well it is an interesting thing, in an open system it is likely not of consequence because the wavelength (the scrunching up) radiates away from the object causing the scrunching, however it is why the lanes in a swimming pool closest to the sides are slower because the waves returning cause further resistance, it is also why a shallow swimming pool is slower than a deep swimming pool. Both situations are a closed system, the waves can not exit the boundary and are reflected back. In an open system (ie not building or rigid structures to reflect the waves back)I don’t think it matters much if at all.
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I don’t know about the specific numbers you are quoting, but the concept is as follows:
Lowering CdA (all else equal) by 10% will lower the aero force by 10%. Which for a given power output allows you to go faster. But - as speed increases, the power required to overcome aero drag increases with the cube of speed. Hence, you won’t go 10% faster.
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I do think there is something wrong with that calculator. Testing limits near zero kph it gives wrong answers. To control you backwards roll when facing a headwind does not require negative power, and i’m sure it would require more than 10w to move at all into a 40kph headwind.
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Making a few assumptions about air density and CdA, the force being applied by a 40kph headwind is ~24N. to overcome this force and have zero ground speed would also require a force of ~24N. If you applied an electric motor to the rear wheel it would require ~269W to keep the bike from rolling backwards or rolling forwards and balance with the force of the headwind.
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Interesting replies, thank you all!
I’m still not sure about the Force x distance or Force x speed part, if that distance should be relative to air or ground. I guess it might be complex because the general assumption is that the wind speed is relatively low, so most of the apparent wind is from the speed of the bike?
Rough diagram with the boat responding to the Watts to stand still: if you didn’t have the friction of the tyres resisting the force of the wind, you would need to do power to have a 0 velocity relative to the ground. Consider the boat on a river, it might do 10kph up stream in a 10kph flow and have a speed of 0 relative to the ground, but still be doing the same watts as doing 10kph up stream in a 0kph flow and have a 10kph speed relative to the ground.
The bike is more complex! Some things are definitely vs. ground, others vs. air, or not? Still confused!
If you stuck the bike on a stand, does that stand require 269W of power to keep the bike from moving?
No, of course not.
If you stick a bike in a wind tunnel at 40kph, does it require zero force to stay stationary?
To answer OPs original question. The forces acting on a cyclist can be summarized by:
F_total = F_mechanical + F_aero, mechanical forces being pretty much drivetrain and tire related.
The power required can be summarized by:
P_total = F_mechanical * mechanical speed + F_aero*wind speed
The aero power required is the same if you’re going 80kph with a 40kph tailwind or 20kph with a 20kph headwind. Nowhere in the aero calculation is ground speed considered (1/2 air density * CdA * air speed ^3)
The power required to overcome drivetrain/tire losses is mostly linear with ground speed and is not impacted by air speed.
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I don’t think this is right.
If it was, riding at 1mph into a 19mph headwind would require the same power to overcome aero drag as riding at 20mph in still air. Go try that and you’ll see this is clearly not the case.
In P = F * V, the aero force is proportional to airspeed squared. But the V is the speed at which that force is moving which is the ground speed of the rider.
This is an interesting discussion, but the aero drag is the same regardless of the ground speed. A stationary (ground speed) object in 40 kph wind and and a 40 kph ground speed in still air is exactly the same aero drag.
It is the “other” drag that changes (drivetrain efficiency, rolling resistance, etc).
An airplane that needs 150 kts of airspeed to generate the lift required to fly, can get that in any combination of ground speed and “wind”. There is no difference in lift, drag, or engine power required (to achieve a steady state flight once airborne).
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This.
It’s the same problem as an airplane on a treadmill. You can put an airplane with a takeoff speed of 100mph on a treadmill doing 100mph opposite the direction of takeoff, and when the plane reaches an airspeed of 100mph, it will take off. The apparent speed of the treadmill (ground) to the plane’s wheels is going to be 200mph, but who cares, ground speed plays no role in takeoff speed.
If an airplane is flying into a 100 mph headwind, with an airspeed of 100mph but a ground speed of zero, does it not require power to do so?
-Tim
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You’re down the exact same mistaken rabbit hole I went down. Force and torque are similar concepts, except torque is used to describe the amount of force required to rotate an object, i.e. your cranks, where force is used for linear motion. Power on the other had is the rate at which you apply force. In cycling terms, Power is the rate at which you apply torque, i.e. torque * cadence. So if wheels aren’t turning, neither is the crank, which means power is 0W. You are expending energy as you fight the wind to stay stationary, but since you are not rotating the cranks, your power remains at 0W. Hopefully my Google-fu is strong and I don’t need to be corrected (much) again 
ETA: I was also thinking force but power was being referred to, whether on purpose or otherwise.
Air resistance is calculated from net velocity of the rider with respect to air, and not relative to ground speed. The formula takes the squared sum of velocity (ground speed) + wind speed relative to the cyclist. Air resistance is only one of the variables, and as the speed goes down, it contributes proportionally less to the overall calculation. Other inputs are rolling resistance, wheel rotation, bearing losses, etc, which are directly unaffected by wind but proportional to ground speed. As ground speed decreases, air resistance contribution from the wind becomes a smaller factor when compared to the remainder of the formula.
Fully agree that airspeed is the right speed for calculating F for aero drag.
In the P = F * V formula, what is the right velocity (V) to use for a bicycle? Airspeed or ground speed?
Think about this: I’m going to go outside on a day with a 20mph wind. I’m going to get on my bike, point into the wind, and trackstand (using force on the pedals, not brakes) for 1 hr and not move.
Is that the same workout as riding for one hour into a 20mph headwind? No, and the difference isn’t just the added rolling resistance from riding at 20mph.
Go to any cycling power calculator online, and repeat what the OP did on Bike Calculator. They all show the same result (in concept, not necessarily the same exact numbers).
Here’s an example from https://wattscalculator.com/
Scenario A.
Ground speed: 10mph
Headwind: 10mph
Power = 85 watts
Scenario B
Ground speed: 20mph
Headwind: 0
Power = 170 watts
Scenario C
Ground speed: 20mph
Tailwind: 10mph
Power = 60 watts
Scenario D
Ground speed: 10mph
Tailwind: 0mph
Power = 30 watts
The power needed in scenario A (85 watts) is less than B (170 watts).
Yes, in A, the riding speed is only 10mph, so there is lower power needed to overcome rolling resistance, so let’s adjust for that.
The power required to overcome rolling resistance to go from 10mph to 20mph is 30 watts (scenario C minus D). So add 30 watts to the 85 from scenario A and get 115 watts. Still less than the 170 watts in scenario B.
I’m way too tired to think about remainder of your post, and way out of my element, though this thread has captured my attention quite well. If you’re relying strictly on the calculator to estimate your values though, you need to stop and rethink your approach.
For example, there is no universe where I am required to do 754W into a 23kmh headwind to move at 42.6kmh:
I picked that exact speed/wind because yesterday I encountered those exact conditions. Here’s a 45s pull I took into basically a straight headwind that required roughly half of the power:
To make matters worse for the calculator, the speed when I started the pull was 37.5kmh, peak was 34.3 and my pull ended at 42.6kmh. That means power was needed to accelerate ~5kmh. That’s yet another ignored factor in equations, with the assumptions that speed is to remain constant.
Back to the calculator - using the slider to get to 376W, I end with 30.9kmh.
Perhaps bikecalculator is more accurate:
No, it most definitely is not, it has me at 27kmh given the same conditions.
Side note - bikecalculator was within 1-2% when estimating climbing time for 6-10% steady grades and no/low wind. I used it to plan a handful of climbs, and it was accurate as long as there was no flat section/descent; those threw a wrench into everything.
Yeah, I can’t vouch for how accurate these calculators are. I just assume they’re correct based on the input assumptions they use.
Curious, how do you measure airspeed when you’re riding?
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I don’t - I’ve no tool available to me to do so. I estimate based on wind speed/direction from the closest weather station(s). It’s a theoretical exercise that holds no value to me in daily riding.
Last night we had wind speeds reported anywhere from 20-35kmh. Based on feel, it was pretty close to a straight head wind, though not 100%, so a sail effect did happen to a smaller extent. Still, there is no accounting for the 2x power requirement in the calculators.
Even going on gut feel alone - I know what a 20-25kmh wind feels like. And I know I can ride 40-45kmh into it, albeit not for a very long time. Yet I do NOT know what 750W for 45-60s feels like as I’ve never been able to hold that power.
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Sure short efforts but the former I can maintain for multiple hours on end. If the requirement was the same you’d be able to hold 41km/h for hours on end, not minutes.
As soon as aero drag comes in, it will never be very accurate unless your CDA is exactly the one the calculator uses. They don’t know if your 1.65 with a narrow build or 1.90 with broad shoulders… If you ride with 44cm bars or 36cm etc… Wind outside is also not really constant so the drag fluctuates and therefore it gets even more complicated…
Bike calculator seems to be pretty accurate ok climbs where aero drag isn’t much of a concern…
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Still trying to get my head around this, found a similar thread if people want to go deep haha
https://forum.cyclinguk.org/viewtopic.php?t=153856
I think one common mistake people are making is the edge case of 0 velocity, which does take no power even in a headwind. This doesn’t prove that the velocity is 0 in the power equation, only that the net forces on the cyclist are zero. The drag force is still present, but there is a force applied through the tyres and brakes in the opposite direction to the wind pushing. This is the same as any object stationary on a slope: gravity pulls it down the slope, and a “normal force” of friction resists the acceleration and applies a force against gravity (or wind).
When you release the brakes and don’t apply torque on the pedals, the bike would be pushed backwards. Now the cyclist is applying a force (like a brake) in connecting the bike to the world through tyres, and again net forces are zero so no power is used (even though a force is being applied by the cyclist).
I’m still not sure what the ground-on-bike force does about the power needed to overcome a headwind. Is it velocity-vs-ground that’s used (what the calculators do) because the remaining force is provided as a normal force, or is it velocity-vs-air that’s used (and the calculators are wrong) because once the bike gets rolling it’s effectively frictionless?