Here's the deal. I have three gallons of fuel I would like to reformulate. I would appreciate verification or correction of my notions below.

Given: 1 gallon of 29% all castor oil, 10% nitro.

Idea: add to 1/3 gallon of the 29% fuel (above) 15+ oz. Klotz, 10 oz. nitro, and 59 oz. methanol

I believe this will yield: 1 gallon of 22% oil (45/55 castor/Klotz), 11% nitro fuel

I arrived at this by computing the number of ounces of castor in 29% fuel (37 oz.), 10% nitro (13 oz.), and methanol (78 oz.), then figured how much Klotz (83 oz.), nitro (43 oz.), and methanol (255 oz.) it would take to make 3 gallons of fuel with a 22/11/67% ratio, then divided the amounts by three.

These numbers are approximations, and I rounded off to make it easy on myself for measuring purposes without getting too far off.

Do you think this is "good to go," or have I miscalculated in some major way.

Thanking you in advance for your consideration, Dzl

To get 10% castor, 12% Klotz, and 11% nitro I figure you would add 9.3 oz nitro, 14.8 oz Klotz, and 56.9 oz of methanol to that 1/3 gallon of the roiginal fuel for a total of 123.7 oz of the mixture. I'd say your calculations are well within the ballpark.