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"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 =
introduce an intake vacuum leak the idle speed goes up. This may be =
included in your 'other forces' above.
Whist there may well be a most efficient point to start combustion and =
the 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 the volume =
available is reducing, which tends to counteract the pressure increase. =
There is also the effect of leverage i.e. the angle the con rod 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 -----=20
From: Richard Lindsay via Mgs=20
To: mgs@autox.team.net List=20
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 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 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 0.26 =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-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, 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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Mgs@autox.team.net
Donate: http://www.team.net/donate.html
Suggested annual donation $12.75
Archive: http://www.team.net/pipermail/mgs =
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<HTML><HEAD>
<META content=3D"text/html; charset=3Dutf-8" http-equiv=3DContent-Type>
<META name=3DGENERATOR content=3D"MSHTML 8.00.6001.23588">
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<BODY bgColor=3D#ffffff>
<DIV><FONT size=3D2>"</FONT><FONT size=3D3>at idle 100% of the available =
power is=20
used to overcome the friction and other forces that exist at idle=20
speed"</FONT></DIV>
<DIV><FONT size=3D3></FONT> </DIV>
<DIV><FONT size=3D2>Most of what the engine is doing at idle is as a =
vacuum pump,=20
generating about 16 in Hg. or so in the intake manifold, and is why when =
you=20
introduce an intake vacuum leak the idle speed goes up. This =
may be=20
included in your 'other forces' above.</FONT></DIV>
<DIV><FONT size=3D2></FONT> </DIV>
<DIV><FONT size=3D2>Whist there may well be a most efficient point =
to start=20
combustion and the flame front, the prime consideration has to be =
avoiding=20
spontaneous combustion at any point, i.e. pinking or =
detonation. As=20
the flame front travels pressure inside the engine is rising, but =
after TDC=20
the volume available is reducing, which tends to counteract the pressure =
increase. There is also the effect of leverage i.e. the angle =
the con=20
rod makes relative to the piston.</FONT></DIV>
<DIV><FONT size=3D2></FONT> </DIV>
<DIV><FONT size=3D2>I'm certainly not going to check your maths, a =
specific engine=20
is what it is, and the timing has to be set taking those =
specifics=20
into account plus other factors like fuel grade and =
type.</FONT></DIV>
<DIV><FONT size=3D2></FONT> </DIV>
<DIV><FONT size=3D2>PaulH.</FONT></DIV>
<DIV><FONT size=3D2></FONT> </DIV>
<DIV><FONT size=3D2></FONT> </DIV>
<DIV><FONT size=3D3>----- Original Message ----- </FONT></DIV>
<BLOCKQUOTE=20
style=3D"BORDER-LEFT: #000000 2px solid; PADDING-LEFT: 5px; =
PADDING-RIGHT: 0px; MARGIN-LEFT: 5px; MARGIN-RIGHT: 0px">
<DIV=20
style=3D"FONT: 10pt arial; BACKGROUND: #e4e4e4; font-color: =
black"><B>From:</B>=20
<A title=3Dmgs@autox.team.net =
href=3D"mailto:mgs@autox.team.net";>Richard Lindsay=20
via Mgs</A> </DIV>
<DIV style=3D"FONT: 10pt arial"><B>To:</B> <A =
title=3Dmgs@autox.team.net=20
href=3D"mailto:mgs@autox.team.net List">mgs@autox.team.net List</A> =
</DIV>
<DIV style=3D"FONT: 10pt arial"><B>Sent:</B> Saturday, March 28, 2020 =
12:54=20
PM</DIV>
<DIV style=3D"FONT: 10pt arial"><B>Subject:</B> [Mgs] Engine =
maths...and spare=20
time</DIV>
<DIV><BR></DIV>
<DIV dir=3Dauto>Hello friends,
<DIV dir=3Dauto><BR>
<DIV dir=3Dauto> When one is a geek, one thinks of geeky =
things. 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=3Dauto><BR></DIV>
<DIV dir=3Dauto> We know that the charge (fuel plus air) =
in a=20
cylinder doesn't burn instantly, despite our perception to the =
contrary.=20
Rather, it takes a finite length of time from the occurance of the =
'spark',=20
the flame front to cross the combustion chamber, and to raise the MEP =
(Mean=20
Effective Pressure) to a maximum - the point where it does the most =
work. But=20
how much time?</DIV>
<DIV dir=3Dauto> Physics problems always start by listing =
the=20
'known' and the property to 'find'. So in this case,</DIV>
<DIV dir=3Dauto><BR></DIV>
<DIV dir=3Dauto>KNOWN:</DIV>
<DIV dir=3Dauto> Idle speed: 900rpm</DIV>
<DIV dir=3Dauto> Idle timing advance: 4=C2=B0 BTDC</DIV>
<DIV dir=3Dauto> Speed at maximum advance: 3500rpm</DIV>
<DIV dir=3Dauto> Maximum timing advance: 32=C2=B0 =
BTDC</DIV>
<DIV dir=3Dauto><BR></DIV>
<DIV dir=3Dauto>FIND:</DIV>
<DIV dir=3Dauto> Time from spark to MEP</DIV>
<DIV dir=3Dauto><BR></DIV>
<DIV dir=3Dauto> The 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 =
available=20
power is used to overcome the friction and other forces that exist at =
idle=20
speed. Stated another way: Idle speed is the fastest the engine can =
achieve=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 back=20
to the problem.</DIV>
<DIV dir=3Dauto><BR></DIV>
<DIV dir=3Dauto> Because the goal at idle is smooth =
running and=20
progression off of idle (e.g. speeding up), not maximum power, the =
calculated=20
wavefront speed may not be correct at idle. But let's see.</DIV>
<DIV dir=3Dauto><BR></DIV>
<DIV dir=3Dauto> At idle speed, 900rpm in this MG TD =
example, the=20
XPAG engine is turning 900rpm or 900rpm / 60mps =3D 15rps =
(revolutions per=20
second).</DIV>
<DIV dir=3Dauto><BR></DIV>
<DIV dir=3Dauto> Distributor speed is 1/2 engine speed so =
at idle=20
the distributor is turning only 7.5 revolutions per second. But timing =
numbers=20
are specified in degrees of crank rotations so we will stick with =
15rps.</DIV>
<DIV dir=3Dauto><BR></DIV>
<DIV dir=3Dauto> We don't know how fast the flame front =
travels=20
across the combustion chamber but we do know that maximum work occurs =
when 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 =
last=20
past that point. Lots of unknowns and theory doesn't always work in =
practice.=20
But if we use the average piston position at half-way down the bore, =
where=20
most work is most effective, and the MEP (Mean Effective Pressure), =
since Mean=20
is average, calculations begin.</DIV>
<DIV dir=3Dauto><BR></DIV>
<DIV dir=3Dauto> A single revolution is 360=C2=B0 so =
half-way down the=20
power stroke is 90=C2=B0. Add the idle spark timing of 4=C2=B0 BTDC =
(Before Top Dead=20
Center) and we get 94=C2=B0 of crank rotation from spark to MEP at =
half-way down.=20
That's 94/360 or about 0.26 of an engine revolution. And the engine is =
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.</DIV>
<DIV dir=3Dauto><BR></DIV>
<DIV dir=3Dauto> If we repeat the calculations for =
operating engine=20
speed and at maximum advance, we get 3500rpm / 60mps =3D 58rps =
(revolutions=20
per second). Maximum advance is 32=C2=B0 BTDC so 90=C2=B0 + =
32=C2=B0 =3D 122=C2=B0, spark to=20
MEP or 122=C2=B0/360=C2=B0 =3D 0.34 of a revolution</DIV>
<DIV dir=3Dauto>
<DIV dir=3Dauto><BR></DIV>
<DIV dir=3Dauto> 58rps is 17ms/r so 17ms/r x 0.34r =3D =
5.78ms from=20
spark to MEP at half-way down the power stroke. This is a more =
representative=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 =
mean at=20
idle, the pressure at idle becomes most effective 5.6ms before =
half-way and=20
for another 5.6ms after midpoint. Interesting that the idle pressure=20
application time is about the same as the maximum pressure application =
time,=20
or is that circular logic?</DIV>
<DIV dir=3Dauto><BR></DIV>
<DIV dir=3Dauto> Yes everything above is ripe with =
assumptions and=20
perhaps even apocryphal and resplendent with errors, but it is only =
7am after=20
all.</DIV>
<DIV dir=3Dauto><BR></DIV>
<DIV dir=3Dauto> Anyone with extra house-bound time on =
their 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=3Dauto><BR></DIV>
<DIV dir=3Dauto>Rick</DIV></DIV></DIV></DIV>
<P>
<HR>
=
<P></P>_______________________________________________<BR><BR>Mgs@autox.t=
eam.net<BR>Donate:=20
http://www.team.net/donate.html<BR>Suggested annual donation =20
$12.75<BR><BR>Archive: http://www.team.net/pipermail/mgs=20
http://autox.team.net/archive<BR><BR>Unsubscribe:=20
=
http://autox.team.net/mailman/options/mgs/paulhunt73@virginmedia.com<BR><=
/BLOCKQUOTE></BODY></HTML>
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