# Inertia and bicycle wheels related to wattage and time

by E190
Tags: bicycle, inertia, time, wattage, wheels
 P: 4 In the bicycle industry, there is a lot of marketing saying that rotational weight is more important than static weight when it comes to accelerations and competition. I understand I=mr^2 but how do you compare differences. Keeping the the hub/spokes constant, how much advantage would you get from removing 150 grams from each rim over a 2hr race. For this problem, r=31.1cm Here is what I "think" I know. It only counts for acceleration and has little to no affect on keeping a steady speed, so lets assume the race is constantly accelerating and decelerating like a criterium or mountain bike race. Am I wrong saying that watts = I/s^3 ? If that is the case, then the following problem would break down as follows I = 150g x 31.1^2 = 145,081.5g/cm^2 to convert to kg/m^2 you multiply by 10 = 1,450,815 kg/m^2 now over a 2hr race divide by 7200^3 Would that mean the savings would only be .00000388 Watts! And if so, is that per second savings or total savings over the entire race? Either way, it seems basically negligible. Or am I way off and did this wrong?
 Engineering Sci Advisor HW Helper Thanks P: 7,293 Most of yuur math is completely wrong, but the marketing guys are right to some extent. The translational kinetic energy of a mass on the rim (i.e. the energy if the wheel was sliding not rotating) = ##mv^2/2##. The rotational KE = ##I\omega^2/2##. ##I = mr^2## and ##\omega = v/r##, so the rotational KE also = ##mv^2/2##. So, in terms effort required to accelerate the bike, a mass on the wheel rim needs twice as much energy as the same mass on the bike frame. However this is mostly marketing hype, because a mass of 0.15kg is very small compared with the mass of the bike rider. The amount of energy to accelerate a 50kg cyclist compared with an mass of 0.15kg for the wheel rim would be 50/(2 x 0.15) = 167 times bigger. FWIW the idea behind this is much more important for the rotating parts of an engine in a race car or bike, because the RPM of the engine is much higher than RPM of the wheels.
 P: 4 Thanks for the help. I understand that accelerating the rider is much more work than accelerating the added wheel weight, how does the decrease in rim weight translate to watts and/or time savings. Lets say I can hold 250 watts over 40km. Using this website (bikecalculator.com) a 300 gram reduction in bike weight saves me only 3 seconds. How much additional time would that save me if I were to reduce that weight from the rim instead of the frame. In theory, this would also apply to shoes and pedals at the end of the crank arms. I am very interested in learning the math/physics behind this to see exactly how much it could save. Aerodynamics and drag are a big factor, but keeping that constant, I would like to learn how weight alone affects the outcome.
 Sci Advisor Thanks PF Gold P: 12,269 Inertia and bicycle wheels related to wattage and time Are there not many other factors involved here? If you reduce the thickness of your rim then won't it flex and cause you even more energy loss, at least a much as the possible difference in Kinetic Energy? In any case, you don't lose all that KE; it's returned when you slow down. Also, a 'special' thin will cost more for a start and then need replacing sooner as it will be more vulnerable. Unless you are v. rich, this is also a factor.
 P: 4 There are tons of factors here. But I am mainly interested in seeing if there is any real life measurable advantage to decreasing inertia. Keeping everything constant (stiffness, thickness, price, etc), is there an advantage if I build up a wheel using rim A at 545 grams, rim B at 425 grams, and rim C at 385 grams. Decreasing the overall weight by x will increase your power to weight ratio. Keeping power output the same, your time would decrease. But that is for static weight, is there a way to quantify weight savings in inertia over static weight. If the wheels are the same price, quality, stiffness, etc, how would going from a 545g rim to a 385g rim benefit me beyond a total static weight savings of 320g. I'm 61kg, bike is 9.9kg. At 250 watts my p/w ratio is 3.526 w/kg. If I get new rims my p/w ratio goes up to 3.542 w/kg. Is there a measurable increase by having the weight savings in inertia? And how do I go about calculating this? Seconds matter, it could mean the difference between a paycheck or a pat on the back.
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P: 7,293
 Quote by E190 Using this website (bikecalculator.com) a 300 gram reduction in bike weight saves me only 3 seconds. How much additional time would that save me if I were to reduce that weight from the rim instead of the frame.
So far as I can tell, "bikecalculator" doesn't include any acceleration effects. If that is correct, it's irrelevant to your question. the "3 seconds" presumably comes from lower rolling resistance of the tires, because of the smaller weight on them.

As a ballpark figure, your acceleration rate will increase by about 0.5%. In other words, if it took you 10 seconds to accelerate through some speed range, you will reduce that by about 0.05 seconds for the same effort.

The effect for pedal cranks etc will be smaller, in proportion to the gearing (i.e. the number of revs of the boke wheel for one rev of the pedals)

 In any case, you don't lose all that KE; it's returned when you slow down.
Returned to where? If cyclists can change the way their leg muscles work so they restore energy to the body through regenerative braking, then I'm seriously impresed. (And maybe the drug testing agencies ought to be told how they do it as well )
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P: 12,269
 Quote by AlephZero Returned to where? (i.e. K.E) If cyclists can change the way their leg muscles work so they restore energy to the body through regenerative braking, then I'm seriously impresed. (And maybe the drug testing agencies ought to be told how they do it as well )
The KE is returned in as far as the cyclist can put less energy into the last few metres of pedalling. The KE (the few mJ) offset some of the resistive losses when there is deceleration.