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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 <barneymg@mgaguru.com> wrote=
:
> 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
> >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=
.
> >
> >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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<div dir=3D"ltr">Barney, I appreciate your words and critique. Same with Pa=
ulH'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 <<a href=3D"mailto:barneymg@mgaguru.com";>barneymg@mg=
aguru.com</a>> 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'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. "we do know that maximum work <br>
occurs when the piston is half way down the <br>
cylinder", 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, "how much time" 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 "MEP" in that entire discussion.<=
br>
<br>
Barney<br>
<br>
<br>
At 08:54 AM 3/28/2020, you wrote:<br>
>Hello friends,<br>
><br>
>When one is a geek, one thinks of geeky things. <br>
>I am a geek and this house-bound morning I woke <br>
>up thinking about ignition timing. Here are the details.<br>
><br>
>We know that the charge (fuel plus air) in a <br>
>cylinder doesn't burn instantly, despite our <br>
>perception to the contrary. Rather, it takes a <br>
>finite length of time from the occurance of the <br>
>'spark', the flame front to cross the combustion <br>
>chamber, and to raise the MEP (Mean Effective <br>
>Pressure) to a maximum - the point where it does <br>
>the most work. But how much time?<br>
><br>
>Physics problems always start by listing the <br>
>'known' and the property to 'find'. So in this case,<br=
>
><br>
>KNOWN:<br>
>Idle speed: 900rpm<br>
>Idle timing advance: 4=C2=B0 BTDC<br>
>Speed at maximum advance: 3500rpm<br>
>Maximum timing advance: 32=C2=B0 BTDC<br>
><br>
>FIND:<br>
>Time from spark to MEP<br>
><br>
>The first thing one might know is that the goal <br>
>at idle is not to produce maximum power. In <br>
>fact, at idle 100% of the available power is <br>
>used to overcome the friction and other forces <br>
>that exist at idle speed. Stated another way: <br>
>Idle speed is the fastest the engine can achieve <br>
>given the available charge. That fact is evident <br>
>(with carbureted engines) when one notices that <br>
>engine speed gradually increases, even for a <br>
>fixed throttle setting, as the engine warms and <br>
>friction forces decrease. But back to the problem.<br>
><br>
>Because the goal at idle is smooth running and <br>
>progression off of idle (e.g. speeding up), not <br>
>maximum power, the calculated wavefront speed <br>
>may not be correct at idle. But let's see.<br>
><br>
>At idle speed, 900rpm in this MG TD example, the <br>
>XPAG engine is turning 900rpm or 900rpm / 60mps <br>
>=3D 15rps (revolutions per second).<br>
><br>
>Distributor speed is 1/2 engine speed so at idle <br>
>the distributor is turning only 7.5 revolutions <br>
>per second. But timing numbers are specified in <br>
>degrees of crank rotations so we will stick with 15rps.<br>
><br>
>We don't know how fast the flame front travels <br>
>across the combustion chamber but we do know <br>
>that maximum work occurs when the piston is half <br>
>way down the cylinder. And we also know that <br>
>work isn't an instantaneous parameter so it must <br>
>begin before the half way point and last past <br>
>that point. Lots of unknowns and theory doesn't <br>
>always work in practice. But if we use the <br>
>average piston position at half-way down the <br>
>bore, where most work is most effective, and the <br>
>MEP (Mean Effective Pressure), since Mean is average, calculations begi=
n.<br>
><br>
>single revolution is 360=C2=B0 so half-way down the <br>
>power stroke is 90=C2=B0. Add the idle spark timing <br>
>of 4=C2=B0 BTDC (Before Top Dead Center) and we get <br>
>94=C2=B0 of crank rotation from spark to MEP at <br>
>half-way down. That's 94/360 or about 0.26 of an <br>
>engine revolution. And the engine is turning 15 <br>
>revolutions per second or 67ms (milliseconds) <br>
>per revolution. So 67 x 0.26 =3D 17ms from spark <br>
>to MEP at half-way down the power stroke, at idle.<br>
><br>
>If we repeat the calculations for operating <br>
>engine speed and at maximum advance, we get <br>
>3500rpm / 60mps =3D 58rps (revolutions per <br>
>second). Maximum advance is 32=C2=B0 BTDC so 90=C2=B0 + <br>
>32=C2=B0 =3D 122=C2=B0, spark to MEP or 122=C2=B0/360=C2=B0 =3D 0.34 of=
a revolution<br>
><br>
>58rps is 17ms/r so 17ms/r x 0.34r =3D 5.78ms from <br>
>spark to MEP at half-way down the power stroke. <br>
>This is a more representative number than the <br>
>17ms at idle. One might even divide the idle <br>
>elapsed time minus the optimal time across the <br>
>strike's midpoint. Doing so would mean at idle, <br>
>the pressure at idle becomes most effective <br>
>5.6ms before half-way and for another 5.6ms <br>
>after midpoint. Interesting that the idle <br>
>pressure application time is about the same as <br>
>the maximum pressure application time, or is that circular logic?<br>
><br>
>Yes everything above is ripe with assumptions <br>
>and perhaps even apocryphal and resplendent with <br>
>errors, but it is only 7am after all.<br>
><br>
>Anyone with extra house-bound time on their <br>
>hands, please check my maths and share your <br>
>corrections, including the logic of the whole <br>
>experiment...or perhaps even why geeks think these ways!<br>
><br>
>Rick<br>
<br>
</blockquote></div>
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