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  1. #1
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    Dumb (but legit) hypathetical Q about rotating weight

    I just wanted to ask a question about rotational weight. This is hypothetical so bare with me. Lets say we have 3 wheelsets with the same hubs and spokes:

    1) 26" wheels with 300gm rims and 800gm tires

    2) 650B wheels with 400gm rims and 700gm tires

    3) 29er wheels with 500gm rims and 600gm tires

    Would these all "spin up" the same given that the wheels all weigh about 1100gm?

    Anything else to be pointed out?

    Any and all comments welcome.

  2. #2
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    Give all these weigh the same you are really only changing one thing: The distance from the center of the wheel to it's outer most edge. That will ultimately affect the spin up. As you get father and farther from the center the rotating weight increases(as it spins), even if the original amount is the same...
    it's easier to imagine this if you goto extremes; think: 12" wheel to a 36" wheel.

  3. #3
    mnt bike laws of physics
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    I realize this is probably more than you want to know, but you are asking about the moment of inertia. Inertia is the property of matter that resists change. For a rotating body, inertia is now refered to as moment of inertia and it is: the sum of (the individual masses x their distances from the center of rotation^squared). So, is you have a tire, you can assume for simplicity, all the mass is concentrated at one distance from the center. For a 650b i'll assume the mass is 335mm from the center. So, in your example the moment of inertia is: 0.7 x 0.335^2 = 0.0786 kg-m^2
    The hub's mass is so close to the center in all 3 situations that it can be neglected from the calculation. Try it if you don't believe.

  4. #4
    nothing relevant to say
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    isn't that what I said...?

  5. #5
    Mr.650b - Mr.27-5
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    Cane Creek has done some interesting work with their wheels.

    Check out what the nipple weight alone does to the moment of inertia (MOI) at the rim when compared to CC's design which uses nipples at the hub.

    http://www.canecreek.com/track_wheels.html (scroll to the bottom).

    Cheers,

    KP

    YP, How did you come up with 335mm for the 650B wheel?
    ďThose that say it canít be done should get out of the way of those doing it.Ē

    Pacenti Cycle Design

  6. #6
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    So 650B rims with 740gm NeoMotos will make me look and feel faster? Cool. I need all the help I can get.

  7. #7
    www.derbyrims.com
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    Quote Originally Posted by yogiprophet
    I realize this is probably more than you want to know, but you are asking about the moment of inertia. Inertia is the property of matter that resists change. For a rotating body, inertia is now refered to as moment of inertia and it is: the sum of (the individual masses x their distances from the center of rotation^squared). So, is you have a tire, you can assume for simplicity, all the mass is concentrated at one distance from the center. For a 650b i'll assume the mass is 335mm from the center. So, in your example the moment of inertia is: 0.7 x 0.335^2 = 0.0786 kg-m^2
    The hub's mass is so close to the center in all 3 situations that it can be neglected from the calculation. Try it if you don't believe.
    26: 1100gm x (26.5/2)^2 = ??? gm-inches^2

    650b: 1100gm x (27.5.5/2)^2 = ??? gm-inches^2

    29: 1100gm x (29/2)^2 = ??? gm-inches^2

    It can be seen that as the wheel diameter increases the same weight spread around a larger circumference for each would be slower to accelerate by a square of the radius.

    However, to accelerate the bike you need to accelerate the axle. And you don't spin a 29'er or 650b wheel up as fast as a 26 wheel to obtain the same ground speed at the axle. The net of the above equations would need to reduce the wheel acceleration force by the difference in circumference.

    If I remember correctly the 650b is about 4.2% greater in circumference than a 26 wheelís, the 29 wheel is about 9.8% greater in circumference.

    So to accelerate the axle with different wheel sizes:
    26 : (??? gm-inches^2 ) * 1 = moment of inertia to accelerate the axle

    650b: (??? gm-inches^2 ) * 1 - 4.2% = moment of inertia to accelerate the axle

    29: (??? gm-inches^2 ) * 1 - 9.8% = moment of inertia to accelerate the axle

    The larger wheels still require more force to accelerate the axle speed.

    Yogi does that make sense?

  8. #8
    The Bubble Wrap Hysteria
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    You should be able to solve the problem and post your results.

    http://en.wikipedia.org/wiki/Moment_of_inertia

    http://en.wikipedia.org/wiki/Rotational_energy
    Last edited by mtnbiker4life; 10-29-2008 at 07:33 PM.

  9. #9
    mnt bike laws of physics
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    Quote Originally Posted by derby
    The larger wheels still require more force to accelerate the axle speed.

    Yogi does that make sense?
    I'll calculate the acceleration of a front wheel.

    acceleration at the outer most edge of the tire (IOW, the acceleration of the bike) = a
    forward force on the axle = F
    moment of inertia = I
    outer tire radius = r
    mass of the whole wheel + tire = m

    a = F/(translational inertia + rotational inertia)

    a = F/(m + I/r^2)

    Wait a minute!!! Do you see what I see? The acceleration of the front wheel/tire system basically does not rely on the radius of the wheel (remember I = m*r^2). The r in I is not exactly the same r in the formula (which as i stated, is the outer radius of the tire) but it is close enough to disregard. WOW! this is evidence that there is too much engineering in bicycle design and not enough physics. The following formula is not exact, but is a close approximation for the front wheel:

    a = F/2*m

    So, when comparing the acceleration of front wheel/tire systems, besides the amount of force you can apply to the axle via the torque that you apply to the pedals.....the only consideration is the mass of the wheel/tire system.
    Why is this so? Because a larger wheel has a higher axle which gives the force applied to the axle through the frame more torque – that takes care of one r. The other r is taken care of because a bigger wheel has to accelerate slower to get to the same speed as a smaller wheel. Get it!

    The rear wheel is torqued (instead of pushed along like the front wheel). Surely it must be radius dependent.

    If you use a larger rear cog to compensate for the larger rear wheel (or use a smaller front ring to create more force on the chain – note: not the same force as in the formulas), the torque at the rear axle will increase but the speed and the torque at the tire patch will be the same as a smaller wheel because of the larger radius wheel. With this larger torque (at the axle) that maintains the same speed as a smaller wheel, F is the same and so acceleration is the same as a smaller wheel if the mass is the same. Acceleration really only comes down to mass of the wheel/tire system whether for the front or rear wheel. Holy Sheite!

    Again, how is this so? The larger wheel does not have to have the same angular acceleration (dw/dt) as does a smaller wheel to achieve the same speed at the tire patch. Angular velocity (w) is also smaller for a larger wheel so even though its I is larger, its w is smaller at the same outer speed. Also, it takes a larger rear cog to maintain the same speed (or to accelerate it to the same outer speed) – with the same chain force – so the rider can apply more torque at the axle in getting the wheel to accelerate without losing speed to the smaller wheeled rider.

    Now, most 29er tires are heavier than their 26er brethren, so the 26er wheel will still in most cases out accelerate the 29er (especially since you have 2 wheels), but the additional mass is not that much, and the rolling resistance (when you do get up to speed) is going to be a little less for a bigger wheel. It is not all about acceleration. Personally, with the 29ers I’ve ridden, I didn’t notice any lag in acceleration, so these calculation are confirmation of my experiences. I had just written it off as being a bigger than average, strong rider, so I just had assumed it wasn’t noticeable for me.

    As you stated Derby, when calculating the acceleration for the wheel/bike system, you have to consider that for a larger radius wheel, dw/dt is smaller for the same translational acceleration of a smaller wheeled bike - and translational acceleration is our ultimate goal in mountain biking - not rotational acceleration. You need to include the rotational acceleration of the wheel/tire, but ultimately, the final answer to ones acceleration question is one of forward motion of the bike/rider system.

    NOTE: the approximation formulas assume that most of the mass is near the outer portion of the wheel/tire system (which it is). I am in no way saying that you can build a wheel however you want and as long as they weigh the same, they will accelerate the same. The same rules apply that it is better to focus the mass towards the center as best as can be done such as with Cane Creek using the spoke nipples at the hub.

    As far as Angular Momentum goes, if the mass is the same, so will be the angular momentum. A bigger wheel is rotating slower to maintain the same speed. The same argument applies here. The angular momentum of a larger wheel is larger only if the mass is larger, which it usually is.
    Last edited by yogiprophet; 10-30-2008 at 08:04 AM.

  10. #10
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    Quote Originally Posted by yogiprophet
    I'll calculate the acceleration of a front wheel.

    acceleration at the outer most edge of the tire (IOW, the acceleration of the bike) = a
    forward force on the axle = F
    moment of inertia = I
    outer tire radius = r
    mass of the whole wheel + tire = m

    a = translational acceleration + rotational acceleration

    a = F/m + F*r^2/I

    Wait a minute!!! Do you see what I see? The acceleration of the front wheel/tire system basically does not rely on the radius of the wheel (remember I = m*r^2). The r in I is not exactly the same r in the formula (which as i stated, is the outer radius of the tire) but it is close enough to disregard. WOW! this is evidence that there is too much engineering in bicycle design and not enough physics. The following formula is not exact, but is a close approximation for the front wheel:

    a = 2*F/m

    So, when comparing the acceleration of front wheel/tire systems, besides the amount of force you can apply to the axle via the torque that you apply to the pedals.....the only consideration is the mass of the wheel/tire system.
    Why is this so? Because a larger wheel has a higher axle which gives the force applied to the axle through the frame more torque – that takes care of one r. The other r is taken care of because a bigger wheel has to accelerate slower to get to the same speed as a smaller wheel. Get it!

    The rear wheel is torqued (instead of pushed along like the front wheel). Surely it must be radius dependent.

    T = torque at the rear axle (which is the force on the chain * the radius of the rear cog)

    a = r*T/I + T/r*m

    If you assume that the radius in I is the same as r (it is slightly smaller as stated in my previous post but using the full mass of the wheel system as in the above approximation, it balances out the approximation), the approximate acceleration of the rear wheel is:

    a = 2*T/r*m

    But, since T = F*r, the above formula does come out to be the same as for the front wheel: a = 2*F/m
    If you use a larger rear cog to compensate for the larger rear wheel (or use a smaller front ring to create more force on the chain – note: not the same force as in the formulas), the torque at the rear axle will increase but the speed and the torque at the tire patch will be the same as a smaller wheel because of the larger radius wheel. With this larger torque (at the axle) that maintains the same speed as a smaller wheel, F is the same and so acceleration is the same as a smaller wheel if the mass is the same. Acceleration really only comes down to mass of the wheel/tire system whether for the front or rear wheel. Holy Sheite!

    Again, how is this so? The larger wheel does not have to have the same angular acceleration (dw/dt) as does a smaller wheel to achieve the same speed at the tire patch. Angular velocity (w) is also smaller for a larger wheel so even though its I is larger, its w is smaller at the same outer speed. Also, it takes a larger rear cog to maintain the same speed (or to accelerate it to the same outer speed) – with the same chain force – so the rider can apply more torque at the axle in getting the wheel to accelerate without losing speed to the smaller wheeled rider.

    Now, most 29er tires are heavier than their 26er brethren, so the 26er wheel will still in most cases out accelerate the 29er (especially since you have 2 wheels), but the additional mass is not that much, and the rolling resistance (when you do get up to speed) is going to be a little less for a bigger wheel. It is not all about acceleration. Personally, with the 29ers I’ve ridden, I didn’t notice any lag in acceleration, so these calculation are confirmation of my experiences. I had just written it off as being a bigger than average, strong rider, so I just had assumed it wasn’t noticeable for me.

    As you stated Derby, when calculating the acceleration for the wheel/bike system, you have to consider that for a larger radius wheel, dw/dt is smaller for the same translational acceleration of a smaller wheeled bike - and translational acceleration is our ultimate goal in mountain biking - not rotational acceleration. You need to include the rotational acceleration of the wheel/tire, but ultimately, the final answer to ones acceleration question is one of forward motion of the bike/rider system.

    NOTE: the approximation formulas assume that most of the mass is near the outer portion of the wheel/tire system (which it is). I am in no way saying that you can build a wheel however you want and as long as they weigh the same, they will accelerate the same. The same rules apply that it is better to focus the mass towards the center as best as can be done such as with Cane Creek using the spoke nipples at the hub.

    As far as Angular Momentum goes, if the mass is the same, so will be the angular momentum. A bigger wheel is rotating slower to maintain the same speed. The same argument applies here. The angular momentum of a larger wheel is larger only if the mass is larger, which it usually is.
    Fantastic . Now I am going to go jab my eyes out with an ice pick.

    Too much thinky...not enough drinky.
    Extreme stationary biker.

  11. #11
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    Quote Originally Posted by Kirk Pacenti
    Cane Creek has done some interesting work with their wheels.

    Check out what the nipple weight alone does to the moment of inertia (MOI) at the rim when compared to CC's design which uses nipples at the hub.

    http://www.canecreek.com/track_wheels.html (scroll to the bottom).

    Comming from CC site (see link above) "Given a nipple weight of 0.27 grams, the effect on the wheel’s MOI would be the same as if we’d replaced them with nipples weighing 48.55 grams."

    So that would mean that if the hub & spokes weighed the same that a CC rim could weigh 1544.96g more (based on 32 spokes) than another rim and have the same MOI. Interisting! So does that mean a CC wheel could weight 1544.96g more than another build and still not require any more wattage to spin up or down? That's a LOT since my wheelset (not just wheel) if about that weight. Please correct me if I'm wrong!



    OK, maybe I see where I translated it wrong. Maybe they mean 8.64g (0.27x32spokes) compared to 48.55g (1.5171gx32 spokes). That sounds more realistic.

  12. #12
    www.derbyrims.com
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    Quote Originally Posted by derby
    The larger wheels still require more force to accelerate the axle speed.
    So I was wrong? (Not the first time!)

    Quote Originally Posted by yogiprophet
    Now, most 29er tires are heavier than their 26er brethren, so the 26er wheel will still in most cases out accelerate the 29er (especially since you have 2 wheels), but the additional mass is not that much, and the rolling resistance (when you do get up to speed) is going to be a little less for a bigger wheel. It is not all about acceleration. Personally, with the 29ers Iíve ridden, I didnít notice any lag in acceleration, so these calculation are confirmation of my experiences. I had just written it off as being a bigger than average, strong rider, so I just had assumed it wasnít noticeable for me.
    Thanks for the expertise!

    I guess I wasnít completely wrong because, even with gear ratio corrections to equalize the torque at the rear tire patch, most of what we feel as slower acceleration on a 29íer compared to smaller wheeled bikes is heavier tires/rims, and the larger (heavier) frame required for the larger wheels. The added overall bike weight (mass) adds up to a slower accelerating and slower handling feel than a smaller wheeled bike.

    Iím bookmarking your calculations!

  13. #13
    mnt bike laws of physics
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    Quote Originally Posted by derby
    So I was wrong? (Not the first time!)


    Thanks for the expertise!

    I guess I wasnít completely wrong because, even with gear ratio corrections to equalize the torque at the rear tire patch, most of what we feel as slower acceleration on a 29íer compared to smaller wheeled bikes is heavier tires/rims, and the larger (heavier) frame required for the larger wheels. The added overall bike weight (mass) adds up to a slower accelerating and slower handling feel than a smaller wheeled bike.

    Iím bookmarking your calculations!
    Yes and there is still the design issues associated with a 29er, and the wheels are only two parts of a very complex system.
    Note, I corrected the formulas. The 2 is in the denominator - not in the numerator. I will do the torque calculations later.

  14. #14
    NedwannaB
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    Best response yet!

    Quote Originally Posted by Wish I Were Riding
    So 650B rims with 740gm NeoMotos will make me look and feel faster? Cool. I need all the help I can get.
    Simple. Go Neo/Quasi Motos!

  15. #15
    Lionel Hutz, Esq.
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    Quote Originally Posted by yogiprophet
    I'll calculate the acceleration of a front wheel.

    acceleration at the outer most edge of the tire (IOW, the acceleration of the bike) = a
    forward force on the axle = F
    moment of inertia = I
    outer tire radius = r
    mass of the whole wheel + tire = m
    I'll preface this by saying that I stopped reading right about here because of my confusion. My question is about the aforementioned confusion.

    Would you mind including units of measure here? Are we using newtons or foot-pounds? Are we going with inches? Grams or pounds? Just so those of us following along at home can know. I realize that it all might cancel out in the end, I just want to make sure I'm plugging in the right numbers. NASA screwed up a Mars probe because they didn't get their units straight. I don't want to screw up my impressions of rotating weight. Thanks.
    2007 Trek Fuel EX 8
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  16. #16
    mnt bike laws of physics
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    Quote Originally Posted by Thirdrawn
    I'll preface this by saying that I stopped reading right about here because of my confusion. My question is about the aforementioned confusion.

    Would you mind including units of measure here? Are we using newtons or foot-pounds? Are we going with inches? Grams or pounds? Just so those of us following along at home can know. I realize that it all might cancel out in the end, I just want to make sure I'm plugging in the right numbers. NASA screwed up a Mars probe because they didn't get their units straight. I don't want to screw up my impressions of rotating weight. Thanks.
    You didn't get very far
    just fucin with ya.

    If you are going for metric (SI), use Newtons(N) for force (1N is = 1kg-m/s^2 - that is a newton is equal to one kilogram-meter per seconds squared). Newtons are measurements of force and a force is m*a - mass times acceleration - kg is a metric mass unit and acceleration is measured in meters per seconds squared. IOW, if you are accelerating, you are increasing your velocity (which is meters per second) so you are increasing your meters per second at some amount per second. I won't go into pounds and slugs.
    So for mass, use kg and for lengths use meters. You use these because the newton consists of kilograms and meters as I pointed out above. This keeps everything consistant.

    I should point out that the acceleration formulas do not consider rolling resistance - this would have to be exerimentally derived and would of course change for every situation.

    Why would you want to calculate this anyway? My purpose for deriving these formulas was for comparing different wheel sizes of similar type wheels. Just know that a wheel's mass is going to have about twice as much inertia (resistance to acceleration) as the rider or frame per unit of mass because it is having to rotate.

  17. #17
    Lionel Hutz, Esq.
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    Quote Originally Posted by yogiprophet
    Why would you want to calculate this anyway? My purpose for deriving these formulas was for comparing different wheel sizes of similar type wheels. Just know that a wheel's mass is going to have about twice as much inertia (resistance to acceleration) as the rider or frame per unit of mass because it is having to rotate.
    To my credit, I actually did read the whole thing. Interesting stuff. I just paused my first time through - wanted to make sure I knew what units to cancel out or square or whatever. Haven't really done much complex math since college. I thought I'd start out understanding the units I was working with before understanding the equations I'm plugging them into. Baby steps.

    I wanted to go through the numbers and see it for myself. I thought it was interesting to the point that I wanted to bring it up next time I'm around some other riders. I wanted to make sure I understood it enough to work it out before I tried to explain it to someone else.
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