Windy. All the time. Is it the same for you ?

paul2718

ChrisAOnABike wrote:

paul2718 wrote:

ChrisAOnABike wrote:

Although, if you insist on being serious, do please explain the difference in experienced airspeed between cycling at 20 mph in nil wind, and cycling at 10 mph into a 10 mph headwind.

If riding at 20mph in nil wind requires 200W, then riding at 10mph into a 10mph headwind only requires 100W. OTOH riding at 20mph into a 10mph headwind requires 450W.

Do please explain why this is the case.

I take it we're agreed on an approximation that the only force to be overcome is air resistance, right? Rolling resistance is well known to be only a small proportion of the total at all non-negligible speeds.

Maybe 20 mph is too slow for the air resistance to completely dominate - but I'd still like to see a justification for the assertion.

It's fairly basic Newtonian stuff. Hope I've got it right.

Power is drag times velocity. Drag is proportional to square of velocity. The first velocity is bike relative to the ground, the second bike relative to the air. No power is required to stand still in a gale, although a lot of force might be involved.

So if 20mph in the still requires 200W (which is somewhere in the zone, but quite dependent on the individual) then 10mph into a 10mph headwind gives you the same drag force, but at half the ground velocity, so half the power, 100W. 20mph into a 10mph headwind requires a 30mph force, at 20mph ground speed. Force is proportional to relative speed squared, so (3/2)^2 = 9/4. Ground speed is the same so, 9*200/4 = 450Watts. 10mph in the still is therefore only 50W, FWIW.

Aero drag is a shocker. The difference between hoods, drops, elbows bent on the hoods, is quite noticeable. Sitting on a wheel can drop the power from 250 to 150.

Paul

chrisaonabike

paul2718 wrote:

It's fairly basic Newtonian stuff. Hope I've got it right.

'Fraid not!

Power is drag times velocity. Drag is proportional to square of velocity.

^^ This bit is right.

However, more specifically, power is drag times the velocity through the thing that's causing the drag. In this case, it's the velocity through the air, which is 20 mph in both cases.

Leaving out the rolling resistance, which is almost the same in both cases, and small compared with the resistance of the air, the power applied via the force on the pedals has to overcome the same air resistance in both cases, and nothing else.

No power is required to stand still in a gale, although a lot of force might be involved.

^^ This is also true. But if you're standing still in a gale, the force that counteracts the force from the gale is the friction between the sole of your shoe and the ground. If you're standing on ice, then you would slide downwind unless you provided a force equal and opposite to the force of the wind, and power would be required to provide this force.

Similarly, suppose you can swim at 1 mph. If you swim upstream at 1 mph, relative to water in a river flowing at 1 mph, your speed over the ground is zero, but you're still developing power to overcome the resistance of the water.

A final example - an aircraft flying at an airspeed of 300 knots requires the same power regardless of the direction and speed of the wind. The wind makes a difference to the ground speed, of course, but not to the power required for that given airspeed.

The mistake is to treat the speed of the bike over the ground as relevant to the power required - it isn't, except insofar as we're approximating by ignoring the rolling resistance.

Cycling into headwinds is hard because we have a mental picture of a speed we want to go at - over the ground. And as you say, Power is Drag x Velocity, and Drag is proportional to the square of the velocity. So power required to overcome air resistance is proportional to the cube of the speed through the air. Double your speed (through the air) and you need eight (yes, 8!!) times the power!

One other thing that sometimes confuses people: it certainly is true that more work is required initially to accelerate the bike to a higher speed over the ground. Kinetic energy is proportional to the mass and to the square of the speed, so to accelerate a given mass from rest to 20 mph requires 4x as much energy as is needed to accelerate it to 10 mph. If that work is done in the same time, then the power required during acceleration to 20 mph is 4x the power required to accelerate to 10 mph.

However, once you reach a steady speed, the only force required is for overcoming drag. If the drag is the same due to moving at the same speed through the air, the force (and hence the power) is the same.

paul2718

Nope. You're wrong. Well, not about everything of course.

Replace the drag force with that necessary to ascend a gradient. Force is force. So the 10mph into a 10mph wind and the 20mph into no wind are equivalent to the same gradient, same force. The power is then related to the rate of ascent, 10mph and 20mph, so double the power to go double the speed.

Work is Force x Displacement, Power Force x Velocity. The force is between the rear wheel and the ground, the velocity of interest therefore is between the bike and the ground.

I think the big thing about the wind is how steady it is. Gusty wind is always hard work, steady is just slower.

Paul

bristolpete

Anyway, 51 solid miles logged today, less windy but pouring down. Nice to be out though.

chrisaonabike

paul2718 wrote:

Nope. You're wrong. Well, not about everything of course.

Replace the drag force with that necessary to ascend a gradient. Force is force. So the 10mph into a 10mph wind and the 20mph into no wind are equivalent to the same gradient, same force. The power is then related to the rate of ascent, 10mph and 20mph, so double the power to go double the speed.

Well I'm not going to argue. But ascending a gradient is different and it just confuses the issue to make a comparison with it.

On a gradient, you're doing work against gravity as well as against air resistance, and since you're going slower, the work done against gravity will tend to dominate, the more so the steeper the gradient.

So on a hill, the work done is mostly related to the gain of potential energy, and certainly if you go up a hill twice as fast, you're gaining potential energy twice as fast, and therefore you have to develop twice as much power to achieve that. At very slow speeds, there is almost no air resistance, so the power required to gain potential energy is proportional to the speed. As you go faster up the hill, air resistance increases and you get a component of additional work due to that.

But on the flat, as I say, the work you do against the pedals is pretty much all used in overcoming air resistance, which is due to the resistance, erm, of the air.

Incidentally:

The force is between the rear wheel and the ground, the velocity of interest therefore is between the bike and the ground.

This does not follow. There is a friction force between the rear wheel and the ground, certainly. But the force that it's equal and opposite to (since there's no acceleration) is the force due to the air resistance, which is the same in the nil wind 20mph case and in the 10 mph into a 10 mph headwind case.

Anyway, feel free to believe you're right and that it's ground speed that dictates the power required. Best avoid the aviation industry if you're convinced about that.

paul2718

OK. Going up a hill in a vacuum. I guess you might be the kind of chap who believes aeroplanes have trouble taking off from moving conveyor belts?

Anyway wind seems to have dropped, but it's raining. One day soon it will be about right for an hour.

Paul

FWIW,

This does not follow. There is a friction force between the rear wheel and the ground, certainly. But the force that it's equal and opposite to (since there's no acceleration) is the force due to the air resistance, which is the same in the nil wind 20mph case and in the 10 mph into a 10 mph headwind case.

How much work is being done in the two cases? You can see it if you consider the gearing. Ride at 50rpm/10mph into a 10mph headwind. We agree the force at the wheel is rear equivalent to the drag, ignoring the other boring stuff. So that gives you a torque at the crank to produce that force at the wheel. Now find a still day and ride at 20mph in the same gear. Same force, therefore same torque, but double the rpm, so double power.

ForumNewbie

paul2718 wrote:

OK. Going up a hill in a vacuum. I guess you might be the kind of chap who believes aeroplanes have trouble taking off from moving conveyor belts?

Anyway wind seems to have dropped, but it's raining. One day soon it will be about right for an hour.

Paul

FWIW,

This does not follow. There is a friction force between the rear wheel and the ground, certainly. But the force that it's equal and opposite to (since there's no acceleration) is the force due to the air resistance, which is the same in the nil wind 20mph case and in the 10 mph into a 10 mph headwind case.

How much work is being done in the two cases? You can see it if you consider the gearing. Ride at 50rpm/10mph into a 10mph headwind. We agree the force at the wheel is rear equivalent to the drag, ignoring the other boring stuff. So that gives you a torque at the crank to produce that force at the wheel. Now find a still day and ride at 20mph in the same gear. Same force, therefore same torque, but double the rpm, so double power.

I've been following this discussion, but a bit technical for me so not sure who is right. However looking at the gearing and the 50rpm/10mph into a 10mph headwind example, I agree that on the flat road with no wind you would be need to doing 100 rpm to ride at 20 mph in the same gear. However I would have thought a power meter would show roughly the same power used for both scenarios? If that is not the case, what would your cadence and speed drop to from 100rpm and 20mph, if you moved from no wind into a 10mph headwind, but continued in the same gear putting in the same amount of power?