Finding the orbits from the focus given

PROPOSITION 53 PROBLEM 35

Granting the quadratures of curvilinear figures, find the forces with which bodies moving in given curve can perform their oscillations in equal times.

Let the body T oscillate in any curve line STRQ,, whose axis is AR passing through the centre of force C.

Draw TX any place of the body T, and in that tangent arc TR. The length of that arc is TX touching TX take TY known from the axis is AR that curve in equal to the common methods used191 OF NATURAL PHILOSOPHY. SEC. X. Y From the point for the quadratures of figures. draw the right line perpendicular to the tangent. YZ CT Draw meeting that perpendicular in Z, and the to the right line centripetal force will be proportional TZ.

For if the force with which the body attracted is TZ from T body in the direction of the length of the thread C be expressed by the right line taken proportional to it, that force will be resolved into two forces TY, YZ, of which drawing the towards YZ PT, whereas the other docs not at all change its motion force directly accelerates or retards its mction in the curve ; TY Wherefore since that force is as the space to be described STRQ. the acceler TR, ations or retardations of the body in describing two proportional parts (u and greater arid a less) of two oscillations, will be always as those parts, But bodies which therefore will cause those parts to be described together. continually describe together parts proportional to the wholes, will describe the wholes together also.

1. Hence if the body T, hanging by a rectilinear thread from the centre A, describe the circular arc STRQ,, COR. AT and in the mean time be acted on by any force tending downwards with parallel directions, which is to the uni form force of gravity as the arc TR to its sine TN, the For because times of the several oscillations will be equal. are parallel, the triangles ATN, TZ, fore AR TZ will ZTY be to AT as TY to TN ; that is, are similar if the ; and there uniform force of gravity be expressed by the given length AT, the force TZ. by which the oscillations become isochronous, will be to the force of gravity AT, as the sine of arc to to the that is arc. equal TY TR

Corollary 2

TN And therefore in clocks, if forces were impressed by some ma the pendulum which preserves the motion, and so compounded chine upon with the force of gravity that the whole force tending downwards should be always as a line produced by applying the rectangle under the arc TR and the radius AR to the sine TN, all the oscillations will become isochronous.