vds c third order equation dsddy = cd 3 y .

Corollary 2.

  1. Let the sine of the angle, that the curve at A makes with the axis AP be equal to μ, and the cosine of this is equal to ( 1 − μ 2 ) = v and the height corresponding to the speed at A is equal to b. Therefore by taking s = 0 and ds : dx = 1 : v there becomes v = b; s therefore there is had from e c v = ads2 this equation v 2b = a , thus recognised to be a 2 dx constant a.

Corollary 3.

  1. Again in the final equation of the curve by taking s = 0 and ds : dx = 1 : v and 1+ μ ds : dy = 1 : μ the constant is found : C = gc + μb + ν 2b l v . Whereby for the curve described there arises this equation : Indeed in order that the speed can be found, this equation is brought into use:

Corollary 4.

  1. If D is the maximum point, then there dx = ds and dy = 0. Therefore: From which the equation is found [p. 378] Truly the height corresponding to the speed that the body has at D, is equal to Corollary 5.

  2. If the curve at B is put to have the same inclination to the axis AP, that is has at A, then the ratio is ds : dx : dy = 1 : ν : − μ . Hence therefore this equation comes about : from which there is produced: Corollary 6.

  3. From the construction of the curve it is also easy to deduce from the equation : dsddy = cd 3 y . For on putting dy = pdx , there arises dpdx ( 1 + pp ) = cddp . Again there dp becomes dx = q , since ddx = 0 dpdq ddp = q , hence this equation is produced: dp ( 1 + pp ) = cdq and q = 1c dp ( 1 + pp ) . ∫ Therefore on taking the abscissa : then we haveEULER’S MECHANICA VOL. 1. Chapter Six (part a). Translated and annotated by Ian Bruce. page 552 And to these there corresponds : Hence it is understood that a negative quantity has to be taken for g.

Corollary 7.

  1. If the body at A is projected along the direction of AP (Fig.83) and the force acts downwards, [p. 379] then the whole curve AM described by the body falls below AP and y or PM becomes negative ; and since μ = 0 and ν = 1 , this equation is had for the curve AM (877) : Corollary 8.

  2. Therefore given the tangent at some point M for the inclination to the line AP or the ratio between ds, dy, and dx, it is possible from this equation to find the length of the arc AM.

Scholium.

  1. Experiments have shown that air resists bodies as the square of the speed. Therefore since the force of gravity is uniform and air at not very great heights keeps almost the same density, the case of bodies projected in air is first referred to this proposition. We therefore determine first the curve that a ball describes projected by a gun or cannon, or by any other means. Commonly a parabola is taken for this curve, which clearly is the trajectory in a vacuum, and the air is supposed to be so fine a fluid that it does not merit to be included in the calculation. Indeed, at any rate, a large body projected with a small speed is insensitive to the resistance of the air. But the trajectory departs the greatest distance from a parabola, if a very small body is projected by a large force. Moreover in these cases, even if here the trajectory has been correctly assigned, it is very painful, as the equation is so involved that hardly any practical use can be extracted from it. [p. 380] Newton in the Phil. Princ. did not touch this problem1 and no one after him attempted it, then John Keil [Sav. Prof. Astronomy Oxford at the time] challenged Johan Bernoulli to this problem, although he had not been able to find a solution himself. Moreover a solution was given, not only by Johan Bernoulli2 in Acta Eruditorum Lips. 1719 in the May issue, but also at nearly the same time Jacob Hermann3 had inserted a solution in his Phoronomiae. But the following problem, in which the resistance is put proportional to the speeds, had been solved, first by Newton4 in Phil. Princ. and then by Huygens in Tractatu de causa gravitatis.

[1. Indeed it was presented by Newton, in Philosophiae naturalis, Book II, Section 2, Concerning the motion of a body, in which the resistance is proportional to the square of the speed of the body, but the problem in which the orbit was to be found, had been omitted.

Ioh. Bernoulli, Responsio ad nonneminis provocationem eiusque solutio quaestionis ipsi ad eodem proposititae de invenienda linea curva, quam describit proiectile in medio resistente, Acta erud. 1719, p. 216; Opera omnia, Tom. 2, Lausannae et Genevae 1742, p. 393. See also Acta erud. 1721, p. 228; Opera omnia, Tom. 2, p. 513; and the dissertation in the 1742 edition Problema ballisticum; Opera omnia, Tom. 4, p. 354. 3.

Iac. Hermann, Phoronomia seu de viribus et motibus corporum solidorum et fluidorum, Amstelodami 1716. 4. I. Newton, Philosophiae naturalis principia mathematica, Londini 1687, Lib. II Sectio 1; Concerning the motion of bodies, in which there is resistance in the ratio of the speed. 5. Chr. Huygens, Traite de la lumiere, avec un discours sur la cause de la pesanteur, Leyden 1691. References supplied by Paul Stackel.]

PROPOSITION 108. PROBLEM

  1. If the resistance of the medium were as the speed of the body and the direction of the force along MP (Fig.82), and in addition the force is uniform as is the resistance, it is required to determine the curve that the body describes and the speed at individual points. SOLUTION. With the force put uniform and equal to g as in the preceding proposition, with the exponent of the medium resistance equal to c, with the height corresponding to the speed at M equal to v, AP = x, PM = y and the arc AM = s , the normal force is as before : gds 2 + 2vddy = 0, with dx taken as constant. For the tangential force, on account of the resistance truly in this case equal to v , this equation c is had : dv = − gdy − ds v . [p. 381] c With these equations solved together as in (872), where we put q = c and m = 12 , it is found that : and hence :

Thus for the curve described there comes about this equation : the integral of which is and again by integration : Thus likewise v is agreed upon from the equation :

Corollary 1.

  1. This differential equation of the third order can immediately be produced for the gds 2 curve described, if from the equation v = − 2ddy , and with the differential of this : the values are substituted in the equation dv = − gdy − ds v . For there arises : c

Corollary 2.

  1. If the body at A is projected with a speed corresponding to the height b and the sine of the angle that the tangent at A makes with AP is equal to μ, and the cosine of this is ( 1 − μ 2 ) = ν , then at the point A, b = a 2 , or a = 4v 2b , from which the indefinite 4v constant a is determined.

Corollary 3.

  1. Again the equation [p. 382] translated to the point A givesEULER’S MECHANICA VOL. 1. Chapter Six (part a). Translated and annotated by Ian Bruce. page 555 as a = 4v 2b. Hence the indefinite quantity is found :

Corollary 4.

  1. Therefore this differential equation is found for the curve described : From which is deduced : Whereby, since there is finally found :

Corollary 5.

  1. Moreover the equation of the integral for the curve sought is : From which the construction of the curve by the logarithm is easily completed. Corollary 6.
  2. Also the time, in which the arc AM is completed, is easily defined. For since ds = 2 dx c , the time for the arc AM is equal to : v 2v bc − x

Corollary 7.

  1. It is also apparent from the equation for the curve to have that asymptote AM . For since x is unable to be greater than 2ν bc , if we take AE = 2ν bc , the applied line [i. e. the y co-ordinate] at E = − ∝ and likewise this is taken as the asymptote of the curve AMDB. This is also understood from the time, [p. 383] since it will be made infinite, before the body arrives at the perpendicular drawn through E.

Corollary 8.

  1. The maximum point D is found by making dy = 0. Moreover, then it is found that Then, since at D, ds = dx and 2ν bc − x = speed at the point is equal to 2vgc b , the height corresponding to the μ b+g c v 2 g 2bc and the speed at D is equal to ( μ b + g c )2 Corollary 9.
  2. Truly the applied line CD of the distance of the point D from the axis AP is equal to and the time, in which the body arrives at D from A, is equal to Scholium.
  3. Therefore it is possible to reduce the case in which the body is projected obliquely from A, to the case, in which it is projected normally to the force from D. For with the speed known, with which the body is projected from A, and from the direction of projection, the point D can be found at which the tangent is parallel to AC, and the speed of the body at D. Whereby to improve our understanding of this motion, it is expedient to consider the motion as beginning at D, that we have added as the following last proposition. [p. 384]EULER’S MECHANICA VOL. 1. Chapter Six (part a). Translated and annotated by Ian Bruce. page 557 PROPOSITION 109. PROBLEM.
  4. If the body is everywhere attracted downwards equally, and it is projected along the horizontal direction at A (Fig.83) with a given velocity in a uniform medium that offers resistance in the simple ratio of the speed, then to determine the curve AM that the body describes and to find the motion of the body on this curve. SOLUTION. Since this proposition is a special case of the preceding, all the derivations remain as before. Moreover y or the applied line PM becomes negative, since the curve AM falls below AP, and we have μ = 0 and ν = 1. Therefore with the force g acting, with c the exponent of the resistance, with b the height corresponding to the speed at A and AP = x and AM = s, this differential equation is found for the curve AM : (888) and this integral: (888). And the time, in which the arc AM is traversed, is equal to (890). Which equations determine the curve AM, and also the motion on the curve. Q.E.I.EULER’S MECHANICA VOL. 1. Chapter Six (part a). page 558 Translated and annotated by Ian Bruce. Corollary 1.
  5. If l 2 bc is converted into a series, it gives : 2 bc − x On account of which, we have [p. 385] and the time, in which the arc AM is traversed, is equal to Corollary 2.
  6. In a vacuum therefore, when c is made infinitely great, the equation becomes gx 2 y = 4b ; in which case therefore the curve AM goes into a parabola, the parameter of which is 4gb , and the time in which the arc AM is completed is equal to x and b gx 2 v = b + 4b = b + gy , as it is clear to recollect from Prop. 72 (564). Corollary 3.
  7. By taking AE = 2 bc the vertical line EF is the asymptote of the curve AM. Whereby the perpendicular MQ is sent from M to EF, and we have MQ = PE = 2 bc − x and EQ = y . On putting MQ = z, we have and the time in which the arc AM is traversed is equal to 2 c l 2 zbc .

Corollary 4.

  1. Therefore the point E through which the asymptote EF passes is as far as the body that has been sent from the point A can reach, if there is no aid to the force g acting, before all the motion has been made available. And likewise in a similar way it is apparent that the time to pass through AM is equal to the time to pass along AP with the force g vanishing. [p. 386] Hence it is understood from this that g is not present in the expression for the time.EULER’S MECHANICA VOL. 1. Chapter Six (part a). Translated and annotated by Ian Bruce. page 559 Corollary 5.
  2. The tangent MT drawn from M is given by Through M draw MR making an angle with EF, the tangent of which is equal to b . g c With which accomplished, QR = gz c and thus RT = 2 gc. b Corollary 6.
  3. Therefore if MR is considered as an oblique applied line [y-axis] of the curve AM to the axis EF, then on account of the constant sub tangent RT, the curve AM is logarithmic with the oblique-angled sub tangent equal to 2gc, and the tangent of the angle of inclination of the applied line MR to the asymptote EF = angle MRQ = z/RQ = b . [i. e. the tangent of the g c b , which is constant, and hence MR can be considered as an g c oblique axis; as this axis slides along EF as the point M varies, the length RT remains constant. A curve that has, no doubt, other interesting properties.] Scholium.
  4. The trajectory in this resisting medium and with the force acting under this hypothesis can be constructed not only with the aid of logarithms, but has been examined by Johan Bernoulli in Act. Erud. Lips. 1719 to be an oblique-angled logarithmic curve, the solution of which agrees uncommonly well with our solution. (See note 2 above.)

PROPOSITION 110. PROBLEM.

  1. With the absolute uniform force put acting along the vertical direction MP (Fig.82) and the medium, that is also put as uniform, [p. 387] with the resistance in some ratio of the multiple of the speeds, to determine the curve AM described by the projected body. SOLUTION. With AP = x, PM = y, Mn = ds, the speed at M = v , and the force equal to g as before, with the exponent of the resisting medium equal to c, with the medium resisting in the ratio of the 2m th power of the speeds, and given by m R = v m and P = g (870). Whereby these equations are to c be had : from which the curve AM as well as the motion of the body on the curve can be gds 2 determined. Moreover, the equation v = − 2ddy gives Hence with v eliminated, this equation is arrived at for the nature of the curve : For the construction of the curve, put dy = pdx and there arises : From which on substituting there becomes dp Again put dx = q and thenEULER’S MECHANICA VOL. 1. Chapter Six (part a). Translated and annotated by Ian Bruce. page 561 Hence there arises and on integrating [p. 388] From which equation q is given in terms of p, from which it is found on taking the abscissa x = dp pdp ∫ q , there corresponds the applied line y = ∫ q . And with the height corresponding to the height and the time in which the arc AM is completed, i. e. ∫ dsv , is equal to Q.E.I.

Corollary 1.

  1. It is evident that whenever 2m is either a positive or negative odd number, the value of q can be shown algebraically in terms of p. Corollary 2.
  2. If the resistance is constant or m = 0 and the body initially at A is projected with a speed b along the horizontal AP, the applied line PM or y likewise taken as negative, has these equations Hence this equation is produced : Corollary 3.
  3. Moreover this case is easier to handle if m = 0 in the differential equation of the 2ddy 2 third order, for it produces gd 3 y = − ds or on substitution in terms of p and q this equation is made : the integral of which isEULER’S MECHANICA VOL. 1. Chapter Six (part a). Translated and annotated by Ian Bruce. page 562 or from the noted homogeneity [p. 389] Hence there arises Which again on integration gives : ∫ And hence there is found y = pdx and this is completed on integration to give : Therefore it is apparent that this curve is algebraic only if g is 1 or 2. Scholium.
  4. And equally general to our solution is the solution given in the Acta Erud. Lipt. in May 1719 by Johan Bernoulli1 on trajectories in resistive media, where the general construction of these curves was given. But before we leave the constant force hypothesis, we must solve the inverse problems, in which we determine the resistance that is effective, in order that the body describes a given curve acted on by a constant downwards force according to the hypothesis. For this matter has been treated several times, first by Newton2 in the Phil. Princ. then again by Johan Bernoulli3 in the Act. Lips, A, 1713: where the sharpest of men have noted many interesting things. [p. 390] [Thus, from the observed curve, one can determine whether or not the resistance has a particular form.] [1 Previous note 2;
  5. I. Newton. Philosophae naturalis principia mathematica, Londini 1687, Lib. II Sectiones 1, 2, 3; 3 Joh. Bernoulli, De motu corporum gravium, pendulorum et proiectium, Acta erud. 1713, p. 77; Opera omnia, Tom. 1, p. 514.]

PROPOSITION 111. PROBLEM.

  1. With an absolute uniform force put in place acting downwards, to determine the resistance at the individual places M (Fig.84) which can be put in place, in order that the body describes the given curve BAM. SOLUTION. As before, put AP = QM = x, PM =AQ = y, and the element of ace AM= ds. Then the speed at M is equal to v , and the resistance at M is equal to R. Therefore with these compared with Prop. 106 (870) we have P = g and y must be taken as negative; and there is produced with dx taken as constant (871) : dv = gdy − Rds (870). From this equation, moreover, we have Therefore with these equations solved there is produced : Whereby, since the curve is given, from this equation the value of R is finally found, and thus the resistance becomes known. Q.E.I. Corollary 1.
  2. Therefore the force of the resistance at M to the force acting g is in the ratio ds , we ds .d 3 y to 2ddy 2 . Or, with the radius of curvature at M set equal to r, since r = dxddy 3 have: Whereby the ratio becomes: Corollary 2.
  3. The height giving rise to the speed at M, surely v, can be determined from this curve; gd 2 s d s = rdx it for it is [p. 391] v = 2ddy . Or by introducing the radius of curvature r from ddy ds grdx becomes : v = 2ds . 2

Corollary 3.

  1. If the resistance is put in the ratio of the square of the speed, the unknown exponent of the resistance taken as q, then R = medium q = gdsd 2 y , and the exponent of the resistance of the 2 ddy 2 dsddy . And in a like manner the exponents can be found for the other d3y hypotheses of the mediums resistances. Corollary 4.
  2. By introducing the radius of osculation r to the determination of q, we have It consequently becomes : Therefore with a known radius of curvature r then R as well as q can be found from a first order differential equation.EULER’S MECHANICA VOL. 1. Chapter Six (part a). Translated and annotated by Ian Bruce. page 565 Corollary 5.
  3. If the curve AM is a parabola, the axis of which is the vertical AC, since in that case d 3 y = 0 , the resistance produced R = 0. From which it is understood that a parabola can be described in a vacuum by a constant force acting downwards, as it agrees with this. Corollary 6.
  4. If the curve is a parabola of some higher order, thus such that is given by a n −1 y = x n , then and And again on account of the constant dx : Whereby the resistance becomes And with the resistance put as in proportion to the square of the speed , then the exponent of the resistance becomes :

Scholium.

  1. I will not add other examples of curvature here which can be put in place of AM; but I am resolved to examine these of merit more carefully in the following propositions. Moreover I intend to consider in particular the circle and the hyperbola, that these men cited have treated especially.

PROPOSITION 112. PROBLEM.

  1. With the absolute force g constant and always acting downwards, to determine the resistance which must be put in place in order that the body is free to move along the periphery of the circle BAMD (Fig.84). [p. 393] SOLUTION. Let the radius of the circle AC = a, then it is the circle x 2 = 2ay − y 2 and the radius of osculation at M = a = MC. Whereby since dr = 0 we have the ratio g : R = 2ds : 3dy . Indeed ds : dy = a : x = AC : QM . On this account, 3 g .QM g : R = 2 AC : 3QM or R = 2 AC . Indeed the height producing the speed at M is equal to gadx g .QC = 2 . But if the resistance is put as the square of 2 ds the speed, then the exponent of the resistance Q.E.I. Corollary 1.
  2. While the body ascends through the arc BA, on account of QM then being negative, the resistance along BA is also negative; or the motion of the body along BA is 3 g .QM accelerated by the medium by the tangential force 2 AC . Corollary 2.
  3. Truly the resistance at the point A is zero, since QM vanishes ; therefore the body at A moves as if in a vacuum. Indeed at the point D the resistance is in the three on two ratio to the force g. Indeed at B the force is acting up by the same amount. Corollary 3. [p. 394]
  4. Therefore since at B and D direction of the resistive force agrees with the direction of the force g, the body at B is urged up by a force equal to 12 g ; at A it is pulled downwards by the force g and at D up again by the force 12 g .EULER’S MECHANICA VOL. 1. Chapter Six (part a). page 567 Translated and annotated by Ian Bruce. Corollary 4. QM
  5. Since AC expresses the sine of the angle ACM with the radius put equal to 1, then the resistance at M = 32 g sin .ACM or the resistance is everywhere as the sine of the angle ACM, by which the body declines from A.

Corollary 5.

  1. Again a .QC is the tangent of the arc MD. Whereby with the resistance put in AM proportion to the square of the speed, the exponent of the resistance q is equal to the third part of the cotangent of the arc AM.

Corollary 6.

  1. Since the height corresponding to the speed at M is equal to g .QC , then the speed at 2 g . AC ; and B and D the speed is equal to zero. Therefore since the body since the 2 g body at B is actually forced up again by the force 2 , it is not a wonder that the body at B A= begins to move up.

Corollary 7.

  1. Therefore the body, so in the quadrant BA as in the quadrant AD, at places equally separated from the point A, has equal speeds. [p. 395] Truly the body is not able to reach places below the horizontal BD on account of QC being negative, and in which case the speed would be imaginary.

Scholium.

  1. But the direction in which the body progresses when it arrives at D can be easily gathered from the reported events. For when the speed at D is equal to 0 and the body at g D is urged up by the force 2 , it is evident that the body must again begin to move up. Moreover, it ascends again through the arc DMA in the same manner, in which in the first g place it ascends through BA, since at D as also at B, it is urged up by the force 2 .

Truly this wonderful thing happens in this motion, since the body is at rest at B and is urged up again and nevertheless moves on the curve, even if it seems to be unaided by any force, since it is able to change direction as the body accepts to go up at B. But to this argument I respond to this argument that the force at B is not in a perfectly vertical direction, but strays an infinitely small amount from the true vertical, since that is sufficient for the production of the oblique motion. For the direction of the resistance force at B or rather the accelerating force is present on the element of the periphery of the circle at B, which is not a perfectly vertical straight line, but inclined at an infinitely small angle towards BC.

Moreover these results of ours agree exceedingly well with these, which the celebrated Bernoulli gave in the Act. Lips. A. 1713 and which are present in the later editions of Newton’s Princ. Phil. [Book II, Sect. 2, Prop. X.]

In the first edition an error has crept into the solution [p. 396], from which it was concluded that the ratio of g to R was established to be equal to the ratio AC to QM. But a word of caution from Nicolas Bernoulli corrected this lapse in the following editions.

Motion in General
The Curvilinear Motion Of A Free Point In A Resistive Medium

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