A study investigating the effective inertial weight of wheels. Warning: technical in nature.

Toy_Guy · 11-15-2002, 11:38 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:

E_kinetic = ½ * (M_total*V² + 4*I_wheel*w²)

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

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

Combining, we find:

E_kinetic = ½ * (M_total + 4*I_wheel / R_tread²) * V²

So the effective additional mass added to the vehicle due to the wheels rotation = 4*I_wheel / R_tread²

Now, I_wheel = M_wheel * R_effective²

Where
M_wheel = mass of the wheel/tire combination
R_effective = radius of gyration (which is always less than R_tread )

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 R_effective = 0.707 x R_disk

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

M_{due to rotation} = 4 * M_wheel * (R_effective / R_tread)²

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

M_{wheel, effective} = M_wheel * (1 + (R_effective / R_tread)² )

R_effective / R_tread 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, M_{wheel, effective} = M_wheel * 1.64

The absolute maximum (impossible) case would be M_{wheel, effective} = M_wheel * 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.

edited 11/16/02

mpwolf · 11-15-2002, 11:41 AM

Toy_Guy · 11-15-2002, 11:43 AM

BiturboBoy_V2.0_S3-- · 11-15-2002, 11:45 AM

don't make me check your work over mister! and you know i'm going to now....over the weekend.

George @ BostonAudi · 11-15-2002, 11:55 AM

Keep in mind, wheels don't have an even weight distribution from the hub to their outer surface... most of the weight is on the very outer part of the wheel.

I didn't go through your math completely, but I think there's something else off with it -- vehicle weight shouldn't be a factor at all. The reason why reducing unsprung and rotational mass helps the car isn't because it makes the car lighter, but because that weight (both unsprung, and more importantly, the higher mass at a distance from the hub) impacts significantly the drive train loss. (Some percentage of your total engine output has to be used to both pass energy across your suspension to the body of the car, and to accelerate the mass of the wheels)

LCP · 11-15-2002, 11:56 AM

BiturboBoy_V2.0_S3-- · 11-15-2002, 12:02 PM

out at the edges of the tire...in other words max possible contribution from wheel weight....but i agree it loooks suspect...i'll take a look at it this weeekend...

George @ BostonAudi · 11-15-2002, 12:04 PM

Sorry, I wasn't clear. I meant to say thats why I think he got 6x not 4x.

~~Mike@VAST~~ · 11-15-2002, 12:09 PM

~~Mike@VAST~~ · 11-15-2002, 12:12 PM

thanks Craig!

11-15-2002, 11:38 AM	#1
Toy_Guy Member Account #: 10011 Join Date: Sep 2000 Posts: 2,636	A study investigating the effective inertial weight of wheels. Warning: technical in nature. 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: E_kinetic = ½ * (M_totalV² + 4I_wheelw²) Where M_total = Total vehicle mass including wheels I_wheel = moment of inertia of a single wheel/tire combination V = Velocity of vehicle w = rotational velocity of the wheel (in radians/sec) Now, V = R_tread w where R_tread = effective radius of the tire's tread Combining, we find: E_kinetic = ½ * (M_total + 4I_wheel / R_tread²) V² So the effective additional mass added to the vehicle due to the wheels rotation = 4I_wheel / R_tread² Now, I_wheel = M_wheel R_effective² Where M_wheel = mass of the wheel/tire combination R_effective = radius of gyration (which is always less than R_tread ) 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 R_effective = 0.707 x R_disk Now, finally, the effective additional mass added to the vehicle due to the wheels rotation is: M_{due to rotation} = 4 * M_wheel * (R_effective / R_tread)² On a per wheel basis, the EFFECTIVE TOTAL wheel mass is given by: *M_{wheel, effective} = M_wheel (1 + (R_effective / R_tread)² )** R_effective / R_tread 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, M_{wheel, effective} = M_wheel * 1.64 *The absolute maximum (impossible) case would be M_{wheel, effective} = M_wheel 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. edited 11/16/02

11-15-2002, 11:41 AM	#2
mpwolf Elder Member Account #: 26345 Join Date: Oct 2001 Posts: 30,421	You could be totally wrong and I would have no clue, but it sounds good =) __________________ \| 1989 BMW M3 \| TMS J-Stock \| Bimmerworld Cage \| iigomotiv \| OMP \| \| Borla \| UUC \| Cobalt Friction \| Kosei \| Toyo \| BBS \| Bridgestone \| V1 \| \| 2005 Kawasaki Ninja ZX-6R \| Pearl Magma Red \| \| 8 Throttle Bodies...B!TCH AIM: mpw8888 MSN Messenger: mpwolf@hotmail.com Yahoo!: mw007e

11-15-2002, 11:43 AM	#3
Toy_Guy Member Account #: 10011 Join Date: Sep 2000 Posts: 2,636	It's not wrong, trust me :-)

11-15-2002, 11:45 AM	#4
BiturboBoy_V2.0_S3-- Elder Member Account #: 13456 Join Date: Dec 2000 Posts: 12,499	damn, engineering is hard to read without subscripts.... don't make me check your work over mister! and you know i'm going to now....over the weekend. __________________ AWESOME post, funny and informative for newbs...its not all right but i suggest you click!

11-15-2002, 11:55 AM	#5
George @ BostonAudi Elder Member Account #: 189 Join Date: Mar 2000 Posts: 15,756	Thats close, but not quite right... Keep in mind, wheels don't have an even weight distribution from the hub to their outer surface... most of the weight is on the very outer part of the wheel. I didn't go through your math completely, but I think there's something else off with it -- vehicle weight shouldn't be a factor at all. The reason why reducing unsprung and rotational mass helps the car isn't because it makes the car lighter, but because that weight (both unsprung, and more importantly, the higher mass at a distance from the hub) impacts significantly the drive train loss. (Some percentage of your total engine output has to be used to both pass energy across your suspension to the body of the car, and to accelerate the mass of the wheels) __________________ Boston Audi Group Website

11-15-2002, 11:56 AM	#6
LCP Member Account #: 125 Join Date: Mar 2000 Posts: 26,556	Please edit your post to include subscripts. It's permitted HTML. See Forum FAQ. __________________ "I can't stop apexing" 2000 S4 :: Nogaro Blue :: 6-spd :: AWE RSK04 :: AP Racing 13" 4-pots STaSIS Tracksports 550f/600r + 19mm rear sway bar :: APR Bipipe RS4 ICs :: Forge BPVs :: Autospeed DPs :: Borla :: Aquamist 1s UUC 40% short shifter w/spacer mod :: AWE center vent boost guage Lightweight flywheel :: 18x8 Ronal LZ's :: 235/40-18 Toyo T1R's 2002 allroad :: Atlas Grey :: Tip :: 245/50-17 Dunlop SP5000's Looking for tires? See My review of all of the tires I've had on my S4 Need info? See Everything anyone needs to know about their S4 Help our troops protect their heads: Donate to Operation Helmet Rest In Peace Matthew Ammon a/k/a O.W. Kenobi. May the Force be with you as you keep on apexing. Read his funniest post here.

11-15-2002, 12:02 PM	#7
BiturboBoy_V2.0_S3-- Elder Member Account #: 13456 Join Date: Dec 2000 Posts: 12,499	george, he's calculating it as if all the weight was all the way out at the edges of the tire...in other words max possible contribution from wheel weight....but i agree it loooks suspect...i'll take a look at it this weeekend... __________________ AWESOME post, funny and informative for newbs...its not all right but i suggest you click!

11-15-2002, 12:04 PM	#8
George @ BostonAudi Elder Member Account #: 189 Join Date: Mar 2000 Posts: 15,756	Right... I've always seen approx 4x equivalence... Sorry, I wasn't clear. I meant to say thats why I think he got 6x not 4x. __________________ Boston Audi Group Website

11-15-2002, 12:09 PM	#9
~~Mike@VAST~~ Banned Account #: 21977 Join Date: Jul 2001 Posts: 19,481	Toy Guy did enough as it is!

11-15-2002, 12:12 PM	#10
~~Mike@VAST~~ Banned Account #: 21977 Join Date: Jul 2001 Posts: 19,481	so my effective weight loss, if i bought lighter wheels, doesn't justify my effective wallet burn. thanks Craig!

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