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

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

9 = the acceleration of gravity (ft./sec. $ ^{2} $ )

$ A_{0} $ area of the opening (ft.2)

h = head of fluid causing flow (ft.)

since $ h=\frac{1 4 4 \Delta p}{d_{q}} $ , we have:

$$
\Delta p = \frac {d g}{2 8 8 g} \left(\frac {1}{C _ {d} A _ {a}} \cdot \frac {d V}{d t}\right) ^ {2} \quad p s i
$$

where $ d_{g} $ is density of the lifting gas (1b./ft. $ ^{3} $ ).

From equation (7) we have:

$$
\frac {d V}{d t} = \frac {d z}{d t} \frac {V}{2 7 8 0 0} f t ^ {3} / \sec
$$

therefore:

$$
\Delta p = \frac {d g}{2 8 8 g} \left(\frac {1}{C _ {d} A _ {a}} \frac {d z}{d t} \cdot \frac {V}{2 7 8 0 0}\right) ^ {2}
$$

psi

Comparing equations (10) and (13) we see that if the equations are equal:

$$
\frac {1}{2 8 8 g C _ {d} ^ {2}} = \frac {1 4 . 7}{1 3 3 3 ^ {2} d _ {a 0}}
$$

If we let $ C_{d}=.975 $ , a reasonable value for the relatively low velocity flow of gas through the appendix, we have:

$$
\frac {1}{2 8 8 g C _ {d} ^ {2}} = 1 1 3. 5 \times 1 0 ^ {- 6} f t - \sec^ {2} / \ln^ {2}
$$

$$
\frac {1 4 . 7}{1 3 3 3 ^ {2} d _ {a 0}} = 1 1 4. 8 \times 1 0 ^ {- 6} f t - \sec^ {2} / \mathrm {i n} ^ {2}
$$

Therefore, the equations (10) and (13) are equal and interchangeable.

It may be noted from equations (10) and (13) that for any given balloon, appendix area and balloon volume are fixed, and the related variables are lifting gas density, rate of rise, and allowable back pressure. For any given allowable back pressure greater rates of rise are allowable at higher altitudes (where $ \mathbf{d}_{g} $ is lower).

Once a floating altitude has been decided upon or it has been decided to carry a given load as part of the balloon system, we can find a maximum allowable rate of rise. We must consider