The Mathematical Principles of Natural Philosophy (1729)/Appendix

​ I.o find the force whereby a sphere (${\displaystyle AdBg}$,) on the diameter ${\displaystyle AB}$, attracts the body ${\displaystyle P}$. (Pl. 19. Fig. 1.)

Let ${\displaystyle SA=SB=r,PS=d,PE=x,PB=a=d+r,PA=\alpha =d-r}$; Theref. ${\displaystyle a\alpha =dd-rr}$; also ${\displaystyle a+a=2d,a-\alpha =2r}$; Therefore ${\displaystyle aa-\alpha \alpha =4dr}$: And ${\displaystyle SE=d-x,AE=x-a,BE=a-x}$.

Now the force whereby the circle, whose radius is ${\displaystyle Ed}$, attracts the body ${\displaystyle P}$, is as ${\displaystyle 1-{\frac {PE}{Pd}}}$ (by Cor. I. Prop. 90.)

​And . Also : Th. . Therefore is the fluxion of the attractive force of the sphere on the body ${\displaystyle P}$, or the ordinate of a curve whose area represents that force. But the fluent of ${\displaystyle {\dot {x}}}$ is ${\displaystyle x}$; and the fluent of is (by Tab. 1. Form 4. Cas. 2 Quadr. of Curv.) Therefore is the general expression of the area of the curve. Now let ${\displaystyle x=a}$, then area = Also let ${\displaystyle x=\alpha }$, then area = And the force whereby the sphere attracts the body ${\displaystyle P}$ is as (${\displaystyle A-B}$ or as .

​2. The force whereby the spheroid ${\displaystyle ADBG}$, attracts the body ${\displaystyle P}$, may, in the same manner, be found thus. Let ${\displaystyle SC=c}$,

The force of a circle whose radius is ${\displaystyle ED}$, to attract ${\displaystyle P}$, is as ${\displaystyle 1-{\frac {PE}{PD}}}$, (by Cor. 1. Prop. 90.) Now (by the Conics;) and Therefore , or) is the fluxion of the attractive force of the spheroid on the body ${\displaystyle P}$, or the ordinate of a curve whose area is the measure of that force. Now the fluent of ${\displaystyle {\dot {x}}}$ is ${\displaystyle x}$; and (by Cas. 2. Form 8. Tab. 2. Quad. Cur.) the fluent of is ​. Therefore is the general expression for the area of the curve. But is an ordinate to a conic section, whose abscissa is x; and's, σ, the areas NMB, NKA, adjacent to the ordinates BM, AK: Put D=-0.

Let x = a, or PE=PB=BM; then v=6, daa-ann-ds, or PD =4. And let x=α, or PE-PA-AK; then v«, or PD =PB = BM, and the area =at •cc + dd―rr daḍ aɑɑ 2 do =PA=AK, and the area = a + =B. cc+dd-rr

And the attractive force of the spheroid on P, is as (A-B±a-α But 2d = (aα=) BM +AK, therefore 2dr= trapezium ABMK; and D= (5-6=) area AKRMB; therefore D-2dr = mixtilinear area KRMLK C; consequently 2dr-D=-C; therefore 2dx 2 dr -D=-2dC; therefore the attractive force of the spheroid on P, is as 2 AS × S C -2 PSX KRMX 2 SC-PS-AS 2rcc 2dC ccdd—rr Confequently, the attractive force of the spheroid upon the body P will be to the attractive force of a sphere, whole diameter is AB, upon the same body P, as ASXSC²- RSK RMK rcc-dC CC-\-derr AS 3 to 83 or SC² +PS² —AS to 3PS

​

For let it be proposed to find the vertex of the cone, a frustum of which has the describ'd property.

Let ${\displaystyle CFGB}$ be the frustum, and ${\displaystyle S}$ the vertex required. (Pl. 19. Fig. 2.)

Now conceive the medium to consist of particles which strike the surface of a body (moving in it) in a direction opposite to that of the motion; then the resistance will be the force which is made up of the efficacy of the forces of all the strokes.

In any line ${\displaystyle Pp}$, parallel to the axis of the cone, and meeting its surface in ${\displaystyle p}$, take ${\displaystyle pm}$ of a given length, for the space describ'd by each point of the cone in a given time: Draw ${\displaystyle mq}$ perpendicular to the side ${\displaystyle (CF)}$ of the cone, and ${\displaystyle qn}$ perpendicular to ${\displaystyle pm}$.

Therefore the line ${\displaystyle pm}$ will represent the velocity, or force, with which a particle of the medium strikes the surface of the cone obliquely in ${\displaystyle p}$.

But the force ${\displaystyle mp}$ is equivalent to two forces, the one ${\displaystyle (mq)}$ perpendicular, the other ${\displaystyle (pq)}$ parallel to the side of the cone; which last is therefore of no effect.

And the perpendicular force ${\displaystyle mq}$ is equivalent to two forces, the one ${\displaystyle (mn)}$ parallel to the axis of the cone, the other ${\displaystyle (qn)}$ perpendicular to it; which also is destroy'd by the contrary action of another particle on the opposite side of the cone.

There remains only the force ${\displaystyle mn}$, which has any effect in resisting or moving the cone in the direction of its axis.

Therefore the whole force of a single particle, or the effect of the perpendicular stroke of a particle, upon the base of a circumscribing cylinder, is to the effect of the oblique stroke upon the surface of the cone (in ${\displaystyle p}$) as ${\displaystyle mp}$ to ${\displaystyle mn}$, or as to , or as to .

​Now the number of particles striking in a parallel direction on any surface, is as the area of a plane figure perpendicular to that direction, and that would just receive those strokes.

Therefore, the number of particles striking against the frustum, that is, against the surfaces describ'd by the rotation of ${\displaystyle FD}$, and ${\displaystyle CF}$, each particle with the forces ${\displaystyle mp}$, and ${\displaystyle mn}$ respectively, is as the circle describ'd by (${\displaystyle FD}$ or) ${\displaystyle OH}$, and the annulus described by ${\displaystyle CH}$, that is, as to .

But the whole force of the medium in resisting, is the sum of the forces of the several particles.

Therefore, the resistance of the medium, or the whole efficacy of the force of all the strokes against the end ${\displaystyle FG}$ of the frustum, is to the resistance against the convex surface thereof, as ( to or as to or as) to . Theref. the whole resistance of the medium against the frustum may be represented by which call ${\displaystyle z}$; that is, (putting ${\displaystyle OC=r,OD=2a,OS=y}$, then , and ; therefore : Consequently ​${\displaystyle 2r^{2}yy-4ar^{2}y=2yzy+y^{2}+y^{2}z+r^{2}z}$; But ${\displaystyle z}$ is a minimum; therefore ${\displaystyle rry-2arr=zy}$; consequently . Hence ${\displaystyle yy-2ay=rr}$; and making ${\displaystyle OQ=QD=a}$; then . On the right-line ${\displaystyle BC}$, (Pl. 19. Fig 3.) suppose the parallelograms ${\displaystyle BG\gamma b,MN\nu m}$, of the least breadth, to be erected, whose hights ${\displaystyle BG,MN}$, their distance ${\displaystyle Mb}$, and half the sum of their bases , are given: Let half the difference of the bases be called ${\displaystyle x}$: Let ${\displaystyle G}$ and ${\displaystyle N}$ be points in the curve ${\displaystyle GND}$; and producing ${\displaystyle b\gamma }$, and ${\displaystyle m\nu }$ to ${\displaystyle g}$ and ${\displaystyle n}$, (so that ${\displaystyle \gamma g=\nu n=b}$.) the points ${\displaystyle g}$ and ${\displaystyle n}$ may also be in the same curve. Now if the figure ${\displaystyle CDNGB}$, revolving about the axis ${\displaystyle BC}$, generates a solid, and that solid moves forwards in a rare and elastic medium from ${\displaystyle C}$ towards ${\displaystyle B}$, (the position of the right-line ${\displaystyle BC}$ remaining the same;) then will the sum of the resistances against the surfaces generated by the lineolæ ${\displaystyle Gg,Nn}$, be the least possible, when is to as ${\displaystyle BG\times Bb}$ to ${\displaystyle MN\times Mm}$. For the force of a particle on ${\displaystyle Gg}$ and ${\displaystyle Nn}$, to move them in the direction ${\displaystyle BC}$, is as and ; and the number of particles that strike in the same time on the surfaces generated by ${\displaystyle Gg}$ and ${\displaystyle Nn}$, are as (the annuli describ'd by ${\displaystyle g\gamma }$ and ${\displaystyle n\nu }$, that is, as ${\displaystyle BG\times g\gamma }$ and ${\displaystyle MN\times n\nu }$, or as) ${\displaystyle BG}$ and ${\displaystyle MN}$; therefore the resistances against those surfaces are as to , that is (putting ${\displaystyle y}$ for , and ${\displaystyle z}$ for ,) as ${\displaystyle {\frac {BG}{y}}}$ to ${\displaystyle {\frac {MN}{Z}}}$.

​But the sum of these resistances ${\displaystyle \left({\frac {BG}{Y}}+{\frac {MN}{Z}}\right)}$ is a minimum. Therefore ${\displaystyle -BGx{\frac {y}{yy}}-MNx{\frac {z}{zz}}=0}$ ,or ${\displaystyle MNx{\frac {z}{zz}}=-BGx{\frac {y}{yy}}}$: But Failed to parse (syntax error): {\displaystyle y=\left ( \overline{Gg}^2 = \overline{Bb}^2 + \overline{yg}^2= \right aa +2ax + xx +bb ) and z=\left ( \overline{Nn}^2 = \overline{Mm}^2 + \overline {vn}^2 = \right ) aa + 2ax + xx + bb;}

therefore ${\displaystyle y=2xx-2ax}$, and ${\displaystyle z=2ax+2xx}$

consequently ; or Therefore

Consequently, that the sum of the resistances against the surfaces generated by the lineola ${\displaystyle Gg}$ and ${\displaystyle Nn}$, may be the

least possible, must be to as ${\displaystyle GBb}$ to ${\displaystyle NMm}$. Wherefore, if ${\displaystyle \gamma g}$ be made equal to ${\displaystyle \gamma G}$, so that the angle ${\displaystyle \gamma Gg}$ may be 45°, and the angle ${\displaystyle BGg}$ 135°; also , and ; then ; and since ${\displaystyle GR}$ is parallel to ${\displaystyle Nn}$, and ${\displaystyle BG,BR}$ parallel to ${\displaystyle n\nu ,N\nu }$; also ; it follows that ; therefore ; also . Consequently . Therefore is to as ${\displaystyle GR}$ to ${\displaystyle MN}$.