Fig. 26.
Let ac, bc, [Fig. 26], be two lines meeting at c, the plane passing through them being the plane of the paper, and let them be viewed by the eyes successively placed at e‴, e″, e′, and e, at different heights in a plane, gmn, perpendicular to the plane of the paper. Let r be the right eye, and l the left eye, and when at e‴, let them be strained so as to unite the points a, b. The united image of these points will be seen at the binocular centre d‴, and the united lines ac, bc, will have the position d‴c. In like manner, when the eye descends to e″, e′, e, the united image d‴c will rise and diminish, taking the positions d″c, d′c, dc, till it disappears on the line cm, when the eyes reach m. If the eye deviates from the vertical plane gmn, the united image will also deviate from it, and is always in a plane passing through the common axis of the two eyes and the line gm.
If at any altitude em, the eye advances towards acb in the line eg, the binocular centre d will also advance towards acb in the line eg, and the image dc will rise, and become shorter as its extremity d moves along dg, and, after passing the perpendicular to ge, it will increase in length. If the eye, on the other hand, recedes from acb in the line ge, the binocular centre d will also recede, and the image dc will descend to the plane cm, and increase in length.
Fig. 27.
The preceding diagram is, for the purpose of illustration, drawn in a sort of perspective, and therefore does not give the true positions and lengths of the united images. This defect, however, is remedied in [Fig. 27], where e, e′, e″, e‴ is the middle point between the two eyes, the plane gmn being, as before, perpendicular to the plane passing through acb. Now, as the distance of the eye from g is supposed to be the same, and as ab is invariable as well as the distance between the eyes, the distance of the binocular centres oO, d, d′, d″, d‴, p from g, will also be invariable, and lie in a circle odp, whose centre is g, and whose radius is go, the point o being determined by the formula
| go = gd = | gm × ab | . |
| ab + rl |
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.