# Moment of inertia of a large flwheel

I am wanting to build an engine inertia dyno. Finding someone capable of machining and balancing a large diameter flywheel has been difficult. Can anyone tell me if the moment of inertia for a larger flywheel be duplicated using multipe smaller diameter flywheels connected together on the same shaft. Or is there a formula to use to calculate how many and what size I would need to use to duplicate the larger flywheel

Well, yes you can replace a large flywheel with several smaller ones, BUT you need to keep in mind that the relationship isn't linear, so you can't just take 3 1/3 sized ones to replace the large one.

This is for a rectangular cross-section (for easy explanation) - here the height is h^3 (whats the english word for this?)

Ix = b h^3 / 12

where

b = width

h = height

More here http://www.engineeringtoolbox.com/area-moment-inertia-d_1328.html :)

I have not been able to find much information regarding the proper size of flywheel I will need for the dyno

can someone show me the formulas I would need to use to calculate the moment of inertia required of a rotating flywheel driven by a 2 cycle engine rated at 28.2 hp and 15.5 ft/lbs of torque.through a gear reduction of 6:1 (like to keep rpm of flywheel below 2500 rpm )
Accelerate the engine from appprox 3,000 rpm through 14,200 rpm in approx 10 seconds

28.2 hp @ 11,500 rpm
15.5 ft/lb @ 8,750 rpm

Contrary to what Claws has said, the relationship is simply additive. Thus, if you put three small flywheels on a common shaft, their combined inertia is
Jeff = J1 + J2 + J3
so the combination is indeed linear when they are all on a common shaft.

dyno1, you ask for the means to calculate the moment of inertia (MOI) required for a 2-cycle engine operating over your specified speed range. At the upper end of that range, the answer is almost certainly that no MOI is required because the internal MOI of the engine parts is sufficient to enable the engine to compress the charge and continue operation. The only reason a flywheel is really required (it may be desirable for other reasons, but the only reason it is absolutely required), is to store enough kinetic energy to enable the system to compress the charge on the upstroke and come to top dead center when it can fire again. This is a problem only at low crank speeds, so at higher speeds, the engine can operate without a flywheel. A flywheel is often advantageous in order to give a smoother rotational speed, but this is simply to reduce fatigue in the other parts of the machine train.

Contrary to what Dr. D has said, Claws statement is correct. Claws was referring to three flywheels all 1/3 the diameter of the original. Dr. D is referring to three flywheels all 1/3 the moment of inertia of the original.

minger
Correct Mike and Claws. The equation for mass moment of inertia of a flywheel will simply be that of a cylinder:
$$I = \frac{mR^2}{2}$$
However, mass itself is a function of R, so:
$$I = \frac{\pi \rho t R^4}{2}$$

It is a rare flywheel that has a rectangular section, so the calculation given by minger, which is apparently the basis for the thinking behind the statement given by Claws, is all simply a very special case.

The original statement simply asked about using three flywheels of a small diameter on a shaft to replicate the effect of a single large flywheel. The answer remains that the net MMOI is simply the sum of the individual MMOI. There is no nonlinearity involved at all.

The nonlinearlity that minger points out quite clearly is the nonlinear relation between MMOI for a single flywheel and the radius of that flywheel, assuming that the flywheel is a flat disk. This all goes out the window when a different cross section is used as is quite common in actual applications.

It is a rare flywheel that has a rectangular section, so the calculation given by minger, which is apparently the basis for the thinking behind the statement given by Claws, is all simply a very special case.

The original statement simply asked about using three flywheels of a small diameter on a shaft to replicate the effect of a single large flywheel. The answer remains that the net MMOI is simply the sum of the individual MMOI. There is no nonlinearity involved at all.

The nonlinearlity that minger points out quite clearly is the nonlinear relation between MMOI for a single flywheel and the radius of that flywheel, assuming that the flywheel is a flat disk. This all goes out the window when a different cross section is used as is quite common in actual applications.
Everything is answered in the thread - read it carefully. Nobody mentions a flywheel with a rectangular cross section - I was in a hurry when I answered the question with the example of a rec. cross sec. It still stands. What are you trying to salvage?