$$
\text {S i n c e a t T}, \left(\frac {\mathrm {d p}}{\mathrm {d} Z}\right) _ {3} = \left(\frac {\mathrm {d p}}{\mathrm {d} Z}\right) _ {5}
$$
$$
m = - \frac {K}{2} \frac {(D - 2 \Delta Z)}{(D \Delta Z - \Delta Z ^ {2}) ^ {3 / 2}}
$$
and, since at T, $ \Delta p_{z3}=\Delta p_{z5} $
$$
m \Delta z _ {T} = - \frac {K \Delta z _ {T}}{2} \frac {(D - 2 \Delta z _ {T})}{(D \Delta z _ {T} - \Delta Z _ {T} ^ {2}) ^ {3 / 2}} = \frac {K}{(D \Delta z _ {T} - \Delta Z _ {T} ^ {2}) ^ {1 / 2}}
$$
and:
$$
\Delta Z _ {T} \left(2 \Delta Z _ {T} - D\right) = 2 \left(D \Delta Z _ {T} - \Delta Z _ {T} ^ {2}\right)
$$
$$
\Delta Z = \frac {3}{4} D
$$
Then:
$$
\Delta p _ {T} = \frac {K}{\left(\frac {3}{4} D ^ {2} - \frac {9}{1 6} D ^ {2}\right) ^ {1 / 2}} = \frac {K}{\sqrt {\frac {3}{4}} D}
$$
$$
m = \left(\frac {d p}{d z}\right) _ {\mathrm {a i r}} (1 - B) = \frac {K \left(2 \cdot \frac {3}{4} D - D\right)}{2 \left(\frac {3}{4} D ^ {2} - \frac {9}{1 6} D ^ {2}\right) ^ {3 / 2}} = \frac {1 6 K}{3 \sqrt {3} D ^ {2}}
$$
Allowable :
$$
\left(\frac {d p}{d z}\right) _ {\mathrm {a i r}} = \frac {1 6 K}{3 \sqrt {3} D ^ {2}} \cdot \frac {1}{1 - B}
$$
For the example above,
$$
D = 3 0 ^ {\prime}, \quad S _ {f} = \frac {9 0 0}{2}, \quad t = . 0 0 1 \mathrm {i n .}, \quad B = \frac {5 3 . 3}{3 8 6} = 0. 1 3 8
$$
Then:
$$
\left(\frac {\mathrm {d p}}{\mathrm {d z}}\right) _ {\mathrm {a i r}} = \frac {1 6}{3 \sqrt {3}} \cdot \frac {4}{2} \cdot \frac {9 0 0}{2} \cdot \frac {. 0 0 1}{(3 0) ^ {2} \cdot 1 2} \cdot \frac {1}{(1 - 0 . 1 3 8)} \quad \mathrm {p s i} / \mathrm {f t}
$$
Allowable
$$
\begin{array}{l} \left(\frac {d p}{d z}\right) _ {\mathrm {a i r}} = 0. 2 9 8 \cdot 1 0 ^ {- 3} \mathrm {p s i} / \mathrm {f t} \\ = 2 0. 5 5 \mathrm {m b} / \mathrm {f t} \\ \end{array}
$$
This is comparable to an altitude of approximately 18,200 ft. Thus the maximum allowable buoyancy for a 30' diameter, .001" thick polyethylene balloon filled with helium is 440 lb.
(2) Appendix-Opening Considerations
As an open-apperdix, constant-volume balloon ascends the lifting gas will expand due to the decrease in the pressure