Sponsor Message:
Aviation Technical / Operations Forum
My Starred Topics | Profile | New Topic | Forum Index | Help | Search 
Would This Case Let Drag Equal Lift?  
User currently offlineLehpron From United States of America, joined Jul 2001, 7028 posts, RR: 21
Posted (13 years 3 days 16 hours ago) and read 1955 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?


The meaning of life is curiosity; we were put on this planet to explore opportunities.
4 replies: All unread, jump to last
 
User currently offlineMinuteman From United States of America, joined Aug 2000, 271 posts, RR: 0
Reply 1, posted (13 years 3 days 14 hours ago) and read 1918 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.

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

Minuteman


User currently offlineLAPA_SAAB340 From Spain, joined Aug 2001, 390 posts, RR: 4
Reply 2, posted (13 years 3 days 13 hours ago) and read 1912 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.


User currently offlineMinuteman From United States of America, joined Aug 2000, 271 posts, RR: 0
Reply 3, posted (13 years 3 days 13 hours ago) and read 1903 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  Smile

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...


User currently offlineLehpron From United States of America, joined Jul 2001, 7028 posts, RR: 21
Reply 4, posted (13 years 2 days 23 hours ago) and read 1874 times:

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?


The meaning of life is curiosity; we were put on this planet to explore opportunities.
Top Of Page
Forum Index

Reply To This Topic Would This Case Let Drag Equal Lift?
Username:
No username? Sign up now!
Password: 


Forgot Password? Be reminded.
Remember me on this computer (uses cookies)
  • Tech/Ops related posts only!
  • Not Tech/Ops related? Use the other forums
  • No adverts of any kind. This includes web pages.
  • No hostile language or criticizing of others.
  • Do not post copyright protected material.
  • Use relevant and describing topics.
  • Check if your post already been discussed.
  • Check your spelling!
  • DETAILED RULES
Add Images Add SmiliesPosting Help

Please check your spelling (press "Check Spelling" above)


Similar topics:More similar topics...
Would This Be One Of The Best Ways Of... posted Mon Oct 10 2005 19:24:50 by Airfly
Would This Warrant An Emergency Descent? posted Tue Apr 19 2005 12:42:07 by AvroArrow
Would This Be Possible (idea For Taxi Power) posted Fri Mar 11 2005 02:56:59 by JumboJim747
If An Airbus, Would This Flight Have Crashed? posted Sat Nov 2 2002 16:03:08 by Rick767
What A300 Flap Setting Would This Be? posted Wed Aug 21 2002 21:08:40 by Mr Spaceman
Would This Be Possible? posted Thu Sep 20 2001 10:17:45 by Victor Hotel
How Would ATC Issue This Clearance posted Thu Jun 15 2006 04:26:36 by Corey07850
Lift-to-Drag Ratio Vs. Velocity For An SST...? posted Tue Oct 4 2005 03:36:06 by Lehpron
Won't This Cause Drag posted Fri Feb 13 2004 15:18:58 by HAWK21M
Tough One! Lift/Drag Vs. Distance/Altitude posted Tue May 14 2002 06:18:04 by Jzucker

Sponsor Message:
Printer friendly format