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#1 |
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Join Date: May 2010
Location: Long Island, NY
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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 |
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#2 |
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Join Date: Apr 2016
Location: Bellingham, WA
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Distributor. 40 degrees at the crank.
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#3 |
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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 |
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#4 |
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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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