Book 2.3

With the centre C, and the rectangular asymptotes CADd and CH, describe an hyperbola BEe, and let AB, DE, de, be parallel to the asymptote CH. In the asymptote CD let A, G be given points; and if the time be expounded by the hyperbolic area ABED uniformly increasing, I say, that the velocity may be expressed by the length DF, whose reciprocal GD, together with the given line CG, compose the length CD increasing in a geometrical progression.

For let the areola DEed be the least given increment of the time, and Dd will be reciprocally as DE, and therefore directly as CD. Therefore the decrement of 1
GD, which (by Lem. II. Book II) is Dd
GD², will be also as CD
GD² or CG+GD
GD², that is, as 1
GD + CG
GD² . Therefore the time ABED uniformly increasing by the addition of the given particles EDde, it follows that 1
GD decreases in the same ratio with the velocity. For the decrement of the velocity is as the resistance, that is (by the supposition), as the sum of two quantities, whereof one is as the velocity, and the other as the square of the velocity; and the decrement of 1
GD is as the sum of the quantities 1
GD and CG
GD², whereof the first is 1
GD itself, and the last CG
GD² is as 1
GD² : therefore 1
GD is as the velocity, the decrements of both being analogous. And if the quantity GD reciprocally proportional to 1
GD, be augmented by the given quantity CG; the sum CD, the time ABED uniformly increasing, will increase in a geometrical progression.   Q.E.D.

Cor. 1. Therefore, if, having the points A and G given, the time be expounded by the hyperbolic area ABED, the velocity may be expounded by 1
GD the reciprocal of GD.

Cor. 2. And by taking GA to GD as the reciprocal of the velocity at the beginning to the reciprocal of the velocity at the end of any time ABED, the point G will be found. And that point being found the velocity may be found from any other time given.

In the asymptote CD let there be given the point R, and, erecting the perpendicular RS meeting the hyperbola in S, let the space described be expounded by the hyperbolic area RSED; and the velocity will be as the length GD, which, together with the given line CG, composes a length CD decreasing in a geometrical progression, while the space RSED increases in an arithmetical progression.

For, because the increment EDde of the space is given, the lineola Dd, which is the decrement of GD, will be reciprocally as ED, and therefore directly as CD; that is, as the sum of the same GD and the given length CG. But the decrement of the velocity, in a time reciprocally proportional thereto, in which the given particle of space DdeE is described, is as the resistance and the time conjunctly, that is, directly as the sum of two quantities, whereof one is as the velocity, the other as the square of the velocity, and inversely as the velocity; and therefore directly as the sum of two quantities, one of which is given, the other is as the velocity. Therefore the decrement both of the velocity and the line GD is as a given quantity and a decreasing quantity conjunctly; and, because the decrements are analogous, the decreasing quantities will always be analogous; viz., the velocity, and the line GD.   Q.E.D.

Cor. 1. If the velocity be expounded by the length GD, the space described will be as the hyperbolic area DESR.

Cor. 2. And if the point R be assumed any how, the point G will be found, by taking GR to GD as the velocity at the beginning to the velocity after any space RSED is described. The point G being given, the space is given from the given velocity: and the contrary.

Cor. 3. Whence since (by Prop. XI) the velocity is given from the given time, and (by this Prop.) the space is given from the given velocity; the space will be given from the given time: and the contrary.

Case 1. Suppose first that the body is ascending, and from the centre D, with any semi-diameter DB, describe a quadrant BETF of a circle, and through the end B of the semi-diameter DB draw the indefinite line BAP, parallel to the semi-diameter DF. In that line let there be given the point A, and take the segment AP proportional to the velocity. And since one part of the resistance is as the velocity, and another part as the square of the velocity, let the whole resistance be as AP² + 2BAP. Join DA, DP, cutting the circle in E and T, and let the gravity be expounded by DA², so that the gravity shall be to the resistance in P as DA² to AP² + 2BAP; and the time of the whole ascent will be as the sector EDT of the circle.

For draw DVQ, cutting off the moment PQ of the velocity AP, and the moment DTV of the sector DET answering to a given moment of time; and that decrement PQ of the velocity will be as the sum of the forces of gravity DA² and of resistance AP² + 2BAP, that is (by Prop. XII Book II, Elem.), as DP². Then the area DPQ, which is proportional to PQ, is as DP², and the area DTV, which is to the area DPQ as DT² to DP², is as the given quantity DT². Therefore the area EDT decreases uniformly according to the rate of the future time, by subduction of given particles DTV, and is therefore proportional to the time of the whole ascent.   Q.E.D.

Case 2. If the velocity in the ascent of the body be expounded by the length AP as before, and the resistance be made as AP² + 2BAP, and if the force of gravity be less than can be expressed by DA²; take BD of such a length, that AB² − BD² maybe proportional to the gravity, and let DF be perpendicular and equal to DB, and through the vertex F describe the hyperbola FTVE, whose conjugate semi-diameters are DB and DF, and which cuts DA in E, and DP, DQ in T and V; and the time of the whole ascent will be as the hyperbolic sector TDE.

For the decrement PQ of the velocity, produced in a given particle of time, is as the sum of the resistance AP² + 2BAP and of the gravity AB² − BD², that is, as BP² − BD². But the area DTV is to the area DPQ as DT² to DP²; and, therefore, if GT be drawn perpendicular to DF, as GT² or GD² − DF² to BD², and as GD² to BP², and, by division, as DF² to BP² − BD². Therefore since the area DPQ is as PQ, that is, as BP² − BD², the area DTV will be as the given quantity DF². Therefore the area EDT decreases uniformly in each of the equal particles of time, by the subduction of so many given particles DTV, and therefore is proportional to the time.   Q.E.D.

Case 3. Let AP be the velocity in the descent of the body, and AP² + 2BAP the force of resistance, and BD² − AB² the force of gravity, the angle DBA being a right one. And if with the centre D, and the principal vertex B, there be described a rectangular hyperbola BETV cutting DA, DP, and DQ produced in E, T, and V; the sector DET of this hyperbola will be as the whole time of descent.

For the increment PQ of the velocity, and the area DPQ proportional to it, is as the excess of the gravity above the resistance, that is, as BD² − AB² − 2BAxAP − AP² or BD² − BP². And the area DTV is to the area DPQ as DT² to DP²; and therefore as GT² or GD² − BD² to BP², and as GD² to BD², and, by division, as BD² to BD² − BP². Therefore since the area DPQ is as BD² − BP², the area DTV will be as the given quantity BD². Therefore the area EDT increases uniformly in the several equal particles of time by the addition of as many given particles DTV, and therefore is proportional to the time of the descent.   Q.E.D.

Cor. If with the centre D and the semi-diameter DA there be drawn through the vertex A an arc At similar to the arc ET, and similarly subtending the angle ADT, the velocity AP will be to the velocity which the body in the time EDT, in a non−resisting space, can lose in its ascent, or acquire in its descent, as the area of the triangle DAP to the area of the sector DAt; and therefore is given from the time given. For the velocity in a non-resisting medium is proportional to the time, and therefore to this sector; in a resisting medium, it is as the triangle; and in both mediums, where it is least, it approaches to the ratio of equality, as the sector and triangle do.

One may demonstrate also that case in the ascent of the body, where the force of gravity is less than can be expressed by DA² or AB² + BD², and greater than can be expressed by AB² − DB², and must be expressed by AB². But I hasten to other things.

Take AC (in these three figures) proportional to the gravity, and AK to the resistance; but take them on the same side of the point A, if the body is descending, otherwise on the contrary. Erect Ab, which make to DB as DB² to 4BAC: and to the rectangular asymptotes CK, CH, describe the hyperbola bN; and, erecting KN perpendicular to CK, the area AbNK will be augmented or diminished in an arithmetical progression, while the forces CK are taken in a geometrical progression. I say, therefore, that the distance of the body from its greatest altitude is as the excess of the area AbNK above the area DET.

For since AK is as the resistance, that is, as AP² x 2BAP; assume any given quantity Z, and put AK equal to AP²+2BAP
Z; then (by Lem. II of this Book) the moment KL of AK will be equal to 2APQ + 2BA x PQ
Z or 2BPQ
Z, and the moment KLON of the area AbNK will be equal to 2BPQ x LO
Z or BPQ x BD³
2Z x CK x AB.

Case 1. Now if the body ascends, and the gravity be as AB² + BD², BET being a circle, the line AC, which is proportional to the gravity, will be AB²+BD²
Z, and DP² or AP² + 2BAP + AB² + BD² will be AK x Z + AC x Z or CK x Z; and therefore the area DTV will be to the area DPQ as DT² or DB² to CK x Z.

Case 2. If the body ascends, and the gravity be as AB² − BD², the line AC will be AB²+BD²
Z, and DT² will be to DP² as DF² or DB² to BP² − BD² or AP² + 2BAP + AB² − BD², that is, to AK x Z + AC x Z or CK x Z. And therefore the area DTV will be to the area DPQ as DB² to CK x Z.

Case 3. And by the same reasoning, if the body descends, and therefore the gravity is as BD² - AB², and the line AC becomes equal to BD²-AB²
Z; the area DTV will be to the area DPQ, as DB² to CK x Z: as above.

Since, therefore, these areas are always in this ratio, if for the area DTV, by which the moment of the time, always equal to itself, is expressed, there be put any determinate rectangle, as BD x m, the area DPQ, that is, ½BD x PQ, will be to BD x m as CK x Z to BD². And thence PQ x BD³ becomes equal to 2BD x m x CK x Z, and the moment KLON of the area AbNK, found before, becomes BP x BD x m
AB. From the area DET subduct its moment DTV or BD x m, and there will remain AP x BD x m
AB. Therefore the difference of the moments, that is, the moment of the difference of the areas, is equal to AP x BD x m
AB; and therefore (because of the given quantity BD x m
AB ) as the velocity AP; that is, as the moment of the space which the body describes in its ascent or descent. And therefore the difference of the areas, and that space, increasing or decreasing by proportional moments, and beginning together or vanishing together, are proportional.   Q.E.D.

Cor. If the length, which arises by applying the area DET to the line BD, be called M; and another length V be taken in that ratio to the length M, which the line DA has to the line DE; the space which a body, in a resisting medium, describes in its whole ascent or descent, will be to the space which a body, in a non-resisting medium, falling from rest, can describe in the same time, as the difference of the aforesaid areas to BD x V²
AB; and therefore is given from the time given. For the space in a non-resisting medium is in a duplicate ratio of the time, or as V²; and, because BD and AB are given, as BD x V²
AB. This area is equal to the area DA² x BD x M²
DE² x AB and the moment of M is m; and therefore the moment ot this area is DA² x BD x 2M x m
DE² x AB. But this moment is to the moment of the difference of the aforesaid areas DET and AbNK, viz., to AB x BD x m
AB, as DA² x BD x M
DE² to ½BD x AP, or as DA²
DE² into DET to DAP; and, therefore, when the areas DET and DAP are least, in the ratio of equality. Therefore the area BD x V²
AB and the difference of the areas DET and AbNK, when all these areas are least, have equal moments; and are therefore equal. Therefore since the velocities, and therefore also the spaces in both mediums described together, in the beginning of the descent, or the end of the ascent, approach to equality, and therefore are then one to another as the area BD x V²
AB, and the difference of the areas DET and AbNK; and moreover since the space, in a non-resisting medium, is perpetually as BD x V²
AB, and the space, in a resisting medium, is perpetually as the difference of the areas DET and AbNK; it necessarily follows, that the spaces, in both mediums, described in any equal times, are one to another as that area BD x V²
AB, and the difference of the areas DET and AbNK.   Q.E.D.

The resistance of spherical bodies in fluids arises partly from the tenacity, partly from the attrition, and partly from the density of the medium. And that part of the resistance which arises from the density of the fluid is, as I said, in a duplicate ratio of the velocity; the other part, which arises from the tenacity of the fluid, is uniform, or as the moment of the time; and, therefore, we might now proceed to the motion of bodies, which are resisted partly by an uniform force, or in the ratio of the moments of the time, and partly in the duplicate ratio of the velocity. But it is sufficient to have cleared the way to this speculation in Prop. VIII and IX foregoing, and their Corollaries. For in those Propositions, instead of the uniform resistance made to an ascending body arising from its gravity, one may substitute the uniform resistance which arises from the tenacity of the medium, when the body moves by its vis insita alone; and when the body ascends in a right line, add this uniform resistance to the force of gravity, and subduct it when the body descends in a right line. One might also go on to the motion of bodies which are resisted in part uniformly, in part in the ratio of the velocity, and in part in the duplicate ratio of the same velocity. And I have opened a way to this in Prop. XIII and XIV foregoing, in which the uniform resistance arising from the tenacity of the medium may be substituted for the force of gravity, or be compounded with it as before. But I hasten to other things.