<div dir="auto">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.<div dir="auto"><br></div><div dir="auto">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.</div><div dir="auto"><br></div><div dir="auto">Again, thank you Paul, <i>et al.</i>, for indulging my house-bound mental exercises.</div><div dir="auto"><br></div><div dir="auto">Rick</div></div><br><div class="gmail_quote"><div dir="ltr" class="gmail_attr">On Sat, Mar 28, 2020, 8:54 AM PaulHunt73 <<a href="mailto:paulhunt73@virginmedia.com">paulhunt73@virginmedia.com</a>> wrote:<br></div><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><u></u>
<div bgcolor="#ffffff">
<div><font size="2">"</font><font size="3">at idle 100% of the available power is
used to overcome the friction and other forces that exist at idle
speed"</font></div>
<div><font size="3"></font> </div>
<div><font size="2">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.</font></div>
<div><font size="2"></font> </div>
<div><font size="2">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.</font></div>
<div><font size="2"></font> </div>
<div><font size="2">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.</font></div>
<div><font size="2"></font> </div>
<div><font size="2">PaulH.</font></div>
<div><font size="2"></font> </div>
<div><font size="2"></font> </div>
<div><font size="3">----- Original Message ----- </font></div>
<blockquote style="BORDER-LEFT:#000000 2px solid;PADDING-LEFT:5px;PADDING-RIGHT:0px;MARGIN-LEFT:5px;MARGIN-RIGHT:0px">
<div style="FONT:10pt arial;BACKGROUND:#e4e4e4"><b>From:</b>
<a title="mgs@autox.team.net" href="mailto:mgs@autox.team.net" target="_blank" rel="noreferrer">Richard Lindsay
via Mgs</a> </div>
<div style="FONT:10pt arial"><b>To:</b> <a title="mgs@autox.team.net" href="mailto:mgs@autox.team.net+List" target="_blank" rel="noreferrer">mgs@autox.team.net List</a> </div>
<div style="FONT:10pt arial"><b>Sent:</b> Saturday, March 28, 2020 12:54
PM</div>
<div style="FONT:10pt arial"><b>Subject:</b> [Mgs] Engine maths...and spare
time</div>
<div><br></div>
<div dir="auto">Hello friends,
<div dir="auto"><br>
<div dir="auto"> 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.</div>
<div dir="auto"><br></div>
<div dir="auto"> 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?</div>
<div dir="auto"> Physics problems always start by listing the
'known' and the property to 'find'. So in this case,</div>
<div dir="auto"><br></div>
<div dir="auto">KNOWN:</div>
<div dir="auto"> Idle speed: 900rpm</div>
<div dir="auto"> Idle timing advance: 4° BTDC</div>
<div dir="auto"> Speed at maximum advance: 3500rpm</div>
<div dir="auto"> Maximum timing advance: 32° BTDC</div>
<div dir="auto"><br></div>
<div dir="auto">FIND:</div>
<div dir="auto"> Time from spark to MEP</div>
<div dir="auto"><br></div>
<div dir="auto"> 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.</div>
<div dir="auto"><br></div>
<div dir="auto"> 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.</div>
<div dir="auto"><br></div>
<div dir="auto"> At idle speed, 900rpm in this MG TD example, the
XPAG engine is turning 900rpm or 900rpm / 60mps = 15rps (revolutions per
second).</div>
<div dir="auto"><br></div>
<div dir="auto"> 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.</div>
<div dir="auto"><br></div>
<div dir="auto"> 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.</div>
<div dir="auto"><br></div>
<div dir="auto"> A single revolution is 360° so half-way down the
power stroke is 90°. Add the idle spark timing of 4° BTDC (Before Top Dead
Center) and we get 94° 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
= 17ms from spark to MEP at half-way down the power stroke, at idle.</div>
<div dir="auto"><br></div>
<div dir="auto"> If we repeat the calculations for operating engine
speed and at maximum advance, we get 3500rpm / 60mps = 58rps (revolutions
per second). Maximum advance is 32° BTDC so 90° + 32° = 122°, spark to
MEP or 122°/360° = 0.34 of a revolution</div>
<div dir="auto">
<div dir="auto"><br></div>
<div dir="auto"> 58rps is 17ms/r so 17ms/r x 0.34r = 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?</div>
<div dir="auto"><br></div>
<div dir="auto"> Yes everything above is ripe with assumptions and
perhaps even apocryphal and resplendent with errors, but it is only 7am after
all.</div>
<div dir="auto"><br></div>
<div dir="auto"> 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!</div>
<div dir="auto"><br></div>
<div dir="auto">Rick</div></div></div></div>
<p>
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