Busher

Hi there !!!

I have been trying to work out what the finished percentages are when I add oil to a Ready mixed commercial brew. If the 4 litre brew contains 17% oil, 10% nitro, and 73% Methanol,
And I add 3% oil to bring it up to 20% oil, I have now affected the other ingredients ratio's what formula do you use to calculate what the results will be. I have tried allsorts of different calculations non of which seem to give me the correct answer. My Mathematics is not the greatest but ratio's and percentages sometimes get the better of me.

Can anybody guide me along to the right method of calculating this correctly..

Thanks

Busher

downunder

Use 1 litre of your original fuel as a starting point (it's easier to do the numbers ).
17% oil is 170cc
10% nitro is 100cc
73% methanol is 730cc
Add it all up and it makes 1000cc or 1 litre.

If the extra 3% oil added is calculated on 1 litre then you'd add 30cc of oil. So now you have 200cc of oil in that fuel. But you now have a total of 1030cc of fuel. The actual % of oil in that fuel is now (200/1030)*100=19.4%
200cc oil (19.4%)
100cc nitro (9.7%)
730cc methanol (70.8%)

But don't be concerned if it looks like you're now down in both nitro and methanol % because that's just the total fuel content. The actual fuel part of the mixture still has the same ratio of nitro to methanol and that's what the needle valve tunes for.

Red B.

Enclosed below you will find an Excel spreadsheet that will help you with your calculations.
In order to upload it I had to change the file extension to .jpg
Download the file (right click and Save File As), rename the file Homebrewer.xls and open the file in Excel.
The spreadsheet works with arbitrary units and does not contain any macros. Enjoy!

/Red B.

Firewalkerd1

My Feedback: (1)

I am also trying to up the oil content of some fuel that I won at a fly in. The fuel is 15 % nitro, 18 % synthetic oil. I would like to run a little castor. If I add 2 oz. of castor would this bring the fuel up to 20 % oil ? I'm I correct to say 1 oz. = 1 cc of oil. Will adding the castor change the running of the fuel as for as power?

Red B.

We need to know the volume of the fuel you have in order to do any calculations. For the example I assume you have 1 US gallon of fuel = 3.875 liters and that the percentages refers to volume.

The amount of oil in the fuel is 18% of 3.875 liters = 0.6975 litres
The remainder is "fuel" = 3.875 - 0.6975 = 3.1775 litres
Assuming that you want to up the oil content to 20%, the equation you need to solve is

X / (X+3.1775) = 0.20 ==>
X = 0.2 x 3.1775 / (1-0,2) = 0.6355 / 0.80 = 0.7944

i.e., you need 0.7944 liters of oil in the fuel, but you already have 0.6975 liters.
The difference 0.7944-0.6975 = 0.0969 liters should be added to your fuel.

0.0969 liters = 96.9 milliliters = 3.28 fl.oz. (US)

Some conversion factors:
1 oz = 28.34952 g (SI)
1 fl oz (US) = 29.57353 ml (SI)
1 fl oz (UK) = 28.41307 ml (SI)
1 ml (milliliter) = 1 cm3 (cubic centimeter) = 1 cc = 0.001 liter

/Red B.


edit: included some conversion factors

downunder

No, it won't make any difference to the power.

Busher

Thanks for all your Help, I will try to make sense of all the calculations and apply it to my mixes.

Red b

Thanks for the calculator It works a treat..

Busher