How much oil to add to a gallon of fuel to get a specific percentage - RC Groups
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Oct 28, 2016, 03:04 PM
A man with a plan
Balsaworkbench's Avatar

How much oil to add to a gallon of fuel to get a specific percentage

This question comes up from time to time, and it's a bit of a head scratcher. For anybody who wants to know how much oil to add to a gallon to get a specific percentage of oil in the final mix (which will be more than a gallon), here's a formula.

A= ounces of additional oil
L = original lube percentage expressed in a decimal. ( 17% is .17)
F = final lube percentage expressed as a decimal.

128L + A = F (128 + A)

Or more simply, A = F (128 + A) - 128L

If you have a gallon (128 oz) of fuel with 17% lube, how much oil do you add to get 20% oil in your final mix, which will be more than a gallon after you add the oil?

A = .2 (128 + A) -(128 x .17)

A = 25.6 + .2A -21.76

A = 3.84 + .2A

.8A = 3.84

A = 4.8

Add 4.8 ounces to get 132.8 ounces of fuel with a total of 20% oil.
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Oct 28, 2016, 03:55 PM
the great Gassif´er
Great formula!

Not to be pedantic, but just as an additional:

For the people working in metric, they would use litres and ml, and the formula would be a lot easier to work with if it is rewritten to this form:

(1-F)A=F-L or even simpler : A=(F-L)/(1-F)

For the same values of initial L=17%, desired F=20%

0,8 x A=0,2-0,17 or: A=(0,2-0,17)/0,8

A= 0.0375 or in plain language, 3,75% of the original total volume, to be added in oil. That way it does not matter if you have a batch of a quart, 2,5 litres, a barrel or whatever amount of fuel you have....

(4,8 ounces is 0.0375 gallon: 0,0375 x 128=4.8)

As a check: 17% fuel would contain (duh!) 170 ml oil
1,0375 L of fuel containing 207,5 ml of oil: (207,5/1037,5)=0,2... Check!
Last edited by Brutus1967; Oct 28, 2016 at 04:15 PM. Reason: Stupid writing error, some additional things
Oct 29, 2016, 12:59 AM
Registered User
gerryndennis's Avatar
So what you both are saying is;

If I have 128 ounces of 17% fuel then 83% of it is methanol, so 128*.83=106.24 of methanol in that gallon.

But I want that 106.24 of methanol to be 80% of the total fuel (ie 20% oil) so 106.24\.8=132.8 ounces of fuel.

So I need to add 132.8-128=4.8 ounces of oil to make the 132.8 total?

Sounds about right.

Dave H
Oct 29, 2016, 01:13 AM
Registered User
gerryndennis's Avatar
I also use my method if I want convert a three (or more) component fuel to something else eg a gallon of 17% oil, 10% nitro fuel to 17% synthetic, 3% castor, 15% nitro fuel.

It also works if I want to end up with a different volume as well.

Eg if you wanted to convert that original gallon of 17% to 2 gallons of 20%.

I just find it easier this way, but I'm sure a formula could be worked out as well.

Dave H
Last edited by gerryndennis; Oct 29, 2016 at 01:19 AM.
Oct 29, 2016, 02:11 AM
the great Gassif´er
Originally Posted by gerryndennis
So what you both are saying is;
Yup... that is what we both are saying....

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