What's the linear distance between two teeth on the belt? That's the number you need to figure out how much offset you'll introduce by decking (or raising) the head.

 

From some generic list of replacement belts
1160mm length
145 teeth

i.e. 8mm/tooth

So if it would have been simple math and 90 degrees, 0.8mm decking would result in 0.1 tooth = 1.5 degrees.

But as Sav mentioned, it's at an angle towards the idler pulley (not 90 degree towards the deck), but not enough angle to translate 0.8mm deck to 6.6mm belt (looks to be less than 45 degrees from vertical).
I'm still confused, but I know how my engine behaves.

 

I think 2.5 degrees is right.

Suppose each tooth is around 5mm and 5mm = 15.65 degrees of timing.
Then (.8mm (around 0.030") × 15.65)/(5mm)= 2.5 degrees

 

I digged my old 2002 HG from the trash and it measures 0.032" so, Emilio was right. The Felpro one is about 0.030" with the same caliper.

 

Nevermind, but it's less than .5 cam degrees per 10 thousandths of an inch shaved.

 

^ This.



This thread makes me sad. It's one of those ClubRoadster moments where we're getting all hung up about a phenomenon which, while calculable on paper, is utterly insignificant in real-world application. In a dynamic (running engine) context, 0.5° of deviation in cam timing literally cannot be measured; it's less than the jitter inherent in the flexation of the cam belt.

Stop worrying about it, and move on. There are more important things in life.

 

Yeah, I did the math a few months ago when trying to figure out why my motor was pinging so much more. It's inconsequential.

--Ian