sids
20+ Year Contributor
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- Jun 24, 2002
Guys,
I need the engineers to check out my math, I think its pretty sound for the paper i'm doing, basically the gist of its to prove that bigger wheels and tires do not neccessairly mean increase braking effort
Effects of Plus Sizing Wheels on Vehicle Braking Systems
There has been much debate of late on the effects of upgrading stock wheels with larger aftermarket wheels and tires. Common misconceptions include:
1. Increase wheel momentum
2. Increase braking requirement
3. Excessive straining of brake systems
For this treatise, we will compare the dynamic forces of a stock Toyota Fortuner wheel/tire package with popular upgrades such as 20 and 22 wheels/tires.
Rotational Inertia and Torque Comparisons
For simplicity, we will treat each wheel and tire combo as a solid disc with rotational inertia given by the formula:
I = MR2
2
where M = mass (kg)
R = radius squared
We physically weighed the stock wheel, a size 20 and a size 22 with tires
16 = 30kg
20 = 36kg
22 = 38kg
Overall radius is the same for all 3 wheels at 0.77m
We then compute torque (turning force) using the following formula:
T = IA
Where I = Rotatinal Inertia
A = Angular acceleration
Since all three wheels will be subjected to the same angular acceleration, we can assume this quantity is constant. The result will give us torque in Newton meters.
T16 = (30kg)(0.77m2)(Am) = 8.9Nm
2s2
T20 = (36kg)(0.77m2)(Am) = 10.67 Nm
2s2
T22 = (38kg)(0.77m2)(Am) = 11.265 Nm
2s2
The quantity derived is actually in joules, the unit for work (using one Newton of work to push and object one meter). The vehicles braking system thus has to do additional work:
16 to 20 = 1.77 joules
16 to 22 = 2.365 joules
This additional work done by the braking system manifests itself in the increased heat generated by the pads on the rotors. We can convert joules to calories (amount of heat required to raise 1 gram of water by 1 degree Celsius) to determine the increase in rotor temperature.
1 calorie = 4.1855 joules
Increase 16 to 20 = 0.42 cal
Increase from 16 to 22 = 0.565 cal
The increase is rotor heat is very minimal with the upsized wheels. Therefore we can conclude that the braking system doesnt have to work significantly harder to slow down a vehicle with upsized wheels.
Momentum & Kinetic Energy Comparison
Vehicle braking systems act directly on the wheels and tires but brake systems have to carry the entire momentum of the vehicle including passengers.
Brakes are energy conversion devices, converting kinetic (moving energy) into heat energy. The total energy of a vehicle can be expressed by the following formula:
K = MV2
2
M = Mass of vehicle
V = Velocity
We shall assume the vehicle (Fortuner) has a mass of 2 tons or 2000 kg with one driver. The total kinetic energy of the vehicle traveling at 40km/h is expressed by:
K= (2000kg)(11m/s)2
2
K = 121,000 Nm
Adding a passenger who weighs 70kg will result in the following kinetic energy of the vehicle:
K= (2070kg)(11m/s)2
2
K = 125,200 Nm
Upsized wheels will add 24kg and 32kg for 20 and 22 respectively, giving the following kinetic energy of the vehicle:
K20 = 122,452 Nm
K22 = 122,936 Nm
Both figures of which yield a lower kinetic energy than one passenger on the vehicles. For al intents and purposes, a vehicles braking system only stops the unsprung weight (weight of vehicle without suspension components), so technically the additional weight of upsized and wheels and tires have no effect on a vehicles kinetic energy.
I need the engineers to check out my math, I think its pretty sound for the paper i'm doing, basically the gist of its to prove that bigger wheels and tires do not neccessairly mean increase braking effort
Effects of Plus Sizing Wheels on Vehicle Braking Systems
There has been much debate of late on the effects of upgrading stock wheels with larger aftermarket wheels and tires. Common misconceptions include:
1. Increase wheel momentum
2. Increase braking requirement
3. Excessive straining of brake systems
For this treatise, we will compare the dynamic forces of a stock Toyota Fortuner wheel/tire package with popular upgrades such as 20 and 22 wheels/tires.
Rotational Inertia and Torque Comparisons
For simplicity, we will treat each wheel and tire combo as a solid disc with rotational inertia given by the formula:
I = MR2
2
where M = mass (kg)
R = radius squared
We physically weighed the stock wheel, a size 20 and a size 22 with tires
16 = 30kg
20 = 36kg
22 = 38kg
Overall radius is the same for all 3 wheels at 0.77m
We then compute torque (turning force) using the following formula:
T = IA
Where I = Rotatinal Inertia
A = Angular acceleration
Since all three wheels will be subjected to the same angular acceleration, we can assume this quantity is constant. The result will give us torque in Newton meters.
T16 = (30kg)(0.77m2)(Am) = 8.9Nm
2s2
T20 = (36kg)(0.77m2)(Am) = 10.67 Nm
2s2
T22 = (38kg)(0.77m2)(Am) = 11.265 Nm
2s2
The quantity derived is actually in joules, the unit for work (using one Newton of work to push and object one meter). The vehicles braking system thus has to do additional work:
16 to 20 = 1.77 joules
16 to 22 = 2.365 joules
This additional work done by the braking system manifests itself in the increased heat generated by the pads on the rotors. We can convert joules to calories (amount of heat required to raise 1 gram of water by 1 degree Celsius) to determine the increase in rotor temperature.
1 calorie = 4.1855 joules
Increase 16 to 20 = 0.42 cal
Increase from 16 to 22 = 0.565 cal
The increase is rotor heat is very minimal with the upsized wheels. Therefore we can conclude that the braking system doesnt have to work significantly harder to slow down a vehicle with upsized wheels.
Momentum & Kinetic Energy Comparison
Vehicle braking systems act directly on the wheels and tires but brake systems have to carry the entire momentum of the vehicle including passengers.
Brakes are energy conversion devices, converting kinetic (moving energy) into heat energy. The total energy of a vehicle can be expressed by the following formula:
K = MV2
2
M = Mass of vehicle
V = Velocity
We shall assume the vehicle (Fortuner) has a mass of 2 tons or 2000 kg with one driver. The total kinetic energy of the vehicle traveling at 40km/h is expressed by:
K= (2000kg)(11m/s)2
2
K = 121,000 Nm
Adding a passenger who weighs 70kg will result in the following kinetic energy of the vehicle:
K= (2070kg)(11m/s)2
2
K = 125,200 Nm
Upsized wheels will add 24kg and 32kg for 20 and 22 respectively, giving the following kinetic energy of the vehicle:
K20 = 122,452 Nm
K22 = 122,936 Nm
Both figures of which yield a lower kinetic energy than one passenger on the vehicles. For al intents and purposes, a vehicles braking system only stops the unsprung weight (weight of vehicle without suspension components), so technically the additional weight of upsized and wheels and tires have no effect on a vehicles kinetic energy.
