basic physics and Di

[sigurd]

I was trying to understand how induced drag relates to speed with a constant lift, by assuming that at twice the speed each volume of water is affected by the same downward force at half the time, so it should be accelerated to half the downward speed.
Can you tell whether I got the stuff right so far?

[jehardiman]

I think you are making this much too complicated, i.e. "ghost lift"?...See Marchaj, Aero-Hydrodynamics of Sailing, part 2D or Hoerner; Fluid Dynamic Drag, Chapt 7.

But as to downwash being half because the time is halfed?...yeah, but that is not a good way to look at it at all. The downwash is haved and therefor the Di is 1/4 because the angle of incidence is quartered to reduce the lift by the speed squared.

L =m *2w
m = rho*A*Vo
2w=Vo*2alphai

therefor

rho *A* Vo^2*2alphao = rho * A*(2*Vo)^2 * 2alpha2
2alphao=4*2alpha2

then

2w(o) = 4*Vo*2alpha2
2w(2) = 2*Vo*2alpha2

and

Di(Vo) = L *(4*2alpha2)/2 = 4*L*alpha2
Di(2) = L*alpha2

This falls apart BTW if viscous effects are considered.

[Rick Willoughby]

To understand "Induced Drag" you need to consider the foil in 3D. Induced drag is related to longitudinal circulation over the ends of the foils so is a function of foil aspect ratio, plan form and lift coefficient. This explains it:
http://selair.selkirk.bc.ca/aerodyna...rag/Page6.html

This site is helpful in getting the lift for a foil:
http://www.lerc.nasa.gov/WWW/K-12/airplane/foil4.html
Unfortunately it does not provide drag.

I have a series of polar curves for high L/D foils to get accurate coefficients of lift and drag. These coefficients are highly dependent on Reynolds number in my area of interest. You can produce some useful foil data using this site for a reasonable range of foils:
http://www.mh-aerotools.de/airfoils/javafoil.htm
(It can take a while to load)

[tspeer]

Originally Posted by sigurd

I was trying to understand how induced drag relates to speed with a constant lift, by assuming that at twice the speed each volume of water is affected by the same downward force at half the time, so it should be accelerated to half the downward speed.
Can you tell whether I got the stuff right so far?

That's correct. Lift = density * velocity * average circulation * span, and the induced velocity is proportional to the circulation, so twice the speed means half the circulation.

Since the induced angle of attack is proportional to circulation / velocity, this means the induced drag at constant lift varies as 1/velocity^2.