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Chapter 6a

The Curvilinear Motion Of A Free Point In A Resistive Medium

by Euler
March 20, 2025 31 minutes  • 6516 words

PROPOSITION 113. PROBLEM.

  1. As before, with the uniform absolute force g put in place and pulling downwards, to find the force of the resistance which is effective in order that the body can move along the hyperbola NAM (Fig.85) freely, with the axis CAQ vertical.

SOLUTION.

Let C be the centre of the hyperbola and the transverse semi-axis AC is equal to a; and the conjugate semi-axis is equal to c. Putting CQ = t and QM = AP = x then from the nature of the hyperbola : c 2t 2 = a 2 x 2 + a 2c 2 .

Moreover taking PM = AQ = y then y = t − a and dy = dt , d 2 y = d 2t and d 3 y = d 3t . Truly from the equation we have: and hence Again there arises : From which there becomes : Consequently the resistance becomes [from (908)]:

Truly with the resistance made equal to the square of the velocity the exponent of the resistance is given by :

Corollary 1.

  1. With the tangent MT drawn, there arises : Consequently there is produced : [p. 397]

Corollary 2.

  1. Since the resistance R is found to be negative, from this it is indicated that it is not possible for a body to descend along the hyperbola AM in a resisting medium, for it is required to be moved by the medium. But while the body ascends the arc NA, since ds becomes negative, then the resistance R is positive. On account of this, if the body is at N, the resistance is given by R = 3 g .CQ .NT . 2 AC 2

Corollary 3.

  1. From the properties of the hyperbola, we have CQ : AC = AC : CT . Thus the 3 g .NT resistance at N, or R, = 2CT . Or the resistance R to the absolute force g as 3NT to 2CT. At the vertex A the resistance hence disappears and then increases, as N is more distant from A. [p. 398]

Corollary 4.

  1. The height corresponding to the speed of the body at M or N is equal to g .CQ 2 .MT 2 , 2 AC 2 .QT as can be easily deduced from the value of v and from the properties of the hyperbola. 2 2 2 AC . Moreover since MT = NT and CQ .NT = 2 R3.AC , we have v = 29Rg ..QT g

Corollary 5.

  1. If the resistance is put in proportion to the speeds of the body and the exponent of 2 . the resistance is q, then R = v and v = R 2q. Whereby it is found that q = 92gAC .QT q Thereby by this hypothesis the exponent varies inversely as the subtangents QT.

Corollary 6.

  1. But if the resistance is put in proportion to the square of the speeds and the exponent NT .CQ of the resistance is q, with the body present at N there is the exponent q = 3TQ . Or with CR drawn parallel to the tangent NT from C, crossing the applied line QN produced . at R, then q = CR 3

Scholium.

  1. Since previously with the circle and now with the hyperbola we have noted that the resistance in the one arc to be made positive and in the other arc to be negative, that is obtained for all curves put in place with two equal arcs AN and AM around the maximum point A. [p. 399] Since in general for the arc AM the resistance is given by R = since in the arc AN ds is negative then at N resistance is given by R = − gdsd 3 y , 2ddy 2 gdsd 3 y , thus in 2ddy 2 order that the resistance at N is the negative of the resistance at M. Whereby, when by the nature of things being such that the resistance cannot be made negative, by which the body is accelerated, then it cannot happen that the body describes a curve in a medium with resistance, which has two similar and equal branches around the maximum point A. gds 2

Truly the height generating the speed at M is the same as at N; for the value of this, 2ddy is not changed, even if ds becomes negative. Therefore with curves of this kind thus considered and due to Newton, so that some might be elicited for which the density of the resisting medium does not vary much, or which can be treated by our method in which the exponent of the resistance has everywhere almost the same value and such a curve can be taken for the trajectory in a medium of uniform resistance without sensible error; in addition we consider other curves without the vertical diameter, also provided by Newton, and hyperbolas having vertical asymptotes are curves of this kind, and which clearly fall closer to the logarithmic curve which is described by a body in a uniform medium with the resistance in the simple ratio of the speed. In other cases indeed Newton could not determine the trajectories from the simple hypotheses of resistance, but was content to assign approximate values. Which arrangement we follow, since the true trajectories to be given by us are so complicated that hardly any can be deduced in practice. [p. 400]

[Newton’s propositions that correspond to those of Euler, and others, are set out in Book II, Sections 2 and 3 of the Principia.]

PROPOSITIO 114. PROBLEM.

  1. Let the curve NM (Fig.86) be a hyperbola of any other kind having a vertical asymptote CP, and it is required to determine the resistance which is effective, in order that the body, always acted on by a downwards force, moves on this hyperbola.

SOLUTION.

The asymptote CP is considered as the axis to the hyperbola, and from M the normal MP is drawn. By taking CP = y and MP = x, which we have generally taken to be the case above, then these are in position if dx is taken as negative everywhere. Let RC be the other asymptote and C the centre of the hyperbola, and the sine of the angle RCP = α and the cosine of this angle is ( 1 − α 2 ) = β . Therefore with PM produced at R, αy y PR = β and CR = β . From M the line MQ is drawn parallel to the asymptote CR, and αy - β x βx MQ = αx and PQ = α , and hence CQ = α . But from the nature of hyperbolas, and hence xdy βx With the tangent MT drawn, we have PT = − dx = − α + ( n −1 )α n−1a n , MT = xds , dx x n−1 [p. 401] thus in order that ds = MTx.dx . This is indeed the case, in which x is taken in the other part, if MT is negative.

Again on account of dx being constant, From these there is produced which with MT made negative is equal to the resistance R. Thus the resistance is given by (908) and the height corresponding to the speed at M, or v, is equal toEULER’S MECHANICA VOL. 1. Chapter Six (part b). page 592 Translated and annotated by Ian Bruce. If the medium is put to resist in the ratio of the 2mth multiple of the speeds, [i. e. the mth m multiple of v] and the exponent of the resistance is q, then R = v m and thus q = v1 . q Rm From which it becomes or Q.E.I.

Corollary 1.

  1. Since x n −1 = ααy −aβx , the resistance is given by n n Hence the resistance R to the force g is as ( n + 1 )MT to 2n( n − 1 )CQ. [p. 402] In a similar manner with this value in place of x n −1 there is produced : Corollary 2.

  2. With the body descending to infinity, x = 0 and with x vanishing. Therefore in the infinite depths, R = magnitude, but v = g ( n +1 ) and thus this is of finite 2n g ( n −1 )α n−1a n . Whereby, since by necessity n > 1, with x vanishing the 2nx n−1 speed of the body becomes infinitely great. Corollary 3. m

  3. Therefore by putting the resistance R = v m in the infinite depths should also make q q infinitely great; and thus with these in place the body will be moving in a vacuum. From which it follows, by however more the body descends, the resistance acting on it becomes smaller or rather the medium becomes rarer.EULER’S MECHANICA VOL. 1. Chapter Six (part b). page 593 Translated and annotated by Ian Bruce.

Corollary 4.

  1. In the Apollonian hyperbola n = 2. Therefore for this curve it is found that [p. 403]

Corollary 5.

  1. If the resistance is put to be in proportion to the speeds, in all these hyperbolas the exponent of the resistance varies directly as CQ, since m = 12 in this case. Therefore for this hypothesis of the resistance, the body is free to describe all the hyperbolas.

Corollary 6.

  1. If the resistance is put in the square ratio of the speeds, since m = 1, then the . From which therefore the more MT is varied, the exponent of the resistance at M = MT n +1 more also the medium is different. Corollary 7.

  2. In addition the time in which the element Mm is described, or ds , is equal to v 2ddy . g Therefore the time taken for the body to reach M, is as Whereby the time, in which the body reaches M from N, is as [Again, the section numbers have increased by 10 in the original. One can presume that Euler decided to remove some sections for some reason at a late stage.] Corollary 8. [p. 404]

  3. Therefore in the Apollonian hyperbola, in which n = 2, the time taken for the body to travel from N to M, is as NP − MP since NP.MP is constant, clearly equal to aα . β 2

Scholium.

  1. From these examples it is clear that a body cannot describe hyperbolas in a medium with a uniform resistance of this kind, since the exponent of the resistance varies too much, and which clearly is finally indefinitely large. On account of which Newton’s custom cannot be approved, in which he desired to substitute these hyperbolas in place of the true trajectories in mediums with uniform resistances. Furthermore, in a medium with the resistance following the square of the speeds, the exponent varies as the tangent MT, which on descending as far as the point N is to be returned as strongly varying according to the calculation. Also this other inconvenience must be understood for the speed, since on descending it becomes infinitely large, when however in a uniform medium it cannot increase beyond a given final speed. Besides clearly, it is not resolved for the hypothesis with the resistance proportional to the square of the speed, how the curve described can have a vertical asymptote, as happens in the case of a medium with the resistance in the simple ratio of the speed. For this resistance, even if the body has no force acting on it, the whole motion is along a finite line; which indeed becomes infinite if the resistance is put as the square of the speed. [p. 405] From which also it follows that the trajectory in this medium does not have an asymptote. It is not even certain that this curve has a hyperbolic asymptote [i. e. becomes hyperbolic for large values of the abscissa.]. Yet meanwhile it can have a vertical asymptote of the other parabolic kind of curve, which is determined from the quadrature of the curve by rectification of the given parabola. But leaving the hypothesis of uniform force we may go on to variable force, yet the direction of this is still everywhere parallel to itself. Indeed we do not investigate the curve described by a given curve with resistance, as this can now be done by Prop. 106 (870) ; but with a given curve and with one of the force, the resistance, or the speed given, we can determine the other quantities.

PROPOSITION 115. PROBLEM.

  1. Let some variable absolute force that acts downwards along MP be given (Fig.87); to determine the required resistance for this, in order that the body moves along the given curve AM.

SOLUTION.

Let AP = x, PM = y and the element of the arc AM = ds; while the force shall be P by which the body at M is acted upon, and the height corresponding to the speed at M is v, and the resistance at M is equal to R. With these put in place, we have (870) dv = − Pdy − Rds and 2 v = − 2Pds (871) on taking dx constant. [p. 406] From ddy this equation there arises : and thus it produces [recall that dsdds = dyddy as ddx = 0] Therefore both v and R can be expressed in terms of the given quantities P, x and y. Q.E.I.

Corollary 1.

  1. With the radius of osculation at M = r there is and hence

Corollary 2.

  1. If the law of the resistance is in the ratio of the square of the speed, and the exponent is equal to q, then R = qv and q = Rv . On which account we have

Corollary 3. y

  1. If the force P to gravity 1 is as y to f, then P = f and thus and

Corollary 4.

  1. If the curve AM is the circle of radius AC = a, then Py There is hence produced : v = 2 and the resistance [p. 407]EULER’S MECHANICA VOL. 1. Chapter Six (part b). page 596 Translated and annotated by Ian Bruce.

Example 1.

  1. Let the curve AM be a circle, the centre of which is C and the radius AC = a. Moreover the body is always attracted to the axis AC in the ratio of the distances, thus so y2 y that P = f ; then v = 2 f and thus the speed at M is as the applied line MP. Then the resistance R becomes equal to Moreover the speed at the point A is equal to 0 and the resistance, while the body ascends in the quadrant, is negative or the body is accelerated by the medium. The time, in which travels from A to M, is indefinitely large; for it becomes This which can also be understood by itself; for since the y speed at A is equal to 0 and here as the force f is acting while the force of the medium vanishes, the body must perpetually remain at A. Example 2.

  2. With the circle AM remaining, if the absolute force varied inversely as the distance f f PM or P = y , then v = 2 . Whereby the speed of the body is the same everywhere, or the body is carried around the periphery of the circle in an equal motion, and the time in which some arc AM is completed, is as the arc AM itself. But the resistance at M is equal to : Therefore the resistance is negative [p. 408] while the body ascends the quadrant ; moreover while it descends the following quadrant, the resistance becomes positive or is .PM . Therefore at the true resistance. Again from the resistance it is found that q = AC 2CP point A the force of the resistance moving forwards is indefinitely large in order that it is equal to the force acting.

Example 3.

  1. With the circle AM remaining, the force acting shall be as some power of the distance MP or P = yn ny n−2 ( a − x )dx . Therefore there is n ; from which [by (956)] dP = f fn produced from these, v = y n+1 and the resistance 2fn Thus the ratio is formed : Whereby, if n = –3 or the force P varies inversely as the cube of the distance MP, then the resistance R vanishes and the body from this force acting in a vacuum is able to move in the circle AM. Then if n + 3 is a positive number, the ascending resistance in the quadrant is negative. But if n + 3 is a negative number, then the resistance in this quadrant is positive. Example 4.
  2. If the curve AMB (Fig. 88) is of such a kind that the radius of osculation at M varies as the reciprocal of the applied line PM, [p. 409] that all elastic curves agree on, then Therefore we have and ds = a and on integrating 2a 2dx = ds( y 2 + b 2 ) , we have Moreover since − dxddy y 3 2 Hence v= And Now P( y 2 + b 2 ) . 4y

Thus R : P = −3dy : 2ds. Therefore as the body ascends for a long time, the resistance is negative, and it is positive when it descends.

PROPOSITION 116. PROBLEM.

  1. If AM is the given curve (Fig.87) and the resistance is given by quantities relating to the curve, to find the absolute force P always acting normal to the axis AC, which can be made, so that the body is free to move in this curve,

SOLUTION.

Let AP = x, PM = y and the element of the curve is equal to ds. Then let the resistance at M be equal to R, which is therefore given in terms of x, y and s ; the force sought is equal to P and the speed at M corresponds to the height v. With these in place, we have (871) (cit.). [p. 410] From this equation as dx is constant, it is found by integration that Moreover with v found, P can become known from the equation P = − 2vddy . Q.E.I. ds 2 Corollary 1. 962. Therefore we have P can then be determined by separating the quantities. Corollary 2. 963. Since a constant can be added as you wish, and this can be determined so that the body at A or at some other given place has a given speed.EULER’S MECHANICA VOL. 1. Chapter Six (part b). page 599 Translated and annotated by Ian Bruce. Corollary 3.

  1. If the resistance is put in proportion to the squares of the speeds, and the exponent of the resistance q, then we have :

Corollary 4.

  1. The time, in which the arc AM is completed, is given by ∫ dxv ; if the value of v is substituted into this expression, the time to traverse the arc AM is equal to Corollary 5. [p. 411]
  2. If the resistance is put to the force of gravity 1 as the tangent at M to the sub tangent or as ds to dx, there arises and the time to complete the arc AM is equal to But again, we have Example.
  3. Let the curve be a circle, the radius of which is AC = b, and y = ( 2bx − xx ) then From these we have With the resistance put equal to ds or R = b , then v = b 2 ( a − x ), and P = 2b 2 ( a − x ), and q = b ( a − x ). The time taken, in dx y y y3 y3 which the body arrives at M from A, is equal to 2 a − 2 ( a − x ) . If further, we let b = a, then we have :EULER’S MECHANICA VOL. 1. Chapter Six (part b). Translated and annotated by Ian Bruce. page 600 Therefore the speed of the body at the highest point of the circle is equal to zero, and the force P vanishes at that place. Moreover the body is not able to progress beyond this point, since otherwise the speed would become imaginary ; thence therefore it retreats to the point A, as it accelerates due to the negative resistance. Moreover while it arrives at A, since here the speed is infinitely great, with this motion of its own it describes the quadrant below AC, in which it is it is acted on by the negative force P upwards. [p. 412]

PROPOSITION 117. PROBLEM.

  1. If the medium is uniform and the resistance in the ratio of the square of the velocity, to determine the absolute force acting downwards which can be made, in order that the body in the medium with this resistance describes the given curve AM (Fig.87).

SOLUTION.

By putting AP = x, PM = y, with the element of the arc AM = ds, with the speed at M = v , with the exponent of the resisting medium equal to c and the absolute force equal to P, the resistance R = vc . With these in place there are the equations : (871). Whereby we have : and on integrating : Therefore with the value of v found, there is Therefore from the given curve, v and then P can be found. Q.E.I.EULER’S MECHANICA VOL. 1. Chapter Six (part b). page 601 Translated and annotated by Ian Bruce. Corollary 1. 969. The time in which the body can complete the arc AM or ∫ dsv , is equal to Therefore with the curve given or the equation between s and x, [p. 413] the time can also be obtained, even by quadrature. Therefore these curves are the most convenient to solve, for which there is a given equation between s and x. Corollary 2. 970. Since a can be taken as a constant of the integration taken as you please, it can be used to determine either the speed at a given point on the curve or the force acting. Corollary 3. 971. If the curve AM is concave towards the axis AP, then ddy is negative; therefore in these cases the force draws the body towards the axis AP. But if the curve is convex towards AP, since then ddy is positive, then the force P is negative, or the body is repelled from the axis AP. Example 1.

  1. Let the curve AM be a parabola having the axis standing normally to the right line AP, of such a kind that is described by a body projected obliquely in a vacuum from A, it is given by by = fx − x 2 and thus dy = b − 2 xdx , and ddy = − 2dx . With these put in b b place, the force acting is given by : fdx 2 From which it is understood, that the longer the motion is continued, [p. 414] there the force P decreases more. Moreover with c made indefinitely large, which is the case in a s vacuum, then e c = 1 and the force P = 4ba is thus constant. Example 2.

  2. If the curve AM is such that its equation is given by y = αx − βx 2 − γx3 , then dy = αdx − 2βxdx − 3γx 2dx and d 2 y = −2β dx 2 − 6γxdx 2 . Hence we have : At the point A we have:EULER’S MECHANICA VOL. 1. Chapter Six (part b). page 602 Translated and annotated by Ian Bruce. 2 Therefore the height corresponding to the initial speed at A is a( 1 + α ) ; which if called b, then a = b 2 , and α is the tangent of the angle below which the body is projected 1+α from A. Then from the value of dy , it is found that Therefore with the remaining terms ignores the arc length becomes : and which boundaries can also be pushed back if c is made very large. Whereby, when P is made approximately constant, truly equal to g, it is the case that And if this equation is assumed : y = αx − β x 2 − γx3 − δx 4 , it produces Therefore we have : [p. 415] Therefore this curve of the fourth order is very close to the trajectory in a very rare uniform medium, which has resistance in the ratio of the square of the velocity, and with a uniform force g acting downwards. Corollary 4.

  3. Since the air resistance is proportional to the square of the speeds, if in air composed of heavy globules [Euler had the belief at this time that air consisted of globules : see E002 De Sono in these translations] and the body is projected with a great force, then b et c are maximum quantities. Whereby it is necessary to take this equation for the projection of this body : since the curve differs little from the true trajectory.

Corollary 5.

  1. Let AMDB (Fig. 89) be this trajectory ; in which in order that the point B can be found, in which the projected body is incident on the horizontal AB, I put y = 0 and the equation becomes [p. 416] Therefore the distance thrown can become known from the given initial speed and the inclination.

Corollary 6.

  1. The maximum height of the trajectory D can be found by making dy = 0. Moreover it becomes : or And from this equation

Corollary 7.

  1. The longest throw, which is produced with the same initial speed the tangent α of the angle of inclination is found from this equation : b is produced if or from this : or thus more simply Corollary 8. [p. 417]

  2. If the sine of the angle, that the curve makes at A with the horizontal AC is set equal to ε with the whole sine equal to 1, then the equation can be written as : From which equation the value of ε elicited gives the direction for the longest throw. Moreover from this equation it is found as an approximation [original formula corrected by Paul St.] Corollary 9.

  3. Therefore the angle, which produces the longest throw, is a little less than half a right angle, which in a vacuum it makes satisfactorily. For if it becomes then it produces ε = 12 and thus the half right angle. But as here we have only 5b in the numerator, the distance becomes less for a short while.EULER’S MECHANICA VOL. 1. Chapter Six (part b). Translated and annotated by Ian Bruce. page 605 Corollary 10.

  4. If the body at A (Fig. 90) is projected horizontally with a speed a , then α = 0 and y becomes negative. On this account on putting AP = x and PM = y the kind of this trajectory can be expressed by this equation : Moreover for the curve AN, in which the body rises, is given by:

Corollary 11.

  1. If more than four terms are taken, this equation is produced for the curve AM : [p. 418] which terms are in agreement with the summable series, as it differs little from the true sum, if y is put equal to the sum of this series. Moreover, it becomes : Truly for the ascending arc AN it becomes :

Corollary 12.

  1. The time in which the arc AM is traversed, is equal to as g = 2vddy , the above equation becomes : ds 2 Therefore we have :EULER’S MECHANICA VOL. 1. Chapter Six (part b). Translated and annotated by Ian Bruce. page 606 And if b and c are expressed in scruples of Rhenish feet, then by (222) the time to traverse AM is equal to Scholium.
  2. Therefore from this reasoning we have determined approximately the trajectory described in air for projected bodies, which can be put in place without too much difficulty rather than the parabola, which generally is accustomed to be used. Indeed we can deduce this same equation from the true equation dsddy = cd 3 y (875) of these trajectories found above. [p. 419] But since there this reduction was omitted, and here we have preferred to present this material, particularly since here it is evident that the later terms of the equation are strongly decreasing. Finally in a like manner too, the curves for trajectories in mediums with other hypothesis of resistance can be conveniently found; but since these other hypotheses do not find a place in the world, we will not tarry here about finding them.

PROPOSITION 118. PROBLEM.

  1. To find the resistance at individual points M (Fig.87) for an absolute force acting downwards along MP, which can be put in place in order that a body can move on a given curve AM, and can move with a given speed at the individual points M.

SOLUTION.

Putting in place as before : AP = x, PM = y, with the element of arc AM = ds, and with the height corresponding to the speed at M equal to v, which are therefore all given. Then let the force acting downwards on the body at M be equal to P and the resistance is equal to R. With these in place, P is immediately found from the equation : (871). But the resistance R is found from the equation dv = 2vdyddy − Rds(cit.). Whereby ds 2 − Rds it is given by : [p. 420] If the law of the resistance is put in the square ratio of the speed, and the exponent of this is equal to q, then we have R = qv , from which there is produced :EULER’S MECHANICA VOL. 1. Chapter Six (part b). page 607 Translated and annotated by Ian Bruce. with dx taken as a constant. Q.E.I. Corollary 1. 985. Since dx is constant, then we have dyddy = dsdds . On account of this, we have And hence, Corollary 2. 986. If the body must be carried by a uniform motion along the curve AM, thus as v = b, it arises that Corollary 3. 987. Therefore in uniform motion, while the body ascends the curve AM, the resistance R is always negative or the motion of the body accelerates. But when the body descends again, the medium actually resists. Corollary 4. 2 2 ds , then ddy = − ds . 988. With the radius of osculation at M equal to r, since r = − dxddy rdx On account of this it is found that Corollary 5. [p. 421]

  1. Again putting v = b and dv = 0 there arises Therefore at the maximum point, since we make dy = 0 and ds = dx, then P = 2rb and the resistance vanishes there, if perhaps the curvature is infinitely large there or r = 0.EULER’S MECHANICA VOL. 1. Chapter Six (part b). Translated and annotated by Ian Bruce. page 608 Example.

  2. Let the curve AM be a circle, the centre of which is at C, which must be described with a uniform speed b . With the radius put as AC = a we have : From these is found the absolute force acting downwards, P = 2rb or which varies inversely as the distance PM. Therefore the resistance is equal to 2b( a − x ) . Therefore ay while the body ascends, the resistance is negative proportional to the reciprocal of the tangent of the arc AM. And if the resistance is proportional to the square of the speed, then the exponent of this is given by : Thus q is negative and equal to half the tangent of the arc AM. Moreover when the body approaches the horizontal AC, then the resistance R and q become positive, or the medium actually offers resistance.

PROPOSITION 119. PROBLEM.

  1. If the medium is uniform and offers resistance in the ratio of some multiple of the speed, and it is given besides that the body progresses uniformly along the horizontal AP at a constant speed (Fig.90), to find the force acting downwards and the curve that the body describes. [p. 422]

SOLUTION.

With AP = x, PM = y, with the arc AM = s, let the horizontal speed of the body while it is at M correspond to the height u; the height corresponding to the true speed of the body at M is equal to uds 2 = v . Again let the exponent of the resistance of dx 2 the medium be equal to c and the ratio law of the 2m-multiple of the speed is the resistance

With these in place we have : (871); for I put ddy in place of –ddy, since in our case y is falling. Then we have : (cit.). But it is the case that : Hence on account of dsdds = dyddy we have : Moreover u is given in terms of x and on account of this ds can be determined from x only; clearly it is given by : Thus it is permitted to find the equation for the curve to be found. With this found, likewise P can become known from the equation : P = 2uddy . Q.E.I. [p. 423] dx 2 Corollary 1. 992. Therefore the horizontal motion cannot be uniform; for on account of du = 0 , ds = 0, except in a vacuum when c =∝ , and where it is always by necessity uniform. The acceleration of the horizontal motion must also be much less also in a resisting medium; for then ds is either given a negative or imaginary value, which is absurd in each case. Therefore the horizontal motion must be one of deceleration, by which du is made negative. Corollary 2. 993. If the resistance is proportional to the speed itself, then we have m = 12 and the equation for the curve becomes du c + dx u = 0. Moreover which, since it does not contain s or y, cannot pertain to the curve. Moreover this equation itself determines the horizontal motion. It is evident in this hypothesis of the resistance that not any horizontal motion can be taken as you please, but by necessity that has to be accepted, which is determined by this equation.

Corollary 3.

  1. Moreover that horizontal motion agrees with the horizontal motion along AP in the same resisting medium, but with no force acting. From which it is understood, if there were any force acting on the body, then the direction of this force is everywhere in the downwards direction only [p. 424], in a medium with the resistance in the simple ratio of the speed always to be the same. Concerning which motion in this hypothesis of the resistance is similar to the motion in a vacuum, in which the horizontal motion is always uniform, however the downwards force acting may vary. Corollary 4.
  2. Therefore with an assumed value for this horizontal motion, the other equation P= 2uddy can determine the described curve, in which for P we are allowed to assume dx 2 some quantity. Therefore in this case of the hypothesis of the resistance, this problem is generally soluble : in order that the curve is found, as the body acted upon by some downwards force describes. Corollary 5.
  3. Moreover since − du c = dx , then 2 bc − 2 cu = x with the initial speed at A put u equal to b . Therefore the equation becomes : Hence for the curve described this equation is found : Corollary 6.
  4. Therefore all these curves have a vertical asymptote at the distance 2 bc from the vertex A. For x cannot be greater than 2 bc , and when the body moves horizontally [p. 425] is unable to progress beyond this boundary, also the body in the wide part of the curve cannot progress beyond this line.

Corollary 7.

  1. If in other hypotheses of the resistance too, the horizontal motion along the curve AM is taken to agree with the horizontal motion along the line AP with the same hypothesis of resistance, thus in order that : and this equation ds = dx is produced for the curve AM, taking P = 0. Therefore that agreement does not therefore find a place with other hypotheses of resistance.

Corollary 8.

  1. Therefore let the resistance vary as the square of the speed, or m = 1. Whereby ds = − cdu and by integrating, with the height b put to correspond to the speed at A, u s = c l ub . Therefore in this hypothesis of the resistance, the arc AM divided by 2c is equal to the logarithm of the horizontal speeds at A and M.

Corollary 9.

  1. Therefore in this hypothesis of the resistance, we have u = e − c b. From which there is produced : s , this becomes [p. 426] Or since ds = − cdu u But Hence we have : Consequently we have:

Corollary 10.

  1. If this equation is taken between u and x then With these substituted, there is produced in the medium with the resistance varying in the ratio of the square of the speed :

Moreover, this equation is obtained for the curve sought :

Corollary 11.

  1. Therefore in this case u 2n − 2 must be greater than f 2n or u greater than c2 Whereby, if the horizontal motion is able to become less than this quantity, the motion on the curve does not correspond to the total horizontal motion given. For if the curve is stretched further, then it could be that dy as well as P become imaginary.

Corollary 12.

  1. To avoid this inconvenience n must be made smaller than unity; therefore making −1 n − 1 = − k or n = 1 − k . With this in place, we have u k = b k − f k kx. Truly with AP the tangent to the curve at A here we have ds = dx, and there u = b. Hence we have Moreover again we obtain

Scholium.

  1. Therefore none of the hypotheses of this kind of the horizontal motion of trajectories in fluids can be integrated. Indeed whatever the value of k, the force at A, where we put u = b, is indefinitely great; then truly is decreases for ever. All these curves also have vertical tangents, where x = kc , which is an asymptote to the curve. Moreover we assign this problem to the end of the first part of the tract, in which we have put the direction of the force to be always parallel among themselves, and we progress to examination of centripetal forces that must soon be considered, in what manner the resisting medium disturbs the motion of bodies attracted to a fixed point.
The Curvilinear Motion Of A Free Point In A Resistive Medium
THE CURVILINEAR MOTION OF A FREE POINT IN A RESISTIVE MEDIUM

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