Use a degree wheel to measure how much you've allowed it to rotate without the crank moving. 6° is usually a good safe number. I don't advance all saws though.
timing advance?
- Thread starter C SAW 090
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Chris-PA
Where the Wild Things Are
Well, if you want to calculate how much to remove for a given number of degrees, it is approximately:
(Crank Diameter)/2 * sin(Advance Angle)
So if your crank is 10mm and you want 3deg advance, then: 5 * sin(3) = 0.26mm = 0.010"
The other way to calculate it is:
(Crank Diameter) * PI * (Advance Angle) / 360
So for 10mm crank: 10 * 3.14 * 3deg / 360 = 0.26mm = 0.010"
I would go easy with it. If it revs up nice with good throttle response it's probably good. I've gotten decent gains from 3deg, and one saw is at 6.
(Crank Diameter)/2 * sin(Advance Angle)
So if your crank is 10mm and you want 3deg advance, then: 5 * sin(3) = 0.26mm = 0.010"
The other way to calculate it is:
(Crank Diameter) * PI * (Advance Angle) / 360
So for 10mm crank: 10 * 3.14 * 3deg / 360 = 0.26mm = 0.010"
I would go easy with it. If it revs up nice with good throttle response it's probably good. I've gotten decent gains from 3deg, and one saw is at 6.
The Ripper
ArboristSite Member
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Thanks for the advance formula,actually on my saw it only has 110 PSI,so with this I just advanced the crap out it and it starts as usual on the third pull without any kick back,etc. Had a fast idle which I corrected,throttle response is excellent and it pulls real hard without heating up. If this where a pro saw with heaps of compression,then I doubt seriously increasing the ignition lead would of paid off. Again thanks for the formula,I'll use it for my pro saw.
one.man.band
ArboristSite Guru
i like numbers:
example saw: 65cc
stroke: 36mm = 1.4173228 inch
connecting rod length: 73mm = 2.8346457 inch
piston travel (movement in height per degree): *= degrees
1* = 0.003417mm = 0.000135 inch
2* = 0.013668mm = 0.000538 inch
3* = 0.030747mm = 0.001211 inch Edit: it would help if i typed in the correct numbers.
4* = 0.054646mm = 0.002151 inch
5* = 0.085355mm = 0.003360 inch
6* = 0.122857mm = 0.004837 inch
continued next post
example saw: 65cc
stroke: 36mm = 1.4173228 inch
connecting rod length: 73mm = 2.8346457 inch
piston travel (movement in height per degree): *= degrees
1* = 0.003417mm = 0.000135 inch
2* = 0.013668mm = 0.000538 inch
3* = 0.030747mm = 0.001211 inch Edit: it would help if i typed in the correct numbers.
4* = 0.054646mm = 0.002151 inch
5* = 0.085355mm = 0.003360 inch
6* = 0.122857mm = 0.004837 inch
continued next post
Last edited:
one.man.band
ArboristSite Guru
....slow typer, didnt want it to time out.
theres two things going on. the formula chris posted is related to the "key" position inregard to the magnets on the flywheel.
what i posted is how much travel the piston makes as the crankshaft is turned.
with different stroke/connecting rod lengths, the amount of piston travel per degree, will change as well.
timing of the magnets, has a relationship with piston position, only to motors of the same: stroke and connecting rod length. in a nutshell, 6* on one motor does not necessarily relate to another motor with a different stroke or conn rod length. as long as apples to apples, this is fine.
this is of course assuming that the magnets are indexed correctly to begin with. i for one, would be hesistant to recommend any "blanket" rule of thumb timing numbers due to this and the above example variances.
regards
-joe
edit: added "blanket"
theres two things going on. the formula chris posted is related to the "key" position inregard to the magnets on the flywheel.
what i posted is how much travel the piston makes as the crankshaft is turned.
with different stroke/connecting rod lengths, the amount of piston travel per degree, will change as well.
timing of the magnets, has a relationship with piston position, only to motors of the same: stroke and connecting rod length. in a nutshell, 6* on one motor does not necessarily relate to another motor with a different stroke or conn rod length. as long as apples to apples, this is fine.
this is of course assuming that the magnets are indexed correctly to begin with. i for one, would be hesistant to recommend any "blanket" rule of thumb timing numbers due to this and the above example variances.
regards
-joe
edit: added "blanket"
Last edited:
Chris-PA
Where the Wild Things Are
For a 10mm crank I get 0.021" for 6deg. Not sure what crank diameter you used or where the numbers come from, but 0.004" is likely about the keyway clearance.
XSKIER
Addicted to ArboristSite
Inches of piston travel is not the same as inches off the key stock. But I agree, .004" is very little.
Mastermind
Work Saw Specialist
I take .020 off the key on the 026.
.004" won't get you any where near 6* advance.
one.man.band
ArboristSite Guru
... i wasn't clear. my numbers are not showing how much to take off the "key."
was trying to illustrate the idea of degrees of dwell time of the piston from a span of 1* to 6* for a motor with a 36mm stroke and 73mm rod.
my motor backround probably is different than a lot of folks. mainly dealt with removable heads. standard was to time 2ST bike motors using a dial indicator. enough about that.
maybe this example will illustrate the point i was trying to make better by comparison.
4ST stroker motor i built in the late '80's.
compare the degrees to the example posted above.
continued on next post.
was trying to illustrate the idea of degrees of dwell time of the piston from a span of 1* to 6* for a motor with a 36mm stroke and 73mm rod.
my motor backround probably is different than a lot of folks. mainly dealt with removable heads. standard was to time 2ST bike motors using a dial indicator. enough about that.
maybe this example will illustrate the point i was trying to make better by comparison.
4ST stroker motor i built in the late '80's.
compare the degrees to the example posted above.
continued on next post.
one.man.band
ArboristSite Guru
(using this motor, because it is extreme case, btw).
stroke: 4-5/8" = 117.25mm
conn rod length = 5.440" = 138.25mm ((shorter than a stock 5.7 gm rod, cannot remember exactly)
1* = 0.01275mm = 0.000502 inch
2* = 0.050994mm = 0.002008 inch
3* = 0.114711mm = 0.004516 inch
4* = 0.203867mm = 0.008026 inch
5* = 0.318417mm = 0.012536 inch
6* = 0.456298mm = 0.018043 inch
look at the difference in 6* of rotataion of the piston in comparison to the 65cc example. the 4-5/8" stroke motor's piston travels further in 1* than the 65cc motors piston. therefore, the timing advance has less of an effect (on the 4-5/8" motor) as far as piston position is concerned.
the 4-5/8" motor can take much more timing advance than the 65cc. although, this is not taking into account other factors such as bore size, compression, etc. .....this probably goes against what some have thought, but assure you that this is the way it is. imagine if the burn rate is equal, static compression ratio equal, bore size equal,....the 4-5/8" motor's piston exposes more chamber volume faster than the 65cc motor does. this varies between the two examples, because of stroke and rod lengths.
-joe
stroke: 4-5/8" = 117.25mm
conn rod length = 5.440" = 138.25mm ((shorter than a stock 5.7 gm rod, cannot remember exactly)
1* = 0.01275mm = 0.000502 inch
2* = 0.050994mm = 0.002008 inch
3* = 0.114711mm = 0.004516 inch
4* = 0.203867mm = 0.008026 inch
5* = 0.318417mm = 0.012536 inch
6* = 0.456298mm = 0.018043 inch
look at the difference in 6* of rotataion of the piston in comparison to the 65cc example. the 4-5/8" stroke motor's piston travels further in 1* than the 65cc motors piston. therefore, the timing advance has less of an effect (on the 4-5/8" motor) as far as piston position is concerned.
the 4-5/8" motor can take much more timing advance than the 65cc. although, this is not taking into account other factors such as bore size, compression, etc. .....this probably goes against what some have thought, but assure you that this is the way it is. imagine if the burn rate is equal, static compression ratio equal, bore size equal,....the 4-5/8" motor's piston exposes more chamber volume faster than the 65cc motor does. this varies between the two examples, because of stroke and rod lengths.
-joe
one.man.band
ArboristSite Guru
...didn't mention piston velocity effect on dwell, because i do not have concrete numbers to post.
strokers and long rod motors get their increased dwell based on piston velocity decrease. learned this just recently. don't want anyone to take me word for anything. try to always show basis for my responses here. this stuff interests me at least.
anyhow, learned this from an animation, my son had told me about. like 150 says, never too old to learn. seems easier for me to understand things by seeing them first.
http://demonstrations.wolfram.com/SliderAndCrankMechanism/
a big program needs to be downloaded first, then the file i linked. but the user can fool around with changing stroke and/or rod lengths. as it spins, there are piston velocity graphs below the animation that show changes with different stroke and or rod lengths chosen. if its bothersome to download the program, you can see from the graphs on the links posted without doing so.
hope it helps.
-joe
strokers and long rod motors get their increased dwell based on piston velocity decrease. learned this just recently. don't want anyone to take me word for anything. try to always show basis for my responses here. this stuff interests me at least.
anyhow, learned this from an animation, my son had told me about. like 150 says, never too old to learn. seems easier for me to understand things by seeing them first.
http://demonstrations.wolfram.com/SliderAndCrankMechanism/
a big program needs to be downloaded first, then the file i linked. but the user can fool around with changing stroke and/or rod lengths. as it spins, there are piston velocity graphs below the animation that show changes with different stroke and or rod lengths chosen. if its bothersome to download the program, you can see from the graphs on the links posted without doing so.
hope it helps.
-joe
anderson3754
Playing the ultimate game every day
Would someone else look at the first formula 5 mm x the 3 degree advance angle =15 . Am I missing a number ?? to use for this formula. Were, how did the 15 become 0.26mm.
The second formula is accurate
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