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Shadowlife25
03-30-2007, 07:22 AM
People have claimed that lowering the weight of the wheel/tire by say 10 lbs per wheel will have an 8-fold effect at each wheel, effectively lightening the vehicle by 320 lbs. I smelled a rat here and decided to perform a little physical analysis.

The answer?

A 10 lb reduction of wheel/tire weight will result in a total overall apparent effect of reducing vehicle weight by about 65 lbs and at absolute maximum, 80 lbs.

The analysis follows:


Question: What is the effective additional mass of wheels/tires due to their rotation?

For the vehicle in motion, the kinetic energy is given by:

Ekinetic = � * (Mtotal*V2 + 4*Iwheel*w2)

Where
Mtotal = Total vehicle mass including wheels
Iwheel = moment of inertia of a single wheel/tire combination
V = Velocity of vehicle
w = rotational velocity of the wheel (in radians/sec)

Now, V = Rtread * w where Rtread = effective radius of the tire's tread

Combining, we find:

Ekinetic = � * (Mtotal + 4*Iwheel / Rtread2) * V2

So the effective additional mass added to the vehicle due to the wheels rotation = 4*Iwheel / Rtread2

Now, Iwheel = Mwheel * Reffective2

Where
Mwheel = mass of the wheel/tire combination
Reffective = radius of gyration (which is always less than Rtread )

What is this radius of gyration (otherwise known as the radius of inertia)? It is the radius at which an infinitely thin hoop of material of identical mass would have the same moment of intertia as the body in question. It is a mathematical abstraction, but can be calculated for any object.

For example, a disk of uniform density would have an Reffective = 0.707 x Rdisk

Now, finally, the effective additional mass added to the vehicle due to the wheels rotation is:

Mdue to rotation = 4 * Mwheel * (Reffective / Rtread)2

On a per wheel basis, the EFFECTIVE TOTAL wheel mass is given by:

Mwheel, effective = Mwheel * (1 + (Reffective / Rtread)2 )

Reffective / Rtread is always less than 1 and probably somwhere around 80% by my guess. This ratio is a function of wheel and tire weight disribution.

So in this case, Mwheel, effective = Mwheel * 1.64

The absolute maximum (impossible) case would be Mwheel, effective = Mwheel * 2

Example: New wheels and tires are fitted, lowering the weight of each wheel / tire combo by 10 lbs. Assuming the radius of inertia of the wheel and tire combination are 80% of the outer radius of the tire, the apparent effect is to lower the weight of the car by 65.6 lbs (16.4 lbs per wheel).

One further note:
I used an energy analysis because it was more convenient for me. The exact same conclusion results if the analysis is performed using Newton's Second Law (F=M*a). This is not a steady-state analysis: it applies to all linear acceleration and deceleration conditions as long as traction is maintained.


from toyguymod back in '02

Slider
03-30-2007, 07:38 AM
Nice post :bigthumbu helps clear things up

david in germany
03-30-2007, 07:49 AM
Ban for using big words and stuff...


Good post man!

partyball
03-30-2007, 07:58 AM
I find this very interesting

Shadowlife25
03-30-2007, 08:38 AM
Enjoy!
Yeah, I know it's a bit wordy, but it is a concise description. I hope you all find this useful.

grayscale
03-30-2007, 09:17 AM
Sooo.....hm. Wish I could put that to good use. :brick:

1990CelicaGT-S
03-30-2007, 11:18 AM
Thats very interesting. Good job on your research.

celica9303
03-30-2007, 04:57 PM
too many big words....whats the answer? lol
nice write up

Cosco
03-30-2007, 08:58 PM
too many big words....whats the answer? lol
nice write up

I believe this is the answer...


People have claimed that lowering the weight of the wheel/tire by say 10 lbs per wheel will have an 8-fold effect at each wheel, effectively lightening the vehicle by 320 lbs. I smelled a rat here and decided to perform a little physical analysis.

The answer?

A 10 lb reduction of wheel/tire weight will result in a total overall apparent effect of reducing vehicle weight by about 65 lbs and at absolute maximum, 80 lbs.

Shadowlife25
03-31-2007, 01:17 AM
^^







;)

you got it buddy. the rest is merely an analysis and mathematical breakdown of WHY.

CHRiS'_CeLiCa
03-31-2007, 01:24 AM
i actually think that it has to do more with physics but technically when you look at it, it doesnt really matter because physics and math are one in the same, so technically, i really didnt need to make this post but im going to click the post quick reply button now because i dont want to backspace all that i have just typed :)

Galcobar
03-31-2007, 03:08 AM
Only two points I note -- the relative effect on handling of sprung versus unsprung weight, and the effect of moving a greater percentage of a wheel/tire's weight to the outer edge (as occurs when you plus-size your wheels, even if the total weight does not change) are not addressed.

The rule of thumb I've worked with is a 5:1 ratio of unsprung, rotating weight versus sprung, static weight when it comes to handling and acceleration. Not having the physics/engineering background to express the effect mathematically, I leave it to someone else to quantify the actual effect.

Shadowlife25
03-31-2007, 03:59 AM
It does address both sprung and unsprung weight, but only as applied in the context of a body in motion.

The necessary energy to move the wheel is increased because of it's position from center or zero(stock), which must now be accounted for. This is like leverage, insofar as you are moving the focal point of mass further out, thus increasing the distance and therefore ENERGY required to move it. This counters to some degree the real and perceived lightening effect of a wheel with less mass.

Moving the wheel to a further point toward the outer edge WILL however, increase the Track, which is nice :hehe:

Remember though, using a lighter wheel while still significantly increasing the diameter can only be brought so far effectively. Note that most professional race teams don't tend to use anything above a 16" wheel.

Adequate cooling of braking systems...
The cost of PROPERLY upgrading a braking system to match the increased rotating mass of a larger wheel... All things to take into account.

If anyone has something to add or address, please do so. I think this might provide a useful and informative tool that will offer some insight.
If I am wrong on any of the previously mentioned information, PLEASE CORRECT ME. :)
We learn something new every day right?

Shadowlife25
04-01-2007, 09:39 PM
Anyone?

alltracman78
04-02-2007, 03:40 AM
hm
I dropped about 20lbs per wheel/tire when I went to the wheels in my sig.

It's a huge difference on the butt dyno.





















:D

Conrad_Turbo
04-02-2007, 03:49 AM
Those forumlas are good "rule of thumb" ideas, such as 1psi=8hp type of mentality. However a wheel/tire assuming the same MOI (which never happens) that is say 5lbs lighter overall will show a more noticeable gain in 1st but the gains will reduce when the speeds increase. This is due to the fact that the car accelerates slower the faster it goes, thus the gains become smaller and smaller. So saying taking 5lbs off each wheel is like taking 50lbs off the car is a very vauge statement...but puts in perspective how important wheel/tire weight is. And I agree it is very important...the race wheels on the Starlet weight 7lbs each and the Hoosier slicks weigh in at around 3-4lbs. :D

Shadowlife25
04-02-2007, 08:03 AM
Hey Jeremy, how did you manage to shave 20lbs. per wheel/tire? What wheel /tire combo were you using to start, and what do you have now? (brand,size,offset,profile..)

Conrad - What you mention is also quite valid, but as you well know, shaving those few extra pounds can make all the difference in a race environment. ;)

Shadowlife25
05-31-2007, 11:48 AM
Bump due to recent interest in wheel/tire sizes and weights. Just something to think about :D

Rix86
05-31-2007, 03:42 PM
real world tests have shown the difference.

balagast
05-31-2007, 06:36 PM
Being a mechanical engineer this post had me really interested as I began to read through it today at work during my lunch. I had never heard anyone say that the lighter tire improves the effective weight so much, but I can believe that someone would. So... i got intrigued as to what the Inertia of wheels are approximately, so I decided to make a couple quick models in solidworks (3D modeling CAD software) one out of aluminum and the other out of something heavy like iron. here were the results:

1. Aluminum: weight 13.1 lbs, Inertia 490.9 lb*in^2
2. Iron: weight 35.4 lbs, Inertia 1327.3 lb*in^2

From above the effective weight of an individual wheel is:
(Nominal Wheel Weight) + (Wheel Inertia/ (Wheel Radius)^2)

So for our two examples: (note 15 inch rims so radius = 7.5)
1. 13.1 + (490.9/56.25) = 21.827 lb
2. 35.4 + (1327.3/56.25) = 58.996 lb

The difference in Nominal Weight = 35.4-13.1 = 22.3 lb
vs
The difference in Effective Weight = 58.996-21.827 = 37.133 lb

Thus with a loss of 22.3 pounds the effect is as if there were 37.133 pounds less at each wheel which is an extra 66.5 percent.

This pretty much goes along with what was said before, but I thought it was an interesting little experiment, and I hope some of you other guys thought so to. Or maybe I'm just a bit too nerdy and need to find a better way to waste my lunch at work :)

Sean
05-31-2007, 06:44 PM
Cool thread guys :)

Trance4c
05-31-2007, 07:19 PM
hrmm.. so should we all be measuring weights of our rims and tires to start to compare?

balagast
05-31-2007, 07:28 PM
hrmm.. so should we all be measuring weights of our rims and tires to start to compare?

That would be interesting, but the differences; in terms of how much "extra" weight is lost; differs based upon the geometry of the rims themselves. If the shape and dimensions of an object are the same, the inertia will vary linearly with respect to its mass (only variance being the density of the material).

Shadowlife25
05-31-2007, 07:46 PM
That would be interesting, but the differences; in terms of how much "extra" weight is lost; differs based upon the geometry of the rims themselves. If the shape and dimensions of an object are the same, the inertia will vary linearly with respect to its mass (only variance being the density of the material).



Thank You for contributing Balagast. Geeks FTW ! :woot:

Fuelish
05-31-2007, 08:36 PM
Sounds about right.... although weight reduction benefits in wheels/tires are much more readily apparent when applied to, oh, say, the modern mountain bike :cool: Switching from factory rims/tires/tubes to lighter versions is MUCH more noticeable, when said vehicle only weighs in at around 26 lbs ;)
I love my Trek Fuel :cool:

PhillyDRFT
06-01-2007, 04:46 AM
Is it just me or is anyone else wondering what alien metal wheels conrad dug up to make 7lb wheels for the starlet?

Shadowlife25
06-01-2007, 08:07 AM
Is it just me or is anyone else wondering what alien metal wheels conrad dug up to make 7lb wheels for the starlet?


He used unobtainium :hehe:

alltracman78
06-01-2007, 07:14 PM
Is it just me or is anyone else wondering what alien metal wheels conrad dug up to make 7lb wheels for the starlet?

Magnesium?
Aircraft aluminum?

My wheels are 11lbs each.
They're 16 x 7.
And I believe they're cheap knockoffs of something else, so they're probably not a very expensive alloy.

7lbs for racing wheels isn't all that shocking.