Pedal-force for bicycle @ 43.2mph ?

[DannoXYZ]

Pedal-force for bicycle @ 43.2mph ?

I heard that bike-racers can put put 1500 watts in sprints. Using this site: http://www.noping.net/english and entering the following:

- Racing bike - hands on the drops
- weight = 170 lbs
- power = 1500 watts

Then hit CALCULATE button and I get a speed of 43.2mph. I came up with this diagram to figure out all the forces and torques in order to come up with pedal-force:

0.336m = Wr = Wheel Radius
0.170m = CL = Crank length

Given 52x13t gear ratio of 4:1, Tc = 4*Tw; torque at crank is 4x higher than torque at wheels.

Then using this site: http://www.analyticcycling.com/GearCadenceSpeed_Page.html, I enter:
speed = 43.2mph
chainring = 53t
cog = 13t
wheel-diameter = 0.672m

Click on RUN MODEL and I get 135 rpms at the crank.

What stops me next is how to figure out what PF - pedal force is in order to generate 1500w and go 43.2mph @ 135rpms. ?

Thanks.

 

[jack action]

The fastest way would be to identified your foot' speed.

v = 135 rpm * (pi/30 rad/s/rpm) * 0.170 m

v = 2.4 m/s

Then the force is:

F = P/v = 1500 W / 2.4 m/s

F = 624 N (= 140 lb)

Longer way:

The tire friction force is:

F_t = 1500 W / (43.2 mph * 0.44704 m/s/mph)

F_t = 77.67 N

The torque at the wheel:

T_w = 77.67 N * 0.336 m

T_w = 26.1 N.m

The torque at the pedal:

T_p = 26.1 N.m * 53/13

T_p = 106.4 N.m

Force at the pedal:

F = 106.4 N.m / 0.170 m

F = 625.9 N

The difference only lies in the rounding-off error (the actual crank rpm is 134.6 rpm).

 

[K^2]

This is a little high, as it should be.

If you burn 1500J/s, that doesn't mean you are doing 1500W of work on the pedals. A lot of this energy is being lost elsewhere.

 

[DannoXYZ]

Thanks Jack, the 1st way really was simple. :)

It may appear pedal-force is high, top-sprinters can exert quite a bit of force on their pedals. Due to the alternating motion, it's only coming mostly from one leg at a time. And we're calculating perfectly symmetric motion, but I suspect the up-stroke has minimal power. So the downstroke would have to be even higher than the numbers we're coming up with.

Chris Hoy has been measured at 2300 watts max and he can sustain 1000w for just over 60-seconds... wow...

 

[jack action]

I was just watching a TV science/game show the other day. The contestants had to guess what type of electrical appliance they could turn on with a continuous effort: a 100 W light bulb, a 350 W mixer or a 1500 W iron. Once they chose, they went on a bike connected to a generator and tried to turn on the appliances.

The woman reached a peak (not an average) of 200 W and the man went up to 300 W. These were un-prepared average joe. So, yes, a 1000 W for 60 s sounds pretty impressive to me!

Furthermore, 1 hp = 746 W, hence that means that he is stronger than an average horse for about 1 min, think about it ... that is freaky!

 

[dbelfordb]

Limited by Force on Pedal

I am a bike rider with a background in engineering physics. I took measurements from my mountain bike and wondered where the reports of 600 W power exerted by a Tour racer came from. What are they measuring to make the calculation and what limits are there to what a rider can exert?

Since forces from wind, grade, and acceleration are variable it seems like the force at the drive wheel or on the pedals should be the sum of all demands on the rider at any instant. Since the direct measure of force at the drive wheel tire surface would be more difficult to measure, the force at the pedal must be more easily come by.

The force stroke at the crank (up (toe clips)+ down, Fo) may be assumed constant for this estimate, but the definition of work is the scalar product of the distance moved and the vector of the force in the direction of motion. To simplify, the vector of force = Fo * sinθ , where θ is the angle of the crank from the positive vertical. To get the work done through the first pi/2 radians, the expression must be integrated between these limits, giving: Fo*Rp = Work Done through pi/2 radians, Rp being the radius of the crank (170 mm for my bike).

To get work done for a complete cycle of the crank multiply by 4. And to get the power (rate of work) multiply by the RPS (revolutions/sec) applied at the crank.

I assumed that the maximum force I could exert would be my full weight alternately applied to each pedal during 1 cycle of the crank. I assumed a 'reasonable max' of 2 RPS under these conditions and calculated just under 1100 W. Riding your bike may convince you that this is an unattainable max with rider weight and crank radius being limits.

 

[VanDerRider]

Can a rider create additional pedal force by pulling up on handle bar while pushing down on pedal? Would allow a force greater than rider weight?

 

[jack action]

Can a rider create additional pedal force by pulling up on handle bar while pushing down on pedal? Would allow a force greater than rider weight?

Yes.

Just imagine putting springs under your shoes and resting on the floor inside your house. Your spring will collapse under your weight. Now put your hands on the ceiling and push: The springs collapse even more. In this case, the house represents your bicycle.

If you pull up on the bike and the bike doesn't move, it means that you were pushing down on it somehow, i.e on the brake pedal with your foot (or it might of been on the seat with your butt! ).