Rebound from air friction

DivisionByZero

I believe that the common drag equation is this:

Fdrag= 1/2*rho*velocity^2*Cross section*Constant

D=1/2rv^2*s*c

Please correct me if this is wrong.

Anyway, I was wondering, what if the drag exceeded the velocity?? the Object would rebound, but certaintly this is not possible. what if density was .2,s and c were 1? then it would be v^2*.1 then, what if velocity was 20? 400*.1=40. Can somthing rebound with twice the speed? I think not.

My equation must be wrong.

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Integral

Staff Emeritus
Gold Member
Isn't drag a force? How do can you say a force is greater then a velocity? They do not have the same units, therefore they cannot be compared in that manner.

turin

Homework Helper
I agree with integral. I would like to add that this equation is an estimation. It is only valid when it is being used in its appropriate range. So, if you realized that it was returning nonsense, then it wouldn't be valid because you wouldn't be using it in the appropriate range. I suppose you are just now realizing that equations are not fundamental laws? I have a general rule of thumb that I go by: The more complicated the equation, the less likely it will be valid in the range that I want to use it.

DivisionByZero

Ah. I see your points. also noted that the equation is also not an absolute yet an aproxamate.

You're right, drag is a force. F=ma so lets say that mass=1. so now we have an acceleration which reverses the veloctity. Taking into acount turin's suggestion/explanation it makes since, this equation is a estimate.

joc

argh, no, the equation doesn't screw up in this case. it's exact for the model we're considering.

eqn for drag force:

Fd = -kv
or
Fd = -kv^2 (for high velocities)

where k is a constant and v is the velocity. the negative sign indicates that the drag acts opposite to the direction of the velocity. let's take the case of high velocities (since low velocities will most likely become high anyway in the course of a long fall).

the only other force of relevance acting on the falling object is the weight, which equals mg.

the NET FORCE acting on the object, Fnet, is given by

Fnet = mg + Fd
Fnet = mg - kv^2

furthermore, Newton's 2nd Law says that

Fnet = ma

where a is the acceleration of the object.

therefore,

ma = mg - kv^2
a = g - kv^2/m

at the start of its fall, the object has 0 velocity, so a = g. immediately after that, the velocity starts increasing pretty quickly, so the kv^2/m term starts increasing as well. this means that the acceleration, a, starts to fall (since a = g - kv^2/m).

eventually, a point comes when kv^2/m exactly equals g. this means that a = g - kv^2/m = 0. if acceleration is 0, the object will be falling at a constant speed, which is known as the terminal velocity. if there's no further change in velocity, it obviously cannot turn negative. negative velocity is equivalent to 'rebounding', and so rebounding never occurs.

a = g - kv^2/m is actually a simple differential equation. a = dv/dt. acceleration DEPENDS on velocity.

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turin

Homework Helper
Originally posted by joc
argh, no, the equation doesn't screw up in this case. it's exact for the model we're considering.
If this is in response to my comment, then I basically agree with you.

But I think the concern was really about what is going on for extremely high velocities. The fact that there is a square term in the equation should be a good indication that it came from some approximation. As we go to higher and higher velocities, it may turn out that we must include higher order terms for the sake of accuracy. Of course, even if we are negligent, there still won't be a nonsensical result, just an inaccurate one.

Let me give an example of where this equation does NOT work:

We have been doing the Milikan Oil Drop experiment in my lab for a few weeks now. For those of you who don't know, this experiment involves measuring the velocities of VERY VERY tiny particles. Then, we apply an equation that takes into acount the drag force to calculate the mass of the particle. This equation basically relates drag force to velocity, but, if you really believe the equation to be absolutely correct, you may be rather perplexed by the large number of particles that don't fall at all (they are so small that you can see them being buffeted around by the air molecules). For particles just larger than this, their drag force becomes rather statistical, since you can almost see each interaction with the air molecules. The equation deals in aggregates.

I know that this is taking the arguement in another direction, but I thought it might be a nice concrete example of the applicability of the drag force equation. I haven't done any experiments in which I observed/noticed the drag force "breaking the rules" for very large velocities, but I know it can happen. For instance, spacecraft on reentry already exceed their terminal velocity. They hit the atmosphere and get really hot. That's why you have to plate them with ceramic. They settle down to their normal terminal velocity, and I know this doesn't show how the equation is violated, but I am imagining that it does get violated.

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russ_watters

Mentor
Originally posted by DivisionByZero
You're right, drag is a force. F=ma so lets say that mass=1. so now we have an acceleration which reverses the veloctity. Taking into acount turin's suggestion/explanation it makes since, this equation is a estimate.
The acceleration will not reverse the velocity because as the velocity decreases, so does the drag force. Eventually the object will just stop.

DivisionByZero

Thanks for everyone's help on the subject.

joc:

Nice equation (a=g-kv^2/m) I'm probably going to directly apply it to my physics system. I don't really agree on the acceleration depending on velocity. True, acceleration is measured as the difference in velocity over time, and is the derivative of velocity. Yet in my imagination of the workings of the universe (disgarding quantum physics and relativity) , it is force imparted on objects which changes the acceleration, which changes velocity, which changes position. For instance, a ball is given a force(throw), this is really an acceleration (the force/mass), which changes the velocity, which changes the position. Perhaps in physics texts, acceleration is dependant on velocity, but I wouldn't know.

I haven't taken any physics classes yet. I have however read "Mathamatics and the physical world" and a book on quantum mechanics. I also read an article on gamasutra on game physics. it was very easily implemented, at first. Now my physics engine is quite advanced(from my view). I'm just saying, I don't have that strong of a background on physics terms.

russ_watters

Mentor
Originally posted by DivisionByZero
I don't really agree on the acceleration depending on velocity. True, acceleration is measured as the difference in velocity over time, and is the derivative of velocity. Yet in my imagination of the workings of the universe (disgarding quantum physics and relativity) , it is force imparted on objects which changes the acceleration, which changes velocity, which changes position.
You are correct: force causes acceleration which changes velocity. But what if force is tied to velocity? Then velocity causes force which causes acceleration which changes velocity (and round and round we go). Thats how drag works. Heck, YOU posted the equation!

joc

russ_watters put it nicely. acceleration is a physical result of force, but via that equation it is tied to velocity. sometimes in physics calculations, what directly causes a phenomenon is not as important as its relationship with various quantitative variables. anyway good for ya that you've got your game physics right :)

turin - i guess if you want to observe extreme/limiting cases like extremely small oil drops or extremely high velocities, the equation will have to be modified to factor in the additional considerations. but i think very accurate descriptions of even these extreme cases can be obtained as a result...we have the mathematical sophistication to do it.

turin

Homework Helper
Originally posted by DivisionByZero
I don't really agree on the acceleration depending on velocity.
Well, it does. How would you explain terminal velocity? Think of it this way: the faster an object moves through a fluid, the more often it will have a collision with a molecule of the fluid. More collisions means more impulse. Therefore, more collisions in some amount of time means more impulse per time means more force.

Originally posted by DivisionByZero
True, acceleration is measured as the difference in velocity over time, and is the derivative of velocity.
If you wanted to measure it directly, then it's the first thing you said. If you want to define it (instantaneously), then it's the second thing you said. Be careful not to get a definition confused with a statement about the physical interactions.

Originally posted by DivisionByZero
Yet in my imagination of the workings of the universe (disgarding quantum physics and relativity) , it is force imparted on objects which changes the acceleration, which changes velocity, which changes position.
First, I don't see why you are mentioning QM or relativity. Force, by definition, changes momentum. If you want to define some quantity that changes the accleration, then you probably don't want to call it force (at least, not on a physics forum), or you will get a lot of confused replies. Again, it is important to distinguish between definitions and statements about physical interactions.

Originally posted by DivisionByZero
For instance, a ball is given a force(throw), this is really an acceleration (the force/mass), which changes the velocity, which changes the position.
I'm confused. Did you mean to say earlier that force changes the acceleration?

Originally posted by DivisionByZero
Perhaps in physics texts, acceleration is dependant on velocity, but I wouldn't know.
I'm sure it is, if it is in the context of damped motion. The applied force is the driving term. The acceleration that depends on velocity comes from the damping force. And, again, the force is not DEFINED to be dependent on velocity, it is OBSERVED to be.

Originally posted by joc
... if you want to observe extreme/limiting cases like extremely small oil drops or extremely high velocities, the equation will have to be modified to factor in the additional considerations.
I don't care one way or the other, but I thought that Divisionbyzero was interested in what happens in the extreme case.

Originally posted by joc
... i think very accurate descriptions of even these extreme cases can be obtained as a result...we have the mathematical sophistication to do it.
Ya, I wasn't really arguing with you, and I know that there is an equation that is supposed to describe the drag for the extremely small. I just thought it would be edifying to show a concrete example of how an equation breaks down.

DivisionByZero

I am interested in the extremes, although mostly out of curiosity, as it most likely wouldn't be put to much use. Although there is the posibility of objects getting caught in a loop where they jitter back and forth due to this equation. (in my physics engine).

I do understand the concepts of drag, etc. to explain terminal velocity using velocity depending on acceleration, let's look at an example in which you shove an object into the water that sinks into the water. The object has a large cross section (towards it's velocity). For a moment (not instantanious) your shove applies an acceleration(force). This force effects the object's velocity, and once in the water, the drag created by the water will cause an acceleration in the opposite direction of the velocity, depending on the velocities magnitude. anyway, in this case, lets say the object's velocity is greater than it's terminal velocity in the water. it slows down rapidly, and reaches it's approximate terminal velocity. finally, it will gradually drift down. Now, what's happening here (as you know), is that the velocity is causing an acceleration, which should be exactly the inverse of the gravity(or bouyancy in this case). This will cause it to move at a constant velocity.

I mentioned QM and relativity simply because I wanted to make sure no one posts saying "Well according to the theory of relativity...." lol. In retrospect, it really doesn't make any since.

About the whole force changes acceleration, well no, that was just a screwup on my part. I'm simply confused on the matter of force because to me it seems to mean exactly the same as an acceleration, except that with objects that have more mass, the 'energy', or substantialness of it's movement is greater.

The whole velocity causing acceleration and acceleration causing velocity is confusing, but I think I got it now. With physics in general(to me), acceleration causes or modifies velocity. With the particular physics in drag, it is a feed-back situation, in which the velocity causes an acceleration which changes the velocity.

anyway, I guess I got that worked out now.

Just thought of somthing..... perhaps an example of rebound due to somthing like this might be say, a brick wall. true, many objects will rebound off the wall, but this is simply a function of the object itself. The object cannot push through the brick wall, and the impulse of the collision would cause a rebound, or at least a velocity of zero. I guese in this case, k is very high causing the rebound effect.

Possibly, probably, I'll post a topic in a little while about bouyancy... This time with a better understanding of how clear I have to be, and a better understanding of bouyancy in general. who whoulda guesed.... k*(enviornment density-object density).....

turin

Homework Helper
Originally posted by DivisionByZero
(1)
... lets say the object's velocity is greater than it's terminal velocity in the water. it slows down rapidly, and reaches it's approximate terminal velocity.

(2)
I'm simply confused on the matter of force because to me it seems to mean exactly the same as an acceleration, except that with objects that have more mass, the 'energy', or substantialness of it's movement is greater.

(3)
With physics in general(to me), acceleration causes or modifies velocity.

(4)
With the particular physics in drag, it is a feed-back situation, in which the velocity causes an acceleration which changes the velocity.

(5)
... perhaps an example of rebound due to somthing like this might be say, a brick wall.
(1) I'm assuming that you mean right after you're done pushing down on the object.

(2) From this statement, you don't seem to be confused at all. In fact, in case you don't know, this is basically Newton's second law (which is really just a definition).

(3) It's actually more general than physics. It is a mathematical definition. To declare something to the contrary would be nonsensical, with or without any physics.

(4) This sounds like a good way to think about it. I would more specifically say "negative feedback," but I suppose that is generally implied.

(5) This sounds like an interesting way of thinking. Are you suggesting that the rebound from a brick wall could be modeled as a caused by a damping force with an extremely high damping coefficient? I think there might be something wrong with that, but I can't put my finger on it right now. I'll have to give it more thought and then get back to you, that is, if you're interested in hearing what I have to say.

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DivisionByZero

Sure, why not?

I agree, the brick wall thin doesn't seem right. The greatest problem I see with it is with slow-moving objects. according to this equation, the drag(or rebound!) would be much less (v^2). The drag might not be signifigant enough to rebound or even stop motion. I believe that mathamatically you can prove that as long as k is finite, there would be a none zero number that would produce negetive acceleration less than the original velocity. Another thing is that rebound is mostly a product of the elasticity of the rebounding object.

russ_watters

Mentor
Originally posted by DivisionByZero
About the whole force changes acceleration, well no, that was just a screwup on my part. I'm simply confused on the matter of force because to me it seems to mean exactly the same as an acceleration, except that with objects that have more mass, the 'energy', or substantialness of it's movement is greater.
Well, acceleration is proportional to force and the proportionality constant is mass: f=ma

The best way I can think of to help you understand this stuff is to start combining related equations.

if a=f/m
and f=Cd*V^2 (not exactly, but sorta)
then you can combine the two to figure out how velocity and acceleration (deceleration due to drag) relate:
a=(Cd*V^2)/m
negetive acceleration less than the original velocity
No. Velocity and acceleration are not the same thing and can't be compared in that way: v=a*t

Maybe this is the problem: deceleration due to drag as calculated in those equations is an INSTANTANEOUS thing. It is only valid for the instant it was calculated. As soon as the velocity changes a little (because the object is decelerating) then you need to plug that new velocity back into the drag equation to find a new drag force and new deceleration. Do this over and over again with small enough time steps (or simply solve the equation through calculus) and eventually the drag force goes to zero as the velocity goes to zero. Or in the case of terminal velocity, drag force becomes equal to weight and velocity becomes constant.

S = k log w

I have a question. At very small diameters water looses it's surface tension. This may change the resistance with air (or other media).
How small is this "drop" of water? At what point does it become a gas?

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