I think the calculator must be wrong, and for the simple reason that the calculator suggests if you have a large amount of tail-wind, you need to apply a larger breaking force to go faster, which can’t be accurate? I think this is because the power is being calculated from ground speed, so as you go faster the apparent power of the wind seems higher, but actually the air speed is lower and so power is lower (and so less braking, not more, is required to go faster)
I can’t tell from the screenshot - but does the model say you need a larger braking force to go faster, or power? I’d be surprised if it showed the former. It does seem to show the latter. Latter seems possible given power increases with speed.
The question is about power, not force. The force is non-zero. The power is zero.
Um, close, but no. This is incorrect. For a bicycle that travels on the ground,
P_total = (F_total) * ground_speed.
Your mistake is thinking that F_aero doesn’t include air speed. It does. But that’s force, not power. In terms of units, force is measured in Newtons, so F_total is measured in Newtons. A watt is a Newton*meter/sec, so to get watts from Newtons, you multiply by velocity (i.e., change in position) in meters/sec.
and “A” is?
Frontal Area
For those who are really interested: Riding against the wind: a review of competition cycling aerodynamics
As others have mentioned, aero drag should only be a function of the velocity difference between the object and the medium it’s in (air).
So there should be no difference in drag between riding 40kph on a calm day, and riding 20kph into a 20kph headwind.
However, as many people will note from personal experience, the amount of power required to go 40kph on a calm day is larger than the amount of power required to go 20kph into a 20kph head wind.
The difference here is mostly due to rolling resistance itself (Crr) increasing with increased ground speed.
You can see that the force required to overcome rolling resistance roughly doubles between 20 and 40kph, though this will be a function of the specific tire.
This could easily workout to a >30w difference in power required to go 40kph on a calm day, vs to go 20kph into a 20kph headwind, despite the fact that there is no difference in drag between these two scenarios.
I’m not so sure about this. The paper does not explicitly discuss air speed vs ground speed, and in a scenario with still air, they are the same. So who’s right - this paper, or all the online calculators?
Look at this another way, based on pedaling torque and cadence.
We know the aero force at 40kph in still air is the same as the aero force at 20kph with a 20kph headwind.
So the torque on the pedals to overcome that force will be the same in both scenarios (for the same gear). But - the cadence at 40kph will be twice that at 20kph.
Power = torque * cadence, so power to overcome aero drag at 40kph = 2x that of the power to overcome aero drag at 20kph + 20kph headwind.
Here’s another thought experiment on this topic. Any maybe a bit of a mind bender 
Let’s say you have 2 bikes - a normal bike where the pedals drive the wheels, and a propeller bike, where the pedals drive a propeller. Let’s assume the propeller is very efficient, and can match the efficiency of a normal bicycle drivetrain.
How much power (think human exertion) does it take to hold each bike stationary at zero speed facing into a 20mph headwind? Let’s assume the rider’s sense of balance is really good and they can easily keep the bike upright in each case.
On the normal bike, the rider just does a trackstand. Not a lot of exertion here.
On the propeller bike - well - the rider is going to have to pedal a good amount to create forward thrust to keep the bike from being blown backwards by the 20mph headwind.
There’s a clear difference in the amount of power required by the rider in each case - even though both bikes are subject to the exact same aerodynamic drag force.
For the normal bike, I think the correct velocity to use when calculating power to overcome aero drag force is ground speed. In this example, zero speed, hence zero power.
For the propeller bike, the correct velocity for calculating power is airspeed. In this case, 20mph, and the rider is having to generate a good amount of power to remain stationary.
Well, we all ride normal bikes, so I think the right speed to use when calculating power is ground speed. And another reason I think the online calculators have it right.
Oh lordy. this thread has become unmanageable and the mods should lock it down. Lock it down, delete the entire thread, and either turn back time or else hand out amnesia drugs so that the collective memory of this thread can be wiped.
You’re not comparing equivalancies here. (DaveWH).
If a person could hold a perfect trackstand without using brakes or pedalling forward, they would be quickly accelerated backwards in a 40 kph wind.
And trying to equate the efficiency of pedal powering a propeller vs a bicycle drivetrain isn’t a useful comparison. Trying to go forward in a no wind condition by pedalling a propellor would also take far more work.
And I think you’re saying to overcome aero drag only requires a certain torque, but that if your going faster you have to pedal faster, so that you would have twice as much power? So are you saying that it is a specific torque that is required to overcome drag, and not a specific power?
Nah. If you look at my posts, I’m trying to say in several different ways that the online calculators are right in using ground speed for V in the P = F*V calculation for power to overcome aero drag.
And answering the OP question - that 40kph in still air does indeed require more power to overcome aero drag than 20kph ground speed + 20kph headwind.
The key thing I realised from that paper is that while the power to overcome drag on the rider and frame is proportional to the cube of air speed, because the wheels are rotating at ground speed not airspeed, the drag on a stationary wheel in a headwind is less than a wheel spinning at speed in a headwind. This explains why cycling into a 20km/h headwind at 20km/h requires less power than riding at 40km/h with no wind.
I’m not sure we’re applying the frames of reference consistently.
For example, I think if you had a treadmill under the bike, as you increased the incline (and so increased the gravitational force on the rider) you’d have to do more power to not roll off the back, even though the velocity is effectively 0.
I think the wind/drag force can be considered a bit like the gravitational force?
The velocity isn’t zero in this case, it’s equal to the treadmill speed. This illustrates one of the complexities in physics: choosing the correct frame of reference.
I know the difference between force and power, i didn’t say anything about turning the cranks. I understand that if the cranks aren’t moving than power through them is zero. Thats why i said apply an electric motor to the rear wheel. I feel like that sheds more light on OPs question. If you lock your cranks stationary then you’d be putting zero power through the cranks, I 100% agree with you, but if you want to move at 0.1kph you’d need ~270W which I think is what OP was really wanting to know.
Moving 0.1kph into a 40kmh headwind will most definitely not require 270W. Source: go out and try it for yourself.