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Hello Richard,
One must take into account the retrograde wave peak when reaching a
maximum pressure in the cylinder. Another consideration is the
displacement of the primary components when the pressure wave is at its
maximum value. Using the Laws of Charles Breindigger then one has to
assume that the MEP has peaked at top dead centre and thus decline
occurs in burn speed but reduced in proportion to the ambient
temperature of the incoming air.  Failing considerable testing it must
be assumed that the burn will reduce to zero when mechanical components
phase into total inactivity. But then this is just thinking off the top
of my head.
Cheers - just a thought:-)
Barrie
On 3/28/2020 9:48 AM, Richard Lindsay wrote:
> Too fun Barrie! Thanks. The biggest unknown is dynamics of pressure
> rise versus piston position.
>
> On Sat, Mar 28, 2020, 8:38 AM Barrie Robinson <barrob@bell.net
>
> Hello Richard,
>
> My oath, I have just got out of bed - so pre-breakfast & coffee. Â
> I read your dissertation with bleary eyes and a befuddled brain
> but I followed your augment but just down to the last
> paragraphs.  So decided that breakfast was more important than
> erudite discussions - thus am going to put down my teddy bear and
> discard my night cap and make breakfast with the help of this
> chick who slept with me last night .
>
> So cheers everyone and I look forward to academic messaging.
>
> Barrie
>
> On 3/28/2020 8:54 AM, Richard Lindsay via Mgs wrote:
>> Hello friends,
>>
>> Â Â 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.
>>
>> Â Â 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?
>> Â Â Physics problems always start by listing the 'known' and the
>> property to 'find'. So in this case,
>>
>> KNOWN:
>> Â Â Idle speed: 900rpm
>>   Idle timing advance: 4° BTDC
>> Â Â Speed at maximum advance: 3500rpm
>>   Maximum timing advance: 32° BTDC
>>
>> FIND:
>> Â Â Time from spark to MEP
>>
>> Â Â 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.
>>
>> Â Â 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.
>>
>> Â Â At idle speed, 900rpm in this MG TD example, the XPAG engine
>> is turning 900rpm or 900rpm / 60mps = 15rps (revolutions per second).
>>
>> Â Â 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.
>>
>> Â Â 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° 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.
>>
>> Â Â 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
>>
>> Â Â 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?
>>
>> Â Â Yes everything above is ripe with assumptions and perhaps even
>> apocryphal and resplendent with errors, but it is only 7am after all.
>>
>> Â Â 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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>>
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>
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<font size="-1"><font face="Arial">Hello Richard,<br>
<br>
One must take into account the retrograde wave peak when
reaching a maximum pressure in the cylinder. Another
consideration is the displacement of the primary components when
the pressure wave is at its maximum value. Using the Laws of
Charles Breindigger then one has to assume that the MEP has
peaked at top dead centre and thus decline occurs in burn speed
but reduced in proportion to the ambient temperature of the
incoming air.  Failing considerable testing it must be assumed
that the burn will reduce to zero when mechanical components
phase into total inactivity. But then this is just thinking off
the top of my head.<br>
<br>
Cheers - just a thought<span
class="moz-smiley-s1"><span>:-)</span></span><br>
Barrie<br>
</font></font><br>
<br>
<div class="moz-cite-prefix">On 3/28/2020 9:48 AM, Richard Lindsay
wrote:<br>
</div>
<blockquote type="cite"
cite="mid:CAOc+-dwiUvTmwLwJNPHPvsTEVSrw7J_6utAfYiFij11ArThbeA@mail.gmail.com">
<meta http-equiv="content-type" content="text/html; charset=UTF-8">
<div dir="auto">Too fun Barrie! Thanks. The biggest unknown is
dynamics of pressure rise versus piston position. </div>
<br>
<div class="gmail_quote">
<div dir="ltr" class="gmail_attr">On Sat, Mar 28, 2020, 8:38 AM
Barrie Robinson <<a href="mailto:barrob@bell.net";
moz-do-not-send="true">barrob@bell.net</a>> wrote:<br>
</div>
<blockquote class="gmail_quote" style="margin:0 0 0
.8ex;border-left:1px #ccc solid;padding-left:1ex">
<div> <font size="-1"><font face="Arial">Hello Richard,<br>
<br>
My oath, I have just got out of bed - so pre-breakfast
& coffee.  I read your dissertation with bleary
eyes and a befuddled brain but I followed your augment
but just down to the last paragraphs.  So decided that
breakfast was more important than erudite discussions -
thus am going to put down my teddy bear and discard my
night cap and make breakfast with the help of this chick
who slept with me last night .<br>
<br>
So cheers everyone and I look forward to academic
messaging.<br>
<br>
Barrie<br>
</font></font><br>
<div>On 3/28/2020 8:54 AM, Richard Lindsay via Mgs wrote:<br>
</div>
<blockquote type="cite">
<div dir="auto">Hello friends,
<div dir="auto"><br>
<div dir="auto">Â Â 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">Â Â 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">Â Â 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">Â Â Idle speed: 900rpm</div>
<div dir="auto">  Idle timing advance: 4° BTDC</div>
<div dir="auto">Â Â 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">Â Â Time from spark to MEP</div>
<div dir="auto"><br>
</div>
<div dir="auto">Â Â 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">Â Â 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">Â Â 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">Â Â 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">Â Â 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">Â Â 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">Â Â 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">Â Â 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">Â Â 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>
<br>
<fieldset></fieldset>
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