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  1. #1
    Senior Member Speedo's Avatar
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    Rotating vs non-rotating mass.

    The issue of rotating vs. non-rotating mass came up in the Downtube thread. I posted about it in the Swift thread once. It's an interesting topic. I haven't found a thread devoted to it; why not in Folding Bikes?

    I looked at this when James_Swift quoted the old saw about a pound of rotating weight being the same as a pound on the bike. I wasn't sure if I believed it, and did the following calculation:

    The energy required to raise a mass M to a speed V is the familiar:

    Et=0.5*M*V^2

    The energy required to rotate an object with moment of inertia I to a rotaional rate omega is:

    Er=0.5*I*omega^2

    If the mass M is distributed at the radius R, the the moment of inertia is:

    I=M*R^2

    For a wheel moving forward at speed V, the rotation rate is:

    omega=V/R

    so Er=0.5*M*R^2*V^2/R^2=0.5*M*V^2

    So a non-rotating mass requires energy

    Et=0.5*M*V^2

    While a rotating mass requires both the translational energy and the rotational energy to get up to speed.

    Et+Er=0.5*M*V^2+0.5*M*V^2 = M*V^2; twice the non-rotating energy.

    So, there is some truth to the pound of rotating = 2 pounds on the bike. I was a little surprised at the time that the result is independent of the radius of the wheel. Prior to the calculation I would have guessed that small wheels get a bit of a free ride.

    This result only applies in accelerating the bike to speed. Once at speed, on a level road, you have to apply a torque to overcome the friction in the bearings (Edit: and rolling resistance and aerodynamic drag). That torque is not a function of the rotating mass. (Edit: What ultimatly limits your speed is mostly aerodynamic drag.)

    So, while I agree in general that lighter is better than heavier, unless quick acceleration is critical for you, there's no particular reason to fuss over rotating vs non-rotating mass.

    I know that there are other opinions out there!

    Speedo
    Last edited by Speedo; 02-10-07 at 07:10 AM.

  2. #2
    Radfahrer Rincewind8's Avatar
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    Quote Originally Posted by Speedo
    I was a little surprised at the time that the result is independent of the radius of the wheel. Prior to the calculation I would have guessed that small wheels get a bit of a free ride.
    The difference usually is that larger diameter wheels (and tires and tubes) are heavier. But in the end they have some drawbacks that make the difference negligible.

    See:
    http://hea-www.harvard.edu/~fine/opi...wheelsize.html
    TH 1.81 (133kg*62)

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    hmmm, omega=V/(2*pi*R), no? So it is much less than double the energy: about 108% rather than 200%. Also, you have to remember that it's only 108% of the energy for the wheel, but the energy put into the wheel is only part of the total energy required to move the bike.

    Furthermore, all else being equal (tire width, spoke material, etc) the smaller wheel will have a smaller mass and thus a smaller moment of inertia. Since most of the rotating mass is due to the tire, let's assume the entire rotating mass is cocentrated at the circumference. Let the mass per unit circumference be m0, then the total rotational mass is 2*pi*R*m0.

    Combining all of the above refinements we obtain the following expression for the energy required to move the wheel:
    Ew=(R*m0*V^2)*pi+(R*m0*V^2)/4*pi

    and for the entire bike:
    E=Ef+2*Ew=(M/2+R*m0*2*pi+R*m0/2*pi)*V^2, where
    R is the radius of the wheels,
    2*pi*R*m0 is the weight of the spokes, tires, and rims of each wheel,
    and M is the weight of the rest of the components (hubs, frame, etc)

    Anyone want to plug some realistic numbers into this equation?
    Last edited by makeinu; 02-09-07 at 01:07 PM.

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    Quote Originally Posted by Rincewind8
    The difference usually is that larger diameter wheels (and tires and tubes) are heavier. But in the end they have some drawbacks that make the difference negligible.

    See:
    http://hea-www.harvard.edu/~fine/opinions/wheelsize.html
    Thanks for the link. He derived an equation similar to mine and when he plugged in some realistic numbers he calculated that less than a 1% difference in energy can be achieved due to shrinking the wheels. So I guess that settles it.

  5. #5
    Senior Member Speedo's Avatar
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    Quote Originally Posted by makeinu
    hmmm, omega=V/(2*pi*R), no?
    No. omega=V/R. Let T be the period of rotation, then omega = 2*pi/T. V=2*pi*R/T, divide both sides of the equation by R, and you get V/R = 2*pi/T = omega.

    Quote Originally Posted by makeinu
    Also, you have to remember that it's only 108% (edit 200%)of the energy for the wheel, but the energy put into the wheel is only part of the total energy required to move the bike.
    I haven't forgotten that. The analysis seeks to give you a feel for the impact of rotating vs non-rotating mass. Maybe we are agreeing that the mass of the complete package, your M, which really has to include the bike and the payload (rider and any junk they are carrying) swamps everything else.

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    Quote Originally Posted by Speedo
    No. omega=V/R. Let T be the period of rotation, then omega = 2*pi/T. V=2*pi*R/T, divide both sides of the equation by R, and you get V/R = 2*pi/T = omega.



    I haven't forgotten that. The analysis seeks to give you a feel for the impact of rotating vs non-rotating mass. Maybe we are agreeing that the mass of the complete package, your M, which really has to include the bike and the payload (rider and any junk they are carrying) swamps everything else.
    Ok.

  7. #7
    jur
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    *applauds*

    I LOVE it when myth gets debunked by solid reasoning and even better, by math proof.

    SO the equations show that to get up to speed, the approx. 1.5-2kg of wheel mass would be equivalent to 3-4kg of "plain" mass. That still is perhaps only 3-4% of total bike+rider mass, so unless you are racing, that won't make any difference. And even when you're racing, hardly any time is spent accelerating. When Robbie McEwen dashes for the finish line at the end, then it would come into play, but perhaps a stronger wheel would be more important to withstand those huge torques that the back wheel has to withstand. I remember from the 2005 TdF that Rasmussen (?) on the last TT day wrecked back wheel after back wheel trying to get up to speed after taking a tumble.
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    jur
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    Further to this, there are those who have somehow been convinced that riding uphill, rotating mass has to be "lifted twice" and therefore is doubly important to get wheel mass down. I found that little gem at the Audax Australia site.

    I suppose those people don't ride their trainers on anything less than a concrete floor for fear of crashing through the floor when spinning that wheel uber-fast.
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  9. #9
    jur
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    From the DT thread:
    Quote Originally Posted by makeinu
    Well, the spinning wheel does affect translational motion because the bicycle (by its very purpose) couples the rotational motion of the wheel to the translational motion of the bike. I know you want to consider the physics of the translational motion by itself, but the translational motion does not exist by itself with a bicycle.
    You are quite correct; however it is possible (and in fact simpler) to treat the 2 different kind of motions entirely separately and add the effects at the end, like in the OP. E.g., you can picture a moving bike as a bike with stationary wheels sliding along, plus a stationary bike with spinning wheels.

    Quote Originally Posted by makeinu
    Also, I think it is a little misleading to say that only the acceleration is affected by the rotational mass. In an ideal physical sense this is true, but in an ideal physical sense all the energy put into the pedals will result in acceleration. However, the bike will slow to a halt if the rider stops pedaling. So it is necessary to "reaccelerate" the wheels to compensate for deceleration due to other factors. I'm pretty sure the smaller wheels will increase the efficiency of this process, even though the actual speed of the bike is not increasing, but I'm not that kind of engineer, so perhaps you can clarify.
    Only a tiny tiny bit misleading: If a wheel rotational speed is constant, then there is no additional effect and I am 100% correct. However, as you pedal, the torque is not constant so there is in fact a small variation in speed, which is a small positive followed by a small negative amount of acceleration. This effect is larger when you ride up a very steep hill. So in that case I am indeed a tiny bit misleading.

    I suppose this could be the reason why spinning with as constant a torque as possible is harped on a being best - you minimise those tiny accelerations.
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  10. #10
    Senior Member Speedo's Avatar
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    Quote Originally Posted by jur
    *applauds*
    My blushes, thank you.

    Quote Originally Posted by jur
    ... so unless you are racing, that won't make any difference. And even when you're racing, hardly any time is spent accelerating. When Robbie McEwen dashes for the finish line at the end, then it would come into play, ...
    I wouldn't go quite that far. Even people who are recreational riders and like to ride fast in pace lines may care. The ability to jump fast can make the difference between hanging on to the wheel in front of you, and dropping off the back. I have plenty of experience being dropped off the back!

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    Small Dia wheels rotate faster, larger have more mass at a further distance. So the angular momentum comes out almost the same.
    2000 Montague CX, I do not recommend it, but still ride it.
    Strida 3, I recommend it for rides < 10mi wo steep hills.
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    Wieleder CARiBIKE (folding), decent frame.

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    You know, I've seen endless thrashes about this, and I really don't think it matters. You're not accelerating wheels, you're accelerating (and decelerating) the combined mass of bike and rider, of which the wheels represent a fairly small fraction of both the total mass and the rotating mass.

    In terms of how fast you go for the amount of energy you put in, the only thing that really matters is the energy that's lost to friction of various kinds. So... forget masses and distances, let's see someone prove the issue in terms of thermodynamics.

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    So are you guys saying I would be wasting 400 bucks on a set of Velocity Tracian wheels, as far as improving my climbing and average speed while riding my RANS Stratus, or any other bike for that matter?

    Thanks guys, you saved me 400 bucks. That will buy me a DT 8H.

    Al

  14. #14
    Neat - w/ ice on the side dalmore's Avatar
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    As someone who flunked more college calculus classes than I passed (4 to 3), I feel like I'm bringing a knife to a gunfight.

    Let's try looking at this from a different perspective. Consider that all the power to propel a bike forward is applied at the hub of the wheel. This power is transfered by the spokes through the tire to the road surface. Essentially we are looking at a long lever (spokes) with the fulcrum at one end (hub) and the load at the other end (tire).

    The mass on that lever and it's distribution along the length of that lever will have a direct influence on the power transferred from your feet to the road service. More mass - particularly mass nearer the load end (rim/tire) of the lever - will require more power to move the lever. The power required to move the lever is power that is not transferred to the road surface and used to propel the bike forward. So a heavier wheel and tire will be diverting power from the process of moving the bike forward.

    Having documented my lack of math skills, I'll leave it to someone else to do the math.
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  15. #15
    Senior Member Speedo's Avatar
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    Quote Originally Posted by dalmore
    ... So a heavier wheel and tire will be diverting power from the process of moving the bike forward.

    Having documented my lack of math skills, I'll leave it to someone else to do the math.
    Sort of, but not quite. Initially some of the engine's (that's you) power will have to do toward spinning the wheel up, but at speed you only have to add power to overcome friction, rolling resistance and aero drag.

    Here's a thought experiment for you. What if, instead of really light wheels, you have really, really, really heavy wheels. Titanium spokes with a rim and rolling surface make of spent uranium. These wheels are so heavy you can't lift them, and can barely provide the energy to start them rolling. Now, the claim above is that the problem with heavy wheels is that they slow your acceleration, but don't appreciably limit your speed. What if we were able to magically skip the acceleration step. Somehow you get started at 20 MPH on a level road with these mondo heavy wheels. What will happen?

    If the claim is true that the heavy wheels slow you down, then you would expect that you would slow and stop immediately. But these wheels act like flywheels, they are really energy storage devices. Even just coasting the energy stored in the rotational momentum of the wheel will easily overcome the rolling resistance, bearing friction, and aero drag. You will gradually slow down, but it would take much much longer to slow down than it would for light wheels, which would have stored less energy.

    I hate to do this to you, but here's an expression to expand your science education:

    Lagrangian Mechanics

    In Lagrangian Mechanics you analyze the motion of a system by following the energy. You can think about what happens when you ride a bike by thinking about where the energy goes. The engine (that's still you) puts energy into the system. From a start (on level ground) that energy goes into translational kinetic energy, rotational kinetic energy, bearing friction (heat), rolling resistance (heat), aerodynamic drag (still more heat). The system will accelerate until the input (you) is balanced by the rest of the system. When you are no longer accelerating none of your input power is going into accelarating the system (duh). It is going into friction (heat), rolling resistance (heat) and aero drag (heat). Most of it will go to aero drag. The power that goes into overcoming aero drag is proportional to the cube of your speed(!!). You haven't lost the energy that went into accelerating the system. It is stored as translational kinetic energy, rotational kinetic energy (and if you climbed a hill, potential energy). If you start to coast then the stored energy, gets turned to heat, and you will slow down and eventually stop.

    The exclamation points on the aero drag comment are to point out that the cube of your speed is a very very steep curve. It's aero drag that will limit your speed on level ground. You want to go faster? Before you spend $400 on some fancy wheels, spend $40 on aero bars.

    Speedo
    Last edited by Speedo; 02-13-07 at 07:25 AM.

  16. #16
    Neat - w/ ice on the side dalmore's Avatar
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    Quote Originally Posted by Speedo
    Sort of, but not quite. Initially some of the engine's (that's you) power will have to do toward spinning the wheel up, but at speed you only have to add power to overcome friction, rolling resistance and aero drag.

    ...

    Before you spend $400 on some fancy wheels, spend $30 on aero bars.

    Speedo
    While I don't find any fault with the logic you present, it doesn't explain the physical observations I have made. Try this physical experiment for yourself. Get up to speed on your bike using any means you desire. Without changing your front chainring, shift to your largest rear cog and gauge how much resistance you find on the pedal. Then while still at speed and again without changing your front chainring, shift to your smallest rear cog and gauge how much resistance you find on the pedal. There was more resistance when you were on the smaller cog yes? Friction, aero drag and rolling resistance are practically unchanged. I argue this supports my theory of the wheel as a lever. (If you are going to argue that it's the effects of gearing - look at how gearing works and see if that brings right back to the lever.) The larger cog is applying the force is a position that is more advantageous effectively shortening the lever. So if the lever theory holds to this degree, why does it suddenly not apply when trying to calculate the effects of varying the load at the end of the lever - ie more mass at the rim and tire?

    As far as upgrades ... I for one can not magically reach 20 mph where the rotational mass of the tire become a flywheel and aerodynamic drag is my biggest enemy... I have to reach that 20 mph point first by exerting energy. And with the rolling terrain around here, I spend more time accelerating up hills than I do gliding down them in a tuck position. So for me, $400 on lighter wheels to get better acceleration might be a more cost effective upgrade than $30 on aero bars.
    Last edited by dalmore; 02-13-07 at 08:06 AM.
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  17. #17
    Senior Member Speedo's Avatar
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    Quote Originally Posted by dalmore
    Try this physical experiment for yourself. Get up to speed on your bike using any means you desire. Without changing your front chainring, shift to your largest rear cog and gauge how much resistance you find on the pedal. Then while still at speed and again without changing your front chainring, shift to your smallest rear cog and gauge how much resistance you find on the pedal. There was more resistance when you were on the smaller cog yes?
    Yes, you apply more torque, but at a lower rotation rate, so the power is the same. The energy you put into the bike is the same in either case. Your engine might run more efficiently at a particular torque/rate combination, so you will find you prefer a gear, but the energy you put in is the same in either case.

    The wheel is a "lever" as you put it. It factors into the gear sense that you experience. That's why Gear Inches includes the diameter of the wheel. But don't confuse the mechanical advantage of gearing with where the energy goes when you put it into the bike.

    Quote Originally Posted by dalmore
    As far as upgrades ... I for one can not magically reach 20 mph where the rotational mass of the tire become a flywheel and aerodynamic drag is my biggest enemy... I have to reach that 20 mph point first by exerting energy. And with the rolling terrain around here, I spend more time accelerating up hills than I do gliding down them in a tuck position. So for me, $400 on lighter wheels to get better acceleration might be a more cost effective upgrade than $30 on aero bars.
    When you go up a hill you, the engine, are increasing your potential energy. Light wheels, being lighter, are easier to carry up the hill. But as you crank up the hill, even the little changes in speed from the crank pulses are a subtle dance between the energy you put in, the energy stored as translational, and rotational kinetic energy, friction and drag, and potential energy. The energy that goes into spinning up a wheel, you get back; it is conserved.

    Look, I've always conceded that lighter is better. But the point is to understand what you are, and what you are not, buying. I'm someone who has bike priorities, and believes that money spent on bikes is money well spent. If you want to spend $400 ($500, $1000!) on wheels and your only reason is that they look cool and make you feel fast, that's a good enough reason in my book. If you have a set of modern alloy rim wheels, with good quality hubs then, upgrading for lightness is not going to dramatically improve your overall speed.

    Speedo

  18. #18
    Seņor Mambo
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    Quote Originally Posted by Speedo
    =The energy that goes into spinning up a wheel, you get back; it is conserved.
    I guess it depends how you define "conserved" because someone else has argued that this is not accurate. He says the bike is a conservative system, but muscles are not. Thus you lose energy every time you pedal; energy does not all of a sudden go back into your leg. The discussion is in terms of frame flex, but it does get to this issue in a roundabout way.

    http://bikeforums.net/showthread.php?t=268395

  19. #19
    Senior Member Speedo's Avatar
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    Quote Originally Posted by spambait11
    I guess it depends how you define "conserved" because someone else has argued that this is not accurate. He says the bike is a conservative system, but muscles are not. Thus you lose energy every time you pedal; energy does not all of a sudden go back into your leg. The discussion is in terms of frame flex, but it does get to this issue in a roundabout way.
    No, but the energy does go toward motion. See the example of the heavy flywheel wheel. The energy is stored in the wheel and goes toward overcoming friction, drag, and rolling resistance. In a normal bicycle wheel, the amount of energy stored is small, but that's the point.

    If you have to hit brakes, it will go toward heating the rims.

    Speedo
    Last edited by Speedo; 02-13-07 at 12:41 PM.

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    Quote Originally Posted by Speedo
    No, but the energy does go toward motion. See the example of the heavy flywheel wheel. The energy is stored in the wheel and goes toward overcoming friction, drag, and rolling resistance. In a normal bicycle wheel, the amount of energy stored is small, but that's the point.

    If you have to hit brakes, it will go toward heating the rims.

    Speedo
    Yeah, but the same holds true apart from the rotating wheel considerations. For example, you won't lose any energy riding a 100 pound bike instead of a 30 pound bike. However, a 100 pound bike is obviously undesirable.

    The important quantity here is the power required, not the energy.

  21. #21
    Senior Member Speedo's Avatar
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    Quote Originally Posted by makeinu
    Yeah, but the same holds true apart from the rotating wheel considerations. For example, you won't lose any energy riding a 100 pound bike instead of a 30 pound bike. However, a 100 pound bike is obviously undesirable.

    The important quantity here is the power required, not the energy.
    Looking at the energy is useful because, regardless of the absolute value, it reveals interrelationships between the mechanical components. I started down the energy path because some posters seemed to be arguing that rotating wheels were a constant sink of energy, and hence consumed an inordinate amount of the available power. Following the energy reveals that that is not the case.

    I'm always happy to concede that light is better than heavy. My only claims are that:

    1) When accelerating the bike it is, in fact, true that a pound in the wheels is like two on the bike.
    2) After the bike is up to speed, there is no continuing penalty due to the fact that the mass of the wheel is rotating.

    Speedo
    Last edited by Speedo; 02-13-07 at 07:10 PM.

  22. #22
    Neat - w/ ice on the side dalmore's Avatar
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    Quote Originally Posted by Speedo
    Looking at the energy is useful because, regardless of the absolute value, it reveals interrelationships between the mechanical components. I started down the energy path because some posters seemed to be arguing that rotating wheels were a constant sink of energy, and hence consumed an inordinate amount of the available power. Following the energy reveals that that is not the case.

    I'm always happy to concede that light is better than heavy. My only claims are that:

    1) When accelerating the bike it is, in fact, true that a pound in the wheels is like two on the bike.
    2) After the bike is up to speed, there is no continuing penalty due to the fact that the mass of the wheel is rotating.

    Speedo
    Sorry to have taken so long to reply - been busy. Looking at energy can also be misleading. Consider that whether a biker rides up a hill or pushes the bike up the hill, the energy is the same but the experience is quite different.

    I think the weight of wheels and tires have a constant effect on bikes that is greater than the effect that same amount of weight would have elsewhere on the bike. I tried to explain the idea with the example of a lever and failed. (I still maintain that's valid but I'll drop that for now.) Let me try again with a different aproach.

    We all agree that rotating mass has an effect on acceleration to a greater degree than non rotating mass. We all agree that the forces of friction (from various sources) and aerodynamic drag are acting to prohibit forward movement of the bike. These forces are constant - if you stop applying power, these forces cause a bike to slow down.

    Now consider that power from the biker is not being applied constantly like gravity or a rocket engine. Instead power is being applied in pulses. So the bike is in a constant cycle of acceleration and deceleration.

    Perhaps you think the pulses are so short that they don't matter. Well regardless of the cadence, a bike with a standard crank and plaform pedals is only under power about half the time. 1/4 of the time from the left foot pushing down on the pedal from about 90 to 180 degrees and 1/4 of the time from the right foot pushing down on the pedal from about 90 to 180. I think that does matter.
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  23. #23
    jur
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    Quote Originally Posted by dalmore
    Perhaps you think the pulses are so short that they don't matter. Well regardless of the cadence, a bike with a standard crank and plaform pedals is only under power about half the time. 1/4 of the time from the left foot pushing down on the pedal from about 90 to 180 degrees and 1/4 of the time from the right foot pushing down on the pedal from about 90 to 180. I think that does matter.
    I wonder if this doesn't come out in the wash: Yes you are correct that there is small variations in speed due to varying pedalling torque in a full revolution (see post #9); so a wheel does indeed have to be re-accelerated. However, a heavy wheel's bigger angular momentum will reduce the slowing by a small amount compared to a lighter wheel, so has to be re-accelerated by less.
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    Neat - w/ ice on the side dalmore's Avatar
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    Quote Originally Posted by jur
    I wonder if this doesn't come out in the wash: Yes you are correct that there is small variations in speed due to varying pedalling torque in a full revolution (see post #9); so a wheel does indeed have to be re-accelerated. However, a heavy wheel's bigger angular momentum will reduce the slowing by a small amount compared to a lighter wheel, so has to be re-accelerated by less.
    Perhaps but I don't think so. Angular momentum is considered to be a small percentage of the force at work here with most of the force at work being aero drag at speed. Since aero drag continues to get greater with more speed, I think any advantage a heavier wheel gains from angular momentum during the deceleration phase would get smaller and smaller as speed increases.
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    Quote Originally Posted by jur
    I wonder if this doesn't come out in the wash: Yes you are correct that there is small variations in speed due to varying pedalling torque in a full revolution (see post #9); so a wheel does indeed have to be re-accelerated. However, a heavy wheel's bigger angular momentum will reduce the slowing by a small amount compared to a lighter wheel, so has to be re-accelerated by less.
    Agree with all this except "re-accelerated less". In the case of pulses of speed as you piston crank up a hill, with each up pulse you accelerate the wheel adding energy, but as it slows that energy is given back. With each piston stroke you add add kinetic energy (translational and rotational) and as you rotate the pedal (without putting a lot of power in) preparing for the next big down stroke that kinetic energy is transformed into potential energy. i.e. you went a little further up the hill.

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