We see that the maximum allowable $(dp/dz)_a (1-B)$ for a 30' diameter, .001" thick polyethylene balloon is $256 \times 10^{-3}$ psi/ft. Dividing by $(1-B)$ we have the maximum allowable:
$$
(dp/dz)_a = \frac{256 \times 10^{-3}}{1 - .138} = .300 \times 10 \quad \text{psi/ft}^{-3}
$$
$$
= 20.7 \times 10^{-3} \quad \text{mb/ft}
$$
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:
$$
\Delta p_z = \Delta z \left( \frac{dp}{dz} \right)_{air} (1-B)
$$
$$
\Delta p_z = \frac{4Sft}{2(D\Delta z - \Delta z^2)^{1/2}}
$$
at the point of tangency of these curves (T in Figure 26):
$$
\Delta p_{T3} = \Delta p_{T5} \quad \text{and} \quad \left( \frac{dp}{dz} \right)_{T3} = \left( \frac{dp}{dz} \right)_{T5}
$$
in equation (5), making $\frac{4Sft}{2} = K$ and in equation (3), making $(dp/dz)_a (1-B) = m$ , 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)
$$
\frac{dp}{dz} = -\frac{K}{2} \frac{(D - 2\Delta z)}{(D\Delta z - \Delta z^2)}
$$
(3b)
$$
\frac{dp}{dz} = m
$$
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