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the equation 2500 ft.3jlb.·mol = ae50b where 50 is the expression for altitude in thousands of feet. Similarly, at 70,000 feet~ 6450 =ae70b, and by solving to eliminate a, we find 2.58 = e Ob or 20b = .95, and the constant b is equal to .0475. Thus, the equation may be written: (3) y was originally defined as the molar volume, equal (for 98% helium) to: Balloon Volume x 24.4 Gross Load K =-w K _ 08.o475Z In turn, w-in thousands of feet. W, the gross load. , where z is the expression for altitude From this relationehip, we may solve for (4) w = (5) In{~) = -.0475 i! or: (6) In W + In f = - .0 4 7 5 ~ Differentiating with respect to W: (7) dz _ 21. 052 ft crw-- w 1b 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 var/ with time. Included is the assumption that over a short period of ti~e buoyancy of lifti~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 31,400 ft./lb. VI A plot of altitude sensit.ivity against load is shown on page 109 of the "Operations" section (Part II of this technical repo~. -32-