Lehpron From United States of America, joined Jul 2001, 7028 posts, RR: 22 Posted (12 years 4 days 2 hours ago) and read 1660 times:
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?
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Minuteman From United States of America, joined Aug 2000, 271 posts, RR: 0 Reply 1, posted (12 years 4 days ago) and read 1623 times:
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.
LAPA_SAAB340 From Spain, joined Aug 2001, 389 posts, RR: 5 Reply 2, posted (12 years 4 days ago) and read 1617 times:
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 From United States of America, joined Aug 2000, 271 posts, RR: 0 Reply 3, posted (12 years 3 days 23 hours ago) and read 1608 times:
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...