Leg strength requirement for cycling?

[SolarEnergy]

Hi,

I was reading the big post about cycling and weight lifting.

I found this statement :

.......
"at the power that e.g., Armstrong is likely to generate (~ 400 W) at 90 revs/min up a long alpine pass, the force requirements assuming 170-mm cranks is 249 Newtons (~ 25 kg between each leg).".......

Anyone has an idea of the forumals involved to get that number (25kg)?

Well, I think I am comfortable with Newtons to kg conversion. But how do you come up with 249 Newtons?

Can this number still be considered as being accurate given that LA probably pushes (100 to 160 degrees) / 360 on each pedal stroke? (the rest being the leg recovery)

Thanks

 

[SATX Devil]

Hi,

I was reading the big post about cycling and weight lifting.

I found this statement posted by Ric Stern:

"at the power that e.g., Armstrong is likely to generate (~ 400 W) at 90 revs/min up a long alpine pass, the force requirements assuming 170-mm cranks is 249 Newtons (~ 25 kg between each leg)."

Anyone has an idea of the forumals involved to get that number (25kg)?

Thanks

I am sure it involves something that I will never understand

 

[whoawhoa]

Hi,

I was reading the big post about cycling and weight lifting.

I found this statement :
Anyone has an idea of the forumals involved to get that number (25kg)?

Well, I think I am comfortable with Newtons to kg conversion. But how do you come up with 249 Newtons?

Can this number still be considered as being accurate given that LA probably pushes (100 to 160 degrees) / 360 on each pedal stroke? (the rest being the leg recovery)

Thanks

Go to analyticcycling.com and plug in some numbers on the "power given speed" page. The 250 newtons is total pedal force I think, the calculator on AC has an input for effective pedal range that will give you the force over whatever duration you input.

 

[frenchyge]

Anyone has an idea of the forumals involved to get that number (25kg)?

Power = torque x angular velocity

torque = force (the value you're solving for) x crank arm length

angular velocity = cadence (rpm) x 3.14 (pi) x 2 x crank arm length / 60

Then divide the force between two legs to determine force being applied to each pedal.

I believe Ric's force figures use the above equations which essentially means that equal force is being applied through the entire pedal stroke. The tools at Analytic Cycling will let you play with different stroke ranges to get more realistic force values.

 

[SolarEnergy]

Power = torque x angular velocity

torque = force (the value you're solving for) x crank arm length

angular velocity = cadence (rpm) x 3.14 (pi) x 2 x crank arm length / 60

Then divide the force between two legs to determine force being applied to each pedal.

I believe Ric's force figures use the above equations which essentially means that equal force is being applied through the entire pedal stroke. The tools at Analytic Cycling will let you play with different stroke ranges to get more realistic force values.

Thanks Frenchyge!

Analytic Cycling?

"angular velocity = cadence (rpm) x 3.14 (pi) x 2 x crank arm length / 60"
Can you explain me a bit more? The arm length is expressed in... mm?

 

[RapDaddyo]

Analytic Cycling?

http://www.analyticcycling.com/

 

[frenchyge]

Analytic Cycling?

www.analyticcycling.com It's a fantastic site!

"angular velocity = cadence (rpm) x 3.14 (pi) x 2 x crank arm length / 60"
Can you explain me a bit more? The arm length is expressed in... mm?

Sorry. I posted before going down to dinner and then thought "I hope people don't get confused by my lack of units there."

To get power, in watts, you need torque (in N-m) times angular velocity in (radians/sec). Crank lengths need to be expressed in *meters.* Angular velocity is the rotational speed of the pedals, so take cadence (Rev/min) and convert to radians/sec by knowing that 1 revolution = 2 x pi radians, and 1 minute = 60 seconds. Then you need to overlook the error in my formula above which includes crank length again.

So 400w @ 90 rpm and 170mm crank means,

400w = Force x (.170m) x 90 rpm x 2pi / 60

Force = 249N and divide by 9.8 to get kg ==> ~25kg

...and looking at Ric's quote again, he doesn't divide the force between the 2 legs, so he's using a 180-deg pedal stroke per leg rather than the 360 that I mentioned above. -again.

 

[SolarEnergy]

www.analyticcycling.com It's a fantastic site!

Sorry. I posted before going down to dinner and then thought "I hope people don't get confused by my lack of units there."

To get power, in watts, you need torque (in N-m) times angular velocity in (radians/sec). Crank lengths need to be expressed in *meters.* Angular velocity is the rotational speed of the pedals, so take cadence (Rev/min) and convert to radians/sec by knowing that 1 revolution = 2 x pi radians, and 1 minute = 60 seconds. Then you need to overlook the error in my formula above which includes crank length again.

So 400w @ 90 rpm and 170mm crank means,

400w = Force x (.170m) x 90 rpm x 2pi / 60

Force = 249N and divide by 9.8 to get kg ==> ~25kg

...and looking at Ric's quote again, he doesn't divide the force between the 2 legs, so he's using a 180-deg pedal stroke per leg rather than the 360 that I mentioned above. -again.

Fantastic site indeed. I'm gonna need time to get familiar with all this, but still, looking for an answer, got it!


But you see Frenchyge, I don't understand why we should divide 25kg by two, 90 rpm refers to the pedal stroke rate for one leg. Both legs are actually moving at 180rpm em I wrong?

 

[SolarEnergy]

Given that the default value, found on analitic cycling, for effective pedalling range 70 (instead of 360) is realistic, I applied a factor of 2.57

So 25 would become ~65.5

Do you think it make sense?

 

[frenchyge]

But you see Frenchyge, I don't understand why we should divide 25kg by two, 90 rpm refers to the pedal stroke rate for one leg. Both legs are actually moving at 180rpm em I wrong?

Divide it by 2 only if you pull on the upstroke. If you're not pulling on the upstroke then you're only applying the force on half the circle or effectively pedalling with one leg....hence, 90 rpm.

Given that the default value, found on analitic cycling, for effective pedalling range 70 (instead of 360) is realistic, I applied a factor of 2.57

So 25 would become ~65.5

Do you think it make sense?

Yep. I'm pretty sure that's the answer AC.com will give you as well.

To be clear, 249N (weight of 25 kg) is the *total* force on the cranks. You can do it all with one leg by pedalling a complete circle and applying 249N of tangential force at all points of the circle (push down & pull up with the same force), or you can push down with each leg for half the stroke and unweight the upstroke, or you can push and pull with *both* legs. The pedal force required for the first 2 modes will be the same (249N or weight of 25kg), and the pedal force for the third mode would be half of that. When you're pushing and pulling at the same time, the torque arm length is effectively doubled, which means same torque for half the force.

 

[Carrera]

I'm afraid this is all rocket science to me so I'll leave all the equations to the experts.
Myself I approached this from a far cruder angle and studied my pedal strokes and angles of contraction as I was climbing a hill. Seems like I myself do very short range contactions to turn the crank.
Obviously, no strength is required on the flat, especially as speed increases.
You also have to consider the effect your body weight has on pedal strokes as you can use your weight to power the pedals too.

 

[SolarEnergy]

I'm afraid this is all rocket science to me...

Yeah I know. The understanding of these things does not garantie better performances. And it is better for a rider to feel these things, rather than understanding per se.

I'm just a bit curious

Yep. I'm pretty sure that's the answer AC.com will give you as well.

Thanks again Frenchge. I already took a look at ac.com
The best model I found for this, is under Static Forces on Rider - Power Given Speed.

But I can not understand an important value : Average Pedal Force (expressed in kg m/s 2)

For example : 156kg m/s2

Any Idea what it means?

 

[beerco]

Given that the default value, found on analitic cycling, for effective pedalling range 70 (instead of 360) is realistic, I applied a factor of 2.57

So 25 would become ~65.5

Do you think it make sense?

We had this discussion a while back on wattage and came to the conclusion that 70 is probably a bit small. The reason being that recorded data has shown that peak pedal forces are right around 2x average pedal forces. So, 65.5kg is close, but it's probably a little high. Peak forces would probably be around 50kg or so.

Even considering this, peak forces are still not very high. Don't forget that every time you walk up steps you cary your whole body's weight with one leg. Unfortunately for me, that's more kgs than I care to admit online

 

[SolarEnergy]

We had this discussion a while back on wattage and came to the conclusion that 70 is probably a bit small. The reason being that recorded data has shown that peak pedal forces are right around 2x average pedal forces. So, 65.5kg is close, but it's probably a little high. Peak forces would probably be around 50kg or so.

Even considering this, peak forces are still not very high. Don't forget that every time you walk up steps you cary your whole body's weight with one leg. Unfortunately for me, that's more kgs than I care to admit online

Sounds good, but I don't understand . What is average pedal force?

 

[beerco]

Sounds good, but I don't understand . What is average pedal force?

Average pedal force is the constant force required to produce a particular power. i.e. if you applied constant force (which we don't) around the whole circle.

However, we push down really hard on the down stroke and not at all anywhere else so what ends up really happening is something that looks like a sine wave as far as torque application goes.

Get it?

 

[asgelle]

Obviously, no strength is required on the flat, especially as speed increases.

I'm not sure what you mean here, but I believe you still need to push on the pedals on the flat, and push harder as speed increases.

 

[frenchyge]

But I can not understand an important value : Average Pedal Force (expressed in kg m/s 2)

For example : 156kg m/s2

Any Idea what it means?

Believe Avg Pedal Force assumes equal tangential force over the full 180-deg downstroke. Effective Pedal Force uses the Effective Pedal Stroke angle that you input and assumes that force is only being applied during that portion of the stroke (and no force applied elsewhere).

 

[SolarEnergy]

Believe Avg Pedal Force assumes equal tangential force over the full 180-deg downstroke. Effective Pedal Force uses the Effective Pedal Stroke angle that you input and assumes that force is only being applied during that portion of the stroke (and no force applied elsewhere).

Ok all right. But how does 156kg m/s 2 translate into kg per downstroke? In other words, on Analytic Cycling, I found no mention of kg per downstroke, only (kg m/s 2).
- Why?
- And how does this translate into kg per downstroke?

 

[Squint]

Ok all right. But how does 156kg m/s 2 translate into kg per downstroke? In other words, on Analytic Cycling, I found no mention of kg per downstroke, only (kg m/s 2).
- Why?
- And how does this translate into kg per downstroke?

kg*(m/s^2) sounds like force

kg is mass

After all, you apply force to move objects, not necessarily mass. You might be able to "convert" using g as 9.8 m/s^2.

 

[frenchyge]

kg*(m/s^2) sounds like force

kg is mass

After all, you apply force to move objects, not necessarily mass. You might be able to "convert" using g as 9.8 m/s^2.

That's right. kg-m/s^2 is a fancy way of describing a Newton of force. Converting Newtons to kg is only useful if you're unfamiliar with Newtons, or you live in America and know that 1 kg "weighs" 2.2 pounds.

To convert Newtons to "Weight of a kilogram on earth" you would divide by 9.81, and then you could convert to pounds of force by multiplying by 2.2.