How to: compute Watts and Calories for for a bicycle, up a hill

  • Thread starter wmazz
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At 6'5", 300 lbs. and old as dirt, I am a walking and rolling testimate to F=ma. This part of my project is difficult for me, because I have never taken a physics class. All the examples I have in books and PF, deal mostly with force moving down a hill. I am trying to compute Watts and Calories, while riding at any measured angle.

I am using a BS2p basic stamp reading an 2 axis accelerometer, 2 hall sensors and a gear tooth sensor. A math coprocessor handles the math. With my current set-up the BS2p determines what gear I am in, at any time, or if I am coasting. Every calculation is started after the front wheel completes a rotation, but I have no idea if it is accurate.

Watts = ma*gr*pcl*s*sin(theta) + Rr (so long as theta is never zero) If theta = 0, then Watts = ma*gr*pcl + Rr

Calories = watts*0.2388459


m = Mass = 145kg
a = Pedal Speed = 22pi/9 rad/sec
gr = Gear Ratio = 44 : 18 (2.44 to 1) (This is gear ratio I use the most)
pcl = Pedal Crank Length = .175m
s = Front tire circumference = 2.2m
Rr = Rolling Resistance ??

I left gravity out, because I felt it would be part of Rolling resistance. Once the initial equation is working I figured on gathering Rr by comparing computed and measured data

thanks

Bill M.
 
Last edited:

Answers and Replies

  • #2
Born2bwire
Science Advisor
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There isn't a conversion between Calories and watts. Watts is power, Calories are energy. You want to know a rough estimate of how many Calories you burn going up a hill? Just find the potential energy from the change in height: m*g*h. This will be in joules though so you will have to convert to Calories which is 4184 J/Cal. So:

145 kg * 9.806 m/s^2 * h (m) / 4184 J/Cal = 0.34 Cal/m * h (m)

This value is going to be less than your actual work done due to the losses in the system, but it could be used as a sanity check for the equations that you have been trying to work with.
 

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