Hi Brains Trust. I was wondering. Assuming everything else is held constant and only the size of a wing is changed. Will a 3m wing have a wind range twice that of a 6m? For example, if a 6m wing has a comfortable wind range of 8-16 kts then will a 3m of the same design have a comfortable wind range of 16-32 kts. I think spending too long in lockdown has muddled my mind. ?????

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ap73 said..
Hi Brains Trust. I was wondering. Assuming everything else is held constant and only the size of a wing is changed. Will a 3m wing have a wind range twice that of a 6m? For example, if a 6m wing has a comfortable wind range of 8-16 kts then will a 3m of the same design have a comfortable wind range of 16-32 kts. I think spending too long in lockdown has muddled my mind. ?????

Roughly,yes.
Lift= (WindSpeed x WindSpeed) x Sail Size

So those two sizes would give approximately the same low end power at 8&16 kt.
EDIT: Got the math wrong,it is 3m/11kt

But Wings are very difficult to "scale", meaning that the same design in different sizes is going to behave differently,bigger the gap bigger the difference.

Ps: if you are thinking of a quiver of 2 i would definitely go for a smaller gap, i ride 3.5 and 5m and that is gappy enough for me :)

Ooo no sure about that ....double the velocity and the force is 4x not 2x ....it's a squared relationship.

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kobo said..
Ooo no sure about that ....double the velocity and the force is 4x not 2x ....it's a squared relationship.

I think we are saying thevsame thing.
(WindSpeed x WindSpeed)= Speed squared

What i meant is he will get the same power from a 6m/8kt and a 3m/16kt.
EDIT: got the math wrong,it is 3m/11kt .

In theory,in practice it is not that simple.

You got the principle right, but calculated the math wrong. All else equal, scaling just area and calculating wind speed from 6m/8knt, then it will be 3m/11kn. Squared relationship means if you halve the area wind speed increases by sqrt(2), not two.

As noted, all else is not in fact equal. Lift coefficients vary across wing sizes and even across wind ranges for the same wing (pesky flexible lifting generating devices), the rider and the board in the air quickly become the dominant source of drag as apparent wind increases, etc etc.

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Svendson said..
You got the principle right, but calculated the math wrong. All else equal, scaling just area and calculating wind speed from 6m/8knt, then it will be 3m/11kn. Squared relationship means if you halve the area wind speed increases by sqrt(2), not two.

As noted, all else is not in fact equal. Lift coefficients vary across wing sizes and even across wind ranges for the same wing (pesky flexible lifting generating devices), the rider and the board in the air quickly become the dominant source of drag as apparent wind increases, etc etc.

You are correct :)

Thanks, that makes sense