232 Of the SOLIDS 



a rectangle, of which the breadth is 2 BC, and the height 2 BL, 

 the attraction of that plane, or of the thin folid, having that 

 plane for its bafe, and n, for its thicknefs, is 4 n . (p. Now, <p, 

 which is thus proportional to the attraction of the plane, is aL 

 fo proportional to the fpherical furface, or the angular fpace, 

 fubtended by the plane at the centre A. 



For fuppofe PSQ^(Fig. 12.) and OQ^to be two quadrants of 

 great circles of a fphere, cutting one another at right angles in 

 Qj let QS = E, and QR = z. Through S, and O the pole of 

 PSQ^ draw the great circle OST, and through P and R, the 

 great circle PTR, interfering OS in T. The fpherical quadri- 

 lateral SQRT, is that which the re&angle CL (Fig. 11.) would 

 fubtend, if the fphere had its centre at A, if the point Q^was 

 in the line AB, and the circle PQ^ in the vertical plane ABL. 



Now, in the fpherical triangle PST, right angled at S, cof T 

 rzcofPS X fin SPT = fin QS x fin QR = fin E X fin z. But 

 this is alfo the value of fin <p, and therefore <p is the complement 

 of the angle T, or <p n 00 — T. 



But the area of the triangle PQR, in which both Q^and R 

 are right angles, is equal to the rectangle under the arch QR, 

 which meafures the angle QPR, and the radius of the fphere. 

 Alfo the area SPT = arc . (S + T + P — 180 ) r ; that is, be- 

 caufe S is a right angle, — arc . (T + P — 90) X r zm 

 arc . (T-f-QR — 90) X r ; and taking this away from the triangle 

 PQR, there remains the area QSTR = arc . (QR — T — QR 

 + 90 °) X r — (90 — T) r — $Xr. The arch <p t therefore, mul- 

 tiplied into the radius, is equal to the fpherical quadrilateral 

 QSTR, fubtended by the redangle BD. 



This proposition is evidently applicable to all rectangles 

 whatfoever. For when the point B, where the perpendicular 

 from A meets the plane of the rectangle, falls anywhere, as in 

 Fig. 1 5. then it may be fhewn of each of the four rectangles 



BD. 



