What is the optimal angle for a go-cart windshield to decrease wind drag?

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In summary, the graduate of a mechanical engineering degree has a problem that they would like to be able to answer, but after a couple of days they are still stumped. They believe that reducing wind resistance by choosing angles alpha and beta would be helpful. They also note that fluid mechanics can be a difficult subject.
  • #1
fpjeepy05
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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
2002_Jeep_Roadster__Yellow___Black___Black.jpg

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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  • #2
I believe you can minimize air resistance by choosing alpha = beta = 90o

:smile:
 
  • #3
NascentOxygen said:
I believe you can minimize air resistance by choosing alpha = beta = 90o

:smile:

I bet that would reduce wind resistance. :smile:

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
 
  • #4
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
 
  • #5
CJSGrailKnigh said:
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?
 
  • #6
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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  • #7
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.
 

1. What is the "Go Cart Windshield Problem"?

The "Go Cart Windshield Problem" refers to the issue of windshields on go carts cracking or breaking during use, potentially causing harm to the driver and passengers.

2. What causes the windshield on a go cart to crack or break?

There are several potential causes for a go cart windshield to crack or break, including impact from debris or other objects, poor quality materials, or excessive force from wind or other elements.

3. How can the "Go Cart Windshield Problem" be prevented?

To prevent the "Go Cart Windshield Problem", it is important to regularly inspect the windshield for any cracks or damage, and to replace it if necessary. Additionally, using a high-quality windshield and proper installation can help prevent breakage.

4. Are there any safety precautions that can be taken to reduce the risk of windshield breakage on go carts?

Some safety precautions that can be taken include wearing protective gear, such as a helmet and goggles, while operating a go cart, avoiding rough terrain or debris, and driving at safe speeds to reduce the risk of impact or excessive force on the windshield.

5. What should I do if my go cart windshield breaks while I am using it?

If your go cart windshield breaks while in use, it is important to stop the vehicle immediately and safely exit. Seek medical attention if necessary, and have the windshield replaced or repaired before operating the go cart again.

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