Project Blue Book File on Tremonton Film, Utah, 1952

Text

A machine read this text from the scan. When the text and the image disagree, the image is right.

Appendix I

Calculation of Angular Velocity of Single Aerial Object:

Pass No. 2: 48 spot positions are plotted, representing an elapsed time interval of 47 frames. At 16 frames per second, this gives:

$$
\frac{47}{16} = 3.0 \text{ seconds elapsed time}
$$

The measured distance on the screen is 15.1 inches. This gives approximately 15.1/3.0 or 5.0 inches per second for image velocity on the screen.

Pass No. 3: 44 spot positions are plotted, for 43 frames of elapsed time. This time is $\frac{43}{16} = 2.7$ seconds.

The distance travelled is 15.7 inches. This gives $\frac{15.7}{2.7} = 5.8$ inches per second

on the screen. Average of these 2 values is 5.4 inches per second; with an estimated precision of ±1.5 per cent.

Conversion to Angular Velocity:

Distance from projector lens to screen = 100 inches

Focal length of projector lens = 2 inches

Enlargement of film on screen = 50 times

$$
\text{Image velocity on film} = \frac{5.4}{50} = .11 \text{ inches per second}
$$

Focal length of camera taking lens = 3 inches

It is adequately precise to divide the image velocity by the focal length to get the angular velocity in radians per second, then to convert to degrees. We have:

$$
\text{Angular velocity} = \frac{.11 \times 57.3}{3} = 2.1 \text{ degrees per second}
$$

The precision of this value is also conservatively set at ±1.5% or ±0.3 degrees per second.

DECLASSIFIED Authority NND 923007