# Go Cart Windshield Problem

I am a graduate of a 4 year university in Mechanical Engineering... But I graduated 3 years ago now and I feel like I'm slipping. I have a problem that I have come up with that I would like to be able to answer, but after a couple of days now I'm still stumped. A little help would be greatly appreciated.

Supposing you were going to make a windshield for a go-cart. In order to decrease wind drag you would want to angle it. Supposing you chose to actually have two pieces of windshield that had a seam down the middle, so you could angle the windshield both upwards and outwards.
Kinda like this

Simply put I would like to know that angle which the wind sees the windshield provided that the windshield is slicked back by angle alpha and inclined back angle beta from the plane contains the z axis and the line which lays on angle alpha. I'm not sure why I choose those to be the angles, but I guess because both of these angles will be less than the angle that the wind sees.

I think I've got thee points that describe the plane, but I'm not sure I went through it the best way. ( [cos(a), sin(a), 0] [sin(a)sin(b), cos(a)sin(b), cos(b)] [0,0,0,] ) I guess you could do a cross product now to get a normal vector then a dot product to find the angle between that normal vector and the normal to the yz plane. If anyone has any suggestions It would be greatly appreciated. Thanks

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NascentOxygen
Staff Emeritus
I believe you can minimize air resistance by choosing alpha = beta = 90o

I believe you can minimize air resistance by choosing alpha = beta = 90o

I bet that would reduce wind resistance.

But honestly know one has any ideas? No Help? And why did this get categorized as "Precalculus" That makes me feel even worst. If an 8th grader can solve this and I can't. \$5 to the first 8th grader that does. haha

Now assuming this flow can be characterized as either a turbulent or laminar flow (and unless you put scramjets on, no shocks)

F = 0.5*rho*C*V^2*A

the area you choose is based on the shape you choose to model the coefficient with. ie it may be frontal area, projected area, or wetted area. So the short answer is yes you are probably going to have to take the projected area of each onto a plane.

Now I remember why I hate fluid mechanics

Now assuming this flow can be characterized as either a turbulent or laminar flow (and unless you put scramjets on, no shocks)

F = 0.5*rho*C*V^2*A

the area you choose is based on the shape you choose to model the coefficient with. ie it may be frontal area, projected area, or wetted area. So the short answer is yes you are probably going to have to take the projected area of each onto a plane.

Now I remember why I hate fluid mechanics
Agreed. So are those three points I chose okay? so take the cross product them the dot product with the normal of the xz plane and then multiply by area?

So I simplified it. The projection onto the xz plane would be cos(a) if it was inclined in only one direction. And it would be cos(b) if it was inclined in the other direction so the compound projection would be cos(a)cos(b) correct? I like it. It might not be right... but I'll take it. By this a windshield inclined at 21.45degrees about both x and z axis would have the same projection as a windshield incline by 30degrees in only one direction. Reducing frontal area (and a proponent of drag) by 14% but not your windshield is also 14% small in a scene. So I'm not sure if that formula is perfect for this problem. A flat windshield at 0 degrees and of area A should not have the same resistance as a windshield at a 60 degree angle with an area of 2A.
:/

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To tell you the truth I deal with upper atmosphere where the continuum assumption isn't valid (>200km) so I'm not actually sure. I do know that the above should be a decent first order approximation.