Silly force question?

[postal_bag]

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?

 

[acoggan]

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

 

[bikeguy]

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

 

[Roadie_scum]

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...

 

[frenchyge]

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.

... 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!

 

[bikeguy]

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.

 

[rmur17]

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

 

[acoggan]

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

 

[sglasgow]

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