36 PHOTOGRAPHIC FACTS AND FORMULAS 
that in the second case. Example: D = 62.7 in.; d= 72.2 in.; 
R =4and r= 5; then 72.2 — 62.7 = 9.5; then (5 X 5+ 1) 
~+5=—26/5 and (4X4+1)+4=17/4; then 26/5 — 
17/4 = 19/20; and 9.5 ~ 19/20 = 10 in., the equivalent focus 
of the lens. 
Another Method:—This method can be carried out in any 
room. First focus on a very distant object, and mark the 
extension of the camera; then focus on a comparatively near 
object, that is to say, one in the length of an ordinary room, 
again mark the position of the camera, and call the distance 
between the two marks x. Again focus on a still nearer 
object, again measure the distance beyond the infinity mark, 
and call this distance y. Let B be the distance between the 
two objects, then the focus f = V~Bry+y—-x. Suppose 
the distance of one object is 144 in. and that of the other 
96 in., then B = 144— 96 = 48. And suppose the extension 
of the camera beyond the infinity or distant mark for 
the object at 12 ft. was 1 in. and at 8 ft., 1% in., then 
f=V48XK1XK14=1%4%—1= V72=-%=v14=12. 
In this method, if the lens is moved in focussing, the distance 
between the object at the two positions, or B, must be ascer- 
tained by measuring from some part of the camera front. 
If, on the other hand, the focussing screen is moved and the 
lens remains stationary, we need only measure the distance 
between the two positions of the object. 
Another Method.—Set up a foot rule on a wall, and shift 
the camera until an image is obtained on the ground glass 
that is exactly the same size as the rule; naturally, how much 
of the rule is included depends on the size of the ground 
glass. Then measure the distance between the rule and the 
image, divide this by 4, and the result will be approximately 
the equivalent focus. 
Or set up the foot rule as in the previous method, and 
