Bicycle wheel inertia

[Vectra01]

There is a saying in cycling circles that 10g off the weight of a wheel is equivalent to 100g off the frame.
Assuming you add the 10g you deduct from the wheels to the frame so the total mass stays the same,
is the statement true? In other words does the total inertia of the complete cycle decrease with
lighter wheels, given the same total weight. What is the maths of the problem?
Typical values are:
frame and components (inc wheels) 12-14kg,
light weight wheels (inc tyres etc) 3.5 kg,
heavyish wheels (inc tyres) 4.5 kg.
Ignore rider weight and rolling resistance etc.

 

[SKillen]

Sorry typo * between I and W terms should be +
sk

 

[SKillen]

Please disregard last comment.

 

[SKillen]

OK pulled the book out and I think we're both right - kinda:

You've modeled the wheel as a thin shell therfore I=mr^2
I modeled it as a thin disc with so ariving at I=mr^2/2
Where what may be more appropriate but overcomplicating things for the purpose of proving (or disproving as the case may be) the theorem is to consider the wheel as a hollow cylinder therfore using I=m/2*(ro^2-ri^2) where ro is outside radius and ri is inside radius.

It is of course posible to get an acurate "I" through considering each material density and it's geometry but in practice it could easier to use an experimental method such as Tri-Filar suspension.

Suggest we argee it is somewhere inbetween 2:1 and 1.5:1 depending on mass distribution...?

Regards
Sean.

 

[electricpete]

Good point.

We assumed worst case mass at the extreme radius which exaggerates the rotational inertia.

You assumed mass spread uniformly over disk which underestimates the rotational inertia.

Real answer is somewhere between (as you say 1.5-2.0)

 

[EnglishMuffin]

Yes, that is correct - I believe I made the point that the 2:1 result stemmed from the approximate assumption that all the wheel mass was concentrated in the rim in my first post, and that the 2:1 was consequently an upper limit in my third. I think it's a reasonable assumption, but my guess is that the truth is probably about 1.9:1.

 

[TMSengineer]

Hey guys,

As most people realize on automotive wheels, the more total mass reduced and the more moved toward the center the better. Our race team is going through wheel options and my boss would like to see "real" numbers, not just "engineering" numbers. I have a short program written to find the total KE of the wheel based on your former replies and my own research. Is there a way to convert from KE to show a difference in acceleration or force to give him an idea of how much a difference more expensive and lighter wheels will make? I should know this...but I guess it's been a while.

Thanks
Ben

 

[EnglishMuffin]

TMSengineer:
I fear all this talk about kinetic energy has confused you.
What it all amounts to is that the effective mass of the car including the wheels is increased by 4*I/r^2, where I is the moment of inertia of a wheel and r is its radius. Of course, this assumes that all the wheels are the same size, which on cars such as the C5 and C6 Corvette is not the case - but I expect you get the idea. You can use this modified mass to figure the actual acceleration, using whatever methods you currently employ. There is no relation between kinetic energy and acceleration, since kinetic energy depends on velocity relative to an inertial reference frame and consequently can have any arbitrary value one chooses depending on the choice of frame, but acceleration is absolute relative to such a frame.
I would like to think that engineering numbers are "real" most of the time, in the sense that they mean something. If they do not, then we are all going to be in big trouble sooner or later!

 

[cab1990]

Helloooo Guys,

I fear something is missing here in this thread. The missing multiplier is in the turning of rotating mass. I will not provide the calculations because it is late and I am lazy. I leave that to those who will come after.

Imagine if you will; you are holding a static wheel in your hand - horizontal from your body by a length of rod welded to the axle. Try to lift it over your head. No problem.

Now, imagine the same wheel rotating at 2000 rpm. Try to lift it over your head. You cannot. The speed, while not changing in magnitude, changes direction. This rotational speed, and moreover, the momentum, is a vector quantity and must be overcome when a changes in direction occur. The change of speed and momentum of the frame is minimal in comparison since it is only a sin change of the translational angle( I'm guessing, 1/10th..because the the men who have the real experience know). Therein lies the saying "10g off the weight of a wheel is equivalent to 100g off the frame"

cab

 

[EnglishMuffin]

Well, I for one have always been careful in this thread to refer specifically to linear acceleration of the bicycle, which depends only on the translational and rotational inertia, and should not be affected at all by the value of the instantaneous angular momentum. If the bicycle happens to be executing a gentle turn, there is a good possibility that the rate of free gyroscopic precession would closely match the forced precession rate dictated by the turning radius, so even in that case it is not necessarily obvious that the wheel inertia would have much noticeable effect on "feel". However, when a rider alternately applies his full weight alternately to the pedals, when hill climbing for example, and rocks the cycle rapidly from side to side to maintain his balance, there is certainly the possibility that he might notice the different gyroscopic resistive torques exhibited by wheels having disparate moments of inertia.

 

[electricpete]

EM is correct, the gyroscopic effect appears when you try to rotate the axis of rotation. But when you just move the axis parallel to the original axis, gyroscopic action is not important.

You should not see any extra resistance lifting the bicycle wheel over your head. But if you try to change the axis of rotation (ie by moving your left hand and not your right), you will see a force perpendicular to your applied force.

Eng-tips forums: The best place on the web for engineering discussions.

 

[EnglishMuffin]

On a somewhat off-topic general note, the effect of gyroscopic action is often exaggerated in simplistic explanations of bicycle behavior. In the case of a simple rolling wheel, it is of undoubted importance, since as the center of gravity of the wheel starts to fall, a gravitational torque is exerted which causes the wheel rotation axis to precess. This in turn causes the wheel to roll in a circle, and centrifugal force then exerts a torque which counteracts the gravitational torque, thus imparting stability. However, in the case of a bicycle, although the same effect occurs, it is not vital to it's stability. This was first conclusively demonstrated in the early 1970's by an Englishman (I forget his name), who constructed a bicycle which exhibited no gyroscopic behavior whatsoever. He did this by mounting contra-rotating wheels alongside the main ones, thus cancelling out all angular momentum. And he discovered that the bicycle was just about as easy to ride as before. It turns out that the kinematics of the front forks alone is such that the turning behavior of the front wheel when the bicycle leans over, mimics that of the gyroscopic effect, as one can observe with a completely stationary bicycle, or with a very slow moving one when pushing it with the saddle. That is the reason that the turning axis of the front forks has to be angled from the vertical. It is also interesting to note that this guy was a chemical engineer, not a mechanical one. It is also the case that I first had this explained to me by a guy in the 1960's who had attended a lecture on the subject while a student at Cambridge University, so this must have been understood in some circles in earlier times, at least in theory.