You're right. I was working backward reading yours and cplager's posts, and clicked on the wrong quote. I'll go back and edit the reference.

I like your idea of drawing a free body diagram.

If we're ignoring rolling resistance (which is already unphysical), then there is no force that depends on speed relative to the ground (consider a hovercraft instead if you prefer).

According to your diagram, what is the difference of the forces acting on a bicycle between (A) a bicycle moving at 30 mph with no wind and (B) a bicycle moving at 10 mph heading into a 20 MPH wind? In both cases, there is a single resistive force of the aerodynamic drag of 30mph acting on a bicycle. In both cases, it takes the same exact effort of the rider.

In reality, rolling resistance is not negligible (and does grow with speed). But even with rolling resistance the OPs numbers do not make sense:

* Into 20 mph wind, riding at 10 mph
* With 20 mph wind, riding at 25 mph

Assuming same aerodynamic positioning, this either suggests that rolling resistance* is much bigger than any of us is considering, or that, well, it wasn't a 20 mph wind.

Cheers,
Charles

* I'm throwing in drive train losses with rolling resistance here.

Let's concentrate on this, and hopefully all the other tings fall into place.

Where you are mixing everything up is that the force in Prathmann's equation is not the force applied by the rider on the pedals. It is the force applied by the back wheels on the ground, that is pushing the bike forward.

And in your example those forces between back wheel and ground are different for each example. So for example when going at 30 mph the force is 2 times smaller than at 15 mph. Multiplying that with the speed gets you to the same power.

Hi,

There is no difference at all in the force diagram. A point repeatedly
made. So why are you repeating the already stated as a question ?

rgds, sreten.

Yes, the motive force required is the same in both the still air and 20mph wind examples. But the distance is different. Work = force x distance, so the issue is what distance you use. If you can't decide whether it's air distance or ground disance, ignor both and consider the force and distance that the rider's legs move.

A bicycle is a machine in the classic sense, so the rider can set his pedaling force to what ever he wants. However the work, force X distance will be the same as the bike's. (not adjusting for frictional losses) So for the same driving force, regardless of gearing the rider's legs will produce the same force through 1/3rd the distance, or he can gear down and produce 1/3rd the force for the same distance. Either way he's producing less work.

No, you're wrong, and everyone else agrees on it. Really, why don't we just drop it, you keep believing what you say, and we what we say, and we'll agree to disagree. If we keep going, it'll get to the point where you realize you're wrong, but you're too far into the "debate" to agree you're wrong, so you'll keep trying to rig the Physics. So let's not.

But as for why you're wrong (if I didn't explain, you'd be sure to ask, and the argument wouldn't be dropped), it's because of what work is, which you've seemed to overlook.

The situation CANNOT be considered a change of reference frame, as you claim, and the bike cannot be replaced by an airplane or a hovercraft, unless you complicate the situation further, which is simply counterintuitive. So, lets stick with the bike.

W=Fd. I think we can agree on this. But that d is the distance the bike travels with reference to the ground, NOT the air/wind around it, which you seem to have claimed (correct me if I'm wrong, your explanations are a bit hard for me to follow). I mean think about it. If you're not moving at all with respect to the ground, just standing there, in a 20mph headwind, surely that's not the same as moving at 20mph at no wind when it comes to "effort," work, which is what matters. The work in one situation in the first situation is 0, in the second much more. And if you even remotely try to hint that the two are the same, I'm done with the conversation. No point to keep going.
And so, if moving at 0mph in 20mph headwind is not the same as moving at 20mph at no wind (with respect to the ground), then moving at 10mph at 20mph headwind is not the same as moving at 30mph at 0mph wind. And since the first is true, so is the second.

Now, case closed? Please?

Seconded. Any counter-argument at this point is basically arguing against the laws of physics...

Hi,

Your saying my reported real world numbers don't make sense (to you).
Well you clearly don't know physics or science, where your job is to make
sense of empirical evidence, not complain the evidence must be wrong.

Take a single gear bike at 10 mph and no wind, and 30mph with a 20mph
tailwind. For the latter pedalling at double rpm does nothing. You have
to pedal at triple the rpm of 10 mph against the same pedal resistance
(If you can at 3 times the 10mph cadence) to do 30mph with the
20mph wind, outputting 3 times as much power as 10mph no wind.

I'll let you work out why my real world numbers actually do make
sense, and why your attitude to analysing a problem basically
sucks, ignoring the correct analyses already given in the thread.

rgds, sreten.

I did posit why can I do ~ 10mph into a ~ 20mph wind and only
~ 25 mph with the same ~ 20mph wind as a tailwind, what was
confusing me ? That was answered properly quite promptly.

My thinking on this had been the same as cplager's, even though I realize that empirical experience doesn't fit with the only-air-resistance-matters theory (e.g., I can't ride 40mph but I can ride - slowly - into a 40mph wind).

This bit of clarification - that the rider is propelling himself forward from the ground - is the key to differentiating a cyclist from airplane. Thanks for helping me understand it. My intuitive-physics mind still wants to hold the airplane analogy, but now I have the rational explanation to know why it doesn't actually work that way.

Hi,

Anything on any legs would be in a lot of trouble if that was true.

Spare a thought for small insects, air drag to weight ratio is phenomenal.
The reason they can cling onto walls and ceilings is simply they need to
be able to cling onto the ground at relative G forces we cannot imagine.

The physics of the small compared to large is fascinating.

rgds, sreten.

Here's a bit of a different scenario:
Our city is mostly East/West and given the choice I usually opt to ride into a strong wind and have it on my back for the return trip. I don't use a computer, but rather a stopwatch for given distances. I'm talking about a 22 km route, mostly on MUP.
Surprisingly my times have been slightly better (3 - 5 min) with the wind on my back going out and fighting it on the return trip. My rationale is that I am stronger starting out and therefore take more advantage of the tail wind, than being tired from bucking the wind. Not being as tired for the return trip may give me more energy while going into the wind.

Hi,

If you do it all in one go that makes a lot of sense.

You warm up with the wind and hammer the return against.

rgds, sreten.

You will never get any sort of sensible returns in time
hammering downhill and/or with the wind, compared to
hammering uphill and/or into the wind, aerodynamics.

Loving the positive feedback. :) I wish more people would be open-minded enough to consider all explanations and follow the evidence/logic, and then admit if/when they're wrong. Glad to see that.

Newton's Third Law FTW!