Wind force in Kilos on a Kite

Dylan123

Question: If you have a kite with a projected area of 6.1m2 and a gust of wind hits the kite at 5omph how many kg of pressure would this exert? The more wind exerted on the kite the harder it is to pull the safety. If there is 200kg of pressure it takes 7kg of pressure to pull the safety - how many kgs of pressure would it take to release the safety at 50 mph

Also wonder how it is possible to calculate this when the shape of the kite will depend on several factors: including that its shape is not a flat surface like a sail on a yacht or windsurefer. (even though these sails are not 100% flat like an airplane wing)

(1) There are two 25m lines connected to the front and two at the back
(2) The lines at the back can be extended by approx 8cm (which has the same effect as the wing on an airplane)

The manufactures state the max wind speed is 30 knots of wind for a 12m kite which actually only has 6.1m2 area.. (no idea why they call it a 12 m if its only 6.1m2 - thats besides the point though)- i am wondering what will happen when the wind doubles or tripples in a gust?
The reason for the question is because the technology is relatively new (every year its updated). Even a slight idea of the pressure we are dealing with would give me a good idea how the safety system operates under massive load! many thanks i know its an awkward question.

ApeXaviour

Dylan123 wrote: »

Question: If you have a kite with a projected area of 6.1m2 and a gust of wind hits the kite at 5omph how many kg of pressure would this exert?

Right I'm gonna really simplify this, assume the kite is a flat plate shape, and is stationary with respect to the wind.

Drag equation is F = (1/2)(p)(v^2)(C)(A)
p = density of air (~1.2 kg per metre cubed)
v = velocity of wind (with respect to object)
C = drag coefficient (~1.28 for flat shape perpendicular to wind direction)
A = Surface area of kite

If I understand it correctly the drag coefficient I chose is for total surface area and not just the side facing the wind... so 12.2m^2. Also 50mph = 80kmh = 22.2m/s

so 0.5*(22.2^2)*1.2*12.2 ~ 3600. Units are newtons. So that's ~360kg equivalent force.

Dylan123 wrote: »

The more wind exerted on the kite the harder it is to pull the safety. If there is 200kg of pressure it takes 7kg of pressure to pull the safety - how many kgs of pressure would it take to release the safety at 50 mph

Pressure is different, but nevermind. I have absolutely no idea how this safety latch works... but if the numbers you say are true, and they increase linearly. 200:7 or ~30:1 ratio then 360kg equivalent force would require 12kg to release the safety. But please acknowledge this is pure conjecture.

Dylan123 wrote: »

Also wonder how it is possible to calculate as this when shape of the kite will depend on several factors:

(1) There are two 25m lines connected to the front and two at the back
(2) The lines at the back can be extended by approx 8cm (which has the same effect as an airplane wing )

The shape of the kite changes the drag coefficient significantly, see: http://en.wikipedia.org/wiki/Drag_coefficient. Fluid dynamics has a lot of variables, I'm only giving you the roughest of approximations... not least of which I'm assuming your kite and person attached are not stationary with respect to the wind... So at any one time, if the kite/person are moving at a constant velocity, the kite will only be exerting the same force as the person is on the kite (coefficient of weight and drag).

Dylan123 wrote: »

The manufactures state the max wind speed is 30 knots of wind - i am wondering what will happen when the wind doubles or tripples in a gust?

Now this where it gets unpredictable, and why the wee equation above is probably of little use in anything but a wind tunnel. A sudden gust would conceivably accelerate a kite (and its passenger) briefly but possibly with a lot of force (force=mass*acceleration). Think about if you were in a fast car. When itis traveling at a constant top speed you're sitting pretty, but if goes full welt from 0 to that speed you're glued to your seat. So depending on the "suddenness", magnitude and direction it hits the kite it could conceivably exert an unmanageable, not to mention unpredictable amount of force.

I was in Achill on the weekend and the power went out all over the island... I'd heard a kite went into it The wind is unpredictable, I hope the person was okay.

SOL

As a less analytical aside, it is worth noting, and VERY noticable if you do and wind based activities, that the power in the wind goes up with the square of the speed of the wind.

So if you are out in a Force 8, the 40 knots you feel has 4 times the energy as a you would feel in a Force 5 at 20 knots...

Dylan123

I posted this on the Mathmatics forum and i think it is better placed here. Fairly technical stuff!!! great to get the advise from such knowledgeable people as this is something i doubt any wind based sport persons would understand! (of course i may be wrong - it wouldnt be the first time)

"This isn't really a maths question, it's a question about a very specific branch of engineering.

Look at some of the replies on this page to see how complicated the calculations are: there are tables and tables for the pressure on different shapes of sail.

In general, however, the pressure varies with the square of the wind speed. If the wind blows twice as hard, there will be four times as much pressure.

This site gives a rough calculator for the pressure on a sail (motivated by predicting the size of the winch you need to turn it). If you try 65 square feet and 65 mph, as you want, you'll get about 1200lb of force (550kg). I wouldn't even consider using this info for what you suggest are life and death matters, but it's probably right to within a factor of two.
http://www.sailingusa.info/cal_wind_load.htm
It's not clear what you mean with your other factors. If the kite is tied down by unbreakable 25m wires, then it doesn't matter how strong the wind is. Changing the shape of the kite will have a big difference on the effective force it feels, though."

Dylan123

What amazes me is the fact that double the wind speed will quadruple the force.

If you have 10 mph wind on an articulated lorry surely 20 mph of wind will exert double the load? and 20 mph will have double the chance of blowing the lorry over opposed to four times.

So if the wind is blowing at 20 knots and i reach a speed across the water of 15 mph. If the wind increases to 40 knots I will reach 60mph..... hmmm that would be faster than the wind itself im missing the point... force and speed have a different relational proximity

ApeXaviour

Dylan123 wrote: »

So if the wind is blowing at 20 knots and i reach a speed across the water of 15 mph. If the wind increases to 40 knots I will reach 60mph..... hmmm that would be faster than the wind itself im missing the point... force and speed have a different relational proximity

Yes indeed, you're not going to outrun the wind. Force is directly proportional to acceleration, not speed. The point is if you were anchored, the force you would feel from the kite at 40 knots would be 4 times what you would feel at 20.

kinetic energy (same with power as SOL mentioned) is proportional to velocity squared... so similarly if you hit an immovable brick wall at 30mph, the impact would be 4 times as powerful than if you hit it at 15.

The same if got caught in a squall, it would cause your kite to accelerate, significantly increasing the force with which the kite is pulling you.

Dylan123

Again a very strange question but relevant: if you don not know there is no panic and i really appreciate your knowldege sharing so far its very complex and highly interesting!

If the kite goes into an uncontrolable loop (perhaps because the left or the right lines break) i suppose it is impossible to work out the added force in kg of pressure on the lines and safety system is it? The kite would end up directly down wind of you in the most powerful wind zone! this i know creates an allmighty amount of force when its not windy! I wonder what the statistics look like for really windy conditions such as:

- 6.1m2 kite
- 42 knots of wind
- Already might have 350kg of pressure as standard

Would u work this out as an estimate by the circumference of a circle multiplied by the downward force and acceleration??? you need a brain the size of a planet to work this stuff out!

krd

ApeXaviour wrote: »

If I understand it correctly the drag coefficient I chose is for total surface area and not just the side facing the wind... so 12.2m^2. Also 50mph = 80kmh = 22.2m/s

I think you're wrong. I think for that equation you would only take into account the surface area facing directly into the wind.

The drag coefficient would also vary from material to material. A really kites change shape with the wind.

krd

Dylan123 wrote: »

Would u work this out as an estimate by the circumference of a circle multiplied by the downward force and acceleration??? you need a brain the size of a planet to work this stuff out!

You would need to use some aerodynamic modelling software to get an idea.

Or build an expensive wind tunnel to get another idea.

You could get a crude estimation

Or you could crudely estimate the maximum force that can be exerted if one of the lines breaks. The Drag equation is F = (1/2)(p)(v^2)(C)(A) looks like it gives the maximum force on the kite. but kites change shape - the sails of a kite or a ship change shape to maximise the force of the wind - they have a concave shape when they're catching the wind. It's actually really complicated when you think of a sail in the real world.

There may be another more complicated equation, that takes into account the shape of the sail. That's where a wind tunnel would come in handy - you could measure the specific drag coefficient for the shape of a sail, without having to do some very ugly looking maths. They may not be that ugly the formulas may have reduced nicely and you may not have to worry about irregular shapes.

There are also probably some very specific formulas for sails.