Ignition Advance vs Crankshaft Rotation

[Bob Bidonde]

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?

[Dick Steinkamp]

Distributor. 40 degrees at the crank.

[Bob Bidonde]

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.

[Synchro909]

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I'd calculate the distance this way but first, we need to know the length of the con rod, centre to centre. I'll call that distance "c" and the distance from TDC , I'll call "x"

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.