Calc (inside) shows very little extra power/torque required for 12 lb heavier wheels
05-30-2008, 01:40 PM
#1
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Calc (inside) shows very little extra power/torque required for 12 lb heavier wheels
I've heard for sometime that extra weight of wheels/tires is very noticeable in acceleration of our cars. I've always understood rationale, but being a mechanical engineer I thought it would be "fun" to actually calculate it.
My detailed calcs are as follows:
My tire/wheel combo weighs 53 lbs (found from STFA). I looked at tire rack and a 235/45/17 tire weighs 22 lbs like my 235/40/18. So my wheel weighs about 31 lbs.
BD has lightwieght Volks wheels that weigh 18.5 lbs (per a previous post by him), therefore I could save about 12 lbs by changing wheels and little/no weight by changing tires.
I estimated that the center of mass of the extra weight of my wheels "acts" at a radius of 2/3 of the wheel radius or about 6" from the centerline.
The formula for the moment of interia (I) is 0.5*m*r*r where m is weight in lbs (12 lbs) and r is 0.5 ft (6 inches). This gives a moment of inertia of 1.5 lb-ft-ft.
Torque (T)is T=I*alpha/g where alpha is angular acceleration in radians/sec/sec and g is acceleration of gravity or 32.2 ft/sec/sec.
If you're still reading, then hopefully you are still following this.
Angular acceleration is calculated as follows.
Initially, it's zero at stop.
A 235/40/18 tire has approx 800 rev/mile, so at 60 miles per hour, it has 48000 rev/hour or 13.3 rev/sec.
If my car can accelerate from 0 to 60 mph in 6 seconds, then the angular acceleration is 13.3 rev/sec divided by 6 sec or 2.22 rev/sec/sec (assuming constant acceleration).
There are 2*PI radians in one revolution, so angular acceleration is 2*PI*2.22 = 14 rad/sec/sec.
Therefore the torque required to spin 12 lbs located 6 inches from axle from stop to 60 mph in 6 sec is
(1.5 lb/ft/ft)*(14 rad/sec/sec)/32.2 ft/sec/sec which equals 0.65 ft-lb. Multiplying by 4 wheels gives 2.6 extra ft-lb of torque.
To translate into horsepower, we need to know the average angular speed of the wheel. If acceleration is constant from 0 to 60 mph, then the average angular speed of the wheel occurs at 3 secs (half way between 0 and 6 sec) and is
(14 rad/sec/sec)(3 sec) or 42 rad/sec.
HP = Torque*angular speed/550 or
HP = (2.6 ft-lb)*(42 rad/sec)/550 = 0.2 HP
Any (constructive) comments?
My detailed calcs are as follows:
My tire/wheel combo weighs 53 lbs (found from STFA). I looked at tire rack and a 235/45/17 tire weighs 22 lbs like my 235/40/18. So my wheel weighs about 31 lbs.
BD has lightwieght Volks wheels that weigh 18.5 lbs (per a previous post by him), therefore I could save about 12 lbs by changing wheels and little/no weight by changing tires.
I estimated that the center of mass of the extra weight of my wheels "acts" at a radius of 2/3 of the wheel radius or about 6" from the centerline.
The formula for the moment of interia (I) is 0.5*m*r*r where m is weight in lbs (12 lbs) and r is 0.5 ft (6 inches). This gives a moment of inertia of 1.5 lb-ft-ft.
Torque (T)is T=I*alpha/g where alpha is angular acceleration in radians/sec/sec and g is acceleration of gravity or 32.2 ft/sec/sec.
If you're still reading, then hopefully you are still following this.
Angular acceleration is calculated as follows.
Initially, it's zero at stop.
A 235/40/18 tire has approx 800 rev/mile, so at 60 miles per hour, it has 48000 rev/hour or 13.3 rev/sec.
If my car can accelerate from 0 to 60 mph in 6 seconds, then the angular acceleration is 13.3 rev/sec divided by 6 sec or 2.22 rev/sec/sec (assuming constant acceleration).
There are 2*PI radians in one revolution, so angular acceleration is 2*PI*2.22 = 14 rad/sec/sec.
Therefore the torque required to spin 12 lbs located 6 inches from axle from stop to 60 mph in 6 sec is
(1.5 lb/ft/ft)*(14 rad/sec/sec)/32.2 ft/sec/sec which equals 0.65 ft-lb. Multiplying by 4 wheels gives 2.6 extra ft-lb of torque.
To translate into horsepower, we need to know the average angular speed of the wheel. If acceleration is constant from 0 to 60 mph, then the average angular speed of the wheel occurs at 3 secs (half way between 0 and 6 sec) and is
(14 rad/sec/sec)(3 sec) or 42 rad/sec.
HP = Torque*angular speed/550 or
HP = (2.6 ft-lb)*(42 rad/sec)/550 = 0.2 HP
Any (constructive) comments?
05-30-2008, 02:55 PM
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with the very distinct possibility of showing my ignorance, I'll ask: would you not
need to multiply the results by four, since you have 4 wheels?
I'm more concerned about torque than HP as that's what makes the car giddy-up.
So, total torque increae req'd would be 2.6 lb-ft x 4 = 10.4 lb ft. For my car (2.7T) that is roughly 3.8% of torque available.
Did I figure this right?
I'm more concerned about torque than HP as that's what makes the car giddy-up.
So, total torque increae req'd would be 2.6 lb-ft x 4 = 10.4 lb ft. For my car (2.7T) that is roughly 3.8% of torque available.
Did I figure this right?
05-30-2008, 04:00 PM
#6
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Onyl that I believe the center of mass is way out near the edge....
..probably somewhere in the region of the tire sidewall, while, if I read right, you had it in the spokes.
The tire mass > wheel mass. And most of the wheel mass is in the rim itself, and the hub, or so I presume from appearances. So I'd place it much further out.
Did you do a delta vs standard wheels/tires? I didn't see it in my scan. So you got the power to drive the wheels, but not the change vs a different wheel/tire. What you have is vs no tires at all, which i would predict to generate less forward motion - call me crazy.
Nice job. You do in fact have too much time on your hands.
G
The tire mass > wheel mass. And most of the wheel mass is in the rim itself, and the hub, or so I presume from appearances. So I'd place it much further out.
Did you do a delta vs standard wheels/tires? I didn't see it in my scan. So you got the power to drive the wheels, but not the change vs a different wheel/tire. What you have is vs no tires at all, which i would predict to generate less forward motion - call me crazy.
Nice job. You do in fact have too much time on your hands.
G
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05-30-2008, 04:05 PM
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Wouldn't the extra mass and subsequent inertial moment help counter rolling resistance?
Thereby freeing up some torque at speed? You can recapture some of the energy wasted while accelerating (spinning up the rims) by coasting down in neutral 