--===============5533120239897141665==
--0000000000004e05d805a1f29b57
Content-Transfer-Encoding: quoted-printable
Excellent. Thank you.
On Sat, Mar 28, 2020, 6:15 PM Barney Gaylord <barneymg@mgaguru.com> wrote:
> 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. 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.
>
> 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 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
>
>
> 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 <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.=C3=82 I thought you were asking how much
> time to complete combustion to get to maximum
> pressure.=C3=82 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.=C3=82 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.=C3=82 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.=C3=82 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.=C3=82 And 900 rpm idle speed will be
> exactly 1/4 of road speed.=C3=82 If you make spark at
> 32d BTDC at road speed, it takes 1.5-ms to reach
> TDC.=C3=82 10d ATDC is at 2-ms, and 30d ATDC would be
> 3-ms (after spark).=C3=82 So the flame front
> propagation to complete combustion is in the 2 to
> 3-ms range.=C3=82 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.=C3=82 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=C3=82=C2=B0 BTDC
> >Speed at maximum advance: 3500rpm
> >Maximum timing advance: 32=C3=82=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=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
> >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=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
> >
> >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
>
>
--0000000000004e05d805a1f29b57
Content-Transfer-Encoding: quoted-printable
<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 <<a href=3D"mailto:barneymg@mgaguru.com";>barneymg@mgaguru.com</=
a>> 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'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'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
<<a href=3D"mailto:barneymg@mgaguru.com"; target=3D"_blank" rel=3D"norefe=
rrer">barneymg@mgaguru.com</a>>
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.=C3=82=C2=A0 I thought you were asking how much <br>
<dd>time to complete combustion to get to maximum <br>
<dd>pressure.=C3=82=C2=A0 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. "we do know that maximum work <br>
<dd>occurs when the piston is half way down the <br>
<dd>cylinder", which is not true.=C3=82=C2=A0 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.=C3=82=C2=A0 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.=C3=82=C2=A0 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.=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>32d BTDC at road speed, it takes 1.5-ms to reach <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>propagation to complete combustion is in the 2 to <br>
<dd>3-ms range.=C3=82=C2=A0 I suppose this is the answer to your <br>
<dd>question, "how much time" 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.=C3=82=C2=A0 And you <br>
<dd>night notice that I did not use "MEP" in that entire
discussion.<br><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>>Hello friends,<br>
<dd>><br>
<dd>>When one is a geek, one thinks of geeky things. <br>
<dd>>I am a geek and this house-bound morning I woke <br>
<dd>>up thinking about ignition timing. Here are the details.<br>
<dd>><br>
<dd>>We know that the charge (fuel plus air) in a <br>
<dd>>cylinder doesn't burn instantly, despite our <br>
<dd>>perception to the contrary. Rather, it takes a <br>
<dd>>finite length of time from the occurance of the <br>
<dd>>'spark', the flame front to cross the combustion <br>
<dd>>chamber, and to raise the MEP (Mean Effective <br>
<dd>>Pressure) to a maximum - the point where it does <br>
<dd>>the most work. But how much time?<br>
<dd>><br>
<dd>>Physics problems always start by listing the <br>
<dd>>'known' and the property to 'find'. So in this case=
,<br>
<dd>><br>
<dd>>KNOWN:<br>
<dd>>Idle speed: 900rpm<br>
<dd>>Idle timing advance: 4=C3=82=C2=B0 BTDC<br>
<dd>>Speed at maximum advance: 3500rpm<br>
<dd>>Maximum timing advance: 32=C3=82=C2=B0 BTDC<br>
<dd>><br>
<dd>>FIND:<br>
<dd>>Time from spark to MEP<br>
<dd>><br>
<dd>>The first thing one might know is that the goal <br>
<dd>>at idle is not to produce maximum power. In <br>
<dd>>fact, at idle 100% of the available power is <br>
<dd>>used to overcome the friction and other forces <br>
<dd>>that exist at idle speed. Stated another way: <br>
<dd>>Idle speed is the fastest the engine can achieve <br>
<dd>>given the available charge. That fact is evident <br>
<dd>>(with carbureted engines) when one notices that <br>
<dd>>engine speed gradually increases, even for a <br>
<dd>>fixed throttle setting, as the engine warms and <br>
<dd>>friction forces decrease. But back to the problem.<br>
<dd>><br>
<dd>>Because the goal at idle is smooth running and <br>
<dd>>progression off of idle (e.g. speeding up), not <br>
<dd>>maximum power, the calculated wavefront speed <br>
<dd>>may not be correct at idle. But let's see.<br>
<dd>><br>
<dd>>At idle speed, 900rpm in this MG TD example, the <br>
<dd>>XPAG engine is turning 900rpm or 900rpm / 60mps <br>
<dd>>=3D 15rps (revolutions per second).<br>
<dd>><br>
<dd>>Distributor speed is 1/2 engine speed so at idle <br>
<dd>>the distributor is turning only 7.5 revolutions <br>
<dd>>per second. But timing numbers are specified in <br>
<dd>>degrees of crank rotations so we will stick with 15rps.<br>
<dd>><br>
<dd>>We don't know how fast the flame front travels <br>
<dd>>across the combustion chamber but we do know <br>
<dd>>that maximum work occurs when the piston is half <br>
<dd>>way down the cylinder. And we also know that <br>
<dd>>work isn't an instantaneous parameter so it must <br>
<dd>>begin before the half way point and last past <br>
<dd>>that point. Lots of unknowns and theory doesn't <br>
<dd>>always work in practice. But if we use the <br>
<dd>>average piston position at half-way down the <br>
<dd>>bore, where most work is most effective, and the <br>
<dd>>MEP (Mean Effective Pressure), since Mean is average,
calculations begin.<br>
<dd>><br>
<dd>>single revolution is 360=C3=82=C2=B0 so half-way down the <br>
<dd>>power stroke is 90=C3=82=C2=B0. Add the idle spark timing <br>
<dd>>of 4=C3=82=C2=B0 BTDC (Before Top Dead Center) and we get <br>
<dd>>94=C3=82=C2=B0 of crank rotation from spark to MEP at <br>
<dd>>half-way down. That's 94/360 or about 0.26 of an <br>
<dd>>engine revolution. And the engine is turning 15 <br>
<dd>>revolutions per second or 67ms (milliseconds) <br>
<dd>>per revolution. So 67 x 0.26 =3D 17ms from spark <br>
<dd>>to MEP at half-way down the power stroke, at idle.<br>
<dd>><br>
<dd>>If we repeat the calculations for operating <br>
<dd>>engine speed and at maximum advance, we get <br>
<dd>>3500rpm / 60mps =3D 58rps (revolutions per <br>
<dd>>second). Maximum advance is 32=C3=82=C2=B0 BTDC so 90=C3=82=C2=B0 +=
<br>
<dd>>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>><br>
<dd>>58rps is 17ms/r so 17ms/r x 0.34r =3D 5.78ms from <br>
<dd>>spark to MEP at half-way down the power stroke. <br>
<dd>>This is a more representative number than the <br>
<dd>>17ms at idle. One might even divide the idle <br>
<dd>>elapsed time minus the optimal time across the <br>
<dd>>strike's midpoint. Doing so would mean at idle, <br>
<dd>>the pressure at idle becomes most effective <br>
<dd>>5.6ms before half-way and for another 5.6ms <br>
<dd>>after midpoint. Interesting that the idle <br>
<dd>>pressure application time is about the same as <br>
<dd>>the maximum pressure application time, or is that circular
logic?<br>
<dd>><br>
<dd>>Yes everything above is ripe with assumptions <br>
<dd>>and perhaps even apocryphal and resplendent with <br>
<dd>>errors, but it is only 7am after all.<br>
<dd>><br>
<dd>>Anyone with extra house-bound time on their <br>
<dd>>hands, please check my maths and share your <br>
<dd>>corrections, including the logic of the whole <br>
<dd>>experiment...or perhaps even why geeks think these ways!<br>
<dd>><br>
<dd>>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>
--0000000000004e05d805a1f29b57--
--===============5533120239897141665==
Content-Transfer-Encoding: 7bit
Content-Disposition: inline
_______________________________________________
Mgs@autox.team.net
Archive: http://www.team.net/pipermail/mgs http://autox.team.net/archive
Unsubscribe: http://autox.team.net/mailman/options/mgs/mharc@autox.team.net
--===============5533120239897141665==--
|