I interested in getting something of an idea about what difference there is between a 205/55-16 (O.D. 24.9") tire weighing 23 lbs. and a 205/50-16 (O.D. 24.1") tire weighing 20 lbs.

I was thinking I could estimate the radius at which the rotational mass was centered and multiply it by the weight. Perhaps someone here knows the correct way to go about this. If so, I'd appreciate the help.

 

Well, moment of inertia can be calculated using the integral of (r^2)*dm, of course that is obviously not practical for the real world unless you want to cut you wheel into little sections or something.

I might have something in my notes on ways to figure it out experimentally for a mass that isnt something simple like a disc or a ring, but I cant think of it off of my head. I bet DR. Obnxs or K-heuvo would know, they are like engineers or something. I'm only about to start my second year of ME... However if you can come up with a rig that can measure angular accleration and then apply a known torque force to your object then you can just for moment of inertia with T=IA (T:torque, I:moment of inertia, A:Angular acceleration),

 

I don't know if this is the correct way of doing it, but I'm a physicist by training, this is the sort of thinking we do:

A simple approximation would be an uniform annular disc for the side walls and a uniform band representing the treads.

A little googling tells me the moment of intertia of a disk is 1/2 MR^2, the mass of a disk is pi*p*R^2 (where p is the areal density if the material). So overall you have 1/2*pi*p*R^4.

The moment of inertia of the annulus is is the moment of inertia of a disk of the outer radius, minus the moment of interia of a disk of the inner radius. Which gives you: 1/2*pi*p*(Ro^4-Ri^4)

The moment of intertia of the treads is MR^2. But in this case, M is pi*Ro*pt*w (where pt is the areal density of the treads, assumed to be different from the sidewalls). (Edit not forgetting the width (w) term.)

The only thing you don't know in those equations is p and pt, however:

You also know that the mass of the tire is the mass of the sidewalls + the mass of the treads. If you assume the areal densitys are the same for both tires (ie the tire material is equally think in both), you have:

pi*p*(Ro1^2-Ri1^2) + 2*pi*Ro1*pt*w= m1
pi*p*(Ro2^2-Ri2^2) + 2*pi*Ro2*pt*w= m2

You have 2 unknowns and 2 equations you can solve for. Ri is the wheel diameter, Ro is the outer diameter.

Edit: Got my Ri and Ro terms the wrong way around. And remember the 2.

 

I can't make my idea work with the numbers given, but it does look like most of the mass of the tire is in the tread, not in the sidewall. So you could probably get a reasonable estimate by assuming all the mass of the tire is in the treads. In that case the the moments of inertia would be the outer diamter squared multiplied by the mass (in weird units). You can take the ratio of those so the weird units don't matter.

The larger tire has 23% more moment of inertia by that approximation.

 

This looks good to my semi trained eye. The reason I avoided doing this was because I was lazy but also because I wanted to avoid using the 1/2MR^2 which would not have appropriately approximated for the sidewall by itself. Your way get around this by removing the "air" and should work as RME (rotational moment of inertia) is additive.

 

I don't think you need to go into that much detail on trying to find the actual numbers...what your really looking for is the delta. You don't need densities of the material to do that.

There will be some assumptions to do it though. You can assume the tread will make up a certain percentage of the total weight of the tire (say 80 or 90%). You also need to assume the thickness of the tread will be the same between both tires (say .5"). The lip where it seals against the wheel will also be the same between both tires.

You know the OD and weights of the tires in question so you can use
I=1/2m(r1^2+r2^2) for the tread. r1 can be the OD/2 and r2 can be that value minus .5. The mass will be .8 or .9 (whatever you choose) * the weight of the tire.

The sidewalls will use the same equation except r1 is the OD/2-.5, r2 is the set radius of the lip, and the mass is .05 or .1 * the weight of the wheel. Whatever number you get for the sidewall multiply by 2 to get both sides.

 

Thanks all. This is a lot more complex than I was hoping. I was just looking for a rough idea. Why not just estimate that the mass of the tire is centered a specific radius -- say 1.5" less than tire O.D.?

 

You could do that. It's just too hard to tell how accurate any of this will be without knowing the weight cross section of a tire is. Using your assumptions, I get somewhere around a 15% reduction in rotational moment.

 

Thanks very much.

 

Robin and All,
The rotational moment of inertia is nothing more than a measure of the distribution of the mass about the axis of interest. To calculate it is not difficult, just tedious. In the ship model testing business we have to find this quantity for our ship models all of the time. There is a neat way to find it experimentally. Then you can do this for the wheels themselves, or just the tires. Then you can merely add up the components.
For the experiment you need two equal lengths of wire, a tape measure, a scale, and a stopwatch. Weigh the wheel and tire. Record as W. Run one piece of wire through a hub bolt hole and secure the end to the wheel. Repeat with the other wire on the opposite bolt hole (works well on the Mini because of the four bolt holes). Now suspend the wheel and tire from the overhead so that the hub axis is vertical and the wires are the same length. You need to measure the lengths of the wire (call the length L). Measure the distance between the wires at the hub (call it 2R and it should be the same as the distance between the wires at the top). Now grab the stopwatch . Twist the tire/wheel about the hub axis maybe 10 degrees. Let go, pick up the stop watch and time 10 oscillations. You want the wheel twisting about the hub axis with as little swing as you can get. It might take a few tries to get it right.
Now for the calculation! Divide the time on the stopwatch by the number of oscillations to get the average period of oscillation (the natural vibration period, T). The mass moment of inertia about the hub axis is then

I = (W x R^2 x T^2)/(4 x Pi^2 x L) (units are slugs-ft^2)

R = half of the distance 2R (in feet)
W = weight in lbs
T = average period in seconds
L = wire length in feet

We engineers usually end up with a shorthand version called radius of gyration (K, in feet) where
K = sqrt(I x g/ W)
The interesting thing is that the radius of gyration as a percentage of diameter will be pretty constant for object of similar construction even with different sizes. So you could estimate the effect of the different tire diameter using percentages.
This is the longest post I have ever tried. Sorry if I bored anyone - but this is kind of my area of expertise.

Cheers,
Greg

 

Good insight Greg. If there's one thing you learn as an Engineer, it's that you don't necessarily need to be the best at anything, you just have to know the right person to go to when the problem exists. And there's nobody better than the person who does it all the time...