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

From: PaulHunt73 via Mgs <[email-masked]>
Date: Sat, 28 Mar 2020 13:36:37 -0000 Ej+6x25zJ+b1sD63Vx/a10XTFY/5WZPr8U15AzNAA4kcRWQaqvd8Y9dTveLeYb5pb/imaStSQ89VLd6RVmEGf/f2TR1rJu7NGjE=

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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: [email-masked] 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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<META name=3DGENERATOR content=3D"MSHTML 8.00.6001.23588">
<STYLE></STYLE>
</HEAD>
<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>&nbsp;</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&nbsp;vacuum leak the idle speed goes up.&nbsp; This =
may be=20
included in your 'other forces' above.</FONT></DIV>
<DIV><FONT size=3D2></FONT>&nbsp;</DIV>
<DIV><FONT size=3D2>Whist there may well be a most efficient&nbsp;point =
to start=20
combustion and the flame front, the prime consideration has to be =
avoiding=20
spontaneous&nbsp;combustion at any point, i.e. pinking or =
detonation.&nbsp; As=20
the flame front travels pressure inside the engine is rising,&nbsp;but =
after TDC=20
the volume available is reducing, which tends to counteract the pressure =

increase.&nbsp; There is also the&nbsp;effect of leverage i.e. the angle =
the con=20
rod makes relative to the piston.</FONT></DIV>
<DIV><FONT size=3D2></FONT>&nbsp;</DIV>
<DIV><FONT size=3D2>I'm certainly not going to check your maths, a =
specific engine=20
is what&nbsp;it is, and the timing has to be set&nbsp;taking those =
specifics=20
into account&nbsp;plus other factors like fuel grade and =
type.</FONT></DIV>
<DIV><FONT size=3D2></FONT>&nbsp;</DIV>
<DIV><FONT size=3D2>PaulH.</FONT></DIV>
<DIV><FONT size=3D2></FONT>&nbsp;</DIV>
<DIV><FONT size=3D2></FONT>&nbsp;</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=[email-masked] =
href=3D"mailto:[email-masked]">Richard Lindsay=20
  via Mgs</A> </DIV>
  <DIV style=3D"FONT: 10pt arial"><B>To:</B> <A =
title=[email-masked]=20
  href=3D"mailto:[email-masked] List">[email-masked] 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>&nbsp; &nbsp;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>&nbsp; &nbsp;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>&nbsp; &nbsp;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>&nbsp; &nbsp;Idle speed: 900rpm</DIV>
  <DIV dir=3Dauto>&nbsp; &nbsp;Idle timing advance: 4=C2=B0 BTDC</DIV>
  <DIV dir=3Dauto>&nbsp; &nbsp;Speed at maximum advance: 3500rpm</DIV>
  <DIV dir=3Dauto>&nbsp; &nbsp;Maximum timing advance: 32=C2=B0 =
BTDC</DIV>
  <DIV dir=3Dauto><BR></DIV>
  <DIV dir=3Dauto>FIND:</DIV>
  <DIV dir=3Dauto>&nbsp; &nbsp;Time from spark to MEP</DIV>
  <DIV dir=3Dauto><BR></DIV>
  <DIV dir=3Dauto>&nbsp; &nbsp;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>&nbsp; &nbsp;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>&nbsp; &nbsp;At idle speed, 900rpm in this MG TD =
example, the=20
  XPAG engine is turning 900rpm or&nbsp;900rpm / 60mps =3D 15rps =
(revolutions per=20
  second).</DIV>
  <DIV dir=3Dauto><BR></DIV>
  <DIV dir=3Dauto>&nbsp; &nbsp;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>&nbsp; &nbsp;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>&nbsp; &nbsp;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>&nbsp; &nbsp;If we repeat the calculations for =
operating engine=20
  speed and at maximum advance, we get&nbsp;3500rpm / 60mps =3D 58rps =
(revolutions=20
  per second). Maximum advance is 32=C2=B0 BTDC so&nbsp;90=C2=B0 + =
32=C2=B0 =3D 122=C2=B0, spark to=20
  MEP or&nbsp;122=C2=B0/360=C2=B0 =3D 0.34 of a revolution</DIV>
  <DIV dir=3Dauto>
  <DIV dir=3Dauto><BR></DIV>
  <DIV dir=3Dauto>&nbsp; &nbsp;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>&nbsp; &nbsp;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>&nbsp; &nbsp;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>[email-masked]=
eam.net<BR>Donate:=20
  http://www.team.net/donate.html<BR>Suggested annual donation&nbsp;=20
  $12.75<BR><BR>Archive: http://www.team.net/pipermail/mgs=20
  http://autox.team.net/archive<BR><BR>Unsubscribe:=20
  =
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From: Barrie Robinson via Mgs <[email-masked]>
Date: Sat, 28 Mar 2020 10:02:33 -0400

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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 <[email-masked] 
>
>     Hello Richard,
>
>     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 .
>
>     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,
>>
>>        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
>>        Idle timing advance: 4° BTDC
>>        Maximum timing advance: 32° 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 = 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° 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.
>>
>>        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
>>
>>        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?
>>
>>        Yes everything above is ripe with assumptions and perhaps even
>>        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
>>
>>     Donate:http://www.team.net/donate.html
>>     Archive:http://www.team.net/pipermail/mgs  http://autox.team.net/archive
>>
>>     Unsubscribe:http://autox.team.net/mailman/options/mgs/[email-masked]
>

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<html>
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    <meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
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  <body>
    <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:[email-masked]">
      <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 &lt;<a href="mailto:[email-masked]"
            moz-do-not-send="true">[email-masked]</a>&gt; 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
                &amp; 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">   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>
              <br>
              <fieldset></fieldset>
              <pre>_______________________________________________

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</pre>
            </blockquote>
            <br>
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        </blockquote>
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[email-masked]

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From: Richard Lindsay via Mgs <[email-masked]>
Date: Sat, 28 Mar 2020 09:17:45 -0500

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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 <[email-masked]> 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 <[email-masked]>
> *To:* [email-masked] 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
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
>    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
>    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
>    Anyone with extra house-bound time on their hands, please check my
> maths and share your corrections, including the logic of the whole
> ------------------------------
>
> Unsubscribe:

--0000000000007b66ac05a1eae2a6
Content-Transfer-Encoding: quoted-printable

<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&#39;t keep up with practice. The factory=
&#39;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&#39;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 &lt;<a href=3D"mailto:=
[email-masked]">[email-masked]</a>&gt; 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">&quot;</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&quot;</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 &#39;other forces&#39; 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&#39;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"[email-masked]" href=3D"mailto:[email-masked]" 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"[email-masked]"=
 href=3D"mailto:[email-masked]+List" target=3D"_blank" rel=3D"noreferre=
r">[email-masked] 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&#39;t burn instantly, despite our perception to the contra=
ry.=20
  Rather, it takes a finite length of time from the occurance of the &#39;s=
park&#39;,=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
  &#39;known&#39; and the property to &#39;find&#39;. 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&#39;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&#39;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&#39;=
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&#39;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&#39;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&#39;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:[email-masked]" target=3D"_blank" rel=3D"noreferrer">[email-masked]=
eam.net</a><br>Donate:=20
  <a href=3D"http://www.team.net/donate.html" target=3D"_blank" rel=3D"nore=
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
  <a href=3D"http://autox.team.net/archive" target=3D"_blank" rel=3D"norefe=
rrer">http://autox.team.net/archive</a><br><br>Unsubscribe:=20
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options/mgs/[email-masked]</a><br></p></blockquote></div>
</blockquote></div>

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[email-masked]

--===============6478281743309314924==--

---

From: Richard Lindsay via Mgs <[email-masked]>
Date: Sat, 28 Mar 2020 09:23:30 -0500

--===============4243197126574771395==

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As I have said before, I don't drive much but I do enjoy understanding and
building engines. Anyone interested is said topic, especially LBC engines,
consider buying and ingesting this book.

https://www.amazon.com/sports-car-engine-tuning-modification/dp/0837600448/=
ref=3Dsr_1_1?keywords=3Dthe+sports+car+engine+its&qid=3D1585403679&sr=3D8-1

Did I say 'consider'? I strongly recommend buying it. Complex, dynamic
engine processes explained in common language.

Rick

On Sat, Mar 28, 2020, 9:17 AM Richard Lindsay <[email-masked]>
wrote:

> And of course, tuning for maximum power is not the only maker's goal. Fue=
l
> On Sat, Mar 28, 2020, 8:54 AM PaulHunt73 <[email-masked]>
> wrote:
>
>> "at idle 100% of the available power is used to overcome the friction
>> about 16 in Hg. or so in the intake manifold, and is why when you introd=
uce
>> an intake vacuum leak the idle speed goes up.  This may be included in y=
our
>> '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
>> flame front travels pressure inside the engine is rising, but after TDC =
the
>> increase.  There is also the effect of leverage i.e. the angle the con r=
od
>> I'm certainly not going to check your maths, a specific engine is what i=
t
>> is, and the timing has to be set taking those specifics into account plu=
s
>> house-bound morning I woke up thinking about ignition timing. Here are t=
he
>> finite length of time from the occurance of the 'spark', the flame front=
 to
>> produce maximum power. In fact, at idle 100% of the available power is u=
sed
>> to overcome the friction and other forces that exist at idle speed. Stat=
ed
>> available charge. That fact is evident (with carbureted engines) when on=
e
>> setting, as the engine warms and friction forces decrease. But back to t=
he
>> problem.
>>
>>    Because the goal at idle is smooth running and progression off of idl=
e
>> (e.g. speeding up), not maximum power, the calculated wavefront speed ma=
y
>>    At idle speed, 900rpm in this MG TD example, the XPAG engine is
>> turning only 7.5 revolutions per second. But timing numbers are specifie=
d
>> chamber but we do know that maximum work occurs when the piston is half =
way
>> point. Lots of unknowns and theory doesn't always work in practice. But =
if
>> work is most effective, and the MEP (Mean Effective Pressure), since Mea=
n
>>    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 w=
e get
>> 94=C2=B0 of crank rotation from spark to MEP at half-way down. That's 94=
/360 or
>> revolutions per second or 67ms (milliseconds) per revolution. So 67 x 0.=
26
>>    If we repeat the calculations for operating engine speed and at
>> maximum advance, we get 3500rpm / 60mps =3D 58rps (revolutions per secon=
d).
>> Maximum advance is 32=C2=B0 BTDC so 90=C2=B0 + 32=C2=B0 =3D 122=C2=B0, s=
park to MEP or 122=C2=B0/360=C2=B0
>> =3D 0.34 of a revolution
>> half-way down the power stroke. This is a more representative number tha=
n
>> optimal time across the strike's midpoint. Doing so would mean at idle, =
the
>> experiment...or perhaps even why geeks think these ways!

--000000000000028a8805a1eaf7a2
Content-Transfer-Encoding: quoted-printable

<div dir=3D"auto"><div dir=3D"auto" style=3D"font-family:sans-serif">As I h=
ave said before, I don&#39;t drive much but I do enjoy understanding and bu=
ilding engines. Anyone interested is said topic, especially LBC engines, co=
nsider buying and ingesting this book.</div><div dir=3D"auto" style=3D"font=
-family:sans-serif"><br></div><div dir=3D"auto" style=3D"font-family:sans-s=
erif"><a href=3D"https://www.amazon.com/sports-car-engine-tuning-modificati=
on/dp/0837600448/ref=3Dsr_1_1?keywords=3Dthe+sports+car+engine+its&amp;qid=
=3D1585403679&amp;sr=3D8-1">https://www.amazon.com/sports-car-engine-tuning=
-modification/dp/0837600448/ref=3Dsr_1_1?keywords=3Dthe+sports+car+engine+i=
ts&amp;qid=3D1585403679&amp;sr=3D8-1</a><br></div><div dir=3D"auto" style=
=3D"font-family:sans-serif"><br></div><div dir=3D"auto" style=3D"font-famil=
y:sans-serif">Did I say &#39;consider&#39;? I strongly recommend buying it.=
 Complex, dynamic engine processes explained in common language.</div><div =
dir=3D"auto" style=3D"font-family:sans-serif"><br></div><div dir=3D"auto" s=
tyle=3D"font-family:sans-serif">Rick</div></div><br><div class=3D"gmail_quo=
te"><div dir=3D"ltr" class=3D"gmail_attr">On Sat, Mar 28, 2020, 9:17 AM Ric=
hard Lindsay &lt;<a href=3D"mailto:[email-masked]">richardolinds=
[email-masked]</a>&gt; wrote:<br></div><blockquote class=3D"gmail_quote" styl=
e=3D"margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><div di=
r=3D"auto">Thank you Paul. As you know, I have always appreciated your dept=
h of knowledge and experience. In fact, as you point out, there are SO many=
 variables. Theory just can&#39;t keep up with practice. The factory&#39;s =
tuning recommendations for our beloved engines are almost certainly empiric=
al. That is, decided upon by trial and error.<div dir=3D"auto"><br></div><d=
iv dir=3D"auto">And of course, tuning for maximum power is not the only mak=
er&#39;s goal. Fuel economy, emissions, build cost, and a plethora of 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 house-bound m=
ental exercises.</div><div dir=3D"auto"><br></div><div dir=3D"auto">Rick</d=
iv></div><br><div class=3D"gmail_quote"><div dir=3D"ltr" class=3D"gmail_att=
r">On Sat, Mar 28, 2020, 8:54 AM PaulHunt73 &lt;<a href=3D"mailto:paulhunt7=
[email-masked]" target=3D"_blank" rel=3D"noreferrer">paulhunt73@virginme=
dia.com</a>&gt; wrote:<br></div><blockquote class=3D"gmail_quote" style=3D"=
margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><u></u>

  <a title=3D"[email-masked]" href=3D"mailto:[email-masked]" rel=3D=
"noreferrer noreferrer" target=3D"_blank">Richard Lindsay=20
 href=3D"mailto:[email-masked]+List" rel=3D"noreferrer noreferrer" targ=
et=3D"_blank">[email-masked] List</a> </div>

  <p></p>_______________________________________________<br><br><a href=3D"=
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</blockquote></div>
</blockquote></div>

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[email-masked]

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---

From: Barney Gaylord via Mgs <[email-masked]>
Date: Sat, 28 Mar 2020 14:56:43 -0400

Rick, ---- Okay, time to spare, so I'll bite.

You do a good job of calculating time from spark 
event to half stroke (about 6-ms at road speed), 
but I think you were asking a different 
question.  I thought you were asking how much 
time to complete combustion to get to maximum 
pressure.  That is, how much time for the flame 
front to propagate all the way across the combustion chamber?

And you also said. "we do know that maximum work 
occurs when the piston is half way down the 
cylinder", which is not true.  Most of the work 
has already been done before the piston gets half 
way down, and maximum torque on the crankshaft 
happens significantly higher in the stroke before 
pressure is lost to expansion.

For best power and efficiency, combustion should 
be completed at or slightly after TDC.  But since 
there is very little motion of the piston 
immediately after TDC, It works just about as 
well if max pressure comes just a little later, 
like maybe 10 to 20d ATDC.  That little delay can 
allow use of higher compression ratio for better power and efficiency.

I like to use 3600 rpm for road speed, because it 
divides evenly into 360 degrees rotation for nice 
round numbers.  And 900 rpm idle speed will be 
exactly 1/4 of road speed.  If you make spark at 
32d BTDC at road speed, it takes 1.5-ms to reach 
TDC.  10d ATDC is at 2-ms, and 30d ATDC would be 
3-ms (after spark).  So the flame front 
propagation to complete combustion is in the 2 to 
3-ms range.  I suppose this is the answer to your 
question, "how much time" for the flame front to cross the combustion chamber.

Distance from the spark plug to far side of the 
combustion chamber is about 2-1/2 inches, which 
it does in about 2-1/2 ms, so flame front speed 
is about 1 inch per ms, or 1000 inches per 
second, which is fairly close to 60-mph.  And you 
night notice that I did not use "MEP" in that entire discussion.

Barney

At 08:54 AM 3/28/2020, you wrote:
>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 
>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
>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 
>= 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 
>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.
>
>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 
>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.
>
>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
>
>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?
>
>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 

[email-masked]

---

From: Richard Lindsay via Mgs <[email-masked]>
Date: Sat, 28 Mar 2020 14:32:11 -0500

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...from Campbell's book.

On Sat, Mar 28, 2020, 2:07 PM Barney Gaylord <[email-masked]> wrote:

> >=3D 15rps (revolutions per second).
> >
> >MEP (Mean Effective Pressure), since Mean is average, calculations begin=
.
> >
> >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
> >per revolution. So 67 x 0.26 =3D 17ms from spark
> >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

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<div dir=3D"auto">...from Campbell&#39;s book.</div><br><div class=3D"gmail=
_quote"><div dir=3D"ltr" class=3D"gmail_attr">On Sat, Mar 28, 2020, 2:07 PM=
 Barney Gaylord &lt;<a href=3D"mailto:[email-masked]">barneymg@mgagur=
u.com</a>&gt; wrote:<br></div><blockquote class=3D"gmail_quote" style=3D"ma=
rgin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">Rick, ---- Oka=
y, time to spare, so I&#39;ll bite.<br>
<br>
You do a good job of calculating time from spark <br>
event to half stroke (about 6-ms at road speed), <br>
but I think you were asking a different <br>
question.=C2=A0 I thought you were asking how much <br>
time to complete combustion to get to maximum <br>
pressure.=C2=A0 That is, how much time for the flame <br>
front to propagate all the way across the combustion chamber?<br>
<br>
And you also said. &quot;we do know that maximum work <br>
occurs when the piston is half way down the <br>
cylinder&quot;, which is not true.=C2=A0 Most of the work <br>
has already been done before the piston gets half <br>
way down, and maximum torque on the crankshaft <br>
happens significantly higher in the stroke before <br>
pressure is lost to expansion.<br>
<br>
For best power and efficiency, combustion should <br>
be completed at or slightly after TDC.=C2=A0 But since <br>
there is very little motion of the piston <br>
immediately after TDC, It works just about as <br>
well if max pressure comes just a little later, <br>
like maybe 10 to 20d ATDC.=C2=A0 That little delay can <br>
allow use of higher compression ratio for better power and efficiency.<br>
<br>
I like to use 3600 rpm for road speed, because it <br>
divides evenly into 360 degrees rotation for nice <br>
round numbers.=C2=A0 And 900 rpm idle speed will be <br>
exactly 1/4 of road speed.=C2=A0 If you make spark at <br>
32d BTDC at road speed, it takes 1.5-ms to reach <br>
TDC.=C2=A0 10d ATDC is at 2-ms, and 30d ATDC would be <br>
3-ms (after spark).=C2=A0 So the flame front <br>
propagation to complete combustion is in the 2 to <br>
3-ms range.=C2=A0 I suppose this is the answer to your <br>
question, &quot;how much time&quot; for the flame front to cross the combus=
tion chamber.<br>
<br>
Distance from the spark plug to far side of the <br>
combustion chamber is about 2-1/2 inches, which <br>
it does in about 2-1/2 ms, so flame front speed <br>
is about 1 inch per ms, or 1000 inches per <br>
second, which is fairly close to 60-mph.=C2=A0 And you <br>
night notice that I did not use &quot;MEP&quot; in that entire discussion.<=
br>
<br>
Barney<br>
<br>
<br>
At 08:54 AM 3/28/2020, you wrote:<br>
&gt;Hello friends,<br>
&gt;<br>
&gt;When one is a geek, one thinks of geeky things. <br>
&gt;I am a geek and this house-bound morning I woke <br>
&gt;up thinking about ignition timing. Here are the details.<br>
&gt;<br>
&gt;We know that the charge (fuel plus air) in a <br>
&gt;cylinder doesn&#39;t burn instantly, despite our <br>
&gt;perception to the contrary. Rather, it takes a <br>
&gt;finite length of time from the occurance of the <br>
&gt;&#39;spark&#39;, the flame front to cross the combustion <br>
&gt;chamber, and to raise the MEP (Mean Effective <br>
&gt;Pressure) to a maximum - the point where it does <br>
&gt;the most work. But how much time?<br>
&gt;<br>
&gt;Physics problems always start by listing the <br>
&gt;&#39;known&#39; and the property to &#39;find&#39;. So in this case,<br=
&gt;<br>
&gt;KNOWN:<br>
&gt;Idle speed: 900rpm<br>
&gt;Idle timing advance: 4=C2=B0 BTDC<br>
&gt;Speed at maximum advance: 3500rpm<br>
&gt;Maximum timing advance: 32=C2=B0 BTDC<br>
&gt;<br>
&gt;FIND:<br>
&gt;Time from spark to MEP<br>
&gt;<br>
&gt;The first thing one might know is that the goal <br>
&gt;at idle is not to produce maximum power. In <br>
&gt;fact, at idle 100% of the available power is <br>
&gt;used to overcome the friction and other forces <br>
&gt;that exist at idle speed. Stated another way: <br>
&gt;Idle speed is the fastest the engine can achieve <br>
&gt;given the available charge. That fact is evident <br>
&gt;(with carbureted engines) when one notices that <br>
&gt;engine speed gradually increases, even for a <br>
&gt;fixed throttle setting, as the engine warms and <br>
&gt;friction forces decrease. But back to the problem.<br>
&gt;<br>
&gt;Because the goal at idle is smooth running and <br>
&gt;progression off of idle (e.g. speeding up), not <br>
&gt;maximum power, the calculated wavefront speed <br>
&gt;may not be correct at idle. But let&#39;s see.<br>
&gt;<br>
&gt;At idle speed, 900rpm in this MG TD example, the <br>
&gt;XPAG engine is turning 900rpm or 900rpm / 60mps <br>
&gt;=3D 15rps (revolutions per second).<br>
&gt;<br>
&gt;Distributor speed is 1/2 engine speed so at idle <br>
&gt;the distributor is turning only 7.5 revolutions <br>
&gt;per second. But timing numbers are specified in <br>
&gt;degrees of crank rotations so we will stick with 15rps.<br>
&gt;<br>
&gt;We don&#39;t know how fast the flame front travels <br>
&gt;across the combustion chamber but we do know <br>
&gt;that maximum work occurs when the piston is half <br>
&gt;way down the cylinder. And we also know that <br>
&gt;work isn&#39;t an instantaneous parameter so it must <br>
&gt;begin before the half way point and last past <br>
&gt;that point. Lots of unknowns and theory doesn&#39;t <br>
&gt;always work in practice. But if we use the <br>
&gt;average piston position at half-way down the <br>
&gt;bore, where most work is most effective, and the <br>
&gt;MEP (Mean Effective Pressure), since Mean is average, calculations begi=
n.<br>
&gt;<br>
&gt;single revolution is 360=C2=B0 so half-way down the <br>
&gt;power stroke is 90=C2=B0. Add the idle spark timing <br>
&gt;of 4=C2=B0 BTDC (Before Top Dead Center) and we get <br>
&gt;94=C2=B0 of crank rotation from spark to MEP at <br>
&gt;half-way down. That&#39;s 94/360 or about 0.26 of an <br>
&gt;engine revolution. And the engine is turning 15 <br>
&gt;revolutions per second or 67ms (milliseconds) <br>
&gt;per revolution. So 67 x 0.26 =3D 17ms from spark <br>
&gt;to MEP at half-way down the power stroke, at idle.<br>
&gt;<br>
&gt;If we repeat the calculations for operating <br>
&gt;engine speed and at maximum advance, we get <br>
&gt;3500rpm / 60mps =3D 58rps (revolutions per <br>
&gt;second). Maximum advance is 32=C2=B0 BTDC so 90=C2=B0 + <br>
&gt;32=C2=B0 =3D 122=C2=B0, spark to MEP or 122=C2=B0/360=C2=B0 =3D 0.34 of=
 a revolution<br>
&gt;<br>
&gt;58rps is 17ms/r so 17ms/r x 0.34r =3D 5.78ms from <br>
&gt;spark to MEP at half-way down the power stroke. <br>
&gt;This is a more representative number than the <br>
&gt;17ms at idle. One might even divide the idle <br>
&gt;elapsed time minus the optimal time across the <br>
&gt;strike&#39;s midpoint. Doing so would mean at idle, <br>
&gt;the pressure at idle becomes most effective <br>
&gt;5.6ms before half-way and for another 5.6ms <br>
&gt;after midpoint. Interesting that the idle <br>
&gt;pressure application time is about the same as <br>
&gt;the maximum pressure application time, or is that circular logic?<br>
&gt;<br>
&gt;Yes everything above is ripe with assumptions <br>
&gt;and perhaps even apocryphal and resplendent with <br>
&gt;errors, but it is only 7am after all.<br>
&gt;<br>
&gt;Anyone with extra house-bound time on their <br>
&gt;hands, please check my maths and share your <br>
&gt;corrections, including the logic of the whole <br>
&gt;experiment...or perhaps even why geeks think these ways!<br>
&gt;<br>
&gt;Rick<br>
<br>
</blockquote></div>

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[email-masked]

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From: Richard Lindsay via Mgs <[email-masked]>
Date: Sat, 28 Mar 2020 14:43:09 -0500

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Barney, I appreciate your words and critique. Same with PaulH's words. We
are such techno-motor-heads!

On Sat, Mar 28, 2020 at 2:07 PM Barney Gaylord <[email-masked]> wrote=

.
a revolution
> >experiment...or perhaps even why geeks think these ways!

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<div dir=3D"ltr">Barney, I appreciate your words and critique. Same with Pa=
ulH&#39;s words. We are such techno-motor-heads!</div><br><div class=3D"gma=
il_quote"><div dir=3D"ltr" class=3D"gmail_attr">On Sat, Mar 28, 2020 at 2:0=
7 PM Barney Gaylord &lt;<a href=3D"mailto:[email-masked]">barneymg@mg=
aguru.com</a>&gt; wrote:<br></div><blockquote class=3D"gmail_quote" style=
=3D"margin:0px 0px 0px 0.8ex;border-left:1px solid rgb(204,204,204);padding=
-left:1ex">Rick, ---- Okay, time to spare, so I&#39;ll bite.<br>
<br>
&gt;<br>
&gt;KNOWN:<br>

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[email-masked]

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From: Barney Gaylord via Mgs <[email-masked]>
Date: Sat, 28 Mar 2020 19:14:07 -0400

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I like the chart.  Pressure peaks at 12d ATDC, and by 90d ATDC it is 90%=
 gone.

If you multiply sine of the angle by the pressure=20
all alog the curve (after TDC), you get another=20
curve representing progression of torque.  Not a=20
lot of torque yet at 10d, pretty good torque by=20
20d, peaking around 30d ATDC.  Still a fair=20
amount of torque at 60d, but by 90d (half stroke)=20
the torque is nearly gone along with the pressure.

Work being done would be represented by the area=20
under the torque curve (not on the chart).  Since=20
both pressure and torque are nearly exhausted by=20
90d ATDC, there is very little energy (work) left=20
to be harvested after mid stroke.  That's why we=20
like to open the exhaust valve early, to let what=20
remains of the mostly useless fumes out of the=20
cylinder, encouraging best intake of fresh air and fuel mix half a turn=
 later.

Barney

At 03:32 PM 3/28/2020, Richard Lindsay wrote:
>...from Campbell's book.
>....
>Attachment: 20200328_113435.jpg

>On Sat, Mar 28, 2020, 2:07 PM Barney Gaylord=20
>question.=C2  I thought you were asking how much
>pressure.=C2  That is, how much time for the flame
>cylinder", which is not true.=C2  Most of the work
>be completed at or slightly after TDC.=C2  But since
>like maybe 10 to 20d ATDC.=C2  That little delay can
>round numbers.=C2  And 900 rpm idle speed will be
>exactly 1/4 of road speed.=C2  If you make spark at
>TDC.=C2  10d ATDC is at 2-ms, and 30d ATDC would be
>3-ms (after spark).=C2  So the flame front
>3-ms range.=C2  I suppose this is the answer to your
>question, "how much time" for the flame front to cross the combustion=
 chamber.
>second, which is fairly close to 60-mph.=C2  And you

> >32=C2=B0 =3D 122=C2=B0, spark to MEP or 122=C2=B0/360=C2=B0 =3D 0.34 of a=
 revolution
> >experiment...or perhaps even why geeks think these ways!

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<html>
<body>
I like the chart.&nbsp; Pressure peaks at 12d ATDC, and by 90d ATDC it is
90% gone.<br><br>
If you multiply sine of the angle by the pressure all alog the curve
(after TDC), you get another curve representing progression of
torque.&nbsp; Not a lot of torque yet at 10d, pretty good torque by 20d,
peaking around 30d ATDC.&nbsp; Still a fair amount of torque at 60d, but
by 90d (half stroke) the torque is nearly gone along with the
pressure.<br><br>
Work being done would be represented by the area under the torque curve
(not on the chart).&nbsp; Since both pressure and torque are nearly
exhausted by 90d ATDC, there is very little energy (work) left to be
harvested after mid stroke.&nbsp; That's why we like to open the exhaust
valve early, to let what remains of the mostly useless fumes out of the
cylinder, encouraging best intake of fresh air and fuel mix half a turn
later.<br><br>
Barney<br><br>
<br>
At 03:32 PM 3/28/2020, Richard Lindsay wrote:<br>
<blockquote type=3Dcite class=3Dcite cite=3D"">...from Campbell's book.<br>
....<br>
Attachment: 20200328_113435.jpg</blockquote><br><br>
<blockquote type=3Dcite class=3Dcite cite=3D"">On Sat, Mar 28, 2020, 2:07 PM
Barney Gaylord
&lt;<a href=3D"mailto:[email-masked]">[email-masked]</a>&gt;
wrote:<br>

<dl>
<dd>Rick, ---- Okay, time to spare, so I'll bite.<br><br>

<dd>You do a good job of calculating time from spark <br>

<dd>event to half stroke (about 6-ms at road speed), <br>

<dd>but I think you were asking a different <br>

<dd>question.=C2&nbsp; I thought you were asking how much <br>

<dd>time to complete combustion to get to maximum <br>

<dd>pressure.=C2&nbsp; That is, how much time for the flame <br>

<dd>front to propagate all the way across the combustion
chamber?<br><br>

<dd>And you also said. &quot;we do know that maximum work <br>

<dd>occurs when the piston is half way down the <br>

<dd>cylinder&quot;, which is not true.=C2&nbsp; Most of the work <br>

<dd>has already been done before the piston gets half <br>

<dd>way down, and maximum torque on the crankshaft <br>

<dd>happens significantly higher in the stroke before <br>

<dd>pressure is lost to expansion.<br><br>

<dd>For best power and efficiency, combustion should <br>

<dd>be completed at or slightly after TDC.=C2&nbsp; But since <br>

<dd>there is very little motion of the piston <br>

<dd>immediately after TDC, It works just about as <br>

<dd>well if max pressure comes just a little later, <br>

<dd>like maybe 10 to 20d ATDC.=C2&nbsp; That little delay can <br>

<dd>allow use of higher compression ratio for better power and
efficiency.<br><br>

<dd>I like to use 3600 rpm for road speed, because it <br>

<dd>divides evenly into 360 degrees rotation for nice <br>

<dd>round numbers.=C2&nbsp; And 900 rpm idle speed will be <br>

<dd>exactly 1/4 of road speed.=C2&nbsp; If you make spark at <br>

<dd>32d BTDC at road speed, it takes 1.5-ms to reach <br>

<dd>TDC.=C2&nbsp; 10d ATDC is at 2-ms, and 30d ATDC would be <br>

<dd>3-ms (after spark).=C2&nbsp; So the flame front <br>

<dd>propagation to complete combustion is in the 2 to <br>

<dd>3-ms range.=C2&nbsp; I suppose this is the answer to your <br>

<dd>question, &quot;how much time&quot; for the flame front to cross the
combustion chamber.<br><br>

<dd>Distance from the spark plug to far side of the <br>

<dd>combustion chamber is about 2-1/2 inches, which <br>

<dd>it does in about 2-1/2 ms, so flame front speed <br>

<dd>is about 1 inch per ms, or 1000 inches per <br>

<dd>second, which is fairly close to 60-mph.=C2&nbsp; And you <br>

<dd>night notice that I did not use &quot;MEP&quot; in that entire
discussion.<br><br>

<dd>Barney</blockquote>
</dl><br><br>
<blockquote type=3Dcite class=3Dcite cite=3D"">
<dl>
<dd>At 08:54 AM 3/28/2020, you wrote:<br>

<dd>&gt;Hello friends,<br>

<dd>&gt;<br>

<dd>&gt;When one is a geek, one thinks of geeky things. <br>

<dd>&gt;I am a geek and this house-bound morning I woke <br>

<dd>&gt;up thinking about ignition timing. Here are the details.<br>

<dd>&gt;<br>

<dd>&gt;We know that the charge (fuel plus air) in a <br>

<dd>&gt;cylinder doesn't burn instantly, despite our <br>

<dd>&gt;perception to the contrary. Rather, it takes a <br>

<dd>&gt;finite length of time from the occurance of the <br>

<dd>&gt;'spark', the flame front to cross the combustion <br>

<dd>&gt;chamber, and to raise the MEP (Mean Effective <br>

<dd>&gt;Pressure) to a maximum - the point where it does <br>

<dd>&gt;the most work. But how much time?<br>

<dd>&gt;<br>

<dd>&gt;Physics problems always start by listing the <br>

<dd>&gt;'known' and the property to 'find'. So in this case,<br>

<dd>&gt;<br>

<dd>&gt;KNOWN:<br>

<dd>&gt;Idle speed: 900rpm<br>

<dd>&gt;Idle timing advance: 4=C2=B0 BTDC<br>

<dd>&gt;Speed at maximum advance: 3500rpm<br>

<dd>&gt;Maximum timing advance: 32=C2=B0 BTDC<br>

<dd>&gt;<br>

<dd>&gt;FIND:<br>

<dd>&gt;Time from spark to MEP<br>

<dd>&gt;<br>

<dd>&gt;The first thing one might know is that the goal <br>

<dd>&gt;at idle is not to produce maximum power. In <br>

<dd>&gt;fact, at idle 100% of the available power is <br>

<dd>&gt;used to overcome the friction and other forces <br>

<dd>&gt;that exist at idle speed. Stated another way: <br>

<dd>&gt;Idle speed is the fastest the engine can achieve <br>

<dd>&gt;given the available charge. That fact is evident <br>

<dd>&gt;(with carbureted engines) when one notices that <br>

<dd>&gt;engine speed gradually increases, even for a <br>

<dd>&gt;fixed throttle setting, as the engine warms and <br>

<dd>&gt;friction forces decrease. But back to the problem.<br>

<dd>&gt;<br>

<dd>&gt;Because the goal at idle is smooth running and <br>

<dd>&gt;progression off of idle (e.g. speeding up), not <br>

<dd>&gt;maximum power, the calculated wavefront speed <br>

<dd>&gt;may not be correct at idle. But let's see.<br>

<dd>&gt;<br>

<dd>&gt;At idle speed, 900rpm in this MG TD example, the <br>

<dd>&gt;XPAG engine is turning 900rpm or 900rpm / 60mps <br>

<dd>&gt;=3D 15rps (revolutions per second).<br>

<dd>&gt;<br>

<dd>&gt;Distributor speed is 1/2 engine speed so at idle <br>

<dd>&gt;the distributor is turning only 7.5 revolutions <br>

<dd>&gt;per second. But timing numbers are specified in <br>

<dd>&gt;degrees of crank rotations so we will stick with 15rps.<br>

<dd>&gt;<br>

<dd>&gt;We don't know how fast the flame front travels <br>

<dd>&gt;across the combustion chamber but we do know <br>

<dd>&gt;that maximum work occurs when the piston is half <br>

<dd>&gt;way down the cylinder. And we also know that <br>

<dd>&gt;work isn't an instantaneous parameter so it must <br>

<dd>&gt;begin before the half way point and last past <br>

<dd>&gt;that point. Lots of unknowns and theory doesn't <br>

<dd>&gt;always work in practice. But if we use the <br>

<dd>&gt;average piston position at half-way down the <br>

<dd>&gt;bore, where most work is most effective, and the <br>

<dd>&gt;MEP (Mean Effective Pressure), since Mean is average,
calculations begin.<br>

<dd>&gt;<br>

<dd>&gt;single revolution is 360=C2=B0 so half-way down the <br>

<dd>&gt;power stroke is 90=C2=B0. Add the idle spark timing <br>

<dd>&gt;of 4=C2=B0 BTDC (Before Top Dead Center) and we get <br>

<dd>&gt;94=C2=B0 of crank rotation from spark to MEP at <br>

<dd>&gt;half-way down. That's 94/360 or about 0.26 of an <br>

<dd>&gt;engine revolution. And the engine is turning 15 <br>

<dd>&gt;revolutions per second or 67ms (milliseconds) <br>

<dd>&gt;per revolution. So 67 x 0.26 =3D 17ms from spark <br>

<dd>&gt;to MEP at half-way down the power stroke, at idle.<br>

<dd>&gt;<br>

<dd>&gt;If we repeat the calculations for operating <br>

<dd>&gt;engine speed and at maximum advance, we get <br>

<dd>&gt;3500rpm / 60mps =3D 58rps (revolutions per <br>

<dd>&gt;second). Maximum advance is 32=C2=B0 BTDC so 90=C2=B0 + <br>

<dd>&gt;32=C2=B0 =3D 122=C2=B0, spark to MEP or 122=C2=B0/360=C2=B0 =3D 0.34=
 of a
revolution<br>

<dd>&gt;<br>

<dd>&gt;58rps is 17ms/r so 17ms/r x 0.34r =3D 5.78ms from <br>

<dd>&gt;spark to MEP at half-way down the power stroke. <br>

<dd>&gt;This is a more representative number than the <br>

<dd>&gt;17ms at idle. One might even divide the idle <br>

<dd>&gt;elapsed time minus the optimal time across the <br>

<dd>&gt;strike's midpoint. Doing so would mean at idle, <br>

<dd>&gt;the pressure at idle becomes most effective <br>

<dd>&gt;5.6ms before half-way and for another 5.6ms <br>

<dd>&gt;after midpoint. Interesting that the idle <br>

<dd>&gt;pressure application time is about the same as <br>

<dd>&gt;the maximum pressure application time, or is that circular
logic?<br>

<dd>&gt;<br>

<dd>&gt;Yes everything above is ripe with assumptions <br>

<dd>&gt;and perhaps even apocryphal and resplendent with <br>

<dd>&gt;errors, but it is only 7am after all.<br>

<dd>&gt;<br>

<dd>&gt;Anyone with extra house-bound time on their <br>

<dd>&gt;hands, please check my maths and share your <br>

<dd>&gt;corrections, including the logic of the whole <br>

<dd>&gt;experiment...or perhaps even why geeks think these ways!<br>

<dd>&gt;<br>

<dd>&gt;Rick
</dl></blockquote></body>
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[email-masked]

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From: Richard Lindsay via Mgs <[email-masked]>
Date: Sat, 28 Mar 2020 18:30:31 -0500

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Excellent. Thank you.

On Sat, Mar 28, 2020, 6:15 PM Barney Gaylord <[email-masked]> wrote:

> I like the chart.  Pressure peaks at 12d ATDC, and by 90d ATDC it is 90%
> gone.
>
> (after TDC), you get another curve representing progression of torque.  N=
ot
> a lot of torque yet at 10d, pretty good torque by 20d, peaking around 30d
> ATDC.  Still a fair amount of torque at 60d, but by 90d (half stroke) the
> torque is nearly gone along with the pressure.
>
> (not on the chart).  Since both pressure and torque are nearly exhausted =
by
> 90d ATDC, there is very little energy (work) left to be harvested after m=
id
> stroke.  That's why we like to open the exhaust valve early, to let what
> remains of the mostly useless fumes out of the cylinder, encouraging best
> intake of fresh air and fuel mix half a turn later.
>
> Barney
>
>
> On Sat, Mar 28, 2020, 2:07 PM Barney Gaylord <[email-masked]> wrote=
:
> question.=C3=82  I thought you were asking how much
> pressure.=C3=82  That is, how much time for the flame
> cylinder", which is not true.=C3=82  Most of the work
> be completed at or slightly after TDC.=C3=82  But since
> like maybe 10 to 20d ATDC.=C3=82  That little delay can
> round numbers.=C3=82  And 900 rpm idle speed will be
> exactly 1/4 of road speed.=C3=82  If you make spark at
> TDC.=C3=82  10d ATDC is at 2-ms, and 30d ATDC would be
> 3-ms (after spark).=C3=82  So the flame front
> 3-ms range.=C3=82  I suppose this is the answer to your
> second, which is fairly close to 60-mph.=C3=82  And you
> >Idle timing advance: 4=C3=82=C2=B0 BTDC
> >Maximum timing advance: 32=C3=82=C2=B0 BTDC
> >
> >FIND:
> >Time from spark to MEP
> >
.
> >
> >single revolution is 360=C3=82=C2=B0 so half-way down the
> >power stroke is 90=C3=82=C2=B0. Add the idle spark timing
> >of 4=C3=82=C2=B0 BTDC (Before Top Dead Center) and we get
> >94=C3=82=C2=B0 of crank rotation from spark to MEP at
> >second). Maximum advance is 32=C3=82=C2=B0 BTDC so 90=C3=82=C2=B0 +
> >32=C3=82=C2=B0 =3D 122=C3=82=C2=B0, spark to MEP or 122=C3=82=C2=B0/360=
=C3=82=C2=B0 =3D 0.34 of a revolution
> >experiment...or perhaps even why geeks think these ways!

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<div dir=3D"auto">Excellent. Thank you.</div><br><div class=3D"gmail_quote"=
><div dir=3D"ltr" class=3D"gmail_attr">On Sat, Mar 28, 2020, 6:15 PM Barney=
 Gaylord &lt;<a href=3D"mailto:[email-masked]">[email-masked]</=
a>&gt; wrote:<br></div><blockquote class=3D"gmail_quote" style=3D"margin:0 =
0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">
<div>
I like the chart.=C2=A0 Pressure peaks at 12d ATDC, and by 90d ATDC it is
90% gone.<br><br>
If you multiply sine of the angle by the pressure all alog the curve
(after TDC), you get another curve representing progression of
torque.=C2=A0 Not a lot of torque yet at 10d, pretty good torque by 20d,
peaking around 30d ATDC.=C2=A0 Still a fair amount of torque at 60d, but
by 90d (half stroke) the torque is nearly gone along with the
pressure.<br><br>
Work being done would be represented by the area under the torque curve
(not on the chart).=C2=A0 Since both pressure and torque are nearly
exhausted by 90d ATDC, there is very little energy (work) left to be
harvested after mid stroke.=C2=A0 That&#39;s why we like to open the exhaus=
t
valve early, to let what remains of the mostly useless fumes out of the
cylinder, encouraging best intake of fresh air and fuel mix half a turn
later.<br><br>
Barney<br><br>
<br>
At 03:32 PM 3/28/2020, Richard Lindsay wrote:<br>
<blockquote type=3D"cite">...from Campbell&#39;s book.<br>
....<br>
Attachment: 20200328_113435.jpg</blockquote><br><br>
<blockquote type=3D"cite">On Sat, Mar 28, 2020, 2:07 PM
Barney Gaylord
&lt;<a href=3D"mailto:[email-masked]" target=3D"_blank" rel=3D"norefe=
rrer">[email-masked]</a>&gt;
wrote:<br>

<dl>
<dd>Rick, ---- Okay, time to spare, so I&#39;ll bite.<br><br>

<dd>question.=C3=82=C2=A0 I thought you were asking how much <br>

<dd>pressure.=C3=82=C2=A0 That is, how much time for the flame <br>

<dd>cylinder&quot;, which is not true.=C3=82=C2=A0 Most of the work <br>

<dd>be completed at or slightly after TDC.=C3=82=C2=A0 But since <br>

<dd>like maybe 10 to 20d ATDC.=C3=82=C2=A0 That little delay can <br>

<dd>round numbers.=C3=82=C2=A0 And 900 rpm idle speed will be <br>

<dd>exactly 1/4 of road speed.=C3=82=C2=A0 If you make spark at <br>

<dd>TDC.=C3=82=C2=A0 10d ATDC is at 2-ms, and 30d ATDC would be <br>

<dd>3-ms (after spark).=C3=82=C2=A0 So the flame front <br>

<dd>3-ms range.=C3=82=C2=A0 I suppose this is the answer to your <br>

<dd>second, which is fairly close to 60-mph.=C3=82=C2=A0 And you <br>

<dd>Barney</dd></dd></dd></dd></dd></dd></dd></dd></dd></dd></dd></dd></dd>=
</dd></dd></dd></dd></dd></dd></dd></dd></dd></dd></dd></dd></dd></dd></dd>=
</dd></dd></dd></dd></dd></dd></dd></dd></dd></dd></dd></dl></blockquote>
<br><br>
<blockquote type=3D"cite">
<dl>
<dd>At 08:54 AM 3/28/2020, you wrote:<br>

<dd>&gt;<br>

<dd>&gt;<br>

<dd>&gt;cylinder doesn&#39;t burn instantly, despite our <br>

<dd>&gt;&#39;spark&#39;, the flame front to cross the combustion <br>

<dd>&gt;<br>

<dd>&gt;&#39;known&#39; and the property to &#39;find&#39;. So in this case=
,<br>

<dd>&gt;<br>

<dd>&gt;KNOWN:<br>

<dd>&gt;Idle timing advance: 4=C3=82=C2=B0 BTDC<br>

<dd>&gt;Maximum timing advance: 32=C3=82=C2=B0 BTDC<br>

<dd>&gt;<br>

<dd>&gt;FIND:<br>

<dd>&gt;<br>

<dd>&gt;<br>

<dd>&gt;may not be correct at idle. But let&#39;s see.<br>

<dd>&gt;<br>

<dd>&gt;<br>

<dd>&gt;<br>

<dd>&gt;We don&#39;t know how fast the flame front travels <br>

<dd>&gt;work isn&#39;t an instantaneous parameter so it must <br>

<dd>&gt;that point. Lots of unknowns and theory doesn&#39;t <br>

<dd>&gt;<br>

<dd>&gt;single revolution is 360=C3=82=C2=B0 so half-way down the <br>

<dd>&gt;power stroke is 90=C3=82=C2=B0. Add the idle spark timing <br>

<dd>&gt;of 4=C3=82=C2=B0 BTDC (Before Top Dead Center) and we get <br>

<dd>&gt;94=C3=82=C2=B0 of crank rotation from spark to MEP at <br>

<dd>&gt;half-way down. That&#39;s 94/360 or about 0.26 of an <br>

<dd>&gt;<br>

<dd>&gt;second). Maximum advance is 32=C3=82=C2=B0 BTDC so 90=C3=82=C2=B0 +=
 <br>

<dd>&gt;32=C3=82=C2=B0 =3D 122=C3=82=C2=B0, spark to MEP or 122=C3=82=C2=B0=
/360=C3=82=C2=B0 =3D 0.34 of a
revolution<br>

<dd>&gt;<br>

<dd>&gt;strike&#39;s midpoint. Doing so would mean at idle, <br>

<dd>&gt;<br>

<dd>&gt;<br>

<dd>&gt;<br>

<dd>&gt;Rick
</dd></dd></dd></dd></dd></dd></dd></dd></dd></dd></dd></dd></dd></dd></dd>=
</dd></dd></dd></dd></dd></dd></dd></dd></dd></dd></dd></dd></dd></dd></dd>=
</dd></dd></dd></dd></dd></dd></dd></dd></dd></dd></dd></dd></dd></dd></dd>=
</dd></dd></dd></dd></dd></dd></dd></dd></dd></dd></dd></dd></dd></dd></dd>=
</dd></dd></dd></dd></dd></dd></dd></dd></dd></dd></dd></dd></dd></dd></dd>=
</dd></dd></dd></dd></dd></dd></dd></dd></dd></dd></dd></dd></dd></dd></dd>=
</dd></dd></dd></dd></dd></dd></dd></dd></dd></dd></dd></dd></dd></dd></dl>=
</blockquote></div>

</blockquote></div>

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From: Barrie Robinson via Mgs <[email-masked]>
Date: Sun, 29 Mar 2020 11:21:34 -0400

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Hello folks,

I am really surprised that no one has blasted me for sending out a 
message that was all (deliberately) gobbledygook.     I had at least 
expected an enquiry as to who Charles Breindigger was.

Cheers
Barrie

On 3/28/2020 7:30 PM, Richard Lindsay via Mgs wrote:
> Excellent. Thank you.
>
> On Sat, Mar 28, 2020, 6:15 PM Barney Gaylord <[email-masked] 
>
>     I like the chart.  Pressure peaks at 12d ATDC, and by 90d ATDC it
>     is 90% gone.
>
>     If you multiply sine of the angle by the pressure all alog the
>     curve (after TDC), you get another curve representing progression
>     of torque.  Not a lot of torque yet at 10d, pretty good torque by
>     20d, peaking around 30d ATDC.  Still a fair amount of torque at
>     60d, but by 90d (half stroke) the torque is nearly gone along with
>     the pressure.
>
>     Work being done would be represented by the area under the torque
>     curve (not on the chart).  Since both pressure and torque are
>     nearly exhausted by 90d ATDC, there is very little energy (work)
>     left to be harvested after mid stroke. That's why we like to open
>     the exhaust valve early, to let what remains of the mostly useless
>     fumes out of the cylinder, encouraging best intake of fresh air
>     and fuel mix half a turn later.
>
>     Barney
>
>
>>     On Sat, Mar 28, 2020, 2:07 PM Barney Gaylord
>>
>>         question.Â  I thought you were asking how much
>>         pressure.Â  That is, how much time for the flame
>>         cylinder", which is not true.Â  Most of the work
>>         be completed at or slightly after TDC.Â  But since
>>         like maybe 10 to 20d ATDC.Â  That little delay can
>>         round numbers.Â  And 900 rpm idle speed will be
>>         exactly 1/4 of road speed.Â  If you make spark at
>>         TDC.Â  10d ATDC is at 2-ms, and 30d ATDC would be
>>         3-ms (after spark).Â  So the flame front
>>         3-ms range.Â  I suppose this is the answer to your
>>         second, which is fairly close to 60-mph.Â  And you
>>         >Idle timing advance: 4Â° BTDC
>>         >Maximum timing advance: 32Â° BTDC
>>         >
>>         >FIND:
>>         >Time from spark to MEP
>>         >
>>         >single revolution is 360Â° so half-way down the
>>         >power stroke is 90Â°. Add the idle spark timing
>>         >of 4Â° BTDC (Before Top Dead Center) and we get
>>         >94Â° of crank rotation from spark to MEP at
>>         >second). Maximum advance is 32Â° BTDC so 90Â° +
>>         >32Â° = 122Â°, spark to MEP or 122Â°/360Â° = 0.34 of a revolution
>>         >

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    <font size="-1"><font face="Arial">Hello folks,<br>
      </font></font><br>
    <font size="-1"><font face="Arial">I am really surprised that no one
        has blasted me for sending out a message that was all
        (deliberately) gobbledygook.     I had at least expected an
        enquiry as to who </font></font><font size="-1"><font
        face="Arial"><font size="-1"><font face="Arial">Charles
            Breindigger was.<br>
            <br>
            Cheers<br>
            Barrie<br>
            <br>
            <br>
          </font></font> </font></font>
    <div class="moz-cite-prefix">On 3/28/2020 7:30 PM, Richard Lindsay
      via Mgs wrote:<br>
    </div>
    <blockquote type="cite"
cite="mid:[email-masked]">
      <meta http-equiv="content-type" content="text/html; charset=UTF-8">
      <div dir="auto">Excellent. Thank you.</div>
      <br>
      <div class="gmail_quote">
        <div dir="ltr" class="gmail_attr">On Sat, Mar 28, 2020, 6:15 PM
          Barney Gaylord &lt;<a href="mailto:[email-masked]"
            moz-do-not-send="true">[email-masked]</a>&gt; wrote:<br>
        </div>
        <blockquote class="gmail_quote" style="margin:0 0 0
          .8ex;border-left:1px #ccc solid;padding-left:1ex">
          <div> I like the chart.  Pressure peaks at 12d ATDC, and by
            90d ATDC it is 90% gone.<br>
            <br>
            If you multiply sine of the angle by the pressure all alog
            the curve (after TDC), you get another curve representing
            progression of torque.  Not a lot of torque yet at 10d,
            pretty good torque by 20d, peaking around 30d ATDC.  Still a
            fair amount of torque at 60d, but by 90d (half stroke) the
            torque is nearly gone along with the pressure.<br>
            <br>
            Work being done would be represented by the area under the
            torque curve (not on the chart).  Since both pressure and
            torque are nearly exhausted by 90d ATDC, there is very
            little energy (work) left to be harvested after mid stroke. 
            That's why we like to open the exhaust valve early, to let
            what remains of the mostly useless fumes out of the
            cylinder, encouraging best intake of fresh air and fuel mix
            half a turn later.<br>
            <br>
            Barney<br>
            <br>
            <br>
            At 03:32 PM 3/28/2020, Richard Lindsay wrote:<br>
            <blockquote type="cite">...from Campbell's book.<br>
              ....<br>
              Attachment: 20200328_113435.jpg</blockquote>
            <br>
            <br>
            <blockquote type="cite">On Sat, Mar 28, 2020, 2:07 PM Barney
              Gaylord &lt;<a href="mailto:[email-masked]"
                target="_blank" rel="noreferrer" moz-do-not-send="true">[email-masked]</a>&gt;
              wrote:<br>
              <dl>
                <dd>Rick, ---- Okay, time to spare, so I'll bite.<br>
                  <br>
                </dd>
                <dd>You do a good job of calculating time from spark <br>
                </dd>
                <dd>event to half stroke (about 6-ms at road speed), <br>
                </dd>
                <dd>but I think you were asking a different <br>
                </dd>
                <dd>question.Â  I thought you were asking how much <br>
                </dd>
                <dd>time to complete combustion to get to maximum <br>
                </dd>
                <dd>pressure.Â  That is, how much time for the flame <br>
                </dd>
                <dd>front to propagate all the way across the combustion
                  chamber?<br>
                  <br>
                </dd>
                <dd>And you also said. "we do know that maximum work <br>
                </dd>
                <dd>occurs when the piston is half way down the <br>
                </dd>
                <dd>cylinder", which is not true.Â  Most of the work <br>
                </dd>
                <dd>has already been done before the piston gets half <br>
                </dd>
                <dd>way down, and maximum torque on the crankshaft <br>
                </dd>
                <dd>happens significantly higher in the stroke before <br>
                </dd>
                <dd>pressure is lost to expansion.<br>
                  <br>
                </dd>
                <dd>For best power and efficiency, combustion should <br>
                </dd>
                <dd>be completed at or slightly after TDC.Â  But since <br>
                </dd>
                <dd>there is very little motion of the piston <br>
                </dd>
                <dd>immediately after TDC, It works just about as <br>
                </dd>
                <dd>well if max pressure comes just a little later, <br>
                </dd>
                <dd>like maybe 10 to 20d ATDC.Â  That little delay can <br>
                </dd>
                <dd>allow use of higher compression ratio for better
                  power and efficiency.<br>
                  <br>
                </dd>
                <dd>I like to use 3600 rpm for road speed, because it <br>
                </dd>
                <dd>divides evenly into 360 degrees rotation for nice <br>
                </dd>
                <dd>round numbers.Â  And 900 rpm idle speed will be <br>
                </dd>
                <dd>exactly 1/4 of road speed.Â  If you make spark at <br>
                </dd>
                <dd>32d BTDC at road speed, it takes 1.5-ms to reach <br>
                </dd>
                <dd>TDC.Â  10d ATDC is at 2-ms, and 30d ATDC would be <br>
                </dd>
                <dd>3-ms (after spark).Â  So the flame front <br>
                </dd>
                <dd>propagation to complete combustion is in the 2 to <br>
                </dd>
                <dd>3-ms range.Â  I suppose this is the answer to your <br>
                </dd>
                <dd>question, "how much time" for the flame front to
                  cross the combustion chamber.<br>
                  <br>
                </dd>
                <dd>Distance from the spark plug to far side of the <br>
                </dd>
                <dd>combustion chamber is about 2-1/2 inches, which <br>
                </dd>
                <dd>it does in about 2-1/2 ms, so flame front speed <br>
                </dd>
                <dd>is about 1 inch per ms, or 1000 inches per <br>
                </dd>
                <dd>second, which is fairly close to 60-mph.Â  And you <br>
                </dd>
                <dd>night notice that I did not use "MEP" in that entire
                  discussion.<br>
                  <br>
                </dd>
                <dd>Barney</dd>
              </dl>
            </blockquote>
            <br>
            <br>
            <blockquote type="cite">
              <dl>
                <dd>At 08:54 AM 3/28/2020, you wrote:<br>
                </dd>
                <dd>&gt;Hello friends,<br>
                </dd>
                <dd>&gt;<br>
                </dd>
                <dd>&gt;When one is a geek, one thinks of geeky things.
                  <br>
                </dd>
                <dd>&gt;I am a geek and this house-bound morning I woke
                  <br>
                </dd>
                <dd>&gt;up thinking about ignition timing. Here are the
                  details.<br>
                </dd>
                <dd>&gt;<br>
                </dd>
                <dd>&gt;We know that the charge (fuel plus air) in a <br>
                </dd>
                <dd>&gt;cylinder doesn't burn instantly, despite our <br>
                </dd>
                <dd>&gt;perception to the contrary. Rather, it takes a <br>
                </dd>
                <dd>&gt;finite length of time from the occurance of the
                  <br>
                </dd>
                <dd>&gt;'spark', the flame front to cross the combustion
                  <br>
                </dd>
                <dd>&gt;chamber, and to raise the MEP (Mean Effective <br>
                </dd>
                <dd>&gt;Pressure) to a maximum - the point where it does
                  <br>
                </dd>
                <dd>&gt;the most work. But how much time?<br>
                </dd>
                <dd>&gt;<br>
                </dd>
                <dd>&gt;Physics problems always start by listing the <br>
                </dd>
                <dd>&gt;'known' and the property to 'find'. So in this
                  case,<br>
                </dd>
                <dd>&gt;<br>
                </dd>
                <dd>&gt;KNOWN:<br>
                </dd>
                <dd>&gt;Idle speed: 900rpm<br>
                </dd>
                <dd>&gt;Idle timing advance: 4Â° BTDC<br>
                </dd>
                <dd>&gt;Speed at maximum advance: 3500rpm<br>
                </dd>
                <dd>&gt;Maximum timing advance: 32Â° BTDC<br>
                </dd>
                <dd>&gt;<br>
                </dd>
                <dd>&gt;FIND:<br>
                </dd>
                <dd>&gt;Time from spark to MEP<br>
                </dd>
                <dd>&gt;<br>
                </dd>
                <dd>&gt;The first thing one might know is that the goal
                  <br>
                </dd>
                <dd>&gt;at idle is not to produce maximum power. In <br>
                </dd>
                <dd>&gt;fact, at idle 100% of the available power is <br>
                </dd>
                <dd>&gt;used to overcome the friction and other forces <br>
                </dd>
                <dd>&gt;that exist at idle speed. Stated another way: <br>
                </dd>
                <dd>&gt;Idle speed is the fastest the engine can achieve
                  <br>
                </dd>
                <dd>&gt;given the available charge. That fact is evident
                  <br>
                </dd>
                <dd>&gt;(with carbureted engines) when one notices that
                  <br>
                </dd>
                <dd>&gt;engine speed gradually increases, even for a <br>
                </dd>
                <dd>&gt;fixed throttle setting, as the engine warms and
                  <br>
                </dd>
                <dd>&gt;friction forces decrease. But back to the
                  problem.<br>
                </dd>
                <dd>&gt;<br>
                </dd>
                <dd>&gt;Because the goal at idle is smooth running and <br>
                </dd>
                <dd>&gt;progression off of idle (e.g. speeding up), not
                  <br>
                </dd>
                <dd>&gt;maximum power, the calculated wavefront speed <br>
                </dd>
                <dd>&gt;may not be correct at idle. But let's see.<br>
                </dd>
                <dd>&gt;<br>
                </dd>
                <dd>&gt;At idle speed, 900rpm in this MG TD example, the
                  <br>
                </dd>
                <dd>&gt;XPAG engine is turning 900rpm or 900rpm / 60mps
                  <br>
                </dd>
                <dd>&gt;= 15rps (revolutions per second).<br>
                </dd>
                <dd>&gt;<br>
                </dd>
                <dd>&gt;Distributor speed is 1/2 engine speed so at idle
                  <br>
                </dd>
                <dd>&gt;the distributor is turning only 7.5 revolutions
                  <br>
                </dd>
                <dd>&gt;per second. But timing numbers are specified in
                  <br>
                </dd>
                <dd>&gt;degrees of crank rotations so we will stick with
                  15rps.<br>
                </dd>
                <dd>&gt;<br>
                </dd>
                <dd>&gt;We don't know how fast the flame front travels <br>
                </dd>
                <dd>&gt;across the combustion chamber but we do know <br>
                </dd>
                <dd>&gt;that maximum work occurs when the piston is half
                  <br>
                </dd>
                <dd>&gt;way down the cylinder. And we also know that <br>
                </dd>
                <dd>&gt;work isn't an instantaneous parameter so it must
                  <br>
                </dd>
                <dd>&gt;begin before the half way point and last past <br>
                </dd>
                <dd>&gt;that point. Lots of unknowns and theory doesn't
                  <br>
                </dd>
                <dd>&gt;always work in practice. But if we use the <br>
                </dd>
                <dd>&gt;average piston position at half-way down the <br>
                </dd>
                <dd>&gt;bore, where most work is most effective, and the
                  <br>
                </dd>
                <dd>&gt;MEP (Mean Effective Pressure), since Mean is
                  average, calculations begin.<br>
                </dd>
                <dd>&gt;<br>
                </dd>
                <dd>&gt;single revolution is 360Â° so half-way down the
                  <br>
                </dd>
                <dd>&gt;power stroke is 90Â°. Add the idle spark timing
                  <br>
                </dd>
                <dd>&gt;of 4Â° BTDC (Before Top Dead Center) and we get
                  <br>
                </dd>
                <dd>&gt;94Â° of crank rotation from spark to MEP at <br>
                </dd>
                <dd>&gt;half-way down. That's 94/360 or about 0.26 of an
                  <br>
                </dd>
                <dd>&gt;engine revolution. And the engine is turning 15
                  <br>
                </dd>
                <dd>&gt;revolutions per second or 67ms (milliseconds) <br>
                </dd>
                <dd>&gt;per revolution. So 67 x 0.26 = 17ms from spark <br>
                </dd>
                <dd>&gt;to MEP at half-way down the power stroke, at
                  idle.<br>
                </dd>
                <dd>&gt;<br>
                </dd>
                <dd>&gt;If we repeat the calculations for operating <br>
                </dd>
                <dd>&gt;engine speed and at maximum advance, we get <br>
                </dd>
                <dd>&gt;3500rpm / 60mps = 58rps (revolutions per <br>
                </dd>
                <dd>&gt;second). Maximum advance is 32Â° BTDC so 90Â° +
                  <br>
                </dd>
                <dd>&gt;32Â° = 122Â°, spark to MEP or 122Â°/360Â° = 0.34
                  of a revolution<br>
                </dd>
                <dd>&gt;<br>
                </dd>
                <dd>&gt;58rps is 17ms/r so 17ms/r x 0.34r = 5.78ms from
                  <br>
                </dd>
                <dd>&gt;spark to MEP at half-way down the power stroke.
                  <br>
                </dd>
                <dd>&gt;This is a more representative number than the <br>
                </dd>
                <dd>&gt;17ms at idle. One might even divide the idle <br>
                </dd>
                <dd>&gt;elapsed time minus the optimal time across the <br>
                </dd>
                <dd>&gt;strike's midpoint. Doing so would mean at idle,
                  <br>
                </dd>
                <dd>&gt;the pressure at idle becomes most effective <br>
                </dd>
                <dd>&gt;5.6ms before half-way and for another 5.6ms <br>
                </dd>
                <dd>&gt;after midpoint. Interesting that the idle <br>
                </dd>
                <dd>&gt;pressure application time is about the same as <br>
                </dd>
                <dd>&gt;the maximum pressure application time, or is
                  that circular logic?<br>
                </dd>
                <dd>&gt;<br>
                </dd>
                <dd>&gt;Yes everything above is ripe with assumptions <br>
                </dd>
                <dd>&gt;and perhaps even apocryphal and resplendent with
                  <br>
                </dd>
                <dd>&gt;errors, but it is only 7am after all.<br>
                </dd>
                <dd>&gt;<br>
                </dd>
                <dd>&gt;Anyone with extra house-bound time on their <br>
                </dd>
                <dd>&gt;hands, please check my maths and share your <br>
                </dd>
                <dd>&gt;corrections, including the logic of the whole <br>
                </dd>
                <dd>&gt;experiment...or perhaps even why geeks think
                  these ways!<br>
                </dd>
                <dd>&gt;<br>
                </dd>
                <dd>&gt;Rick </dd>
              </dl>
            </blockquote>
          </div>
        </blockquote>
      </div>
      <br>
      <fieldset class="mimeAttachmentHeader"></fieldset>
      <pre class="moz-quote-pre" wrap="">_______________________________________________

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[email-masked]

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---

From: Dan DiBiase via Mgs <[email-masked]>
Date: Sun, 29 Mar 2020 11:38:18 -0400

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It's all Greek to me anyway, so I just put it in the same category as the
others writings.... ;-)

Dan D
'76B '65B
Central NJ USA
http://dandibiase.cbintouch.com/

On Sun, Mar 29, 2020 at 11:21 AM Barrie Robinson via Mgs <[email-masked]=
t>
wrote:

> Hello folks,
>
> I am really surprised that no one has blasted me for sending out a messag=
e
> that was all (deliberately) gobbledygook.     I had at least expected an
> enquiry as to who Charles Breindigger was.
>
> Cheers
> Barrie
>
>
> On Sat, Mar 28, 2020, 6:15 PM Barney Gaylord <[email-masked]> wrote=
:
>> (after TDC), you get another curve representing progression of torque.  =
Not
>> a lot of torque yet at 10d, pretty good torque by 20d, peaking around 30=
d
>> ATDC.  Still a fair amount of torque at 60d, but by 90d (half stroke) th=
e
>> (not on the chart).  Since both pressure and torque are nearly exhausted=
 by
>> 90d ATDC, there is very little energy (work) left to be harvested after =
mid
>> remains of the mostly useless fumes out of the cylinder, encouraging bes=
t
>> >MEP (Mean Effective Pressure), since Mean is average, calculations begi=
n.
> Unsubscribe: http://autox.team.net/mailman/options/mgs/[email-masked]
>
>
> http://autox.team.net/mailman/options/mgs/[email-masked]
>

--000000000000d9475e05a200212d
Content-Transfer-Encoding: quoted-printable

<div dir=3D"ltr">It&#39;s all Greek to me anyway, so I just put it in the s=
ame category as the others writings.... ;-)<div><br></div><div>Dan D</div><=
div>&#39;76B &#39;65B</div><div>Central NJ USA<br clear=3D"all"><div><div d=
ir=3D"ltr" class=3D"gmail_signature" data-smartmail=3D"gmail_signature"><di=
v dir=3D"ltr"><a href=3D"http://dandibiase.cbintouch.com/" target=3D"_blank=
">http://dandibiase.cbintouch.com/</a></div></div></div><br></div></div><br=
><div class=3D"gmail_quote"><div dir=3D"ltr" class=3D"gmail_attr">On Sun, M=
ar 29, 2020 at 11:21 AM Barrie Robinson via Mgs &lt;<a href=3D"mailto:mgs@a=
utox.team.net">[email-masked]</a>&gt; wrote:<br></div><blockquote class=
=3D"gmail_quote" style=3D"margin:0px 0px 0px 0.8ex;border-left:1px solid rg=
b(204,204,204);padding-left:1ex">
 =20
   =20
 =20
  <div>
    <font size=3D"-1"><font face=3D"Arial">Hello folks,<br>
      </font></font><br>
    <font size=3D"-1"><font face=3D"Arial">I am really surprised that no on=
e
        has blasted me for sending out a message that was all
        (deliberately) gobbledygook. =C2=A0 =C2=A0 I had at least expected =
an
        enquiry as to who </font></font><font size=3D"-1"><font face=3D"Ari=
al"><font size=3D"-1"><font face=3D"Arial">Charles
            Breindigger was.<br>
            <br>
            Cheers<br>
            Barrie<br>
            <br>
            <br>
          </font></font> </font></font>
    <div>On 3/28/2020 7:30 PM, Richard Lindsay
      via Mgs wrote:<br>
    </div>
    <blockquote type=3D"cite">
     =20
      <div dir=3D"auto">Excellent. Thank you.</div>
      <br>
      <div class=3D"gmail_quote">
        <div dir=3D"ltr" class=3D"gmail_attr">On Sat, Mar 28, 2020, 6:15 PM
          Barney Gaylord &lt;<a href=3D"mailto:[email-masked]" target=
=3D"_blank">[email-masked]</a>&gt; wrote:<br>
        </div>
        <blockquote class=3D"gmail_quote" style=3D"margin:0px 0px 0px 0.8ex=
;border-left:1px solid rgb(204,204,204);padding-left:1ex">
          <div> I like the chart.=C2=A0 Pressure peaks at 12d ATDC, and by
            90d ATDC it is 90% gone.<br>
            <br>
            If you multiply sine of the angle by the pressure all alog
            the curve (after TDC), you get another curve representing
            progression of torque.=C2=A0 Not a lot of torque yet at 10d,
            pretty good torque by 20d, peaking around 30d ATDC.=C2=A0 Still=
 a
            fair amount of torque at 60d, but by 90d (half stroke) the
            torque is nearly gone along with the pressure.<br>
            <br>
            Work being done would be represented by the area under the
            torque curve (not on the chart).=C2=A0 Since both pressure and
            torque are nearly exhausted by 90d ATDC, there is very
            little energy (work) left to be harvested after mid stroke.=C2=
=A0
            That&#39;s why we like to open the exhaust valve early, to let
            what remains of the mostly useless fumes out of the
            cylinder, encouraging best intake of fresh air and fuel mix
            half a turn later.<br>
            <br>
            Barney<br>
            <br>
            <br>
            At 03:32 PM 3/28/2020, Richard Lindsay wrote:<br>
            <blockquote type=3D"cite">...from Campbell&#39;s book.<br>
              ....<br>
              Attachment: 20200328_113435.jpg</blockquote>
            <br>
            <br>
            <blockquote type=3D"cite">On Sat, Mar 28, 2020, 2:07 PM Barney
              Gaylord &lt;<a href=3D"mailto:[email-masked]" rel=3D"no=
referrer" target=3D"_blank">[email-masked]</a>&gt;
              wrote:<br>
              <dl>
                <dd>Rick, ---- Okay, time to spare, so I&#39;ll bite.<br>
                  <br>
                </dd>
                <dd>You do a good job of calculating time from spark <br>
                </dd>
                <dd>event to half stroke (about 6-ms at road speed), <br>
                </dd>
                <dd>but I think you were asking a different <br>
                </dd>
                <dd>question.=C3=82=C2=A0 I thought you were asking how muc=
h <br>
                </dd>
                <dd>time to complete combustion to get to maximum <br>
                </dd>
                <dd>pressure.=C3=82=C2=A0 That is, how much time for the fl=
ame <br>
                </dd>
                <dd>front to propagate all the way across the combustion
                  chamber?<br>
                  <br>
                </dd>
                <dd>And you also said. &quot;we do know that maximum work <=
br>
                </dd>
                <dd>occurs when the piston is half way down the <br>
                </dd>
                <dd>cylinder&quot;, which is not true.=C3=82=C2=A0 Most of =
the work <br>
                </dd>
                <dd>has already been done before the piston gets half <br>
                </dd>
                <dd>way down, and maximum torque on the crankshaft <br>
                </dd>
                <dd>happens significantly higher in the stroke before <br>
                </dd>
                <dd>pressure is lost to expansion.<br>
                  <br>
                </dd>
                <dd>For best power and efficiency, combustion should <br>
                </dd>
                <dd>be completed at or slightly after TDC.=C3=82=C2=A0 But =
since <br>
                </dd>
                <dd>there is very little motion of the piston <br>
                </dd>
                <dd>immediately after TDC, It works just about as <br>
                </dd>
                <dd>well if max pressure comes just a little later, <br>
                </dd>
                <dd>like maybe 10 to 20d ATDC.=C3=82=C2=A0 That little dela=
y can <br>
                </dd>
                <dd>allow use of higher compression ratio for better
                  power and efficiency.<br>
                  <br>
                </dd>
                <dd>I like to use 3600 rpm for road speed, because it <br>
                </dd>
                <dd>divides evenly into 360 degrees rotation for nice <br>
                </dd>
                <dd>round numbers.=C3=82=C2=A0 And 900 rpm idle speed will =
be <br>
                </dd>
                <dd>exactly 1/4 of road speed.=C3=82=C2=A0 If you make spar=
k at <br>
                </dd>
                <dd>32d BTDC at road speed, it takes 1.5-ms to reach <br>
                </dd>
                <dd>TDC.=C3=82=C2=A0 10d ATDC is at 2-ms, and 30d ATDC woul=
d be <br>
                </dd>
                <dd>3-ms (after spark).=C3=82=C2=A0 So the flame front <br>
                </dd>
                <dd>propagation to complete combustion is in the 2 to <br>
                </dd>
                <dd>3-ms range.=C3=82=C2=A0 I suppose this is the answer to=
 your <br>
                </dd>
                <dd>question, &quot;how much time&quot; for the flame front=
 to
                  cross the combustion chamber.<br>
                  <br>
                </dd>
                <dd>Distance from the spark plug to far side of the <br>
                </dd>
                <dd>combustion chamber is about 2-1/2 inches, which <br>
                </dd>
                <dd>it does in about 2-1/2 ms, so flame front speed <br>
                </dd>
                <dd>is about 1 inch per ms, or 1000 inches per <br>
                </dd>
                <dd>second, which is fairly close to 60-mph.=C3=82=C2=A0 An=
d you <br>
                </dd>
                <dd>night notice that I did not use &quot;MEP&quot; in that=
 entire
                  discussion.<br>
                  <br>
                </dd>
                <dd>Barney</dd>
              </dl>
            </blockquote>
            <br>
            <br>
            <blockquote type=3D"cite">
              <dl>
                <dd>At 08:54 AM 3/28/2020, you wrote:<br>
                </dd>
                <dd>&gt;Hello friends,<br>
                </dd>
                <dd>&gt;<br>
                </dd>
                <dd>&gt;When one is a geek, one thinks of geeky things.
                  <br>
                </dd>
                <dd>&gt;I am a geek and this house-bound morning I woke
                  <br>
                </dd>
                <dd>&gt;up thinking about ignition timing. Here are the
                  details.<br>
                </dd>
                <dd>&gt;<br>
                </dd>
                <dd>&gt;We know that the charge (fuel plus air) in a <br>
                </dd>
                <dd>&gt;cylinder doesn&#39;t burn instantly, despite our <b=
r>
                </dd>
                <dd>&gt;perception to the contrary. Rather, it takes a <br>
                </dd>
                <dd>&gt;finite length of time from the occurance of the
                  <br>
                </dd>
                <dd>&gt;&#39;spark&#39;, the flame front to cross the combu=
stion
                  <br>
                </dd>
                <dd>&gt;chamber, and to raise the MEP (Mean Effective <br>
                </dd>
                <dd>&gt;Pressure) to a maximum - the point where it does
                  <br>
                </dd>
                <dd>&gt;the most work. But how much time?<br>
                </dd>
                <dd>&gt;<br>
                </dd>
                <dd>&gt;Physics problems always start by listing the <br>
                </dd>
                <dd>&gt;&#39;known&#39; and the property to &#39;find&#39;.=
 So in this
                  case,<br>
                </dd>
                <dd>&gt;<br>
                </dd>
                <dd>&gt;KNOWN:<br>
                </dd>
                <dd>&gt;Idle speed: 900rpm<br>
                </dd>
                <dd>&gt;Idle timing advance: 4=C3=82=C2=B0 BTDC<br>
                </dd>
                <dd>&gt;Speed at maximum advance: 3500rpm<br>
                </dd>
                <dd>&gt;Maximum timing advance: 32=C3=82=C2=B0 BTDC<br>
                </dd>
                <dd>&gt;<br>
                </dd>
                <dd>&gt;FIND:<br>
                </dd>
                <dd>&gt;Time from spark to MEP<br>
                </dd>
                <dd>&gt;<br>
                </dd>
                <dd>&gt;The first thing one might know is that the goal
                  <br>
                </dd>
                <dd>&gt;at idle is not to produce maximum power. In <br>
                </dd>
                <dd>&gt;fact, at idle 100% of the available power is <br>
                </dd>
                <dd>&gt;used to overcome the friction and other forces <br>
                </dd>
                <dd>&gt;that exist at idle speed. Stated another way: <br>
                </dd>
                <dd>&gt;Idle speed is the fastest the engine can achieve
                  <br>
                </dd>
                <dd>&gt;given the available charge. That fact is evident
                  <br>
                </dd>
                <dd>&gt;(with carbureted engines) when one notices that
                  <br>
                </dd>
                <dd>&gt;engine speed gradually increases, even for a <br>
                </dd>
                <dd>&gt;fixed throttle setting, as the engine warms and
                  <br>
                </dd>
                <dd>&gt;friction forces decrease. But back to the
                  problem.<br>
                </dd>
                <dd>&gt;<br>
                </dd>
                <dd>&gt;Because the goal at idle is smooth running and <br>
                </dd>
                <dd>&gt;progression off of idle (e.g. speeding up), not
                  <br>
                </dd>
                <dd>&gt;maximum power, the calculated wavefront speed <br>
                </dd>
                <dd>&gt;may not be correct at idle. But let&#39;s see.<br>
                </dd>
                <dd>&gt;<br>
                </dd>
                <dd>&gt;At idle speed, 900rpm in this MG TD example, the
                  <br>
                </dd>
                <dd>&gt;XPAG engine is turning 900rpm or 900rpm / 60mps
                  <br>
                </dd>
                <dd>&gt;=3D 15rps (revolutions per second).<br>
                </dd>
                <dd>&gt;<br>
                </dd>
                <dd>&gt;Distributor speed is 1/2 engine speed so at idle
                  <br>
                </dd>
                <dd>&gt;the distributor is turning only 7.5 revolutions
                  <br>
                </dd>
                <dd>&gt;per second. But timing numbers are specified in
                  <br>
                </dd>
                <dd>&gt;degrees of crank rotations so we will stick with
                  15rps.<br>
                </dd>
                <dd>&gt;<br>
                </dd>
                <dd>&gt;We don&#39;t know how fast the flame front travels =
<br>
                </dd>
                <dd>&gt;across the combustion chamber but we do know <br>
                </dd>
                <dd>&gt;that maximum work occurs when the piston is half
                  <br>
                </dd>
                <dd>&gt;way down the cylinder. And we also know that <br>
                </dd>
                <dd>&gt;work isn&#39;t an instantaneous parameter so it mus=
t
                  <br>
                </dd>
                <dd>&gt;begin before the half way point and last past <br>
                </dd>
                <dd>&gt;that point. Lots of unknowns and theory doesn&#39;t
                  <br>
                </dd>
                <dd>&gt;always work in practice. But if we use the <br>
                </dd>
                <dd>&gt;average piston position at half-way down the <br>
                </dd>
                <dd>&gt;bore, where most work is most effective, and the
                  <br>
                </dd>
                <dd>&gt;MEP (Mean Effective Pressure), since Mean is
                  average, calculations begin.<br>
                </dd>
                <dd>&gt;<br>
                </dd>
                <dd>&gt;single revolution is 360=C3=82=C2=B0 so half-way do=
wn the
                  <br>
                </dd>
                <dd>&gt;power stroke is 90=C3=82=C2=B0. Add the idle spark =
timing
                  <br>
                </dd>
                <dd>&gt;of 4=C3=82=C2=B0 BTDC (Before Top Dead Center) and =
we get
                  <br>
                </dd>
                <dd>&gt;94=C3=82=C2=B0 of crank rotation from spark to MEP =
at <br>
                </dd>
                <dd>&gt;half-way down. That&#39;s 94/360 or about 0.26 of a=
n
                  <br>
                </dd>
                <dd>&gt;engine revolution. And the engine is turning 15
                  <br>
                </dd>
                <dd>&gt;revolutions per second or 67ms (milliseconds) <br>
                </dd>
                <dd>&gt;per revolution. So 67 x 0.26 =3D 17ms from spark <b=
r>
                </dd>
                <dd>&gt;to MEP at half-way down the power stroke, at
                  idle.<br>
                </dd>
                <dd>&gt;<br>
                </dd>
                <dd>&gt;If we repeat the calculations for operating <br>
                </dd>
                <dd>&gt;engine speed and at maximum advance, we get <br>
                </dd>
                <dd>&gt;3500rpm / 60mps =3D 58rps (revolutions per <br>
                </dd>
                <dd>&gt;second). Maximum advance is 32=C3=82=C2=B0 BTDC so =
90=C3=82=C2=B0 +
                  <br>
                </dd>
                <dd>&gt;32=C3=82=C2=B0 =3D 122=C3=82=C2=B0, spark to MEP or=
 122=C3=82=C2=B0/360=C3=82=C2=B0 =3D 0.34
                  of a revolution<br>
                </dd>
                <dd>&gt;<br>
                </dd>
                <dd>&gt;58rps is 17ms/r so 17ms/r x 0.34r =3D 5.78ms from
                  <br>
                </dd>
                <dd>&gt;spark to MEP at half-way down the power stroke.
                  <br>
                </dd>
                <dd>&gt;This is a more representative number than the <br>
                </dd>
                <dd>&gt;17ms at idle. One might even divide the idle <br>
                </dd>
                <dd>&gt;elapsed time minus the optimal time across the <br>
                </dd>
                <dd>&gt;strike&#39;s midpoint. Doing so would mean at idle,
                  <br>
                </dd>
                <dd>&gt;the pressure at idle becomes most effective <br>
                </dd>
                <dd>&gt;5.6ms before half-way and for another 5.6ms <br>
                </dd>
                <dd>&gt;after midpoint. Interesting that the idle <br>
                </dd>
                <dd>&gt;pressure application time is about the same as <br>
                </dd>
                <dd>&gt;the maximum pressure application time, or is
                  that circular logic?<br>
                </dd>
                <dd>&gt;<br>
                </dd>
                <dd>&gt;Yes everything above is ripe with assumptions <br>
                </dd>
                <dd>&gt;and perhaps even apocryphal and resplendent with
                  <br>
                </dd>
                <dd>&gt;errors, but it is only 7am after all.<br>
                </dd>
                <dd>&gt;<br>
                </dd>
                <dd>&gt;Anyone with extra house-bound time on their <br>
                </dd>
                <dd>&gt;hands, please check my maths and share your <br>
                </dd>
                <dd>&gt;corrections, including the logic of the whole <br>
                </dd>
                <dd>&gt;experiment...or perhaps even why geeks think
                  these ways!<br>
                </dd>
                <dd>&gt;<br>
                </dd>
                <dd>&gt;Rick </dd>
              </dl>
            </blockquote>
          </div>
        </blockquote>
      </div>
      <br>
      <fieldset></fieldset>
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---

From: Richard Lindsay via Mgs <[email-masked]>
Date: Sun, 29 Mar 2020 12:24:44 -0500

--===============3149900410903260634==

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I am Mr. Gobbledygook while you and PaulH are DOCTOR Gobbledygook winners.
Thank you!

Some of us drive 'em. Some race 'em. Some just study 'em. And silly guys
like me restore then and try to understand them. Honestly, I don't drive
much.  :-o

rick

On Sun, Mar 29, 2020 at 10:21 AM Barrie Robinson via Mgs <[email-masked]=
t>

e
Not
d
e
 by
mid
t
n.
> http://autox.team.net/mailman/options/mgs/[email-masked]
>

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<div dir=3D"ltr"><div dir=3D"ltr">I am Mr. Gobbledygook while you and PaulH=
 are DOCTOR Gobbledygook winners. Thank you!</div><div dir=3D"ltr"><br></di=
v><div>Some of us drive &#39;em. Some race &#39;em. Some just study &#39;em=
. And silly guys like me restore then and try to understand them. Honestly,=
 I don&#39;t drive much.=C2=A0 :-o</div><div><br></div><div>rick</div><br><=
div class=3D"gmail_quote"><div dir=3D"ltr" class=3D"gmail_attr">On Sun, Mar=
 29, 2020 at 10:21 AM Barrie Robinson via Mgs &lt;<a href=3D"mailto:mgs@aut=
ox.team.net">[email-masked]</a>&gt; wrote:<br></div><blockquote class=

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