Physics and Accident Reconstruction: Solving the Drag Equation

[Brett R.]

HI all,
I am new to the forum and have a real question for you physics buffs. It has to do with drag. For a mathematics paper, I am exploring the use of physics in accident reconstruction. I want to use calculus to create differential equations for the drag equation .5p(Cd)(A)v^2... I want a drag equation that can be used for a car's motion. However, sources are scarce. I am having trouble with the integration and trying to get the equation into differential form. I am a BC calc student, but am new to physics and the application is a bit tricky for me. Help would be so very much needed!

 

[brainpushups]

Hi,
You will want to start with Newton's second law ##ΣF_x=m*dv_x/dt## where the subscripts indicate that the forces are along the x direction as I assume you'll be working on a horizontal surface. Is friction present in your model as well or is this an exercise just in using the drag equation? Regardless of the answer to that question it is advisable that you collect all of the constants and just label them as a single constant for the purposes of the integration. For example just call ##1/2 C_dρA=β## or something.

 

[Brett R.]

Well I am using a Skid mark equation √μgdcosθ. So a car is traveling along the x and skids to a stop. I want to compare the various effects of velocity and frontal area on drag. I want to know if drag will have a major effect on the calculated velocity based on a skid mark length. So there is friction. I think I get what youre saying. So -μmg-βv² = dv/dt then integrate to find the equation?

 

[brainpushups]

So -μmg-βv2 = dv/dt then integrate to find the equation?

Yup. The integral isn't bad. The constants make it seem more complex than it is. If it gives you trouble I can make another suggestion about how to rename everything so that it might look more familiar.

 

[Brett R.]

how do I integrate -μg-βv² = dv/dt

 

[brainpushups]

Ok. First you want to get all terms with t and dt on one side and all terms with v and dv on the other. This is called 'separation of variables.' Then you will integrate both sides with the appropriate limits. First, let see the separated form of the equation (also note that you have a typo - a missing m)

 

[Brett R.]

I divided both sides by m to get the m on the right side to go away. My problem is I was having trouble getting the v^2 term to the right

 

[Brett R.]

 

[brainpushups]

I divided both sides by m to get the m on the right side to go away

Uh oh. You better check your algebra; m is not a common factor for ALL of the terms so it cannot be eliminated from the equation.

My problem is I was having trouble getting the v^2 term to the right

How did you get dt to one side?

 

[Brett R.]

Uh oh. You better check your algebra; m is not a common factor for ALL of the terms so it cannot be eliminated from the equation.

I know but if you divide by both sides, you are left with an m under the second term. Then I substituted out.

As for dt, if I multiply across I can not get the v^2 to the right since it would be trapped by the dt.

 

[brainpushups]

Oh. Thanks for the attachment - the algebra is ok - you've absorbed m into your constant β. So from here

-μg-βv²=dv/dt

how can you get dt on one side and and v and dv on the other?

 

[brainpushups]

As for dt, if I multiply across I can not get the v^2 to the right since it would be trapped by the dt.

What if I write it like this:

$(-μg-βv^2)dt=dv$

Does that help?

 

[Brett R.]

that would leave it in the form. (-μg)dt + (-βv²) dt=dv This is where I was stuck. I do not know how to go farther without further complicating the terms.

 

[brainpushups]

Don't distribute the dt. Just divide by the entire factor in parentheses.

 

[Brett R.]

OHH I think i got it. Because nothing here but velocity varies with time. So if I just divide, I can integrate dt on the left and 1/the rest on the right

 

[brainpushups]

Bingo. The last 'trick' is to get the integral involving velocity into a form that is recognizable. You may have seen certain integrals where the denominator has a term that includes x². See what you can do with it and check back if you need another hint.

 

[Brett R.]

So it is arc tan. i manipulate to get ∫dt= (1/μg)[1/(1+(β/μg)v²)]dv

 

[Brett R.]

Sorry for all the questions, but my math skills have not gotten me to the point where I can take all the information I learn and apply it to physics equations. I really appreciate your help.

 

[Brett R.]

However, I need the form 1/1+v². I have 1/(1+ (β/μg)v²). How do I get that other part out.

 

[brainpushups]

So it is arc tan

Yes.

Sorry for all the questions, but my math skills have not gotten me to the point where I can take all the information I learn and apply it to physics equations. I really appreciate your help.

No problem. This stuff takes practice.

However, I need the form 1/1+v2. I have 1/(1+ (β/μg)v2). How do I get that other part out.

There is nothing preventing you from performing the integration as it stands. You know that the function should involve arctangent and the only difference between the familiar 1/1+v² and what you have is a constant (you may as well call β/μg = k or something just to make things neater; in fact you could call β/μg = k² to make things even nicer) .

My suggestion is to make a guess at what you think the integral will be and then differentiate your answer. If you end up with the original thing you were trying to integrate then you have succeeded. If you end up with something else upon differentiating your guess I bet you'll see what needs to change. For example, what if you just guess that arctan(k v) is the solution (where k² is β/μg)? If this is the 'guess' then what is the derivative of this guess?

I hope that wasn't too verbose...

 

[Brett R.]

So... I think I got it. So the integral is arctan(kv) if you set β/μg = k² which makes things a lot easier. Then you substitute back in for k and you get arctan(βv/μg)=0 right?

 

[Brett R.]

However, I want to know how drag changes with velocity. How will this help me with that?

 

[brainpushups]

So... I think I got it. So the integral is arctan(kv) if you set β/μg = k2 which makes things a lot easier. Then you substitute back in for k and you get arctan(βv/μg)=0 right?

Not quite. Try differentiating arctan(k v). What is the derivative?

However, I want to know how drag changes with velocity. How will this help me with that?

You already know how drag changes with velocity drag = 1/2CpAv²

I thought your goal was to see if the drag had a significant affect on the motion...

 

[Brett R.]

Is it 1/1+(kv)²?
That is my goal, however I don't think this equation will tell me that goal anymore.

 

[brainpushups]

Is it 1/1+(kv)2?

No. You need to apply the chain rule.

Once you have the integral evaluated you can solve for v as a function of time. This will allow you to compare, for example, the time it takes for the object to come to rest with and without drag. Perhaps this is too detailed. Another thing you can do to help decide whether drag is significant is to compare the maximum force of drag to the constant force of kinetic friction. This requires no calculus.

 

[brainpushups]

Looking at your original post I think you DO want the equation for the velocity. Once you have this you can separate variables and integrate again for the position as a function of time.

 

[mp3car]

Yes, drag will obviously have some effect on velocity while stopping, which would mean values obtained from skid-based formulas without factoring in drag would not be accurate. However, my initial response is that drag is not usually accounted for when stopping or skidding is because it is extremely negligible compared to the braking force (well, the tire/road interaction). Just think about how long you coast if you put it in neutral, compared to how short of a distance it takes to stop when braking, even in a skid. My previous vehicle had a 0.29 Cd and with no wind, it could coast in neutral from a mile or more when beginning at highway speeds, which tells me the drag force is very minimal when compared to braking (even including a skid, you still would stop MUCH sooner than coasting to a stop).

I am interested to see some examples once you have it figured out! I will be very surprised if the difference by taking drag into effect makes any bigger of a difference than real life variability of other assumptions that are being made, such as the friction between tires and the road. There are so many variables with that interaction that the assumptions made there would probably play more into the end result than drag... Such as age and wear on the concrete or asphalt, temperature of the surface and the tire, tire composition (e.g. a harder economy tire vs. a soft high performance tire), did you hit a little bit of gravel or slightly sandy spots during the skid, etc... AND - was there wind? If there was a "tailwind" of 30mph, then at 30mph, there would be very little drag force at 30mph. If it was a headwind of 30mph, then at 30mph, the drag force is roughly the same as it would be at 60mph.

It's been a long time since I've done motion problems or even advanced math, so maybe I'm wrong and drag plays a contributing role, but my gut tells me it's negligible... Keep at it because I am curious, maybe my hunch is wrong! :)

 

[mp3car]

Figure out the impact of drag yet? (with respect to its impact when braking/skid)