Table of Contents

PROPOSITION 99. PROBLEM

To determine the inclination of the plane, in which the elements Mm and mμ (Fig.

described by a body have been placed, relative to the fixed plane APQ and to find the line of intersection of the two planes. [p. 344]

SOLUTION

In the plane, of which we seek the inclination, the three points M, m and μ are given; therefore in this plane a certain line is placed of which the points pass through both planes. [This is the element mM ; the plane APQ can be considered as horizontal while the eye looks down and sees the plane containing QM, the z coordinate, rising vertically, as in the previous proposition. The point μ arises as previously due to the actions of the forces M and N, and the lines mM and mμ define a plane, which is tangential to the surface on which the body moves.]

Whereby if the line mM is produced, then it meets the line qQ [both in the plane APQ] produced in S, while the point S then lies in the plane Mmμ as well as in the fixed plane APQ; therefore the line formed by the intersection of the two planes passes through S. Therefore with the lengths remaining as before: AP = x, PQ = y and QM = z and with the elements of the abscissa Pp and pπ equal to each other, then qm − QM : Qq = QM : QS and hence : For the position of the line QS is known from the angle PQS, the sine of which is equal to …

In addition, the point n is also situated in the plane Mmμ ; on account of this the line passing through n and μ or drawn parallel to this passes through M extant in the same plane. Moreover this line crosses the plane APQ in the point R of the line QP produced, and QR is known from this ratio rn − ρμ : rρ = QM : QR ; hence we have …

[For rn − rμ = nμ = −ddz ; rρ = −ddy , QM = z, giving QR as shown, etc.]

For Qq : Pp = QS : PT with ST drawn perpendicular to AP. From which there arises PT = zdx . dz Furthermore, Pp : ( pq − PQ ) = PT : ( PQ + ST ) and thus zdy PQ + ST = dz and ST = zdy − y. dz

The line RS produced cuts the axis AP in O and hence PR − ST : PT = PR : PO , from which it is found that:

and [p. 345]

Again from these, the tangent of the angle POR is equal to [Note that the numerator of PO can be factored to give dx( zddy − yddz ) , which cancels with the numerator of PR above to give the simpler expression shown]; thus the position of the intersection RO of the plane Mmμ with the fixed plane APQ is known.[On the annotated diagram, the distance AO along the fixed axis AP is known, and the angle θ to AP has been found.] Moreover, the mutual inclination of these planes is found by sending the perpendicular QV from Q to RS ; for then the tangent of the angle ..

of inclination is equal to QV . Indeed [as the right triangles RVQ and RPO are similar] this becomes : and thus from which the angle of the inclination between the two planes Mmμ and APQ is determined. Q.E.I

Corollary 1

If the angle POR is always the same, then and the tangent of the angle is equal to α. This equation on integration gives αdx + dy + βdz = 0 and αx + y + βz = f .

[Note that β is defined in the next cor.] From which equation it is known that the whole orbit described by the body lies in the same plane in this case.

Corollary 2

If indeed it is the case that αx + y + β z = f , then αdx + dy + βdz = 0 and ddy + βddz = 0. Hence we have : ddy

[on substituting β = − ddz into the equation for PO above, and simplifying,] and because − dy − βdz = αdx , then and likewise AO is constant.

Corollary 3

Again with the angle POR remaining constant or αx + y + β z = f , then the tangent of the angle of inclination of the planes Mmμ and APQ is equal to : Whereby this angle itself is constant.

[This is readily seen on substituting β = − ddz into and using the result of Cor. 2.]

Corollary 4

For neither can the point of intersection O be put as invariable, unless likewise the orbit described by the body becomes a plane. For let AO = f and putting x − f = t [= OP] and dx = dt , then: [from the expression set equal to t, we have] and hence

This is multiplied by t, from which it is found that : Which equation can be integrated, in account of dt being constant; for it becomes

This divided by tt and integrated gives : y αz = t + β or y = az + βx − βf . t Which is seen to be a plane surface. [The first result follows by setting ..

= tdy − ydt and likewise tdztddz = tdz − zdt and noting that ddt = 0; while the tdy − ydt − zdt second case follows directly from tdy − ydt α ( tdz − zdt )

on integrating by parts, where t2 t2 integrals cancel.]

Corollary 5

But if the tangent of the angle of inclination of the planes Mmμ and APQ is constant, an equation of the kind αx + y + βz = f is not produced. And elsewhere it has been shown that the orbit described by the body is then not by necessity a plane.

Corollary 6

Whereby, with the curve described by the body not being planar, then neither the point O nor the angle POR can be take as invariable. Moreover if these are variables, then no more can the angle between the planes Mmμ and APQ be considered as constant. [p. 347]

Corollary 7

The line of intersection RO, which is called the line of the nodes in astronomy, if it does not have a constant position, turns about the point S. For the element mMS has been put in the plane of the elements Mm from the preceding argument. Whereby the intersection of RO and the preceding line cross over each other at S.

Corollary 8

Therefore this point S is at that place, where [ AP – TP = ] …

From which the position of the point S is known.

Corollary 9

If the point S is put invariable, then xdz − zdx = adz and zdy − ydz = bdz , hence it is found that x − a = α z and y − b = β z . Therefore in this case the orbit described by the body is not only planar, but also it is a straight line, since the projection of Qqρ gives a straight line in the x-y plane, and since y − b = βz , Mmμ is also a straight line in the y-z plane.

Scholion

Now from the principles established, which are concerned with the non planar motion of a body on a surface, the description itself can be divided into two parts as before with regard to motion in a plane. In the first of these we instruct how to find the curve described by a body from given forces, and in the second truly it is shown, if the curve is given which the body describes, what kind of forces are required to do this. Here only the curve itself needs to be given, or likewise also the speed of the body at individual points.

Proposition 99