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Thank you Paul. As you know, I have always appreciated your depth of
knowledge and experience. In fact, as you point out, there are SO many
variables. Theory just can't keep up with practice. The factory's tuning
recommendations for our beloved engines are almost certainly empirical.
That is, decided upon by trial and error.
And of course, tuning for maximum power is not the only maker's goal. Fuel
economy, emissions, build cost, and a plethora of other constraints
contribute.
Again, thank you Paul, *et al.*, for indulging my house-bound mental
exercises.
Rick
On Sat, Mar 28, 2020, 8:54 AM PaulHunt73 <paulhunt73@virginmedia.com> wrote=
:
> "at idle 100% of the available power is used to overcome the friction and
> other forces that exist at idle speed"
>
> Most of what the engine is doing at idle is as a vacuum pump, generating
> about 16 in Hg. or so in the intake manifold, and is why when you introdu=
ce
> an intake vacuum leak the idle speed goes up. This may be included in yo=
ur
> 'other forces' above.
>
> Whist there may well be a most efficient point to start combustion and th=
e
> flame front, the prime consideration has to be avoiding
> spontaneous combustion at any point, i.e. pinking or detonation. As the
> flame front travels pressure inside the engine is rising, but after TDC t=
he
> volume available is reducing, which tends to counteract the pressure
> increase. There is also the effect of leverage i.e. the angle the con ro=
d
> makes relative to the piston.
>
> I'm certainly not going to check your maths, a specific engine is what it
> is, and the timing has to be set taking those specifics into account plus
> other factors like fuel grade and type.
>
> PaulH.
>
>
> ----- Original Message -----
>
> *From:* Richard Lindsay via Mgs <mgs@autox.team.net>
> *To:* mgs@autox.team.net List
> *Sent:* Saturday, March 28, 2020 12:54 PM
> *Subject:* [Mgs] Engine maths...and spare time
>
> 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/paulhunt73@virginmedia.com
>
>
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<div dir=3D"auto">Thank you Paul. As you know, I have always appreciated yo=
ur depth of knowledge and experience. In fact, as you point out, there are =
SO many variables. Theory just can't keep up with practice. The factory=
's tuning recommendations for our beloved engines are almost certainly =
empirical. That is, decided upon by trial and error.<div dir=3D"auto"><br><=
/div><div dir=3D"auto">And of course, tuning for maximum power is not the o=
nly maker's goal. Fuel economy, emissions, build cost, and a plethora o=
f other constraints contribute.</div><div dir=3D"auto"><br></div><div dir=
=3D"auto">Again, thank you Paul,=C2=A0 <i>et al.</i>, for indulging my hous=
e-bound mental exercises.</div><div dir=3D"auto"><br></div><div dir=3D"auto=
">Rick</div></div><br><div class=3D"gmail_quote"><div dir=3D"ltr" class=3D"=
gmail_attr">On Sat, Mar 28, 2020, 8:54 AM PaulHunt73 <<a href=3D"mailto:=
paulhunt73@virginmedia.com">paulhunt73@virginmedia.com</a>> wrote:<br></=
div><blockquote class=3D"gmail_quote" style=3D"margin:0 0 0 .8ex;border-lef=
t:1px #ccc solid;padding-left:1ex"><u></u>
<div bgcolor=3D"#ffffff">
<div><font size=3D"2">"</font><font size=3D"3">at idle 100% of the ava=
ilable power is=20
used to overcome the friction and other forces that exist at idle=20
speed"</font></div>
<div><font size=3D"3"></font>=C2=A0</div>
<div><font size=3D"2">Most of what the engine is doing at idle is as a vacu=
um pump,=20
generating about 16 in Hg. or so in the intake manifold, and is why when yo=
u=20
introduce an intake=C2=A0vacuum leak the idle speed goes up.=C2=A0 This may=
be=20
included in your 'other forces' above.</font></div>
<div><font size=3D"2"></font>=C2=A0</div>
<div><font size=3D"2">Whist there may well be a most efficient=C2=A0point t=
o start=20
combustion and the flame front, the prime consideration has to be avoiding=
=20
spontaneous=C2=A0combustion at any point, i.e. pinking or detonation.=C2=A0=
As=20
the flame front travels pressure inside the engine is rising,=C2=A0but afte=
r TDC=20
the volume available is reducing, which tends to counteract the pressure=20
increase.=C2=A0 There is also the=C2=A0effect of leverage i.e. the angle th=
e con=20
rod makes relative to the piston.</font></div>
<div><font size=3D"2"></font>=C2=A0</div>
<div><font size=3D"2">I'm certainly not going to check your maths, a sp=
ecific engine=20
is what=C2=A0it is, and the timing has to be set=C2=A0taking those specific=
s=20
into account=C2=A0plus other factors like fuel grade and type.</font></div>
<div><font size=3D"2"></font>=C2=A0</div>
<div><font size=3D"2">PaulH.</font></div>
<div><font size=3D"2"></font>=C2=A0</div>
<div><font size=3D"2"></font>=C2=A0</div>
<div><font size=3D"3">----- Original Message ----- </font></div>
<blockquote style=3D"BORDER-LEFT:#000000 2px solid;PADDING-LEFT:5px;PADDING=
-RIGHT:0px;MARGIN-LEFT:5px;MARGIN-RIGHT:0px">
<div style=3D"FONT:10pt arial;BACKGROUND:#e4e4e4"><b>From:</b>=20
<a title=3D"mgs@autox.team.net" href=3D"mailto:mgs@autox.team.net"; target=
=3D"_blank" rel=3D"noreferrer">Richard Lindsay=20
via Mgs</a> </div>
<div style=3D"FONT:10pt arial"><b>To:</b> <a title=3D"mgs@autox.team.net"=
href=3D"mailto:mgs@autox.team.net+List"; target=3D"_blank" rel=3D"noreferre=
r">mgs@autox.team.net List</a> </div>
<div style=3D"FONT:10pt arial"><b>Sent:</b> Saturday, March 28, 2020 12:5=
4=20
PM</div>
<div style=3D"FONT:10pt arial"><b>Subject:</b> [Mgs] Engine maths...and s=
pare=20
time</div>
<div><br></div>
<div dir=3D"auto">Hello friends,
<div dir=3D"auto"><br>
<div dir=3D"auto">=C2=A0 =C2=A0When one is a geek, one thinks of geeky th=
ings. I=20
am a geek and this house-bound morning I woke up thinking about ignition=
=20
timing. Here are the details.</div>
<div dir=3D"auto"><br></div>
<div dir=3D"auto">=C2=A0 =C2=A0We know that the charge (fuel plus air) in=
a=20
cylinder doesn't burn instantly, despite our perception to the contra=
ry.=20
Rather, it takes a finite length of time from the occurance of the 's=
park',=20
the flame front to cross the combustion chamber, and to raise the MEP (Me=
an=20
Effective Pressure) to a maximum - the point where it does the most work.=
But=20
how much time?</div>
<div dir=3D"auto">=C2=A0 =C2=A0Physics problems always start by listing t=
he=20
'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 speed: 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: 3500rpm</div>
<div dir=3D"auto">=C2=A0 =C2=A0Maximum timing advance: 32=C2=B0 BTDC</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 the=
goal at=20
idle is not to produce maximum power. In fact, at idle 100% of the availa=
ble=20
power is used to overcome the friction and other forces that exist at idl=
e=20
speed. Stated another way: Idle speed is the fastest the engine can achie=
ve=20
given the available charge. That fact is evident (with carbureted engines=
)=20
when one notices that engine speed gradually increases, even for a fixed=
=20
throttle setting, as the engine warms and friction forces decrease. But b=
ack=20
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=20
progression off of idle (e.g. speeding up), not maximum power, the calcul=
ated=20
wavefront speed may not be correct at idle. But let's see.</div>
<div dir=3D"auto"><br></div>
<div dir=3D"auto">=C2=A0 =C2=A0At idle speed, 900rpm in this MG TD exampl=
e, the=20
XPAG engine is turning 900rpm or=C2=A0900rpm / 60mps =3D 15rps (revolutio=
ns per=20
second).</div>
<div dir=3D"auto"><br></div>
<div dir=3D"auto">=C2=A0 =C2=A0Distributor speed is 1/2 engine speed so a=
t idle=20
the distributor is turning only 7.5 revolutions per second. But timing nu=
mbers=20
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=20
across the combustion chamber but we do know that maximum work occurs whe=
n the=20
piston is half way down the cylinder. And we also know that work isn'=
t an=20
instantaneous parameter so it must begin before the half way point and la=
st=20
past that point. Lots of unknowns and theory doesn't always work in p=
ractice.=20
But if we use the average piston position at half-way down the bore, wher=
e=20
most work is most effective, and the MEP (Mean Effective Pressure), since=
Mean=20
is average, calculations begin.</div>
<div dir=3D"auto"><br></div>
<div dir=3D"auto">=C2=A0 =C2=A0A single revolution is 360=C2=B0 so half-w=
ay down the=20
power stroke is 90=C2=B0. Add the idle spark timing of 4=C2=B0 BTDC (Befo=
re Top Dead=20
Center) and we get 94=C2=B0 of crank rotation from spark to MEP at half-w=
ay down.=20
That's 94/360 or about 0.26 of an engine revolution. And the engine i=
s turning=20
15 revolutions per second or 67ms (milliseconds) per revolution. So 67 x =
0.26=20
=3D 17ms from spark to MEP at half-way down the power stroke, at idle.</d=
iv>
<div dir=3D"auto"><br></div>
<div dir=3D"auto">=C2=A0 =C2=A0If we repeat the calculations for operatin=
g engine=20
speed and at maximum advance, we get=C2=A03500rpm / 60mps =3D 58rps (revo=
lutions=20
per second). Maximum advance is 32=C2=B0 BTDC so=C2=A090=C2=B0 + 32=C2=B0=
=3D 122=C2=B0, spark to=20
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 5.78=
ms from=20
spark to MEP at half-way down the power stroke. This is a more representa=
tive=20
number than the 17ms at idle. One might even divide the idle elapsed time=
=20
minus the optimal time across the strike's midpoint. Doing so would m=
ean at=20
idle, the pressure at idle becomes most effective 5.6ms before half-way a=
nd=20
for another 5.6ms after midpoint. Interesting that the idle pressure=20
application time is about the same as the maximum pressure application ti=
me,=20
or is that circular logic?</div>
<div dir=3D"auto"><br></div>
<div dir=3D"auto">=C2=A0 =C2=A0Yes everything above is ripe with assumpti=
ons and=20
perhaps even apocryphal and resplendent with errors, but it is only 7am a=
fter=20
all.</div>
<div dir=3D"auto"><br></div>
<div dir=3D"auto">=C2=A0 =C2=A0Anyone with extra house-bound time on thei=
r hands,=20
please check my maths and share your corrections, including the logic of =
the=20
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>
<p>
<hr>
<p></p>_______________________________________________<br><br><a href=3D"=
mailto:Mgs@autox.team.net"; target=3D"_blank" rel=3D"noreferrer">Mgs@autox.t=
eam.net</a><br>Donate:=20
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ferrer">http://www.team.net/donate.html</a><br>Suggested annual donation=C2=
=A0=20
$12.75<br><br>Archive: <a href=3D"http://www.team.net/pipermail/mgs"; targ=
et=3D"_blank" rel=3D"noreferrer">http://www.team.net/pipermail/mgs</a>=20
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rrer">http://autox.team.net/archive</a><br><br>Unsubscribe:=20
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options/mgs/paulhunt73@virginmedia.com</a><br></p></blockquote></div>
</blockquote></div>
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