Finding distance travelled with drag

[SirHall]

I've been working on a little project recently and I've been scratching my head over this one problem, is there a way to calculate how far an object will travel while drag is taking effect. Assuming there is no gravity or any other forces acting on the object, if we take it's velocity, mass, air density, drag coefficient and so on, is it possible to find how far it will travel before completely stopped by the drag?
Yeah sorry Google didn't help me out much.

 

[Simon Bridge]

Yes it is - generally drag force would be proportional to the square of the velocity, and it is the only force on the object: it's Newton's second law and some calculus.
The distance it travels is the area under the velocity-time graph.

 

[SirHall]

*facepalm* Just realized all I have to do is find how much distance would be required to equal the energy of the moving object without any drag. Therefore just find how much energy is required to stop the moving object, and that'll lead to the distance covered. Correct me if I'm wrong.
By the way, thanks for the advice.

 

[Simon Bridge]

Sure - work against drag will equal the initial kinetic energy ... however, the force of drag depends on the square of the velocity.
model $\vec F = -\sigma \vec v^2$ where $\sigma$ is the result of air density and so on.

Then by work-energy theorem:
$\frac{1}{2}mu^2 = \int_{x_0}^x F(x')\; dx' = \sigma \int_{x_0}^x v^2(x)\; dx'$
... if you are comfortable doing that integral then more power to you.

By Newton:
$-\sigma v^2 = m\dot v$ with initial value $v(0)=u$
... maybe use relation $\dot v = vv'$ and solve for v(x)

A problem should appear about now :)

 

[SirHall]

Okay, just did some working out and I think I have the answer.
Firstly I got the drag equation, and swapped out the density for mass/volume (because my use requires that) and replaced velocity for distance/time(as that is where I can get a distance variable for the equation). This gave me $F\ =\ C\ \cdot \ \frac{\left(\frac{m}{v}\cdot \left(\frac{d}{t}\right)^2\right)}{2}A$ . So I simply solved this for the distance and got $d\ =\ \sqrt{\frac{\left(2ft^2v\right)}{acm}}$
Sorry but I really don't want to type up every step in-between, it's just easy algebra.

 

[Simon Bridge]

This is not correct ...
1. your equation for d has two unknowns on the rhs... t and f... you need d in terms of things you know.
2. d/t is for average velocity over a time interval, not the instantanious velocity, so the substitution is not valid the way you are doing this. Putting distance as x, the correct substitution is: v=dx/dt (the time derivative of the position.)

Aside: comparing, my posts put $\sigma = \frac{1}{2}\rho CA$

 

[SirHall]

This is not correct ... d/t is for average velocity over a time interval, not the instantanious velocity.
Putting distance as x, then v=dx/dt

Aside: comparing, my posts put $\sigma = \frac{1}{2}\rho CA$

Hmm, I remember something along those lines from back in high school.
While sort of a cheap solution, since dt = t2 - t1, I'd assume it's safe to say: $dx\ =\ \sqrt{\frac{\left(2fdt^2v\right)}{acm}}$

 

[Simon Bridge]

That is not safe.

 

[SirHall]

That is not safe.

*Shot down*

 

[Simon Bridge]

It suffers the same two problems already pointed out.
Lets say you know m=1kg, V=1m^3, C=0.5, A=1m^2, and the initial speed u=1m/s
Use the equation you derived to find the stopping distance.

Thats point 1.

To approximate dx/dt as $\Delta x/ \Delta t$ you need the drag force to be a constant over the entire time. Is it?

 

[SirHall]

It suffers the same two problems already pointed out.
Lets say you know m, V, C, A, and the initial speed u.
Use the equation you derived to find the stopping distance.

In stopping distance I'd assume you mean $dx\ =\ x_1-x_2$ where $x_2$ is the stopping distance.
Sorry if I'm coming a little slow to this, but I do appreciate actually being led to the solution.

 

[Simon Bridge]

The stopping distance is what you wanted to calculate in post #1.
Its what you asked about. Its what the equation is supposed to be for right?

Note: $\Delta x = x_2 - x_1$ ... change = final minus initial.
$$\frac{dx}{dt} = \lim_{\Delta t \to \infty} \frac{x(t+\Delta t) - x(t)}{\Delta t}$$

You can use average velocities if you divide the time to stop into lots of very small intervals where the drag is aproximately constant and add up the distances traveled in each interval, but it is better to just use calculus.

 

[SirHall]

Yes, I'm still trying to find the stopping distance. But now I'm sort of at a loss. Would I simply throw that equation into the original to replace the dx/dt?

 

[Simon Bridge]

Yes, I'm still trying to find the stopping distance. But now I'm sort of at a loss. Would I simply throw that equation into the original to replace the dx/dt?

I don't know, it's your equation. You are the one who has said that it works (post #5). (How were you expecting to use it?)

I disputed that the equation works - you do not seem to understand the objections I have raised, so I am trying to get you to demonstrate how you would go about using your equation to find the stopping distance. That way, I hoped, you would come to understand the 1st objection I raised.

Lets just make sure I have properly understood you - back in post #5 you wrote:

Okay, just did some working out and I think I have the answer. ...

... you basically did this:

Drag force: $F_d = \frac{1}{2}\rho AC v^2 = \frac{m}{2V}AC(\frac{d}{t})^2$ then you did some algebra to make d the subject:

and got $d =\sqrt{\frac{\left(2ft^2v\right)}{acm}}$

Tidying that up a bit: $$d= \sqrt{ \frac{2V Ft^2}{ACm} }$$ ... all right. If I am following you, then "d" in that equation is the stopping distance (the distance it takes the object to come to rest under drag alone).
A is the crossection area, C is the drag coefficient, V is the volume of something (what?), m is the mass of something (what?), F is the drag force (how do you determine that?), and t is some time interval (for what?).

Have I understood you correctly?

My first and strongest objection to that equation is that you have the thing you want to find in terms of some stuff you know (V, A, C, m) and some stuff you don't know (F, t). This means that you have not, in fact, done what you set out to do in post #1.

Do you follow me so far?

 

[SirHall]

I don't know, it's your equation. You are the one who has said that it works (post #5). (How were you expecting to use it?)

I don't recall stating that it worked, more just asking if or whether it did indeed work.

I disputed that the equation works - you do not seem to understand the objections I have raised, so I am trying to get you to demonstrate how you would go about using your equation to find the stopping distance. That way, I hoped, you would come to understand the 1st objection I raised.

I'd assume you were talking about:

Yes it is - generally drag force would be proportional to the square of the velocity, and it is the only force on the object: it's Newton's second law and some calculus.
The distance it travels is the area under the velocity-time graph.

Yes, I understand what you said, which is why I used the original drag force equation which does indeed use velocity squared.

... all right. If I am following you, then "d" in that equation is the stopping distance (the distance it takes the object to come to rest under drag alone).
A is the crossection area, C is the drag coefficient, V is the volume of something (what?), m is the mass of something (what?), F is the drag force (how do you determine that?), and t is some time interval (for what?).

d = Distance it takes for the moving object to go from initial velocity to final velocity.
V = Volume of the moving object (This came from the density variable which was replaced with Mass/Volume)
M = Mass of the moving object
F = Drag force (Admittedly there is a problem here...)
T = Time Interval (Well I simply got Distance/Time from the Velocity variable in the original equation. So I'd assume that you'd get the initial velocity of the object and divide that by... Wait, I can't divide it by the distance covered as that is what we're trying to find out. Either the entire idea behind the formula is wrong, or I'm forgetting something really simple.

Have I understood you correctly?

My first and strongest objection to that equation is that you have the thing you want to find in terms of some stuff you know (V, A, C, m) and some stuff you don't know (F, t). This means that you have not, in fact, done what you set out to do in post #1.

Do you follow me so far?

Yes, I see your points.
I see I have not achieved what I wanted back in post #1 but I can assure you that is still my aim.

 

[Simon Bridge]

Either the entire idea behind the formula is wrong, or I'm forgetting something really simple.

... well done: the entire idea behind the formula is wrong.
More accurately, the reasoning that lead you to the formula was flawed. Happens to the best of us.
Best to start again. Let me know where I lose you...

You want to find the distance that an object stops in, given:
m: mass of the object
u: initial speed
A: cross-sectional area
C: drag coefficient
$\rho$: air/fluid density

It is important to understand the equations you use.

1. The proper drag force equation is: $\vec F= -\rho CA v\vec v$
The minus sign means that the force is in the opposite the direction to the velocity.
The magnitude of this force is $F=\rho CA v^2$ ... which is the equation you used.
In that equation:

$\rho$ is the density of the air, not of the object that is moving;
$v(t)=dx(t)/dt$ is the instantaneous velocity at the instant of time $t$.​

... this is very important because the velocity is always changing, and not in a simple way.
This means that simple averages like you tried are not going to be very good.
This also means that the drag force is not a constant.

Note: it is inconvenient to use "d" for "distance" because I want to be able to use it for calculus notation.
So let's call the stopping distance "b" .. (a "d" backwards).

2. Work: $W=-\int_C \vec F(\vec s)\cdot d\vec s$ ... this is the full equation for work moving against the force $\vec F$.
It is what happens to "force times distance" when the force is not a constant, and may point in any direction wrt the displacement.
Here $\int_C$ means you integrate over the path taken, $\vec s$ is the displacement vector in general coordinates, and the dot is the vector dot product.
In your case, this simplifies to: $W=\rho AC \int_0^b v^2(x)\; dx$ ... which means the object starts at x=0 and ends at x=b and has to push against F.

3. Conservation of energy:
The loss in kinetic energy is equal to the work done pushing through the air (work against drag).

$\frac{1}{2}mu^2 = \rho CA \int_0^bv^2(x)\; dx$​

... and this is the equation you have to solve to find b.

Alternatively:
4. Newtons Laws:
$\sum \vec F = m\vec a$ ... this is actually a differential equation, because $\vec a = \frac{d}{dt}\vec v$
We have one force so: $-\rho AC v^2 = m\frac{d}{dt}v$ you also have initial conditions $x(0)=0$ and $v(0)=u$
(again the minus sign is because the force is opposite the velocity)
... solve this for $v(t)$, then find $T: v(T)=0$, then $b=\int_0^T v(t)\; dt$ or something.

How good is your maths?

 

[SirHall]

... well done: the entire idea behind the formula is wrong.
More accurately, the reasoning that lead you to the formula was flawed. Happens to the best of us.
Best to start again. Let me know where I lose you...

... Taunt appreciated :)

How good is your maths?

Great question!

Well aren't I damned, you pointed out three to four things I forgot which were trivial, made three to four steps which I needed some researching to get a basic understanding of, and to top it all off, I'm not entirely even sure exactly how to script this.

Well then, time to get cracking.

By the way, thanks for the in depth explanation.

 

[SirHall]

Okay, (yes, still trying to figure this out) I've re-learnt derivatives and integrals.
I think I've figured out the answer to :

3. Conservation of energy:
The loss in kinetic energy is equal to the work done pushing through the air (work against drag).
12mu2=ρCA∫b0v2(x)dx12mu2=ρCA∫0bv2(x)dx\frac{1}{2}mu^2 = \rho CA \int_0^bv^2(x)\; dx... and this is the equation you have to solve to find b.

But, before I come to any conclusions I was wondering, what is V? You have u for initial velocity, and there isn't any t if I was supposed to find the instantaneous velocity. I also don't think I could use:

By Newton:
−σv2=m˙v−σv2=mv˙-\sigma v^2 = m\dot v with initial value v(0)=uv(0)=uv(0)=u
... maybe use relation ˙v=vv′v˙=vv′\dot v = vv' and solve for v(x)

A problem should appear about now :)

I assume this was for something slightly different. However, if I am supposed to use this to find V, then what exactly is V in this sense, and what are those pretty little dots and apostrophes floating around them?(I'd assume the apostrophe's are not referring to derivatives.)
By the way, thanks for the help so far!