I've been searching for this formula for quite a while now, with no results...

So simply put:

Anyone can give me the formula for the force of air friction? My guess is that lots of variables are involved here, and I would expect the formula to be beyond my comprehension, but I'm willing to give it a try.

Thanks for any input.

Dear A.M.,
You have asked very tough question.
I could mention Stokes formula
F = akSv
where a is some numerical coefficient (usually close to 1), k is medium’s (air) viscosity, S is normal cross section of body and v is its velocity in respect to medium. But this formula works only for well-streamlined bodies (like ball) and at v small enough in comparison with so called the Raynolds first critical velociyu. For example, this formula very well works for balls dropping in oil.
I could mention Karman formula which works at creation of so called “Karman chain” of vortices behind streamlined body.
I could mention my own formula which works at creation of so called “vortical boiling” (when surface of the streamlined body creates vortices and/or rotons; is used for golf balls, in blades of jet engines, uranium cassettes in reactors, and so on). Etc, etc….
The problem is that no one of these formulas will be useful in your case of an arbitrary body and arbitrary regime of streamlining. There is no such formula in Physics – too complex phenomena are happened at such streamlining. It is why physicists build Air-tubes – to recreate the real conditions of streamlining and simply measure the friction force

So, my advice is: If you really want know value of this force for your device, go in library and find reports on streamlining experiments with bodies like yours and in range of velocities like you are expecting to have.
And good luck with this…..

The only real way to do it is numerical simulation on a computer, everything else is a rough estimate.

Simulation of what?

Air flow.... FEA for fluids basically.

How you will simulate turbulent effects, if nobody knows how to describe them, at all? What equations, models and conditions of initiating of different possible turbulence you will be programming at Computer simulation?

Turbulent effects are the result of small surface/mass changes. The simulations take this into account.

Please, show us the simpleast procedure of simulation of simpliest type of turbulence you know

You're looking for simple? You're going to be disappointed...

The simplest way is to use the navier-stokes equations and just treat the result as an average. That's cheating though, and you don't really 'see' the turbulance.

To get slightly more complicated you can introduce the reynolds stress (tied in with roughness of object) into the navier-stokes equations.

To get anywhere near reality you move onto computers. You break the object and air into an adaptive mesh for FEA. You introduce friction where two meshes meet and there is a different velocity/material. Depending on the resolution of your mesh you can go all the way from currents (small matrix) to detailed turbulance (large matrix).

Since this is remarkably slow, people are currently researching a method of a 'threaded' numerical simulation. Basically saying that you have cells which are under the same basic condition... calculate once and applying to all increase the speed. I haven't seen much on this lately though.


The first two are modeling the results of turbulance. The last two are actually modeling the turbulance... but to varying degrees of accuracy. The field is still a very hot research topic.

As you understand, you have described well known principles of simulation of ... the turbulence in so called the Reynolds stress model, which has nothing to do with turbulence caused by phenomena of vortex creation, for example (or with instabilities of flows, like Garter’s, or Hamlet’s, or any torn off Prandtl layers, and so on , and so on). But all types of known and even unknown yet mechanisms of turbulence will take place in case of any real body of general shape streamlined by air. So, there is no practical use in advice: go, do simulation on PC. BTW: the numerical simulation of turbulence for today is the most complex and complicated problem known in whole Cybernetics. It was why your advice so amazed me…

As you understand, you have described well known principles of simulation of ... the turbulence in so called the Reynolds stress model, which has nothing to do with turbulence caused by phenomena of vortex creation, for example (or with instabilities of flows, like Garter’s, or Hamlet’s, or any torn off Prandtl layers, and so on , and so on).Agreed.
But all types of known and even unknown yet mechanisms of turbulence will take place in case of any real body of general shape streamlined by air.Obviously.
So, there is no practical use in advice: go, do simulation on PC. BTW: the numerical simulation for today is the most complex and complicated problem known in whole Cybernetics. It was why your advise so amazed me…I disagree with you here. For the majority of problems (especially if you are only concerned with drag) the computer simulations work fine. It's not exact, and I don't expect it to be... but it does give you a close estimate of the stress, force and drag on an object.

Your only other real option is to make a scale model and test it (which introduces other problems depending on your surface - magnus effect, etc).

I'm assuming that he doesn't wants something more than the stokes forumla, but that accuracy is not his number one concern. The PC simulation is ideal for this.

We really need to know why he wants to know.

I agree.

Well guys, all of this has been very interesting to read. But quite frankly, it is beyond me, as I have feared at first. What I need is a way to compare the air friction of a parachute + person vs. the air friction of a person. So I guess my only way now is to simply search for these values, since creating my own computer simulation is out of the question... Thanks though, lots of useful input there guys.

Oh, well that's not to complicated. I'll try and get back to you soon, but dinner just came.

Sorry for the extra stuff...

Dear A.M.,
Now, when you formulated your problem, everything came to order:
1. First of all, "parachute" problem has nothing to do with air friction force - so, you should find data on air-resistance force for parachutes (it strongly depends on shape of parachute). Such data you can find only in special books or sites: by my best knowledge there is no physical theories of that force (mostly because experimental data can be easily obtained and are much more accurate then any possible theoretical results)
2. If you need that data for serious purposes (I am scared with necessity of comparison “parachute + body” and "body ... without parachute") you should pay attention at what load these parachute data was obtained. Because changing the shape of load you can get dramatically different data for the same parachute.
So, good luck...

F = ½ *(rho)*Cd*A*v^2
rho = 1.22 kg/m3
Cd = drag coefficient
A = area of the chute
v = velocity

You'll have to look up Cd depending on the shape/make of the parachute.

The force on a person alone will be equal to their weight during most of the trip. A person will accelerate to terminal velocity and then the force down (wieght) will be equal to force up (the air resistance).
(I am scared with necessity of comparison “parachute + body” and "body ... without parachute")Hehe.

Sorry for the extra stuff... Actually that was very interesting and informative. Thank you :)

(I am scared with necessity of comparison “parachute + body” and "body ... without parachute") Not to worry, I am not about to commit siucide just yet.

Now regarding the pretty formula of Persol here. What exactly is rho? And the units for that is kg/m^3, I assume? The rest seems pretty clear. Thanks very, very much.

rho is density of air.

Thanks! I will try this in the context of my problem. I'll let you guys know tomorrow how this works sout.

I've been searching for this formula for quite a while now, with no results...

So simply put:

Anyone can give me the formula for the force of air friction? My guess is that lots of variables are involved here, and I would expect the formula to be beyond my comprehension, but I'm willing to give it a try.

Thanks for any input.

You might try searching on 'wing design in aeronautical engineering', 'viscouis flow dynamics' etc,

The formula given by Persol above is correct for an object falling through a fluid, such as a person or a parachute.

For shear in a pipe, one can calculate dP from one point to another, and that pressure times the cross sectional area of the pipe is equal to the shear drag on the walls between the two points. One doesn't need to resort to Navier Stokes equations, computational fluid dynamics or anything so exotic to calculate shear forces on pipe walls.

You don't need any math besides addition for that. Pipe losses are generally in precalculated tables.

Actually, engineers don't use tables for pressure drop. There are equations for all of that.

Lol, I'm not sure what enigineers you hang around with.

This may seem extremely stupid for some of you... but here is a formula I have developed to determine the amount of total acceleration due to gravity acting on a 65-kg free faling person.

a=9.8-0.0036v^2

I did this based on the assumption that this body would achieve a terminal velocity of 54m/s. I also know that the greater the speed, the greater the force of friction, so I squared it. This may seem totally childish... but it does seem to work quite well.

What would you guys say about this?

It is very believable that "good" formula should be something like
a = g - bv^2 - cv
where b and c are some coefficients,
but I am very in doubt that your "b" is right: it is hardly to believe that parachutist is landing at 54m/sec! "b" should be much bigger.

No, the formula I came up with is for a person with no parachute.

I understand the part of your formula where a = g - bv^2

But where did you get a = g - bv^2 - cv from?

In this case it looks like a right answer, but .... without parachute? Again? You really scare me....

This may seem extremely stupid for some of you... but here is a formula I have developed to determine the amount of total acceleration due to gravity acting on a 65-kg free faling person.

a=9.8-0.0036v^2

Well:
F =ma= ½ *(rho)*Cd*A*v^2
a = ½ *(rho)*Cd*A*v^2/m

So the your constant would equal rho*Cd*A/2*m
Is that the way you did it?

If so, what did you use for Cd?

Your answer sounds right.. and that's the reason we need parachutes:)

Actually what I did to come up with this is:

I researched the value of 54m/s as the terminal velocity for the average person with no parachute. So, since at terminal velocity a = 0, 0 = 9.8 - air drag.

I know that the higher the speed, the higher the drag, so V should be squared to achieve a pretty close approximation.

Thus: 0 = 9.8 - V^2 Obviously this lacked something... so I inserted a value into the equation to make it true for v=54. 0 = 9.8 - XV^2 From here I just solved for X, substituting 54 for V.

Oh and one more thing: do you guys know what is the safe landing velocity for a person? I could use that to develop a similar formula which would apply to a person + a parachute. If you guys don't know this right off, it's ok, I'll rsearch this too.

Persol,
this last post needs your comments: I do not have approptiate words....

AM, your equation is correct, it's a matter of determining the coefficients.

The way one might do this is to do sum forces in the verticle direction.

The force down due to weight is m*g
Where m=mass
g=acceleration due to gravity

The force up due to drag is .5*rho*A*Cd*V^2
Where rho = fluid density
A = Cross sectional area in direction of motion
Cd = Coefficient of Drag (I've seen 0.8 to 1.3+ used for a parachute)
V = Velocity

Doing a force balance:
m*a = m*g - .5*rho*A*Cd*V^2

So the acceleration down (or up) is given by dividing through by mass:

a = g - rho*A*Cd*V^2/(2*m)



Persol - perhaps you're thinking of tables used for L/D and K as found in Crane paper 410? But those are just constants such as drag coefficient. Crane 410 is the bible for fluid flow, I've been using it for 15 years as a professional engineer.

Q_Goest,

That formula is not very useful in the context of my problem, and besides, my peers at school wouldn't exactly follow that. I'm just happy you think my formula works out :)

The formula I got for the person with a parachute, assuming that the terminal velocity of a person with a parachute is 6.2m/s, is a = g - 0.2567v^2.

Does that seem reasonable?

Thus: 0 = 9.8 - V^2 Obviously this lacked something... so I inserted a value into the equation to make it true for v=54. 0 = 9.8 - XV^2 From here I just solved for X, substituting 54 for V.This is actually a very nice way of doing it. While it doesn't tell you what the coefficient actually is, you get an equation that works.
Oh and one more thing: do you guys know what is the safe landing velocity for a person?Now you have me scared....

10-20mph seems to be about the limit. That's assuming a good 'collapse' upon impact.
Persol - perhaps you're thinking of tables used for L/D and K as found in Crane paper 410? But those are just constants such as drag coefficient. Crane 410 is the bible for fluid flow, I've been using it for 15 years as a professional engineer.I was thinking about calculating the head losses using L/D, completely forgetting about flow. My mistake.

AM, the 6.2 m/s terminal velocity sounds about right. I'm sure that will vary depending on the type of parachute and weight of the person. Your equation is identical to what I provided, so it tells you that the value for X you provide is equal to:
rho*A*Cd/(2*m)

If you know terminal velocity, you can do what you did, and just put the value in that equates to the other variables.

Note that terminal velocity is also dependant on mass (ie: 1/m). So you could set X = Z/m and say that Z=m*X. Then put Z back into the equation and divide by mass to get terminal velocity for people of different mass (ie: weight). Obviously, a lighter person will have a lower terminal velocity, and a heavier person will have a higher terminal velocity, so putting mass back into the equation will correct for that.

As for how fast someone hits the ground, I've heard paratroopers hit the ground at about 20 mph. I suspect a search of the net could turn up some good values for how fast other types of parachutists hit the ground.

Agreed 100%. However, my equation suits its purpose perfectly well, so I'd rather keep it as simple as possible. In my case I'm dealing with only one mass, so there's no need to include it in the formula. rho*A*Cd/(2*m) makes sense though, so thanks for teaching me that part :)

Anyway I'm done my little project already. If it interests you guys at all, I'll let you know what mark I got :)

Note that terminal velocity is also dependant on mass (ie: 1/m). So you could set X = Z/m and say that Z=m*X. Then put Z back into the equation and divide by mass to get terminal velocity for people of different mass (ie: weight). Obviously, a lighter person will have a lower terminal velocity, and a heavier person will have a higher terminal velocity, so putting mass back into the equation will correct for that.I wouldn't be 100% confident about this. A larger person is going to have higher drag as well.... which will somewhat offset the mass increase.
Anyway I'm done my little project already. If it interests you guys at all, I'll let you know what mark I gotGood luck.