Quote:
Originally Posted by ZDan
Tire weight affects rotational inertia a lot more than wheel weight. However, it's still only on the order of 1.5x non-rotating mass (i.e, save 4 lb on tire weight and acceleration is improved as if you saved 6 lb).
But other tire qualities might be more important to you..
Continental tires are very lightweight, but they are in general also poorly rated for steering responsiveness and cornering stability.
https://www.tirerack.com/tires/tests...y.jsp?ttid=190
Conti DW:
"What We'd Improve: More precise steering response and handling"
I've only had one set of Conti tires. They were lightweight for sure. But the sidewalls were limp and fun-to-driveness went right out the window... |
This. Sometimes even the faster things don't feel as fast or as good to drive.
Also, if anyone has any leads on a constant of rotational moment of inertia of an actual car wheel and tire and not the theoretical stuff that google keeps bringing up, I'd love that info. I had heard weight savings in the wheel/tire package is like saving 150% before and I'd also heard it is like 1000% and I got bored so I plugged through it.
Power required to get a non-rotating item to a certain linear velocity in a certain time is:
P=E/t=0.5*m*(v^2)/t
Where:
E= Linear kinetic energy of the item
m = mass of the item
v = linear velocity of the item (equal to car's velocity)
t = time spent accelerating
Rotational kinetic energy of the rotating wheel tire = 0.5*k*m*(R^2)*(w^2) = 0.5*k*m*(v^2) since v=R*w
w=rotational velocity (radians/sec)
R=radius of wheel/tire
m = mass of the item
v = linear velocity of the item (equal to car's velocity)
So,
Power to get a wheel/tire assembly to a certain linear velocity in a certain time equals:
P=linear kinetic energy/t + rotational kinetic energy/t = [0.5*m*(v^2)/t]+[0.5*k*m*(v^2)/t]
m = mass of the wheel/tire
v = linear velocity of the wheel (equal to car's velocity)
t = time spent accelerating
k = inertial constant of rotation for wheel/tire setup
Notice how the first half of that is the same formula as for the power required to move a non-rotating object, such as a passenger in the cabin, or the exact same wheel in the trunk. The latter part is the contribution due to the wheel rotating.
So,
Total power required for wheel assembly = power required if wheel was sitting in the trunk + [0.5*k*m*(v^2)/t]
We can choose a speed and estimated acceleration time for our cars, and a decent wheel/tire mass guestimate:
v= 60mph = 27 meters/second
t = 7 seconds (rough 0-60 estimate for an NA twin)
m = 55 lbs = 25 kg
P for non rotating = [0.5*35 kg*(27(m/s)^2)/7 sec] = 1822 watts = 2.44 horsepower is required to accelerate 25 kg to 60 mph in 7 seconds
P for rotating wheel/tire = 1822 watts + [0.5*k*35 kg*(27(m/s)^2)/7 sec] = 1822 watts + 1822*k watts = 2.44 horsepower + 2.44*k horsepower
Get your k for the wheel/tire setup and you've got your answer as to how much the rotational part adds.