0
$\begingroup$

Hey there Physics friends!

I've been struggling with what seems to be a very basic physics question. Let's say that I have wind vector for a given altitude and location, and I want to calculate the flight path of a hot air balloon.

So once the balloon hits that wind, would you calculate forward distance based on the wind speed (so would the balloon speed be exactly the average wind speed) or would you use the standard drag equation to calculate the force of the wind on the balloon to determine the distance traveled by the balloon in a given time period?

$\endgroup$
1
$\begingroup$

Yes you would have to use the standard drag equation. Imagine the balloon is at rest (or moving with a different velocity from the wind) initially. As the balloon hits the wind it can't suddenly reach the same velocity as the balloon, as this would require an infinite acceleration (which is impossible). Instead the wind applies a force to the balloon. This force will increase the balloon's velocity until it does match that of the wind (drag force is proportional to the relative speed between the balloons), at which point the velocity of the balloon will match that of the wind. If you can assume that the time it takes for the balloon to reach the velocity of the wind is negligible (such as in the case of a large drag coefficient or cross sectional area of the balloon) then the velocity of the balloon could be assumed to be the same as that of the wind at all times.

$\endgroup$
  • 1
    $\begingroup$ Thanks so much for your answer. This really breaks down the problem very nicely! I'm just looking at coding out a predictor so I believe for my purposes (hot air balloon with large drag coefficient and people likely won't take one up in crazy wind shear), the balloon speed could equal wind speed, but this wouldn't be the case for something with a different shape. $\endgroup$ – Kyle Hotchkiss Feb 27 '16 at 16:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.