Hence, in order to find the binocular centres D, D′, D″, D‴, &c., at any altitude, E, E′, &c., we have only to join EG, E′G, &c., and the points of intersection D, D′, &c., will be the binocular centres, and the lines DC, D′C, &c., drawn to C, will be the real lengths and inclinations of the united images of the lines AC, BC. When GO is greater than GC there is obviously some angle A, or E″GM, at which D″C is perpendicular to GC. This takes place when GC Cos. A = ————. GO When O coincides with C, the images CD, CD′, &c., will have the same positions and magnitudes as the chords of the altitudes A of the eyes above the plane GC. In this case the raised or united images will just reach the perpendicular when the eye is in the plane GCM, for since GC = GO, Cos. A = 1 and A = 0. When the eye at any position, E″ for example, sees the points A and B united at D″, it sees also the whole lines AC, BC forming the image D″C. The binocular centre must, therefore, run rapidly along the line D″C; that is, the inclination of the optic axes must gradually diminish till the binocular centre reaches C, when all strain is removed. The vision of the image D″C, however, is carried on so rapidly that the binocular centre returns to D″ without the eye being sensible of the removal and resumption of the strain which is required in maintaining a view of the united image D″C. If we now suppose AB to diminish, the binocular centre will advance towards G, and the length and inclination of the united images DC, D′C, &c., will diminish also, and _vice versa_. If the distance RL (Fig. 26) between the eyes diminishes, the binocular centre will retire towards E, and the length and inclination of the images will increase. Hence persons with eyes more or less distant will see the united images in different places and of different sizes, though the quantities A and AB be invariable. While the eyes at E″ are running along the lines AC, BC, let us suppose them to rest upon the points _ab_ equidistant from C. Join _ab_, and from the point _g_, where _ab_ intersects GC, draw the line _g_E″, and find the point _d″_ from the formula _g_E″ × _ab_ _gd″_ = ———————————-. _ab_ + RL Hence the two points _a_, _b_ will be united at _d″_, and when the angle E″GC is such that the line joining D and C is perpendicular to GC, the line joining _d″_C will also be perpendicular to GC, the loci of the points D″_d″_, &c., will be in that perpendicular, and the image DC, seen by successive movements of the binocular centre from D″ to C, will be a straight line.