--===============3149900410903260634==
--000000000000f306ea05a2019c2a
Content-Transfer-Encoding: quoted-printable
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 <mgs@autox.team.ne=
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 3/28/2020 7:30 PM, Richard Lindsay via Mgs wrote:
>
> 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. =
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
>> 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 bes=
t
>> 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 begi=
n.
>> >
>> >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
>>
>>
> _______________________________________________
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> _______________________________________________
>
> Mgs@autox.team.net
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--000000000000f306ea05a2019c2a
Content-Transfer-Encoding: quoted-printable
<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 '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.=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 <<a href=3D"mailto:mgs@aut=
ox.team.net">mgs@autox.team.net</a>> 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 <<a href=3D"mailto:barneymg@mgaguru.com"; target=
=3D"_blank">barneymg@mgaguru.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">
<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 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'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"; rel=3D"no=
referrer" target=3D"_blank">barneymg@mgaguru.com</a>>
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.=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. "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.=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, "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.=C3=82=C2=A0 An=
d 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=3D"cite">
<dl>
<dd>At 08:54 AM 3/28/2020, you wrote:<br>
</dd>
<dd>>Hello friends,<br>
</dd>
<dd>><br>
</dd>
<dd>>When one is a geek, one thinks of geeky things.
<br>
</dd>
<dd>>I am a geek and this house-bound morning I woke
<br>
</dd>
<dd>>up thinking about ignition timing. Here are the
details.<br>
</dd>
<dd>><br>
</dd>
<dd>>We know that the charge (fuel plus air) in a <br>
</dd>
<dd>>cylinder doesn't burn instantly, despite our <b=
r>
</dd>
<dd>>perception to the contrary. Rather, it takes a <br>
</dd>
<dd>>finite length of time from the occurance of the
<br>
</dd>
<dd>>'spark', the flame front to cross the combu=
stion
<br>
</dd>
<dd>>chamber, and to raise the MEP (Mean Effective <br>
</dd>
<dd>>Pressure) to a maximum - the point where it does
<br>
</dd>
<dd>>the most work. But how much time?<br>
</dd>
<dd>><br>
</dd>
<dd>>Physics problems always start by listing the <br>
</dd>
<dd>>'known' and the property to 'find'.=
So in this
case,<br>
</dd>
<dd>><br>
</dd>
<dd>>KNOWN:<br>
</dd>
<dd>>Idle speed: 900rpm<br>
</dd>
<dd>>Idle timing advance: 4=C3=82=C2=B0 BTDC<br>
</dd>
<dd>>Speed at maximum advance: 3500rpm<br>
</dd>
<dd>>Maximum timing advance: 32=C3=82=C2=B0 BTDC<br>
</dd>
<dd>><br>
</dd>
<dd>>FIND:<br>
</dd>
<dd>>Time from spark to MEP<br>
</dd>
<dd>><br>
</dd>
<dd>>The first thing one might know is that the goal
<br>
</dd>
<dd>>at idle is not to produce maximum power. In <br>
</dd>
<dd>>fact, at idle 100% of the available power is <br>
</dd>
<dd>>used to overcome the friction and other forces <br>
</dd>
<dd>>that exist at idle speed. Stated another way: <br>
</dd>
<dd>>Idle speed is the fastest the engine can achieve
<br>
</dd>
<dd>>given the available charge. That fact is evident
<br>
</dd>
<dd>>(with carbureted engines) when one notices that
<br>
</dd>
<dd>>engine speed gradually increases, even for a <br>
</dd>
<dd>>fixed throttle setting, as the engine warms and
<br>
</dd>
<dd>>friction forces decrease. But back to the
problem.<br>
</dd>
<dd>><br>
</dd>
<dd>>Because the goal at idle is smooth running and <br>
</dd>
<dd>>progression off of idle (e.g. speeding up), not
<br>
</dd>
<dd>>maximum power, the calculated wavefront speed <br>
</dd>
<dd>>may not be correct at idle. But let's see.<br>
</dd>
<dd>><br>
</dd>
<dd>>At idle speed, 900rpm in this MG TD example, the
<br>
</dd>
<dd>>XPAG engine is turning 900rpm or 900rpm / 60mps
<br>
</dd>
<dd>>=3D 15rps (revolutions per second).<br>
</dd>
<dd>><br>
</dd>
<dd>>Distributor speed is 1/2 engine speed so at idle
<br>
</dd>
<dd>>the distributor is turning only 7.5 revolutions
<br>
</dd>
<dd>>per second. But timing numbers are specified in
<br>
</dd>
<dd>>degrees of crank rotations so we will stick with
15rps.<br>
</dd>
<dd>><br>
</dd>
<dd>>We don't know how fast the flame front travels =
<br>
</dd>
<dd>>across the combustion chamber but we do know <br>
</dd>
<dd>>that maximum work occurs when the piston is half
<br>
</dd>
<dd>>way down the cylinder. And we also know that <br>
</dd>
<dd>>work isn't an instantaneous parameter so it mus=
t
<br>
</dd>
<dd>>begin before the half way point and last past <br>
</dd>
<dd>>that point. Lots of unknowns and theory doesn't
<br>
</dd>
<dd>>always work in practice. But if we use the <br>
</dd>
<dd>>average piston position at half-way down the <br>
</dd>
<dd>>bore, where most work is most effective, and the
<br>
</dd>
<dd>>MEP (Mean Effective Pressure), since Mean is
average, calculations begin.<br>
</dd>
<dd>><br>
</dd>
<dd>>single revolution is 360=C3=82=C2=B0 so half-way do=
wn the
<br>
</dd>
<dd>>power stroke is 90=C3=82=C2=B0. Add the idle spark =
timing
<br>
</dd>
<dd>>of 4=C3=82=C2=B0 BTDC (Before Top Dead Center) and =
we get
<br>
</dd>
<dd>>94=C3=82=C2=B0 of crank rotation from spark to MEP =
at <br>
</dd>
<dd>>half-way down. That's 94/360 or about 0.26 of a=
n
<br>
</dd>
<dd>>engine revolution. And the engine is turning 15
<br>
</dd>
<dd>>revolutions per second or 67ms (milliseconds) <br>
</dd>
<dd>>per revolution. So 67 x 0.26 =3D 17ms from spark <b=
r>
</dd>
<dd>>to MEP at half-way down the power stroke, at
idle.<br>
</dd>
<dd>><br>
</dd>
<dd>>If we repeat the calculations for operating <br>
</dd>
<dd>>engine speed and at maximum advance, we get <br>
</dd>
<dd>>3500rpm / 60mps =3D 58rps (revolutions per <br>
</dd>
<dd>>second). Maximum advance is 32=C3=82=C2=B0 BTDC so =
90=C3=82=C2=B0 +
<br>
</dd>
<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>
<dd>><br>
</dd>
<dd>>58rps is 17ms/r so 17ms/r x 0.34r =3D 5.78ms from
<br>
</dd>
<dd>>spark to MEP at half-way down the power stroke.
<br>
</dd>
<dd>>This is a more representative number than the <br>
</dd>
<dd>>17ms at idle. One might even divide the idle <br>
</dd>
<dd>>elapsed time minus the optimal time across the <br>
</dd>
<dd>>strike's midpoint. Doing so would mean at idle,
<br>
</dd>
<dd>>the pressure at idle becomes most effective <br>
</dd>
<dd>>5.6ms before half-way and for another 5.6ms <br>
</dd>
<dd>>after midpoint. Interesting that the idle <br>
</dd>
<dd>>pressure application time is about the same as <br>
</dd>
<dd>>the maximum pressure application time, or is
that circular logic?<br>
</dd>
<dd>><br>
</dd>
<dd>>Yes everything above is ripe with assumptions <br>
</dd>
<dd>>and perhaps even apocryphal and resplendent with
<br>
</dd>
<dd>>errors, but it is only 7am after all.<br>
</dd>
<dd>><br>
</dd>
<dd>>Anyone with extra house-bound time on their <br>
</dd>
<dd>>hands, please check my maths and share your <br>
</dd>
<dd>>corrections, including the logic of the whole <br>
</dd>
<dd>>experiment...or perhaps even why geeks think
these ways!<br>
</dd>
<dd>><br>
</dd>
<dd>>Rick </dd>
</dl>
</blockquote>
</div>
</blockquote>
</div>
<br>
<fieldset></fieldset>
<pre>_______________________________________________
<a href=3D"mailto:Mgs@autox.team.net"; target=3D"_blank">Mgs@autox.team.net<=
/a>
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