Moment of Inertia Calculations

[jayb]

There's a recent message on the FE Forum regarding using an aluminum flywheel versus a steel flywheel in a performance application.  To quantify the performance differences you need to do a torque calculation using something called the moment of inertia of the flywheel.  I thought it might be of general interest here to show this calculation, at least if you are an FE tech geek like me  

For the purposes of an example, let's pretend that we have a choice of two flywheels, a 15 pound aluminum flywheel or a 40 pound steel flywheel.  The 40 pound flywheel has more inertia, and will be harder to accelerate than the 15 pound flywheel.  How much more torque will it take to accelerate the 40 pound flywheel at 1000 RPM per second than the 15 pound flywheel?

First, we have to calculate the center of mass of the flywheels.  This is the imaginary ring, some distance from the center of each flywheel, where if all the weight of the flywheel was concentrated it would behave the same under acceleration as the flywheel itself.  The easiest way to do that is to find the imaginary ring where half the flywheel's weight is outside the ring, and half is inside the ring.  If we approximate the flywheel as a simple disc 13" in diameter, the diameter of the imaginary ring is 0.707 X 13", or 9.19".  We actually will need the radius of this ring, which is 4.595".

Next we have to calculate the moment of inertia of each flywheel, but we need to get the values into a form that we can use.  In English units, this means that we will calculate the moment of inertia in slug-feet.  In order to do this, we have to convert the radius of our imaginary ring to feet, which is 4.595/12, or 0.383 feet.  We also have to calculate the mass of the flywheels, by dividing their weight in pounds by the acceleration due to gravity, 32.2 feet per second per second.  For the aluminum flywheel, this is 15/32.2, or 0.466, and for the steel flywheel this is 40/32.2, or 1.242.  Moment of inertia is the mass times the radius of the ring squared:

Moment of inertia of aluminum flywheel = 0.466 X 0.383 X 0.383 = 0.068

Moment of inertial of the steel flywheel = 1.242 X 0.383 X 0.383 = 0.182

To calculate the torque required, we have to multiply the acceleration rate by the moment of inertia.  We need the acceleration rate in radians per second per second, but we have it in rotations per minute per second.  To convert one rotation to radians, multiply by 6.28, and to convert from rotations per minute to rotations per second, divide by 60.  So, 1000 RPM (Rotations per minute) per second = (1000 X 6.28)/60, or 104.66 radians per second per second.

Torque required to accelerate the 15 pound aluminum flywheel at 1000 RPM per second is 0.068 X 104.66 = 7.12 lb-ft of torque.

Torque required to accelerate the 40 pound steel flywheel at 1000 RPM per second is 0.182 X 104.66 = 19.05 lb-ft of torque.

So, if you switch out the 15 pound aluminum flywheel in your 428CJ Mustang and replace it with a 40 pound steel flywheel, as you accelerate in third gear at 1000 RPM per second, you have lost about 12 lb-ft of torque all along the acceleration curve.  This is because this torque is required to accelerate the additional mass added by the steel flywheel.  In terms of horsepower, at 3000 RPM this is a 6.85 HP loss, and at 7000 RPM it is a 16 HP loss.

This calculation can be applied to any part of the rotational drive components of the vehicle, including the driveshaft, axles, transmission components, and of course the reciprocating assembly.  For anything other than a round disc, the center of mass portion of the calculation is more complex, but an estimate can be made in most cases.

I was able to confirm these general results with some dyno tests comparing steel and aluminum flywheels back in 2006.  I don't have the test results handy, but if I run across them I will post them in this thread.

Hope this is of some general interest  

[My427stang]

Good stuff Jay

First, let me ask you to break the calculator back out, as an MBA and Strategic Intelligence grad, I'll have to lean on you to tell me how to mathematically get to the answer to my question

So, if you look at what you calculated, could you back it into a "resistance to change" calculation?

What I am trying to say is if you had a an engine at 450 ft lb at a steady 2000 RPM with one flywheel, is there enough information to calculate how much difference in torque the motor would have to provide to keep one or the other flywheel at 2000 rpm with an increase in load? 

I assume there would be time and load variables in there too, because once the inertia is "eaten up" the motor would have to work harder to keep the wheel spinning.

Thanks

[jayb]

Ross, that is not an easy one to answer.  The factors that will reduce the speed of the car from a steady state are wind resistance and friction, from internal components and the tires on the road.  I really don't know the answer, but intuitively I think I would make the following argument:  You could lift the throttle at the speed of interest, and determine the rate of the speed decrease of the car.  So, this is the rate of deceleration.  It has to be countered by an equal rate of acceleration in order to keep the car at a constant speed.  

To test out this idea let's say that that the car has a 3.91 gear and is in fourth gear at 2000 RPM with a 26" tire.  This means it is going 39.58 MPH (MPH = [Tire Diameter X RPM]/[336 X Gear Ratio]).  Next, let's say that when the throttle is closed, the car slows down to 30 MPH in 10 seconds.  Using the preceding formula 30 MPH would be equivalent to 1516 RPM.  So, the rate of deceleration would be 484 RPM per 10 seconds, or 48.4 RPM per second, or 5.06 radians per second per second, as calculated using the formulas in my original post.

Now, if we assume that this deceleration must be matched by an acceleration to keep the speed of the car constant, then the torque numbers for the two flywheels work out like this:

Aluminum flywheel:  .068 X 5.06 = 0.344 lb-ft

Steel flywheel:  .182 X 5.06 = 0.92 lb-ft

So the difference is very small, which seems intuitively correct.  But to be honest I have no idea if the assumptions that I based this calculation on are really correct.   I guess I'm not really sure how to analyze that problem

[machoneman]

Nice to see the actual calculations Jay. Have seen something like this before but not nearly as well put.
I'm surprised though that switching to a much smaller 'wheel (more than 50% smaller) only results in such a small proportional loss in torque....thought it would be even more.

Ross has a very good question though.  I'll put it anther way. When I used to race door slammers long ago, the trick was to use a heavy flywheel (think: very high-winding SBCs) and one would sidestep the clutch at launch.  This got out of hand with high class (G/MP, H/MP and up) Modified Production and Gas cars that actually ran 60# wheels in an attempt to improve the launch with relatively heavy cars and avoid smoking what were organic clutch discs. This was a real parts-breaker combo btw.

But this was in the days before more modern slipper clutches came into vogue like the (mainly) Long style clutches, now called Soft-Loks, that have an adjustable pressure plate and/or base spring loads that allow mucho slippage to keep RPM's up with long-wearing sintered metallic discs. Not only did this allow lighter flywheels but also minimized driveline shock breakage to a great extent. Sad that we stopped racing doorslammers just as these modern clutch combos came about as we broke tons of pricey parts racing a 3,600 lb. 427 BBC Camaro leaving at about 7,500 via a clutch dump. Anyway that's all racing stuff.

Street wise and having seen the other replies on the old Forum, unless one is running a slipper clutch on the street and/or is making big hp the heavier steel wheel IMO is the way to go, especially when we mere mortals don't have Jay Brown Engineering hp under the hood...LOL!

Seriously though, I've also dealt much more recently with SCCA corner carvers (think: underpowered Japanese and VW street cars) who switched to very lightweight wheels for gymkhana events and goofing around. They like them for track work but the constant slipping at takeoff (leading to very fast disc wear) and that weird issue of the difficulty in keeping a constant rpm level at cruise speeds both have lead some to switch back.

The moral is in my book ...it depends on just how much inconvenience one will tolerate, how heavy one's ride is and how much hp they are making.  


 


  

  

[jayb]

Those are all good points Bob, and it really does boil down to what you are willing to put up with on your street car.  I guess I'm willing to put up with more than most.  One thing that confuses me though is the comments from you and some others that with an aluminum flywheel you would have a hard time maintaining cruise RPM.  I have never, ever noticed that to be a problem.  Having said that, the cars I have with manual transmissions are not as high horsepower as some of the others...

[My427stang]

Thanks Jay, if the engineers can't do the math, I'll just say its FM LOL

As far as the sustaining cruise RPM, I assume you have seen those 1-lunger farm motors.  They run a heavy flywheel to store energy and the engine fires to keep it going.

If thats one far limit of a heavy wheel and the other limit would be no flywheel at all, I am looking it it as at low/steady RPM the heavy wheel would be more efficient at a steady cruise, especially at the bottom end of the torque curve.  Essentially behaving the same as the small motor/hard launch mentioned above but with constant speed.

There is no doubt the motor could compensate with a little throttle, but my thought is that it would be happier with the big wheel.

The question of course is, how much happier at cruise compared to how much happier during acceleration? 

I still go back to Steve Engberg and his expertise on the the Mach, but a sample size of one guy who didn't like the alum wheel vs one who did doesn't get us very far either

Thanks for taking the time to work the math, its interesting if nothing else

[mmason]

Oh how I wish I could follow along with the math formulas but maybe in another life. Anyway I always thought that the benefits of an aluminum flywheel in a four speed would mostly be in first gear and maybe some in second but hardly any in forth because of the quick acceleration in first and relatively slow acceleration in forth.

I went from a 40# flywheel to a 20# years ago and the only problem I in countered was driving through a parking lot at 3 or 4 miles per hour. I have to keep letting the clutch out to stop the car from lurching.

Michael Mason

[hotrodfeguy]

I think on a BB one should also consider that the crank weighs in a lot more than some 289/302 crank. There for a alloy wheel is just fine. The only car I ever drove I think worthy of a factory wheel was a 69 B2 man that thing needs a hand getting to 3K in first gear LOL