Since at T, $\left(\frac{dp}{dz}\right)_3 = \left(\frac{dp}{dz}\right)_5$ :
$$
m = -\frac{K}{2} \frac{(D - 2\Delta Z)}{(D\Delta Z - \Delta Z^2)^{3/2}}
$$
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:
$$
\Delta Z_T (2\Delta Z_T - D) = 2(D\Delta Z_T - \Delta Z_T^2)
$$
$$
\Delta Z = \frac{3}{4} D
$$
Then:
$$
\Delta p_T = \frac{K}{\left(\frac{3}{4} D^2 - \frac{9}{16} D^2\right)^{1/2}} = \frac{K}{\sqrt{\frac{3}{4}} D}
$$
$$
m = \left(\frac{dp}{dz}\right)_{air} (1 - B) = \frac{K(2 \cdot \frac{3}{4} D - D)}{2 \left(\frac{3}{4} D^2 - \frac{9}{16} D^2\right)^{3/2}} = \frac{16K}{3\sqrt{3} D^2}
$$
Allowable:
$$
\left(\frac{dp}{dz}\right)_{air} = \frac{16K}{3\sqrt{3} D^2} \cdot \frac{1}{1 - B}
$$
For the example above,
$$
D = 30', \quad S_f = \frac{900}{2}, \quad t = .001 \text{ in.}, \quad B = \frac{53.3}{386} = 0.138 \quad \text{(helium)}
$$
Then:
$$
\left(\frac{dp}{dz}\right)_{air} = \frac{16}{3\sqrt{3}} \cdot \frac{4}{2} \cdot \frac{900}{2} \cdot \frac{.001}{(30)^2 \cdot 12} \cdot \frac{1}{(1 - 0.138)} \quad \text{psi/ft}
$$
Allowable $\left(\frac{dp}{dz}\right)_{air} = 0.298 \cdot 10^{-3} \text{ psi/ft}$
$$
= 20.55 \text{ mb/ft}
$$
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-appendix, constant-volume balloon ascends the lifting gas will expand due to the decrease in the pressure
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