The Roswell Report: Fact vs. Fiction in the New Mexico Desert

Read from the scan by GLM-OCR; expect the odd misread word.

If a balloon is teardrop in shape rather than spherical, the curve would be modified so that the value of $ C_{D} $ , for a given Reynolds number would be lower. In this case the sudden drop in $ C_{D} $ as Reynolds number increases (the change from viscous to turbulent flow) would occur at a lower Reynolds number.

We have thus far in our discussion assumed that there is no vertical motion of the air surrounding the balloon system relative to the coordinate z . However, this is not necessarily the case under actual conditions. In many instances vertical air movement is found in the atmosphere. Therefore, we must introduce a term to allow for this vertical air movement. In equation (2) this term was indicated as $ \pm F_{A} $ , the external atmospheric force.

We may consider this vertical air movement in terms of a velocity D $ \zeta $ . Then the vertical velocity of the balloon system relative to the air surrounding the system will be the difference between the velocity of the balloon relative to the absolute altitude Dz and the velocity of the surrounding air relative to the absolute altitude This may be equated as Dz-D $ \zeta $ , where Dz and D $ \zeta $ are both considered positive in the direction of increase of altitude.

The total force due to the drag, or friction will be:

$$
F _ {D} + F _ {A T M} = C _ {D} \frac {\rho}{z} A (D z - D \xi) ^ {2}
$$

where the notations are those used previously, except that now $ N_{R}=\frac{(Dz-D\zeta)d\rho}{\mu} $ . The relationship between $ N_{R} $ and $ C_{D} $ will be those used previously.

The force due to buoyancy of the lifting gas $ F_{b}=V_{b} \left( \dot{\rho}_{a}-\dot{\rho}_{g} \right) $ where:

$$
V _ {b} = \text {b a l l o o n v o l u m e (f t .} ^ {3})
$$

$ \rho_{a}, \rho_{g} $ density of the air and lifting gas, respectively (1b./ft. $ ^{3} $)

This term may also be stated as: F = $ V_{b}\left(\frac{P_{g}}{R_{a}T_{a}}-\frac{P_{g}}{R_{g}T_{g}}\right) $ where:

$ p_{a}, p_{g} = $ pressure of air and lifting gas

$ R_{a}, R_{g} = $ specific gas constant of air and lifting gas

$ T_{a}, T_{g} $ = temperature of air and lifting gas

The changes that will take place in this expression are those due to a temperature difference between the lifting gas and the free air, change in volume of the balloon due to loss of lifting gas, change of the gas constant of the lifting gas due to dilution with air, and (in the case of a balloon that will hold an internal pressure) pressure difference between lifting gas and surrounding air.