center of lateral resistance

[tspeer]

Originally Posted by Raggi_Thor

I just used the formula for a flat plate, Clo = 2 * pi * a
5 degrees = PI * 5/180 ~0.17 Rad
0.17 * 2 * pi = 2.2
Maybe This is not applicable for that angle of attac...

Check your math:
5 deg = 0.087 rad.

Cla * a = 0.54

As a practical matter, the lift curve slope is closer to 0.1 due to boundary layer displacement.

And don't confuse the two-dimensional lift curve slope with the angle of attack required for a three-dimensional planform. Considerably more angle of attack is needed to reach the same level of lift.

The apparent lift curve slope of planform is closer to

a = a0 / (1 + a0 * 57.3 / (pi * AR * e))

a = 3D lift curve slope, per deg
a0 = 2D lift curve slope, per deg
AR = aspect ratio
e = Oswald efficiency factor

[RHough]

Originally Posted by tspeer

The apparent lift curve slope of planform is closer to

a = a0 / (1 + a0 * 57.3 / (pi * AR * e))

a = 3D lift curve slope, per deg
a0 = 2D lift curve slope, per deg
AR = aspect ratio
e = Oswald efficiency factor

Thanks Tom,

What is 57.3?

I've been using .7-.8 for value of e, is that "close enough" for a sailboat?

-Randy

[jam007]

57.3 = 180 / pi
Conversion from degrees to radians.

Anders M

[Raggi_Thor]

I was just too quick to copy paste some formula...

[chandler]

So is that where all you rocket scientists get your formulas, copying and pasting? For a while there I thought you all Knew and understood these formulas? Silly me.

[jam007]

No no Chandler don´t confuse egineers with scientists.

[Raggi_Thor]

Sorry again, Chandler :-)
One thing is copy/paste or look up in a reference. The most embarrassing is not to check the answer, is it reasonable. When dynamic pressure is 0.5xROxV^2, it is not reasonable to have a lift coefficient much larger than 0.5...