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Bicycle wheel inertia

Bicycle wheel inertia

Bicycle wheel inertia

(OP)
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.

RE: Bicycle wheel inertia

From the point of view of linear acceleration, as far as I can see the effective translational inertia of the bicycle should be Mc + 2*Mw where Mc is the mass of the cycle without the wheels and Mw is the combined mass of both wheels. This assumes that the wheel mass is totally concentrated in the rim. It would mean that 10g off the mass of a single wheel would be equivalent to 20g off the frame. So I can't see where the 10:1 ratio comes from, but maybe I've made a mistake and overlooked something, which wouldn't be the first time.

RE: Bicycle wheel inertia

You have forgotten the angular momentum of the wheels.

Cheers

Greg Locock

RE: Bicycle wheel inertia

I was discussing something similar with one of my graduate students who is a mountain bike champion. When you get big air over a jump, can you extend your flight by hitting the brakes hard in mid air?

M

RE: Bicycle wheel inertia

It seems like it should be an easy problem.

But when I try to work it out...I get same answer as EM, even considering angular momentum.

Let's say there is mass M at center of the wheel and mass m rotating at radius r.

Mass M at Center moves at v.  
Mass m at radisu r moves at radian speed w=v/r.

Make the accelerating power of both masses the same (equivalent), and solve for M in terms of m.

F * v = w * T
M*dv/dt * v = J * dw/dt*w

Substitute:
J = m*r^2
w = v/r

M*dv/dt * v = m*r^2 * (dv/dt)/r * v/r = m*dv/dt
M=m

Seems m has same effect as M when m=M.

Or perhaps m has to be counted twice... once for it's rotational acceleration and once for its linear acceleration?

My head hurts. I'm sure I'm missing something. Someone else please help with this simple (?) problem.

RE: Bicycle wheel inertia

I re-read EnglishMuffin's post again and I see he concludes the same as me:  A mass at the radius of the wheel has twice as much effective inertia as a mass at the center.

Like EM I have been wrong many times before and waiting any other explanation.

RE: Bicycle wheel inertia

You get the double whammy from the rotating inertia, which is additional to the linear inertia.

As usual it is simpler to consider the energy equation

the total kinetic energy of the bike is 1/2*M*v^2+1/2*I*w^2

M is the total mass, I is the total rotational inertia of the wheels and w is their angular velocity=v*r

MikeyP Applying the brakes in mid flight will tend to rotate the bike nose down, but will not change the path of the centre of gravity of the complete system, as all forces are internally resolved.

Cheers

Greg Locock

RE: Bicycle wheel inertia

Greg Locock : I did not "forget" the angular momentum of the wheels (That's where the 2 comes from)!. However, as I implied, I assumed the moment of inertia for a wheel was Mw*r^2. Please explain on that basis why you disagree with my equation.

RE: Bicycle wheel inertia

Ooh. Ah, I misread your meaning of Mw, as being the mass of one wheel. What threw me was when you said translational inertia, strictly speaking when you add the rotational component in it is more descriptive to call it referred (or effective) translational inertia.

... and I got the angular velocity wrong in my post. w=v/r

Sorry about that.

Cheers

Greg Locock

RE: Bicycle wheel inertia

GregLocock :Saright! It's refreshing to learn that even the oft celebrated tipster of the week can be wrong sometimes! And I did call it "effective translational inertia"! It still doesn't explain where the 10:1 originated, particularly since the 2:1 is an upper limit, as not all of the wheel mass is concentrated in the rim.

RE: Bicycle wheel inertia

So you did. Ack even the corrections are incorrect!

I'd guess the 10 is an exaggeration, possibly based on comparing a bad steel wheel with a good aluminium one. The subjective difference is quite spectacular, yet the difference in weight is probably not all that great. The better wheel is better mostly because it is stiffer, I suspect.



Cheers

Greg Locock

RE: Bicycle wheel inertia

Well, many years ago, when I used to ride a bike a lot, I got a pair of aluminum wheels (had them made up specially, and they even had aluminum hubs). I remember running the numbers at the time and concluding that they wouldn't make any spectacular improvements to the acceleration. But I really liked them - the difference in "feel" was very marked - maybe something to do with the unsprung weight and the springiness of the front forks. Also, the lower gyroscopic moment might give them a more lively feel. But everyone warned me against them because of their susceptibility to damage (running into curbs etc). As far as I know, that's the only disadvantage. I would imagine that the Tour de France guys all use aluminum wheels since they have instant replacements.

RE: Bicycle wheel inertia

(OP)
Thanks for the feedback guys.
I'm not a mechanical engineer and thought it would be better
to get the answer from the experts. However, I have put
together my own interpretation and will post it on the
cycling forum I normally use (www.cyclingplus.co.uk/forum)
and hope that a consensus will be reached to produce the
definitive answer.

RE: Bicycle wheel inertia

Just to clarify, it appears there is consensus among all responders here that the effective inertia of mass on the rim of the wheel is twice as much as mass not located on the wheel.

RE: Bicycle wheel inertia

If you mean in regard to linear acceleration, then as far as I personally am concerned - yes. But as GregLocock says, and my experience also indicates, there seems to be a very marked difference in feel between aluminum and steel rims. And I think such subjective things, even if they can be shown so to be, are by no means insignificant, especially in competition situations. I'm so out of date on this stuff - I expect all "serious" racing bikes have aluminum rims these days, or carbon fibre or something.

RE: Bicycle wheel inertia

No steel, mostly aluminium deep V rims, some carbon fibre, on reasonably serious roadbikes.

Mountain bikes are off in their own little world, but again, no steel.

Cheers

Greg Locock

RE: Bicycle wheel inertia

I didn't quite follow as to how the figure of 2:1 was arrived at so had a go at the Energy method and came up with a slightly different result:

The change in kinetic energy of the system due to a change in wheel mass "m" at radius "r" on a wheel with a radius of "R" will be equal to: m+(m*R^2)/(2*r^2).

So if the mass is added/subtracted at maximum possible radius, that is R=r, then this will be the same as adding/subtracting 1.5 times the mass to/from the frame.

So the figure of 10:1 is way off!! even more so if the weight is lost from closser to the hub.

Regards,
Sean.

Method:

let mf=mass of frame
let v=velocity
KE frame =(mf*v^2)/2

let inertia of mass a mass at radius "r" be (m*r^2)/2
let angular velocity of wheel radius R be w=v/R
so rotational component of wheel KE is:
KE=(I*w^2)/2
and translational component of wheel KE
KE=(m*v^2)/2

Conservation of energy:
(mf*v^2)/2=((m*r^2)/2*(v/R)^2)/2+(m*v^2)/2
which is solved for mf:
mf=m+(m*R^2)/(2*r^2) as in text above.

RE: Bicycle wheel inertia

I don't understand the statements:

"Conservation of energy:
(mf*v^2)/2=((m*r^2)/2*(v/R)^2)/2+(m*v^2)/2"

I think you are saying mf is the equivalent mass whose linear kinetic energy (LHS) is same as the total linear plus rotational kinetic energy (RHS).  

The RHS kinetic energy is correct, but where did you come up with the rotational kinetic energy ((m*r^2)/2*(v/R)^2)/2

You have two factors of 2 in the denominator and you only need one. If you get rid of one your result will agree and give mf = 2*m for the case r=R.

RE: Bicycle wheel inertia

Using the corrected term for rotational kinetic energy:

Conservation of energy:
(mf*v^2)/2=((m*r^2)*(v/R)^2)/2+(m*v^2)/2

mf = m*(r/R)^2 + m

mf = 2m for r=R

RE: Bicycle wheel inertia

Think we're disagreeing on value for Inertia of a wheel, from memory I though mass moment of inertia of a solid disc about it's origin was  m*r^2/2, memory could be failing have to check that...?
Rgds
Sean

RE: Bicycle wheel inertia

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

RE: Bicycle wheel inertia

Please disregard last comment.

RE: Bicycle wheel inertia

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.

RE: Bicycle wheel inertia

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)

RE: Bicycle wheel inertia

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.

RE: Bicycle wheel inertia

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

RE: Bicycle wheel inertia

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!

RE: Bicycle wheel inertia

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

RE: Bicycle wheel inertia

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.

RE: Bicycle wheel inertia

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.

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RE: Bicycle wheel inertia

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.

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