Fuelman
Posts: 980
Joined: 12/31/2001
From: Jordan,
NY, USA
Status: offline
|
Hauckf;
Lots of numbers, several ways to look at them. You are correct, to be exactly accurate the increase in contents needs to be taken into consideration. Let me try to illustrate what you are refering to for the modelers that might be getting lost with numbers.
If I were to make two gallons of fuel, for example two gallons of 10% nitro, one gallon has 18% oil and 72% methanol the other has 20% oil, 70% methanol, the contents would be as follows:
#1 (nitro to methanol ratio of 1:7.2)
N: 12.8 oz
O: 23.04 oz (18%)
M: 92.16 oz (72%)
#2 (nitro to methanol ratio of 1:7)
N: 12.8 oz
O: 25.6 oz (20%)
M: 89.6 oz (70%)
Now, if I wanted to take the 18% oil jug of fuel and doctor it up so it has around 20% oil in it, I would add 1.28 oz per percentage point increase or 2.56 oz total to the mix., the resulting mixture would be a total of 130.56 oz broken down as follows:
#1+ (130.56 oz, with the added oil), (nitro to methanol ratio 1:7.2, same as it started out)
N: 12.8 oz (9.8%)
O: 25.6 oz (19.6%)
M: 92.16 oz (70.6%)
Now, to make the fuel run exactly like a gallon of 10/20 factory made, you would have to adjust the nitro up 0.2% , the oil 0.4% and that would drop the percentage ( by volume) of the methanol down by 0.6%, making the chemical mixture , by percentage exactly the same as the factory made 10/20 fuel.
OK, lots to soak in. For the purpose and intent of the modeler that has a need to adjust the oil percentage of oil up a few percent, the ballpark 1.28 oz per percentage puts you in the right ballpark. For greater adjustments than just a few percent, a little simple equation is required. For modelers that require exact final proportions (assuming you are measuring by volume), here's the formula that works in english or metric usits of volumetric measure. this formula will not work when calculating by weight measurement since each component has different specific gravities, and it only works with one component alterations.
z(x+y)= x(s) + y
x= starting volume of fuel.
y= volume of oil to be added
z= final percentage of oil
s= initial percentage of oil
To use our example above of 10/18 fuel to start adding to make it a true 20% oil by volume. (You'll still need to add nitro)
z(x+y) = x(s) + y
.20(128+y) = 128(.18) + y
25.6 + .20y = 23.04 + y
25.6 = 23.04 + .80y
2.56 = .80y
3.2 = y
Now, using the formula, here's what we get so you have comparison to the above, lets call it #1B:
#1B (131.2 total ounces of fuel) (nitro to methanol ratio is still 1:7.2)
N: 12.8 oz (9.75%)
O: 26.24 oz (20%)
M: 92.16 oz (70.24%)
You folks can make your own discussion from here, so I hope this is clarified for the purpose of exactness. As I stated previously, the earlier 1.28 per percentage point gets you in the ball park. Now for the average flyer that adjusts fuel oil, lets keep in perspective a couple things; are you able to measure by volume this level of accuracy for the small amounts required in adding to a gallon. and how many engines can tell the difference between 9.75% nitro and 9.8% nitro or between 19.6% oil and 20% oil?
Very observent point Haukf, thank you for your contributions to the forum.
_____________________________
Fuelman
Cooper Fuels LLC
|
Printable Version
All Forums >> Glow Engines, Gas Engines, Fuel & Mfg Support Forums >> RC Fuels >> How much oil needed to bring 16% oil up to 20% 
How much oil needed to bring 16% oil up to 20% - 
Cox Texaco Fuel - 
RE: Changes? - 
RE: Water in fuel -