Tailwind and speed difference

Originally Posted by Fday

What is a problem for me here is the answer has to be the same regardless of the frame of reference it is solved from. If we agree that the energy loss here is entirely from the wind resistance then the energy loss going to heat is the same for the rider going 1 mph into a 49 mph headwind and the rider going 50 mph into zero wind. Then the rider must simply put in the power lost to maintain speed. I am having trouble immediatly understanding what the speed of the massless wheel has to do with anything. But, as I said above, I will cogitate on this awhile and see if I can understand your viewpoint or find the flaw in it.

Originally Posted by dhk2

OK, maybe I see your hangup here. Agree that the airstream is losing energy at the same rate in both cases, but that doesn't dictate that the riders have identical power output. A moving airstream loses velocity (and energy) whenever it encounters obstructions, but nothing says the rider must match or restore that energy.

Consider a slightly more extreme scenerio where the slow bike isn't moving at all but just holding a track-stand, like a tree, in a 50 mph wind. How much power is the rider (or tree) putting out then? It would be zero, because the bike pedals (and bike) isn't moving, so no work is being done. The rider is exerting static force, ie, pressing on the forward pedal to hold the bike in the wind, essentially equal to that required to go 50 mph on a calm day, but there is no movement (velocity) and no work as a result. It's just like pushing your bike into the garage wall; no power is exerted via static pressure on the pedals no matter how hard you push.

Energy is being lost by the moving airstream, but the rider sitting in the gale at 0 mph isn't doing any work to replace it; he just let's the moving air go by, letting it compress and heat at the front and then expand/cool in back of him. No law says he has to restore or replace the energy lost by the airstream.

The speed of the rear wheel is important since that determines the speed of the bike, which is half of the power equation. Again, force x velocity is the definition of power. You can use the pedal forces and velocity to compute the rider's power output, or use the rear wheel force/velocity. Both results will be identical as the ratio of force/velocity is established by the relative lengths of the crank arm and wheel radius as well as the ratio of chainring-to-cog.

Originally Posted by rmur17

honestly, it is just high-school physics. Maybe pre-school physics these days!

Originally Posted by koger

Yes and that's the worst part of it, I already learned this stuff years ago. Now when I finally seems to get interrested in it, the knowledge is gone. I'm sure somebody told me that "you are going to need that someday", shame on me :-)

Originally Posted by dhk2

OK, maybe I see your hangup here. Agree that the airstream is losing energy at the same rate in both cases, but that doesn't dictate that the riders have identical power output. A moving airstream loses velocity (and energy) whenever it encounters obstructions, but nothing says the rider must match or restore that energy.

Originally Posted by koger

Yes and that's the worst part of it, I already learned this stuff years ago. Now when I finally seems to get interrested in it, the knowledge is gone. I'm sure somebody told me that "you are going to need that someday", shame on me :-)

Originally Posted by frenchyge

Or to put it another way, if the spot at the bottom of the wheel (or point of propulsion, let's say) is allowed to move with the air, then the wind *does* do work by moving the rider. Whereas, if the tire is fixed from sliding then the wind is simply another force that either aids or opposes the rider.

Originally Posted by Fday

I think I get it now. What is bothering me now is it would seem that wind would have a bigger effect on time-trial results than it seemingly has. I tried going to analytic cycling dot com to run some projected speed effects on an out and back course to see how bad it might really be and couldn't find a page that took into account wind effects.

It may be that we adjust to different conditions to minimize the effects of the wind on performance. For instance, perhaps we try to maximize aerodynamics when going into the wind even at the expense of a lower power while sitting up to maximize power with a tail-wind when aerodynamics are less important. The people who are winning races are probably better at this than those of us at the back, where the "real" effects of wind may be more obvious. Yet, the times we tend to see are only those of the winners so wind seems to be less of a factor than it should be if we couldn't adapt.

Originally Posted by rmur17

Originally Posted by Fday

I seem to be having a similar problem. Use it or lose it as they say. :-)

Originally Posted by dhk2

Glad the explanations worked for you. Now I'm curious to check out that ACC model for some of the detailed calculations on wind effects.

Originally Posted by Fday

I went there and got this data for a 4k flat course using his standard inputs, 250 watts.

speed in zero wind, 5 m/s head wind, and 5 m/s tail wind respectively is:

10.72m/s, 7.88m/s, 13.88m/s

so the headwind slows the rider 2.83 m/s and the tail wind speeds him up 3.16 m/s

On a 4km out and back TT this wind would supposedly slow the rider about 30 seconds compared to calm conditions.

Originally Posted by rmur17

remember the model is dynamic so the time to get up to ~steady-state at 250W power is included there. I would suspect if you're using distance/duratioin or avg. speed that longer legs would allow the true differential to show up somewhat higher than that (-2.9 m/s and + 3.4 m/s) ... but in any case you can play around with that.