4 ms·

g (9.8 m/s^2) is an acceleration, not a speed. It's right there in the units. The actual terminal velocity of an object is meet when its wind resistance -- whic

g (9.8 m/s^2) is an acceleration, not a speed. It's right there in the units. The actual terminal velocity of an object is meet when its wind resistance -- which is a force (acceleration with mass) with negative magnitude -- matches g. Until then, it will accelerate. So, generally, the larger the surface-area-to-mass ratio of an object, the lower its terminal velocity. Aerodynamics also plays a part, of course. So, a lead feather will have a much higher terminal velocity than a real feather, due to the extra mass that the wind resistance is fighting against. Meanwhile, g is an acceleration so it affects all mass equally. Here's a really cool video of a feather and bowling ball falling in vacuum together: https://youtu.be/frZ9dN_ATew https://youtu.be/frZ9dN_ATew Since aircraft rely on lift, which is also a force, to overcome gravity, minimizing weight is ideal. Gravity will affect all the mass equally. But the same lift force with double the mass will only be half as effective.

antoineMoPa 9y ago
The helicopter people on this forum seem to conclude that the fall speed of heavier helicopters is slower: https://www.pprune.org/rotorheads/484972-weight-vs-rod-autorotation.html https://www.pprune.org/rotorheads/484972-weight-vs-rod-autor...
- jdmichal 9y ago
  Ah, I see where you were going now. That is interesting; never heard of it before. I agree with you, that at first glance it seems like the gain in potential energy would be countered by the increase in the lift force necessary. Maybe there's some sort of flat efficiency loss going on, that is overcome by the higher energies present in the heavier scenarios? It does seem that everyone is in agreement that the heavier helicoptor would require more applied pitch. Which may not actually be desirable, given my impression that a pilot would attempt to burn the kinetic energy in the rotors to lift via pitching the blades right before landing.

antoineMoPa 9y ago

The helicopter people on this forum seem to conclude that the fall speed of heavier helicopters is slower: https://www.pprune.org/rotorheads/484972-weight-vs-rod-autorotation.html https://www.pprune.org/rotorheads/484972-weight-vs-rod-autor...

jdmichal 9y ago
Ah, I see where you were going now. That is interesting; never heard of it before. I agree with you, that at first glance it seems like the gain in potential energy would be countered by the increase in the lift force necessary. Maybe there's some sort of flat efficiency loss going on, that is overcome by the higher energies present in the heavier scenarios? It does seem that everyone is in agreement that the heavier helicoptor would require more applied pitch. Which may not actually be desirable, given my impression that a pilot would attempt to burn the kinetic energy in the rotors to lift via pitching the blades right before landing.