I understand how to get the vertical (descent speed) of a parachute and its payload, but how could one find the horizontal speed/velocity of this parachute depending on the speed of the wind? (yeah I understand the higher the wind is, the smaller it's influence is as well)

I'm having a hard time trying to figure out what the winds influence on a descending balloon is. I know it would push the parachute, but how much would that matter? I know it would push the payload, which trying to figure out a CD for would be rather hard!


yeah I understand the higher the wind is, the smaller it's influence is as well

do you mean the higher wind speed? or that the parachute is higher? if you mean a higher wind speed, then you have it backwards. give it a thought. if higher (stronger wind) had less influence (force) then slower wind would have more, and no wind would have the most. that makes no sense.

but on to the question. the answer is that the speed of the balloon horizontally asymptotically approaches the speed of the wind. think of it this way, there is some relationship between windspeed and force on the parachute, in so much as when the speed increases, the force does too, and when the speed decreases, the force does, too. in fact the relationship is that Force = Const*Speed^2, we just dont know what that constant is, just that it is positive.

so. we have that relationship and your starting point. a parachute with 0 velocity (in the horizontal, we're just gonna work there) and a large windspeed, relative to the parachute. that means the force is high to begin with, and the parachute accelerates in the same direction as the wind is blowing. but now the relative windspeed is lower, so the force decreases, and this keeps happening. the parachut keeps going faster and faster in the direction of the wind, but the force reduces, and so does the acceleration, until the force is nearly 0 because the parachute is moving at almost the speed of the wind.

  • $\begingroup$ Higher in altitude (less pressure = pushing power) =) Thanks for the awesome description... after thinking a bit, it makes sense that it could travel near to the wind speed until the pressure drops to a certain point. $\endgroup$ – Kyle Hotchkiss Jan 8 '12 at 19:59
  • $\begingroup$ PS welcome to physics/stackexchange!! $\endgroup$ – Kyle Hotchkiss Jan 8 '12 at 20:01

Start by assuming a uniform, unsteerable canopy (think big rount chutes from WWII movies).

If you can assume a reasonably thick, uniform mass of moving air for the wind, the horizontal speed can't really be anything except the speed of the wind.

Of course, that's probably not a really great description of the real world, so now you want to figure how long the parachutist would be in the final wind layer to start getting an idea of who much of that velocity would be imparted.

Already the problem is getting pretty hard for a Back of the Envelope calculation.

But most chutes are more complicated then that: the user can pull the shrouds to dump air out in various direction giving some control, so the user can (and I presume should) steer into the ground level wind to reduce their horizontal speed and landing.

At this point I think we're stuck on BotE work: we'll have to hope someone know the actual performance characteristics of typical examples.

  • $\begingroup$ In this situation there are no pull cords, etc because the parachute is just one for a balloon payload. $\endgroup$ – Kyle Hotchkiss Jan 7 '12 at 4:10
  • $\begingroup$ @Kyle: Well, that balloon is going to drift with the wind as well, so the first approach sketched in this answer will work. $\endgroup$ – MSalters Jan 9 '12 at 13:43

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