Fuel Formulation

Oct 23, 2005 3 Replies

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pcoopy 20 years ago

I know I had a formula written down somewhere (probably in 8th grade about 1953) to calculate how much castor to add to a given volume of fuel, with a given amount of lube already in it, to raise it to another level (%by volume).

For example: I want to raise 1 gallon (128 oz.) of fuel, which has about 18% (23.04 oz.) lube in it, up to 28% lube.

Considering that whatever you add will raise the total volume which makes the total% a moving target, there was a specific formula that I had at one time.

How do I calculate how much much castor, by volume, to add?

Phil AMA609

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Joe Bill 20 years ago

I'm no math whiz, but playing around I came up with the following:

23.04oz oil + ??? oz oil

-------------------------------------------- = 0.28 (28%)

128 oz total volume + ??? oz oil

I'm gonna forgo units cause it only confuses me...

23.04oz oil + ??? oz oil = 0.28 x (128 oz + ??? oz oil) 23.04 + ??? oz oil = 35.84 + 0.28 x ??? oz oil 0.72 x ??? oz oil = 12.8 12.8 / 0.72 = 17.7778

So ??? = 17.7778 oz oil to be added to make 28% oil, new total volume =

145.7778 oz

To check:

23.04 oz originally + 17.7778 oz added 40.8178 oz oil

----------------------------------------------------- --------------------------------- = 0.279996 (27.9996%) 145.7778 new total volume 145.7778 total volume

Hopefully there's an easier way, but I think this'll work if you can follow the algebra.

Jim

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Paul McIntosh 20 years ago

It really isn't that critical and I doubt that you would be able to accurately measure those tiny increments anyhow.

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pcoopy 20 years ago

Jim Thanks for the start, I dusted off my old algebra gears and came up with this:

V(N%-O%) X = ----------- 1 - N%

Where: V=starting volume, N%=new percentage, O%=old percentage, x= amt to add

So 128(.28-.18) X = ------------ 1-.28

12.8 X = -------- .72

X = 17.7777777778 = amt to add to 1 gal of 18% lube to get to 28% lube.

All this to mix FOX 35 fuel!

Phil AMA609

Joe Bill wrote:

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