Flywheel weight theory?

[Kerry j]

OK, what is the theory on flywheel weight? I see there are 40, 28 & 14 lb options for my FE projects. I'll be putting them in pretty heavy cars; a 61 Starliner and a 64 Galaxie; both have 4 speed wide ratio toploaders and 3.91 rear gears. The toploaders will both have the lower 2.90 first gear as well.

Just a bit confused about why the different weights and what their advantages and disadvantages are.

[rcodecj]

The heavier the flywheel the more energy it stores, which when you take off from a dead stop helps to get a heavier car going.
Lighter cars can use a lighter flywheel and it won't affect them as much, because they don't have as much mass to get moving.  I've seen the real light ones used in road racing. I broke my engine in without putting the torque converter on and you could sure tell when you revved the engine that it picked up quicker. Of course that's without trying to take off in a car, just revving it in park.

[jayb]

In addition, heavier flywheels take power to accelerate.  This means that the car will accelerate more quickly with a lighter flywheel.  To put an actual number on this, you need to do a moment of inertia calculation.  Let's take a couple of examples, one with your 14 pound flywheel and one with your 40 pound flywheel. 

First, you have to calculate the center of mass of the flywheel.  For a disc like a flywheel, this is .707 multiplied by the radius.  Let's assume the flywheel diameter is 13".  Radius is 6.5", so the center of mass will be .707 * 6.5 = 4.59".  We will need this in feet, so 4.59" is about 0.38 feet (4.59/12).

Next you have to calculate the Gs of the flywheel, which is just its weight divided by 32.2 ft/sec^2.  For the 14 pound flywheel, this is .434 slug-feet, and for the 40 pound flywheel this is 1.242 slug-feet.

Now you can calculate the moment of inertia, which is just the weight in Gs multiplied by the center of mass radius squared.  So, for the 14 pound flywheel, the moment of inertia is .434 * .38 * .38, or 0.0626.  For the 40 pound flywheel, this calculation is 1.242 * .38 * .38 = 0.179

Finally, to determine how much torque it will take to accelerate each flywheel, we have to decide on an acceleration rate.  Let's pick 1000 RPM per second to make it easy.  We need to convert this acceleration rate into radians per second squared, so we need to divide by 60, then multiply by 2pi (6.28).  So, 1000 RPM per second is the same as 104.6 radians per second squared. 

Torque equals Moment of Inertia multiplied by Angular Acceleration.  For the 14 pound flywheel, a torque of 104.6 * 0.0626, or 6.5 lb-ft, is required to accelerate the flywheel at 1000 RPM per second.  For the 40 pound flywheel, a torque of 104.6 * 0.179, or 18.7 lb-ft, is required for the same acceleration rate.

The engine has to deliver this torque to the flywheel in order to accelerate the car, so that means that the 40 lb flywheel will result in a reduction of 12.2 lb-ft of torque going to the transmission input compared to the 14 lb flywheel. 

The morale of the story is use a light flywheel if you can.  A heavy flywheel will steal torque from the rest of the drivetrain.

[rcodecj]

and its effect on acceleration. It also covers wheels, driveshafts, and other rotational mass.

http://www.popularhotrodding.com/tech/0602phr_rear_wheel_power_increase/index.html

[Qikbbstang]

http://www.knfilters.com/news/news.aspx?id=3107

[TorinoBP88]

On a street car with less then ideal traction a heavier flywheel can be a good thing to help reduce wheel spin on launch allowing for a smoother acceleration.

Depends on the engine combination. As seen in Jays book, some intake/engine combos have a really steep torque rise, that combined with an advanced cam and a very light drive line could lead a Mustang straight into the ditch... making it not as drivable.