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

PROPOSITION 58. Problem.

«««< HEAD 505. According to the hypothesis of uniform gravity g, and in a medium with some uniform resistance, to determine the motion of the body with the initial speed of ascent given at A (Fig.62) on the straight line AB inclined at some angle to the horizontal.

Solution.

With the line AC drawn to the horizontal and with the perpendicular MP drawn to that from M , call PM = x, and let AM = nx. Let the height corresponding to the initial speed at A be equal to b and the height corresponding to the speed at M be equal to v; now the resistance at M is equal to VK . With these in place, there is the equation :

======= 505. According to the hypothesis of uniform gravity g, and in a medium with some uniform resistance, to determine the motion of the body with the initial speed of ascent given at A (Fig.62) on the straight line AB inclined at some angle to the horizontal.

Solution.

nV gK + nV

with this integral thus taken, so that it vanishes on putting v = b. If then we put v = 0, then there is produced x = BC, where the body has lost all the speed at the point B. Now the time, in which the body ascends along AM, is equal to ∫ [p. 261] ∫ ( gK−+Kdv nV ) v with this integral also taken so that it vanishes on putting v = b; in which if again we put v = 0, the time for the whole ascent along AMB is found. Moreover the force sustained by the line AMB is everywhere constant, and equal to the normal force g ( n 2 −1) n . Q.E.I.

Corollary 1.

If the line AMB is horizontal, then with the angle BAC vanishing, n becomes ∝ . Therefore on putting AM = z = nz , there is found «««< HEAD

=======

Corollary 2.

======= Corollary 2.

If the resistance is as some power 2m of the speeds, then we have V = v m and K = k m . Therefore in this case, there becomes : x= ∫ − k m dv gk m + nv m and the time to pass along AM is equal to : ∫ ( gk −+knvdv) v . m m m «««< HEAD

Corollary 3.

======= Corollary 3.

This expression and the other converted into series give and the time to pass along AM is equal to : [p. 262] On account of which, on putting v = 0, the series becomes and the time of the whole ascent along AB is equal to :

Example 1.

Let the resistance be proportional to the speed, then m = 12 and Hence the total height BC, to which the body is able to reach,

Now the time in which it rises along AM is equal to : Whereby the time of the whole ascent along AMB is equal to : Therefore if the body descends on the inclined line AC (Fig. 63) and with the speed acquired at C ascends on CB as far B and let AC = N .AD and BC = n.BE and the speed at C corresponds to the height b, then (486) And the descent time along AC = (cit.) and the ascent time along CB = [p. 263] Hence the descent and the ascent on the straight lines can be compared with each other.

Corollary 4.

If these logarithms are expressed in series, it is clear that it is not possible that BE = AD ; for according to any hypothesis of the resistance, as it is understood from the series (508 and 488), that we have BE < bg and AD > bg . But it can happen that AC = BC. «««< HEAD

Corollary 5.

======= Corollary 5.

Moreover it is easily demonstrated that the descent time along AC is equal to the ascent time along CB. Clearly this must become ng k = Ng k + Nn b or n = Ng k g k −N b «««< HEAD

Hence we have n > N or angle BCE < angle ACD. Moreover, the relation between N and n depends on the speed at the point C.

Corollary 6.

But if the angle BCE is equal to the angle ACD or N = n, the ascent time along BC is less than the descent time along AC. And this generally is the position for any hypothesis of the resistance; for the time of the descent along CB < 2ng b , as is apparent from the series given above ((488) and (508)).

Example 2.

======= Hence we have n > N or angle BCE < angle ACD. Moreover, the relation between N and n depends on the speed at the point C. Corollary 6. 512. But if the angle BCE is equal to the angle ACD or N = n, the ascent time along BC is less than the descent time along AC. And this generally is the position for any hypothesis of the resistance; for the time of the descent along CB < 2ng b , as is apparent from the series given above ((488) and (508)). [p. 264] Example 2.

The medium resists in the ratio of the square of the speeds; then m = 1 . Whereby there is obtained : − kdv gk + nb x= = kn l gk + nv gk + nv and (Fig. 62) ∫ gk + nb gk + nb BC = kn l gk and AB = kl gk Now the time of the descent along AM is equal to . with the radius equal to 1 and with A denoting the arc of the circle. Hence the ascent time along AB is equal to If now the body descends along the inclined line AC (Fig. 63) and with the speed at C acquired, which corresponds to the height b, it ascends along CB again where «««< HEAD AC = N .AD and BC = n.BE , then we have :

======= AC = N .AD and BC = n.BE , then we have :EULER’S MECHANICA VOL. 2. Chapter 3b. Translated and annotated by Ian Bruce.

Corollary 7.

======= (407). Now again we have : and the time of the ascent along CB is equal to : Corollary 7.

In this hypothesis of the resistance it is convenient to put AC = BC; for it must become Ngk ngk = Ngk + Nnb or n = gk − Nb . Hence we have n > N, and hence the angle BCE < the angle ACD. «««< HEAD

Example 3.

======= Example 3.

Let the resistance be taken as very small and proportional to the 2m th power of the speed; then k is a very large quantity. [p. 265] If therefore the speed at C corresponds to the height b, AC = N .AD and BC = n.BE , and the body descends on the line AC and ascends on the line CB , then and the time of descent along AC is equal to : (488). Now for the ascent, and the time to pass along CB is equal to : «««< HEAD (508). If therefore it is to be brought about that AC = BC, then it is required that thus this becomes : since k is a very large quantity. But where the time of the descent along AC is equal to the time of the ascent along CB, it must be the case that : or Scholium 1.

======= (508). If therefore it is to be brought about that AC = BC, then it is required thatEULER’S MECHANICA VOL. 2. Chapter 3b. Translated and annotated by Ian Bruce. page 414 thus this becomes : since k is a very large quantity. But where the time of the descent along AC is equal to the time of the ascent along CB, it must be the case that : or Scholium 1. [p. 266]

In the case of this example, where the resistance is very small, the curve AMD (Fig.

can be determined by this property, as the body by ascending from C with a speed corresponding to the height b upon some line CM reaches the curve AMD. For on putting CM = z and MP = x then n = xz and Whereby this equation is obtained : Let CP = y and let f be written in place of b m+1 , ( m +1) gk m and there arises : or b− f If we put y = 0, then both x = 0 and x = g = CA. Hence the curve also passes through the point C, which moreover ceases to satisfy the part of this question on account of the following neglected terms, which have been wrongly ignored, if n or xz is also made very large. Now the equation gives a curve in the form of an ellipse with the maximum length described about the minor axis AC. But the true curve has the form AMD, the asymptote of which is the horizontal line CE, if indeed m > 1 , the equation of this curve is obtained «««< HEAD with all the terms taken, which is :

======= with all the terms taken, which is :EULER’S MECHANICA VOL. 2. Chapter 3b. Translated and annotated by Ian Bruce. page 415

Scholium 2.

======= Scholium 2. [p. 267]

If the resistance of the medium is not uniform, then the line is not straight upon which the motion can be most easily determined; the same too is to be observed, if the force acting is not uniform. As the force is set equal to P and the resistance equal to VQ , where Q is such a function of the exponent of the variable resistance q, as V is of v; with these put in place the motion of the body upon any curve is expressed by this equation : i.e. Hence this is the equation of this curve, upon which the motion is most easily defined : PQdx = Ads , and from which the following arises : «««< HEAD

=======

PROPOSITION 59. Problem.

======= dimension, that which arises if the resistance is proportional to the square of the speed.EULER’S MECHANICA VOL. 2. Chapter 3b. Translated and annotated by Ian Bruce. page 416 PROPOSITION 59. Problem.

According to the hypothesis of uniform gravity g and with the resistance proportional to the square of the speed, the body descends on some curve AMB (Fig.57); to determine the motion of this body, and the force sustained by the curve at individual points. [p. 268] «««< HEAD

Solution.

======= Solution.

Corollary 1. [p. 269]

======= Q.E.I.EULER’S MECHANICA VOL. 2. Chapter 3b. page 417 Translated and annotated by Ian Bruce. Corollary 1. [p. 269]

The force that the curve sustains can be changed into this form : s which, upon integrating e k dx for the given curve, is more convenient to apply to any case. Corollary 2. gkdx «««< HEAD

=======

The body on descending has the maximum speed where v = ds . Now this comes about, where ∫ s s e k kdx = ds e k dx or where ∫ s d .l e k dx = ds ; k at which point it is clear that the tangent is not horizontal. «««< HEAD

Corollary 3.

======= Corollary 3.

If it should be that then and Whereby if this equation expresses the nature of the curve sought, then v= gs n a n−1 and the time to traverse AM is equal to [p. 270] if indeed n is less than two; for if n = 2 or n > 2, then the curve has a horizontal tangent at A and the body remains there permanently.EULER’S MECHANICA VOL. 2. Chapter 3b. Translated and annotated by Ian Bruce. page 418 «««< HEAD

Corollary 4

======= Corollary 4.

In a similar manner it is also evident, if x is some power of s or a sum of powers of s this kind, then it is always possible to integrate e k dx and thus the terminal speed can be shown. «««< HEAD

Corollary 5

======= Corollary 5.

Moreover if the abscissae are taken on the vertical axis BQ and the speed that the body has at B corresponds to the height b, and besides calling BQ = x and BM = s, then dv = − gdx − vds k the integral of this equation is : e − k v = b − g e − k dx ∫ s s clearly with the integral e − k dx thus taken, in order that it vanishes on putting x = 0. On ∫ s this account we have : v = be k − ge k e − k dx s s ∫ s and the time, in which the arc descended MB is completed, is equal to : ∫e ds s 2k −s . (b − g ∫ e k dx ) «««< HEAD

Corollary 6

======= Corollary 6. [p. 271]

Therefore if the speed is given at the point B, clearly b , it is possible to find the point A on the curve BMA, from which the body always begins its descent and has the speed equal to zero. Whereby this is clearly the place where ∫ e − k dx = bg . s And also the expression for the time : ∫e ds s 2k −s (b − g ∫ e k dx ) gives the time of the whole descent along AMB, if we put after the integration, − ks ∫ e dx = bg . «««< HEAD

Scholium

======= Scholium .

Thus we have presented two ways in which the motion can be investigated : as it can be adapted to the descent made from a given point, or the descent as far as a given point is considered, as is usually the case in oscillatory motion.EULER’S MECHANICA VOL. 2. Chapter 3b. page 419 Translated and annotated by Ian Bruce. «««< HEAD

PROPOSITION 60. Problem.

======= PROPOSITION 60. Problem.

With the force arising acting uniformly and with a uniform medium with resistance in the square ratio of the speed, to determine the motion of the body ascending on the given curve AMD (Fig.58) and the force pressing on the curve sustained at individual points M. «««< HEAD

Solution.

======= Solution.

Hence on putting v = 0 the equation becomes : ∫ s g e k dx = b , from which equation the point D is obtained, and to which the body is able to ascend ∫ s from A. And the time of the whole ascent along AMD can be obtained, if g e k dx = b is «««< HEAD put in the expression for the time.

Corollary 2

If, in the formula showing the force pressing, in place of v this value found is substituted, then there is obtained : Which it is possible to change into this form :

Corollary 3. [p. 273]

Now for the descent, if the speed at A also corresponds to the height b, then the pressing force that the curve sustains at M along the normal MN is equal to :

Corollary 4.

======= put in the expression for the time.EULER’S MECHANICA VOL. 2. Chapter 3b. Translated and annotated by Ian Bruce. page 420 Corollary 2. 528. If, in the formula showing the force pressing, in place of v this value found is substituted, then there is obtained : Which it is possible to change into this form : Corollary 3. [p. 273] 529. Now for the descent, if the speed at A also corresponds to the height b, then the pressing force that the curve sustains at M along the normal MN is equal to : Corollary 4.

If therefore the ascent as well as the descent are defined with respect to the axis AP, the equation determining the ascent can be changed into the equation for the descent by writing – k in place of k in turn. Whereby if the descent has been determined on the curve AM, then the ascent can also be determined in turn. «««< HEAD

Scholium

Since the formulas determining the ascent and the descent have so much in common, the ascents and the descents can easily be compared with each other, and thus the oscillations on a given curve can be determined. We present this in the following proposition, as generally as it can be done.

PROPOSITION 61. Problem

======= Scholium. 531. Since the formulas determining the ascent and the descent have so much in common, the ascents and the descents can easily be compared with each other, and thus the oscillations on a given curve can be determined. We present this in the following proposition, as generally as it can be done. [p. 274]EULER’S MECHANICA VOL. 2. Chapter 3b. Translated and annotated by Ian Bruce. page 421 PROPOSITION 61. Problem.

Let whatever curves MA and NA (Fig.65) be joined at the lowest point A and the body descends on the curve [MA and ascends on the curve] AN in a medium with uniform resistance following the square of the speeds; to be compares between themselves are the descends on the curve MA and the ascents on the curve AN. «««< HEAD

Solution.

======= Solution.

======= AMEULER’S MECHANICA VOL. 2. Chapter 3b. Translated and annotated by Ian Bruce.

Therefore if MAN and mAn are two arcs described by neighbouring oscillations, then always Qq < Pp and from this Qq is less than Pp, in which the sum of the arcs AM + AN is greater. Therefore also, AQ shall always be less than AP. Corollary 2. AM + AN k
In vacuo, since k =∝ , then e = 1 ; and the above equation becomes Qq = Pp and hence AQ = AP. Whereby the body oscillating in vacuo ascends to as great a height, as that from which it descends. Corollary 3.
If the resistance is very small and thus k extremely large, then e AM + AN k = 1 + AM k+ AN . Whereby in this case the ratio is : Qq + MAN .Qq Pp ( k − MAN ) = Pp and Qq = k k Corollary 4. [p. 276]
If the point O is the place, at which the descending body has the maximum speed, and there put AO = s, AS = x, then s s gkdx − ks k k = e b − ge e dx ds ∫ or [corrected in O. O.] b = g e − k dx + ds . Whereby if the body descends from m, the point of the maximum speed arising at o ∫ s gkdx db = ge − k dx + p , with dy = ov and p radius of osculation at the point O, or s gkdy db = e − AO k Ss + k .pov . gEULER’S MECHANICA VOL. 2. Chapter 3b. Translated and annotated by Ian Bruce. page 423 Corollary 5.
The time of a single motion along MAN is found, if in the sum of the integration, there is inserted And thus the time of a single semi–oscillation is produced. Corollary 6.
From what has been said another elegant property follows, for if the body ascends from A with the speed b to N and from N it again falls through NA and the speed that it then has at A corresponds to the height c, [p. 277] then from A it can reach n with a speed corresponding to the height b + db, hence again on descending to A it acquires a speed corresponding to the height c + dc. Hence and Qq 2 = db.2dc g and also db = e 2 AN k . dc Scholion.
It remains, that we adapt these generalisations to examples or given curves, where the use of these will there become more apparent. Moreover we take only the cycloid for ∫ s the given curve, as there the integral e k dx can be easily demonstrated and also the equation between s and x is algebraic. Hence we will examine both descents made on cycloids as well as oscillations, from which it is apparent to what extent the oscillations made on the cycloid depart from being isochronous, clearly in vacuo all have been shown «««< HEAD to be completed in the same time.

PROPOSITION 62. Problem.

======= to be completed in the same time. [(191)].EULER’S MECHANICA VOL. 2. Chapter 3b. Translated and annotated by Ian Bruce. page 424 PROPOSITION 62. Problem.

Let the given curve be the cycloid ACB (Fig.66) described by the circle of diameter CD rolling along upon the given horizontal base AB, and the body descends on that cycloid from A in a medium with the resistance proportional to the square of the speed; to determine the motion of the descending body. [p. 278] «««< HEAD

Solution.

======= Solution.

Since s 2 e − k = 1 − 1s.k + s 2 − 1.2.k s3

etc., 1.2.3.k 3 if we put a + k = c or a = c – k, then 2 3 2 3 (c − k )v = − s + 1cs.k − cs 2 + cs 3 − etc. = 1as − cs 2 + cs 3 − etc. gk . k 1.2.k 1.2.3.k 1.2.k 1.2.3.k Whereby gs v = a ( a1 − 1s.2 − 1.as

2.k (a + k )s 2 (a + k ) s 3 −

etc. ) 1.2.3.k 2 1.2.3.4 k 3 Corollary 2.

Therefore in vacuo, where k is infinite, then gs 2 v = gs − 2a = gx , is taken as a constant. But if the resistance is only very small, and therefore k becomes very large, then gs 2 gs 2 gs 3 v = gs − 2a − 2k + 6ak .EULER’S MECHANICA VOL. 2. Chapter 3b. Translated and annotated by Ian Bruce. page 426 Corollary 3.
In vacuo it is apparent that the speed of the body is zero at two points, where s = 0 and s = 2a, i.e. at the two cusps A et B. Now in a resisting medium, the one place is where s = 0; the other must be elicited from the equation : hence it is found that : If indeed k is very large, then as an approximation. «««< HEAD

Corollary 4.

This same series found on substituting a + k in place of c is transformed into this : from which the value of s gives the arc length ACN, by which the body is able to travel so far by its own motion.

Corollary 5.

======= Corollary 4. [p. 280] 544. This same series found on substituting a + k in place of c is transformed into this : from which the value of s gives the arc length ACN, by which the body is able to travel so far by its own motion. Corollary 5.

The arc AO from A as far as to O, where the body has its maximum speed, is equal to : Whereby the arc is given by [the following expressions have been corrected as in O. O. ]: and «««< HEAD

Corollary 6.

Now the speed at the point C is found corresponding to the height, is equal to

======= Corollary 6. 546. Now the speed at the point C is found corresponding to the height, is equal toEULER’S MECHANICA VOL. 2. Chapter 3b. Translated and annotated by Ian Bruce. page 427

The height corresponding to the maximum speed at O is equal to : gk 2 gk − a l a +k k = g ( a2 − 3ak + a 2 − a 3 + etc) 4k 5k Whereby the excess of this height over the height corresponding to the speed at C is equal to 3 g( a 2 − 1.2.4k 2 3 4 -a a 3 − a 4 + etc.) 5a 4 23a 5 k 3 + 4 - etc) = ge ( 1.2.3.5k 1.2.3.4 ,6k 8k 2 24k 3 This expression is certainly obtained if kl a +k k is substituted in place of s in the general value of v, and clearly the arc AO is equal to this quantity. Scholium.
In the solution of this proposition it is required to be examined, since by determining only the formula for the descents we have also derived the ascent of the body on the curve CN ; from which it is possible that doubt arises, or that the true ascent is defined. But this is evidently easy to be derived from the formula for the ascent. Indeed we have been using this formula dv = gdx − Rds , in which with the point M falling beyond the point C as dx has been made negative is changed into : dv = − gdx − Rds , which actually contains the nature of the ascent. From these the continuity between the ascent and the descent is understood, which touch each other with no jump in between. [p. 282] Where indeed the curve itself begins to change direction and rise, where the like formula serving the descent is changed freely into the formula for the ascent. And this junction has a place in a medium with some kind of resistance, as is apparent from the general formulas, which only disagree with the sign of dx . On account of which it is not necessary from the given equation for some curve, that the body ascending or descending on one part of the curve or the other is sought, as either formula can be adapted to the equation and gives the true motion on the proposed curve. Only this has to be understood, that the abscissae are taken on the vertical axis, and that formula either for the ascent or «««< HEAD of the descent is adhered to, which agrees with the initial motion.

PROPOSITION 63. Problem.

======= of the descent is adhered to, which agrees with the initial motion.EULER’S MECHANICA VOL. 2. Chapter 3b. Translated and annotated by Ian Bruce. page 428 PROPOSITION 63. Problem.

Let the given curve ACB (Fig.66) be a cycloid described on the horizontal base AB and considering the body to complete downwards oscillations on that curve in a medium with resistance in the square ratio of the speeds; to determine the motion of the oscillations. «««< HEAD

Solution.

======= Solution.

=======

Therefore if the maximum speed of the descending body corresponds to the height c, since ac CO = gk thenEULER’S MECHANICA VOL. 2. Chapter 3b. Translated and annotated by Ian Bruce. page 431 Now the total ascent CN is equal to : Hence and Corollary 2.
If the whole arc of the descent MC is put equal to E and the following arc of the ascent CN is equal to F and the height corresponding to the height at C = b, then And with k made negative it is found in the same way: From which there becomes : and the height corresponding to the maximum speed is given by : [p. 286] Corollary 3.
Since F is the arc of the ascent in the first half oscillation, the arc F is likewise the arc of the descent in the following oscillation ; therefore as it can be joined with the ascent arc, And in a like manner the following oscillations can be defined, however many it might please to consider.EULER’S MECHANICA VOL. 2. Chapter 3b. Translated and annotated by Ian Bruce. page 432 Corollary 4.
It is apparent from the equation setting out the time, that the time in which the body arrives at O from M, is always less than the time in which the body reaches as far as N from O. In a similar manner also the arc ON is greater than the arc OM, and now the arc CN is less than the arc MC. Corollary 5.
If the oscillations should become infinitely small or c becomes a vanishing quantity, then the oscillations agree with the oscillations made in vacuo ; for in the individual expressions the same terms vanish, which vanish on putting k =∝ . Therefore isochronous with the smallest oscillations of a pendulum of length a in vacuo acted on by a force g , or of a pendulum according to the hypothesis of gravity equal to 1, the length of which is equal to ag . [This agrees with modern analysis, and reinforces the belief considered earlier, that Euler is using a set of units in which the second and the acceleration of gravity are taken as one.] Corollary 6. [p. 287]
But if the oscillations become greater, then the times of the oscillations also become greater; whereby according to the hypothesis of resistance the cycloid is not favoured with the property of tautochronism. For when the maximum speed is made greater in some oscillation, the greater also is the excess of the time of the oscillation of this kind over the time of the smallest oscillation. Scholium 1.
Since we have said that the smallest oscillations agree with oscillations in vacuo, then the situation arises that a and k are quantities of finite size. For if a should be infinitely large or k infinitely small, the following terms express the time : that do not vanish, even if c should be indefinitely small. Moreover, therefore, only the smallest oscillations upon some curve agree with the oscillations in vacuo and in a resisting medium, when neither the radius of osculation of the curve at the lowest point is infinitely large nor the resistance infinitely great. Example.
We explain the case by means of an example, in which the resistance is so small and thus the quantity k so large that the fractions in the denominators of which k has more than two dimensions can be put equal to zero without harm. Therefore with the aforesaid ac , then the arc of height c corresponding to the maximum speed at O, thus so that CO = gk the descent is given by : [p. 288]EULER’S MECHANICA VOL. 2. Chapter 3b. Translated and annotated by Ian Bruce. page 433 and the following arc of the ascent is given by : From these it is found or and Therefore the time of half an oscillation along MCN is equal to : In the following half oscillation the arc of descent is equal to : and the arc of ascent that follows is equal to : and the time of this half oscillation is equal to : where the final term can be neglected on account of k3 in the denominator. In the third half oscillation, the descent arc is equal to : and the ascending arc is equal to :EULER’S MECHANICA VOL. 2. Chapter 3b. Translated and annotated by Ian Bruce. page 434 And generally in that half oscillation, which in indicated by the number n, the arc of descent is equal to : = and the arc of ascent is equal to : On account of which after n half oscillations the body is at a distance from the lowest point C by the arc : which is less than the first arc of descent by the amount : Moreover the time of the half oscillation indicated by the given number n is : But if the whole arc of the first half oscillation MCN is called A, then with the following term vanishing spontaneously. Hence the whole arc described by the oscillation indicated by the number n is equal toEULER’S MECHANICA VOL. 2. Chapter 3b. Translated and annotated by Ian Bruce. page 435 Corollary 7. [p. 289]
If n half oscillations are made, the descending arc of the first oscillation is E, and the final ascending arc is equal to L, then which expression, if taken with the nearby series, almost agrees with the geometric progression of the same starting value, [and ratio 23nE ]and on this account it follows that k Corollary 8.
Hence, for some number of oscillations performed, if the first descending arc of the first oscillation E is given together with the final ascending arc L, it is possible to find the number of oscillations; for it is in fact Corollary 9.
Hence it is clear that the diminution of the arcs does not depend on the length of the pendulum, but from n and E given the same arc L is found, whatever the length a of the pendulum should be. and n is always proportional to L1 − E1 . Scholium 2. [p. 290]
Newton examines experiments of this kind concerning oscillations in a resisting medium in Phil. Book. II, where he observes the first descending arc, the final descending arc, and the number of oscillations for [a pendulum] in air, as in water and mercury. Whereby if the mediums resisted perfectly in the ratio of the square of the speed, there should be agreement with these formulas, thus in order that the decrease of the arc should be proportional to the number of oscillations, and both the first and the final arcs taken together. Since also the situation arises for larger oscillations to be observed, in which the speed is not extremely small. But in the smallest oscillations the greatest aberration from this rule is observed. From which it is gathered, when the speed of the body in the fluid is made greater, the resistance to that falls closer to the ratio of the square of the speeds, but the slowest motion is liable to be affected by other resistances, which vanishes in motions for speeds with the previous resistance, which is proportional to the square of the speeds. Also in these experiments, Newton assumed that the resistance was in part simply proportional to the ratio of the speed, partly as the speed to the three on two power, and partly as the square of the speeds, yet this was not satisfactory for the slowest motions. Now in the final Phil. edition Newton recognises theEULER’S MECHANICA VOL. 2. Chapter 3b. Translated and annotated by Ian Bruce. page 436 insufficiency of his own initial theory and by many other reasons he shows that another resistance of the fluid to be constant, or proportional to the lengths of time, as before he had considered for the proportionality with the speeds. Hence on account of this, in the following proposition we will consider joining that resistance with this, which is proportional to the square of the speed, since the resolution of the equation and the determination of the speed with this addition does not present any more special difficulty. [The interested reader should consult Cohen’s translation of the third ed. of the Principia, where these matters of Newton are discussed at some length. Cohen’s work is remarkable for the complete lack of comment on the follow-up to Newton that we are providing here. One wonders why Newton has been ‘done to death’ as it were by numerous commentaries, while this wonderful work of Euler has been hidden in the closet, as it were, for centuries………] Corollary 10. [p. 291]
Since according to the times of the oscillations and the semi oscillations [the pendulum] attains, it is evident that this decreases when the arc described becomes smaller, and if the arcs completely vanish, then the time of half an oscillation becomes equal to π 2a . g Corollary 11.
Moreover the excess of the time of this half oscillation over the smallest half oscillation in the case of the smallest resistance is πE 2 2a with E denoting the descent of 2 24k g this half oscillation. Whereby the excess itself is in proportion to the square of the arc descended or also to the square of the whole arc of the half oscillation described. Scholium 3.
Therefore the cycloid, that has been suitably described by Huygens according to the production of isochronous pendulums, loses this property on account of the resistance in proportional to the square of the speeds, and hence is not serviceable in air, unless the oscillations are very small or are almost equal to each other. [p. 292] Now from this, since larger oscillations endure a long time, it is possible to gather that the true tautochrone curve according to this hypothesis of the resistance to be more curved than the cycloid. As clearly the cycloid is continued in a circle of the same radius, of which the cycloid is the lowest point, thus also the true tautochrone will be continued in a cycloid and with the curvature decreased more from the lowest point than with the «««< HEAD curvature of the cycloid. ======= curvature of the cycloid.EULER’S MECHANICA VOL. 2. Chapter 3b. Translated and annotated by Ian Bruce. page 437 CAPUT TERTIUM DE MOTU PUNCTI SUPER DATA LINEA IN MEDIO RESISTENTE. [p. 260] PROPOSITIO 58. Problema.
In hypothesi gravitatis uniformis g et medio quocunque resistente uniformi determinare motum corporis data cum celeritate initiali ex A (Fig.62) ascendis super linea recta AB utcunque inclinata ad horizontem. Solutio. Ducta horizontali AC et ex M ad eam perpendiculari MP vocetur PM = x, sitque AM = nx. Sit altitudo debita celeritati initiali in A = b et altitudino debita celeritati in M = v; resistentia vero in M sit = VK . His positus erit dv = − gdx − nVdx K (479), unde habetur dx = − Kdv atque x = gK + nV ∫ gK− Kdv

nV hoc integrali ita accepto, ut evanescat posito v = b. Si deinde ponatur v = 0, prodibit x = BC, ubi in puncto B corpus omnem celeritatem amittit. Tempus vero, quo corpus per AM ascendit, est = − Kdv [p. 261] ∫ ( gK + nV ) v hoc integrali quoque ita accepto, ut evanescat posito v = b; in quo si porro ponatur v = 0, prodibit tempus totius ascensus per AMB. Pressio autem, quam linea AMB sustinet, ubique est constans et aequalis vi normali = g ( n 2 −1) n Q.E.I.EULER’S MECHANICA VOL. 2. Chapter 3b. Translated and annotated by Ian Bruce. page 438 Corollarium 1.

Si linea AMB fit horizontalis, evanescente angulo BAC fiet n =∝ . Posito igitur AM = z = nz erit z = − Kdv V ∫ et tempus, quo per AM progreditur, erit = ∫ −VKdvv . Corollarium 2.
Si resistentia fuerit ut potestas exponentis 2m celeritatum, erit V = v m et K = k m . Hoc ergo casu erit x= ∫ − k m dv gk m + nv m atque tempus per AM = ∫ ( gk −+knvdv) v . m m m Corollarium 3.
Utraque haec expressio in seriem conversa dat atque tempus per AM = [p. 262] Quamobrem posito v = 0 erit atque tempus totius ascensus per AB =EULER’S MECHANICA VOL. 2. Chapter 3b. Translated and annotated by Ian Bruce. page 439 Exemplum 1.
Sit resistentia celeritatibus proportionalis; erit m = 12 atque Hinc erit tota altitudo BC, ad quam corpus pertingere valet, Tempus vero, quo per AM ascendit, est = Quare tempus totius ascensus per AMB erit = Si igitur corpus super linea inclinata AC (Fig. 63) descenderit et celeritate in C acquisita ascendat in CB usque ad B sitque AC = N .AD et BC = n.BE atque celeritas in C debita altitudini b, erit (486) Atque tempus descensus per AC = (cit.) et tempus ascensus per CB = [p. 263] Unde descensus et ascensus super lineis rectis inter se comparari possunt. Corollarium 4.
Si hi logarithmi per series exprimantur, patet fieri non posse, ut sit BE = AD; est enim in quacunque resistentiae hypothesi, ut ex seriebus (508 et 488) intelligitur, BE < bg et AD > bg . Fieri autem potest, ut sit AC = BC.EULER’S MECHANICA VOL. 2. Chapter 3b. Translated and annotated by Ian Bruce. page 440 Corollarium 5.
Effici autem facile potest, ut tempus descensus per AC aequale sit tempori ascensus per CB. Fieri solicet debet Est igitur n > N seu ang. BCE < ang. ACD. Relatio autem inter N et n pendet a celeritate in puncto C. Corollarium 6.
Si autem angulus BCE aequalis fuerit angulo ACD seu N = n, tempus ascensus per BC minus erit tempore descensus per AC. Atque hoc generaliter locum habet in quacunque resistentiae hypothesi; est enim tempus descensus per CB < 2ng b , ut ex seriebus supra datis ((488) et (508)) apparet. [p. 264] Exemplum 2.
Resistat medium in duplicata ratione celeritatum; erit m = 1 . Quare habebitur − kdv gk + nb x= = kl gk + nv n gk + nv atque (Fig. 62) ∫ gk + nb BC = kn l gk Tempus vero ascensus per AM erit = gk + nb et AB = kl gk . existente radio = 1 et A denotante arcum circuli. Erit ergo tempus ascensus per AB = Si nunc corpus super linea inclinata AC (Fig. 63) descenderit et celeritate in C acquisita, quae debita sit altitudini b, rursus ascendat per CB fueritque AC = N .AD et BC = n.BE , erit atque tempus descensus per AC = (407). Porro vero erit et tempus ascensus per CB =EULER’S MECHANICA VOL. 2. Chapter 3b. Translated and annotated by Ian Bruce. page 441 Corollarium 7.
In hac resistentiae hypothesi commode effici potest, ut sit AC = BC; debebit enim esse Ngk ngk = Ngk + Nnb seu n = gk − Nb . Est igitur n > N hincque ang. BCE < ang. ACD. Exemplum 3.
Sit resistentia quam minima et proportionalis potestati 2m celeritatum; erit k quantitas vehementer magna. [p. 265] Si ergo celeritas in C fuerit debita altitudini b et AC = N .AD atque BC = n.BE corpusque super AC descendat et super CB ascendat, erit et tempus descensus per AC = (488). Pro ascensu vero erit et tempus per CB = (508). Si igitur effici debeat, ut sit AC = BC, oportet esse unde fit propter quantitatem k valde magnam. At quo tempus descensus per AC aequale sit tempori ascensus per CB, debet esse seuEULER’S MECHANICA VOL. 2. Chapter 3b. Translated and annotated by Ian Bruce. page 442 Scholion 1. [p. 266]
In casu huius exempli, quo resistentia est valde parva, curva AMD (Fig. 64) potest determinari huius proprietatis, ut corpus ex C celeritate altitudini b debita ascendendo super quavis recta CM ad curvam AMD pertingat. Posita enim CM = z et MP = x erit n = xz atque Quare habebitur ista aequatio Sit CP = y et loco b m+1 ( m +1) gk m scribatur f; orietur seu b− f Si ponatur y = 0, erit et x = 0 et x = g = CA. Curva ergo etiam per punctum C transit, quae autem eius pars quaestioni satisfacere cessat propter sequentes terminos neglectos, qui perperam negliguntur, si n seu xz fit quoque valde magnum. Aequatio vero dat curvam ellipsoformem maxime oblongam circa axem minorem AC descriptam. Vera autem curva habet formam AMD, cuius asymtotos est horizontalis CE, si quidem est m > 1 , eiusque aequatio habetur omnibus sumendis terminis, quae erit Si m < 1, curva non in infinitum progredietur, sed incidet in CE sumendo CE = bk m . (1− m)b m Nam si m < 1, corpus horizontaliter non in infinitum progredi potest, sed in distantia finita omnem amittit celeritatem. Scholion 2. [p. 267]
Si medium resistens non sit uniforme, tum linea non esset recta, super qua motus facillime determinari posset; idem quoque est notandum, si potentia sollicitans non fuerit uniformis. Ut sit potentia = P et resistentia = VQ , ubi Q talis est functio exponentisEULER’S MECHANICA VOL. 2. Chapter 3b. Translated and annotated by Ian Bruce. page 443 resistentiae variabilis q, qualis V est ipsius v; his positis motus corporis super quacunque curva exprimitur hac aequatione Eius ergo curvae, super qua motus facillime definitur, haec erit aequatio PQdx = Ads , ex qua oritur sequens motum super hac curva determinans, in qua indeterminatae sunt a se invicem separatae. Quare si etiam huiusmodi hypotheses persequi vellemus, loco linearum rectarum huiusmodi curvas hac aequatione expressas PQdx = Ads assumere deberemus. Sed quia statuimus hypothesin potentiae sollicitantis et resistentiae uniformis tantum fusius pertractare, missis his ad eos casus progredimur, in quibus v unicam tantum habet dimensionem, id quod evenit, si resistentia proportionalis fuerit quadratis celeritatum PROPOSITIO 59. Problema.
In hypothesi gravitatis uniformis g resistentiae quadratis celeritatum proportionalis descendat corpus super curva quacunque AMB (Fig.57); determinare eius motum et pressionem, quam curvam in singulis punctis sustinet. [p. 268] Solutio. In axe verticali sumatur abscissa AP = x et ponatur arcus AM = s, celeritas in M debita altutudini v et exponens resistentiae k ; erit resistentia = kv . Quamobrem ista habebitur aequatio motum corporis exponens dv = gdx − vds k s (465). Ad hanc integrandam multiplico per e k eritque integralis s ∫ s e k v = e k gdx Ita autem sumi debet hoc integrale, ut posito s = 0 abeat v in altitudinem celeritati initiali s in A debitam. Si igitur descensus ex quiete fieri ponatur e k gdx ita debet integrari, ut evanescat posito s = 0. Hoc itaque facto erit −s ∫ s v = ge k e k dx. Unde tempus per AM erit =EULER’S MECHANICA VOL. 2. Chapter 3b. page 444 Translated and annotated by Ian Bruce. s 2k ∫ ge edxdx . ∫ s k Posito nunc PM = y sumtoque dx constante erit pressio, quam curva in M secundum normalem MN sustinet, = Q.E.I. Corollarium 1. [p. 269]
Pressio, quam curva sustinet, in hanc formam potest transmutari s quae, postquam pro data curva integratum est e k dx , commodius ad quosvis casus accommodatur. Corollarium 2. gkdx
Corpus in descensu maximam habet celeritatem, ubi est v = ds . Hoc vero evenit, ubi est ∫ s s e k kdx = ds e k dx seu ubi ∫ s d .l e k dx = ds ; k in quo puncto patet tangentem non esse horizontalem. Corollarium 3.
Si fuerit erit atque Quare si haec aequatio exprimat curvae quaesitae naturam, erit v= et tempus per AM = gs n a n−1EULER’S MECHANICA VOL. 2. Chapter 3b. Translated and annotated by Ian Bruce. page 445 [p. 270] si quidem n fuerit minor binario; nam si n = 2 vel n > 2, curva in A tangentem horizontalem habebit atque corpus ibi perpetuo permanebit. Corollarium 4.
Simili modo etiam perspicitur, si x fuerit potestas quaecunque ipsius s vel s huiusmodi potestatum aggregatum, semper integrari posse e k dx atque ideo celeritatem terminus finitis exhiberi. Corollarium 5.
Si autem abscissae in axe verticali BQ sumantur et celeritas, quam corpus in B habebit, debita sit altitudini b praetereaque vocetur BQ = x et BM = s, erit dv = − gdx − vds k cuius integralis est e − k v = b − g e − k dx ∫ s s integrali scilicet e − k dx ita accepto, ut evanescat posito x = 0. Hanc ob rem erit ∫ s v = be k − ge k e − k dx s s ∫ s et tempus, quo in descensu arcus MB absolvitur, = ∫e ds s 2k −s k . (b − g ∫ e dx ) Corollarium 6. [p. 271]
Si igitur detur celeritas in puncto B, nempe b , inveniri potest in curva BMA punctum A, ex quo corpus descendere incepit ubique celeritatem habuit = 0. Quari debet scilicet locus, ubi est ∫ e − k dx = bg . s Atque etiam expressio temporis ∫e ds s 2k −s (b − g ∫ e k dx ) dabit tempus totius descensus per AMB, si post integrationem ponatur − ks ∫ e dx = bg . Scholion .
Duplicem hic motum investigandi modum ideo attulimus, ut tam ad descensus ex dato puncto facto quam ad descensus usque ad datum punctum, ut in motu oscillatorio fieri solet, accommodari possit.EULER’S MECHANICA VOL. 2. Chapter 3b. Translated and annotated by Ian Bruce. page 446 PROPOSITIO 60. Problema.
Existente potentia sollicitante uniformi et medio uniformi resistente in duplicata ratione celeritatum, determinare motum corporis ascendentis super data curva AMD (Fig.58) et pressionem, quam curvam sustinet in singulis punctis M. Solutio. In linea verticali AP ponatur abscissa AP = x, the arc AM = s, the speed at A corresponds to the height b and the speed at M to the height v. Let the force acting downwards be equal to g and let the resistance be equal to kv . With these in place, then dv = − gdx − vds k s (475), which on being multiplies by e k , the integral gives ∫ s s e k v = b − g e k dx [p. 272] ∫ s k Thus on taking e dx so that it vanishes on putting x = 0. On this account, this becomes: v = be − k − ge − k e k dx s s ∫ s Hence the time of the ascent along the arc AM is equal to : s ∫ (b− g e dx) e 2 k ds s ∫ k From the speed found there is obtained the pressing force, that is allowed on the curve at M along the normal, equal to gdy 2vdxddy − ds ds 3 (475) on putting PM = y and on taking dx as constant. Q.E.I. Corollarium 1.
Posito ergo v = 0 erit ∫ s g e k dx = b , ex qua aequatione obtinebitur punctum D, quousque corpus ex A ascendere poterit. Atque tempus totius ascensus per AMD habebitur, si in expressione temporis ponatur ∫ s g e k dx = b.

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

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