Silly force question?

postal_bag said:

I understand that the required force is greater at a lower cadence compared to a higher one, when riding at a given power output.

Does anyone know of a graph or spreadsheet that would plot power and cadence, so that I could see what force was required at any combination of the two?

No, but the formulae you need can be found here:

http://home.earthink.net/~acoggan/quadrantanalysis

You can use analyticcycling.com to calculate the force at different wattages and rpms. kg m/s2 = 1 newton , 9.8 newtons = 1 kg, well, on earth anyway.
AA seems to have effective pedal (peak?) force as somewhat more than 2 times the average, that seems about right.

Here's an example:

Forces On Rider

Frontal Area 0.50 m2
Coefficient Wind Drag 0.50 dimensionless
Air Density 1.226 kg/m3
Weight 85.0 kg
Coefficient of Rolling 0.004 dimensionless
Grade 0.000 decimal
Wind Resistance 44.3 kg m/s2
Rolling Resistance 3.3 kg m/s2
Slope Force 0.0 kg m/s2
Cadence 100. rev/min
Crank Length 170. mm
Pedal Speed 1.78 m/s
Average Pedal Force 454.8 kg m/s2 = 45.5 kg roughly, average
Effective Pedaling Range 70. degree
Effective Pedal Force 1169.4 kg m/s2 = 116.9 kg roughly, peak
Speed 17.00 m/s
Power 809.6 watts

-bikeguy

postal_bag said:

I understand that the required force is greater at a lower cadence compared to a higher one, when riding at a given power output.

Does anyone know of a graph or spreadsheet that would plot power and cadence, so that I could see what force was required at any combination of the two?

For some reason AC's link didn't work for me.

Excuse clumsy notation and possible clumsy physics and algebra. I don't know how to get better symbols out of my keyboard, and it's possible I can't remember how to do maths and physics (it's been a while!).

Defn: Power = Torque X Angular Speed

Therefore (I can't get the funky three dots on my keyboard so will go with "tf"😉

tf: Torque = Power(Angular Speed)^-1

Defn: Torque = Force (orthogonal to lever) X Lever Length (that is crank length)

tf: Force = (Power(Angular Speed)^-1)(Crank Length)^-1

Angular speed is radians per second (although what with radians being dimensionless, I guess it's nothings per second). A cadence of 60rpm is 2pi radians/second, so the conversion factor between cadence and radians per second should be 2pi/60

So we end up with:

Force = (Power) X [(2pi/60)XCadence]^-1 X (Crank Length)^-1

So a given power and crank length will yield a nice little constant and Force will vary as 1/Cadence. Yay!

Now will someone please explain to me how I got something wrong? High school physics was a good discipline for me and I need it back... never study law... never... it makes you forget everything else...

Roadie_scum said:

Now will someone please explain to me how I got something wrong?

Nothing wrong there. Power is proportional to Force x Cadence, meaning that it's equal to a constant x Force x Cadence. Figure out the constant without making any math mistakes and you've got it. 🙂

Roadie_scum said:

... never study law... never... it makes you forget everything else...

It's something about the idea that anything can be made to be correct if you argue it well enough. Uck! 😄

Roadie_scum said:

For some reason AC's link didn't work for me.

Excuse clumsy notation and possible clumsy physics and algebra. I don't know how to get better symbols out of my keyboard, and it's possible I can't remember how to do maths and physics (it's been a while!).

Defn: Power = Torque X Angular Speed

Therefore (I can't get the funky three dots on my keyboard so will go with "tf"😉

tf: Torque = Power(Angular Speed)^-1

Defn: Torque = Force (orthogonal to lever) X Lever Length (that is crank length)

tf: Force = (Power(Angular Speed)^-1)(Crank Length)^-1

Angular speed is radians per second (although what with radians being dimensionless, I guess it's nothings per second). A cadence of 60rpm is 2pi radians/second, so the conversion factor between cadence and radians per second should be 2pi/60

So we end up with:

Force = (Power) X [(2pi/60)XCadence]^-1 X (Crank Length)^-1

So a given power and crank length will yield a nice little constant and Force will vary as 1/Cadence. Yay!

Now will someone please explain to me how I got something wrong? High school physics was a good discipline for me and I need it back... never study law... never... it makes you forget everything else...

There isn't anything wrong with your formula.. I get the same answer as below but you're making it needlessly complicated with terms in brackets and raised to powers, which leads to possible errors. I'd use F=P/V

foot velocity on pedal[m/s] = crank length x 2pi x cadence/60 [m/s] Solve for F. i.,e 400 watts at 70 rpm and 175 mm cranks, gives pedal[m/s]=1.282 m/s.
1 W = 1 kg / m^2 s^3
F=P/V; F(newtons)=(400 kg /m^2 s^3) / (1.28 m/s) = 311.8 kg /m s^2
=311.8 N = 31.8 kg average

Peak force will likely be more than double that, so 70 kg according to the AA model. For sprinters I don't know if the model for peak force holds, I know I start pulling up hard if sprinting but I use a slower cadence than most track sprinters.

Roadie_scum said:

For some reason AC's link didn't work for me.

Excuse clumsy notation and possible clumsy physics and algebra. I don't know how to get better symbols out of my keyboard, and it's possible I can't remember how to do maths and physics (it's been a while!).

Defn: Power = Torque X Angular Speed

Therefore (I can't get the funky three dots on my keyboard so will go with "tf"😉

tf: Torque = Power(Angular Speed)^-1

Defn: Torque = Force (orthogonal to lever) X Lever Length (that is crank length)

tf: Force = (Power(Angular Speed)^-1)(Crank Length)^-1

Angular speed is radians per second (although what with radians being dimensionless, I guess it's nothings per second). A cadence of 60rpm is 2pi radians/second, so the conversion factor between cadence and radians per second should be 2pi/60

So we end up with:

Force = (Power) X [(2pi/60)XCadence]^-1 X (Crank Length)^-1

So a given power and crank length will yield a nice little constant and Force will vary as 1/Cadence. Yay!

Now will someone please explain to me how I got something wrong? High school physics was a good discipline for me and I need it back... never study law... never... it makes you forget everything else...

Force = Power / [(2pi/60) x Cadence x Crank_Length]

simplifies it a bit. Force is actually Average Effective Pedal Force (AEPF) around the crank cycle ... versus instantaneous. All units should be metric: Force in Newtons, power in Watts, cadence in rpm, and crank_length in meters.

e.g 370W at 90 rpm and with 175mm cranks yields 217N or ~ 49 lbs-f

Roadie_scum said:

For some reason AC's link didn't work for me.

I made a typo when spelling "earthLink" - it's: http://home.earthlink.net/~acoggan/quadrantanalysis

postal_bag said:

I understand that the required force is greater at a lower cadence compared to a higher one, when riding at a given power output.

Does anyone know of a graph or spreadsheet that would plot power and cadence, so that I could see what force was required at any combination of the two?

Here are easy to understand graphs for 60 and 90 rpms plus a graph on efficient cadence at each wattage

http://www.53x12.com/do/show?page=article&id=27