VOL. XXXII.] PHILOSOPHICAL TRANSACTIONS. Q2g 



inconvenient method of finding the refracting power of any fluid, or indeed of 

 any transparent substance, if it be formed into a spherical or cylindrical figure. 

 For this purpose therefore I have found, that in the latter determination of the 

 first proposition, if the sine of the angle under cmg, fig. 4, be denoted by a, 

 the tangent of the complement of this angle to a right one be denoted by b, 

 and the secant of this complement by c; the root of this equation z^ — 3aaz 

 = 2aa X 2c — a will exceed the sine of the angle under fma, that is the sine 

 of the angle of incidence, by the sine of the angle under cmg ; and the sine 

 of the angle under fmo, which is double the refracted angle, will be the root 

 of this equation x^ + 3aax = 4aab; this angle being acute, when the tangent 

 of the angle under cmg is less than half the radius, or when the angle itself is 

 less than 26° 33' 54" iV", and when this tangent is more than half the radius, 

 the angle under omf is obtuse. 



The roots of these cubic equations are found by seeking the first of two 

 mean proportionals, between each of the versed sines appertaining to the arches 

 CG, AG, and the sine of those arches, counting from the versed sines; for the 

 sum of these two mean proportionals is the root of the former equation, and 

 the difference between them the root of the latter ; as may be collected from 

 Cardan's rules. 



And hence also, if the first and last of the 5 mean proportionals, between 

 the sine and cosine of half the angle under cmg be found, twice the sum of 

 the squares of these mean proportionals applied to the radius, exceeds the sine 

 of the angle of incidence, by the sine of the angle under cmg ; and twice the 

 difference of the squares of the same mean proportionals, applied to the radius, 

 is equal to the sine of double the refracted angle. Further, this double of the 

 refracted angle exceeds the angle of incidence, by the angle under cmg. 



In the latter determination of the 2d proposition, draw ky, fig. 8, and ay 

 being parallel to mn, the angle under cky will be equal to twice the angle 

 under cmn, that is equal to the complement of half the distance of the exterior 

 rainbow from the point opposite to the sun. Then putting a for the radius ak, 

 and b for the sine of the angle under cky, the sine of the angle under akv 

 will be the root of this equation y'' + 4by^ — 8aaby -f 4aabb = O. But the 

 angle of incidence and refraction may also be found as follows: 



Let two mean proportionals between the radius and the sine of the angle 

 under cky be found; then take the angle, whose cosine is the first of these 

 mean proportionals, counting from the radius; and also the angle whose sine, 

 together with the second mean proportional, shall be to the radius, as the cosine 

 of the angle under cky, to the sine of the angle before found. The sum of 



