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I came across a question about wasted spark systems, and I was wondering whether there is a risk of prematurely igniting the incoming fresh mixture charge with that wasted spark.

At the top of the exhaust stroke, the inlet valve is opened already. An inlet valve timing of say, 20 degrees before TDC isn't uncommon, and ignition advance can be less than that. This means that these two overlap.

So with the right conditions concerning engine speed and engine load etc, I guess there's a chance that there's a bit of fresh mixture sucked into the cylinder already, when it isn't at it's TDC, nor at the time of ignition yet. Especially with forced induction engines. The wasted spark then could ignite that incoming charge. In the worst case, the flame could propagate throught the intake port.

These are all just guesses though, but i'm curious why it doesn't happen. Is it that the mixture isn't inflammable yet at this point? Or is it impossible for an intake charge to be forced into the cylinder so early, even when the inletvalve is already opening?

I'm curious about your theories!

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  • To reviewers: please check out this meta post: I certainly believe this question is on-topic. – anonymous2 May 3 '17 at 15:40
  • whatever fuel is present does not ignite because it is not under any compression. – agentp May 3 '17 at 20:14
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If the inlet valve opens 20 deg before TDC, will fresh mixture be going in or not? As there is still pressure in the cylinder - the exhaust leaving is helping to "draw" the fresh mixture into the cylinder, so a wasted spark has no "new" or very little mixture to fire so it doesn't happen.

Also, the engines designed with wasted spark ignition may not have had very high scavenge ratios... That is, overlap of exhaust & inlet valves.

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  • That's what i'm not sure about. The leaving exhaust gases might help drawing mixture inside the cylinder, and a turbo would help even further I guess. So logically speaking, it would be possible for mixture to be present in the cylinder at wasted sparktime. The reason for it not to happen then could indeed be the weakness of the mixture at that time. But that's just a guess. – Bart May 3 '17 at 12:05
  • As per my earlier response the amount present will be really small and not likely to ignite - a mixture that is too weak does not ignite... – Solar Mike May 3 '17 at 12:07
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The wasted spark will likely be 30 degrees or so before TDC. There would be no fresh fuel charge at this point.

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  • Why would the spark be so early? If the engine speed is eg. around 1500 rpm, an ignition advance like 10-12 degrees before TDC could be perfectly normal. – Bart May 3 '17 at 17:12
  • @Bart - You are only considering mechanical advance. There could also be vacuum and electronic advance as well. 30deg BTDC is not uncommon. – Pᴀᴜʟsᴛᴇʀ2♦ May 3 '17 at 17:29
  • Mostly because I’m more familiar with high performance engines. But at low RPM you’re not going to have intake flow during the overlap period. At higher RPM the exhaust velocity will act on the intake as the IV opens. Still, even at 10-12 degrees advance, the wasted spark is complete before the intake charge appears. At higher RPM more advance is needed because of the rather slow travel of the ignition kernel at the spark plug, i.e. conduction of the flame front. The time for fuel mixture ignition is a time domain function of RPM. TomO – TomO May 3 '17 at 17:31
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    @Bart, typically at idle I would be expecting to command about 15-18 degrees timing BTDC 10-12 seems very low, above that it would continue to rise, 20+ at 1500 is what I would normally expect, higher and higher RPM's would go up to 30, 40. I can see what your thinking, on a 4 cylinder for example with C1 coming up to TDC Compression, and C2 coming up to TDC Exhaust, if there was sufficient valve overlap, and good scavenging, some air fuel mix could be pulled in, however this much overlap is unlikely. I will make a little write-up with my thoughts on it. – H. Daun Jun 14 '19 at 1:49

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