Just to broaden our knowledge about windsurfing, short question:

Did somebody try and estimated /calculated what could be our maximum power generated by sail during speed trial ?

If Lauderitz top breaking record require 10 Hp or rather 50 ??

Just to illustrate human body can generate about 200-300 W continuously during excercise and about 1000 W at the peak.

So how much power our sail deliver ?

Using the formula (Mass x Wind Speed)/Sail size = Power

My (100kg x 25kts)/7.8m = 320.5KillerWasps

So watch out come Burrum wind fest.

The lift coefficient could get as low as 0.6 at low angle of attack, sail area say 5.5m,
true wind speed: 45 knots, angle off the wind: 110deg, boardspeed: 50knots --> apparent wind speed: 60 knots, apparent wind angle = 45deg, proportion of lift in board direction = 0.75ish
Power = 1/2*density of air*lift coeff*area*windspeed^2*0.75*boardspeed = 0.5*1.2*0.6*5.5*30^2*0.75*25 = 33 kW = 45 HP, so yeah 50 isn't far off. Comparable to a older, small car that's had a hard life.

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7tim said..
The lift coefficient could get as low as 0.6 at low angle of attack, sail area say 5.5m,
true wind speed: 45 knots, angle off the wind: 110deg, boardspeed: 50knots --> apparent wind speed: 60 knots, apparent wind angle = 45deg, proportion of lift in board direction = 0.75ish
Power = 1/2*density of air*lift coeff*area*windspeed^2*0.75*boardspeed = 0.5*1.2*0.6*5.5*30^2*0.75*25 = 33 kW = 45 HP, so yeah 50 isn't far off. Comparable to a older, small car that's had a hard life.

taking into account the complete rig weight less then 15 kg ( board 6kg. mast 1.5, sail 3, harness 0.5 , fin 0.2 boom 2 kg) , power to mass ratio is quite impressive !
With 3HP per kg it makes much better then superbike or F1

now interesting will be to know what could be force / power equivalent required to pull the same sailor on the same speed board with same fin board ( but without sail/rig) on rope behind speed boat at 50- knots.
That could give us idea about resistance created by water to planing board+fin.
If the sailor is capable of easy hang with his bare hands on the line that give us idea that planing board, fin resistance and air resistance don't amount for more that 1 Hp on straight course ( alternatively equivalent 50 kg or 500N force on his handle bar)

Pretty sure thats a mile off isnt it ?

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Simon100 said..
Pretty sure thats a mile off isnt it ?

Considering a 25hp outboard will push 2 blokes, a dog, and a full esky in a 14 foot long tinnie along at 20knots. Having said that the power required for speed increase is exponential but 50hp seems excessive. The powerboats that run the Avon decent are 8 or 12hp so to think of one of them bolted to the back of one's board should see 40 knots achieved without to much problem.

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7tim said..
The lift coefficient could get as low as 0.6 at low angle of attack, sail area say 5.5m,
true wind speed: 45 knots, angle off the wind: 110deg, boardspeed: 50knots --> apparent wind speed: 60 knots, apparent wind angle = 45deg, proportion of lift in board direction = 0.75ish
Power = 1/2*density of air*lift coeff*area*windspeed^2*0.75*boardspeed = 0.5*1.2*0.6*5.5*30^2*0.75*25 = 33 kW = 45 HP, so yeah 50 isn't far off. Comparable to a older, small car that's had a hard life.

I think 60 knots is a bit of an overestimate for apparent wind. ~ 30 is more like it, at the broad angles sailors need to go to get to 50 knots. That would reduce your calculations to 11 hp or less.

A better approach may be to find the tension in the rope pulling a water skier at 50 knots? That should be easy to measure? A narrow windsurfer won't be far off it. I'd reckon 15 kg tops, you've got to hold on after all, and they do it for hours. Power = force times velocity = 15 * 9.8 * 25 = 3.7 kW ~ 5 hp.

Bicycle power is easier to get a handle on.

Most of the bicycling wind resistance would be the rider, the apparent wind on a water skier at 50 knots is 50 knots, the windsurfer, ~ 30 knots, just on that graph, look where that graph is heading at 50 knots! So I'm thinking the aerodynamic force in the forward direction of a windsurfer at 50 knots may be less than the tension in the rope of a water skier at the same speed.

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Ian K said..

I'd reckon 15 kg tops, you've got to hold on after all, and they do it for hours. Power = force times velocity = 15 * 9.8 * 25 = 3.7 kW ~ 5 hp.

This approach may be close to real value.
But I would rather use now different data to calculate our max output.

1. First example could be approx power during record breaking speeds around 50 knots
Lets assume 110 kg , sailor and rig altogether.
we could also assume that at this speed our sailor is almost levitating

P=F * V
P= ( 110 kg x9.8 ) *( 50 knots x0.514) =27.7 kw ~ 37 Hp

2.Next could be my output when doing my speed record , say 33 knots( assume that I am not the efficient so levitating whole weight too almost)

P=(95kg x9.8) * ( 33 knots x .514) = 15.7 kW ~21 HP

3.I guess 5 HP will be adequate average for low speed comfortable everyday planing.

We still must be realistic ,that our rig and board , fin are not 100% efficient machines and most of that power is wasted,
not simply converted into speed.

Now if all that calculation above are about real. interesting will be to learn what was actually wind power swept by rig during those 3 examples. If somebody want to play here is equotation we may use:
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Hopefully , our calculated HP power for sails doesn't come at higher value that wind power available

Great math Ian. Now how does a facade and air create a greater resistance to structural steel thermal expansion than the reinforced core of a steel skyscraper?

Answer: it doesn't, unless you believe lies.

No really, just kidding.

PS. Oops. Wrong discussion.

I think thats the right formula the force direction isnt in the direction you would typicaly expect it to be but i think it all works out the same , but remember velocity will be apparent wind speed not speed over ground and force is lateral force on the sail id guess only about 30 kg . remmeber we are super efficient compared to almost everything else on the water most motor boats are very inefficent designs.

Dear wind god please send these sailers wind

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Simon100 said..
I think thats the right formula the force direction isnt in the direction you would typicaly expect it to be

I realize that, In fact all force vectors are very complicated.
On another hand , when rig is properly designed we could multiply those forces to exceed our own weight as counter balance.
For example that skipper of sail rocket doesn't really need to do much to achieve this astronomical speeds on the water.


I think that designers of our windsurfing speed gear my one day follow similar path - transferring most of the forces into our gear - rig, board ,fin.At his moment the ultimate restriction are :
-sailors weight
-sailors strength
-Sail size
-wind speed
I imagine that in few year , nearest future of advanced computer design all those becomes irrelevant.
We could design sort of "power steering" where sailor only steer and direct without the need to involve extreme strength and stamina.Lets imagine that one day you will be able to handle and steer 20 m2 sail in 50 knots wind !!! What could be your speed then ???

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Ian K said..

A better approach may be to find the tension in the rope pulling a water skier at 50 knots?

Lets imagine now, Ian that we have exact model of our car pulled /towed on the rope in our lab. Forced measured.
Do we really need 1000 Hp to pull the exact copy by weight and shape on simulated drag race 1/4 mile at exact acceleration and speed ?

BTW. Knowing exact weight of nowadays dragster we could calculate exactly net physical power to accelerate this vehicle.
So interesting question is somebody already did the proper calculation what is actual efficiency of those drag racing machines ?

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Macroscien said..

Ian K said..

A better approach may be to find the tension in the rope pulling a water skier at 50 knots?

Lets imagine now, Ian that we have exact model of our car pulled /towed on the rope in our lab. Forced measured.
Do we really need 1000 Hp to pull the exact copy by weight and shape on simulated drag race 1/4 mile at exact acceleration and speed ?

BTW. Knowing exact weight of nowadays dragster we could calculate exactly net physical power to accelerate this vehicle.
So interesting question is somebody already did the proper calculation what is actual efficiency of those drag racing machines ?

Dragsters aren't a good example, they waste energy spinning tyres, but for a car driven without wheel spin the results of calculating hp on a chassis dynamometer or a piece of string will be the same. (apart from tyre distortion differences, and whether you are able to free wheel out the transmission losses ).

If towing a windsurfer board and rider though you will have to consider if you put the the fin in or not. The fin could be considered part of the sail's HP generating system and is thus not 'taxable". We need an accountant, but once you get that sorted, F=ma always works.

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Ian K said..

Macroscien said..

Ian K said..

A better approach may be to find the tension in the rope pulling a water skier at 50 knots?

Lets imagine now, Ian that we have exact model of our car pulled /towed on the rope in our lab. Forced measured.
Do we really need 1000 Hp to pull the exact copy by weight and shape on simulated drag race 1/4 mile at exact acceleration and speed ?

BTW. Knowing exact weight of nowadays dragster we could calculate exactly net physical power to accelerate this vehicle.
So interesting question is somebody already did the proper calculation what is actual efficiency of those drag racing machines ?

Dragsters aren't a good example, they waste energy spinning tyres, but for a car driven without wheel spin the results of calculating hp on a chassis dynamometer or a piece of string will be the same. (apart from tyre distortion differences, and whether you are able to free wheel out the transmission losses ).

If towing a windsurfer board and rider though you will have to consider if you put the the fin in or not. The fin could be considered part of the sail's HP generating system and is thus not 'taxable". We need an accountant, but once you get that sorted, F=ma always works.

F = ma sure does. Measure the force of a category 5 hurricane on a 110 storey building facade and compare that to a 767. Then do it twice and check probability of complete failure.

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Bonominator said..

F = ma sure does. Measure the force of a category 5 hurricane on a 110 storey building facade and compare that to a 767. Then do it twice and check probability of complete failure.

That's a chalk and cheese comparison? Why bother?

But then again I can't resist a bit of back of the envelope doodling. How quickly do you want me to stop that 767?

(I'm sure they'll let you back in heavy weather if you ask nicely )

Sail rocket runs its main fin on the same angle as the sail completely balancing each other no counter weight is actually needed in its design . if you get have a nice new slalom gear all set up just right the whole lot does look after its self with very little input from the rider it just seems to take ages to figure out how to do nothing for this to happen , I think its increadible how well our sails actually work these days they are almost a rigid wing down the bottom that is asymetric and flips sides and with the out haul you can change the lift caracteristics and active twisting to let out gusts compared to those triangle things they used to have its a completely different thing , im lucky i only started windsurfing when they figured things out a few years ago :)