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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. |
whats there discuss. this is obvious.
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When I first read that, it seemed more than just oversimplified, but rather entirely wrong unless I'm missing something.
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How do powermeters measure power? Dont they measure torque and use that? Or do they know what gear your in?
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Pt has a torque tube in the hub. as well as the Ergemo I believe has one in the BB.
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measure by strain
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my PT had a-torque tube, ergemo i'm not familiar with
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PT measures torque and wheel speed to give you force and distance per time. From that you can calculate work and power.
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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. |
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. |
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. 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. |
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.
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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. |
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.
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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. |
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.
~7% at 250 W for the same rider up a 5% grade. (per analyticcycling.com) |
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) |
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. |
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. |
Originally Posted by haimtoeg
(Post 5816194)
Your power does not vary with the gear if your speed remains constant.
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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.
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Originally Posted by operator
(Post 5816267)
Next to nothing ...
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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.
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.
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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.
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I'm too tired for this.
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