Proposition 15

PROPOSITION 15. Problem.

124. If as before there is an infinite family of similar curves AM, AM, etc. (Fig.18) and the straight line DE in the position given, to find that curve AMN, upon which the body arrives at the line DE in the shortest possible time from A.

Solution

By the preceding proposition, with some curve CMM described cutting the isochronous arcs AM, the tangent GMH is drawn parallel to the given line DE. It is evident that the body is to arrive in the shortest time along the curve AM, which touches the line GMH at the point of contact M, since all other points of the line GMH fall beyond the curve CMM and thus a longer time is needed for the body to reach that line.

Since all the curves [such as CMM] cutting isochronous arcs from the curves AM, AM, are similar to each other, (121), one is taken from these , which the line DE touches;

The point of contact is to be at the point N, in which the line AM drawn through the previous point of contact M crosses the line DE. From this it then follows from the nature of the similarity of the curves CMM with respect to the point A, that also it follows, as the arc AMN is similar to the arc AM and crossed the line DE at the same angle that the curve AM crosses the line GH. Whereby, [p. 59] since the body arrives in the shortest time along AM to GH, it is necessary that it also arrives in the shortest time on the curve AMN to the line DE. Q.E.I.

Corollary 1.

125. From this it is understood, that if the line DE is horizontal, with the descent along the vertical AC , then the body arrives in the shortest possible time, on account of the horizontal tangent to the curve CMM at C ; which is indeed evident by itself.

Corollary 2.

126. If therefore the curves AM, AM are cycloids, as we put in example 3 of the preceding proposition, the body on that cycloid arrives at the line DE the fastest which crosses this line at N at right angles, since the angle that any cycloid makes with the curve CM is right.

Corollary 3.

127. If therefore the line DE is vertical or parallel to AC, the portion of the cycloid AMM is half the cycloid. Whereby the horizontal motion on half the cycloid is the fastest.

Corollary 4

128. If the curves AM, AM are straight lines drawn from the point A to the given line position DE, the body on that line AM (Fig. 19) arrives at DE, which is the chord of the circle passing through A and having the centre in the vertical line AB, and having the tangent line DE (112) [p. 60]

Corollary 5

130. If therefore the angle DEA is n degrees, the angle BAM is 902+ n degrees and the angle AMG is 902− n . Or with AGH drawn the horizontal, and with the angle DGH bisected by the line GF, then the sought line AM is parallel to GF.

Corollary 6

131. Whereby if the line DE is vertical, the body arrives at that line the fastest by descending on the line inclined at 45 degrees to the horizontal [on setting n = 0]. Therefore a body inclined at this angle to the horizontal advances the quickest.

Scholium

132. In a like manner it can be found too, which of an infinite number of similar curves AM, AM (Fig. 18), a body can arrive at a given curve by descending the fastest.

For if the line GMH should be any kind of curve touching the curve CMM at M, the body on this curve AM arrives the quickest at the curve GMH, if indeed the whole of the curve GMH is placed beyond the curve CMM. Also in the same way it might be possible to be determined, if the curves AM, AM are not similar, how the above body can arrive the fastest at a given line GH. Indeed from the infinity of curves CMM the isochronous arc from those being cut is the one sought, which touches the given curve GMH, and it is on that curve AM, which passes through that point of contact, it is that point which is sought. But since in these cases generally the curves CMM are to be found with difficulty and it is much more difficult to determine that curve which is a tangent to the given line, [p. 61] we have restricted the question to similar curves only.

PROPOSITION 16. Theorem.

132. The times of the descent, by which a body placed on the curves traverses the curves AM and Am, etc. (Fig.20) similar and similarly from the point A, are in the ratio of the square root of the homologous sides.

Demonstration.

Since the curves AM, Am are similar, AM:Am, AP:Ap and PM:pm are in a given ratio, clearly that, which the homologous sides hold; let the ratio of the homologous sides be N:n. Because the speed at M is to the speed at m as AP to Ap , the speeds at M and m are in the square root ratio of the homologous sides. Now similar elements are taken from M and m, clearly holding the ratio N to n, are the times in which these two elements are traversed, in the ratio composed from the direction of the elements, i. e. N to n, and to the reciprocals of the speeds, i. e. N : n . From which it follows that the times, in which the homologous elements of the curves AM, Am are traversed, are in the square root ratio of the homologous sides. Whereby, since this ratio is constant, the times in which the whole curves AM and Am are traversed, keep this same ratio. Q.E.D. [p. 62]

Corollary 1.

133. Therefore the times, in which similar and similarly circular arcs put in place for the descent, are in the square root ratio of the radii.

Corollary 2.

134. Therefore pendulums, which describe similar arcs, complete oscillations in times which are in the square root ratio maintained by the lengths of the pendulums.

Corollary 3.

135. The same ratio of the times is kept in place, if the pendulum bodies do not describe circular orbits, but other curves, provided these are similar to each other and they complete similar arcs.

Scholium.

136. Moreover in these all the forces acting we put to be uniform and to be pulling downwards, even if we have disregarded this condition. For we have put this hypothesis in place to be acted on previously, as we are now about to progress to others.