Calculations - the REAL effect of rotating weight

bhuff30 · Jul 8, 2009

There is a common saying for drag racing that says, reducing rotating weight by 10lbs is like removing 100lbs from the car. Well, I wanted to see if this was true or not.

I did a few calculations to see what the actual effect is. The first calculation was a heavier wheel/tire combo. Lets say you bought new wheels and tires and each wheel/tire weighs 10lbs more than the origional combination, and assume that all the extra weight is at the outside, near the tire. In this situation, adding 10lbs of rotating weight has the same effect as adding 20lbs of dead weight. So for the set of 4 wheels and tires, you added an actual 40lbs, but because it is rotating weight, it is as if you added a total of 80 lbs of dead weight.

Then, I thought about the effect of a light weight driveshaft. Lets say an aluminum driveshaft is 10lbs ligher. I assumed the driveshaft is the same diameter of a steel one, and that it is 5" diameter, that 3.73 gears are being used, and that all the weight was at the outside again. In this case, removing 10.00 lbs of rotating weight from the driveshaft has the same effect as removing 26.6 lbs of dead weight from the car. It is interesting that this gain is better than the wheels/tires. The diameter is much smaller, so you'd think the weight has a much smaller effect, but actaully it spins 3-4 times faster than the wheels which makes up for the difference.

So it looks like light weight componets really do make a difference... but no where near the 10x difference in that common saying. Keep in mind, that 100lbs of dead weight is worth approximately .1 seconds in the quarter mile... so your 10lb lighter driveshaft shaves off .026 seconds and your wheel tire combo that weighs 10lbs more per wheel/tire adds nearly .1 seconds to your quarter mile.

sneaky98gt · Jul 8, 2009

Good info.

If you don't mind me asking, how'd you do those "calculations"? Just figure up the rotational inertia of each rotating component? If that's the case, then the radius of each would also have an effect, right? (i.e. larger tire = more dead weight) I'm a physics junkie and I love seeing how physics I learn in the classroom can be used in real life, such as with racing and such.

-Will

bhuff30 · Jul 8, 2009

I just calculated the kenetic energy for a 10lb rotating weight. As you said, the major factors of that kenetic energy is the radius, moment of inertia and the other the rotational velocity (which depends on the velocity of the car). Then, I set the kenetic energy of said rotating 10lb weight equal to the kenetic energy of an unknown mass traveling the same speed (but not rotating). A lot of the terms cancel out, and you can find that unknown mass, which is the equivalent dead weight.

I realized I could do this calculation during my dynamics class. I was a bit bored, so I just started playing with it and made the calcs.

If you are a real physics junkie, try entering calculating average acceleration from your best 60' time. It makes absolutely no sense at all because the numbers don't work. After thinking about why the numbers aren't even close for such a simple calculation, I realized the flaw. The 60' time printed on your timeslip doesn't account for acceleration that occurs before you roll out of the beams. It is very possible the car will reach 5-10mph before your tire completely rolls out of the starting beam. Rolling out of the beams can take several tenths of a second, which in a problem where your 60' time was 1.7 is very significant.

007 · Jul 9, 2009

To lay it out simply for folks... for your chassis the kinetic energy at a given speed (how much energy it took to get it moving this fast) is:

E = 1/2 mv2

For a rotating component its:

E = 1/2 Iw2, where w is the angular velocity and the I is the rotational inertia (mr2)

From the equations above you can see that for your chassis the amount of power needed to accelerate the car increases linearly with mass and with the square of your speed.

For your wheels it also increases linearly with mass but increases with the square of your speed and also the square of the radius of rotation.

007 · Jul 9, 2009

bhuff30 said:
Then, I thought about the effect of a light weight driveshaft. Lets say an aluminum driveshaft is 10lbs ligher. I assumed the driveshaft is the same diameter of a steel one, and that it is 5" diameter, that 3.73 gears are being used, and that all the weight was at the outside again. In this case, removing 10.00 lbs of rotating weight from the driveshaft has the same effect as removing 26.6 lbs of dead weight from the car. It is interesting that this gain is better than the wheels/tires. The diameter is much smaller, so you'd think the weight has a much smaller effect, but actaully it spins 3-4 times faster than the wheels which makes up for the difference.

I get that answer if I use a 5" radius rather than 5" diameter. Its less impressive if you use a 5" diameter (2.5" radius).

This calc is interesting for wheels because you can get an idea of what 15" vs 20" rims do to the cars acceleration. The radius of the center of mass of the wheel increases, but if the final tire diameter stays the same there is no change in the angular velocity. As a result, the rotational chunk of the energy goes up with the square of the radius of the center of mass.

007 · Jul 9, 2009

Here are some calcs showing the effect of 1kg in various places. First is chassis, then a wheel with all the mass at the outside. The next wheels then move the center of mass inward slightly (smaller rim, same final diameter tire), then the 5" diameter driveshaft rotating 3.73 times faster than the wheels.

Chassis Wheel 1 Wheel 2 Wheel 3 Driveshaft
Mass (kg) 1 1 1 1 1
Radius of center of mass (inch) 10 9 8 2.5
Radius (m) 0.2540 0.2286 0.2032 0.0635
Rotational inertia (kg m^2) 0.0645 0.0523 0.0413 0.0040
Velocity (km/h) 100 100 100 100 100
Velocity (m/s) 27.78 27.78 27.78 27.78 27.78
Circumference (m) 1.596
Angular velocity (rads/s) 109 109 109 408
Linear Kinetic Energy (J) 386 386 386 386 386
Rotational Kinetic Energy (J) 386 313 247 335
Total Energy (J) 386 772 698 633 721

007 · Jul 9, 2009

Another interesting result. Since steeper gears spin the driveshaft faster at the same speed, you obviously suck up more energy spinning the driveshaft.

The steeper the gears the more benefit you get from an aluminum driveshaft:

Driveshaft w 3.27s Driveshaft w 3.73s Driveshaft w 4.10s
644 721 791

bhuff30 · Jul 9, 2009

I figured out the problem here. The rotational energy increases with the square of velocity, and because the driveshaft is rotating 3.73 times as fast as the wheels, it is increasing faster. My calculation used a higher speed. I actually used 110mph because that's what my car was trapping. In the wheel example, this velocity completely cancels, but in the driveshaft calculation, it is still present.

007 said:
I get that answer if I use a 5" radius rather than 5" diameter. Its less impressive if you use a 5" diameter (2.5" radius).

Definately an interesting results on the significance of rear end gears. I knew it would be higher with steeper gears, but I didn't immidiately remember it was being squared... intensifying the effect.

96SCstang · Jul 9, 2009

omg my head hurts!! i feel like i just left the 4.6 talk forum and right into the nasa space shuttle forum lol....good info though

Stan Weiss · Jul 9, 2009

bhuff30 said:
I just calculated the kenetic energy for a 10lb rotating weight. As you said, the major factors of that kenetic energy is the radius, moment of inertia and the other the rotational velocity (which depends on the velocity of the car). Then, I set the kenetic energy of said rotating 10lb weight equal to the kenetic energy of an unknown mass traveling the same speed (but not rotating). A lot of the terms cancel out, and you can find that unknown mass, which is the equivalent dead weight.

I realized I could do this calculation during my dynamics class. I was a bit bored, so I just started playing with it and made the calcs.

If you are a real physics junkie, try entering calculating average acceleration from your best 60' time. It makes absolutely no sense at all because the numbers don't work. After thinking about why the numbers aren't even close for such a simple calculation, I realized the flaw. The 60' time printed on your timeslip doesn't account for acceleration that occurs before you roll out of the beams. It is very possible the car will reach 5-10mph before your tire completely rolls out of the starting beam. Rolling out of the beams can take several tenths of a second, which in a problem where your 60' time was 1.7 is very significant.

This is what makes people that drive different class / speed / hp vehicles so impressive to watch. As each vehicle will take a different amount of time to travel the roll out distance. They have to mental adjust to this so as to not red light. Then you can through in deep and sallow staging.

sneaky98gt · Jul 9, 2009

Yea, we did this stuff in my AP Physics course this past year.

bhuff30 said:
If you are a real physics junkie, try entering calculating average acceleration from your best 60' time.

Yea, I just did this real quick, and you're right. Assuming the initial velocity is 0 will get a crazy high number (like what I wish my car would actually do, lol). But as soon as you put in an initial velocity (because of what you described) the acceleration comes back down to a much more reasonable number.

It's cool to see that I'm not the only one who likes seeing how physics and racing (or most anything for that matter) do go hand in hand.

-Will

Stark77 · Jul 10, 2009

my aluminum DS looked much wider (larger diameter) than the stock clunker - not sure on actual specs or dimensions though

Calculations - the REAL effect of rotating weight

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