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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 <richardolindsay@gmail.com>
wrote:
> 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. Fue=
l
> 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 <paulhunt73@virginmedia.com>
> 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 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
>> 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 r=
od
>> makes relative to the piston.
>>
>> 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
>> other factors like fuel grade and type.
>>
>> PaulH.
>>
>>
>> ----- Original Message -----
>>
>> *From:* Richard Lindsay via Mgs <mgs@autox.team.net>
>> *To:* mgs@autox.team.net 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 t=
he
>> 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 u=
sed
>> to overcome the friction and other forces that exist at idle speed. Stat=
ed
>> another way: Idle speed is the fastest the engine can achieve given the
>> available charge. That fact is evident (with carbureted engines) when on=
e
>> notices that engine speed gradually increases, even for a fixed throttle
>> 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
>> 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 specifie=
d
>> 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 Mea=
n
>> 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 w=
e 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 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
>>
>> 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 tha=
n
>> 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
>>
>> ------------------------------
>>
>> _______________________________________________
>>
>> Mgs@autox.team.net
>> Donate: http://www.team.net/donate.html
>> Suggested annual donation $12.75
>>
>> Archive: http://www.team.net/pipermail/mgs http://autox.team.net/archive
>>
>> Unsubscribe:
>> http://autox.team.net/mailman/options/mgs/paulhunt73@virginmedia.com
>>
>>
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<div dir=3D"auto"><div dir=3D"auto" style=3D"font-family:sans-serif">As I h=
ave said before, I don'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&qid=
=3D1585403679&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&qid=3D1585403679&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 'consider'? 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 <<a href=3D"mailto:richardolindsay@gmail.com";>richardolinds=
ay@gmail.com</a>> 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't keep up with practice. The factory'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'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 <<a href=3D"mailto:paulhunt7=
3@virginmedia.com" target=3D"_blank" rel=3D"noreferrer">paulhunt73@virginme=
dia.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"><u></u>
<div bgcolor=3D"#ffffff">
<div><font size=3D"2">"</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"</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 'other forces' 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'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"mgs@autox.team.net" href=3D"mailto:mgs@autox.team.net"; rel=3D=
"noreferrer noreferrer" target=3D"_blank">Richard Lindsay=20
via Mgs</a> </div>
<div style=3D"FONT:10pt arial"><b>To:</b> <a title=3D"mgs@autox.team.net"=
href=3D"mailto:mgs@autox.team.net+List"; rel=3D"noreferrer noreferrer" targ=
et=3D"_blank">mgs@autox.team.net 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't burn instantly, despite our perception to the contra=
ry.=20
Rather, it takes a finite length of time from the occurance of the 's=
park',=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
'known' and the property to 'find'. 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'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'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'=
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'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'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'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:Mgs@autox.team.net"; rel=3D"noreferrer noreferrer" target=3D"_blank">=
Mgs@autox.team.net</a><br>Donate:=20
<a href=3D"http://www.team.net/donate.html"; rel=3D"noreferrer noreferrer"=
target=3D"_blank">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"; rel=
=3D"noreferrer noreferrer" target=3D"_blank">http://www.team.net/pipermail/=
mgs</a>=20
<a href=3D"http://autox.team.net/archive"; rel=3D"noreferrer noreferrer" t=
arget=3D"_blank">http://autox.team.net/archive</a><br><br>Unsubscribe:=20
<a href=3D"http://autox.team.net/mailman/options/mgs/paulhunt73@virginmed=
ia.com" rel=3D"noreferrer noreferrer" target=3D"_blank">http://autox.team.n=
et/mailman/options/mgs/paulhunt73@virginmedia.com</a><br></p></blockquote><=
/div>
</blockquote></div>
</blockquote></div>
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