Air friction formula?

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.

 

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

 

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

 

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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....

 

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

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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

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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?

 

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.

Now you have me scared....

10-20mph seems to be about the limit. That's assuming a good 'collapse' upon impact.

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

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

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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.

Good luck.