| D | dA | ||
| t = | = | ||
| V | aV |
If A = 2000 meters, d = 65 millimeters, and a = 25 centimeters, and if the plane is traveling 200 kilometers per hour, the time interval must be—
| .065 × 2000 × 3600 | |
| = 9.4 seconds | |
| .25 × 200,000 |
At 1000 meters altitude the interval will be half this, and so on in proportion. If the pictures are taken with a 50 centimeter focus camera, and are hence to be viewed at 50 centimeters convergence distance instead of at 25, the time will again be halved. These relations are clearly shown in the diagram (Fig. [156]). Here the left-hand portion shows how to find the stereoscopic base line at each altitude for each focal length; while the right-hand portion shows how to translate this into time interval for any plane velocity. The Burchall slide rule (Fig. [130]) shows another way to arrange these data in form for rapid calculation.
Fig. 156.—Chart for calculating intervals between exposures for stereoscopic pictures.
Plates used for stereoscopic negatives should be at least twice as long as the ocular separation, if correct relief is desired, and the full size of the stereoscope field is to be utilized. This relation follows at once if we consider that we wish to cut from each negative a rectangle 65 millimeters wide, and that the image of the target has shifted 65 millimeters between exposures. If the plate is larger than this there is opportunity to select the view, or to pick several. If the plate is smaller the elements of the stereogram must be narrow strips. This, however, holds only for contact prints.
The ordinary English practice in making stereo negatives is to take successive pictures with an overlap of 60 to 75 per cent. This practice is probably dictated by the 4 × 5 inch plate, since 60 per cent. overlap on 4 inches means a separation of just over an inch and a half instead of 2¾, but it leaves 2½ inches of picture common to the two negatives. With ¾ overlap the common portion is 3 inches, which permits of cutting 2¾ inch prints, and allows some latitude for irregular motion of the plane or for chance error in calculation of intervals. Data on the basis of ¾ overlaps for a 4-inch plate are shown in connection with Fig. [155] which shows in diagrammatic form the variation of exposure interval with height, together with other points of interest.
Elevation Possible to Detect in Stereoscopic Views.—Can the actual difference in elevation be discovered by the use of stereoscopic views? An approximate idea may be obtained from the following considerations: Suppose we have two small point-like objects, one above the other, such as a street lamp globe and the base of the lamp pillar. In a view taken from directly overhead these will be superposed, and so will not be capable of separation. But, as the point of view is shifted sideways, the two objects separate, until a point is reached where they can just be distinguished as double. When this condition holds for either picture of the stereoscopic pair it will be possible to obtain stereoscopic relief.
Now the separation which can just be distinguished is commonly assumed to be one minute of arc. This angle corresponds to about 1
3400 the distance from the eye to the object. If the object is assumed at a distance a from the face, and on a line with one of the eyes, which are separated by the distance d, then (all angles being small) the object must be of height a
d times the horizontal distance which corresponds to one minute. For 25 centimeters' viewing-distance this quantity is about 4, so that the least perceptible elevation is 4
3400 or about 1
900. The stereogram having been made under conditions giving correct relief, this fraction is also the fraction of the altitude of the plane when the photograph was taken which may be detected. An object as high as a man (6 feet) should be visible as a projection in a stereoscopic view taken at 6 × 900 = 5400 feet. This relation—1
900—holds (irrespective of the focal length of the lens), as long as the conditions for correct relief are maintained.