RChung

Hmmm. It does address your original question. Once you know your drag parameters under zero wind you can back-calculate what the "effective" wind would have been to have seen what you saw. But it's become clear that you didn't really want to know, you just wanted opinions. So I'll bow out. Carry on.

f4rrest

Yes, I shouldn't have added the wind speeds assuming same power.

f4rrest

No.

Carbonfiberboy 

To the contrary, I appreciate your input. My original question was, "Where did the watts go?" I wasn't particularly concerned with how many watts, just where they went. I think I have the answer to that. Calculations as to how many watts and thus the average of the wind resistance at various distances from the ground will have to wait until another day, until I can sort out Crr at 25 mph, which should automatically allow for the spoke churning.

Bunyanderman

Moving at at 25 mph with a 25 mph tail wind, you are still fighting wind resistance, just not as much with no tail wind. I believe a bike+rider is more aerodynamic when comparing the back side to the front. The front has a higher drag coefficient when compared to the back.

rpenmanparker 

Yes, I already said that above. Plus the rider himself shields his front from the tailwind leaving air in front of him that he has to move through. Isn't this fun?

achoo

What about physics do you fail to understand?

Power is force times velocity. Period. Applying a force at 12 mph requires 1/3 the POWER if the speed were 36 mph EVEN IF THE FORCE IS THE SAME.

Like when you're riding into a headwind.

Which is why 300W of power into a 25 mph wind results in you going 12 mph, even though with the same relative wind speed you'd need just about 900W to go the same 37 mph through the air were there no wind.

That's because with the 25 mph headwind the force is being applied to the ground, which the bike is moving over at only 12 mph. About 1/3 the speed of the air going past.

achoo

Also, because you're going over the ground a lot faster, it's going to take a lot more power to increase the force driving you through whatever resistance you do get.

Carbonfiberboy 

Pretty easy to see why we picked up a line of shattered singles on the way north, doing 9.7. I came up on a woman about 10' behind her man and commented, "This is what is called a soul-destroying wind." She replied, "Yeah, my soul is destroyed." Her face was quite pale, too. I relayed that info to her man who then dropped back to her.

This was on Hwy. 1 and bizarrely, we saw many loaded tourists heading north at about 5 mph. Some folks must not have gotten the memo.

That F X V thing is not intuitive for some reason. Intuitively, one would think that power required would be proportional to the square of the apparent wind speed. However anyone with a power meter should be able to post power files showing that achoo and physics are correct.

f4rrest

Again, here's the problem. The force is not the same if power is constant and velocity is different. This was the premise in my description of two scenarios: pedaling into a headwind versus pedaling in still wind at the same perceived effort.

If the rider is putting out 300 watts, then the wheel is putting that power into the ground to move forward, regardless of what the ground speed is relative to the wheel.

Intuitively, this makes complete sense. Pedaling into a headwind it gets more difficult, so you select a lower gear to maintain the same power at the pedals. The power at the wheel is the same as the power put into the pedals, but through gear reduction, the angular velocity at the wheel is now lower and the torque at the wheel is now higher.

This is how gears work. As you also know, P=Torque x Angular Velocity. Selecting a lower gear results in a lower angular velocity at the wheel and a higher torque at the wheel for a given power. Now, since the radius of the wheel is constant, force at the contact patch between wheel and ground is necessarily higher if torque is higher, and velocity at the contact patch is lower, since angular velocity of the wheel is lower.

Yes, P=FV, so into a headwind, the force applied by the wheel at 12 mph is greater than the force applied to the ground without the headwind, precisely because V is different and P is the same.

Power is the same; Velocity is lower; Force is higher. Got it?