Subject: [Mgs] Engine maths...and spare time

From: Richard Lindsay via Mgs <[email-masked]>
Date: Sat, 28 Mar 2020 07:54:59 -0500

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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 the
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 used
to overcome the friction and other forces that exist at idle speed. Stated
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 the
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 turning
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 way
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 if
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 g=
et
94=C2=B0 of crank rotation from spark to MEP at half-way down. That's 94/36=
0 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 0.26
=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 maximum
advance, we get 3500rpm / 60mps =3D 58rps (revolutions per second). Maximum
advance is 32=C2=B0 BTDC so 90=C2=B0 + 32=C2=B0 =3D 122=C2=B0, spark to MEP=
 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-w=
ay
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, the 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

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<div dir=3D"auto">Hello friends,<div dir=3D"auto"><br><div dir=3D"auto" sty=
le=3D"">=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" style=3D""><br></div><div di=
r=3D"auto" style=3D"">=C2=A0 =C2=A0We know that the charge (fuel plus air) =
in a cylinder doesn&#39;t burn instantly, despite our perception to the con=
trary. Rather, it takes a finite length of time from the occurance of the &=
#39;spark&#39;, the flame front to cross the combustion chamber, and to rai=
se the MEP (Mean Effective Pressure) to a maximum - the point where it does=
 the most work. But how much time?</div><div dir=3D"auto" style=3D"">=C2=A0=
 =C2=A0Physics problems always start by listing the &#39;known&#39; and the=
 property to &#39;find&#39;. So in this case,</div><div dir=3D"auto" style=
=3D""><br></div><div dir=3D"auto" style=3D"">KNOWN:</div><div dir=3D"auto" =
style=3D"">=C2=A0 =C2=A0Idle speed: 900rpm</div><div dir=3D"auto" style=3D"=
">=C2=A0 =C2=A0Idle timing advance: 4=C2=B0 BTDC</div><div dir=3D"auto" sty=
le=3D"">=C2=A0 =C2=A0Speed at maximum advance: 3500rpm</div><div dir=3D"aut=
o" style=3D"">=C2=A0 =C2=A0Maximum timing advance: 32=C2=B0 BTDC</div><div =
dir=3D"auto" style=3D""><br></div><div dir=3D"auto" style=3D"">FIND:</div><=
div dir=3D"auto" style=3D"">=C2=A0 =C2=A0Time from spark to MEP</div><div d=
ir=3D"auto" style=3D""><br></div><div dir=3D"auto" style=3D"">=C2=A0 =C2=A0=
The first thing one might know is that the goal at idle is not to produce m=
aximum power. In fact, at idle 100% of the available power is used to overc=
ome the friction and other forces that exist at idle speed. Stated another =
way: Idle speed is the fastest the engine can achieve given the available c=
harge. That fact is evident (with carbureted engines) when one notices that=
 engine speed gradually increases, even for a fixed throttle setting, as th=
e engine warms and friction forces decrease. But back to the problem.</div>=
<div dir=3D"auto" style=3D""><br></div><div dir=3D"auto" style=3D"">=C2=A0 =
=C2=A0Because the goal at idle is smooth running and progression off of idl=
e (e.g. speeding up), not maximum power, the calculated wavefront speed may=
 not be correct at idle. But let&#39;s see.</div><div dir=3D"auto" style=3D=
""><br></div><div dir=3D"auto" style=3D"">=C2=A0 =C2=A0At idle speed, 900rp=
m 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" style=3D"=
"><br></div><div dir=3D"auto" style=3D"">=C2=A0 =C2=A0Distributor 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.</div><div dir=3D"auto" style=3D""><br></div><=
div dir=3D"auto" style=3D"">=C2=A0 =C2=A0We don&#39;t know how fast the fla=
me front travels across the combustion chamber but we do know that maximum =
work occurs when the piston is half way down the cylinder. And we also know=
 that work isn&#39;t an instantaneous parameter so it must begin before the=
 half way point and last past that point. Lots of unknowns and theory doesn=
&#39;t always work in practice. But if we use the average piston position a=
t half-way down the bore, where most work is most effective, and the MEP (M=
ean Effective Pressure), since Mean is average, calculations begin.</div><d=
iv dir=3D"auto" style=3D""><br></div><div dir=3D"auto" style=3D"">=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&#39;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" style=3D""><br></div><div dir=3D"auto" styl=
e=3D"">=C2=A0 =C2=A0If we repeat the calculations for operating engine spee=
d and at maximum advance, we get=C2=A03500rpm / 60mps =3D 58rps (revolution=
s 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 revo=
lution</div><div dir=3D"auto" style=3D""><div dir=3D"auto" style=3D""><br><=
/div><div dir=3D"auto" style=3D"">=C2=A0 =C2=A058rps 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 divid=
e the idle elapsed time minus the optimal time across the strike&#39;s midp=
oint. Doing so would mean at idle, the pressure at idle becomes most effect=
ive 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 p=
ressure application time, or is that circular logic?</div><div dir=3D"auto"=
 style=3D""><br></div><div dir=3D"auto" style=3D"">=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" style=
=3D""><br></div><div dir=3D"auto" style=3D"">=C2=A0 =C2=A0Anyone with extra=
 house-bound time on their hands, please check my maths and share your corr=
ections, including the logic of the whole experiment...or perhaps even why =
geeks think these ways!</div><div dir=3D"auto" style=3D""><br></div><div di=
r=3D"auto" style=3D"">Rick</div></div></div></div>

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