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Old 12-06-06, 09:43 AM   #1
oldacura
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Aerodynamics

This is a discussion I had on Tandem@Hobbes a number of years ago with no satisfactory resolution.

A website: http://www.exploratorium.edu/cycling/aerodynamics1.html. provides a simplistic model to show how much power a cyclist produces under given conditions. If one enters their speed & wind velocity, it calculates power produced. It shows that riding 20mph in still air is the same as 10mph into a 10mph headwind. Where it falls apart is that it shows the same power output at 1mph into a 19mph headwind (and even the same standing still with a 20mph headwind). So, the model is obviously flawed.

It is my intuitive experience that power output is not solely dependent on "airspeed". If not, why does is seem to take more power to ride 20mph in still air than 10mph into a 10mph headwind? Where does the power go?

When I think about where power goes when cycling on flat ground, there are really only two sinks: rolling/bearing friction and air drag. I assume that rolling/bearing friction are small and not speed dependent.

The only explanation I can come up with is that wheels spinning at 20mph "churn" more air than wheels spinning at 10mph.

Does anyone have a clear explanation of where the power goes?
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Old 12-06-06, 09:56 AM   #2
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I don't know the exact math...but I do not believe that the relationship between headwind speed & the force it takes to overcome the headwind is linear.

In other words...the wind resistance model you linked to is really faulty (as you mentioned)...because it assumes that 1 mph headwind slows you down by 1 mph...given that you are pushing the pedals at the same effort level.

My example would be, depending on a persons fitness level...riding at 17mph (single bike) is a relatively easy pace to maintain...but increase that to 20 mph...and the average rider cannot maintain that speed for long.

In order to maintain 20 mph...not only do you have to pedal harder...but you also have to overcome the increased wind resistance created by going 20 mph vs. 17 mph.
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Old 12-06-06, 10:13 AM   #3
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Here's a website that I found that explains things a little more scientifically:

http://sheldonbrown.com/brandt/wind.html

Here are some of the "key" statements:

"Wind drag is a nonlinear function of the relative wind. That is, drag does not increase in proportion (linearly) to wind but rather with the square of its speed relative to the rider."

"Although inline winds add or subtract directly to drag, they do not similarly affect the power required to overcome that drag. For instance, standing still in a 15mph wind requires no power because power is the product of inline drag and rider speed."

"Drag is proportional to the square of relative wind-speed while power is the product of the inline portion of that drag and rider speed. Thus, required power (in still air) increases as the cube of rider speed."


As you can see...as soon as you start seeing "square or cube of a rider's speed"...or "drag is proportional to the square...", that wind resistance, drag, or speed are not linear relationships.

Now after reading all of that...you are allowed to let your head explode!!!

- Nick
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Old 12-06-06, 10:31 AM   #4
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Yes - I understand that power to overcome air drag increases as the cube of the relative windspeed. If one is riding 10mph into a 10mph headwind or 20mph in still air, the air is moving past the rider at 20mph.

If it really does require more power to ride 20mph in still air than 10mph in a 10mph headwind, where does this extra power go?

Maybe I should email Sheldon.
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Old 12-06-06, 10:44 AM   #5
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The power transforms into heat and motion from compression and decompression of the air flowing off the rider and equipment. There is a huge high pressure zone in front of the rider and a huge negative pressure zone behind the rider, and it takes a lot of power to maintain these pressure differentials since the air can rapidly escape or enter the zones as fluid dynamics in open space allow.

The air has a mass, and it takes energy to move that mass. The faster you go, the more air you have to move, and faster. As you go through it, the air retains the energy you provided through power. The energy then exists as turbulence (kinetic), heat (thermal), and soundwaves.

Some of the energy turns into heat in your tires and bearings, and is absorbed and dissipated through the road, air, spokes, frame, etc.

Last edited by waterrockets; 12-06-06 at 10:49 AM.
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Old 12-06-06, 10:59 AM   #6
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Originally Posted by oldacura
Yes - I understand that power to overcome air drag increases as the cube of the relative windspeed. If one is riding 10mph into a 10mph headwind or 20mph in still air, the air is moving past the rider at 20mph.

If it really does require more power to ride 20mph in still air than 10mph in a 10mph headwind, where does this extra power go?
Maybe friction of the components/bearings, tire deformation, bumps in the road (if you go 20mph you hit statistically twice as many bumps/pebbles/... than when you go 10mph),...?

I can ride 20mph in still air. Now, I guess according to the claims of that website, if I had a wind at 20mph from exactly behind me, I could go 40mph. Awesome!
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Old 12-06-06, 11:05 AM   #7
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Oldacura,

One thing to consider in your example...

...riding at 10mph into a 10mph headwind vs. riding at 20mph with no headwind...

Rarely do you get a day where you actually get a steady 10mph headwind...and rarely will that 10mph headwind always be an exact headwind. Sometimes it may shift to a cross-wind...or be partially blocked by obstacles such as trees, hills, cars, trucks, and buildings.

When you are riding at 20mph with no winds...in this case you ALWAYS have to overcome the wind resistance at 20mph...because that is the speed you are riding at.

So the actual "real world" experience of riding at 10mph, with a 10mph headwind, may not always be equivalent to riding at 20mph with no wind.

- Nick

Last edited by Pigoo3; 12-06-06 at 11:10 AM.
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Old 12-06-06, 11:06 AM   #8
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I understand that power is converted to relative motion which is then converted to heat.

But this still does not answer my question: If it takes more power to ride 20mph in still air than to ride 10mph into a 10mph headwind, Why? Where does the additional power go? The air is still moving past the rider at 20mph.
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Old 12-06-06, 11:16 AM   #9
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Oldacura,

Remember...you are basing this discussion based on the model in your original post...which I think we agree is faulty.

So if the original model you referenced is faulty...then you cannot believe any of the results that is "spits" out.

So maybe in a perfect world, riding at 20mph with no winds...and riding at 10mph with a "perfect & steady" 10mph headwind...are actually the "similar" efforts.

I think that we also have to assume at both the 10mph & 20mph riding speeds, that we are spinning the pedals at the same rate/cadence.

Not choosing the proper gears for the situation may also modify the perception of pedaling difficulty at the different speeds.

- Nick

Last edited by Pigoo3; 12-06-06 at 11:32 AM.
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Old 12-06-06, 01:06 PM   #10
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Here is a link to a simulator that calculates speed for power imput and other variables including wind speed that may assit you. http://www.kreuzotter.de/english/espeed.htm It includes a tandems as an input and provides the formulas and assumptions. I have used it a few times and the results seem reasonable compared with field results.
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Old 12-06-06, 01:23 PM   #11
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Quote:
Originally Posted by oldacura
If it takes more power to ride 20mph in still air than to ride 10mph into a 10mph headwind, Why? Where does the additional power go? The air is still moving past the rider at 20mph.
  • There's 1/2 as much rolling resistance from tires and wheel bearings
  • Since your wheels are rotating slower, the tops of the wheels and tires are having a much lower impact since at 20mph your wheel tops are going 40mph, where they're only going 20mph if you're traveling 10mph -- so the wheel tops make a significant impact
  • You're not moving the chain as fast through the drivetrain, so it has less resistance itself, as well as less pulley resistance and gear entry/exit resistance
  • Bigger cogs and slower wheel RPM mean a lower chain/gear entry/exit rate
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Old 12-06-06, 01:39 PM   #12
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Power equals force times distance divided by time.

When travelling at 10mph into a 10mph headwind, the aerodynamic drag FORCE (the vast majority of the total drag on a bike) will be the same, but the distance covered will be half as much as riding 20mph in still air. Half the power required.
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Old 12-06-06, 03:26 PM   #13
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Lord of the Rings - thanks for the link. It completely disagrees with the link I provided. It shows that riding 10mph in a 10mph headwind takes about 1/2 the power of riding 20mph in still air.

I think the reason for this is given by Jinker.

However, the equation (Power = force X distance / time) falls apart for a stationary bike. Just because you are not moving are you producing no power?
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Old 12-06-06, 03:43 PM   #14
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Quote:
Originally Posted by oldacura
Lord of the Rings - thanks for the link. It completely disagrees with the link I provided. It shows that riding 10mph in a 10mph headwind takes about 1/2 the power of riding 20mph in still air.

I think the reason for this is given by Jinker.

However, the equation (Power = force X distance / time) falls apart for a stationary bike. Just because you are not moving are you producing no power?
P=Fd assumes perfectly efficient locomotion. Stationary bicycles are perfectly inefficient, so you have to find another way to calculate power. You can look at force on a pedal and how far that pedal moves instantaneously.
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Old 12-06-06, 04:35 PM   #15
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Quote:
Originally Posted by waterrockets
P=Fd assumes perfectly efficient locomotion. Stationary bicycles are perfectly inefficient, so you have to find another way to calculate power. You can look at force on a pedal and how far that pedal moves instantaneously.
Oldacura: Your welcome. I think waterrockets is correct. It is difficult to mathematically model real systems under all operating, startup and transient conditions. It is not surprising that it may not work for some conditions.
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Old 12-06-06, 04:36 PM   #16
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Quote:
Originally Posted by oldacura
Lord of the Rings - thanks for the link. It completely disagrees with the link I provided. It shows that riding 10mph in a 10mph headwind takes about 1/2 the power of riding 20mph in still air.

I think the reason for this is given by Jinker.

However, the equation (Power = force X distance / time) falls apart for a stationary bike. Just because you are not moving are you producing no power?
As waterrockets said... all your power is going into losses on a stationary bike, whatever the resistance system is, air, fluid, magnetic etc.

On a moving road bike (assuming steady speed, no hills), though, the vast majority of the energy fed into the pedals is actually fighting air resistance. You lose a few percent to the drivetrain, maybe a few more to rolling resistance. It looks like around 75% of the drag at 20mph is aero drag.

Playing around with this:
http://www.kreuzotter.de/english/espeed.htm

should give you a good idea of the relationships between speed, power, wind and slopes.
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Old 12-11-06, 03:10 PM   #17
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Quote:
Originally Posted by oldacura
Lord of the Rings - thanks for the link. It completely disagrees with the link I provided. It shows that riding 10mph in a 10mph headwind takes about 1/2 the power of riding 20mph in still air.

I think the reason for this is given by Jinker.

However, the equation (Power = force X distance / time) falls apart for a stationary bike. Just because you are not moving are you producing no power?
The equation is fine for a stationary bike. In a 20 mph head wind you dont need any power to stand still with one foot down on the curb, but you still feel the same wind pressure on your face as riding at 20 mph in still air.
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Old 12-12-06, 05:27 AM   #18
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What I think is you should get off the chair and go ride. All questions will be answered.
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