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.

 

[EnglishMuffin]

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.

 

[GregLocock]

You have forgotten the angular momentum of the wheels.



Cheers

Greg Locock

 

[MikeyP]

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

 

[electricpete]

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.

 

[electricpete]

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.

 

[GregLocock]

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

 

[EnglishMuffin]

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.

 

[GregLocock]

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

 

[EnglishMuffin]

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.

 

[GregLocock]

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

 

[EnglishMuffin]

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.

 

[Vectra01]

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 (

http://www.cyclingplus.co.uk/forum)

and hope that a consensus will be reached to produce the
definitive answer.

 

[electricpete]

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.

 

[EnglishMuffin]

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.

 

[GregLocock]

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

 

[SKillen]

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.

 

[electricpete]

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.

 

[electricpete]

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

 

[SKillen]

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