need second opinion on brake force computation

Guys,

Here's a little paper I'm working on, would appreciate it very much if the engineers here can cross check my computations to see if i've missed anything. The point of the paper is to prove that plus sizing wheels doesn't neccessarily affect braking performance.




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 vehicle’s 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 doesn’t 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 vehicle’s 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 vehicle’s kinetic energy.

It is late, but this is my $0.02…

For simplicity, we will treat each wheel and tire combo as a solid disc with rotational inertia given by the formula:

I realize this is nice for simplification but I personally find this to be a very suspect assumption. The heart of you braking model is the wheel’s moment of inertia (MoI), and it is very sensitive to the mass distribution.

Your assumption is for identically sized discs with different densities, which will result in the same mass distribution. However, in actuality the larger wheels will have their mass distributed farther from the axis of rotation. This is something that should not be overlooked when comparing moments of inertia.



Overall radius is the same for all 3 wheels at 0.77m

Those are some monster wheels!!!


The quantity derived is actually in joules, the unit for work (using one Newton of work to push and object one meter).

I do not agree. Torque and work are not equivalent, and therefore the units are not interchangeable without proper conversion.


For al intents and purposes, a vehicle’s 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 vehicle’s kinetic energy.

Is this a typo? Did you mean to write the sprung weight as that is what you are describing. Either way I disagree as the brakes one way or another stop the entire mass of the vehicle (body, suspension, wheels, etc) other wise parts of the vehicle would continue down the street after hitting the brakes. The distinction is that the wheels are storing rotational motion kinetic energy, and everything else is storing linear motion kinetic energy.

EDIT: In regards to the last sentence, I think technically the wheels would be storing kinetic energy from both rotation and translation.

It is late, but this is my $0.02…



I realize this is nice for simplification but I personally find this to be a very suspect assumption. The heart of you braking model is the wheel’s moment of inertia (MoI), and it is very sensitive to the mass distribution.

Your assumption is for identically sized discs with different densities, which will result in the same mass distribution. However, in actuality the larger wheels will have their mass distributed farther from the axis of rotation. This is something that should not be overlooked when comparing moments of inertia.






I agree...the formular you used is supposed to be for a completely solid disc with the mass of the disc evenly distributed throughout, something that most rims, much less the blingly aftermarket rims I assume you're using here do not come close to having.

I agree...the formular you used is supposed to be for a completely solid disc with the mass of the disc evenly distributed throughout, something that most rims, much less the blingly aftermarket rims I assume you're using here do not come close to having.

A little background on the purpose of the study. We currently supply aftermarket wheels to a number of dealers and automotive dealerships such as toyota. currently the hot car is the Toyota fortuner (asian domestic maket suv, slightly huger than a 4runner) which we fit with 20 and 22 inch wheels.

there was a news item in thailand where a customer was claiming warranty on a set of cracked brake pads. upon taking them to the dealership, they refused his claim and blamed the upsized 20 inch wheels as the source of the pads cracking. The whole point of this paper is to show that upsized wheels do no overstress the braking components even with bigger wheels.

alsator, hmm you do have a point on the whole sprung unsprung weight. thought about it a little bit more and yea the rbakes carry everything moving including the suspension bits.

"The quantity derived is actually in joules, the unit for work (using one Newton of work to push and object one meter).
----------------------------------------------------------------------
I do not agree. Torque and work are not equivalent, and therefore the units are not interchangeable without proper conversion."

I've browsed a couple of sites and they do relate Nm to joules directly

http://www.unc.edu/~rowlett/units/dictN.html

while they have the same units, the application is different, anyway that i can relate the two?

there was a news item in thailand where a customer was claiming warranty on a set of cracked brake pads.

What causes brake pads to crack? I guessing it could be caused by excessive brake temperatures, no?



The whole point of this paper is to show that upsized wheels do no overstress the braking components even with bigger wheels.


I would like to suggest that you be careful on making this kind of a blanket statement if your paper will be used as the verification. In your first post you were comparing the increase in wheel weight with respect to the addition of passenger. This is a fair enough comparison for most daily use.

However, if the original brake system was designed properly it was designed for the worst case. That is for the maximum gross vehicle weight and maximum towing load with the corresponding factors of safety.

So your quoted statement may not always be true, it will be dependent on your ‘failure criteria’. This may be something you want to consider if you are in anyway liable for your results.

Also, note that in the first post when you compared the different KEs using 0.5 * m * v^2 you are not taking in to account the rotational energy stored by the wheels.


while they have the same units, the application is different, anyway that i can relate the two?

Torque is more or less a force applied along the circumference of a circle. The radius of the circle can be assumed to be 1 [unit length] and the magnitude of the force equal to the magnitude of the torque. Then using the equation of work one can calculated the work done by applying a torque along a given distance around a circle.

For example, less say 100 N-m is applied along the total circumference of a circle or 360°:

Torque = 100 N-m

=>
Force = 100 N
Radius = 1 m

=>

Circumference = 2 * Pi m
Work = 200 * Pi joules


You are going to need to determine how many revolutions you are interested in as this will have a large impact on your results. So for example looking at a wheel with a 0.77 m radius and a vehicle traveling at 11 m/s:

First,

A 0.77 m radius implies an overall diameter of over 5 feet, you might want to double check that.

Looking at the work done over a duration of one second would imply the following number of revolutions:

11 m /s * 1 s = 11 m

2 * Pi * 0.77 m = 4.84 m

11 m / 4.84 m = 2.27 revolutions

Good points, so the total energy of the system will have to be the vehicle's kinetic energy plus rotational energy in the wheels.

RE = (I*W*W)/2 interia times angular velocity squared over 2.

since the assumption that all wheels will be subjected to the same forces, W will be the same so the incremental increase will only be due to the difference in moment of inertia.

also keeping all variables the same, the work done (distance x newton) will also be the same save for the difference in torque from the different sizes.

I also like how you took into consideration the vehicles gross weight (will full passengers) and towing capacity. And this is toyota after al so were assuming a properly designed break system

Guys,
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


In addition to all the other problems the others noted, this really strikes me as suspect. These wheels come up to my chin and weight 66-84 pounds each. Consider you have 1.54m high wheels on a vehicle that's only 1.85m high overall. I think you have mistaken radius for diameter, because a wheel diameter of 0.77m (30") seems a lot better.

The observation that all three wheels have the same "radius" is also suspect. Perhaps I'm mistaken, but doesn't "plus size" wheels mean the overall radius increases? If just the rim gets bigger, then the sidewall of the tire should get smaller to maintain the same total radius, and to me the mass would DECREASE as a result, trading a good portion of rubber for a little bit of metal. Please verify this.

I'm also a little skeptical of the masses, but at least they seem reasonable.

----------------

The formula you want to use for calculating the rotational inertia of the wheel would be thus:

It = Mt*((Dt1-Dt2)/4)

Dt1 being the diameter of the whole wheel (0.77m) and Dt2 being the diameter of the inside of the hub. This is slightly wrong, since it will put the mass of the hub "spokes" farther out, but that mass is pretty small compared to the rest of the wheel. If you want a better approximation, find the inertia of the rim and tire separately and add them together (It will still be slightly off, but more accurate.)

----------------

Total energy in the system (the whole vehicle) is:
E = (1/2) * (Mc*Vc^2 + 4*(IWt^2 + MtVc^2))

Where:

Mc = Mass of vehicle MINUS wheels (Since we want to see the effect of changing the wheels, it should be a separate term)

Vc = Forward velocity of the vehicle

It = Inertia of each wheel. (See above)

Wt = Angular velocity of each wheel, in radians per second.

Mt = Mass of each wheel

----------------

If we assume the wheels roll without slipping (And they better be!) we can express Wt in terms of Vc:

Vt = Wt * (Dt1/2)

Vt being the velocity of a point on the edge of the wheel relative to the axis of the wheel. Since the motion of the edge relative to the axis is the same of the axis relative to the edge:

Vc = Vt = Wt * (Dt1/2)

So solving for Wt in terms of Vc:

Wt = 2*Vc/Dt1

----------------

Plug the above equation into our energy formula:

E = (1/2) * (Mc*Vc^2 + 4*(I(2*Vc/Dt1)^2 + 4*MtVc^2))

A little bit of algebra and we can simplify this a bit..

E = 2*(Vc^2) * (Mc/4 + 4*It/(Dt1^2) + Mt)

----------------

Figure out the energy for all three configurations. To do this we need the inner diameter of the rims, which I don't have. You can also assume a vehicle velocity Vc, since it will be the same for all three configurations and all you are interested in is the change that occurs for different wheels.

No matter what, the brakes must dissipate that energy to stop the vehicle. This energy will go completely to heat. Of course, the stopping distance/time will have an effect, but ONLY if the time to stop is long enough to make heat dissipation nontrivial. In other words, we also need to consider how quickly the braking system cools off!

But there's a shortcut! If the total change in energy for each configuration is relatively small, then we can safely assume the net effect is also small, and we can conclude there is no practical impact on braking performance.

If the difference is large, then we'll be left with a big "maybe" and have to go further.
=Smidge=

well, yall could save a bunch of time and headache by just asking techs in the field. while brake performance may not be affected greatly, brake pad and rotor life is shortened. take the stock 16" aluminum wheels off a tahoe and swap them for 22" steelies and watch the pad life be cut in half. some drug dealer at my old GM dealer did this. he was in every 8-10k miles for a front brake job. and he didnt tow anything with it either. i would guess average front brake pad life on those tahoes was about 20k miles, depending on driver habit.

The question was about brake performance, not wear.

If the brake pads cracked because they were worm out then the warranty issue would be moot, so they probably cracked while fairly new. Was it overheating or manufacturer's defect? One is covered by warranty, one isn't.
=Smidge=

The question was about brake performance, not wear.

If the brake pads cracked because they were worm out then the warranty issue would be moot, so they probably cracked while fairly new. Was it overheating or manufacturer's defect? One is covered by warranty, one isn't.
=Smidge=


car was barely 4 months old when it happened and the pads were still very thick. the service department found the cracks during routine check up and blamed the wheel for the cracking.

smidge, that was a lot of help, many thanks