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It takes about 3ms to burn the fuel in a cylinder from when the spark
ignites it. The engine designers want the peak pressure to occur at
between 5-10 ATDC.
At idle speed 3ms is about 15 degrees.
So if you want your peak pressure at 5 ATDC, your base timing is 10 BTDC,
if you want your peak pressure at 10 ATDC your timing is 5 BTDC.
If you look at tune up specs for non smog controlled engines, most have
base timing in the 5-10 BTDC range.
Now if you double the engine speed, it still take 3ms to burn, but to keep
the max pressure at 5-10 ATDC, you have to light the mixture sooner.
You do eventually get to a point where no more advance will help. That
point varies with the engine design, cam profile, combustion chamber
design, headers etc.
Rick
On Sat, Mar 28, 2020 at 5:55 AM Richard Lindsay via Mgs <mgs@autox.team.net=
>
wrote:
> Hello friends,
>
> When one is a geek, one thinks of geeky things. I am a geek and this
> house-bound morning I woke up thinking about ignition timing. Here are th=
e
> details.
>
> We know that the charge (fuel plus air) in a cylinder doesn't burn
> instantly, despite our perception to the contrary. Rather, it takes a
> finite length of time from the occurance of the 'spark', the flame front =
to
> cross the combustion chamber, and to raise the MEP (Mean Effective
> Pressure) to a maximum - the point where it does the most work. But how
> much time?
> Physics problems always start by listing the 'known' and the property
> to 'find'. So in this case,
>
> KNOWN:
> Idle speed: 900rpm
> Idle timing advance: 4=C2=B0 BTDC
> Speed at maximum advance: 3500rpm
> Maximum timing advance: 32=C2=B0 BTDC
>
> FIND:
> Time from spark to MEP
>
> The first thing one might know is that the goal at idle is not to
> produce maximum power. In fact, at idle 100% of the available power is us=
ed
> to overcome the friction and other forces that exist at idle speed. State=
d
> another way: Idle speed is the fastest the engine can achieve given the
> available charge. That fact is evident (with carbureted engines) when one
> notices that engine speed gradually increases, even for a fixed throttle
> setting, as the engine warms and friction forces decrease. But back to th=
e
> problem.
>
> Because the goal at idle is smooth running and progression off of idle
> (e.g. speeding up), not maximum power, the calculated wavefront speed may
> not be correct at idle. But let's see.
>
> At idle speed, 900rpm in this MG TD example, the XPAG engine is turnin=
g
> 900rpm or 900rpm / 60mps =3D 15rps (revolutions per second).
>
> Distributor speed is 1/2 engine speed so at idle the distributor is
> turning only 7.5 revolutions per second. But timing numbers are specified
> in degrees of crank rotations so we will stick with 15rps.
>
> We don't know how fast the flame front travels across the combustion
> chamber but we do know that maximum work occurs when the piston is half w=
ay
> down the cylinder. And we also know that work isn't an instantaneous
> parameter so it must begin before the half way point and last past that
> point. Lots of unknowns and theory doesn't always work in practice. But i=
f
> we use the average piston position at half-way down the bore, where most
> work is most effective, and the MEP (Mean Effective Pressure), since Mean
> is average, calculations begin.
>
> A single revolution is 360=C2=B0 so half-way down the power stroke is =
90=C2=B0.
> Add the idle spark timing of 4=C2=B0 BTDC (Before Top Dead Center) and we=
get
> 94=C2=B0 of crank rotation from spark to MEP at half-way down. That's 94/=
360 or
> about 0.26 of an engine revolution. And the engine is turning 15
> revolutions per second or 67ms (milliseconds) per revolution. So 67 x 0.2=
6
> =3D 17ms from spark to MEP at half-way down the power stroke, at idle.
>
> If we repeat the calculations for operating engine speed and at maximu=
m
> advance, we get 3500rpm / 60mps =3D 58rps (revolutions per second). Maxim=
um
> advance is 32=C2=B0 BTDC so 90=C2=B0 + 32=C2=B0 =3D 122=C2=B0, spark to M=
EP or 122=C2=B0/360=C2=B0 =3D 0.34
> of a revolution
>
> 58rps is 17ms/r so 17ms/r x 0.34r =3D 5.78ms from spark to MEP at
> half-way down the power stroke. This is a more representative number than
> the 17ms at idle. One might even divide the idle elapsed time minus the
> optimal time across the strike's midpoint. Doing so would mean at idle, t=
he
> pressure at idle becomes most effective 5.6ms before half-way and for
> another 5.6ms after midpoint. Interesting that the idle pressure
> application time is about the same as the maximum pressure application
> time, or is that circular logic?
>
> Yes everything above is ripe with assumptions and perhaps even
> apocryphal and resplendent with errors, but it is only 7am after all.
>
> Anyone with extra house-bound time on their hands, please check my
> maths and share your corrections, including the logic of the whole
> experiment...or perhaps even why geeks think these ways!
>
> Rick
> _______________________________________________
>
> Mgs@autox.team.net
> Donate: http://www.team.net/donate.html
> Suggested annual donation $12.75
>
> Archive: http://www.team.net/pipermail/mgs http://autox.team.net/archive
>
> Unsubscribe:
> http://autox.team.net/mailman/options/mgs/richard.ewald@gmail.com
>
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<div dir=3D"ltr">It takes about 3ms to burn the fuel in a cylinder from whe=
n the spark ignites it.=C2=A0 The engine designers=C2=A0want the peak press=
ure to occur at between 5-10 ATDC.<div>At idle speed 3ms is about 15 degree=
s.=C2=A0=C2=A0</div><div>So if you want your peak pressure at 5 ATDC, your =
base timing=C2=A0is 10 BTDC, if=C2=A0 you want=C2=A0 your peak pressure at =
10 ATDC your timing is 5 BTDC.</div><div>If you look at tune up specs for n=
on smog controlled engines, most have base timing in the 5-10 BTDC range.</=
div><div>Now if you double the engine speed, it still take 3ms to burn, but=
to keep the max pressure at 5-10 ATDC, you have to light the mixture soone=
r.</div><div>You do eventually=C2=A0get to a point where no more advance wi=
ll help.=C2=A0 That point varies with the engine=C2=A0design,=C2=A0 cam pro=
file, combustion chamber design, headers etc.=C2=A0 =C2=A0</div><div>Rick</=
div></div><br><div class=3D"gmail_quote"><div dir=3D"ltr" class=3D"gmail_at=
tr">On Sat, Mar 28, 2020 at 5:55 AM Richard Lindsay via Mgs <<a href=3D"=
mailto:mgs@autox.team.net";>mgs@autox.team.net</a>> wrote:<br></div><bloc=
kquote class=3D"gmail_quote" style=3D"margin:0px 0px 0px 0.8ex;border-left:=
1px solid rgb(204,204,204);padding-left:1ex"><div dir=3D"auto">Hello friend=
s,<div dir=3D"auto"><br><div dir=3D"auto">=C2=A0 =C2=A0When one is a geek, =
one thinks of geeky things. I am a geek and this house-bound morning I woke=
up thinking about ignition timing. Here are the details.</div><div dir=3D"=
auto"><br></div><div dir=3D"auto">=C2=A0 =C2=A0We know that the charge (fue=
l plus air) in a cylinder doesn't burn instantly, despite our perceptio=
n to the contrary. Rather, it takes a finite length of time from the occura=
nce of the 'spark', the flame front to cross the combustion chamber=
, and to raise the MEP (Mean Effective Pressure) to a maximum - the point w=
here it does the most work. But how much time?</div><div dir=3D"auto">=C2=
=A0 =C2=A0Physics problems always start by listing the 'known' and =
the property to 'find'. So in this case,</div><div dir=3D"auto"><br=
></div><div dir=3D"auto">KNOWN:</div><div dir=3D"auto">=C2=A0 =C2=A0Idle sp=
eed: 900rpm</div><div dir=3D"auto">=C2=A0 =C2=A0Idle timing advance: 4=C2=
=B0 BTDC</div><div dir=3D"auto">=C2=A0 =C2=A0Speed at maximum advance: 3500=
rpm</div><div dir=3D"auto">=C2=A0 =C2=A0Maximum timing advance: 32=C2=B0 BT=
DC</div><div dir=3D"auto"><br></div><div dir=3D"auto">FIND:</div><div dir=
=3D"auto">=C2=A0 =C2=A0Time from spark to MEP</div><div dir=3D"auto"><br></=
div><div dir=3D"auto">=C2=A0 =C2=A0The first thing one might know is that t=
he goal at idle is not to produce maximum power. In fact, at idle 100% of t=
he available power is used to overcome the friction and other forces that e=
xist at idle speed. Stated another way: Idle speed is the fastest the engin=
e can achieve given the available charge. That fact is evident (with carbur=
eted engines) when one notices that engine speed gradually increases, even =
for a fixed throttle setting, as the engine warms and friction forces decre=
ase. But back to the problem.</div><div dir=3D"auto"><br></div><div dir=3D"=
auto">=C2=A0 =C2=A0Because the goal at idle is smooth running and progressi=
on off of idle (e.g. speeding up), not maximum power, the calculated wavefr=
ont speed may not be correct at idle. But let's see.</div><div dir=3D"a=
uto"><br></div><div dir=3D"auto">=C2=A0 =C2=A0At idle speed, 900rpm in this=
MG TD example, the XPAG engine is turning 900rpm or=C2=A0900rpm / 60mps =
=3D 15rps (revolutions per second).</div><div dir=3D"auto"><br></div><div d=
ir=3D"auto">=C2=A0 =C2=A0Distributor speed is 1/2 engine speed so at idle t=
he distributor is turning only 7.5 revolutions per second. But timing numbe=
rs are specified in degrees of crank rotations so we will stick with 15rps.=
</div><div dir=3D"auto"><br></div><div dir=3D"auto">=C2=A0 =C2=A0We don'=
;t know how fast the flame front travels across the combustion chamber but =
we do know that maximum work occurs when the piston is half way down the cy=
linder. And we also know that work isn't an instantaneous parameter so =
it must begin before the half way point and last past that point. Lots of u=
nknowns and theory doesn't always work in practice. But if we use the a=
verage piston position at half-way down the bore, where most work is most e=
ffective, and the MEP (Mean Effective Pressure), since Mean is average, cal=
culations begin.</div><div dir=3D"auto"><br></div><div dir=3D"auto">=C2=A0 =
=C2=A0A single revolution is 360=C2=B0 so half-way down the power stroke is=
90=C2=B0. Add the idle spark timing of 4=C2=B0 BTDC (Before Top Dead Cente=
r) and we get 94=C2=B0 of crank rotation from spark to MEP at half-way down=
. That's 94/360 or about 0.26 of an engine revolution. And the engine i=
s turning 15 revolutions per second or 67ms (milliseconds) per revolution. =
So 67 x 0.26 =3D 17ms from spark to MEP at half-way down the power stroke, =
at idle.</div><div dir=3D"auto"><br></div><div dir=3D"auto">=C2=A0 =C2=A0If=
we repeat the calculations for operating engine speed and at maximum advan=
ce, we get=C2=A03500rpm / 60mps =3D 58rps (revolutions per second). Maximum=
advance is 32=C2=B0 BTDC so=C2=A090=C2=B0 + 32=C2=B0 =3D 122=C2=B0, spark =
to MEP or=C2=A0122=C2=B0/360=C2=B0 =3D 0.34 of a revolution</div><div dir=
=3D"auto"><div dir=3D"auto"><br></div><div dir=3D"auto">=C2=A0 =C2=A058rps =
is 17ms/r so 17ms/r x 0.34r =3D 5.78ms from spark to MEP at half-way down t=
he power stroke. This is a more representative number than the 17ms at idle=
. One might even divide the idle elapsed time minus the optimal time across=
the strike's midpoint. Doing so would mean at idle, the pressure at id=
le becomes most effective 5.6ms before half-way and for another 5.6ms after=
midpoint. Interesting that the idle pressure application time is about the=
same as the maximum pressure application time, or is that circular logic?<=
/div><div dir=3D"auto"><br></div><div dir=3D"auto">=C2=A0 =C2=A0Yes everyth=
ing above is ripe with assumptions and perhaps even apocryphal and resplend=
ent with errors, but it is only 7am after all.</div><div dir=3D"auto"><br><=
/div><div dir=3D"auto">=C2=A0 =C2=A0Anyone with extra house-bound time on t=
heir hands, please check my maths and share your corrections, including the=
logic of the whole experiment...or perhaps even why geeks think these ways=
!</div><div dir=3D"auto"><br></div><div dir=3D"auto">Rick</div></div></div>=
</div>
_______________________________________________<br>
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</blockquote></div>
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