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.
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
RE: Bicycle wheel inertia
Cheers
Greg Locock
RE: Bicycle wheel inertia
M
RE: Bicycle wheel inertia
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
Like EM I have been wrong many times before and waiting any other explanation.
RE: Bicycle wheel 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
RE: Bicycle wheel inertia
... and I got the angular velocity wrong in my post. w=v/r
Sorry about that.
Cheers
Greg Locock
RE: Bicycle wheel inertia
RE: Bicycle wheel inertia
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
RE: Bicycle wheel inertia
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
RE: Bicycle wheel inertia
RE: Bicycle wheel inertia
Mountain bikes are off in their own little world, but again, no steel.
Cheers
Greg Locock
RE: Bicycle wheel inertia
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
"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
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
Rgds
Sean
RE: Bicycle wheel inertia
sk
RE: Bicycle wheel inertia
RE: Bicycle wheel inertia
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
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
RE: Bicycle wheel inertia
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
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
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
RE: Bicycle wheel inertia
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