Wing Area

HomeBuiltAirplanes.com

Help Support HomeBuiltAirplanes.com:

yankeeclipper

Well-Known Member
Joined
Jun 1, 2009
Messages
47
Location
Northeast.
After a bit of research and playing around with various tools, I find myself just a little confused about wing area. Assuming the following:

-1150 lbs TOW
-.75 Oswald
-23012 airfoil (12% thickness)

it looks to me and FoilSim like a measly 105 ft sq at 2.5 degrees angle of attack are plenty for reasonably short take-off's and even for a well-flapped 50mph landing. But most light aircraft matching the above description end up being invariably close to 130 sq ft. Clearly I'm missing some important factors. Any thoughts on what they might be?
 

Norman

Well-Known Member
Joined
Nov 28, 2003
Messages
2,933
Location
Grand Junction, Colorado
Are you accounting for altitude in your CL calculation? I'm getting 1,071 lbs at cl=1.6 from 105 ft^2 at 50 mph at sea level but only 866 lbs at 7,000 ft. You can't count on always being able to find a sea level airport.;)
 

addaon

Well-Known Member
Joined
Feb 24, 2008
Messages
1,686
Location
San Jose, CA
However, landing requirements (including the LSA requirements) are traditionally given in CAS, not TAS, so altitude would not be a factor.

CL in the 1.65 range is not unreasonable, but it's pretty good. One question is how much leeway you want if you don't hit the CL_max you expected.
 

orion

Well-Known Member
Joined
Mar 2, 2003
Messages
5,800
Location
Western Washington
There's a few more variables to consider when looking at wing design than might be obvious from a cursory examination. Stall speed is of course one so one must be able to account for several performance issues including scaling factors as related to the Rn and section performance, wing performance as a function of three dimensional characteristics, 3-D flap performance and how that performance affects the part of the wing that is not flapped, and on and on.

Another issue to consider is maneuvering - simply said, how many G's can you pull. Cutting it too close due to other considerations like stall speed may result in a wing that might be susceptible to early stall at low speed such as might be encountered in the pattern.

As has been said here many times, design is a practice of compromise so optimizing for one aspect may hurt you in another. For that reason you'll often find physical characteristics on some airplanes the reasons for which might not be obvious at first glance.
 

radfordc

Well-Known Member
Joined
Feb 5, 2008
Messages
1,283
But most light aircraft matching the above description end up being invariably close to 130 sq ft.quote]

Some planes are better than others, here's mine:
Wing Span - 22'
Wing Area - 98 Sq Ft
Airfoil - 64-415
Stall full flap - 40 mph
Stall clean - 46 mph
Va - 125 mph
Takeoff weight - 1150
 

wsimpso1

Super Moderator
Staff member
Log Member
Joined
Oct 18, 2003
Messages
6,290
Location
Saline Michigan
OK, I get a fair amount more wing needed. Let's check our math:

L = rho/2*V^2*Clmax*S

where S is wing area in ft^2, Clmax is max coefficient of lift for the 2D foil, V is airspeed in ft/s, and rho is air density, 0.002378 slug/ft^3 at STP. This works fine in a wind tunnel with the section going to the walls and the boundary layer sucked off of the walls near the foil.

Put the whole thing on an airplane and a couple more effects come in:

First is 3-D flow. Air runs from the high pressure side to the low pressure side, causing tip vortcies and resulting in the lift distribution trailing off towards zero at the tips. Regardess of wing shape, the spanwise lift distribution will look like the classic elliptical shape with the Cl at the root approximating the the lift from the wind tunnel model and at the very tip, lift is zero. So, we have to put PI/4 (as an reasonable approximation) into the equation to get the lift for a real wing;

Then, we have make lift to carry tail down load. A guess is that it will be 5 to 10% of the wing's lift.

So now the lift required of the wing is:

1.05*W = pi/4*rho/2*V^2*Clmax*S so

S = 1.05*W/(pi/4*rho/2*V^2*Clmax)

When I do this, it indicates 150 ft^2 of wing area needed to stall clean at 50 mph.

Billski
 

addaon

Well-Known Member
Joined
Feb 24, 2008
Messages
1,686
Location
San Jose, CA
I thought that the "standard" approximation for, say, a hershey bar wing without twist is that it's halfway between the lift distribution indicated by the chord distribution (linear) and elliptical; so (1+pi/4)/2 = ~0.9 instead of 0.8. Not a big difference, but brings it closer to 130 ft^2.
 
Group Builder
Top