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

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

We see that the maximum allowable $ \mathrm{(d p / d z)}_{\mathrm{q}} $ (1-B) for a 30' diameter, .001" thick polyethylene balloon is $ 2 5 6 \times 1 0^{-3} \mathrm{p s i / f t} $ . Dividing by (1-B) we have the maximum allowable:

$$
\begin{array}{l} \left(\mathrm {d p} / \mathrm {d z}\right) _ {a} = \frac {2 5 6 \times 1 0 ^ {- 3}}{1 - . 1 3 8} = . 3 0 0 \times 1 0 \mathrm {p s i} / \mathrm {f t} ^ {- 3} \\ = 2 0. 7 \times 1 0 ^ {- 3} \mathrm {m b} / \mathrm {f t} \\ \end{array}
$$

This is comparable to an altitude of 18,300 ft. or a gross buoyancy of 450 lb., the maximum allowable inflation of a 30' diameter, .001" thick polyethylene balloon from the standpoint of pressure distribution.

In order to determine mathematically the point of failure due to pressure distribution we may use equations (3) and (5) and their derivatives:

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

$$
\Delta p _ {z} = \frac {4 S f t}{2 \left(D \Delta Z - \Delta Z ^ {2}\right) ^ {1 / 2}}
$$

at the point of tangency of these curves (T in Figure 26):

$$
\Delta p _ {T 3} = \Delta p _ {T 5} \quad \text {a n d} \quad \left(\frac {d p}{d z}\right) _ {T 3} = \left(\frac {d p}{d z}\right) _ {T 5}
$$

in equation (5), making $ \frac{4311}{2}=K $ and in equation (3), making $ (dp/dZ)_{0}(1-B)=m $ , the slope of the line $ \Delta p_{z}=\Delta Z\cdot m $ we have:

(5a)

$$
\Delta p _ {z} = \frac {K}{(D \Delta z - \Delta z ^ {2}) ^ {1 / 2}}
$$

and:

(3a)

$$
\Delta p _ {z} = m \Delta z
$$

differentiating with respect to z :

(5b)

$$
\frac {d p}{d z} = - \frac {K}{2} \frac {(D - 2 \Delta Z)}{(D \Delta Z - \Delta Z ^ {2})}
$$

$$
\frac {d p}{d z} = m
$$

(3b)