that once you put a fresh timing belt on a 2.2 and line up the timing marks and the marks painted on the NEW belt, that the markings on the belt, won't line up again?

 

Of course the timing marks on the pulleys and crankshaft do line up every other rotation, but the belt marks won't.

 

Todd

that once you put a fresh timing belt on a 2.2 and line up the timing marks and the marks painted on the NEW belt, that the markings on the belt, won't line up again?

 

Of course the timing marks on the pulleys and crankshaft do line up every other rotation, but the belt marks won't.

 

Todd

 

It is, as far as it's exactly what happened to me when I replaced my belt. Some members tried to figure out recently when (after how many rotations) the marks on the belt would line up again and, IIRC, they could not figure out the maths necessary to find the exact solution. FWIW.

The marks would line up again if you turn the crankshaft as many times as there are teeth on the timing belt.

Would be fair to say that you most likely won't see them lined up agin. As stated the math hasn't been done to awnser that question, and as far as I know nobody has taking the time the crank the engine over long enough to find out.

The math here isn't rocket science as they say. The required number of revolutions would be less if the tooth counts on the belt and the camshaft sprocket had their largest common multiple greater than one, but that isn't the case. Anyhow, such answer is not practically useful.

It is interesting to ponder Subaru mathmatics!

 

My guess was right. Unless you turn the engine over as many times as there are teeth in the belt, the marks won't line up again. And even then it's questionable.

 

This is good for everybody to know.

 

Todd

when it does come around for lining up, it would be a bad thing if it didint on only one mark... might indicate bad timing..

 

 

look on nasioc, i think this was up there somewhere a few weeks ago, i cant remember how many rotations that came up with though. i cant remember, but i think someone came up with a math formula to figure it out...

.

OK. Think of crank and cam sprockets, and the t. belt as three gears, with tooth counts C, 2*C, and T, respectively. When crankshaft is rotated N times,

the cams will rotate N/2 times, and the belt N*C/T times. We want the smallest N such that all three numbers are whole. N=T will work if T is even which I believe is the case (otherwise N=2*T). A smaller value would work if C and T have a common factor but that shouldn't be so if the design provides for the most even wear of the teeth, like with ring-pinion pairs (37:9 etc).