## Induced Drag Coefficient document.SUBSCRIPTION_OPTIONS = { "thing": "thread", "subscribed": false, "url": "subscribe", "icon": { "css": "fa fa-envelope-o" } };

Developers
2008-05-05
2012-09-26
• Eduardo Borota - 2008-05-05

i´ve designed a horizontal tail and ran vortice lattice method in order to get induced angles on it. Then I corrected the lift coefficient with wind tunnel data (MS excel) and calculated the induced drag coefficient, by having induced angles and the corrected lift coefficients on each station.
The values i obtained are "exactly" double of the ones I got from XFLR5. So I started to look for what made both values different.
I´ve found out that the calculation of the Cdi, when using the whole wing area instead of only half-wing (which seems correct, because I´m calculating only half-wing), returns pretty much the same values of XFLR5.
Is that correct? I´m running XLFR5 4.02Beta.
If that is correct, the only thing that must be changed is the area by which the sum of all spanwise stations induced drag coefficients is divided.
thanks... =)

• Eduardo Borota - 2008-05-11

Andre, i´ve seen it again.
supose you have an induced drag on each section, calculated as fdi=claiSn*q, where "Sn" is the area of each section and "q" is the dinamic pressure of the non-disturbed flow, "cl" is the local lift coefficient and "ai" is the induced angle of attack. In order to obtain the full induced drag force, we´ve got to sum all of then.
If you sum only half wing, because of its symmetry, you have to multiple the sum by two (in order to get the induced drag force of the entire wing).
Then, to obtain the induced drag coefficent, we´ve got to divide by the dinamic pressure of the non-disturbed flow and the area of the wing (which is also the refference area for the airplane).
do u agree with it?

• Anonymous - 2008-05-07

Eduardo,

In the the early days of the development, I have had this issue with the factor 2. However the Induced drag calculated with XFLR5 is now the same as for other codes.
It is also easy to compare your results with theoretical values available for the elliptical wing...

andre

• Anonymous - 2008-05-12

For each section of the wing :

Cdi = Cl ai
Fdi = q S CDi = q S Cl ai

Overall, you should find for a near elliptical planform : Cdi = (Cl x Cl) / (pi AR e)
e is Oswald's efficiency factor, close to 1

For the wing
CDi = 1/S Sum(c cl ai dy)

c cl is the local lift, dy is the width of a span strip.

I checked and everything looks good in the code ?

andre