| 2000 × .18 × .8 × 3600 | ||
| t = | = 10.3 seconds | |
| .5 × 200,000 |
The principle of overlapping map exposures is shown in the accompanying diagram (Fig. [129]), together with data calculated as above for a 4 × 5 inch plate.
It is particularly to be noted that it is the ground speed of the plane that is used. This may be calculated by knowing the air speed and the wind velocity and direction. Fig. [136] shows the method of doing this graphically. First an arrow is drawn representing the direction it is desired to fly. Next a second arrow is drawn of length to represent the wind velocity. This must be inclined toward the first arrow in the direction of the wind, and its head is to touch the head of the first arrow. Then with the farther end of this second arrow as a center, describe a circle of such a length as to represent the air speed of the plane, in the same units as the wind velocity. Connect the point where this circle cuts the arrow of flight direction to the center of the circle by a straight line. This line constitutes the air speed arrow, giving the direction it is necessary to fly, at the given air speed, to make the course desired. The length of the flight direction arrow between its head and its point of intersection with the air speed arrow gives the ground speed.
Fig. 136.—Diagram showing method of calculating ground speed from air speed and wind velocity.
When the wind is ahead or astern this calculation reduces to the simple subtraction or addition of the wind velocity to the air speed of the plane. Whenever possible, mapping should be done up and down the wind (Fig. [137]). If the plane is “crabbing,” the above calculations for overlap are only valid if the camera can be turned normal to the direction of travel over the ground. If the camera cannot be so turned the corners of the successive pictures overlap instead of their sides, with quite unsatisfactory results (Fig. [138]).
Calculation of the distance apart of the parallel flights necessary to make a map of any width is done by the use of a formula similar to the longitudinal overlap formula above, distance figuring instead of time. Using the same symbols, and denoting the distance by D, we have—
| Ad(1 - f) | |
| D = | |
| a |
With the same figures as before, but substituting 24 centimeters for the plate dimension, this relation gives—
| 2000 × .24 × .8 | ||
| D = | = 768 meters | |
| .5 |