The Rectilinear Motion Of A Free Point Acted On By Absolute Forces

THE MOTION OF A POINT ON A GIVEN LINE IN A MEDIUM WITH RESISTANCE.

PROPOSITION 78. Problem.

691. In a medium with some kind of resistance and with some kind of forces acting to find the brachistochrone AM (Fig.78), upon which a descending body arrives the fastest from A to M. Solution.

Let A be the starting point of the motion, through which some straight line AP is drawn being taken for the axis, on which the abscissa AP = x it taken; to which there corresponds the applied line PM = y and the arc AM = s. Again let the speed of the body at M correspond to the height v and the resistance depending somehow on the speed equal to R. Now the body is acted on by some absolute forces act on the body, in place of these two forces can be substituted in the given directions ML and MN, of which the former is parallel to the axis AP, and indeed the latter normal to that axis. Moreover the force acting on the body following ML is equal to P and the force along MN = Q. From these forces, there arises the equation : dv = Pdx + Qdy − Rds. And the nature of the brachistochrone curve gives : 2v = 2vdxddy = Pdy −Qdx r ds ds 3 (673) with r denoting the radius of osculation of the cure at M towards the upper

• ds , since the other direction; hence from which on taking dx as constant we put dxddy 3 − ds . Hence from these two equations : direction must give r = dxddy 3EULER’S MECHANICA VOL. 2. Chapter 3f. Translated and annotated by Ian Bruce. 2vdxddy dv = Pdx + Qdy − Rds and = Pdy − Qdx ds 2 page 558 if v is eliminated, the equation is had for the brachistochrone curve sought ; clearly there is given v= Pdyds 2 −Qdxds 2 ; 2dxddy with the differential of this substituted in place of dv and in place of v itself in the resistance R, the equation is given for the curve sought. Q.E.I. Corollary 1. [p. 377]

692. The equation for the curve, if v is eliminated in the stated manner, becomes a differential of the third order. Whereby if the threefold integration is to be obtained, then also three constants can be added, in which to be effective, so that on x vanishing, likewise both y and s or v vanish and in addition the curve passes through the given point M . Corollary 2.
693. Therefore since the brachistochrone curve can always be shown, which has the starting point A and passes through a given point, an infinite number of brachistochrone curves can be drawn from the point A. Corollary 3.
694. With the equation for the brachistochrone curve AM known, likewise the speed of the descent on that curve can be found at individual points; and as much as v= Pdyds 2 −Qdxds 2 . 2 dxddy Corollary 4.
695. With the speed given, from that the time can be determined in which the body completes the arc AM ; clearly the time to pass along the arc AM is equal to : and that can now be found on account of the equation between x and y, even if it has to be shown by quadrature.

Corollary 5. [p. 378]

696. Therefore if the curve is to be found, which all the brachistochrones drawn from A must cross at right angles, then the construction of this line is obtained, if an amount of the same magnitude is cut off from all the brachistochrones. According to this, the arc of the isochrone is cut off from this infinitude of curves which, since all the curves are brachistochrones, terminate at right–angles to the trajectory. [Thus, the beginnings of sets of orthogonal curves is set out.] Example.
697. Let the resistance be proportional to the square of the resistance and let the exponent of the resistance be some variable q; then R = qv . Hence since dv = Pdx + Qdy − vds , q on integration it becomes : But since v= Pdyds 2 −Qdxds 2 , 2dxddy then the above equation becomes : in which equation v is no longer present. Yet meanwhile this equation becomes an equation of the third order, if the integration signs are removed by differentiation; besides the indeterminate values of P, Q, and q are reasons by which the equation is less able to be prepared for construction. Scholium.
698. Here it is clearly evident, concerning the brachistochrone curves that have been deduced from the two forces P and Q, since however many forces are acting on the body, all of these can be resolved in this manner into two, but only if the directions of all the forces lie in the same plane. [p. 379] On this account also in this proposition, the brachistochrones are held for any hypothesis of the centripetal forces acting, which moreover, since neither well–ordered nor constructible equations are forthcoming, we shall not pursued further. Therefore with these disposed of, in which the speed is prescribed by a certain rule, we progress to the following questions, in which curves are required, which sustain a given force on these from the body in motion.

PROPOSITION 79. Problem. 699. According to the hypothesis of uniform gravity and a uniform medium, which resists according to some ratio of the speeds, to determine the curve of constant pressure AM (Fig.79), which sustains the same pressing force everywhere by a body descending on it. Solution. On putting AP = x, PM = y, AM = s and for the speed at M to correspond to a height equal to v, let the force of the body pulling downwards following ML be equal to g, and the force of the resistance at M m = v m . Therefore, while the body progresses through k the element Mm, m dv = gdx − v mds . k [p. 380] We put the curve to be convex downwards, thus so that MR is the direction of the radius of curvature and the radius of osculation itself is MR = ds 3 ,with dx put constant. Hence the centrifugal dxddy force is in the direction of the normal MN and the size of this is equal to 2vdxddy . Now following the ds 3 same direction, the curve is pressed by a normal force gdy arising from the resolution of the force ML = g , which is equal to ds . Therefore the gdy 2vdxddy total force, by which the curve is pressed along MN , is equal to ds + ds 3 since it has to be constant, that is put equal to αg and there is obtained : αgds 2 = gdyds 2 + 2vdxddy and hence v= αgds 2 − gdyds 2 2 dxddy Let ds = pdx and dy = dx ( p 2 − 1) ; then we have : ddy = pdpdx ( p 2 −1) With these substituted, the equation becomes : . . ; whichEULER’S MECHANICA VOL. 2. Chapter 3f. page 561 Translated and annotated by Ian Bruce. Again, let dx = 2qdp; then or v = Pq on putting Hence this equation becomes m With which values substituted into the equation dv = gdx − v mds , there is produced : k Now it is the case that : dx = 2qdp et ds = pdx = 2pqdp. On account of which, the above equation becomes : which contains only the two variables p and q, since P is given in terms of p. In solving this equation, there is put from which this equation is obtained : [p. 381] dp which multiplied by e 2mg ∫ P and integrated goes into this : Hence therefore there is found u in terms of p, and from u found then q = 1 1 and Pu mEULER’S MECHANICA VOL. 2. Chapter 3f. Translated and annotated by Ian Bruce. page 562 Moreover since we have then and and Now, On account of which [Paul Stackel has corrected the following formulae in the O. O. ], Therefore since u can be found from p in this manner, the solution of the curve sought can be brought about. Q.E.I. Corollary 1. [p. 382] 700. The curve found hence has this property, that at any point M it is pressed towards MN by a constant force, which is in the same ratio to the force of gravity ML = g, as α is to 1. Corollary 2. 701. If a negative number is taken for α , the curve is equally pressed everywhere along MR, in the opposite direction. Therefore in this case the curve must be concave downwards, since the centrifugal force is in the opposite direction and is greater than the normal force, and the direction of this force has always been placed along MN.EULER’S MECHANICA VOL. 2. Chapter 3f. Translated and annotated by Ian Bruce. page 563 Corollary 3. 702. If α = 0 , then the curve is produces that sustains no pressing force from the body. Hence this is the curve that the motion of the body projected freely describes. Corollary 4. 703. If α = 1 or the total pressing force is equal to g, then the curve is convex downwards everywhere. For since the normal force alone is everywhere less than g except in the case that ds = dy, the centrifugal force acts together with that, and thus the radius of osculation falls on the opposite side of MN . Corollary 5. [p. 383] 704. If we put then Hence a special solution is obtained, since in this equation the indeterminates can be separated from each other, on putting P = 0. Moreover this becomes ; or Thus there is produced : αs = y . Hence making a straight line angle to the vertical AP, the sine of which is α on taking 1 for the whole sine. For in this case the centrifugal force vanishes and the normal force becomes equal to αg . Example. 705. Let α = 1 or the curve is sought, which can be pressed everywhere by a force gdp equal to g ; in which case the integration of P emerges simpler; for then it becomes : On this account then,EULER’S MECHANICA VOL. 2. Chapter 3f. Translated and annotated by Ian Bruce. page 564 Which equation, as long as 2m is a whole number, admits to being integrated. For on putting : then and Moreover with this in place : As if m = 12 or the resistance becomes proportional to the speeds, then [p. 384] or modified by the constant β , it is Moreover with this value put in place of r it becomes : Let β = 0; then and on this account, then andEULER’S MECHANICA VOL. 2. Chapter 3f. Translated and annotated by Ian Bruce. page 565 Scholium. 706. The integration of a similar formula to which u is equal, also follows if α = −1 , in which case there is produced a curve concave downwards, in which the centrifugal force is in the opposite direction and is greater than the normal force ; obviously the excess is equal to g. Now the same equation is produced, as for the case α = 1 , except that the sign of ( p 2 − 1) has to be changed. As pertains to the remaining questions touched on so far, in which other laws of the pressing force are proposed, these either can be deduced from exceedingly long calculations, or now have been handled. Indeed we have seen that the curves in which the total pressing force is twice as great as either the centrifugal force alone, or the normal force alone, are the brachistochrones, and curves in which another ratio acts, we have now handled these too, while we should investigate curves on which the motion is accelerated the least. [p. 385] Hence it follows that we proceed to finding curves, upon which many diverse descents and ascents are governed among themselves by given laws, which questions are the most difficult. For it is necessary in solving problems of this kind, that the speed of the body at individual points can be expressed by quantities from which the nature of the curve can be determined. Moreover, since they cannot be completed according to any hypothesis of the resistance, as we have noted above, such questions can only be proposed for special hypothesis of the resistance. Therefore this treatment is especially adapted to resistance in proportion to the squares of the speeds, since in that case the canonical equation, by which the speed is determined, leads to separation of the variables, and thus the speed can be determined. Then also to be considered is the resistance which is proportional to the biquadratic of the speeds, since for that hypothesis the speed can become known in a certain way. And finally for whatever law of the resistance, but only if the resistance becomes very small, questions of this kind emerge with a solution more easily. Now in these problems either the ratio of the speeds which have been acquired in different descents on the same curve is investigated, or the ratio of the times in which different descents or ascents have been completed. And in each kind, either from the ratio of the given times or speeds acquired from different descents, the curves are to be found.EULER’S MECHANICA VOL. 2. Chapter 3f. page 566 Translated and annotated by Ian Bruce. PROPOSITION 80. [p. 386] Problem. 707. In a uniform medium, which resists in the ratio of the squares of the speeds, and with an absolute force acting downwards, to compare the speeds between themselves at the point A (Fig.80), which are acquired in different descents of the body on the curve MA. Solution. Let the speed at A, acquired in one descent, correspond to the height b, and the speed at M correspond to the height v. Put AP = x, AM = s, the force acting at M, which is variable in some way, is equal to P and the exponent of the resistance is equal to k. With these in position, we have : dv = − Pdx + vds , k which equation integrated gives : ∫ −s with the integral e k Pdx thus taken, so that it vanishes on putting x = 0. Now let M be the starting point of the descent, where v = 0; this point is found from the equation ∫ −s b = e k Pdx . Now another descent is put in place to come from a neighbouring point m and the speed acquired at A corresponds to the height b + db. Hence we have ∫ −s b + db = e k Pdx −s k equal to the sum of all the terms e Pdx from A as far as m; now in the first equation, ∫ −s −s b = e k Pdx signifies the sum of all the terms e k Pdx only as far as from A to M . −s Hence the first sum is greater than this sum by the final element e k Pdx , with AM = s and Pp = dx. [p. 387] Hence we have −s db = e k Pdx. From which equation the relation is given between the arc traversed in the descent MA and the speed acquired at the lowest point A. Q.E.I.EULER’S MECHANICA VOL. 2. Chapter 3f. Translated and annotated by Ian Bruce. page 567 Corollary 1. 708. Hence from the given arc of the descent AM = s, the height corresponding to the speed acquired by the speed at A is given by : −s ∫ b = e k Pdx Or if the point M and the speed at A are considered as the only variable quantities, then there is this equation between them : −s db = e k Pdx. Corollary 2. 709. Hence from this equation, if some ratio was proposed between the arc of the descent and the speeds acquired at the point A, the equation is found for the curve AM satisfying the proposed conditions. Corollary 3. 710. If the medium is not uniform, but dissimilar in some way with the exponent of this arising set equal to q, in place of the equation found there is produced this equation : db = e ∫ q Pdx , − s which is of similar use. Corollary 4. [p. 388] 711. Because the value of e is greater than unity, clearly 2,7182818284, then e ∫ q or − ds −s e k is less than unity and on this account db < Pdx. In vacuo now it is the case that db = Pdx. Scholium 1. 712. In a similar manner in ascending it occurs that, when the body rises from A along the arc AM = s to a corresponding height b. Then indeed it becomes : s db = e k Pdx. or in a non-uniform medium : db = e ∫ q Pdx . Which formulas from the descents are found to be made serviceable on putting – s in place of + s ; with which put in place the descent is always changed into the ascent. Hence it is apparent, as for the descent always it is the case that db < Pdx, thus for the ds ascents it is always the case that db > Pdx, since e k or e ∫ q is greater than one. s Corollary 5. ds ∫ 713. Hence in a medium it will not be possible for an ascent of descent if b = Pdx or ∫ s k b = α Pdx ; for then we have e = α or s = const., in which equation no line is continued.EULER’S MECHANICA VOL. 2. Chapter 3f. Translated and annotated by Ian Bruce. page 568 Corollary 6. 714. Neither can a curve be found, for which we can write, either in the ascent or descent, b = Qdx , with Q denoting some function of s and x, unless Q can be ∫ Q prepared thus, so that P becomes equal to 1 on putting s and x = 0. Indeed it happens that e ∫ q P = Q and e ∫ q has the value 1 on putting s = 0. ± ds ± ds Scholium 2. [p. 389] 715. The reason for this is that we have put s to vanish with x vanishing; and on this account the equation db = e ∫ q Pdx ± ds thus must be integrated, so that b vanishes on putting x = 0. Moreover if b is given thus, so that db is expressed in terms of dx , the equation can be divided by dx. On which account it may not be possible to apply this condition to that equation, unless perhaps the equation freely has this pleasing property. But if such a value of b is given, in order that db = Rds or b = Rds with b vanishing on making s = 0, then the ∫ equation for the curve sought becomes : Rds = e ∫ q Pdx , ± ds ∫ which is always for a real curve, provided Rds has a positive value, and gives rise to ds > dx or e ∫ q P > R. ± ds Example 1. 716. Let the uniform force acting or P = g and the medium resists uniformly and the curve MA has this required property, that the body in individual descents, descends from A until it has acquires speeds which are in the square root ratio of the descended arcs traversed. . Hence b is as s or b = αs ; thus the equation becomes : [p. 390] −s s αds = ge k dx or αe k ds = gdx , the integral of which is : s αk (e k − 1) = gx with a constant added, in order that s vanishes on making x = 0. There is hence had s αk + gx e k = αk and ks = l (αk + gx) − lαk .EULER’S MECHANICA VOL. 2. Chapter 3f. page 569 Translated and annotated by Ian Bruce. Which differentiated gives : ds = gdx ; k αk + gx from which it is understood that the curve is tractrix for a thread of length k on a horizontal base at a distance down from the point A by the interval αgk . Hence the curve can be constructed in this manner : upon the horizontal base CE (Fig. 81) and with the string BC = k the tractrix BA is described; then the horizontal line DA is drawn at a distance DC = αgk from CE ; with which done the part of the curve BA sought is satisfied. Moreover we can place the vertical line BC and B the highest point of the tractrix; from which it is understood that by necessity α must be less than g. If indeed it should be greater, then it follows that CD

CB and thus the point A becomes imaginary. But if it is the case that α = g , then the point A falls on B and thus is the only satisfactory point. If it is the case that α = 0 , then the point A is at an infinite distance and the descending body loses all its speed. Therefore since it must be the case that α < g , then b < gs. Example 2.

717. As according to the above hypothesis of resistance and force acting, the curve AMF (Fig. 82) is sought, upon which all the ascents made from the point A thus have in common, that the whole arcs for each of the individual complete ascents are proportional to the squares of the speeds of the motion starting from A. [p. 391] Hence as before we have b = αs and s db = αds . But since for the ascents, db = ge k dx , −s then αe k ds = gdx and on integrating, −s αk (1 − e k ) = gx. Hence therefore it is found that −s αk − gx e k = αk and ds = gdx or ( αk - x) ds = k . k αk − gx g dx From which it is apparent that the satisfying curve is again a tractrix upon the horizontal base CE made by a line of length k, but considered downwards, the cusp of which is at B with BC becoming equal to k. Moreover taking CD = αgk ; and with DA drawn to the horizontal then A is the point, in which all the ascents must begin. HenceEULER’S MECHANICA VOL. 2. Chapter 3f. Translated and annotated by Ian Bruce. page 570 therefore it is also understood that α cannot be greater than g, as previously the point A becomes imaginary. But if α = g or b = gs, then A falls on B and the arc traversed for any ascent is equal to bg . Scholium 3.
718. Many examples of this kind I omit here, since they can be resolved so easily from the general formula ; nor do I report also on questions of this kind for other hypothesis of the resistance, for which the solution of these are able to be found, since such questions are now neither subject to discussion nor are considered curious enough that the solutions of these are required. Therefore we proceed to more worthy problems, in which the curves sought are tautochrones, upon which either all the ascents or descents are completed in equal times. PROPOSITION 81. [p. 392] Problem.
719. According to the hypothesis of a uniform force acting in the downwards direction and with a uniform medium, which resists in the ratio of the square of the speeds, to find the tautochrone curve AM (Fig.80), upon which all the descents as far as the point A are completed in equal times. Solution. Some descent is considered, in which the speed that the body has acquired at the lowest point A corresponds to the height b. Putting AP = x, AM = s, the height corresponding to the speed at M is equal to v and the force acting is equal to g and the exponent of the medium is equal to k, thus so that the resistance at M is to the force of gravity as kv to 1. With these in place, we have : dv = − gdx + vds , k which equation on integration gives : ∫ −s thus with the integral e k gdx taken thus, so that it vanishes on putting x or s = 0. Hence from this equation the start of the descent is found by putting ∫ −s v = 0 or e k gdx = b. Now the time, in which the arc MA is completed, hence is equal to from which the time of the whole descent is produced, if after the integration puttingEULER’S MECHANICA VOL. 2. Chapter 3f. Translated and annotated by Ian Bruce. page 571 −s k ∫ e gdx = b. For the sake of brevity we put [p. 393] −s k ∫ e gdx = t and eds = du , s 2k in order that the time of the whole descent thus becomes equal to ∫ (dub-t ) , on putting t = b after the integration. Now since this expression always gives the same value, it is du is a function of zero dimensions of b and t, so that on putting t necessary that ∫ (b-t ) = b, b vanishes from the formula. On this account du must be a function of dimensions one half only of t, since u cannot depend on b. Therefore it is necessary that du = αdt t with the quantity α arising not containing b. With this in place the time of one descent is equal to α dt , putting t = b after the integration. Or with the ratio of the ∫ (bt-tt ) diameter to the periphery put as 1 : π , then the time of one descent is equal to απ ; which value always remains, in whatever manner b or the start of the descent is changed. Hence the tautochrone curve sought is determined from this equation : du = αdt = dss , the integral of which is : t e 2k clearly with a constant added, which makes t disappear on putting s = 0. But since ∫ −s t = e k gdx , then We put α 2 g = a or α = a ; then we have g the integral of which is : [p. 394] which equations indeed, since the variables s and x have been separated from each other, are sufficient to construct the curve. But if the equation is desired to be free from the exponentials, since from the other equation we have : and with this value substituted in the other equation, it becomes axds + ksds = 2akdx. Q.E.I.EULER’S MECHANICA VOL. 2. Chapter 3f. page 572 Translated and annotated by Ian Bruce. Corollary 1.
720. Since α = a , then the time of one descent is equal to π g a . Moreover in vacuo g and with gravity equal to 1, the time of descent of a pendulum of length f is equal to : π 2f 2 (166). Whereby the length of an isochronous pendulum in vacuo is equal to 2a . g Corollary 2.
721. Therefore if 2ga = 3166 thousandth parts [or scruples] of a Rhenish foot (170), the descent is completed in half a second; this therefore arises, if we put a = 1583g scruples of a Rhenish foot. [If we use the elementary formula for the period of a simple pendulum of length f, T = 2π f , then the time for a quarter period, or one descent, is given by T4 = π2 g f f . g If this time T/4 is put as half a second, then 1 = π g ; if we approximate the Rhenish foot with the English or American foot, then g = 32000 scruples/sec2 approx. In this case the corresponding length f is given by 3240 scruples, while the Euler value for 2a = 3170, which is in approximate agreement. Note that the diameter of the generating circle of the cycloid is the radius of circle for the simple pendulum, and f is approx. 39 inches, or 1 metre. Is it a ‘happy accident’ that π 2 = 9.87 ≈ g ? ] Corollary 3.
722. The height corresponding to the speed at M is v and this is equal to and because of this, thenEULER’S MECHANICA VOL. 2. Chapter 3f. Translated and annotated by Ian Bruce. page 573 Corollary 4. [p. 395]
723. On putting v = 0 the whole of the descent arc is produced from this equation : Hence if the descent arc is put equal to f, then Whereby from the given arc of the descent f , then Corollary 5.
724. The equation for this curve s can be changed into a series of the exponential e 2 k , which is and hence it becomes : or Scholium 1.
725. Here it is appropriate to observe that the curve is expressed by an equation similar to that serving to express the ascent for the above brachistochrone; indeed there it was : (687), which equation only differs from this one of ours found for the tautochrone, as here it is 2a, and there it was a, and the exponent of the resistance of the brachistochrone is twice as larges as the exponent for the resistance of the tautochrone. Hence the brachistochrone curve can be adapted to produce tautochronism also, with the arc of the ascent attributed to the descent in the resisting medium, the exponent of which is twice as small. [p. 396]EULER’S MECHANICA VOL. 2. Chapter 3f. Translated and annotated by Ian Bruce. page 574 Corollary 6.
726. To find the continuation of the curve MA beyond A, s must be made negative, with which done there is obtained : or Which same equation emerges, if we make k negative. But with k made negative the descent is changed into ascent ; on account of which the curve MA continued beyond A serves to ascend, and above that point all the ascents are completed in the same times, clearly π ag . Corollary 7.
727. Hence the same curve continued BMANC (Fig. 83) is a tautochrone for the ascent as well as for the ascent. In fact on the curve BMA all the descents are completed in the same time and on the arc ANC all the ascents. Whereby all the half oscillations which begin on the arc BMA are isochronous with each other, and the time of one semi- oscillation is equal to 2π ga . Corollary 8. [p. 397]
728. If the resistance vanishes, in which case k becomes ∝ , this curve is changed into a cycloid, which is the tautochrone curve in vacuo. This itself indicates equation expressed through the series; for it becomes 2ax = 1s.2 or 4ax = s 2 , the equation for the 2 cycloid. Corollary 9.
729. Hence the curve BMANC exactly as the cycloid has vertical cusps at B and C, and so that these can be found, put dx = ds, and hence for the arc BMA and the height of this BD becomes equal to Now for the arc of the ascent ANC we have :EULER’S MECHANICA VOL. 2. Chapter 3f. Translated and annotated by Ian Bruce. page 575 Or through series we have : and Corollary 10.
730. From these it is evident that the cusp C of the ascending arc is higher than the cusp A of the descending arc. And the arc ANC of the cusps becomes infinite, if k = a; and if a > k, the cusp C becomes imaginary. Likewise it is apparent from the equation that BD as well as CE are diameters of the curve sought. Corollary 11. [p. 398]
731. If the body in the half–oscillation has a speed corresponding to the height b, then the arc of the descent is equal to : (723) or through the series is equal to : and the following arc of the ascent is equal to : Corollary 12.
732. Hence if the descent is made from the point B, thus in order that the arc of the descent is equal to AMB = 2kl a +k k , then now the arc of the following ascent is equal to 2kl a +k k .EULER’S MECHANICA VOL. 2. Chapter 3f. Translated and annotated by Ian Bruce. page 576 Corollary 13.
733. From the equation for this curve it appears that the curve has a horizontal tangent at the point A. Moreover since on putting ds constant, the radius of osculation at M is equal to : since (719) : then the equation becomes ; and and the radius of osculation at M is equal to : Hence on putting s = 0 the radius of osculation at the lowest point A = 2a. Now at B and C the radius of osculation vanishes. Corollary 14. [p. 399]
734. The radius of osculation is not a maximum at the lowest point A; but through the method of maxima the maximum is found in the arc of the ascension, and that is present at the point O : For at this point the radius of osculation is equal to 2ak . From which is inferred (k 2 −a 2 ) that, unless k > a, the curvature of the curve ANC is to be continually diminished and the point O does not exist anywhere.

Scholium 2.

735. Therefore in this problem we have found two curves, upon the first of which all the descents are to be completed, and indeed on the second all the ascents are completed in equal times. And while on the whole curve BAC the whole journeys or semi–oscillations are performed in equal times, if they start from a certain portion of the curve BA, this curve is suitable for isochronous oscillations to be made in fluids, but only if the returns also should be isochronous, concerning which there is no agreement. [As the initial descent-ascent and the following descent-ascent are made in opposite directions; if one is a tautochrone, then the other is not.]Moreover because in fluids besides resistance proportional to the square of the speeds another is observed above, that jointly with that resistance which for proportional or constant moments of time it is probably worth the effort to determine the tautochrone, since ; because that can be easily done from the preceding. For let the constant resistance be equal to h; then for the descent we have : Whereby if in the former operation in place of gdx only, everywhere gdx – hds is substituted, a satisfying tautochrone is obtained in a like manner ; clearly for the descent this curve is obtained : [p. 400] and now for the ascent this equation : which curve constitutes a continuation with the first curve; indeed the one goes into the other on putting s negative. It is to be observed that if k = ah , then the tautochrone g curve is the tractrix BAF (Fig. 82) having the horizontal asymptote CE, which is distant from the point A by the interval AE = 2ak . Moreover the length of the thread , 2 by which the tractrix is described, is equal to 2k. But because above with the curve of this kind never having a horizontal tangent the semi–oscillation can be completed and somewhere a point of equilibrium A exists, it is no wonder as now we have observed above, that in such a hypothesis of the resistance the body is able to stop at some place on the slope. Moreover in these cases, in which the curve is allowed to descend beyond A, there is no return and thus no oscillations can be performed, since the body is able to come to rest somewhere on the inclined plane, yet is unable to ascend above that ; for in some part of the curve AF the body is able to remain at rest.

Scholium 3.

736. The solution of the problem is not made much more difficult, if the force tending downwards is not constant, but P is made variable somehow, and the exponent of the resistance also is put as the variable q . Indeed there is had for the element of time in the descent :EULER’S MECHANICA VOL. 2. Chapter 3f. Translated and annotated by Ian Bruce. page 578 If now there is put as before (719) there also corresponds : Which equation differentiated again on putting ds constant and with this value substituted in place of dt there is given : for the curve having the isochronous descent. And the continuation of this curve beyond A serves as the ascent curve.

Scholium 4.

737. I first gave this tautochrone according to the hypothesis of resistance proportional to the square of the speeds in Book IV of the Commentaries [of the St. Petersburg Ac. of Sc.;E013 in these translations], in which I have used the same method here. Since then also the Cel. Joh. Bernoulli has signified to me by letter that he too has found the same tautochrone for the same hypothesis of the resistance [See the letter from Joh. Bernoulli 27 Dec. 1729 given to Leonhard Euler, that G. Enestrom published in the Biblioth. Mathem. 4, 1903, p.378. Note by P. St. in the O. O.]; the method of which can be seen in the Comm. Acad.Paris. A 1730 [Joh. Bernoulli, Méthode pour trouver les tautochrones, dans les milieaux résistans comme le quarré des vitesses, Mém. de l’acad. d. sc. de Paris 1730, p. 78; Opera Omnia, Book. 3, Lausannae et Genevae 1742, p. 173.]. Now for other hypotheses of the resistance with that exception, in which it is proportional to the speeds, nobody to date as far as I know has determined the tautochrones. For that which pertained to these curves , to which I gave name of tautochrones in the Act. Lips. A. 1726 [E001], did not give a satisfactory answer to this question, as the Cel. Hermann, who first fell upon the same, and I afterwards, have shown. [Jac. Hermann, General theory of the motion, which arises from any forces acting constantly on bodies, Comment. Acad. Sc. Petrop. 2 (1727), p. 139; see in particular p. 158.] But the difficulty of this method of finding tautochrones rests of this, that [p. 402] from other hypotheses of the resistance the speed cannot generally be determined from the canonical equations. Now how nevertheless the tautochrones are able to be investigated for other resistances of the tautochrones, can be gathered from the following proposition, in which it is proposed how for the rarest mediums with the resistance in some ratio of the speeds the tautochrones are to be found.