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Cyclologist 12-15-07 01:19 AM

Power = work/time
 

Originally Posted by The Cyclists Training Bible
power = work/time

At the risk of oversimplification, in cycling "work" is essentially gear size and "time" is cadence. So if gear size is increased and cadence kept steady, power rises. Or, if cadence is increased (time per revolution of the crank is decreased) while using the same gear size, power also rises.

Discuss.

vic32amg 12-15-07 01:30 AM

whats there discuss. this is obvious.

Cyclologist 12-15-07 01:35 AM

When I first read that, it seemed more than just oversimplified, but rather entirely wrong unless I'm missing something.

joecool2727 12-15-07 01:57 AM

How do powermeters measure power? Dont they measure torque and use that? Or do they know what gear your in?

vic32amg 12-15-07 02:00 AM

Pt has a torque tube in the hub. as well as the Ergemo I believe has one in the BB.

CEE 12-15-07 02:00 AM

measure by strain

vic32amg 12-15-07 02:12 AM

my PT had a-torque tube, ergemo i'm not familiar with

waterrockets 12-15-07 07:04 AM

PT measures torque and wheel speed to give you force and distance per time. From that you can calculate work and power.

JayhawKen 12-15-07 07:06 AM


Originally Posted by Cyclologist (Post 5813130)
When I first read that, it seemed more than just oversimplified, but rather entirely inaccurate unless I'm missing something.


I agree - he probably shouldn't have taken the risk. :D

Since work = force x distance, I can see where Friel equates distance with gear size. But reading that text you could infer that he sees it as a linear relationship. I think he's missing (or choosing to overlook for simplification) the relationship that force must increase with increasing proportions of distance over time (i.e. velocity).

It's a good book, but there are a few areas like that where it could have benefited by better editing.

ElJamoquio 12-15-07 07:58 AM


Originally Posted by Friel
Power = Torque * Rotational Speed

At the risk of correct explanation, torque is the amount of force you put on the pedals (multiplied by crank length) and rotational speed is your cadence. So if you put more force on the pedals (say, by going to a higher gear), power rises. Or, if cadence is increased while using the same gear size, power also rises.

I don't know what you guys are *****ing about, it's well explained and makes perfect sense.

Hocam 12-15-07 10:00 AM


Originally Posted by JayhawKen (Post 5813490)
I agree - he probably shouldn't have taken the risk. :D

Since work = force x distance, I can see where Friel equates distance with gear size. But reading that text you could infer that he sees it as a linear relationship. I think he's missing (or choosing to overlook for simplification) the relationship that force must increase with increasing proportions of distance over time (i.e. velocity).

It's a good book, but there are a few areas like that where it could have benefited by better editing.

This is a better way of looking at it:
I can push a 53x13 on a flat surface at 90 rpm with no headwind.

Does my power change if my speed, gearing and cadence stay the same but now I'm going downhill? Or uphill? Or into a headwind? Or into a tailwind?

Obviously yes. There is more to power than gearing and cadence.

Wesmo15 12-15-07 11:02 AM

Power is also going to depend on the weight of the rider (ie. when going uphill). We did power tests in physics, and this simplified explanation of power would only apply in a setting with zero friction from both rolling resistance and wind.

brianappleby 12-15-07 03:09 PM

power = work/time.

change some (fundamental physics)stuff around and you get

Power= torque*angular velocity.

Angular velocity varies directly with cadence, so
Power = torque*cadence to a constant.

This applies both at the cranks and at the hub on the wheel. The gearing system can trade out torque for cadence, but power will be the same under the came conditions (speed, incline, etc) regardless of cadence.

operator 12-15-07 03:10 PM


Originally Posted by Wesmo15 (Post 5814156)
Power is also going to depend on the weight of the rider (ie. when going uphill). We did power tests in physics, and this simplified explanation of power would only apply in a setting with zero friction from both rolling resistance and wind.

So what you're saying is watt/kg is important.

...

brianappleby 12-15-07 04:25 PM

on a hill, watts/kg are going to determine your steady state speed.

On a flat, watts/frontal area are what matters

For sprinting, it's some combo of the 2, but it's mostly positioning.

This is true even with air resistance. Generally it's safe to ignore rolling resistance in this type of discussion, because it's minimal (i would say negligible) compared to air resistance at high speeds, and it's again minimal compared to the force of gravity at low speeds.

Generally a heavier rider will have a higher overall power output, but a lower power/weight ratio.

This is itself a gross oversimplification, and outside of the professional peloton, it's mostly wrong.

asgelle 12-15-07 04:49 PM


Originally Posted by brianappleby (Post 5815349)
Generally it's safe to ignore rolling resistance in this type of discussion, because it's minimal (i would say negligible) compared to air resistance at high speeds, and it's again minimal compared to the force of gravity at low speeds.

~13% of total power at 25 mph for a 75 kg bike + rider on level ground
~7% at 250 W for the same rider up a 5% grade. (per analyticcycling.com)

uberclkgtr 12-15-07 06:41 PM


Originally Posted by asgelle (Post 5815476)
~13% of total power at 25 mph for a 75 kg bike + rider on level ground
~7% at 25 W for the same rider up a 5% grade. (per analyticcycling.com)

yes, rolling resistance is not negligible, as you indicate. it takes significantly more power for me to average 25mph on a still day than it does to average 15mph into a 10mph headwind.

Brian Ratliff 12-15-07 06:53 PM

I'm not sure what there is to discuss.

FWIW, this is how I do trainer and roller interval sessions. I don't use a heart rate monitor or powermeter or anything, I just pick a cadence and gearing that has me feeling the right amount of pain, then use the speedometer to maintain wheel speed, and therefore, my power output, constant over the interval. If my speed starts drooping, then I know that my power output is drooping, and I need to step it up.

haimtoeg 12-15-07 07:04 PM

Your power does not vary with the gear if your speed remains constant.

What varies with gear is the force and the distance equation as jayhawken said. The higher the gear, the more you need more force over less distance, with distance being the circumference of your pedal stroke.

asgelle 12-15-07 07:07 PM


Originally Posted by haimtoeg (Post 5816194)
Your power does not vary with the gear if your speed remains constant.

Of course it does. Drive train losses are proportional to the tension in the chain and diameter of the sprockets. Now, how much of a change that makes is another story.

operator 12-15-07 07:19 PM


Originally Posted by asgelle (Post 5816203)
Of course it does. Drive train losses are proportional to the tension in the chain and diameter of the sprockets. Now, how much of a change that makes is another story.

Next to nothing unless it's so tight that your derailleur breaks off. All you need is a fixed gear and moving the axle/back forth to see the effects yourself.

asgelle 12-15-07 07:32 PM


Originally Posted by operator (Post 5816267)
Next to nothing ...

You mean like rolling resistance? Put a number to it.

waterrockets 12-15-07 10:52 PM


Originally Posted by brianappleby (Post 5815349)
Generally it's safe to ignore rolling resistance in this type of discussion, because it's minimal (i would say negligible) compared to air resistance at high speeds, and it's again minimal compared to the force of gravity at low speeds.

Well, as a guy who drafts trucks at 40mph a couple times/week, I'll tell you that rolling resistance can be quite substantial. With pretty much no air hitting me, I can't hold 40mph on the flats for longer than maybe five minutes (368W 5-minute power).


Originally Posted by asgelle
Of course it does. Drive train losses are proportional to the tension in the chain and diameter of the sprockets. Now, how much of a change that makes is another story.

This is true only for measured power, if it's measured at the hub. Your actual power at the cleat doesn't change with gearing.

asgelle 12-16-07 08:03 AM


Originally Posted by waterrockets (Post 5817217)
This is true only for measured power, if it's measured at the hub. Your actual power at the cleat doesn't change with gearing.

Look at the original statement. Speed remains constant; therefore, power to the road is constant. Since drive train losses change (slightly) with gearing, the only way for speed to stay constant is for power from the rider to change. This has nothing to with how power is measured or even that it be measured at all.

BillyD 12-16-07 10:11 AM

I'm too tired for this.


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