How much pop from a Kitewing?

  • Thread starter Phrak
  • Start date
  • #1
4,239
1

Main Question or Discussion Point

There's a product on the market called a Kitewing. It's a hand held sail for ice, water or land. It can be swung horizontally to lift. I'm curious to know how much mimium wind speed would be needed to get some pop?

There are some utube videos out where some get 5 ft of air from level ground. Suprizing for such a small wing.

The sail is symmetric like a wing, having a 5.5 square meters, with an aspect ration of about 2.2, and a roughly eliptical profile. Wing span and washout aren't given.
 
Last edited:

Answers and Replies

  • #2
4,239
1
So I guess there aren't any AEs around who have any interest in fluid flows without a cabin wraped around them and 500+ horses in the front...
 
  • #3
2,985
13
Your question doesnt leand itself to a simple answer. Aerodynamics is highly experimental. There isnt a magic formula you can use to arrive at the answer you want.
 
  • #4
4,239
1
I've run out of time. I'll have to generate some numbers tomorrow afternoon.
 
  • #5
4,239
1
Your question doesnt leand itself to a simple answer. Aerodynamics is highly experimental. There isnt a magic formula you can use to arrive at the answer you want.
I'm looking for a, not too, rough idea of how well it could perform. So far I have,

Mg = ½CLρV²S

ρO = 1.225 kg/m³, density of air at sea level
ρ = 1.1 kg/m³
CL_MAX=1.2
V = 30 mph = 44 fps = 13.4 m/s
S = 5.5 m²
g = 9.8 m/s², acceleration due to gravity.
M = 67 kg
Mg = 146 lb ...very marginal.

For a sail on land, we could assume a "beta" of at least 30 degrees. This means wind speed is 15 mph to obtain a 30 mph relative wind, V at the most favorable tack. The ground speed would be 30*0.866 mph.

For a wing that is mostly single sided except along the leading edge, I'm assuming a max coefficient of lift of only 1.2.

I still don't know if I'm using the right coefficient of lift. CL_MAX is the lift coefficient of an infinite aspect ration section.
 
  • #6
2,985
13
Sorry, that's way wrong. You want 'not too close' build or use a wind tunnel, or find data generated by someone who has. You've pretty much assumed 99.9% of the problem. Your answer is probably as much as 40% off.
 
  • #7
FredGarvin
Science Advisor
5,066
7
You're estimating on the Cl and treating like an infinite wing. Of course it's going to be off, but, IMO, for a first guess, those numbers don't seem all that bad. At least you aren't in the area of an order of magnitude off. Especially since you are treating Cl as a constant (which it isn't) and who knows just what the relative velocity a wing actually sees is. As a total SWAG I think you're not too bad. Just understand that to be accurate you need way more information.
 
  • #8
4,239
1
You're estimating on the Cl and treating like an infinite wing. Of course it's going to be off, but, IMO, for a first guess, those numbers don't seem all that bad. At least you aren't in the area of an order of magnitude off. Especially since you are treating Cl as a constant (which it isn't) and who knows just what the relative velocity a wing actually sees is. As a total SWAG I think you're not too bad. Just understand that to be accurate you need way more information.
I don't understand the sources of large amounts of error. Is it just CL? But, this is what I was asking. How is CL max obtained from CL max from section data?
 
Last edited:
  • #9
FredGarvin
Science Advisor
5,066
7
I don't understand the sources of large amounts of error. Is it just CL? But, this is what I was asking. How is CL max obtained from CL max from section data?
It's usually not. It is done experimentally...hence Cy's comments.
 
  • #10
4,239
1
It's usually not. It is done experimentally...hence Cy's comments.
I'll read theory of wing sections.
 

Related Threads for: How much pop from a Kitewing?

  • Last Post
Replies
1
Views
3K
Replies
5
Views
3K
Replies
7
Views
3K
Replies
12
Views
6K
Replies
2
Views
8K
Replies
4
Views
7K
Top