lehpron
Topic Author
Posts: 6846
Joined: Tue Jul 10, 2001 3:42 am

### Would This Case Let Drag Equal Lift?

I read in an aerodynamincs book that the coeff of drag against a flat plate area is 1.18. If this plate is rotated then wouldn't the component of drag eventually become a component of lift and the coeff of lift becomes 1.18?
The meaning of life is curiosity; we were put on this planet to explore opportunities.

Minuteman
Posts: 260
Joined: Sat Aug 05, 2000 1:01 am

### RE: Would This Case Let Drag Equal Lift?

Nope, but good idea.

Part of the reason the *total* drag on a flat plate is so much more than that of a cylinder or airfoil with the same frontal area is because of the massive changes in momentum in the fluid required to fill the void behind the plate.

Think about it...the airflow around a cylinder only has to curve around the cirumference, while the air at the edge of the plate has to make almost a 90 degree turn to get behind the plate. So, much more force/energy is required to deflect the airstream.

http://www.grc.nasa.gov/WWW/K-12/airplane/shaped.html

Minuteman

lapa_saab340
Posts: 398
Joined: Tue Aug 07, 2001 8:42 pm

### RE: Would This Case Let Drag Equal Lift?

Actually, the high drag of a flat plate perpendicular to the flow is mainly due to the fact that the flow is brought to a stop at the face of the plate. What Minuteman says about the fluid having to fill that void behind the plate is also true, but it's not the main contributor to the drag in this case. If we take a look at the examples in that link, we'll see that a wedge is indeed an improvement over the flat plate. But if you look at the bullet, it has a lower drag coefficient than the wedge. The cylinder or the sphere can be a bit misleading because the coefficient of drag highly depends on the Reynolds number (As the boundary layer goes turbulent, the flow tends to remain attached to the surface of the cylinder a little longer, and the wake gets smaller). I was just reading that link Minuteman provided, and they do mention that in there.

If you rotated the plate 90 degrees, the only way you'd get lift out of it is if the flow remains perpendicular to the plate. In this case, you'd have a flat plate with the flow blowing into it from below, creating a 'lifting' force. If you only rotate the plate and the flow becomes parallel to the plate, you won't get any lift out of it.

Minuteman
Posts: 260
Joined: Sat Aug 05, 2000 1:01 am

### RE: Would This Case Let Drag Equal Lift?

dhoh!...its been longer than I thought since I've worked with this stuff. You're absolutely right that a cylinder/sphere is a poor comparison because the Cd is so dependent on the Reynolds number. Perhaps I should read the text of pages I link to instead of just looking at the purty pictures

I also agree that the "upstream" shape is more important that the downstream shape in determining the drag coefficient. There's a stagnation "point" at the center of the plate that requires that the fluid makes a roughly 90 degree turn to go rushing out to the edge and then make another roughly 90 degree turn. However, the air won't go rushing to the center of the plate on the back because there's a relatively stagnant wake behind the plate that "reshapes" the tail...hence the bullet which encourages a more gradual change in momentum up front and is the same a a flat plate in back has a much lover Cd. Otherwise, d'Alembert would have been right and cannon balls would cruise much further.

Yeah, its supersonic, but you still get the idea of the wake...

lehpron
Topic Author
Posts: 6846
Joined: Tue Jul 10, 2001 3:42 am

### RE: Would This Case Let Drag Equal Lift?

I understand the wake stuff but let's say the plate is rotated at an angle 45-degrees, would the CL and CD be about 70% in magnitude of original?
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