Headwind and Power
01-02-22, 01:02 PM
#51
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If it has devolved into nonsense, it's only because people are trying to find some explanation that will resonant with you so that you can understand why you are wrong. (And, yes, you are wrong.)
01-02-22, 01:02 PM
#52
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Omg! Just stop :-) This was addressed several times before. Including by me in #13 and Im sure in other posts too since you all seem to throw this at me, even if I made no claim to the contrary.
This thread has devolved in to nonsense and Im out of here as I said in #27
This thread has devolved in to nonsense and Im out of here as I said in #27
01-02-22, 06:07 PM
#53
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so, in the spirit of the season - 'off-season' for some - a thread we can all present our As and Os...
...so, I've been wonderin, for so very long...
given the same rider, on the same bike
if you're riding at 10 mph (16 kph) into a 15 mph headwind (24 kph) , are you using the equal/same 'power' as if you were riding at 25 mph (40 kph) ???
is it an equal relationship, and is it linear ?
what other factors might influence this greatly ? - are rolling resistence and impedance nominal or significant players in this drama ?
So Bill Nyes _ The Science Guys in our midst - what is the reality?
not a scientist ? have an opinion? lets hear it !
My wishes for more power to all for the coming year
...so, I've been wonderin, for so very long...
given the same rider, on the same bike
if you're riding at 10 mph (16 kph) into a 15 mph headwind (24 kph) , are you using the equal/same 'power' as if you were riding at 25 mph (40 kph) ???
is it an equal relationship, and is it linear ?
what other factors might influence this greatly ? - are rolling resistence and impedance nominal or significant players in this drama ?
So Bill Nyes _ The Science Guys in our midst - what is the reality?
not a scientist ? have an opinion? lets hear it !
My wishes for more power to all for the coming year
horse power = [Vg*W(.0053 + %G/100) + .0083(Va^3)]/375
calories/hr = [Vg*W(.0053 + %G/100) + .0083(Va^3)]*7.2
calories/hr = [V*W(.0053 + %G/100) + .0083(V^3)]*7.2
watts = [V*W(.0053 + %G/100) + .0083(V^3)]*2
where Vg is ground speed, Va is speed through the air (includes head/tail winds), W is bike + rider weight in lbs., and %G is grade in per cent. The factors listed here (0.0053 for friction + rolling resistance and 0.0083 for aerodynamic drag) are obviously not absolute. They will vary with efficiency of the tires and drive train, and with the aerodynamics of the bike + rider combination. Both of these assume a racing position on a racing bike. A clunker bike or a more efficient riding position will change these numbers, which are averages anyway. Power to overcome friction and gravity is proportional only to rider weight and ground speed. Power to overcome wind drag is proportional to the cube of the air speed. For reference, 1 hp = 2700 calories (because of human metabolic efficiency of 24%); 1 calorie = 0.276 watts; 1 hp = 746 watts. Here, all calories are kg-calories, or "food calories."
01-02-22, 06:23 PM
#54
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Otto
01-02-22, 08:29 PM
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Yep, I spotted that too. Drag force is proportional to the square of the airspeed and Power is proportional to Drag Force x Vg. When there is no headwind Va = Vg and the Power equation then simplifies to V cubed. It's actually this very fact that causes confusion when you start introducing headwind.
