How does wind speed relate to the force needed to lift a person?

[infraray]

Someone at work had stated that it would take a wind speed of 120mph to lift a person lying on the ground into the air. I was curious and set out to find out if this was accurate. I am having some difficulty. In order to figure this out how does one equate wind speed to force. I am a beginning physics student in college and am trying to exercise my skills. Do I figure this out as a drag coefficient problem? Or perhaps maybe this problem is more complex than I am prepared for, but I am curious. Thanks!

 

[LURCH]

Never thought of trying to solve this for a person lying flat, but the first thing I thought of was to apply a Bernoulli equation. If the man is laying flat, a pocket of air should be trapped underneath him and it should be fairly stationary. If the air above him is building, this should create a differential in pressure, generating lift. When the amount of lift is greater than the weight of the individual, up he comes!

 

[chroot]

As far as I know, the terminal velocity of a person is around 120-130 mph in the horizontal posture.

Terminal velocity is the point at which the upward drag force equals the downward gravitational force -- so it makes sense. It would take a 120 mph or so wind to levitate a person.

- Warren

 

[zoobyshoe]

Lurch,

I just lay down on the floor on
my stomach, and also on my back.

The only possible pocket of air
was the very small one in my
navel.

What are you shaped like?

 

[LURCH]

Originally posted by zoobyshoe
Lurch,

I just lay down on the floor on
my stomach, and also on my back.

The only possible pocket of air
was the very small one in my
navel.

What are you shaped like?

LOL!

But there must be air in there, otherwise you'd be vaccum-sealed to the ground. If you were to lay on the ground and remove all the air from underneath yourself (and I'm not for one instant suggesting that you should), then for every square inch of your "sunny-side", there would be about 14lbs of air pressure pushing down no you. It is very unlikely you would be able to get up again.

 

[zoobyshoe]

Originally posted by LURCH
LOL!

But there must be air in there, otherwise you'd be vaccum-sealed to the ground. If you were to lay on the ground and remove all the air from underneath yourself (and I'm not for one instant suggesting that you should), then for every square inch of your "sunny-side", there would be about 14lbs of air pressure pushing down no you. It is very unlikely you would be able to get up again.

I see what you mean now. It's not
a monocoque pocket, but a million
little pockets that prevent you
from getting stuck. I think wear-
ing clothes helps, too.

 

[russ_watters]

A person isn't a wing. No amount of wind will lift a person off the ground if the wind is parralel to the ground.

 

[Clausius2]

Hi! First of all sorry for my bad english. I am spanish and i am in the "first stages" of learning it.
I am going to answer the problem, but I have to say to the truth that only russ_watters seems to be near of the correct answer.
Well, it is imposible that a lift is created only with a flow of wind parallel to the ground. Of course, it would be possible if there were a gap of air beneath the surface in contact with the ground. Then we could apply Bernoulli's equation in order to evaluate the difference of pressures and calculate the lift force.
But the problem I think it is a person standing on the ground. His shoes are in contact with the ground's surface. Then, two forces are acting over him: friction force (between ground and shoes) and drag force (created by the flow). Well, this is the demostration in which i state that the figure of 100 mph isn't so far of the truth.
Under the assumption of Reynold's Number too large (Re>>>1), the viscous forces can be neglected. Then, the force balance results:
d=density of air (it's of the order of 1,24 Kg/m3
U=velocity of the flow of air (unknown)
M=mass of the man (of the order of 75 Kg)
g=9,8m/s2
k=friction coefficient between shoes and ground (of the order of 0,02)
A= Transversal area of the man (of the order of 50cm2)
As an approximate calculation the pressure (drag) forces are of the order of AdU^2 where ^ means "powered to".
So AdU^2 = kMg to start the dragging movement.
Clarifying the velocity U=sqrt(kMg/Ad)=46 m/s = 104 Mph where sqrt means "square root".
Of course, this calculation is approximate, so to obtain the exact solution it is necessary to know the geometry of the problem (boundary conditions) and the initial conditions, in order to equate and resolve the Momentum Equation of the Fluid Mechanics (it would be too difficult!)

 

[Tyro]

Such a problem involves transient aerodynamics. Using a drag coefficient found for a person (in say, a wind tunnel) or terminal velocity values as some other posters have done invariably refer to a steady-state value.

Transient aerodynamics is not an trivial matter, its modelling is best done computationally. Another example of a transient aerodyamic effect is 'Jump aerodynamics', first discovered by Vietnam helo pilots, who found that they could get their helicopters to lift off even though they exceeded the maximum take-off weight slightly, by giving the collective (not the cyclic!) a shove. Quite a handy thing for an experienced chopper pilot to know, especially if they were medevac'ing one extra injured soldier over the MTOW Ask if you want the explanation, if enough ask I'll explain it.

Paying attention in otherwise [zz)] Helicopter Dynamics lectures helps