This has been a very interesting and educational exercise.
I used the illustration and accompanying table that Ved provided above to get me started. I used the thick ring illustration and formula for my flywheel. I also received confirmation from the Machinerys Handbook that the most inertia will come from the outer most part of the flywheel. Also that, in order to simply the calculations a bit, I ignored the area of the spokes and inner hub. They are inner-most and of the smallest area and therefore provide the least amount of inertia.
Next, I used the Online Metal Store weight calculator to determine that the weight per cubic inch of Brass, Steel, and Aluminum were .307, .283, .098 pounds respectively.
Next, I caculated the mass of the outer "donut" ( new technical term for outside portion of a spoked flywheel ). That was easy - calculate the area of a circle x thickness x weight per C.I. of the metal as noted above. First I calc'd the OD and subtracted the calc'd amount for the ID. That gave me the mass for each metal, OD and ID variation on my table.
Finally, I used the Mass Moment of Inertia ( MMI ) formula for the Thick Ring on the table that Vled provided - A.K.A. the "Donut", using the formula M(r1^2 + r2^2)/2. Substituting the info above, I used Mass x (O.D. squared + I.D. squared ) / 2.
Now back to practicalities. Jan Ridders design, roughly translated from metric to inches is about 6" O.D., 5" I.D and .75" thick. This results in a MMI of 41.95 for steel and 45.50 for brass if my arithmetic and Excel skills are working. I changed the O.D., I.D., thickness, and weight to use aluminum and came up with and O.D. of 7", I.D. of 6", and thickness of .750" resulting in a MMI of 42.53. Sounds just about right.
So, there you have it, at least by my calculations, is the answer. 7" x 6" x 1" using aluminum.
Thank you all for your input and contribution to my continuing education. Please pass the aspirin...