Does anyone know how heavy is the 16" OEM rim for the V6 Mustang. I have a friend who is going to sell me the 18" rim with the correct bolt pattern and offset. The rim weights around 24lbs. Do you think it's too heavy? I am afraid that it's going to be too heavy on the V6.

 

The weight is not an issue in this case but be careful not to increase tire diameter too much it will reduce performance.

kc

 

I am using Michellin Pilot Sport A/S. 235/50/18. Do you think it's okay? I am trying to keep the height the same as the OEM height (27").

 

How narrow is the 18" rim you're getting to get 235's on them?

 

The Rimss are 18x8.

 

When you go to larger diameter rims, even if the rims weigh the same amount, you will be slower with the larger diameter rims because now the weight is concentrated farther from the center of rotation. This means that the larger diameter rims have a larger moment of inertia, which means it now requires more torque to provide the same rotational acceleration to them.

You can aproximate the moment of inertia for a rim be the formula I= 1/2*MR^2 where I is moment of inertia, M is mass, and R is the mass of the rim. As you can see, the R term is squared, which means if you increase the radius of the rim by a certain amount, the moment of inertia increases by that amount squared. In other words, going from a 16" to a 18 " rim, the radius goes from 8" to 9". Let's say that both rims weigh 24 lbs (for comparison's sake.) Lbs aren't really a unit of mass, but of gravitational force so we use 1 lb = 4.45 N to get that the force on these rims if acceleration due to gravity is 9.8 m/s^2 to be 106.8 N. We then use Newton's second law, F=ma and solve for m. m=F/a, so the mass of these rims is about 10.9 kg. Now, 1 m = 39. 4", so the radius of these rims is .203 m and .228 m respectively. Plugging these values into the equation of moment of inertia for the rim gives us I= .449 kg*m for the 16" rim and I = .569 kg*m for the 18" rim. That results in a 26.6 % increase in moment of inertia for the two rims. I could have arived at the same result by simply taking the difference in the squares of the two radii, deviding by the original radius, and multiplying by 100. In other words, % change in I = (RF-RI)/RI*100 where RF is final rim radius and RI is initial rim radius.Newton's second law for rotational motion is T=Ia, where T is torque, I is moment of inertia, and a is rotational acceleration. As you can see, the increase in rotational inertia results in a proportional decrease in acceleration for a given applied torque. In other words, you would accelerate 26.6 % slower with rims 2" greater in diameter and equal weight.

The formula used is for the moment of inertia of a hollow cylinder about a central axis. It obviously is not compleely acurate for a rim because of the spokes and the fact that it is not perfectly cylindrical, but it provides a decent aproxamation and lets the point be seen. Even with equal weight rims, a small increase in diamter (only 2") can result in a large increase in moment of inertia, as well as a proportional decrease in rotational acceleration.

 

Yeah, what he said.