the equation 2500 ft. $ ^{3} / 1 \mathrm{b}. $ mol = ae $ ^{50 b} $ where 50 is the expression for altitude in thousands of feet. Similarly, at 70,000 feet, $ 6 4 5 0=a e^{7 0 b}, $ and by solving to eliminate a, we find $ 2. 5 8=e^{2 0 b} $ or 20b = .95, and the constant b is equal to .0475. Thus, the equation may be written:
(3)
$$
y = a e ^ {. 0 4 7 5 z}
$$
y was originally defined as the molar volume, equal (for 98% helium) to:
$$
\frac {\text {Balloon Volume} \times 2 4 . 4}{\text {Gross Load}} = \frac {K}{W}
$$
In turn, $ \frac{K}{W}=a e^{-0. 4 7 5 z} $ , where z is the expression for altitude in thousands of feet. From this relationship, we may solve for W, the gross load.
(4)
$$
W = \frac {K}{a} e ^ {- 0. 4 7 5 z}
$$
(5)
$$
\ln \left(\frac {W a}{K}\right) = - 0. 4 7 5 z
$$
or:
(6)
$$
\ln W + \ln \frac {a}{K} = - 0. 4 7 5 z
$$
Differentiating with respect to W:
$$
\frac {d z}{d W} = - \frac {2 1 . 0 5 2}{W} \frac {f t}{l b}
$$
where W 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 O and 30,000 feet. A in this range is equal to 31,400 ft./lb.
A plot of altitude sensitivity against load is shown on page 109 of the "Operations" section (Part II of this technical report).