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  #1  
Old 12-13-2005, 06:12 AM
fede fede is offline
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rudder's joessel formula

Hi Guys, I need the Joessel's formula to calculate the force generated by a rudder at different speeds and different angles.
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Old 12-17-2005, 01:58 AM
tspeer tspeer is offline
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Quote:
Originally Posted by fede
Hi Guys, I need the Joessel's formula to calculate the force generated by a rudder at different speeds and different angles.
C = (0.2 +0.3 Sin a) L.
C= The distance from the front edge for the center of pressure.
L= The whole length fore and aft.
a = The angle of incidence.

But since this formula predates the Wright Brothers, is only good for square planforms, and isn't all that good even for them - surely you can do better?
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Old 12-17-2005, 03:13 AM
Karsten Karsten is offline
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The formula today used is

L = cL * Rho / 2 * V^2 *A

L = Lift in Newtons
Rho = Density of fluid in kg/m^3
V = speed in m/s
A = area in m^2

cL is the lift coefficient. For symmetrical rudders with infinite span the relation to the angel of attack is

cL = 2 * PI * alpha

alpha is the angle of attack in radians. This gets inaccurate at greater angles of attack. The maximum cL you can achieve is usually about 1.2 then the foil stalls.
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Old 12-17-2005, 01:46 PM
tspeer tspeer is offline
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Jossel's formula has to do with where the center of pressure is located - in other words, the moment arm.

The equations you present are for the lift - in other words, the force that acts at the center of pressure.

And the formula you give for the lift coefficient is only for surfaces of infinite span. For finite span surfaces, you have to reduce the lift curve slope according to the aspect ratio.
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Old 12-19-2005, 01:19 AM
Karsten Karsten is offline
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Fede was asking for "the Joessel's formula to calculate the force generated by a rudder at different speeds and different angles". Now that I know what the Joessel's formula actually is it is clearly not suitable to calculate lift depending on the speed and angle of attack. The formula I provided gives the lift as requested for an infintie span foil as Tom pointed out.

Isn't the centre of pressure just at 25% of the chord for symmetric foils again with infinite span? It moves aft when the foil starts to stall.
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