Wheel Weight and Rotational Mass
#1
04-23-2009, 12:32 PM
Wheel Weight and Rotational Mass
I've got a question about Wheel Weight and Rotational Mass. This could be too good to be true:
I've been looking at getting some light weight wheels and I was searching online for the calculation for how much lightening of your wheels equates to in lightening of "non-unsprung weight" (in-car weight reduction) and I came across this page on a honda forum:
http://hondaswap.com/general-tech-ar...art-2-a-29058/
If this information is correct, 1 lb. of Wheel Weight= 10 lbs Weight Reduction.
This means if you replace your stock 27lb. Avus' with 17lb. Advans or OZ ULs, this is the equivalent of 400 lbs. of weight reduction.
I've heard many different numbers quoted on forums in the past, but when I searched this was the only calculation I could find. Is this too good to be true? It is from a honda forum.
I've been looking at getting some light weight wheels and I was searching online for the calculation for how much lightening of your wheels equates to in lightening of "non-unsprung weight" (in-car weight reduction) and I came across this page on a honda forum:
http://hondaswap.com/general-tech-ar...art-2-a-29058/
If this information is correct, 1 lb. of Wheel Weight= 10 lbs Weight Reduction.
This means if you replace your stock 27lb. Avus' with 17lb. Advans or OZ ULs, this is the equivalent of 400 lbs. of weight reduction.
I've heard many different numbers quoted on forums in the past, but when I searched this was the only calculation I could find. Is this too good to be true? It is from a honda forum.
#2
04-23-2009, 03:20 PM
The claim of 10:1 improvement is highly exagerated
I've heard that 10:1 bit of wisdom before, but I don't think it stands up to scrutiny. It is absolutely true that a pound saved in the wheel or tire is worth more than a pound saved somewhere else, but not a factor of 10 more. I did a "back of the envelope" calculation using basic physics principles, and what I find is that a 1 pound savings in wheel and/or tire weight is equaivalent to about a 1.5 pound savings in the non-rotating weight of the car. What I did was compare the power needed to accelerate a 1 pound mass in a straight line versus the power needed to increase the rotational veocity of a wheel with a 1 pound mass concentrated at the rim an equivalent amount.
Here's the anaysis. To keep the math simple, let's assume the following:
1. The diameter of the wheel is 18", or in other words the radius of the wheel (r) is 9 inches, or 3/4 ft.
2. The radius of the tire (R) is one foot.
3. All the weight savings for a 1 pound reduction in wheel weight is concentrated at the outer edge of the rim - that is, 3/4 ft from the center. This is a bit unrealistic, but is the most favorable assumption with respect to trying to show the advantage of reducing rotating mass.
The instantaneous power needed to linearly accelerate a given mass (m) is given by P(linear) = m*a*v, where a = the acceleration, v is the instantaneous velocity.
The power needed to increase the rotational velocity of a wheel is P(rotation) = I*alpha*omega, where I = rotational moment of inertia of the wheel & tire, alpha is the rotational acceleration (in radians per sec squared), and omega is the instantaneous rotational velocity (in radians/sec). Alpha is related to the linear acceleration of the car by a = alpha*R, and omega is related to the car's velocity by v = omega*R. So the power required for rotational acceleration can be expressed in terms of a and v by: P=Iav/R^3. As for the value of I: for a 1 pound mass concentrated at a distance r from the center of rotation, I = mr^2. Putting it together: P(rotation) = I*alpha*omega = mr^2av/R^3. Since r = 3/4 ft and R = 1 ft, that means you have P(rotation) = 9/16 * mav.
Note that P(rotation) is 9/16 of the amount of P(linear). Now if you can save that 1 pound of mass in the wheel you save on the power needed to both spin the wheel up to speed and to acelerate on down the road, so the total savings in power because of the lighter wheel is (1+9/16)*mav, or just more than one and a half times the power savings if the weight was some place other than the wheel. So while saving a pound in the wheel is definitely more advantageous than saving a pound of weight in the rest of the car, it's no more than about 50% better (not 900% better as claimed).
Of course this says nothing about the other improvements that come about from a ligher unsprung weight - namely faster damping of vibrations and less transfer of energy from the wheel into the car when you hit a bump.
Here's the anaysis. To keep the math simple, let's assume the following:
1. The diameter of the wheel is 18", or in other words the radius of the wheel (r) is 9 inches, or 3/4 ft.
2. The radius of the tire (R) is one foot.
3. All the weight savings for a 1 pound reduction in wheel weight is concentrated at the outer edge of the rim - that is, 3/4 ft from the center. This is a bit unrealistic, but is the most favorable assumption with respect to trying to show the advantage of reducing rotating mass.
The instantaneous power needed to linearly accelerate a given mass (m) is given by P(linear) = m*a*v, where a = the acceleration, v is the instantaneous velocity.
The power needed to increase the rotational velocity of a wheel is P(rotation) = I*alpha*omega, where I = rotational moment of inertia of the wheel & tire, alpha is the rotational acceleration (in radians per sec squared), and omega is the instantaneous rotational velocity (in radians/sec). Alpha is related to the linear acceleration of the car by a = alpha*R, and omega is related to the car's velocity by v = omega*R. So the power required for rotational acceleration can be expressed in terms of a and v by: P=Iav/R^3. As for the value of I: for a 1 pound mass concentrated at a distance r from the center of rotation, I = mr^2. Putting it together: P(rotation) = I*alpha*omega = mr^2av/R^3. Since r = 3/4 ft and R = 1 ft, that means you have P(rotation) = 9/16 * mav.
Note that P(rotation) is 9/16 of the amount of P(linear). Now if you can save that 1 pound of mass in the wheel you save on the power needed to both spin the wheel up to speed and to acelerate on down the road, so the total savings in power because of the lighter wheel is (1+9/16)*mav, or just more than one and a half times the power savings if the weight was some place other than the wheel. So while saving a pound in the wheel is definitely more advantageous than saving a pound of weight in the rest of the car, it's no more than about 50% better (not 900% better as claimed).
Of course this says nothing about the other improvements that come about from a ligher unsprung weight - namely faster damping of vibrations and less transfer of energy from the wheel into the car when you hit a bump.
#3
04-23-2009, 04:43 PM
About 15 years separate me from my last physics class, thanks, I would have had no chance in figuring that out. This seems to make sense, 1lb:1.5lbs is still a savings of 60lbs. in-car if I bought OZ Ultraleggeras, not bad.
I could swear that I've heard 4-5lbs for every 1lb. saved at the very least, though.
I could swear that I've heard 4-5lbs for every 1lb. saved at the very least, though.
I've heard that 10:1 bit of wisdom before, but I don't think it stands up to scrutiny. It is absolutely true that a pound saved in the wheel or tire is worth more than a pound saved somewhere else, but not a factor of 10 more. I did a "back of the envelope" calculation using basic physics principles, and what I find is that a 1 pound savings in wheel and/or tire weight is equaivalent to about a 1.5 pound savings in the non-rotating weight of the car. What I did was compare the power needed to accelerate a 1 pound mass in a straight line versus the power needed to increase the rotational veocity of a wheel with a 1 pound mass concentrated at the rim an equivalent amount.
Here's the anaysis. To keep the math simple, let's assume the following:
1. The diameter of the wheel is 18", or in other words the radius of the wheel (r) is 9 inches, or 3/4 ft.
2. The radius of the tire (R) is one foot.
3. All the weight savings for a 1 pound reduction in wheel weight is concentrated at the outer edge of the rim - that is, 3/4 ft from the center. This is a bit unrealistic, but is the most favorable assumption with respect to trying to show the advantage of reducing rotating mass.
The instantaneous power needed to linearly accelerate a given mass (m) is given by P(linear) = m*a*v, where a = the acceleration, v is the instantaneous velocity.
The power needed to increase the rotational velocity of a wheel is P(rotation) = I*alpha*omega, where I = rotational moment of inertia of the wheel & tire, alpha is the rotational acceleration (in radians per sec squared), and omega is the instantaneous rotational velocity (in radians/sec). Alpha is related to the linear acceleration of the car by a = alpha*R, and omega is related to the car's velocity by v = omega*R. So the power required for rotational acceleration can be expressed in terms of a and v by: P=Iav/R^3. As for the value of I: for a 1 pound mass concentrated at a distance r from the center of rotation, I = mr^2. Putting it together: P(rotation) = I*alpha*omega = mr^2av/R^3. Since r = 3/4 ft and R = 1 ft, that means you have P(rotation) = 9/16 * mav.
Note that P(rotation) is 9/16 of the amount of P(linear). Now if you can save that 1 pound of mass in the wheel you save on the power needed to both spin the wheel up to speed and to acelerate on down the road, so the total savings in power because of the lighter wheel is (1+9/16)*mav, or just more than one and a half times the power savings if the weight was some place other than the wheel. So while saving a pound in the wheel is definitely more advantageous than saving a pound of weight in the rest of the car, it's no more than about 50% better (not 900% better as claimed).
Of course this says nothing about the other improvements that come about from a ligher unsprung weight - namely faster damping of vibrations and less transfer of energy from the wheel into the car when you hit a bump.
Here's the anaysis. To keep the math simple, let's assume the following:
1. The diameter of the wheel is 18", or in other words the radius of the wheel (r) is 9 inches, or 3/4 ft.
2. The radius of the tire (R) is one foot.
3. All the weight savings for a 1 pound reduction in wheel weight is concentrated at the outer edge of the rim - that is, 3/4 ft from the center. This is a bit unrealistic, but is the most favorable assumption with respect to trying to show the advantage of reducing rotating mass.
The instantaneous power needed to linearly accelerate a given mass (m) is given by P(linear) = m*a*v, where a = the acceleration, v is the instantaneous velocity.
The power needed to increase the rotational velocity of a wheel is P(rotation) = I*alpha*omega, where I = rotational moment of inertia of the wheel & tire, alpha is the rotational acceleration (in radians per sec squared), and omega is the instantaneous rotational velocity (in radians/sec). Alpha is related to the linear acceleration of the car by a = alpha*R, and omega is related to the car's velocity by v = omega*R. So the power required for rotational acceleration can be expressed in terms of a and v by: P=Iav/R^3. As for the value of I: for a 1 pound mass concentrated at a distance r from the center of rotation, I = mr^2. Putting it together: P(rotation) = I*alpha*omega = mr^2av/R^3. Since r = 3/4 ft and R = 1 ft, that means you have P(rotation) = 9/16 * mav.
Note that P(rotation) is 9/16 of the amount of P(linear). Now if you can save that 1 pound of mass in the wheel you save on the power needed to both spin the wheel up to speed and to acelerate on down the road, so the total savings in power because of the lighter wheel is (1+9/16)*mav, or just more than one and a half times the power savings if the weight was some place other than the wheel. So while saving a pound in the wheel is definitely more advantageous than saving a pound of weight in the rest of the car, it's no more than about 50% better (not 900% better as claimed).
Of course this says nothing about the other improvements that come about from a ligher unsprung weight - namely faster damping of vibrations and less transfer of energy from the wheel into the car when you hit a bump.
#4
04-24-2009, 11:33 AM
I found some more information on wheel weights, rotational mass and weight reduction. This website has a spreadsheet with calculations that take EBAINES' formula (thanks again) for rotational mass and acceleration with tires using 3 calculations: one for uniform density (wheel and tire weight is uniform from center to edge of tire,) one for the wheel and tire weights separately, and one assuming more weight at the edge of the tire.
http://www.the-welters.com/racing/rotational.html
Long story short: Using the least advantageous calculation (Wheel and Tire calculated separately,) going from a 23lb. wheel to an 18lb. tire saves you an equivalent 50lbs. of chassis weight.
1 lb. (Wheel) = 2.5 lb. (Chassis)
So at the very least, going from a stock B5 Avus wheel (27lbs.) to a 17lb. wheel would be equivalent to shaving 100 lbs. from the chassis, or 0.1 sec from your 1/4 ET.
http://www.the-welters.com/racing/rotational.html
Long story short: Using the least advantageous calculation (Wheel and Tire calculated separately,) going from a 23lb. wheel to an 18lb. tire saves you an equivalent 50lbs. of chassis weight.
1 lb. (Wheel) = 2.5 lb. (Chassis)
So at the very least, going from a stock B5 Avus wheel (27lbs.) to a 17lb. wheel would be equivalent to shaving 100 lbs. from the chassis, or 0.1 sec from your 1/4 ET.
#5
04-24-2009, 02:18 PM
I think the 10:1 theory takes into account the other benefits of lighter wheels: enhanced braking, improved effectiveness of the suspension due to reduced unsprung weight, improved handling from reduced mass, AND improved acceleration. The combined benefits in those areas of a 1 lb reduction in wheel weight (the theory goes) is equivalent to taking 10 lbs out of the chassis.
That being said, since the origins of this theory come from the Honda/Acura world, it is reasonable to assume that the benefits of reduced wheel weight on a car with 160 hp/110 lb-ft would probably seem greater than they would on a car with 220 hp/235 lb-ft. Also, there is likely a point of diminishing returns, where assuming the 10:1 ratio was valid, that may be true for the first 5 lbs of saving in wheel weight, after which it would probably begin to fall off.
Regardless, a drop in wheel weight can have significant impact on the performance of the car, and it would be relatively higher than a similar reduction in non-rotating mass.
And having just changed wheels on my S4, these stock wheels are heavy mothers compared to what I am used to when changing wheels on the Civics and Miatas that I typically autocross.
That being said, since the origins of this theory come from the Honda/Acura world, it is reasonable to assume that the benefits of reduced wheel weight on a car with 160 hp/110 lb-ft would probably seem greater than they would on a car with 220 hp/235 lb-ft. Also, there is likely a point of diminishing returns, where assuming the 10:1 ratio was valid, that may be true for the first 5 lbs of saving in wheel weight, after which it would probably begin to fall off.
Regardless, a drop in wheel weight can have significant impact on the performance of the car, and it would be relatively higher than a similar reduction in non-rotating mass.
And having just changed wheels on my S4, these stock wheels are heavy mothers compared to what I am used to when changing wheels on the Civics and Miatas that I typically autocross.
Last edited by pms; 04-24-2009 at 02:23 PM.
#6
03-03-2010, 11:04 AM
Weight Ratios
This is one of the few correct approaches to the rotating v/s body mass issue. However, I think a small correction is due. The R^3 term has to be replaced with R^2, as only two terms (a = alpha*R and v = omega*R) create R-divider. Correct me if I misunderstand your thought. The numeric result is not affected and is correct, because of the value of 1 (1^3 = 1 = 1^2)!
.................. The power needed to increase the rotational velocity of a wheel is P(rotation) = I*alpha*omega, where I = rotational moment of inertia of the wheel & tire, alpha is the rotational acceleration (in radians per sec squared), and omega is the instantaneous rotational velocity (in radians/sec). Alpha is related to the linear acceleration of the car by a = alpha*R, and omega is related to the car's velocity by v = omega*R. So the power required for rotational acceleration can be expressed in terms of a and v by: P=Iav/R^3. As for the value of I: for a 1 pound mass concentrated at a distance r from the center of rotation, I = mr^2. Putting it together: P(rotation) = I*alpha*omega = mr^2av/R^3. Since r = 3/4 ft and R = 1 ft, that means you have P(rotation) = 9/16 * mav. ...................................
#7
03-03-2010, 09:02 PM
But can you feel the difference?
My s4 has final outlived its warranty and my need to drive it only in the summer. As such i find myself looking for wheels. Either I buy crap rims/winter tires or nice rims/use stock for winter.
That decision has me wondering. Can you really tell the difference when you are driving with lighter rims?
Or should I just save my $$$ for a chip/exhaust/etc.
That decision has me wondering. Can you really tell the difference when you are driving with lighter rims?
Or should I just save my $$$ for a chip/exhaust/etc.
#8
03-03-2010, 11:18 PM
Just to toss this into the physics soup... When considering wheels, also consider thier diameter. A lighter wheel will of course accelerate faster, but what about a smaller wheel? I don't know the exact calculations to weigh the two but think of gears and mechanical advantage. A 16" rim will feel faster than a 19" rim because of the increased distance needed to travel for 1 full rotation. Maybe someone with more mathematical know how can show the difference in acceleration for difference size wheels.
#9
08-08-2011, 09:56 AM
hey guys, i know this is a old thread, just wanted to add a little detail.
Your diameter of your tire is not only your rim/wheel diameter. Wheel diameter maters by width and profile. For example a 215/55/r16 tire has a greater circumference/diameter (25.3in) compared to a 205/55/r16 (24.8in).
Since this value is squared or cubed it can change your numbers substantially.
it has to do with the gearing of the drivetrain as a member of the forum mentioned.
I usually just use this website: Tire Diameter And Circumference Calculator
or anything similar. I have the formulas somewhere but this is much simpler. Hopes it helps someone.
Your diameter of your tire is not only your rim/wheel diameter. Wheel diameter maters by width and profile. For example a 215/55/r16 tire has a greater circumference/diameter (25.3in) compared to a 205/55/r16 (24.8in).
Since this value is squared or cubed it can change your numbers substantially.
it has to do with the gearing of the drivetrain as a member of the forum mentioned.
I usually just use this website: Tire Diameter And Circumference Calculator
or anything similar. I have the formulas somewhere but this is much simpler. Hopes it helps someone.
#10
08-09-2011, 07:53 AM
Presentation
Drope me a note at Metal Processing & Metal Forming Consulting by IBD Connection, Inc. and I send you an presentation on the subject.
Thread
Thread Starter
Forum
Replies
Last Post
