480 , PHILOSOPHICAL TRANSACTIONS. [aNNO lOgi-S. 



circular base, and covered by a hemispherical dome, having 4 windows in its 

 circumference, with a circular aperture at the top, so combined, that the re- 

 mainder of the hemispherical surface of the dome was quadrable, or equal to a 

 rectilineal figure ; also in the cylindric part of the temple was a door, absolutely 

 quadrable. It is proposed to determine by what geometrical means the architect 

 could accomplish the construction of this monument. This problem being 

 proposed to Dr. Wallis, he resolves it on the following principles, viz. on the 

 relation between a triangle and a portion of a circle in the form of a lune, as 

 determined by Hippocrates of Chios ; and then by the relation between cir- 

 cular spaces and the curve surfaces of spheres and cylinders, as demonstrated by 

 Archimedes. 



Thus, if the semicircle ABD, fig. IQ, pi. 11, be divided into the two 

 quadrants, A CD, BCD, the chord AD of the quandrantal be drawn, and 

 this be bisected in H by the radius CE ; and on the centre H there be described 

 the semicircle AD F : then it will be (because the square of AD is equal to 

 half the square of AB), that the semicircle ADF is half the semicircle 

 ABD, and therefore equal to the quadrant ACD. And hence, by taking away 

 the common segment ADE, the remaining lune AEDF is equal to the triangle 

 ADC ; and consequently 4 such lunes equal to 4 such triangles, that is, to the 

 whole square A DBG inscribed in the circle. 



Again, by what has been demonstrated by Archimedes, the superficies of a 

 sphere, is equal to 4 great circles of the same sphere ; and therefore the surface 

 of the hemisphere equal to 4 such semicircles, and so the surface of a quadrant 

 of the hemisphere equal to one such semicircle. 



Now let the circle ADBG be the base of the hemispherical curve surface; 

 of which surface let P be the pole, fig. 20, the axis CP perpendicular to the 

 plane of the base, and of which also DPA is a quadrant, which is bisected by 

 the plane EPC passing through the axis. — Now, for the easier calculation, put 

 r for the radius of the circle, its diameter d = 2r, its circumference c, and a 

 proposed arc a. And putting the quadrantal arc DEA = a =: ^c; then the 

 semicircle ABD is ^ ar ^ -^cr ; the triangle ADC = -Jr^ = -i-dr ; and, 

 deducting the triangle, the remainder of the semicircle -^icr — -i^dr ; and this 

 taking away from DPA (the quadrant of the hemispherical curve surface, which 

 is equal to the semicircle ABD) the remainder will be equal to the proposed tri- 

 angle ADC. 



There are various ways of performing this ; one is as follows : since the seg- 

 ments of spherical superficies, cut by parallel planes, have the same ratio as 

 the segments of the axis; if there be taken in the axis CP, fig. 21, as the 

 semicircle -^cr, is to the same diminished by the triangle, -^cr ■- -^dr, that 



