BRT505 Posts: 1, Reputation: 1 New Member #1 May 2, 2008, 07:58 AM
Have an object that is 477 square feet and the wind speed is 50 mph. what would be the wind load/force on this object? Also what would be the formula that could be used for other wind speed on this object ?

Thanks
 ebaines Posts: 12,131, Reputation: 1307 Expert #2 May 7, 2008, 07:34 AM
The general formula for drag from wind resistance is:

$
F_d = \frac 1 2 \rho v^2 A C_d
$

where $C_d$ is the object's drag coefficient, which depends on the shape of the object (typically 0.25 - 0.50 for a car, and usually assumed to be about 1.17 for a flat plate), $\rho$ is the density of the medium (air in this case , which is 1.293 kg/m^3), A is the frontal surface area, and v is the velocity of the wind.

See: Drag (physics - Wikipedia, the free encyclopedia) for details.

Putting everything into SI units:
A is 477 ft^2 = 44.3 m^2
v is 50 MPH = 22.35 m/s

So:
$
F_d = \frac 1 2 \rho v^2 A C_d = \frac 1 2 (1.293 kg/m^3) (22.35 m/s)^2 (44.3 m^2) 1.17\\
= 16740 N\\
= 3760\ pounds.
$
 Gordon4900 Posts: 2, Reputation: 1 New Member #3 Feb 18, 2011, 08:35 AM
The force of a water jet (a fluid) with area A and velocity v onto a stationery flat plate is given by the formula:

F = density of water x A x v2

When I compare this with the formula above for a wind (also a fluid) why do we half the force for wind loading? I can be contacted on gmarshall49@hotmail.com if anyone wants to give me an answer.

Thanks
 ebaines Posts: 12,131, Reputation: 1307 Expert #4 Feb 18, 2011, 09:34 AM

Gordon - Your formula assumes that the full momentum of the stream of water is absorbed by the plate, and the velocity of the water once it hits the plate is zero. It is derived from:

$
m\Delta v = F \Delta t
$

For a stream of water of area A and velocity v:

$
\frac m {\Delta t} \Delta v = F \\
(\rho A v )\Delta v = F
$

So if $\Delta v = v$ you get your formula.

For this to be true the plate must be larger in area than the area of the stream of water. However, The aerodynamic drag formula I gave is for a body immersed in the air, so the area of the stream of air is larger than the area of the body, and the wind is deflected around the body. Also note that for the formula I gave A is the area of the body, whereas for your formula A is the area of the stream.
 Gordon4900 Posts: 2, Reputation: 1 New Member #5 Feb 18, 2011, 10:24 PM
Comment on ebaines's post
Ebaines - OK, thanks for this. I understand the requirements for 50% to be used and it makes sense. Regards.
 mayan02 Posts: 1, Reputation: 1 New Member #6 Apr 9, 2011, 05:42 AM
given velocity(V) in meter per second
pressure is equal to 0.0000473V*V
then Force(F) = pressure(area)

for trusses:
normal pressure = P(2sin(angle))/ 1 + square of sin(angle)
horizontal pressure = 0.0000473V*V
same goes in computing for F
 Archon01 Posts: 1, Reputation: 1 New Member #7 Jul 25, 2011, 09:14 AM
An airplane can average 130mph. If a trip takes 2 hours on way and the return takes 1hour and 15 minutes. What is the wind speed assuming it is constant? *** help
 ebaines Posts: 12,131, Reputation: 1307 Expert #8 Jul 25, 2011, 09:28 AM

Please do not tag a new question onto an old thread - it's better to start a new question, especially since the subject matter is so different.

But to get you started - recall that distance = rate times time. Here the plane flies the same distance in both directions, and so you have

$
D = R_1T_1
$

and

$
D = R_2T_2
$

So

$
R_1T_1 = R_2T_2
$

Given a wind speed of W, the rate of travel of the plane is 130 MPH + W in one direction and 130 MPH - W in the other. You've been given values for T_1 and T_2, so can you take it from here?

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