# Trajectory of raindrop with horizontal wind

1. Aug 1, 2015

### koolraj09

Hi all,
This question just occurred to me: Whether a falling raindrop in presence of a uniform horizontal wind (ideally speaking) has a curve trajectory or a straight line?
Assuming the wind force is constant, the velocity of the raindrop in horizontal direction will keep on changing since there is no other force in this direction to bring the raindrop to a terminal velocity in horizontal direction.
In the vertical direction, there is also only one force, gravity. No other force being acted on.
So writing the two equations of motion in horizontal and vertical directions give:
d2x/dt2 = (Fx/M) = constant; and d2y/dt2 = g;
Solving this for x and y with initial conditions set to zero, gives a straight line.
Is this possible that in such a condition the raindrop is falling in an inclined straight line?
Why does then the projectile motion trajectory is a parabola?
Is this model a good approximation (if correct) for a first cut analysis of raindrop trajectories?
Anything I've missed or that can be added to bring the model more close to reality? How do I confirm the prediction of the model?

2. Aug 1, 2015

### tech99

I assume that the vertical velocity will be constant - the terminal velocity. In the horizontal direction the drop will move at the wind speed. So it will travel in a straight, sloping line. You may have seen this beneath clouds when a rain shower occurs. I believe the drop is slightly flattened, sometimes called an oblate spheroid, and is canted so the short axis aligns with the direction of travel. These drops have been studied due to their de-polarizing effect on microwaves.

3. Aug 1, 2015

### nasu

The solution of your equation may be a straight line trajectory.

For projectile motion there is no horizontal force so the equations are not the same.

As for modeling reality, it seems to me that your model is inconsistent. If you neglect air resistance there will be no force in the horizontal direction.
If you don't neglect it, it should act on the vertical too.
Anyway, the horizontal force won't be constant.

4. Aug 2, 2015

### CWatters

That's not correct. The horizontal velocity is limited by the wind speed.

5. Aug 2, 2015

### stedwards

If we assume the raindrops are being carried by the wind, the solution is simple.

Change to the frame of reference where the wind has zero velocity, and the ground is moving. The raindrops have a constant and vertical velocity in this frame.

Change back to the Earth's frame of reference. This just adds a constant horizontal velocity. $\bar{V} = \bar{v_x} + \bar{v_y}$. The velocity is constant.

6. Aug 2, 2015

### koolraj09

Thanks for your replies. I got some thoughts on this today. Description of the whole phenomena according to my understanding is following. Please correct me if you find anything inaccurate.
1. The raindrop forms by condensation of water vapor and starts to fall. Thus as soon as it is formed it has zero initial velocity in both x and y directions. If we place our origin here then xo = yo = 0...the start points of raindrop.
2. The raindrop starts to move towards earth because of gravity. It accelerates towards the earth with 'g'. Similarly in x direction too, due the wind force it starts to accelerate for a particular amount of time. As mentioned above, assuming Fwind = constant ( why can't it be constant is another doubt i have now!), the raindrop also accelerates in x direction and reaches the final speed = wind speed in this direction. In x - direction we have only one force. But if the drop ceases to accelerate in this direction what other force is to make the drop move with constant velocity which equals wind velocity?
The resultant of the Fx and Fy force is some force inclined at some angle to the vertical. The trajectory of the raindrop during it's acceleration in the direction of this force. Since the raindrop is free to move in any direction it will move in a straight line in the direction of the resultant force.
3. Motion continues and the acceleration ceases when in both the cases when both reach the terminal velocities. The vy is limited by drag force and vx is limited by wind velocity.
Now from this point since there is no net force on the raindrop it will continue to follow the straight line direction (as found out above) because of inertia.

I still haven't been able to imagine why won't it go in a curved trajectory? In general for me, this leads to the question, in which cases and how do objects follow curved paths?
The next thing is if the above description is correct, I think i would go on to write the relevant equations and then mathematically find the trajectory. (Just for sense of proof)

7. Aug 2, 2015

### stedwards

Do you think all rain drops begin life with an initial velocity of zero with respect to the earth, or might their initial velocity have something to do with the state of motion of the air from which they condensed?