THE CURVILINEAR MOTION OF A FREE POINT IN A RESISTIVE MEDIUM

by Euler

March 20, 2025 28 minutes • 5764 words

PROPOSITION 126. PROBLEM.

If the curve AM is given (Fig.93), on which the body is moving, and the angular motion about the centre of force C, to find both the centripetal force attracting the body towards C as well as the resistance at individual places.

SOLUTION.

As before by placing CM = y, CT = p, Mm = ds, with the speed at M corresponding to the height v, the centripetal force equal to P and the resistive force equal to R, the periphery ELl of a circle is taken described with centre C and with radius EC = 1 on which the body [p. 455] is carried around C with the same angular motion C as the body is carried around the curve AM. Hence the element Ll is completed in the same time as the element Mm. Now let the speed along the element Ll correspond to the height u; and u is given, since the angular motion is given. And it is found that :

Truly we have : Ll : mr = 1 : y or Ll = mr , again it is the case that y mr : Mm = p : y and likewise mr = and consequently, Ll = On account of this : p .Mm . y2 p .Mm , y

Now from this ratio found for v, it is the case that (1005) : (1007). And if the resistance is itself put proportional to the square of the speeds, and the exponent of the resistance is equal to q, then the exponent becomes : Q.E.I.

Corollary 1.

If the centripetal force P is proportional to dp , and since that happens when the p 3 dy body moves in a vacuum, then u varies inversely as y 4 . Whereby the angular speed then varies inversely as the square of the distance of the body from the centre. Moreover, with y 4u put constant, it is evident that the resistance R vanishes from the other equation.

Corollary 2.

If the body approaches the centre C, thus as y decreases we have: Whereby the resistance becomes :

From which it is understood, if y 4u is the power of this y, [p. 456] of which the exponent is a positive number, then the resistance also becomes positive. But if the exponent of this power of y is negative, then the resistance also is negative. Corollary 3.

If the angular motion is made uniform or u is constant, then du = 0 and likewise

Corollary 4.

Let the angular speed be as the power of the exponent n of the distance y or u= y 2n , then the resistance f 2 n−1 the centripetal force and for the resistance of the medium in the square ratio of the speed, the exponent of the resistance is given by : Example.
Let the curve AM again be the hyperbolic spiral expressed by the equation : and the angular speed is as y n or as before u = y 2n . Moreover since f 2 n−1 then [p. 457] and the centripetal force If the resistance is put proportional to the speed itself, then the exponent of the resistance is equal to [Recall that the exponent of the resistance goes with the corresponding height or the square of the speed] : But on the contrary if the resistance is put proportional to the square of the speed and the exponent of the resistance is q, then we have Which clearly agrees with these which have been treated in the above example (1061).EULER’S MECHANICA VOL. 1. Chapter Six (part d). page 684 Translated and annotated by Ian Bruce. PROPOSITION 127. PROBLEM.
If the resistance is given by some power of the speed and likewise the exponent of the resistance, also in addition the angular motion of the body about the centre C is given (Fig.93), from these to find the curve that the body describes, and the centripetal force pulling towards the centre C. SOLUTION. [p.433] On putting CM = y, CT = p, Mm = ds, with the height corresponding to the speed at m M equal to v, the centripetal force equal to P , the resistance R is given by R = v m , q where q is given in terms of y. Then the angular motion is considered as before [p. 458] as the motion performed by a point on the periphery of the circle ELl, the radius of which CE = 1. Now with the speed put in place in which Ll is describes, corresponding to the height u, we can elucidate as in the preceding proposition, Moreover this last equation, in which u and q are given quantities, expresses the nature of the given curve, for which we therefore have : or Truly from the known nature of the described curve, or from the equation between p and y, the centripetal force becomes known at once, clearly it is Q.E.I. Corollary 1.
If the angular speed is to become constant, which cannot happen in a vacuum unless the body moves in a circle, then du = 0 and this produces the equation for the curve sought : Hence with q given in terms of y, this equation between p and y is integrable, from which the curve can be constructed.EULER’S MECHANICA VOL. 1. Chapter Six (part d). Translated and annotated by Ian Bruce. page 685 Corollary 2.
If u is given in terms of y as u = y 2n , then [p. 459] f 2 n−1 Which equation between y and p is also integrable, and likewise is sufficient for the curve to be constructed. Corollary 3.
If the angular speed is given through the arc EL or du through the element of this (1063), thus in order that there becomes : dy ( y2 − p2 ) = g k ydu . With this value substituted the equation becomes : uk p and From the equation the value of p substituted into the equation du = u k pdy g k y ( y2 − p2 ) determines u in terms of y. Hence also the equation between p and y is obtained. Corollary 4.
If the resistance should be in the simple ratio of the speed or m = 12 , the equation immediately gives u in terms of y. [p. 460] Which value substituted in the equation du = u k pdy g k y ( y2 − p2 ) gives the equation between y and p.EULER’S MECHANICA VOL. 1. Chapter Six (part d). page 686 Translated and annotated by Ian Bruce. Example 1.
The medium offers resistance in the square ratio of the distances and the exponent y of the medium is q = α . Truly the angular motion is taken as uniform or u = b. Then m = 1 and 4 ( y 2 − p 2 ) = αy (1070). Hence there is produced : Whereby the curve described is a logarithmic spiral, in which the sine of the angle that the radius makes with the tangent is centripetal force is equal to ( 16 −α 2 ) and the cosine is equal to α4 . Truly the 4 32by . 16 −α 2 But if the medium is made uniform or q = c, then the equation becomes : Example 2.
The medium offers resistance in the simple ratio of the speed and that is uniform, also the angular speed is made constant; then m = 12 , q = c , u = b. With these substituted we have this equation for the curve described : Which curve is also a logarithmic spiral, in which the sine of the angle of intersection is 4 bc 1 , the cosine is equal to , and the tangent is equal to 4 ( 1+16bc ) ( 1+16bc ) this case the centripetal force is equal to bc . Truly in y( 1+16bc ) . 8c Because in these formulas uniformity of dimensions is not observed, the reason for is because we have put the radius of the circle EC equal to 1. Therefore by this means uniformity with unity is restored. Scholium. [p. 461]
There are many central forces that we do not consider in this chapter, since also in this case as in a vacuum, hardly any can equations can be deduced in order to determine the motion. If a certain centre of force attracts in the simple ratio of the distances, any number greater than one have no difficulty in attracting in the simple ratio of the distances, as we have shown above (702). And this agreement applies equally with a resisting medium in place of the vacuum. On account of which, when now we consider one centre of force attracting in the simple ratio of the distances, there is no need for us toEULER’S MECHANICA VOL. 1. Chapter Six (part d). Translated and annotated by Ian Bruce. page 687 go through the motions of including several more of the same kind. Therefore we progress to the case of the widest applicability, in which all the motions produced in the same plane are dealt with. Obviously we consider two absolute forces the directions of which are mutually perpendicular to each other, and the individual forces making up each kind are parallel to each other. For it is agreed that any forces existing in the same plane can be resolved into two forces of this kind. Besides in this treatment not only do we embrace all the cases of absolute forces, but also it is allowed to observe certain special cases concerning centripetal forces, which were barely evident in the preceding. For here we can immediately reduce the description of the curve to an equation between the orthogonal coordinates, because in this case the distance between the centre and the perpendicular to the tangent has been found for any point on the curve. [p. 462] PROPOSITION 128. PROBLEM.
If the body at M (Fig.94) is acted on by two forces, of which the one has the direction MP normal toe the given line AC, and the other truly has the direction MQ parallel to AC itself or normal to BC, to determine the curve AM which the body describes in any medium with resistance due to action of these forces. SOLUTION. Calling CP = MQ = x, PM = CQ = y, the element Mm = ds; and with drawing mp and mq there is produced Pp = – dx and Qq = dy and ds = ( dx 2 + dy 2 ) . Let the force, by which the body is drawn along MP be equal to P and the force, by which the body is drawn along MQ, be equal to Q, truly the resistance is equal to R and the speed at M corresponds to the height v. Now the forces P and Q are resolved into normal and tangential forces with the help of the perpendiculars sent from P and Q to the tangent Tt ; hence the normal force arising from P is equal to : and the tangential force is equal to : Truly by resolution from the force Q the tangential force is equal to : and the normal force is equal to :EULER’S MECHANICA VOL. 1. Chapter Six (part d). Translated and annotated by Ian Bruce. page 688 Therefore the total normal force is equal to : and the total tangential force emerging is equal to : which force is diminished by the resistance R, and from which total the accelerating force produces the motion. Hence from these with the radius of osculation at M put equal to r , it follows that (866). [p. 463] Hence on eliminating v from these equations the equation arises expressing the nature of the described curve. Q.E.I. Corollary 1.
If the element ds of the curve is placed constant, then the radius of osculation Therefore with this value substituted there is Corollary 2.
From the equations solved together it is found that From which, if the relation between P and Q is given, the equation is immediately obtained, for which the curve is given in terms of v alone. Corollary 3.
If the body is always attracted by some force towards the centre C, then P : Q = y : x. Therefore this equation is then obtained : Which on putting y = px results in this equation :EULER’S MECHANICA VOL. 1. Chapter Six (part d). Translated and annotated by Ian Bruce. page 689 Corollary 4.
In a vacuum, in which R vanishes and the body is attracted towards the centre, this becomes : Besides truly there is : [p. 464] From which this equation is obtained : or by taking Q in place of P this equation : Corollary 5.
If the centripetal force attracting towards C is equal to Therefore in a vacuum this equation is obtained for the described curve : With B put in place for − Af n and with dx constant, this equation arises from differentiation : Truly with dp made constant, this equation is produced : On making x = 1q , this becomes : Of these equations although the integration is not apparent, yet the integral is which was found in the previous chapter.EULER’S MECHANICA VOL. 1. Chapter Six (part d). page 690 [This has been solved by Paul Stackel in the O.O. : With MC = ( xx + yy ) = z the equation is found (601, 685) Translated and annotated by Ian Bruce. Let then we have or On substituting thus in order that : there is made : With which equations solved there is obtained : ] Corollary 6.
Nevertheless although this equation : is of second order differentials, yet it is more convenient than the differential equation of the first order in determining the curves [p. 465], which the projected body describes attracted either in the simple ratio of the distances or inversely as the square of the distances. For in the simple ratio there is n = 1 and 2 Bq3ddq + dp 2 = 0. On making dp = wdq; on dwdq account of dp being constant we have ddq = − w , henceEULER’S MECHANICA VOL. 1. Chapter Six (part d). Translated and annotated by Ian Bruce. page 691 Hence there is obtained : or (with the meaning of the symbols changed) since B is a negative quantity. Corollary 7.
If n = –2 or the body is attracted in the reciprocal ratio of the distances squared, it is in the vacuum the described curve with B taken negative, as is required. Hence on integrating it becomes : and on integrating again : Of which each curve is the section of a cone; that one indeed an ellipse, yet all of these are embraced. Scholium 1.
In the preceding chapter, in which we presented the motion of bodies in a vacuum, we also determined curves which a body described with a centripetal force either proportional to the distances or inversely proportional to the square of the distances; and it was convenient to find these curves from these laws in the given corollaries. [p. 466] Indeed the methods are maximally different; for there we arrived at algebraic equations from the comparison of circular arcs, here truly by integration an algebraic equation between the coordinates is spontaneously given. Truly this method, although it is more convenient in the two cases already given, yet in other cases it is troubled with difficulties. For in other hypotheses of centripetal force this method indeed is unable to give a differential equation for the curve described, that yet can always be done with equal ease by the other direct method to give the curve described. Yet this should be ascribed to defective analysis rather than to the method, when we may know the integral of the second order differentials [This is referred to as a differential of the differential equation in the text] of the equation that isEULER’S MECHANICA VOL. 1. Chapter Six (part d). Translated and annotated by Ian Bruce. page 692 from the method used in the preceding chapter, truly we may not be able to elicit this integral from the second order equation. Scholium 2.
The problems of reciprocal natures that can be proposed around these forces, this has now been solved in Cor. 2, in which from a given curve, with the resistance of the medium and the speed at individual points given also, the forces acting along MP and MQ are sought which produce this motion. As if the curve AMB is a circle with centre at C and having radius AC = a and the resistance is equal to vc and speed is constant, truly v = b, then it is x 2 + y 2 = a 2 and [p. 467] In a similar manner, since there are five things that are arrived at in the consideration : truly the two forces P and Q, thirdly the resistance R, in the fourth place the speed at the individual places or v, and in the fifth place the nature of the curve described or the equation between x and y, always three of these can be taken as given and the two remaining are to be found from these. On this account there are ten problems that can be formed from the number of combinations, in which three are taken from five. But so that we are not detained to any extent in working these out, and from which not much can be deduced that is useful, we treat a single problem, in which the medium offers resistance in the square ratio of the speed and the curve described is sought from a given centripetal force. PROPOSITION 129. PROBLEM.
If a body moves in a medium that resists as the square ratio of the speed, and if the force P is to the force Q as MP to MQ (Fig.94) or, which is the same, if the body is drawn to the centre C by some force, to determine the curve AMB described by the body. SOLUTION. As before on placing CP = x, PM = y, Mm = ds and y = px, let the speed at M correspond to the height v and the exponent of the resistance is q; that is, R = qv . And since there is the ratio P : Q = y : x , then this gives (1080). [p. 468] Which equation divided by vxdp becomes this equation : the integral of which isEULER’S MECHANICA VOL. 1. Chapter Six (part d). page 693 Translated and annotated by Ian Bruce. Truly this equation : ∫ dsq e dx and integrated gives (1079) multiplied by ds 2 in which the value of v found substituted gives : This equation is differentiated with dx placed constant, and it becomes : Which is the equation for the curve sought. Q.E.I. Corollary 1.
This equation for the curve found does not differ from the equation found in a ds Qdx vacuum(1081), except that this has e ∫ q Qdx , when there − Adx 4 2 is equal to ∫ 2 x dp ∫ only. Corollary 2.
If the element dp is assumed for the constant, then this equation is produced : [p. 469] In which if we put x = 1z , there arises : Corollary 3.
If the centripetal force attracting towards C is equal to : Whereby we have this equation for the curve given :EULER’S MECHANICA VOL. 1. Chapter Six (part d). Translated and annotated by Ian Bruce. page 694 Which with the logarithms taken and differentiated , gives Truly this is : Corollary 4.
With everything the same in place : giving Put z = e ∫ udp and this equation is produced : Scholium.
I doubt that this second order differential equation can in any case be reduced to a differential equation of the first order [p. 470]; or yet that which we put in place above, where we considered the customary centripetal forces (1020). Therefore in a medium with resistance the method of working does not yet seem to be useful, as much as it brought in a vacuum, even for the cases in which n is either 1 or –2. On this account, since in this matter hardly anything more can be expected, I leave the motion made in a plane with a resisting medium, and I proceed to consider non–coplanar motion, connecting the body with the absolute forces and the force of resistance acting on it. Where in this business it is alright for a little easy understanding to lead to apparent knowledge, I am content to expound the rules, from which we are able to arrive at an equation for any proposed problem.EULER’S MECHANICA VOL. 1. Chapter Six (part d). Translated and annotated by Ian Bruce. page 695 PROPOSITION 130. PROBLEM.
In a medium with some resistance acting a body is acted on by three forces, of which one is along the tangent, and the remaining two are normal to the direction of the body, and in two planes with the normals between each normal in turn; to determine the motion of the body and the curve which it describes. SOLUTION. From the element Mm (Fig. 95) which the body describes, from the ends M and m the perpendiculars MQ and mq are sent, and from the points Q and q the perpendiculars QP and qp are sent to the fixed axis AP in the fixed plane APQ [p. 471]. Then put AP = x, PQ = y and QM = z, with the height corresponding to the speed at M equal to v. Now let the tangential force be equal to T. The first of the normals , the direction of which lies in the plane Mq, is equal to N and the other, the direction of which is normal to the plane Mq, is equal to M. The force of the resistance is truly equal to V. Moreover since the force of the resistance V does not affect the normal forces, but only has a little effect on the tangential force, the effect of the normal forces N and M remains unchanged, but in effect the tangential force has to be defined by putting T – V in place of T. Whereby, when we determine the effect of these forces now above (809), the same equations prevail given there and here, if in this way T – V is put in place of T. On this account these equations are produced for the resisting medium : and (809). From which with v eliminated there are two equations involving the three coordinates x, y, z, which express the nature of the curve sought. Moreover it is assumed that the element dx is assumed to be constant in these equations. Q.E.I. [p. 472] Corollary 1.
The two latter equations can be solved together in order that v is eliminated and give this equation : Which are equally valid for some medium and for a vacuum (810).EULER’S MECHANICA VOL. 1. Chapter Six (part d). Translated and annotated by Ian Bruce. page 696 Corollary 2.
It is evident from this equation, if N or M vanishes, the motion of the body is of such a kind. For on putting N = 0 it becomes : Truly dz is the tangent of the angle, by which the element Mm is inclined to Qq. ( dx 2 + dy 2 ) Whereby this angle is constant; since QM has a given ratio to the projection BQ of the described curve in the plane APQ. Corollary 3.
If M = 0, then ddy = 0 and thus the projection BQ is a straight line. Therefore the total curve described by the body is put in a plane normal to the plane APQ and cutting the line BQ. Corollary 4.
From the equation (1094) there arises Whereby, since it becomes then Corollary 5.
Whereby, if then the body also moves in a plane, since then ddz = 0 and dz = αdx. For the projection of the described curve is a straight line in the plane normal to the plane [p. 473] APQ with the normal AP.EULER’S MECHANICA VOL. 1. Chapter Six (part d). Translated and annotated by Ian Bruce. page 697 Corollary 6.
Moreover the plane, in which the two elements Mm and mμ have been placed (Fig. 96) which the body describes, is determined in a like manner to that in the vacuum, since the determination of this plane only depends on the coordinates x, y and z. Truly if this plane is SMR, and it cuts the plane APQ in the line OR, then And the tangent of the angle that the plane RMS constitutes with the plane MQ APQ, or QV , is equal to (812). Corollary 7.
Therefore the tangent of the angle, which the plane RMS makes with the plane ddz . APQ , is equal to the secant of the angle POR taken by ddy Corollary 8.
Therefore in the case, in which the force N vanishes, since it is given by : dx or POR = RQS. Therefore it then follows that QV falls the tangent of the angle POR = dy on QS. Truly the tangent of the angle, that RMS makes with RQS , is equal to Whereby this angle is constant, on account of (1095). Truly it is found that :EULER’S MECHANICA VOL. 1. Chapter Six (part d). Translated and annotated by Ian Bruce. page 698 Corollary 9. [p. 474]
As in corollary 1, the ratio is given between ddy and ddz for the normal forces M and N, and if the proportionals are substituted in their place then the position of the plane RMS is determined by a first order differential equation. But all these apply equally well for a vacuum and for a resisting medium. Whereby also this result agrees with these that were presented above in proposition 98 in a straightforward manner. PROPOSITION 131. PROBLEM.
If a body M (Fig.97) in some resisting medium is drawn by three forces, of which the direction of one is Mf parallel to the AP, the direction of another Mg is parallel to the applied line PQ placed in the plane APQ and the direction of the third is MQ sent normally to the plane APQ from M, to find the motion of the body and the line that it describes. SOLUTIO. As before by putting AP = x. PQ = y and QM = z and with the speed at M corresponding to the height v, let the force drawing along Mf be equal to P, the force drawing along Mg be equal to Q, and the force drawing along MQ equal to R, and the force of the resistance at M is equal to V. These three forces can be resolved into three others, the directions of which agree with these in the previous proposition, [p. 475] and the tangential force produced (these are the normal forces) and (823). For here we use the same denominators, as with these in Proposition 99. Therefore with these values substituted in the preceding formulas we have the following three equations :EULER’S MECHANICA VOL. 1. Chapter Six (part d). Translated and annotated by Ian Bruce. page 699 and Which three equations with v eliminated give two equations in the coordinates x, y and z, which express the nature of the curve described. Moreover in these formulae the element dx has been assumed constant. Q.E.I. Corollary 1.
The two last equations agree perfectly with these that we found for the vacuum (823). Whereby the equations which follow from these have a place, as in the vacuum case so in the resistive case. Moreover the whole distinction that lies between the motion in the vacuum case and the motion with resistance, depends on the first equation. Corollary 2.
Moreover from the final two equations solved together there arises this ratio : On account of which in place of second normal equation, which makes up the greater part, this substitution can be made [p. 476] or which does not involve v. Corollary 3. 1105a. With the help of this ratio the determination of the plane RMS is found in terms of first order differentials as follows : the tangent of the angle POR is equal toEULER’S MECHANICA VOL. 1. Chapter Six (part d). Translated and annotated by Ian Bruce. page 700 and the tangent of the angle of inclination of the plane RMS to the fixed plane RQS is equal to (825). Corollary 4.
If two of the forces P, Q, R vanish, the motion by necessity becomes that in a plane. For if P and Q vanish, this makes ddy = 0 ; if P and R vanish, this makes (if Q and R vanish, this makes ddz = 0;) dzddy = dyddz or dz = αdy. Which all indicate that the motion lies in a plane. Corollary 5.
If P, Q and R are proportionals of x, y et z, then the body is always attracted to the point A, and thus the motion of this body becomes that in a plane. The formulae indicate the same; for set AO = 0. But because then it follows that xddy = xdzxddz and on integrating xdy − ydx = αxdz − αzdx. xdy − ydx − zdx Whereby αddz = ddy , thus the proposal is agreed upon. [p. 477]

Corollary 6.

If the force P vanishes, then the ratio becomes ddy : ddz = Q : R and With these values P, Q and R in place in the equation, by which dv is defined, on substitution there arises Where, if the resistance V is put equal to vc and on placing equal to ds, there becomes ( dx 2 + dy 2 + dz 2 ) , or Mm

Corollary 7.

If the force R vanishes, the ratio becomes and Therefore we have Where, if V = vc , there becomes In a like manner, if Q vanishes, there is produced : Scholium.
All the forces can be reduced to these three forces P, Q and R, in whatever way they are able to be devised. On account of which, whatever problem that is proposed, two equations can be elicited that contain the nature of the described curve[p. 478]. Truly of these one is a differential equation of the second degree, and the other a differential equation of the third degree, if indeed the value of v found from the equation is differentiated and with the differential is substituted in place of dv in the equationEULER’S MECHANICA VOL. 1. Chapter Six (part d). Translated and annotated by Ian Bruce. page 702 PROPOSITION 132. PROBLEM.
In a uniform medium, which resists in the simple ratio of the speeds, the body is always attracted normally to the line AP (Fig.97); to define the curve that the body describes projected in any manner. SOLUTION. As before these are put in place : AP = x, PQ = y, QM = z, the speed at M = v the exponent of the resistance is equal to c, the force by which the body at M is drawn along MP , = S. With these in place, the resistance is given by hence the ratio Q:R = y: z. On account of which we have ddy:ddz = y:z and The integral of this equation is ydz − zdy = αdx. Again also, as P = 0 we have this equation : (1108) on placing ds = ( dx 2 + dy 2 + dz 2 ) . The integral of this is With this value substituted, there is produced : Putting z = py; [p. 479] we have the following two equations, from which the nature of the curves described ought to be determined, Q.E.I.EULER’S MECHANICA VOL. 1. Chapter Six (part d). page 703 Translated and annotated by Ian Bruce. Corollary 1. ds ( b − x ), the element of time
Since it is the case that 2 cv = dx ∫ dsv = b− x . 2dx c Therefore the whole time, in which the body is moved horizontally along AP by the motion, is equal to 2 c l b −b x . Therefore the horizontal motion agrees with the motion in the same resisting medium along the line AP with no force acting, with the initial speed at A corresponding to the height bb : 4c.

Corollary 2.

Truly neither does the motion have this special amount of time only in place if the body is drawn along MP or if Q : R = y : z, but it always prevails if P = 0. For this follows from (1108), in which P is put equal to zero.

Corollary 3.

Therefore the progressive motion of the body along AP has been slowed down, and it cannot go beyond the limit, which is x = b. Moreover the time taken is infinitely great for the body to be able to reach this limit. [p. 480] Example.
We put the force by which the body is attracted to the line AP to be in proportion to the distances MP or Therefore in order to determine the curve we have these equations in that equation is put y = e ∫ udx and it becomes q Which equation becomes separable on putting u = b − x ; for it produces Therefore with q given and on also on account of u given in terms of x. Consequently also y in terms of x is known, from which the projection of the curve described in the plane APQ is obtained. Then from the given y in terms of x, also p is given in terms of x on account of dp = αdx 2 , and likewise z in terms of x. On account of which the whole y curve described by the body can be constructed.
If b vanishes, likewise also the progressive motion of the body along AP vanishes and on account of this the body moves in the plane through A normally to AP and is attracted to A in the ratio of the distances. Moreover the curve, which the body describes in this case, can also be constructed (1027) along with the others.