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Perfectly logical. Thank you. My TD is set to 5=C2=B0 BTDC but it might
advance a little smoother at 10=C2=B0 BTDC. I will experiment. Smooth is th=
e
goal.
New parts for the TD distributor arrived today so rebuilding then
retuning the carbs tomorrow.
Today I painted my work truck's replacement left front fender, but
another coat is needed. Yea, house-bound and a little stir crazy.
http://aubard.us/97_Nissan/20200401_135540.jpg
Rick
On Wed, Apr 1, 2020, 6:56 PM Richard Ewald <richard.ewald@gmail.com> wrote:
> It takes about 3ms to burn the fuel in a cylinder from when the spark
> ignites it. The engine designers want the peak pressure to occur at
> between 5-10 ATDC.
> At idle speed 3ms is about 15 degrees.
> So if you want your peak pressure at 5 ATDC, your base timing is 10 BTDC,
> if you want your peak pressure at 10 ATDC your timing is 5 BTDC.
> If you look at tune up specs for non smog controlled engines, most have
> base timing in the 5-10 BTDC range.
> Now if you double the engine speed, it still take 3ms to burn, but to kee=
p
> the max pressure at 5-10 ATDC, you have to light the mixture sooner.
> You do eventually get to a point where no more advance will help. That
> point varies with the engine design, cam profile, combustion chamber
> design, headers etc.
> Rick
>
> On Sat, Mar 28, 2020 at 5:55 AM Richard Lindsay via Mgs <
> mgs@autox.team.net> 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 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/richard.ewald@gmail.com
>>
>
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<div dir=3D"auto"><div>=C2=A0 =C2=A0Perfectly logical. Thank you. My TD is =
set to 5=C2=B0 BTDC but it might advance a little smoother at 10=C2=B0 BTDC=
. I will experiment. Smooth is the goal.=C2=A0<div dir=3D"auto"><br></div><=
div dir=3D"auto">=C2=A0 =C2=A0New parts for the TD distributor arrived toda=
y so rebuilding then retuning the carbs tomorrow.=C2=A0</div><div dir=3D"au=
to"><br></div><div dir=3D"auto">=C2=A0 =C2=A0Today I painted my work truck&=
#39;s replacement left front fender, but another coat is needed. Yea, house=
-bound and a little stir crazy.</div><div dir=3D"auto"><br></div><div dir=
=3D"auto"><a href=3D"http://aubard.us/97_Nissan/20200401_135540.jpg";>http:/=
/aubard.us/97_Nissan/20200401_135540.jpg</a><br></div><div dir=3D"auto"><br=
></div><div dir=3D"auto">Rick</div><br><br><div class=3D"gmail_quote"><div =
dir=3D"ltr" class=3D"gmail_attr">On Wed, Apr 1, 2020, 6:56 PM Richard Ewald=
<<a href=3D"mailto:richard.ewald@gmail.com";>richard.ewald@gmail.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 dir=3D"ltr">It tak=
es about 3ms to burn the fuel in a cylinder from when the spark ignites it.=
=C2=A0 The engine designers=C2=A0want the peak pressure to occur at between=
5-10 ATDC.<div>At idle speed 3ms is about 15 degrees.=C2=A0=C2=A0</div><di=
v>So if you want your peak pressure at 5 ATDC, your base timing=C2=A0is 10 =
BTDC, if=C2=A0 you want=C2=A0 your peak pressure at 10 ATDC your timing is =
5 BTDC.</div><div>If you look at tune up specs for non smog controlled engi=
nes, most have base timing in the 5-10 BTDC range.</div><div>Now if you dou=
ble the engine speed, it still take 3ms to burn, but to keep the max pressu=
re at 5-10 ATDC, you have to light the mixture sooner.</div><div>You do eve=
ntually=C2=A0get to a point where no more advance will help.=C2=A0 That poi=
nt varies with the engine=C2=A0design,=C2=A0 cam profile, combustion chambe=
r design, headers etc.=C2=A0 =C2=A0</div><div>Rick</div></div><br><div clas=
s=3D"gmail_quote"><div dir=3D"ltr" class=3D"gmail_attr">On Sat, Mar 28, 202=
0 at 5:55 AM Richard Lindsay via Mgs <<a href=3D"mailto:mgs@autox.team.n=
et" target=3D"_blank" rel=3D"noreferrer">mgs@autox.team.net</a>> wrote:<=
br></div><blockquote class=3D"gmail_quote" style=3D"margin:0px 0px 0px 0.8e=
x;border-left:1px solid rgb(204,204,204);padding-left:1ex"><div dir=3D"auto=
">Hello friends,<div dir=3D"auto"><br><div dir=3D"auto">=C2=A0 =C2=A0When o=
ne 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.</di=
v><div dir=3D"auto"><br></div><div dir=3D"auto">=C2=A0 =C2=A0We know that t=
he 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 f=
rom the occurance of the 'spark', the flame front to cross the comb=
ustion 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=
=3D"auto">=C2=A0 =C2=A0Physics problems always start by listing the 'kn=
own' 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 ad=
vance: 4=C2=B0 BTDC</div><div dir=3D"auto">=C2=A0 =C2=A0Speed at maximum ad=
vance: 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"aut=
o"><br></div><div dir=3D"auto">=C2=A0 =C2=A0The 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 forc=
es 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 (wi=
th carbureted engines) when one notices that engine speed gradually increas=
es, even for a fixed throttle setting, as the engine warms and friction for=
ces decrease. But back to the problem.</div><div dir=3D"auto"><br></div><di=
v dir=3D"auto">=C2=A0 =C2=A0Because the goal at idle is smooth running and =
progression off of idle (e.g. speeding up), not maximum power, the calculat=
ed 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, 900rp=
m in this MG TD example, the XPAG engine is turning 900rpm or=C2=A0900rpm /=
60mps =3D 15rps (revolutions per second).</div><div dir=3D"auto"><br></div=
><div dir=3D"auto">=C2=A0 =C2=A0Distributor speed is 1/2 engine speed so at=
idle the distributor is turning only 7.5 revolutions per second. But timin=
g numbers 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 across the combustion chamb=
er 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 parame=
ter so it must begin before the half way point and last past that point. Lo=
ts of unknowns and theory doesn't always work in practice. But if we us=
e 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 avera=
ge, 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-way down the power st=
roke is 90=C2=B0. Add the idle spark timing of 4=C2=B0 BTDC (Before Top Dea=
d Center) and we get 94=C2=B0 of crank rotation from spark to MEP at half-w=
ay down. That's 94/360 or about 0.26 of an engine revolution. And the e=
ngine is turning 15 revolutions per second or 67ms (milliseconds) per revol=
ution. So 67 x 0.26 =3D 17ms from spark to MEP at half-way down the power s=
troke, at idle.</div><div dir=3D"auto"><br></div><div dir=3D"auto">=C2=A0 =
=C2=A0If we repeat the calculations for operating engine speed and at maxim=
um advance, we get=C2=A03500rpm / 60mps =3D 58rps (revolutions per second).=
Maximum advance is 32=C2=B0 BTDC so=C2=A090=C2=B0 + 32=C2=B0 =3D 122=C2=B0=
, spark to MEP or=C2=A0122=C2=B0/360=C2=B0 =3D 0.34 of a revolution</div><d=
iv dir=3D"auto"><div dir=3D"auto"><br></div><div dir=3D"auto">=C2=A0 =C2=A0=
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 a=
t 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 abo=
ut the same as the maximum pressure application time, or is that circular l=
ogic?</div><div dir=3D"auto"><br></div><div dir=3D"auto">=C2=A0 =C2=A0Yes e=
verything above is ripe with assumptions and perhaps even apocryphal and re=
splendent with errors, but it is only 7am after all.</div><div dir=3D"auto"=
><br></div><div dir=3D"auto">=C2=A0 =C2=A0Anyone with extra house-bound tim=
e on their hands, please check my maths and share your corrections, includi=
ng the logic of the whole experiment...or perhaps even why geeks think thes=
e ways!</div><div dir=3D"auto"><br></div><div dir=3D"auto">Rick</div></div>=
</div></div>
_______________________________________________<br>
<br>
<a href=3D"mailto:Mgs@autox.team.net"; target=3D"_blank" rel=3D"noreferrer">=
Mgs@autox.team.net</a><br>
errer" target=3D"_blank">http://www.team.net/donate.html</a><br>
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Archive: <a href=3D"http://www.team.net/pipermail/mgs"; rel=3D"noreferrer no=
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</blockquote></div>
</blockquote></div></div></div>
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