Max Force = kite (sq. meters) x rider (lbs) x wind (mph)

[MK]

Is this the equation to guage how much force can be created by a kite?

I don't claim to be able to solve this, just throwing it out as a query for someone to solve. I think there is some variables missing but if the equation is set up I could probably dust off the Dimensional Analysis method for solving equations I used back in Chemistry. Maybe it was some other class? Organic Chem? Whatever it was I probably bombed it

I did a quick search thinking this topic has been discussed before but my keyword was "force" and I uncovered a lot of threads related to: motivation, Star Wars, and Liquid.

The first correct answer gets a free copy of the lastest Kiteboarder Magazine or a chocolate pudding snack pack single (that's what is in front of me, unopened!).

[pepe]

Here is more accurate definition of this question:

A force is pulling a 100 kg kiter across water surface. The force acts at an angle of 45 degrees above the surface. Assuming that the coefficient of friction is 0.5 and that the kiter moves with 10 m/h. How big of a kite he/she needs to have if the wind is 15 m/h.

*The coefficient of friction in this case will depend of on the the size, shape and fins of the board you are riding.
** The vertical component of the pulling force is calculated: F1(y) = F1 x sin φ

[MK]

This response came in via email, Thanks Mark:

The end force equation will be a somewhat deviation of

where
FD is the force of drag, which is by definition the force component in the direction of the flow velocity
ρ is the mass density of the fluid,
u is the velocity of the object relative to the fluid,
A is the reference area, and
CD is the drag coefficient

Anyone care to plug some variables in?

[Kosk]

Your equation: kite area (m^2) X mass (kg) X wind velocity (m/s) ends up with a product with units of (kg*m^3)/(s).

Force is (kg*m)/(s^2)

Your equation appears to be flawed.

I think Lift generated by the kites forward movement plays a critical roll in the total pulling force a kite has, not just the area projected normal to the wind direction. If this were not the case, it would be impossible to move upwind, as the projected area normal to the wind direction can only push the kite in that direction (down wind) and drag you along with it.

Lift is related to the square of velocity. (Twice the wind speed = 4times the lift force). This explains the "apparent wind" phenomenon, the reason some kiters can ride in really light wind and others struggle. It's all about momentum. The more you keep your speed up, the more speed the kite sees, and thus the more lift generated (the harder it pulls).

I have no desire to figure out your equation actually kiteboarding is WAY more fun. But hopefully reading this made your brain hurt a little bit.

[adventure in]

Kosk responded as I wrote this up. I would agree with apparent wind and such. My imagined formula is below.

Force = (Force due to air resistance) + (Force from airfoil shape)

Force due to air resistance = 1/2 * p * Cd * A*v^2
With p = density of air = 1.2 kg/m^3
Cd = Coefficent of drag --> assume flat plate = 1.28
A = Area presented lets assume straight downwind and so you have about 10^2 m of your 12 meter kite
v = wind velocity (will keep as a variable for the graph)
_____________
Fa = 1/2*1.2kg/m^3*1.28*10m^2*v^2=7.68*v^2

Now this should be the force the kite produces when it is sitting leading edge down in the water directly down wind of you. To me these calcs look a little high.

_________________
Now the next part is more tricky to calculate. Since a kite is by no means an ideal airfoil, I do not believe the equations will hold very well. This will most likely cause a great difference kite to kite on force produced. Not to mention this also depends on angle of attack.

If you did want to figure this part out you could construct a curve based on the force the kite produces an nearly 90 to the wind direction at various wind speeds. Again Angle of attack has so much to do with the equation I don't think that would be very accurate either.

[adventure in]

Another Thought...

If you wanted to know how much force it takes to move you through the water you could attach a force gauge on a tow rope and get pulled on your board behind a boat at various speeds.

I predict that you will have a higher force required to keep you up at slower speeds due to you sinking down more.

[MK]

Yes, brain hurts and it is more fun to just go kiting.

I'm totally out of my league here. In trying to teach, I was confronted with this questioin. I'm not sure I could regergitate these factors in an effective way. Maybe I need a good, relevant analogy?

Good stuff. Glad to see the motivation for a Kiteboarding Magazine. Or was it over the pudding?

[JimPat]

Hey Homies!!!!!!!!!
If you get this figured out we can mount force meters on our bars so we can look at the bar while we move the kite through the air.
We will however need some type of warning device to tell us if something is in our path since we are not looking where we are going. I think this guy wants this info to make a new Wii game! Sweet- we can kite in the comfort of our home on the couch with a beer and some chips and pull all sorts of ticks in the air with our game controller. Now if we could do this on a trampoline we would have a great simulation tool! On second thought, keep working on that formula! On the other hand (she had warts) I prefer to get my WEEE's (sometimes woops) out on the water but I do think a kiting game would be super cool!!
Jimster, Wii na be, Wanna be a Weeeeeeeeeeeee Team rider!

[Kosk]

Mike,

If any student asks you this question again, the standard response should be; "Why don't you grab the right side of the bar with both hands, pull really hard, and tell me what you think." Too much chit chat, not enough kiting....

PK