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5

Wikipedia's Drag equation is $$F_D = \frac{1}{2} \rho v^2 C_D A$$ shows drag's $\frac{1}{2} \rho v^2$ dependence you mention, as does @MarkAdler's answer about dynamic pressure or "Q": Max Q is simply the maximum of the dynamic pressure of the external flow, ${1\over 2}\rho v^2$. It has nothing to do with the vehicle, except for the vehicle's ...

2

Firstly, tipping won't be possible because of Mars' low wind speeds It would take wind moving at around 0.7 mach to see any interesting results Ingenuity's long leg span must also be considered and its wide stance There is also a third factor that comes into play..... The heavy batteries at the base give it a low center of gravity ( yes the gravity on ...

6

The wind on Mars simply isn't that powerful. For comparison, the blades, which are rather large, are moving at the tips around 0.7 mach. It would take wind speeds on that order to cause it to have any measureable effects, and the winds aren't anywhere near that strong on Mars.

0

If you consider the grid fins at a small angle of attack, am I right that drag will act in the direction of the flattened axis (thickness of the grid of fins) and that lift will act perpendicular to it? For any body in a flow, you can draw a single net aerodynamic force vector. Conventionally, the component of this vector inline with the vehicles ...

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