We three agree.roubaixtuesday wrote: ↑12 Nov 2022, 6:50pmWe agree.axel_knutt wrote: ↑12 Nov 2022, 6:46pmForce is proportional to square of relative air velocity.roubaixtuesday wrote: ↑12 Nov 2022, 6:13pm
The total energy input is force x distance (this is a completely general equation).
The power is energy/ time.
So power = force x (distance/ time)
(Distance / time) is speed.
The force is proportional to the square of relative wind speed, so the power must be too
Energy is force on the bike multiplied by distance the bike travels along the road surface that the wheels are pushing against, (if the bike's not moving, there's no power even when the wind's blowing a hoolie).
Power is energy/time = force x bike distance/time = force x bike velocity
Power vs Wind.png
How does a headwind affect speed and power output?
- Chris Jeggo
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Re: How does a headwind affect speed and power output?
Re: How does a headwind affect speed and power output?
From a real world scenario, we rolled out about 50 miles last weekend into a nearly continuous stormy headwind, with the odd bit of rain. It was awful and one if the guys we regularly ride with was really struggling. On the return leg, I got what I imagine electric bike owners get: that effortless pedalling sensation with a direct in the tail tailwind.
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Re: How does a headwind affect speed and power output?
It would be interesting to see the energy/work/power relationship given a tailwind.Dingdong wrote: ↑13 Nov 2022, 7:36amFrom a real world scenario, we rolled out about 50 miles last weekend into a nearly continuous stormy headwind, with the odd bit of rain. It was awful and one if the guys we regularly ride with was really struggling. On the return leg, I got what I imagine electric bike owners get: that effortless pedalling sensation with a direct in the tail tailwind.
Re the headwind case, coming from an aeronautical background I'm struggling to see the difference between maintaining a constant speed into a headwind that has got x knots faster, and increasing speed by x knots into a constant headwind.
Must be something to do with the road.
Re: How does a headwind affect speed and power output?
I finally got my mind round the difference between the two formulations. And why working on distance produces it.
And this is the crucial point, and it needs to be discussed as early as possible. By taking the general case it's easier to see that the distance moved though the air is the relevant distance. What other distance could it be?
Jonathan
Agreed. And for any arbitrary object experiencing aerodynamic drag the force is the aerodynamic drag and the distance is the distance that the object moves through the air.roubaixtuesday wrote: ↑12 Nov 2022, 5:51pmThey are not cycling related! Here they are again, with all cyclists removed, and simplified.
Please indicate which you agree with.
The total energy input is force x distance (this is a completely general equation).
And this is the crucial point, and it needs to be discussed as early as possible. By taking the general case it's easier to see that the distance moved though the air is the relevant distance. What other distance could it be?
Jonathan
Re: How does a headwind affect speed and power output?
That smiley is doing a lot of work!DaveReading wrote: ↑13 Nov 2022, 8:11am ...
Re the headwind case, coming from an aeronautical background I'm struggling to see the difference between maintaining a constant speed into a headwind that has got x knots faster, and increasing speed by x knots into a constant headwind.
Must be something to do with the road.
: - )
Jonathan
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Re: How does a headwind affect speed and power output?
It's the distance along the ground.Jdsk wrote: ↑13 Nov 2022, 8:31am I finally got my mind round the difference between the two formulations. And why working on distance produces it.
Agreed. And for any arbitrary object experiencing aerodynamic drag the force is the aerodynamic drag and the distance is the distance that the object moves through the air.roubaixtuesday wrote: ↑12 Nov 2022, 5:51pmThey are not cycling related! Here they are again, with all cyclists removed, and simplified.
Please indicate which you agree with.
The total energy input is force x distance (this is a completely general equation).
And this is the crucial point, and it needs to be discussed as early as possible. By taking the general case it's easier to see that the distance moved though the air is the relevant distance. What other distance could it be?
Jonathan
The source of the retarding force, being aerodynamic, frictional, gravitational or anything else is entirely irrelevant.
Whatever the source of the force, the work done by the cyclist is the distance the cyclist travels multiplied by the force he overcomes.
Think about your claim.
If a cyclist is going uphill, and experiencing an aerodynamic drag equal to the gravitational force, do you
a) calculate the work by distance x (sum of the forces) or
b) calculate the work by somehow weighting according to windspeed?
The object doing the work does the same amount of work regardless of the reason for the retarding force.
[You can do the calculation as you suggest, using a moving frame of reference. But by doing so , you need to calculate using a different distance, as the distance depends on windspeed]
Re: How does a headwind affect speed and power output?
There is no "ground" for an arbitrary object experiencing aerodynamic drag. Just the object and the air.roubaixtuesday wrote: ↑13 Nov 2022, 2:17pmIt's the distance along the ground.Jdsk wrote: ↑13 Nov 2022, 8:31am I finally got my mind round the difference between the two formulations. And why working on distance produces it.Agreed. And for any arbitrary object experiencing aerodynamic drag the force is the aerodynamic drag and the distance is the distance that the object moves through the air.roubaixtuesday wrote: ↑12 Nov 2022, 5:51pm They are not cycling related! Here they are again, with all cyclists removed, and simplified.
Please indicate which you agree with.
The total energy input is force x distance (this is a completely general equation).
And this is the crucial point, and it needs to be discussed as early as possible. By taking the general case it's easier to see that the distance moved though the air is the relevant distance. What other distance could it be?
...
Jonathan
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Re: How does a headwind affect speed and power output?
The speed of the cyclist is defined relative to the ground.Jdsk wrote: ↑13 Nov 2022, 2:58pmThere is no "ground" for an arbitrary object experiencing aerodynamic drag. Just the object and the air.roubaixtuesday wrote: ↑13 Nov 2022, 2:17pmIt's the distance along the ground.Jdsk wrote: ↑13 Nov 2022, 8:31am I finally got my mind round the difference between the two formulations. And why working on distance produces it.
Agreed. And for any arbitrary object experiencing aerodynamic drag the force is the aerodynamic drag and the distance is the distance that the object moves through the air.
And this is the crucial point, and it needs to be discussed as early as possible. By taking the general case it's easier to see that the distance moved though the air is the relevant distance. What other distance could it be?
...
Jonathan
You are doing a calculation applying different frames of reference to different parts of the calculation.
Again, work = force x distance
It matters not a jot what the source of the force is, the law still applies.
Your approach breaks the laws of physics.
Re: How does a headwind affect speed and power output?
We'd just agreed that there wasn't a cyclist:roubaixtuesday wrote: ↑13 Nov 2022, 5:12pmThe speed of the cyclist is defined relative to the ground.
You are doing a calculation applying different frames of reference to different parts of the calculation.
Again, work = force x distance
It matters not a jot what the source of the force is, the law still applies.
Your approach breaks the laws of physics.
Jonathanroubaixtuesday wrote: ↑12 Nov 2022, 5:51pmThey are not cycling related! Here they are again, with all cyclists removed, and simplified.
Re: How does a headwind affect speed and power output?
My approach is the fundamental equation of drag. For an arbitrary object in an airstream:
Force (of the drag) is proportional to the square of the speed of the object through the air.
plus:
Energy (or work) = force x distance.
And distance here is the distance that the object travels through the airstream. What other distance could it be?
Jonathan
Last edited by Jdsk on 13 Nov 2022, 5:26pm, edited 2 times in total.
Re: How does a headwind affect speed and power output?
If at this point any one is interested in why the distance matters I'm happy to explain, but I'd rather work through it one step at a time.
Jonathan
Jonathan
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Re: How does a headwind affect speed and power output?
Here we are discussing an object whose distance is measured vs ground, not vs airstream.Jdsk wrote: ↑13 Nov 2022, 5:24pmMy approach is the fundamental equation of drag. For an arbitrary object in an airstream:
Force (of the drag) is proportional to the square of the speed of the object through the air.
plus:
Energy (or work) = force x distance.
And distance here is the distance that the object travels through the airstream. What other distance could it be?
Jonathan
You can't use the velocity vs airstream to calculate work, and say the velocity vs ground is constant.
That's your fundamental error - you're using two different frames of reference in the same calculation.
If you do the calculation with a single consistent frame of reference you'll get the correct answer.
Try it!
Re: How does a headwind affect speed and power output?
What other frame of reference is there for an arbitrary object moving though an airstream?roubaixtuesday wrote: ↑13 Nov 2022, 5:35pmHere we are discussing an object whose distance is measured vs ground, not vs airstream.Jdsk wrote: ↑13 Nov 2022, 5:24pmMy approach is the fundamental equation of drag. For an arbitrary object in an airstream:
Force (of the drag) is proportional to the square of the speed of the object through the air.
plus:
Energy (or work) = force x distance.
And distance here is the distance that the object travels through the airstream. What other distance could it be?
You can't use the velocity vs airstream to calculate work, and say the velocity vs ground is constant.
That's your fundamental error - you're using two different frames of reference in the same calculation.
If you do the calculation with a single consistent frame of reference you'll get the correct answer.
Try it!
Thanks
Jonathan
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Re: How does a headwind affect speed and power output?
I have offered you a step by step approach already.
You insist that distance must be measured relative to air speed.
If, for instance, the headwind is the same as the object's speed this means you have doubled its speed.
Which means the speed is not constant - which is the question at hand; what is the relationship between power and headwind for an object at constant speed.
Work is force x distance regardless of the force. You are breaking this physical law.
Also, when your answer is different to all the online calculators available, could I suggest asking yourself the question "where have I made my mistake" will get you to an answer.
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Re: How does a headwind affect speed and power output?
Relative to the frame the question itself uses for distance. The ground.Jdsk wrote: ↑13 Nov 2022, 5:37pmWhat other frame of reference is there for an arbitrary object moving though an airstream?roubaixtuesday wrote: ↑13 Nov 2022, 5:35pmHere we are discussing an object whose distance is measured vs ground, not vs airstream.Jdsk wrote: ↑13 Nov 2022, 5:24pm
My approach is the fundamental equation of drag. For an arbitrary object in an airstream:
Force (of the drag) is proportional to the square of the speed of the object through the air.
plus:
Energy (or work) = force x distance.
And distance here is the distance that the object travels through the airstream. What other distance could it be?
You can't use the velocity vs airstream to calculate work, and say the velocity vs ground is constant.
That's your fundamental error - you're using two different frames of reference in the same calculation.
If you do the calculation with a single consistent frame of reference you'll get the correct answer.
Try it!
Thanks
Jonathan
You're using two frames of reference inconsistently.