Ratio between wind speed and force

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lampshuk

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I have it in my head as a "fact" that the force exerted by wind on a sail is proportional to the square of its velocity. I was comforting myself with this thought when I was cycling uphill into a headwind on the return leg of a bike ride.

Then I asked myself "why?" and I realised I didn't know. The ensuing mental gymnastics did at least take my mind off my excruciating unmentionables for the rest of the ride.

I have just done a Google search (admittedly quite cursory) with no satisfactory outcome. Anybody got a good explanation? Or did I make this up? Is it in fact a linear relationship?

Note that this is not a question about real vs apparent wind which, as any fule no, is governed by Murphy's Law of Inconvenient Headwinds.
 
It's just one of those facts in physics that one has to learn and accept. You could work out for yourself the fact that the intensity of radiation is only a quarter at double the distance, but I for one would not expect that the energy of a moving body varies with the square of its speed. Since wind, or air, is a moving object, this is what happens.
 
Complicated answer really as the force exerted by a sail isn't directly proportional to the drag, added to which there's more than one type of drag. In general lift generated will rise exponentially as airspeed increases, all others things remaining constant (which they never do in practice of course). That's one of the biggest problems with wing sails. Keeping it simple and talking about your bike then, yes, the force required to overcome the "wind resistance" of the bike and your body rises as the square of the speed.

https://en.wikipedia.org/wiki/Gliding_flight has some relevant stuff.
 
lampshuk said:
I have it in my head as a "fact" that the force exerted by wind on a sail is proportional to the square of its velocity. I was comforting myself with this thought when I was cycling uphill into a headwind on the return leg of a bike ride.

Then I asked myself "why?" and I realised I didn't know. The ensuing mental gymnastics did at least take my mind off my excruciating unmentionables for the rest of the ride.

I have just done a Google search (admittedly quite cursory) with no satisfactory outcome. Anybody got a good explanation? Or did I make this up? Is it in fact a linear relationship?

Note that this is not a question about real vs apparent wind which, as any fule no, is governed by Murphy's Law of Inconvenient Headwinds.
Click to expand...

To put it as simply as possible: the work done driving the boat comes from loss of kinetic energy of the wind. And kinetic energy is mass x (velocity) squared/2.
 
Thanks, all. Guernseyman, thankyou for reminding me of the MV^2 formula. I had forgotten to apply it, even though I have experienced it - most recently on the rugby pitch.

Johnalison, I'm not sure if this is what you were referring to, but the squared decrease in intensity of radiation with distance is one I could visualize, since the surface of the sphere around an object expands proportionally to its radius. As you say, I should just have remembered the MV^2 relationship.

G_O_G, thanks for that link. There's a lot of useful linkage between wings and sails. Fascinating stuff.

Thanks, Neeve, I think the interesting - and slightly sobering - corollary of your point is that if you're sailing happily at a comfortable angle of heel, and the wind doubles, you need to reduce sailing to a quarter of its original area to maintain the same force on the sail (though there will be many other complicating factors, of course).
 
According to Wiki on Beaufort scale
v = 0.836 B3/2 m/s
Where v is the equivalent wind speed at 10 metres above the sea surface and B is Beaufort scale number. For example, B = 9.5 is related to 24.5 m/s which is equal to the lower limit of "10 Beaufort". Using this formula the highest winds in hurricanes would be 23 in the scale.
Well worth reading the derivation of the scale on various websites. All related to the amount of sail a frigate could carry.
 
To further complicate things, consider that the "force" of the wind is also varied by the density of the air. Thus winter (cold) winds tend to be more destructive.
 
Good point.
Does the same apply to the humidity of air? ie air with a higher moisture content (with the same density and temperature) weighs more per m^3?
 
It's all in Wiki. The simple theory is that the incident wind is completely stopped and its momentum is converted into a stagnation pressure I.e. drag which opposes the forward motion of the body. The momentum is proportional to speed but also the mass of incident air per second is also proportional to speed so you get a square effect.
Re. radiation, it's only an inverse square for a point source (think point at centre of a spherical surface so the radiation is spread over that whole surface). For a linear source it's inversely proportional to distance and, amazingly, for a uniform extended source it's independant of distance ....in this case think that the further away you are the greater area of radiating surface contributes to the intensity at the point where you are located. At Uni. I only heard about inverse square variation and discovered the other cases for myself many years later.
 
lampshuk said:
I have it in my head as a "fact" that the force exerted by wind on a sail is proportional to the square of its velocity. I was comforting myself with this thought when I was cycling uphill into a headwind on the return leg of a bike ride.

Then I asked myself "why?" and I realised I didn't know.
Click to expand...

Force is change of momentum, and momentum is mass times velocity. The more it weighs or the faster it goes, the more force it takes to stop it.

For a fluid, the mass involved is also proportional to velocity. The faster it's going, the more gets to your sail.

So combine these and you get the velocity-squared idea: if the wind doubles in strength then (a) ever kilogramme hitting your sail has twice the effect and (b) twice as many kilogrammes hit your sail in any given time.
 
guernseyman said:
To put it as simply as possible: the work done driving the boat comes from loss of kinetic energy of the wind. And kinetic energy is mass x (velocity) squared/2.
Click to expand...

I'm afraid that it doesn't work like that. For a fluid, kinetic energy flux (the amount passing through or reaching a given area) is proportional to velocity cubed, for exactly the same reason that momentum flux (and therefore force) is proportional to velocity squared: the "mass" term is also proportional to velocity.
 
NormanS said:
To further complicate things, consider that the "force" of the wind is also varied by the density of the air. Thus winter (cold) winds tend to be more destructive.
Click to expand...

There is very little difference in density between warm and cold air. For equal pressures, air at 0oC (273K) is only 7% denser than air at 20oC (293K).

lampshuk said:
Good point.
Does the same apply to the humidity of air? ie air with a higher moisture content (with the same density and temperature) weighs more per m^3?
Click to expand...

Damp air weighs less than dry air at the same pressure and temperature. You are replacing molecules of nitrogen (mass 28) and oxygen (mass 32) with the same number of molecules - remember molar volumes - of H2O (mass 18). It's a while since I did the full calculation for a thread here but I think air with 100% humidity at 20oC is about 3% less dense than dry air.
 
Wow. Lots of Chemistry and physics revision here. Thanks, JD and others.

I think I will have to do like the constipated mathematician and work it out on a piece of paper.
 
JumbleDuck said:
I'm afraid that it doesn't work like that. For a fluid, kinetic energy flux (the amount passing through or reaching a given area) is proportional to velocity cubed, for exactly the same reason that momentum flux (and therefore force) is proportional to velocity squared: the "mass" term is also proportional to velocity.
Click to expand...

You seem to mixing up energy and energy flux, clearly different things.
 
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