Ehh... "Still wind?"
No, but yes, I know what you mean. In all cases, the power is proportional to the difference between your speed and the windspeed squared, times the wind speed. If there is no wind, this simplifies to just your speed cubed.

Hi,

Rolling resistance increases linearly with the road speed moreorless.

The rougher the road surface the higher it is. It has little to do with
variations in tarmac elasticity, the tyre is far more elastic. It has
everything to do with how much the surface is deforming the elastic
tyre. Everyone must have experienced this, the effect of hitting a
section of smooth tarmac after riding some rougher stuff, or less
immediately obvious, being slowed down by a rougher surface.

rgds, sreten.

Prathmann nailed it. It's easy to forget that it's times ground speed for power and not windspeed (or incorrectly V cubed), which completely answers the OP question.

The other factor is power needed to overcome rolling resistance which is usually fudged to be some coefficient times ground velocity.

Yup, case closed.

Very interesting. Reminds me of something one of my professor used to say: common sense falls short.

Now, what if the bicycle is an airplane?

This is a complicated question. You're assuming that the power should be equal in the two cases and that's not true (as it isn't just the power provided by the rider that are setting the speeds).

Instead of looking at the power, go back to force:

An object will move at a constant velocity if there is 0 net force acting upon it (Newton's first law).

I agree that we usually think in terms of power delivered, but one can also think about the average force (or, if you prefer, average torque) with which we can push on our bikes. And for comparing two situations, you can see what happens when these two forces/torques are equal.

In other words, the OP's original formulation was correct: If there is no rolling resistance (which there is) riding 10 mph into a 20 mph headwind should be equivalent to riding 30 mph into no headwind. Alternatively, with no rolling resistance, he should have been absolutely flying with the wind at his back.

First, rolling resistance isn't negligible. Second, I completely agree with you questioning the level of the wind, not only for the reasons you suggested, but also because in order to count it as a 20 mph wind, it would have to be (1) blowing at 20 mph and (2) blowing either parallel or anti-parallel to his velocity. (Not to mention I'm somehow unswayed by the OPs assertion that the wind was "that strong" as he provides no information as to how he knows that).

Cheers, Charles

In terms of the force required, yes. In terms of the power required, no, for the reasons discussed above. Power is what actually matters.

Your effective cross-sectional area is different for headwinds and tailwinds. Unfortunately, it's less in a tailwind than in a headwind.

Uhhhhh, and how is that relevant for this discussion?

You can't use force a bicycle is a geared machine. The rider can trade force for distance simply by using a lower gear. the fact that he's moving slower means that the total pedaling distance in his example will be one third of what it would be at triple the speed and distance. I made an error in my earlier calculation by forgetting to credit the distance traveled and came up with 17.3mph and Prathmann was spot on at roughly 20.8.

Hi,

At ~ 10mph + ~ 20mph headwind I can put the torque required for ~ 30mph no
wind through the back wheel, trouble is I haven't a hope in hell of producing that
torque with the gearing needed to do 30mph, i.e. I can't triple my power output.

The best you can assume is constant power output and that gives ~ 21 mph no wind.
I can't go quite that fast, I assume due to increased rolling resistance, which will be
~ double that doing ~ 10mph, and the actual amount depends on a lot of variables.

I'm also sure I'm trying harder into the headwind, I wouldn't call it cruising, I'd call
it comfortable pushing, same effort no wind my best guess is I'd do ~ 18/19 mph.

rgds, sreten.

The mistake that you are making is that the effort we make is in terms of power not force. So when lifting a 50 lbs weight at a constant speed you need to apply the same force regardless of the speed. But it is much harder to do if the speed is higher.
Same thing for on the bike, the force on the pedals might be the same but if you are spinning faster it is harder for you.

You can get to 30 mph without wind at any number of torque values. It is not torque that gets you to a certain speed, it is power.

Hi,

Your missing the point. You need the same torque through the back
wheel to do 30mph no wind and 10mph into a 20mph wind, aero only.

As I pointed out, the power is very different, but me on my bike and
an effective aero drag of 30mph and my back wheel, only one value
of torque through the wheel relates to 30mph aero only effective drag.

(Torque / wheelsize) * groundspeed = power
Torque / wheelsize = force = aero drag (if only aero is considered)

30mph no wind needs 3 times the power of 10mph into 20 mph
and for both cases the torque through the back wheel is identical.

rgds, sreten.

Ah I see, I was thinking in terms of torque on the pedals.

Yes, that is also a good way of putting it: in both cases you would need to push the bike forward with the same force but going at 3 times the speed would require 3 times the power.

Hi,

It was "that strong" if not worse in sections. Forecast was ~ 25mph and of course
not exactly head on. But cycling on the seafront at the bottom of big cliffs there
is no place to hide with any wind at any reasonably oblique angle to the cliffs.

Simply you don't know if my assertion was reasonable, but there is nothing
reasonable in casting aspersions about something you know nothing about.

If you can't accept assertions at face value, that means think you are cleverer
than other people, I can assure you are not, and continually implying you
know something better about simple statements that are intended to be
taken at face value without pedantic explanations of the context, arrogant.

rgds, sreten.

~ 10 mph into ~ 20mph was ballpark, accurate enough for the question.
Can I do 10mph into a 20mph wind? Yes, certainly. Can I do 30mph
into no wind ? No, nowhere remotely close, can I do 20mph ? Yes.

It's completely irrelevant to the discussion whether the wind was 20mph or not. That's not what he was asking. That was only a hypothetical situation provided to help with the general question of why riding at v speed is not equal to riding at v-V speed at a head wind of V. To this general question he also receiving a very satisfactory answer.

+1, the details don't matter. The general question and the answers are more informative than whether the wind was 20mph of 19mph or head on or at an angle.

Imagine this scenario:

You have a bike and rider, an electric motor on the front wheel and a "power" meter on the crank (which in this case measures average torque assuming it knows the cranklength).

1) For the first run, the rider is going at his "I can keep this up for an hour" effort, pedaling at 90 RPM and goes 15 mph with no power assist.

2) For the second run, the rider is again going at the same intensity and cadence (and as the "power" meter actually measures the force and cadence, it will give report the same value), but this time the front motor is turned on and the rider is cruising at 30 mph. (The bike is in a higher gear).

3) For the third run, the rider is again going at the same intensity and cadence (confirmed by the crankset power meter), but the brakes are rubbing and he is only moving at 7.5 mph.

In all three cases, the rider is putting the same torque into the system and as far as the rider is concerned, they take exactly the same amount of effort. But when you apply the calculation as prathmann showed, you'd end up with the fastest leg having the rider produce twice as much power as the first leg, and the slowest leg having the rider produce half as much power (since he was applying the same force to the bike each time).

Now replace the second scenario with a very (even ridiculously) strong tail wind (or going down hill) and the third scenario with a strong headwind (or going up hill).

I know we talk about power meters and power / weight ratio, but we need to be careful applying the formulas. In all three scenarios, if you do a careful conservation of energy calculation, you'll see that energy is of course conserved, but that the energy involved in the three scenarios are different from one another.

In other words: Consistent effort != same power calculation.

As I said before, one does want to go back to Newton's first law. Another way of looking at this:

If there is no rolling resistance and aerodynamic drag is the only thing that matters, then you can consider this as a change of reference frame. Riding 10 mph faster than the air would be the same, regardless of speed versus the ground, so being able to ride 10 mph with no wind would be identical to being able to ride 30 mph with a 20 mph tail wind. In the case of bicycles this of course isn't true in real life because there is rolling resistance and it is not constant with respect to speed. This is exactly the case with airplanes when calculating drag and fuel consumption.

So the OPs original question (where ignoring rolling resistance) was right:

(Incorrectly) ignoring rolling resistance, if he can do 10 mph into a 20 mph headwind, he ought to be able to ride at 30 mph on the flats. (A) Rolling resistance is important in these calculations and (B) I don't believe his estimates of the wind speeds because the math doesn't bear it out.

Cheers,
Charles

I don't follow your reasoning here. In all 3 cases the rider is producing the same power, that is how you have defined it, same torque and same cadence. The motor is providing the extra power that is required to go at 30 mph and the brakes are providing the extra drag to go at 7.5 at that same power output of the rider. That's all there is to it. How can you conclude that the rider is producing a different power output in each case?

The rider is producing the same (average) force in all three cases. But the calculation used (power = force . velocity) has the power generated is different. When talking about bicycling, we often use power when we should be really talking about force.

The other point I made is important: If aerodynamic drag is the only resistive force (like with airplanes), then only the relative velocity of the rider to the air is important. Any calculation that shows this isn't the case is necessarily flawed.

Cheers,
Charles

You defined your example such that the rider is putting down the same force and same cadence, therefore the same power. The example is not disproving the formulas previously stated in the thread.

You are confused in saying that it is force that matters and not power, effort is measured in power, not force. Consider a weight lifter in the gym that is bench pressing 100lbs, raising the weight at a constant speed. Independent of the speed the force is the same. So are you saying it does not take more effort to raise the weight at a faster speed?

I actually brought up the airplane example a few posts back, riding a bicycle taking only wind resistance into account is not the same as an airplane. The reason is that standing still on the bike with 20 mph wind blowing at you takes almost no power (compare that to a weight lifter that is just holding a heavy weight steady). An airplane with a 20 mph head wind hovering over the ground would still need to put out the power required to go at 20 mph.

I hate trying to convinced the already convinced, but I'll give it a shot. The reason that ground speed must be figured into the solution, is that while the rider is riding against the wind drag, he's doing all his work levering the bike forward against a fulcrum on the ground. This is very different from an airplane that thrusts against the wind alone.

If you take a minute to draw free body diagrams of both an airplane (in flight) and a bicycle you'll see the difference, and understand why ground force and distance (work) come into play.

Hi,

But it's not the same power as defined by the formula. That's the point. (I agree in terms of biomechanics, it is the same power).

As for your weight lifting example, it takes more energy to get the bar moving to a faster speed. Once you have hit a static situation (the bar is moving at a constant speed), it takes the same force to keep the bar moving at that speed (the mass of the bar times the acceleration of gravity). Again, this is why I think trying to do these calculations with power is a mistake.

(And don't confuse biomechanical energy with physics energy: in physics, it takes no energy to hold still a heavy weight off the ground because there is no work being done on that weight. It takes a lot of biomechanical energy, however).

The bicycle and airplane aren't so different (except that, of course, if an airplane isn't moving relative to the air, it falls).

When calculating fuel usage of an airplane, it is (basically) sufficient to know about the relative velocity of the airplane to the surrounding wind. The speed of the airplane to the ground is irrelevant to instantaneous fuel usage calculations.

The same thing is true for a bicycle with no rolling resistance. In this case, with no power applied, the bicycle will go the same speed as the wind (you can think of a trike if you don't want to worry about it blowing over, etc). If you go 10 mph faster than the wind, it doesn't matter how much faster you are going to the ground, it is only the velocity relative to the wind that imparts any force onto the bicycle.

Again, rolling resistance is not negligible. But if you don't like thinking about a rolling resistance free bicycle, think about a hovercraft on a frozen lake.

Cheers,
Charles

I agree, maybe you wrongly quoted me?