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06-14-2019, 09:49 AM | #1 |
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Ignition Advance vs Crankshaft Rotation
The maximum ignition advance for a properly timed Model "A" is 20 degrees. Is the 20 degrees distributor rotation or is it crankshaft rotation BFTDC?
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Bob Bidonde |
06-14-2019, 09:58 AM | #2 |
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Re: Ignition Advance vs Crankshaft Rotation
Distributor. 40 degrees at the crank.
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06-14-2019, 10:18 AM | #3 |
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Re: Ignition Advance vs Crankshaft Rotation
Thanks Dick for confirming my understanding. So at 20 degrees distributor advance, the piston is 40 degrees BFTDC. This puts the piston about 0.92 inches down from TDC.
40/180 = x/ 4.125 where 4.125 is the stroke at 180 degrees BDC of the crankshaft. Therefore x=0.92" BFTDC when the crankshaft is 40 degrees BFTDC, and the distributor is at 20 degrees spark advance.
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Bob Bidonde |
06-14-2019, 06:53 PM | #4 |
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Re: Ignition Advance vs Crankshaft Rotation
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x = c - (c cos 40*) or in other words, c x versine 40* I realise that will only hold true for an engine where the centreline of the crankshaft intersects the centreline of the bores. For engines where they are offset, the game changes.
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I'm part of the only ever generation with an analogue childhood and a digital adulthood. Last edited by Synchro909; 06-20-2019 at 08:29 PM. |
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