The Roswell Report: Fact vs. Fiction in the New Mexico Desert

Read from the scan by GLM-OCR; expect the odd misread word.

Comparing rate of leakage at 40,000 feet with leakage at sea level:

$$
\frac {L _ {4 0}}{L _ {0}} = \sqrt {\frac {2 7}{1 1 2} \cdot \frac {1 8 8}{1 0 1 3} \cdot \frac {2 8 8}{2 1 8}} = 0. 2 4 3
$$

Comparing rate of leakage at 100,000 feet with leakage at sea level:

$$
\frac {L _ {1 0 0}}{L _ {0}} = \sqrt {\frac {2 7}{1 8 8 0} \cdot \frac {1 0 . 9}{1 0 1 3} \cdot \frac {2 8 8}{2 1 8}} = 0. 0 4 4
$$

Therefore, if leakage of a full balloon at sea level is known, it is possible to compute theoretical leakage at any altitude. However, if it is not possible to completely inflate a balloon on the ground in order to make a sea level test (if lift would be great enough to rupture balloon or load lines), a method of comparing full balloon leakage with partially full balloon leakage must be found.

Let us assume that it is possible to obtain results of a leakage test for a balloon inflated to a volume $ \frac{1}{x} $ of fullballoon volume. Again starting with equation (1):

$$
Q = C _ {d} A \sqrt {2 g h}
$$

We see that in this case the total area of openings, A is not constant but is a function of volume. Therefore, we have:

$$
Q \propto A \sqrt {h}
$$

We have shown that:

$$
h = \frac {\Delta p}{d _ {g}} \cdot 1 4 4 = \frac {\Delta z \left(\frac {d p}{d z}\right) _ {\mathrm {a i r} (1 - B)}}{d _ {g}} \cdot 1 4 4
$$

Since we are comparing partially inflated balloon leakage at sea level with full balloon leakage at sea level the variable in the above expression is $ \Delta z $ . This is graphically illustrated in Figure 24.