Lord Oxmantown on a new Reflecting Telescope. 29 



To compute the respective lengths of the curves composing 

 a compound speculum, such as has been described, so that the 

 aberration of the whole speculum may be equally divided be- 

 tween them. 



Let E, Fig. 2, be the centre of the surface. Let a ray pro- 

 ceed from Q and intersect the axis in V. Let q be the geo- 

 metrical focus of rays proceeding from the point Q. 

 Let q represent E Q, Let q x represent E v, 



q' E q, 6 R E A 



■f E F, v ver sin 6 



Then considering q x a function of ver. sin % v being of course 

 — in each coefficient, we have 



which by proper substitutions becomes 



The aberration being q x — <f is evidently represented by this 

 series without its first term, which expressed geometrically 

 amounts to 



QE 2 AN QE 5 AN 2 

 Q F 2 ' T" + QF 3 ' 4~EF + C * 

 which, when the rays are parallel, becomes 

 AN AN 2 



2 + 4 E F C ' 

 It is evident that the first term of this series will afford a 

 sufficiently near approximation for compound specula ; if we 

 therefore represent the ver. sines of the arcs DO, DP, DQ. 

 Fig. 3, by AN, AN/ AN," the problem becomes, Given AN =z 

 AN' — AN = AN" — AN/ the length of the arc DQ and its 

 radius, to find the lengths of the arcs DO and DP. 



Example. — Let a be the arc DQ — 3 inches, r its radius 



= 48 inches, w the seconds in a, then f a? = 206265 - = 

 206265 ~ g =3 12891% which in degrees amounts to 3° 35' 



whose v. s. == .0019550. 



I of which = .0006516, corresponding to arc, 2° 4'= 7440" 

 = 1.77 inches. 



* Coddingtons Optics, + I*ardners Trigonometry. 



