I'm trying to make a basic simulation for horizontal drift of a falling object due to wind. I need the force balance so that I can find out the acceleration of the object from the wind.

The object is initially dropped with zero horizontal velocity with the only force acting on the object being forces applied by the wind which is a constant velocity. I would like to use an observer on the ground as a reference frame. The projectile will be dropped directly above an observer and will accelerate in the direction of the wind until reaching steady state (velocity of the wind).

I want measure how far the projectile moves in the horizontal direction while it falls to the ground. I am getting confused on how forces are applied to the object.

The way I originally thought the only force was wind pressure applied over the surface area of the object.

$F_{wind} = \frac{1}{2}\rho v^2 A$

I was unsure however, if there was any help from the drag force. Does the drag force add to the force from wind pressure? Is the wind force just the drag force with a different name?

EDIT: Adding some clarification to the problem setup.

  • $\begingroup$ If you want to assume that the object is initially released with zero groundspeed, and you are interested in analyzing the initial acceleration, then you need to make that clear. Speaking of "drift" but not "acceleration" tends to suggest you are looking at a steady-state condition; see xxavier answer. $\endgroup$ Oct 24, 2019 at 19:24
  • $\begingroup$ If you do want to consider acceleration, then you need to consider all aerodynamic forces. You can either deal in terms of airspeed, or in terms of ground speed plus wind speed. They are the same. This may possibly be of some use to you: see all associated links as well: physics.stackexchange.com/questions/491600/… $\endgroup$ Oct 24, 2019 at 19:33
  • $\begingroup$ Re initial comment: yes of course you do mention acceleration as well as drift. Need some clarification as to initial conditions, and reference frame. $\endgroup$ Oct 24, 2019 at 20:13
  • $\begingroup$ Cutting to the chase, is the object being dropped from a freely drifting balloon, or from a tower, or from an airplane, or do you wish to be able to consider 3 cases? Actually now that my answer addresses all 3 I guess technically you need to be careful about edits but still curious what you had in mind. $\endgroup$ Oct 24, 2019 at 21:40
  • $\begingroup$ Technically the edit may be construed to invalidate some parts of my answer, but in this case I don't mind and I feel it is a good edit. If someone complains about the edit invalidating the other answer, you (or anyone) could add "and case b, what if the object were dropped from a free balloon". It would seems silly to suggest that a question must be left in an ambiguous state so as not to invalidate an existing answer. $\endgroup$ Oct 24, 2019 at 21:50

2 Answers 2


Additional response to edited question: object is dropped with an initial groundspeed of zero: the only forces present as object falls are gravity and aerodynamic drag. Drag is a function of airspeed, which has a vertical component as well as a horizontal component. If you use the airmass reference frame, the ONLY significance of wind is that 1) the initial condition is that airspeed equals wind speed, and 2) during the time the object is falling there will be some horizontal translation between the airmass reference frame and the ground reference frame, which will affect the final answer when converted back to the ground reference frame. See notes below, and links at end, about not being able to solve for vertical and horizontal accelerations independently of each other.

Read on for more (some content was originally aimed at a broader version of the question) :

Cutting to the chase, is the object being dropped from a freely drifting balloon, or from a tower, or from an airplane? I.e. is the horizontal airspeed initially zero, or not? Dropping from an airplane is more like dropping from a tower than dropping from a freely drifting balloon, because the initial horizontal component of airspeed is not zero.

Once an object has accelerated to freely move with the wind, the horizontal component of airspeed is zero, so the wind exerts no force on the object. This makes calculations easier. In theory, if the initial horizontal component of airspeed is non-zero, it will take a infinite amount of time for the horizontal component of airspeed to fall all the way to zero, just as it takes an infinite amount of time to fully achieve the vertical terminal velocity. In practice, after a few tens of seconds the horizontal airspeed will probably be very close to zero.

If the object is initially released with zero groundspeed, then the horizontal component of the airspeed vector is initially not zero, but rather is equal to the wind speed. It turns out that this has an effect on the object's vertical acceleration as well as the object's horizontal acceleration, even if the body is a perfect sphere! (See links below for more.)

If the object is dropped from an airplane, wind or no wind, the situation is essentially the same as if the object is dropped from a tower in wind. The key point is that the horizontal component of airspeed is not zero at the instant of release.

If horizontal as well as vertical airspeed, force, and acceleration components are present at the instant of release, then you need to take care to properly consider the true magnitude and direction of the net aerodynamic force acting on the object. You need to write your equations in terms of the airspeed vector, which also is equal to the ground speed vector (including the vertical component) plus the wind speed vector. You can't treat drag due to wind speed, drag due to horizontal ground speed, and drag due to vertical falling speed as independent entities, because drag is dependent on airspeed squared, and all these components of motion contribute to airspeed.

Ultimately it's up to you whether you want to work in terms of the airmass reference frame (which moves with the wind) or the ground reference frame, so long as your equations appropriately deal with the x, y, and z components of the airspeed vector. It will generally be much easier to work in terms of the airmass reference frame, and view the ground as moving carpet down below, like one of those airport people-mover beltways. Just bear in mind that IF the object is released with zero groundspeed, then the initial condition is that the horizontal airspeed equals the wind speed. IF the object is dropped from a freely drifting balloon, then the initial condition is that airspeed is zero. IF the object is dropped from an airplane, then the initial condition is that the dropped object's airspeed equals the airplane's airspeed. Regardless of which is true, once the appropriate initial condition is set, forget about the wind, except for checking at the end to see how much the ground moved relative to the airmass reference frame, during the time of fall.

In other words, if the object is released from an aircraft with a known airspeed vector, and the airmass (wind field) is uniform, then you can ignore the wind while you do your calculations in terms of airspeed, and then you can add on the "drift" relative to the ground due to the translation of the airmass reference frame relative to the ground during the time of fall.

It appears from your question that you hoped to be able to compute horizontal acceleration components independent of vertical acceleration components. In any case where the object's initial horizontal airspeed is not zero, that's not going to work. Just as the horizontal airspeed at any instant affects the rate of change of vertical airspeed, so too does the vertical airspeed at any instant affect the rate of change of horizontal airspeed. See links below for more.

On the hand, if you are just dropping objects from a freely drifting balloon in a uniform wind field, then you can ignore most of this answer. In that case, horizontal airspeed, force and acceleration are zero.

These links should be of some use to you: see all associated links as well:



  • $\begingroup$ Thank you for the in depth response. So when I look at the object in the wind frame I am still a little confused on how the force is applied. Is the only force (other than gravity) drag due to wind (in both the x and y axis)? $\endgroup$
    – Boto
    Oct 24, 2019 at 22:10
  • $\begingroup$ See last bit added in italics near end. Not so much wind per se, but airflow. $\endgroup$ Oct 24, 2019 at 22:30
  • $\begingroup$ Now moved to top of answer $\endgroup$ Oct 24, 2019 at 22:42
  • $\begingroup$ Ok, so when I calculate the drag with respect to the airflow am I able to then break each forces into x and y axes at that point? From there I could have the projectile decelerate with respect to the the airflow until the horizontal axis has the same velocity as the wind which in this frame would be 0 velocity and the vertical axis is equal to the force of gravity. Is there still a concept that I am not grasping? $\endgroup$
    – Boto
    Oct 24, 2019 at 22:45
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    $\begingroup$ Sounds basically right except that the vertical SPEED will continue to increase till terminal velocity is reached; the vertical FORCE is equal to weight minus vertical component of drag throughout the whole time. Got to run, consider asking Physics SE for more details? $\endgroup$ Oct 24, 2019 at 22:48

The falling object moves within the mass of air, and suffers no horizontal forces at all... For a ground observer, it drifts with the wind...

  • $\begingroup$ But if the object was dropped from a fixed point, which I think is what the questioner is talking about, it will start out with its relative wind off of perpendicular, starting at 90 degrees at the moment of release and eventually becoming 0 degrees. So it sees a side force that diminishes to 0 once it's moving laterally at the same speed as the wind. The question relates to the lateral acceleration phase. $\endgroup$
    – John K
    Oct 24, 2019 at 19:23
  • $\begingroup$ Downvoted because this is not true. If you dropped a bowling ball from shoulder height would it "drift with the wind" from the perspective of a ground observer? Density, time, windspeed, and possibly other factors all play into how much a falling object deflects from vertical. $\endgroup$ Oct 24, 2019 at 19:50
  • $\begingroup$ @MichaelHall the answer is obviously taking the perspective of dropping a bowling ball from a freely drifting balloon, etc. It could use some clarification as to that point. $\endgroup$ Oct 24, 2019 at 20:03
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    $\begingroup$ @quiet flyer - "Obviously" and "could use some clarification" don't really go together. It is obvious, or it is not. You of course are free to presume what you want, but the question says nothing about the object being dropped from a platform already moving at the same speed as the wind. And if it was, a dense object like a bowling ball would eventually overcome the wind drift of the platform (think hot air balloon) and take a path much closer to vertical so that it would not strike the ground directly below the drifting platform from which it was dropped. $\endgroup$ Oct 24, 2019 at 20:14
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    $\begingroup$ @MichaelHall - point taken about "obviously", but I definitely strongly disagree that a bowling ball dropped from a balloon, in a uniform wind field, will not impact directly below the balloon. A ribbon streamer attached to the ball would stream straight up vertically throughout the fall, as well. The ball will never "feel" any sideways wind at all in such a case. $\endgroup$ Oct 24, 2019 at 20:24

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