that are neither all parallel among term, or to one common point, that the ezera ege feveral lines. (Pl. 13, Fig. 1.)

Let the four right lines ABC. pofition; the first cutting the second n à. the fourth in C; and fuppote a mapema scribed that may be fimilar to the mom

whofe angle f, equal to the given age 3 zur right line ABC; and the other age ! L other given angles, G, H, I, nay

95.

arly

is

EV

BD.

ter

luce

på

ven

BD, CE, refpectively. Join F

and the

as the letters CBDC, and the poem F

as the letters ACEA. Complete me gam cles, and let P be the centre of the i

centre of the fecond FTH. Jan

line PQ, and in it take QR in ne te mall a

has to AB. But QR is to be a

point Q, that the order of me ees?

fame as of the letters A, B, C; and

the interval RF defcribe a fura zm

circle FVI in c. Join Fe catang ne

fecond in b. Draw aG,

be made fimilar to the igre E fghi will be that which was

For let the two iri cres

in K; join PK, QE, RI. 15. ii a to L. The angles FaÃ, FL Piz

are the halves of the anges 21 KA L

tres, and therefore equain 21 2

those angles. Wherefore the igre Y2

ZJ

1

s

HI to FH. But AK, KM, AL, LN are to be drawn towards thofe fides of the lines AD, AK, AL, that the letters CAKMC, ALKA, DALND may be carried round in the fame order as the letters FGHIF; and draw MN meeting the right line CE in i. Make the angle iEP equal to the angle IGF, and let PE be to Ei as FG to GI; and through P draw PQf that may with the right line ADE contain an angle PQE equal to the angle FIG, and may meet the right line AB in f, and join fi. But PE and PQ are to be drawn towards thofe fides of the lines CE, PE, that the circular order of the letters PEIP and PEQP may be the fame as of the letters FGHIF; and if upon the line fi, in the fame order of letters, and fimilar to the trapezium FGHI, a trapezium fghi is conftructed, and a trajectory given in kind is circumfcribed about it, the problem will be folved.

So far concerning the finding of the orbits. It remains that we determine the motions of bodies in the orbits fo found. SECTION VI.

How the motions are to be found in given orbits.

PROPOSITION XXX. PROBLEM XXII. To find at any affigned time the place of a body moving in a given parabolic trajectory.

Let S (Pl. 14, Fig. 1) be the focus, and A the principal vertex of the parabola; and fuppofe 4AS x M equal to the parabolic area to be cut off APS, which either was defcribed by the radius SP, fince the body's departure from the vertex, or is to be defcribed thereby before its arrival there. Now the quantity of that area to be cut off is known from the time which is proportional to it. Bifect AS in G, and erect the perpendicular GH equal to 3M, and a circle defcribed about the centre H, with the interval HS, will cut the parabola in the place P required. For letting fall PO perpendicular on the axis, and drawing PH, there will be AG2+GH2 (=HP2 =AO – AG|* + PO− GH|*) = AO + PO* 2GAO 2GH + PO + AG2 + GH2. Whence 2GH × PO ( = AO2 + PO2GAO) AO2 + PO. For AO write AO X PO2

--

; then dividing all the terms by 3PO, and multiplying

them by 2AS, we fhall have GH × AS (=÷AO × PO

+AS × PO=

AO + 3AS
6

× PO=

4AO - 3SO
6

× PO= to the area APO-SPO)| = to the area APS. But GH was 3M, and therefore 4GH × AS is 4AS × M. Wherefore the area cut off APS is equal to the area that was to be cut off 4AS x M. Q.E.D.

COR. 1. Hence GH is to AS as the time in which the body described the arc AP to the time in which the body defcribed the arc between the vertex A and the perpendicular erected from the focus S upon the axis.

COR. 2. And fuppofing à circle ASP perpetually to pafs through the moving body P, the velocity of the point H is to the velocity which the body had in the vertex A as 3 to 8; and therefore in the fame ratio is the line GH to the right line which the body, in the time of its moving from A to P, would describe with that velocity which it had in the vertex A. COR. 3. Hence alfo, on the other hand, the time may be found in which the body has defcribed any affigned arc AP. Join AP, and on its middle point érect a perpendicular meeting the right line GH in H.

LEMMA XXVIII.

There is no oval figure whofe area, cut off by right lines at pleasure, can be universally found by means of equations of any number of finite terms and dimenfions.

Suppofe that within the oval any point is given, about which as a pole a right line is perpetually revolving with an uniform motion, while in that right line a moveable point going out from the pole moves always forward with a velocity proportional to the fquare of that right line within the oval, By this motion that point will describe a spiral with infinite circumgyrations. Now if a portion of the area of the oval cut off by that right line could be found by a finite equation, the distance of the point from the pole, which is proportional to this area, might be found by the fame equation, and therefore all the points of the spiral might be found by a finite equation also; and therefore the interfection of a right line given in pofition with the fpiral might alfo be found by a VOL. I.

H