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

• Lotto
In summary: I think I cannot solve it, so what is the solution? Because in this problem below, I used exactly that my equation above and I got a correct solution. So doesn't the solution correspond to that...No, the solution is not the same as the equation given.
Lotto
TL;DR Summary: I have a water drop falling with a constant velocity ##v##, since ##mg=\frac 12 C \rho S v^2##. Wind is blowing with a velocity ##u## only in a horizontal direction. What will be its force acting on the drop?

I would use this equation ##F= \frac 12 C\rho S u^2##. When I want to calculate the net velocity of the drop, will it be the sum of ##\vec v## and ##\vec u##? Bacause we assume that the drop is so light that its horizontal velocity will be equal to the wind's one.

So the drop's velocity should be then constant, but when threre is a force ##F## acting on it, how is it possible? Doesn't act on the drop also a force ##-F##, the force caused by a normal air resistance?

I would guess an ansatz for the drag force like
$$\vec{F}_{\text{drag}}=-\alpha |\vec{v}-\vec{u}|(\vec{v}-\vec{u}),$$
where ##\vec{u}## is the velocity field of the air. Of course after a sufficiently long time the drop gets into equilibrium with the flow of the air, and then
$$\vec{F}_{\text{drag}}-m \vec{g}=0,$$
if ##\vec{u}=\text{const}##.

erobz
In the vertical direction the drop is always imparting vertical momentum to the air as it falls. In the horizontal direction the air is imparting momentum to the drop, and in the limit as the drops horizontal velocity approaches ##u## ( the speed of the air mass) the horizontal force goes to zero.

vanhees71
erobz said:
In the vertical direction the drop is always imparting vertical momentum to the air as it falls. In the horizontal direction the air is imparting momentum to the drop, and in the limit as the drops horizontal velocity approaches ##u## ( the speed of the air mass) the horizontal force goes to zero.
And why does it go to zero when it approaches that velocity? Is that because of a drag force in a horizontal direction that is equal to the wind force?

Lotto said:
And why does it go to zero when it approaches that velocity? Is that because of a drag force in a horizontal direction that is equal to the wind force?

Imagine you are running as fast as you can with a friend in front of you also running that fast at an arms length ahead of you ( maybe you pushed them to that speed, maybe you didn’t…it doesn’t matter) You reach out your arm, can you accelerate your friend? Also there is another friend in front of them jogging backwards at an arms ahead of your friend, can they accelerate your friend slowing them down while also moving at that speed?

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Lotto said:
And why does it go to zero when it approaches that velocity? Is that because of a drag force in a horizontal direction that is equal to the wind force?
It is because the horizontal velocity of the droplet is then equal to the wind velocity, so there is no differential velocity to square in the drag equation.

Lotto said:
So the drop's velocity should be then constant, but when threre is a force ##F## acting on it, how is it possible? Doesn't act on the drop also a force ##-F##, the force caused by a normal air resistance?
There is only one force by the air on the drop, and its direction is parallel to the flow of the air relative to the drop (ignoring lift).

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Chestermiller
The drag force is based on the relative velocity of the air with respect to the drop (not the absolute air velocity), and acts in the direction of the air with respect to the drop, where ##V=\sqrt{u^2+v^2}##. So $$F=\frac{1}{2}C\rho SV^2$$The horizontal and vertical components of the drag force are $$F_x=\frac{1}{2}C\rho SuV$$ and $$F_y=\frac{1}{2}C\rho SvV$$

Chestermiller said:
The drag force is based on the relative velocity of the air with respect to the drop (not the absolute air velocity), and acts in the direction of the air with respect to the drop, where ##V=\sqrt{u^2+v^2}##. So $$F=\frac{1}{2}C\rho SV^2$$The horizontal and vertical components of the drag force are $$F_x=\frac{1}{2}C\rho SuV$$ and $$F_y=\frac{1}{2}C\rho SvV$$
So when the wind accelerates the drop in the horizontal direction, can I write for the "final" time ##t##:## \frac 12 C \rho Su^2 t=mu##? That time corresponds to the time when the drop gains the velocity ##u##.

Not even close. The correct differential equation, assuming no gravity, is $$m\frac{du_{drop}}{dt}=\frac{1}{2}C\rho S(u_{air}-u_{drop})^2$$Do you know how to solve this ordinary differential equation?

Chestermiller said:
Not even close. The correct differential equation, assuming no gravity, is $$m\frac{du_{drop}}{dt}=\frac{1}{2}C\rho S(u_{air}-u_{drop})^2$$Do you know how to solve this ordinary differential equation?
I think I cannot solve it, so what is the solution? Because in this problem below,

I used exactly that my equation above and I got a correct solution. So doesn't the solution correspond to that equation?

Lotto said:
So when the wind accelerates the drop in the horizontal direction, can I write for the "final" time
For the problem above you don't need time. Just assume terminal velocity, where drag balances gravity, and horizontal velocity equals wind-speed. In the rest frame of the airmass the drop just falls vertically, so the drag is only vertical.

A.T. said:
For the problem above you don't need time. Just assume terminal velocity, where drag balances gravity, and horizontal velocity equals wind-speed. In the rest frame of the airmass the drop just falls vertically, so the drag is only vertical.
Yes, I solved it this way as well, but then I wanted to solve it using that wind force and I used that equation in #9 and I got a correct solution. So why could I use it when it is wrong?

I would be interested in seeing the derivation of the "accepted solution" to this problem (because I'm not sure it would be consistent with the correct equations I wrote). I would like to evaluate the approximations that were made to arrive at the accepted solution.

Chestermiller said:
I would be interested in seeing the derivation of the "accepted solution" to this problem (because I'm not sure it would be consistent with the correct equations I wrote). I would like to evaluate the approximations that were made to arrive at the accepted solution.
My solution:

The "official" solution:

I have the same result, so how is that possible when my main equation is wrong?

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Lotto said:
My solution:
View attachment 328443
The "official" solution:
View attachment 328445
I have the same result, so how is that possible when my main equation is wrong?
It is hard to say because your handwriting is difficult to read, however it seems likely that you have made a compensating error. Given that the correct equation is ## \frac 12 C \rho Su^2 = mg ## whereas you have ## \frac 12 C \rho Su^2 = m \frac ut ## I suspect that you have somehow managed to use a numerical value for ## t ## that makes ## \frac ut = g ##.

Lotto said:
My solution:
View attachment 328443
The "official" solution:
View attachment 328445
I have the same result, so how is that possible when my main equation is wrong?
What this solution indicates and the problem statement fails to mention is that the droplets are traveling with the air velocity in the horizontal direction, such that $$u=u_{air,x}-u_{drop,x}=0$$Also, the velocity of the air in the vertical direction is zero, such that $$v=0-v_{drop,y}=-v_{drop,y}$$, and $$V=\sqrt{u^2+v^2}=v_{drop,y}^2$$So $$F_x=0$$$$u_{drop,x}=u_{air,x}$$and $$F_y=\frac{1}{2}C\rho Sv_{drop,y}^2=mg$$

Chestermiller said:
The drag force is based on the relative velocity of the air with respect to the drop (not the absolute air velocity), and acts in the direction of the air with respect to the drop, where ##V=\sqrt{u^2+v^2}##. So $$F=\frac{1}{2}C\rho SV^2$$The horizontal and vertical components of the drag force are $$F_x=\frac{1}{2}C\rho SuV$$ and $$F_y=\frac{1}{2}C\rho SvV$$
Forces and velocities are vectors! The correct law for a drag force quadratic in the velocity should read
$$\vec{F}_{\text{drag}}=-\alpha |\vec{v}-\vec{u}| (\vec{v}-\vec{u}),$$
where ##\vec{v}## is the body under consideration and ##\vec{u}## is the fluid flow field.

vanhees71 said:
Forces and velocities are vectors! The correct law for a drag force quadratic in the velocity should read
$$\vec{F}_{\text{drag}}=-\alpha |\vec{v}-\vec{u}| (\vec{v}-\vec{u}),$$
where ##\vec{v}## is the body under consideration and ##\vec{u}## is the fluid flow field.
We have reached equilbrium so ## \vec u = \vec 0 ##.

Chestermiller said:
What this solution indicates and the problem statement fails to mention is that the droplets are traveling with the air velocity in the horizontal direction.
To be fair the OP did mention this:
Lotto said:
[The drop's] horizontal velocity will be equal to the wind's one.

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BvU
vanhees71 said:
Forces and velocities are vectors! The correct law for a drag force quadratic in the velocity should read
$$\vec{F}_{\text{drag}}=-\alpha |\vec{v}-\vec{u}| (\vec{v}-\vec{u}),$$
where ##\vec{v}## is the body under consideration and ##\vec{u}## is the fluid flow field.
I did treat them as vectors. But it is important to recognize that the drag force vector is proportional to the relative velocity vector between the air and the drop. In the OP, the member treats u and v as the absolute air velocities in the heretical and horizontal directions.

A.T. and vanhees71
Chestermiller said:
I did treat them as vectors. But it is important to recognize that the drag force vector is proportional to the relative velocity vector between the air and the drop. In the OP, the member treats u and v as the absolute air velocities in the heretical and horizontal directions.
OK, I understand that this equation for the horizontal velocity ##\frac 12 C \rho S u^2t=mu## is wrong because the velocity in the drag formula is relative and when the drop has a wind's velocity the force is zero, but how to calculate the wind's velocity using the drag force formula for the horizontal direction?

I understand the "official' solution but I wanted to solve it differently. But the steps should be similar, shouldn't they? I need to express the velocity ##u## using time ##t## as I did it in my solution, then calculate the time and by using it express the unknown velocity.

However, I still don't understand how it is possible that I got a correct solution when my equation for the horizont velocity was wrong. Is it only a coincidence? But it is a weird coincidence.

Lotto said:
I need to express the velocity ##u## using time ##t##...
Why do you need to use time here at all? Are you trying to solve a different problem, where the drops are released at rest and then are accelerated by the wind?

A.T. said:
Why do you need to use time here at all? Are you trying to solve a different problem, where the drops are released at rest and then are accelerated by the wind?
As it has been mentioned somewhere above, the wind accelerates the drops until they have the same velocity like the wind, so I wanted to solve it with this assumption. The "official" solution has an asumption that the water drops have been already accelerated to the velocity ##u##.

Lotto said:
OK, I understand that this equation for the horizontal velocity ##\frac 12 C \rho S u^2t=mu## is wrong because the velocity in the drag formula is relative and when the drop has a wind's velocity the force is zero, but how to calculate the wind's velocity using the drag force formula for the horizontal direction?
You can't: there is no horizontal drag force because there is no horizontal relative velocity.

Lotto said:
However, I still don't understand how it is possible that I got a correct solution when my equation for the horizont velocity was wrong. Is it only a coincidence? But it is a weird coincidence.
Not a coincidence, a compensating error. The key to the solution is understanding that the ratio of the horizontal and vertical components of velocity is determined by the fact that the raindrops hit Robo's shoulder, and that the vertical component of velocity is simply terminal velocity. The equation you wrote down incorrectly for horizontal drag force (which is of course 0) was nearly the right one for (vertical) terminal velocity, and the way you used it to calculate time compensated for this.

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Lotto said:
As it has been mentioned somewhere above, the wind accelerates the drops until they have the same velocity like the wind, so I wanted to solve it with this assumption. The "official" solution has an asumption that the water drops have been already accelerated to the velocity ##u##.
But where does time come into this? Do you understand that the drops do not reacu ## u ## in finite time?

Again you have got the right numerical answer because you have calculated ## u = \frac xt ## and ## v = \frac yt ## so your meaningless value of ## t ## has cancelled out of the answer given by ## \frac u v ##

pbuk said:
You can't: there is no horizontal drag force because there is no horizontal relative velocity.Not a coincidence, a compensating error. The key to the solution is understanding that the ratio of the horizontal and vertical components of velocity is determined by the fact that the raindrops hit your shoulder, and that the vertical component of velocity is simply terminal velocity. The equation you wrote down incorrectly for horizontal drag force (which is of course 0) was nearly the right one for (vertical) terminal velocity, and the way you used it to calculate time compensated for this.
When there was no wind, the drop would be falling only in a vertical direction and then its velocity would be determined by using ##\frac 12 C \rho Sv^2=mg##. That I understand. But when the wind will suddenly start to blow with the velocity ##u##, it has to accelerate the drops in a horizontal direction to the velocity ##u##, because until the wind and the drops have the same velocities, the wind force is not zero. That's how I understand it.

So my attempts are to find an equation connecting ##u## and ##t##.

Lotto said:
When there was no wind, the drop would be falling only in a vertical direction and then its velocity would be determined by using ##\frac 12 C \rho Sv^2=mg##. That I understand. But when the wind will suddenly start to blow with the velocity ##u##, it has to accelerate the drops in a horizontal direction to the velocity ##u##, because until the wind and the drops have the same velocities, the wind force is not equal. That's how I understand it.
The wind does not suddenly start to blow, the wind has been blowing the whole time the drop has been falling for 1000s of metres. In this time, the drops have reached a horizontal velocity of (## u - \epsilon) ## where ## \epsilon ## is negligible. We are not concerned with this, we are only concerned with what happens between the roof and Robo's shoulder.

pbuk said:
The wind does not suddenly start to blow, the wind has been blowing the whole time the drop has been falling for 1000s of metres. In this time, the drops have reached a horizontal velocity of (## u - \epsilon) ## where ## \epsilon ## is negligible. We are not concerned with this, we are only concerned with what happens between the roof and your shoulder.
I know that the wind has been already blowing, I just wanted to show what I want to do - to calculate the time ##t## needed to accelerate the drops to the velocity ##u##, because the wind had to accelerate them, no matter what time exactly.

Lotto said:
When there was no wind, the drop would be falling only in a vertical direction and then its velocity would be determined by using ##\frac 12 C \rho Sv^2=mg##.
Thats its terminal velocity. Its only approached in the limit as ##t \to \infty##

In the absence of "wind" ( 1D motion) the velocity as a function of time ##v(t)## is expressed as the solution to the nonlinear ODE:

$$\uparrow^+ \sum F = \frac{1}{2} \rho A C_d v^2 - mg = m \dot v$$

Lotto said:
I just wanted to show what I want to do - to calculate the time ##t## needed to accelerate the drops to the velocity ##u##

pbuk said:
Do you understand that the drops do not reach ## u ## in finite time?

And why don't they reach ##u## in finite time?

Lotto said:
And why don't they reach ##u## in finite time?
To give you a helpful answer to this we need to know a bit more about your background in Physics and what level you are studying at.

Edit: oh and I've just realised this is not in the Homework section where it belongs, I'll get it moved.

pbuk said:
To give you a helpful answer to this we need to know a bit more about your background in Physics and what level you are studying at.

Edit: oh and I've just realised this is not in the Homework section where it belongs, I'll get it moved.
At the beginnig, it wasn't supposed to be a homework, my question was inspired by it.

And I studied at a high school, now I am going to go to university.

pbuk
Lotto said:
And why don't they reach ##u## in finite time?
Because that is a characteristic of the solution to the proposed ODE.

$$\uparrow^+ \sum F = \frac{1}{2} \rho A C_d v^2 - mg = m \dot v$$

$$\implies \dot v = \frac{1}{2 m} \rho A C_d v^2 - g$$

as ##v## increases ## \dot v \to 0 ##. That creates an issue that is circular, because in order for ##v## to increase we need ## \dot v > 0 ##. The acceleration ##\dot v## keeps getting smaller, and the change in velocity keeps getting smaller add infinitum as ##t \to \infty##

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Lotto said:
OK, I understand that this equation for the horizontal velocity ##\frac 12 C \rho S u^2t=mu## is wrong because the velocity in the drag formula is relative and when the drop has a wind's velocity the force is zero, but how to calculate the wind's velocity using the drag force formula for the horizontal direction?
They are telling you that the drop velocity is equal to the air velocity by the time that the air flows under the shelter. That means that the horizontal component of the drag force has dropped to zero by the entrance to the shelter and the drop velocity equals the air velocity. For this problem, the time that a parcel of air with its entrained water drop enters the shelter is taken as t = 0.

Therefore, you have $$u_{air,x}t=R$$and $$v_{drop,y}t=H-h$$where H is the height of the shelter overhang and h is the height of the top of the guy's head. Eliminating t between these equations gives $$u_{air,x}=\frac{R}{H-h}v_{drop,y}$$

pbuk

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