186 Prof. H. Hennessy on Ronayne's Cubes. 



equally inclined planes terminating at the corners of the 

 cube. These knife-edges must manifestly be equal to the side 

 of the cube ; and as the sliding cube on each flange has its 

 side perpendicular to the flange, the two flanges must have 

 their edges a little distant from the diagonal to which each is 

 parallel. The interval secures junctions of the flange with 

 the two prisms. The thickness t of flange downwards must 

 be also secured in order that the flange holds its place. The 

 inclination of the face of the flange will depend upon these 

 two quantities. 



The relations hetween x the distance of the knife-edge 

 from the corner of the cube, t the thickness of the flange at 

 the points of junction with the prisms, and the angle of 

 inclination of the inner face of the flange to the face of the 

 cube can he easily found. As the sliding cube must have one 

 of its sides always perpendicular to the face of the flange, the 

 following equation must subsist : — 



a 



= {a-t) cos0+[pV2-(,£ — ±a)] sin 0. 



As the least thickness of the flange parallel to the long side 

 of the prism may be represented by 



f= (%— l a) tan 0, 

 the above becomes 



(a — 2t) cos 6 +2?^ 2 sin 6=a, . . . . (1) 



a = acosd+[p\/2 — 2(<^» — Jaj] sin 0, 



a — a cos = a(\/2 — 2x) sin 0, ... (2) 



or 



remembering that^ = a I 1 ^=1. 



From (2) the relation between x and gives for 0, 



ginf >_ 2a{a^2-2x) _ 

 3a 2 + 4x 2 -4ax*/2 



From (1) the value of which makes t a maximum can be 

 found by the usual methods, 



2t = a{^2-\) tan0-a (sec 0-1), 

 9 (U _ g(*/2 — 1) asinfl 

 dO " — ^0" cos 2 6' 



dH __ a cos 3 0-2(^2-1) cos sin 0- 2a sin 2 cos 

 2 oW~ cos 4 



