Combating Air Resistance on a Bike

AI Thread Summary
The discussion focuses on calculating the torque and power required to maintain a constant speed of 27.5 mph on a bike, considering factors like air resistance and rolling resistance. The user calculates a total required torque of 28.99 pound-feet based on their leg strength and gear ratio. They express confusion regarding their power calculation, which estimates 133.8 watts, compared to a calculator that suggests 400 watts is necessary for the same speed. The discrepancy arises from misunderstanding the relationship between drag force and power, as the power formula includes additional factors beyond just drag. Clarification is needed on the correct application of the power formula to accurately assess the energy required to sustain the desired speed.
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I'm trying to calculate the torque i would need to pedal when in the highest gear on my bike to sustain a constant speed on level ground. Basically i just want someone to review my calculations and make sure they're correct. I leg press 270 pounds and therefore each of my legs can supply 135 pounds of force. So when i turn the cranks with 135 pound-feet of torque with a gear ratio of 1:4.33 (I have a 52 tooth on the front and 12 on the back) , then the hub of the wheel on my bike in the rear wheel will be turning with 31.1 pound-feet of torque. When pedaling at 80 rpm, I would be going 27.5 mph. Now using the following information i calculate the air resistance:
Fd= 1/2pv^2ACd.
p= 1.3 kg/m^3
v= 13.44 meters/sec.
A= 1.15 meters^2
Cd= 0.9
I get 121.52 Newtons or 27.28 pounds of force for the air resistance.
So then taking into account rolling resistance of 2.352 Newtons or 1.71 pounds, this brings me to a total required torque of 28.99 pound-feet. My process for determining this required torque is correct, right? And if it is correct, then how would i determine the amount of power required to sustain this speed?
 
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You need this:

http://www.kreuzotter.de/english/espeed.htm

Also remember, unless you are on a recumbent, the max force you will be able to apply will be your weight as you "mash" on each pedal during the down-stroke.
 
This calculator calculates the power, however i was correct with my torque calculations right?
And also, couldn't i just calculate the power on my own. Power = Work/Time. Work = Force*Distance. The Crank that I turn has a 175mm diameter, so the circumference is 0.5495 meters, and at 80rpm as the pedaling cadence, this equates to 1.333 revolutions of the cranks per second, so the distance force is being applied is 0.732 meters. The force is 182.9 Newton-Meters, or Joules, so then the power should be 133.8 watts, right?
 
montoyas7940 said:
unless you are on a recumbent, the max force you will be able to apply will be your weight as you "mash" on each pedal during the down-stroke.

If you pull up on the handlebars (without pulling them off, of course), you can push down more on the pedals, by a corresponding amount.
 
I know, I can only apply as much force as my weight while sitting down on the saddle, 135 pounds is as much as i weigh.
 
jtbell said:
If you pull up on the handlebars (without pulling them off, of course), you can push down more on the pedals, by a corresponding amount.

I didn't consider that. My 'bent preference is showing I suppose. :blushing:
 
With pedal clips (toe clips or the lock in shoes), you can pull up on the pedals as well, increasing the maximum force without having to pull up on the handlebars, or at least not as much.
 
I'm still confused about the power though, i calculated out 133.8 watts to sustain 27.5 mph, and the bike velocity and power calculator calculates 400 watts necessary to maintain the speed. I inputed all the correct information to match my bike setup, why are the numbers so drastically different? I took into account air resistance with my power calculation.
 
Here is the Wiki on drag and power to overcome drag:

http://en.wikipedia.org/wiki/Drag_(physics )

The formula you are using is only the drag force. Look at the power formula in the wiki.
 
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