Switch from 50:1 to 60:1?

[andoman]

Calling all mathematicians!

I'll agree that the point is true in theory, but I challenge anyone (even Jake!) to be able to tell the difference at the throttle resulting from the ultra minute mixture change due to going from 50:1 oil ratio to 60:1 oil ratio.
Please read on before throwing all the anecdotes at me.

I'll leave the math to someone smarter than I. However, think about the fact that the ideal air/fuel mixture is 15:1 and were talking about a change from 1 part in 50 to 1 part in 60 in only one part of the original 16. I'm betting that is a change so minute that it couldn't even be accounted for with adjustments so course as jets, clips, and needle taper.

Again, I'll leave the math to the arithmeticians. Perhaps, after lots of mathematics (please show your work!), someone can equate the actual mixture change by changing oil ratios to a common change in jetting. I.e, "going from 50:1 to 60:1 oil ratio is like changing your needle clip 1/20th of a step down on an N3CH taper needle, or going from a 168 main to a 168.28 main at the average fuel volumes burned by a 250cc two stroke in the middle of the RPM range at standard atmosphere."
Until someone does the volumetric comparisons I simply will not be convinced.
The answer to the question as to why one may notice a reduction in spooge when they go down to 60:1 is obvious: were talking about only the percentage of oil and not the ratio of oil to fuel in an overall ratio with air.
A change of 50:1 to 60:1 is a change from approx. 2% to 1.66% oil. Enough to notice a less messy tailpipe.

I intend no disrespect nor am I trolling for a heated debate. I'm truly curious. My Forrest Gump mind tells me the math, if I could do it accurately, wouldn't validate the accepted postulate that a change from 50:1 to 60:1 oil ratio is a significant enough change in mixture to re-jet.

[iamovru]

I used to run my 09 300XC at 50:1. I only used Motul 800 and or Klotz Technoplate. They have a higher flash point but I never had serious spooge issues. I also had the FMF on it. I firmly believe that jetting is the biggest issue when it comes to the mess coming out the tail pipe. The only time that it was an issue, was when the bike was started cold. But that goes for every other 2T I have been around.

[Zman]

I was previously aware of the fuel viscosity change issue and oil displacement of fuel affecting jetting, but I just found out that the increased oil in a lower numeric gasil ratio will also reduce the fuel octane rating. Here are some interesting links.

http://airfoolers.com/?page_id=385
http://articles.superhunky.com/4/194

[Zman]

Math based answer - The difference between 50:1 and 60:1 is less than 1 tablespoon oil difference per gallon of gas.

Gut based answer supported by experience making more severe changes in gasil ratios without perceivable effect on jetting - Seasonal changes in air temperature and humidity will affect your jetting more than this minor change in oil between these two ratios.

[Caravan Monster]

When I clean the bike I wipe the spooge off with a dab of solvent and forget about it. So long as the bike runs well, it's not a big deal. (50:1 fully synthetic)

[noobi]

Originally Posted by andoman

Calling all mathematicians!

I can math.

So lets take a 250cc 2 stroke, running at 5000rpm.

That's cycling 0.25L x 5000/min of air/fuel mix. 1250L/min, or 20.83L/sec.
At 20.83L/sec flowing through a carb of, say, 38mm diameter. Thats 18.37m/s air velocity though the carb. Which comes from 20.83L/s / ((pi*(0.038m/2)^2) /1000L/m^3)

If we assume the engine operates at the 'ideal' 15:1 air/fuel ratio.
Then its 19.52L/s of air and 1.301L/s of fuel. Of which, at 50:1, 98% is gasoline.
So its actually 1.301L of gasoline and 0.025L of oil
Now lets assume that all the fuel is vapourised completely and it all came through the main jet only. This just simplifies my lunchtime calculations.
The vapour density of gasoline is around 3-4 times that of air, which is 1.2kg/m^3, so 3.6-4.8kg/m^3. Lets assume it 0.004kg/L just because.
The liquid density of gasoline is between 0.71-0.77kg/L, lets say 0.75kg/L.

So if we flow 1.325L/s of vapour fuel mix, that is 1.325x0.004 = 0.0053kg/s
Which must of come from 0.053/0.75 = 0.00707L/s of liquid fuel mix.

Now, iv'e assumed that the vapour and liquid densities of the fuel mix are approximately that of pure gasoline, because the density of oil is not hugely different to that of pure gasoline, its around 0.75kg/L for gasoline, and around 0.8-0.9kg/L for oils, and seeing as the're mixed at around 2%, the effect on density is small. 0.98x0.75 + 0.02x0.9 = 0.753 which is ~ 0.75.

So lets consider a main of 160, which is 1.6mm diameter, so the velocity through is must be (0.00707/(0.0008^2*pi))/1000 = 3.516m/s.

Now, lets assume that this engine is in its 'sweet' spot, and the ideal fuel delivery is based on engine dynamics and the flow through the main jet is defined by pressure differences across it, this is actually how it works too by the way.

So lets change the mixture ratio, but the engine dynamics stay the same, so the engine mass flow is the same, and the flow through the 160 main jet is the same. So 0.00707L/s of liquid fuel mixture. Which as before, is 0.0053kg/s fuel mix, thus 1.325L/s fuel vapour
But now lets make is 60:1 instead of 50:1, so its 1.66% oil and 98.33% fuel
So we have 1.303L/s of pure gasoline instead of 1.301L/s. I'm assuming that all the oil falls out of suspension and is not burnt with gasoline.
We're still flowing the same total volume of 20.83L/s, but theres more fuel in it, less oil, the same amount is air. So 19.52L/s air and 1.303L/s fuel vapour, which gives an air/ fuel ratio of 14.98 instead of 15.0038. Based on the rounded numbers I used to start with.
Pretty much no difference.

Now we could have gone straight to that derivation from the 50:1 v 60:1, but you asked about changes in the carb.
So, if we wanted to go back to the 15:1 ratio with the 60:1 fuel / oil mix.
As before, flowing 20.83L/s of fluid, 15:1 ratio air/fuel.
So 19.52L/s of air and 1.301L/s of fuel, 1.301L/s fuel is actually 1.301L/s of gasoline, so it must actually be 1.323L/s gas/oil vapour mix through the carb.
Which is 0.007056L/s gas/oil mix liquid.
However, I set the velocity across the main jet before at 3.516m/s.
So recalculating the main jet size based on flow rate and velocity, 0.007056/3.516/1000 = 0.0000020m^2. Solve for main jet size, 0.0000020 = r^2*pi, r = 0.000799, which gives a mainjet diameter of


1.598mm, equivalent to a 159.8 size mainjet.

So, your gut feeling of a change of 50:1 to 60:1 making sfa difference is correct.

A change to 40:1 gives a mainjet of 160.5
You would have to go to 20:1 before a change in mainjet was needed to accommodate the change is fuel, and even then, its only to a 162 main.



Now take all of this with a pinch of salt, or whatever the stupid expression is, it is very off the cuff as I could think of relating one thing to another. It also assumes steady state in the engine and carb, which is ridiculous of itself, but I didn't want to get involved with differentials.

[iamovru]

Originally Posted by noobi

I can math.

So lets take a 250cc 2 stroke, running at 5000rpm.

That's cycling 0.25L x 5000/min of air/fuel mix. 1250L/min, or 20.83L/sec.
At 20.83L/sec flowing through a carb of, say, 38mm diameter. Thats 18.37m/s air velocity though the carb. Which comes from 20.83L/s / ((pi*(0.038m/2)^2) /1000L/m^3)

If we assume the engine operates at the 'ideal' 15:1 air/fuel ratio.
Then its 19.52L/s of air and 1.301L/s of fuel. Of which, at 50:1, 98% is gasoline.
So its actually 1.301L of gasoline and 0.025L of oil
Now lets assume that all the fuel is vapourised completely and it all came through the main jet only. This just simplifies my lunchtime calculations.
The vapour density of gasoline is around 3-4 times that of air, which is 1.2kg/m^3, so 3.6-4.8kg/m^3. Lets assume it 0.04kg/L just because.
The liquid density of gasoline is between 0.71-0.77kg/L, lets say 0.75kg/L.

So if we flow 1.325L/s of vapour fuel mix, that is 1.325x0.04 = 0.053kg/s
Which must of come from 0.053/0.75 = 0.0707L/s of liquid fuel mix.

Now, iv'e assumed that the vapour and liquid densities of the fuel mix are approximately that of pure gasoline, because the density of oil is not hugely different to that of pure gasoline, its around 0.75kg/L for gasoline, and around 0.8-0.9kg/L for oils, and seeing as the're mixed at around 2%, the effect on density is small. 0.98x0.75 + 0.02x0.9 = 0.753 which is ~ 0.75.

So lets consider a main of 160, which is 1.6mm diameter, so the velocity through is must be (0.0707/(0.0008^2*pi))/1000 = 35.16m/s.

Now, lets assume that this engine is in its 'sweet' spot, and the ideal fuel delivery is based on engine dynamics and the flow through the main jet is defined by pressure differences across it, this is actually how it works too by the way.

So lets change the mixture ratio, but the engine dynamics stay the same, so the engine mass flow is the same, and the flow through the 160 main jet is the same. So 0.0707L/s of liquid fuel mixture. Which as before, is 0.053kg/s fuel mix, thus 1.325L/s fuel vapour
But now lets make is 60:1 instead of 50:1, so its 1.66% oil and 98.33% fuel
So we have 1.303L/s of pure gasoline instead of 1.301L/s. I'm assuming that all the oil falls out of suspension and is not burnt with gasoline.
We're still flowing the same total volume of 20.83L/s, but theres more fuel in it, less oil, the same amount is air. So 19.52L/s air and 1.303L/s fuel vapour, which gives an air/ fuel ratio of 14.98 instead of 15.0038. Based on the rounded numbers I used to start with.
Pretty much no difference.

Now we could have gone straight to that derivation from the 50:1 v 60:1, but you asked about changes in the carb.
So, if we wanted to go back to the 15:1 ratio with the 60:1 fuel / oil mix.
As before, flowing 20.83L/s of fluid, 15:1 ratio air/fuel.
So 19.52L/s of air and 1.301L/s of fuel, 1.301L/s fuel is actually 1.301L/s of gasoline, so it must actually be 13.23L/s gas/oil vapour mix through the carb.
Which is 0.07056L/s gas/oil mix liquid.
However, I set the velocity across the main jet before at 35.16m/s.
So recalculating the main jet size based on flow rate and velocity, 0.07056/35.16/1000 = 0.0000020m^2. Solve for main jet size, 0.0000020 = r^2*pi, r = 0.000799, which gives a mainjet diameter of


1.598mm, equivalent to a 159.8 size mainjet.

So, your gut feeling of a change of 50:1 to 60:1 making sfa difference is correct.

A change to 40:1 gives a mainjet of 160.5
You would have to go to 20:1 before a change in mainjet was needed to accommodate the change is fuel, and even then, its only to a 162 main.



Now take all of this with a pinch of salt, or whatever the stupid expression is, it is very off the cuff as I could think of relating one thing to another. It also assumes steady state in the engine and carb, which is ridiculous of itself, but I didn't want to get involved with differentials.

WOW! That hurt my head and made my eyes bleed

[Zman]

Very Well Done! What are your thoughts on the difference in viscosity?

[bowhunter007]

Originally Posted by iamovru

WOW! That hurt my head and made my eyes bleed

I didn't want to say it...

[andoman]

Noobi, that was awesome! Thanks!

Looking at a density altitude chart, we see that just the change in temperature from the chill in the morning as we unload the bikes to the heat of the afternoon has a much greater effect on jetting needs than changing pre-mix ratios.