# Fluid dynamics kite question

1. Jan 4, 2010

### Marshall10488

just wanted to say hi first as this is my first post.
I will explain the story...
There is an ongowing debate within the Kitesurfing world regarding the aerodynamics of LEI (leading edge inflateable) kites and Foil (like RAM aim parachutes) kites.

LEI Kites:
They have large inflateable 'bladders' that keep the shape of the kite and also make it float. they are single skin and the foil shape comes from the leading edge's cylndrical shape. They have only a few 'bridle' lines on the leading edge with a thickness of about 3-5mm. very curved shape so projected area is not same as surface area
http://www.boardlife.sk/userfiles/16-North_Rebel_2007_12m2_a_North_Rebel_2008_10m2.JPG.311008_122249_11.JPG [Broken]
some links to LEI companies
http://www.northkites.com/public/content/index_eng.html" [Broken]
http://www.naishkites.com/en/index.html" [Broken]

Foil kites:
closed cell RAM air parachutes. twin skin. lots of bridles, the majority of which are about 1-2mm. then some 3-5mm and 2 5-8mm. very flat so projected area is close to surface area. very light.
http://www.flysurfer.com/gallery2_code/d/141624-2/SPEED3+Deluxe+mit+Bridles.png [Broken]
main water re-launchable foil company
http://www.flysurfer.com" [Broken]

Ok so the argument is that some beleave that foil kites are better in low wind as they are lighter, tend to have a higher AR. Others that LEIs are better as less drag etc.
My thinking is: Lift needed to fly it L=ma
Lα1/M
Lα V[squared]
Lα Lift coefficient
L α Area

But i need to add in the drag equation so lift needed to fly:
l=ma + 0.5 Cd ρ V[squared] a (don't know if this is right)
so by equating i get that the velocity to beed gravity and drag
v=Root[(2ma)/(ρ(Cl Ak-Cd Ab)
Where Cl is lift coefficent
Ak is area of kite
Ab is area of bridles
Cd is drag coefficient

knowing that Cl = [2(pi)(AoA)]/[1+(2pi)/AR]
AoA= Angle of attack
AR = Aspect ratio
is there another equation for Cd without simply just rearaging the drag equation?
anyone who knows more about parachute/kite physics/fluid dynamics have a definitive answer to this discussion?
Thanks
Ben

Last edited by a moderator: May 4, 2017
2. Jan 4, 2010

### Andy Resnick

Those are cool!

I suspect your simplified analysis is not worth doing; those look like very complicated shapes (you must consider the full 3-D problem) and the wind conditions are highly variable in time. Wind tunnels are used for a reason.

3. Jan 4, 2010

### Marshall10488

i'm not looking for exact just estimate, to prove either way. the big problem i have is the drag from the bridles due to harmonics.

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