How to calculate the wind force acting on a water drop?

[A.T.]

... the wind accelerates the drops...

Only if the wind suddenly changes. If you have a uniformly moving airmass, then the drops are created within it already moving at the same horizontal velocity as the airmass.

The horizontal acceleration of the drops and the time it takes are completely irrelevant to this problem with uniform wind.

 

[jbriggs444]

What follows will be an attempt to show how it takes infinite time for the drop to reach the horizontal velocity of the wind. This will be done without using differential calculus. I have done this derivation in the past, but goodness knows where it is buried.

We will work in one dimension. The two dimensional situation is rather nasty because vertical relative velocity affects horizontal drag force and horizontal relative velocity affects vertical drag force. Let us take a step back and consider this "two dimensional complication" in greater detail.

1. Constant drag.

This is the situation for standard kinetic friction. Force magnitude (for a given normal force) is a constant.

Possibly you have seen cars out in the parking lot in snowy weather "doing doughnuts". With rear wheel drive automobiles, this is done by stepping on the accelerator so that the back wheels spin rapidly while the car accelerates slowly forward. The driver steers hard in one direction. The front wheels stick while the rear of the car spins around the front. The key here is that the interaction between ice and tires is "constant drag".

The magnitude of the frictional force between rear tire and ice is fixed. At a high relative speed (rapidly spinning tires), the direction of this force is almost purely forward. The lateral component is negligible. So the back end can slip sideways while the front end does not.

With constant drag, a high slip rate in one direction means low drag in the perpendicular direction.

2. Linear drag.

Here, force is directly proportional to relative velocity.

Linear drag occurs when drag is primarily the result of viscosity rather than turbulence and wave effects. This will tend to be the case for very viscous fluids, low speeds and small objects. One can summarize these factors into a number called the "Reynolds Number". If the Reynolds number is low, drag is approximately linear. If the Reynolds number is high, drag tends to be quadratic instead.

With linear drag, horizontal force depends only on horizontal relative velocity and vertical force depends only on vertical relative velocity.

You can solve for the horizontal and vertical components independently. This is convenient.

3. Quadratic drag.

Quadratic drag is the usual situation for human scale interactions. Force is proportional to the square of relative velocity. So we have a force law such as $F=kv^2$ where the $k$ includes the coefficient of drag, the fluid density and any unit conversion factors.

Let us work an example. Set $k=1$ for convenience. We have a raindrop moving with a horizontal velocity of 3 and a vertical velocity of 0. What is the resulting force? $F=kv^2 = 9$. Here, $F$ is in a purely horizontal direction so $F_x = 9$.

Now take the same raindrop but give it a vertical velocity of 4. What is the resulting force now? $F=kv^2 = 25$. This time the force is split into components with $\frac{3}{5}$ in the horizontal direction and $\frac{4}{5}$ in the vertical direction. The horizontal component is $F_x = 25\frac{3}{5} = 15$.

With quadratic drag, a high slip rate in one direction means high drag in the perpendicular direction.

Back to the derivation. Clearly it will be much easier if we stick to one dimension.

Let us use the frame of reference where the air is at rest and the projectile/raindrop is moving. We want to ask "how much time will elapse before the raindrop comes to rest".

Let us say that the raindrop starts with velocity $v$. How much time will elapse before the raindrop is slowed to $\frac{v}{2}$?

Well, the drag force on the raindrop during this time interval will be at least as much as the drag force at relative velocity $\frac{v}{2}$. So we divide that force by the raindrop's mass and get an answer. It does not matter much what that answer is. Call it $\Delta t$.

The time required to cut the raindrop's velocity in half is at least $\Delta t$.

How much time is required to cut the raindrops velocity in half again?

This time we only have to lose half as much speed. If we look at the drag force at the end of the interval, it will be the result of a relative velocity of $\frac{v}{4}$. That's half the velocity to lose and one quarter of the force. It will take twice as long.

The time required to cut the raindrop's velocity in half again is at least $2 \Delta t$.

We can keep going. The total time required to reduce the raindrop's velocity to zero is at least the sum:$$T = \Delta t \times \sum_{i=0 \to \infty}2^i = \infty$$That's infinite. It takes infinite time for the raindrop to come to rest under quadratic drag.

We could also ask about how much distance is covered during the decelleration.

It will turn out that the distance covered for each halving of velocity will be a constant. So we get that the distance is at least:$$S = \Delta s \times \sum_{i=0 \to \infty} 1 = \infty$$So that is also infinite.

If one shifts to linear drag, the situation changes somewhat. It still takes infinite time -- it's the infinite sum of a constant series. That diverges to infinity. But it takes finite distance -- it's the infinite sum of a decaying geometric series. That converges to a finite sum.

With constant drag, the SUVAT equations immediately tell us that both elapsed time and distance travelled are finite.

We are not done yet...

Let us step back to the two dimensional case. This time we will be considering a raindrop that is (at least approximately) at terminal velocity in the vertical direction. It has some non-zero relative velocity $v$ in the horizontal direction.

Clearly, the magnitude of the resultang velocity will quickly (geometrically) converge on terminal velocity with a drag force of approximately $mg$. The horizontal component of drag will depend almost entirely on the angle of descent.

With the small angle approximation, $F_x = mg \sin \theta = mg \frac{v_x}{V}$ where $\theta$ is the angle of descent (measured from the vertical) and $V$ is terminal velocity.

This is effectively linear drag in the horizontal direction.

We've already reasoned above that it takes infinite time and finite horizontal distance to bring the raindrop to horizontal rest relative to the wind.

 

[haruspex]

You seem to have used “slip rate in one direction" differently in these two conclusions.

With constant drag, a high slip rate in one direction means low drag in the perpendicular direction.

I think you mean that if the slip (all of it) is in one direction then there is no drag in the perpendicular direction. Merely having a high slip rate in a given direction does not mean there is no slip (perhaps greater) in another direction.

With quadratic drag, a high slip rate in one direction means high drag in the perpendicular direction.

Here you are assuming there is slip in the perpendicular direction. How about “With quadratic drag, if the slip has components in each of two perpendicular directions then increasing the slip rate in one increases the drag in the other"?

 

[jbriggs444]

Here you are assuming there is slip in the perpendicular direction.

Yes, indeed. The alternative seems to be one dimensional motion which is the easy case.

 

[vanhees71]

The only case you can solve analytically is the "free fall". The equation of motion corresponding component of $v$ (with the gravitational acceleration in the negative direction of the corresponding basis vector) reads (assuming that for the initial condition $v(0)=0$ we always have $v<0$)
$$m \dot{v}=-m\alpha |v| v-g=m \alpha v^2 -m g$$
or
$$\dot{v}=\alpha v^2 -g.$$
This equation can be solved by "separation of variables":
$$\frac{\mathrm{d} v}{\alpha v^2-g}=-\mathrm{d} t.$$
Integrating gives
$$t=-\frac{1}{\sqrt{\alpha g}} \mathrm{artanh} \left (\sqrt{\frac{\alpha}{g}} t \right ),$$
where I've assumed that at $t=0$ we start with $v=0$. Solved for $v$ gives
$$v=-\sqrt{\frac{g}{\alpha}} \tanh \left (\sqrt{\alpha g} t \right).$$
For $t \rightarrow \infty$ we get $v\rightarrow -\sqrt{g/\alpha}$. So the motion goes asymptotically into uniform motion, i.e., the velocity adjust such that the friction force and the gravitational force compensate each other.

Integrating wrt. $t$ assuming the initial condition $x(0)=0$ you get
$$x=-\frac{1}{\alpha} \ln \left [ \cosh(\sqrt{g \alpha} t) \right]$$
For $t \rightarrow \infty$ one has $\cosh(\sqrt{g \alpha} t) \rightarrow \exp(\sqrt{g \alpha} t)/2$ and thus
$$x \simeq -\sqrt{\frac{g}{\alpha}} t + \frac{1}{2 \alpha} \quad \text{for} \quad t \rightarrow \infty.$$