Projecticle

Proposition 9 Theorem 7

Supposing what is above demonstrated, I say, that if the tangents of the angles of the sector of a circle, and of an hyperbola, be taken proportional to the velocities, the radius being of a fit magnitude, all the time of the ascent to the highest place will be as the sector of the circle, and all the time of descending from the highest place as the sector of the hyperbola.

To the right line AC, which expresses the force of gravity, let AD be drawn perpendicular and equal. From the centre D with the semi-diameter AD describe as well the quadrant AtE of a circle, as the rectangular hyperbola AVZ, whose axis is AK, principal vertex A, and asymptote DC. Let Dp, DP be drawn; and the circular sector AtD will be as all the time of the ascent to the highest place; and the hyperbolic sector ATD as all the time of descent from the highest place; if so be that the tangents Ap, AP of those sectors be as the velocities.

Case 1. Draw Dvq cutting off the moments or least particles tDv and qDp, described in the same time, of the sector ADt and of the triangle ADp. Since those particles (because of the common angle D) are in a duplicate ratio of the sides, the particle tDv will be as .. that is (because tD is given), as …

But pD² is AD² + Ap², that is, AD² + AD × Ak, or AD × Ck.

qDp is ½AD × pq.

Therefore tDv, the particle of the sector, is as … that is, as the least decrement pq of the velocity directly, and the force Ck which diminishes the velocity, inversely.

Therefore as the particle of time answering to the decrement of the velocity. And, by composition, the sum of all the particles tDv in the sector ADt will be as the sum of the particles of time answering to each of the lost particles pq of the decreasing velocity Ap, till that velocity, being diminished into nothing, vanishes; that is, the whole sector ADt is as the whole time of ascent to the highest place. Q.E.D.

Case 2

Draw DQV cutting off the least particles TDV and PDQ of the sector DAV, and of the triangle DAQ; and these particles will be to each other as DT² to DP², that is (if TX and AP are parallel), as DX² to DA² or TX² to AP²; and, by division, as DX² - TX² to DA² - AP² . But, from the nature of the hyperbola, DX² - TX² is AD²; and, by the supposition, AP² is AD × AK.

Therefore the particles are to each other as AD² to AD² - AD × AK.

That is, as AD to AD - AK or AC to CK: and therefore the particle TDV of the sector is …

Therefore (because AC and AD are given) as … that is, as the increment of the velocity directly, and as the force generating the increment inversely; and therefore as the particle of the time answering to the increment.

By composition, the sum of the particles of time, in which all the particles PQ of the velocity AP are generated, will be as the sum of the particles of the sector ATD; that is, the whole time will be as the whole sector. Q.E.D.

Corollary 1

Hence if AB be equal to a fourth part of AC, the space which a body will describe by falling in any time will be to the space which the body could describe, by moving uniformly on in the same time with its greatest velocity AC, as the area ABNK, which expresses the space described in falling to the area ATD, which expresses the time.

For since AC is to AP as AP to AK, then (by Cor. 1, Lem. II, of this Book) LK is to PQ as 2AK to AP, that is, as 2AP to AC, and thence LK is to ½PQ as AP to ¼AC or AB; and KN is to AC or AD as AB to CK.

Therefore, ex æquo, LKNO to DPQ as AP to CK. But DPQ was to DTV as CK to AC. Therefore, ex æquo, LKNO is to DTV as AP to AC; that is, as the velocity of the falling body to the greatest velocity which the body by falling can acquire.

Since, therefore, the moments LKNO and DTV of the areas ABNK and ATD are as the velocities, all the parts of those areas generated in the same time will be as the spaces described in the same time; and therefore the whole areas ABNK and ADT, generated from the beginning, will be as the whole spaces described from the beginning of the descent. Q.E.D.

Corollary 2

The same is true also of the space described in the ascent. That is to say, that all that space is to the space described in the same time, with the uniform velocity AC, as the area ABnk is to the sector ADt.

Corollary 3

The velocity of the body, falling in the time ATD, is to the velocity which it would acquire in the same time in a non-resisting space, as the triangle APD to the hyperbolic sector ATD. For the velocity in a non-resisting medium would be as the time ATD, and in a resisting medium is as AP, that is, as the triangle APD. And those velocities, at the beginning of the descent, are equal among themselves, as well as those areas ATD, APD.

Corollary 4

By the same argument, the velocity in the ascent is to the velocity with which the body in the same time, in a non-resisting space, would lose all its motion of ascent, as the triangle ApD to the circular sector AtD; or as the right line Ap to the arc At.

Corollary 5

Therefore the time in which a body, by falling in a resisting medium, would acquire the velocity AP, is to the time in which it would acquire its greatest velocity AC, by falling in a non-resisting space, as the sector ADT to the triangle ADC.

The time in which it would lose its velocity Ap, by ascending in a resisting medium, is to the time in which it would lose the same velocity by ascending in a non-resisting space, as the arc At if to its tangent Ap.

Corollary 6

Hence from the given time there is given the space described in the ascent or descent. For the greatest velocity of a body descending in infinitum is given (by Corol. 2 and 3, Theor. VI, of this Book); and thence the time is given in which a body would acquire that velocity by falling in a non-resisting space.

Taking the sector ADT or ADt to the triangle ADC in the ratio of the given time to the time just now found, there will be given both the velocity AP or Ap, and the area ABNK or ABnk, which is to the sector ADT, or ADt, as the space sought to the space which would, in the given time, be uniformly described with that greatest velocity found just before.

Corollary 7

By going backward, from the given space of ascent or descent ABnk or ABNK, there will be given the time ADt or ADT.