Timing mark from BTDC

[tradosaurus]

I took my timing gun and checked the timing on my car.

I found it to be 10 Deg BTDC and low and behold the Ignition timing at idle is 10 Deg BTDC.

My car is still idling funny once it warms up so I'm wondering if it is still the timing belt being off one tooth?

Could it be a faulty Crankshaft position sensor?

[GasSavers_Erik]

How is the drive-ability and mileage? I think a bad crank sensor would likely set a code or at least result in poor mpg or poor drive-ability.

[tradosaurus]

It appears to drive fine until you come to a stop and the RPM fluctuates.

[cems70]

Originally Posted by tradosaurus

My car is still idling funny once it warms up so I'm wondering if it is still the timing belt being off one tooth?

If the belt were off one tooth in either direction, it would be nearly undriveable.

[kamesama980]

What kind of car might help. If it uses a MAF, vacuum leak would be my first suggestion. cleaning the throttle body and IAC are next either way.

It also decides whether the crank sensor will throw a code (very unlikely to cause a warm idle problem and nothing at any other time though)

If the timing belt slipped a tooth the ignition timing would NOT be correct (except for very few cases) and it would run MUCH worse.

[tradosaurus]

Here is an update. I've rechecked the timing marks and I believe it is dead-on perfect.

I have a 2000 Mazda Protege 1.6L 5 speed.

I've cleaned the MAF sensor, checked intake hose for cracks, tested the resistance of the IAC and throttle position sensor. The IAC was at the maximum resistance allowed and the throttle position sensor was midrange of resistance allowance.

I've replaced the throttle body gasket and verified all vacuum hoses are attached. The previous owner took out the EGR valve and piping so there is no issue there.

My last effort will be to replace the intake manifold gasket and injector O-rings in the hope that there are vacuum leaks at those points. If the gasket and O-rings don't fix it then I will buy a new IAC valve ($$$$).