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Old Aug 8th 2011, 09:20 AM   #1
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speed required to move an object

Hello forum
I have designed a license plate holder that hangs from single pivot point. A question has come up that it might swing upward in such a manner as to not allow law enforcement to see the plate. Below are some details of the holder.
Area = 90 sq in
weight = 2.7 lbs.
So I guess that it would take a certain velocity just to get it to move and then a sustained velocity to get it to move from the vertical to the horizontal position. And I guess objects in front of like a spare tire and the rear drive shaft differential would also play a factor in the airflow. So for sake of argument lets assume it is hanging in clear air.

I would appreciate any and all help. Thanks in advance.
Bill
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Old Aug 9th 2011, 09:40 AM   #2
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If the plate is vertical initially and is 'fixed' against a vertical support, I don't see how it can more up... or is the support here at an inclination?

A sketch would greatly help, I think, to show your problem.
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Old Aug 9th 2011, 09:52 AM   #3
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How about a picture of the holder. It hangs from the tow pintle of a truck. It is made fron .250 2024-T3 aluminum plate. The contact area is approx 2.0" x .250" the contact area is not rounded. Also you can see from the picture that the holder cannot swing forward as it contacts the edge of the bumper. Hopes this helps.
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Old Aug 9th 2011, 11:38 AM   #4
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Hm.. interesting Thanks for the picture!

But another thing to note, it needs a constant acceleration for the plate to remain horizontal after getting there. So it's always acceleration. But of course, it might be like 1 degree lower than horizontal and from that point, it's not horizontal, though practically, it is. So, it'd be better if a certain range was assumed where the plate is considered horizontal.

I'll use base units to make it simpler for me.

The mass becomes 1.13 kg
The area becomes 0.5776 m^2

Also, I think that the height of the plate is required too, so that its centre of gravity can be estimated.

From that, assuming that the air remains still (no breeze from underneath, besides the one created by the motion of the vehicle), I can estimate the force that the air exerts on the plate, using

F = dm/dt * v

F = d(density*volume)/dt * v

F = d(density*length*area)/dt * v

F = density*area*v^2

Taking air density as 1.225 kg/m^3, the area as 0.5776, the force becomes F = 1.225Av^2

The area now varies with the angle which the plate makes with the horizontal, by the equation A = Ao cos(x) where Ao = 0.5776 m^2 and x is the angle.

F = 1.225(0.5776)cos(x) v^2

F = 0.7076v^2 cos(x)

The force that the plate exerts against the breeze is also dependent on the angle;

Fg = mg tan(x) = 11.058 tan(x)

So when the angle is 90 degrees with the vertical, and combining the two equations, we should get the speed required, but as I said earlier, that would be impossible. Let's take 80 degrees with the vertical to consider it horizontal.

F = 0.7076v^2 cos(x)
Fg = 11.058 tan(x)

0.7076v^2 cos(80) = 11.058 tan(80)

0.7076v^2 * 0.174 = 11.058*5.671

we get v = 509 m/s

or about 1100 mile/hr

That's huge, lol

Of course, acceleration now would put another variable into the game
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Old Aug 9th 2011, 11:43 AM   #5
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WOW! a supersonic diesel truck, and a new land speed record.
My friend Unknown008 you rock. Thank you very much.

bill
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Old Aug 9th 2011, 11:53 AM   #6
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Yes that's because as you go faster, the part being striken by the 'breeze' gets smaller, but as I said there are more variables to be considered. But those will be out of my league

Like for example, when a wind current goes past an edge, it tends to create wind vortexes which 'suck' the plate upwards and makes it reach that faster.

Anyway, that's it for a high school level analysis =P

Take care!
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