<div dir="auto"><div> Perfectly logical. Thank you. My TD is set to 5° BTDC but it might advance a little smoother at 10° BTDC. I will experiment. Smooth is the goal. <div dir="auto"><br></div><div dir="auto"> New parts for the TD distributor arrived today so rebuilding then retuning the carbs tomorrow. </div><div dir="auto"><br></div><div dir="auto"> Today I painted my work truck's replacement left front fender, but another coat is needed. Yea, house-bound and a little stir crazy.</div><div dir="auto"><br></div><div dir="auto"><a href="http://aubard.us/97_Nissan/20200401_135540.jpg">http://aubard.us/97_Nissan/20200401_135540.jpg</a><br></div><div dir="auto"><br></div><div dir="auto">Rick</div><br><br><div class="gmail_quote"><div dir="ltr" class="gmail_attr">On Wed, Apr 1, 2020, 6:56 PM Richard Ewald <<a href="mailto:richard.ewald@gmail.com">richard.ewald@gmail.com</a>> wrote:<br></div><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><div dir="ltr">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.<div>At idle speed 3ms is about 15 degrees. </div><div>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.</div><div>If you look at tune up specs for non smog controlled engines, most have base timing in the 5-10 BTDC range.</div><div>Now if you double the engine speed, it still take 3ms to burn, but to keep the max pressure at 5-10 ATDC, you have to light the mixture sooner.</div><div>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. </div><div>Rick</div></div><br><div class="gmail_quote"><div dir="ltr" class="gmail_attr">On Sat, Mar 28, 2020 at 5:55 AM Richard Lindsay via Mgs <<a href="mailto:mgs@autox.team.net" target="_blank" rel="noreferrer">mgs@autox.team.net</a>> wrote:<br></div><blockquote class="gmail_quote" style="margin:0px 0px 0px 0.8ex;border-left:1px solid rgb(204,204,204);padding-left:1ex"><div dir="auto">Hello friends,<div dir="auto"><br><div dir="auto"> 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.</div><div dir="auto"><br></div><div dir="auto"> 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?</div><div dir="auto"> Physics problems always start by listing the 'known' and the property to 'find'. So in this case,</div><div dir="auto"><br></div><div dir="auto">KNOWN:</div><div dir="auto"> Idle speed: 900rpm</div><div dir="auto"> Idle timing advance: 4° BTDC</div><div dir="auto"> Speed at maximum advance: 3500rpm</div><div dir="auto"> Maximum timing advance: 32° BTDC</div><div dir="auto"><br></div><div dir="auto">FIND:</div><div dir="auto"> Time from spark to MEP</div><div dir="auto"><br></div><div dir="auto"> 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.</div><div dir="auto"><br></div><div dir="auto"> 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.</div><div dir="auto"><br></div><div dir="auto"> At idle speed, 900rpm in this MG TD example, the XPAG engine is turning 900rpm or 900rpm / 60mps = 15rps (revolutions per second).</div><div dir="auto"><br></div><div dir="auto"> 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.</div><div dir="auto"><br></div><div dir="auto"> 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.</div><div dir="auto"><br></div><div dir="auto"> A single revolution is 360° so half-way down the power stroke is 90°. Add the idle spark timing of 4° BTDC (Before Top Dead Center) and we get 94° 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 = 17ms from spark to MEP at half-way down the power stroke, at idle.</div><div dir="auto"><br></div><div dir="auto"> If we repeat the calculations for operating engine speed and at maximum advance, we get 3500rpm / 60mps = 58rps (revolutions per second). Maximum advance is 32° BTDC so 90° + 32° = 122°, spark to MEP or 122°/360° = 0.34 of a revolution</div><div dir="auto"><div dir="auto"><br></div><div dir="auto"> 58rps is 17ms/r so 17ms/r x 0.34r = 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?</div><div dir="auto"><br></div><div dir="auto"> Yes everything above is ripe with assumptions and perhaps even apocryphal and resplendent with errors, but it is only 7am after all.</div><div dir="auto"><br></div><div dir="auto"> 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!</div><div dir="auto"><br></div><div dir="auto">Rick</div></div></div></div>
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