Timing a 78 911-SC (euro)

There's no vacuum retard or advance on my
distributor. Is it timed at 5 degrees BTDC
@ 900 RPM? Is there anything special (other
than setting the RPM) I need to do to check
timing? Does anyone know what the specs are for the rest of the advance curve for a euro 911, and how they are verified?

Thanks again.

Anyone ????????????

Beuller ?????????????

That is pretty funny.

And surely we are all old enough to remember as well.

Cam

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Cameron Baudinet
1975 911S

From the spec book, '78 911SC ignition timing is 5 BTDC at 900+-50 and 24-31 BTDC at 6000. Both measurements with the vacuum control detached. (You might not have a vacuum control unit NOW, but it came from the factory with one!)

Brian,
Thanks for the specs. Are they also applicable for the euro cars? One other thing. I took my distributor out (it's a Boch unit) and couldn't find a spot where the vacuum unit would have been attached. I'll call the dealer with the part number and ask him.

Thanks, Leon

Those are ROW specs ("Rest of World"), which includes Euro. (Other specs "USA", "JAPAN" ...). The USA advance spec is 24-28 degrees BTDC at 6000 and the same 5 degrees BTDC at 850-950., again with the vacuum detached.

For the centrifugal advance curve for the 1978 USA 911SC, plot the following points on a graph. These graph points are in degrees of advance AT THE DISTRIBUTOR at RPM. 4@2000 would be 4 degrees AT THE DISTRIBUTOR at 2000 engine RPM.

0@600, 9@1200,8.5@2200, 12@2900, 11.5@4000

0@800, 7@1200, 6.5@2400, 9.5@3050, 9.2@4000

That give you two roughly parallel lines depicting the maximum and minimum advance at each RPM. Note that this is advance AT THE DISTRIBUTOR. Advance measured at the CRANKSHAFT would be DOUBLE the amount at the DISTRIBUTOR.

There is a separate curve for the vacuum advance. But the vacuum advance is DETACHED for the test, so its curve does not apply. If you are curious, the two lines are (given in degrees advance AT THE DISTRIBUTOR against TORR of vacuum):

0.5@0-80 torr, 3.4@>=140 torr
-0.5@0-120 torr, 1.4@>=160 torr

That gives you two horizonal lines on the graph. Connect the adjacent ends of each pair of lines to complete the picture.


[This message has been edited by Brian K. Haggard (edited 07-15-2000).]

Thanks so Brian