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Mathinline
bodyF_{D, main} = \frac{1}{2}*1.225 \frac{kg}{m^3}* 1.44 * (0.96*(1.4478 m)^2 * \pi) *V^2 = 40.75 kg * 9.81 \frac{m}{s^2}

V 8.47 m/s = 28 ft/s

This is is acceptable per Requirement 3.1.2 of the DTEG.

2 Because Raziel separated in two locations rather than just one, this number may vary from what will be seen on Hermes, which has a single separation point and a different height. We're using it here as an approximation for right now. One idea is to use a pitot tube to test for wind speed and rig the parachute up on a windy day with a fish scale or load cell. If a convenient opportunity during spring semester presents itself for this, we can do that. Or we can just use new flight test data from when we fly it (assuming it works...)

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Because the CdS of the drogue is 2 orders of magnitude greater than that on the mission package, we expect there to be a good deal of tension in the line connecting the drogue and the main. At sea level, there willwe can calculate approximately how many pounds of force will be available to extract the deployment bag from the cup (assuming it hasn't fallen out prematurely) and then the main from the deployment bag.

Drogue to

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Booster

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As previously mentioned, the variability in falling orientation makes it difficult to predict the relative ratio of CdS between the drogue and the booster. Given our current estimations, the drogue could see anywhere between 1.25x and 45x the amount of drag on the booster (if they are going the same velocity).

dd

Resources

[1] DRAG OF CIRCULAR CYLINDERS FOR A WIDE RANGE OF REYNOLDS NUMBERS AND MACH NUMBERS

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