the equation $2500 \text{ ft.}^3/\text{lb. mol} = ae^{50b}$ where 50 is the expression for altitude in thousands of feet. Similarly, at 70,000 feet, $6450 = ae^{70b}$ , and by solving to eliminate $a$ , we find $2.58 = e^{20b}$ or $20b = .95$ , and the constant $b$ is equal to $.0475$ . Thus, the equation may be written:
(3)
$$
y = ae^{.0475z}
$$
$y$ was originally defined as the molar volume, equal (for 98% helium) to:
$$
\frac{\text{Balloon Volume} \times 24.4}{\text{Gross Load}} = \frac{K}{W}
$$
In turn, $\frac{K}{W} = ae^{.0475z}$ , where $z$ is the expression for altitude in thousands of feet. From this relationship, we may solve for $W$ , the gross load.
(4)
$$
W = \frac{K}{a} e^{-.0475z}
$$
(5)
$$
\ln\left(\frac{Wa}{K}\right) = -.0475z
$$
or:
(6)
$$
\ln W + \ln \frac{a}{K} = -.0475z
$$
Differentiating with respect to $W$ :
(7)
$$
\frac{dz}{dW} = -\frac{21.052}{W} \frac{\text{ft}}{\text{lb}} \quad \text{where } W \text{ is gross load in lb.}
$$
We see that the value of the constant $a$ is unimportant here, and the expression is independent of balloon volume, as long as it does not vary with time. Included is the assumption that over a short period of time buoyancy of lifting gas does not change.
Thus, we have an expression for $A$ , the altitude sensitivity, which is valid between 40,000 and 105,000 feet. Similarly, it is possible to evaluate altitude sensitivity for operation between 0 and 30,000 feet. $A$ in this range is equal to $\frac{31,400 \text{ ft.}/\text{lb.}}{W}$
A plot of altitude sensitivity against load is shown on page 109 of the "Operations" section (Part II of this technical report).
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