I ordered a 2# upper downs pulley. Now I want to get a 4# lower pulley to make 6#'s Does the math work like that or will I mess up the whole ratio and get 10 lbs. Want to make sure before I make a real mistake.

Any math wizzards?

 

this may help
http://www.ncs-stl.com/Files/BlowerRPM.xls

 

That was interesting. But the real question I have I guess is... If you have a smaller than stock on the top and you have something that spins it faster on the bottom, then you will have an accelerated motion/spin on the top (even more so than they would expect). It would make it more like a 7# or 8# pully, no?

Lets pretend that octane isn't a problem and I can adjust it at will without retarding it. Can I run more boost with the stock eaton supercharger? Will it even be nominal the extra spin that I get from the "accelerated motion"?

I knew I shouldn't have slept in trig.

 

I would assume that the person that creates the say 2lb upper pulley has already taken the lower stock ratio aspect into consideration.

So I would say that you simply need to figure the stock blower rpm to PSI rations per size of the stock pullies to start with.

Next calulate the rpms of say a +2lb pulley and calculate the new rpms to make again +2lbs. YOu can also play with the lower +2lb pulley sizes.

I'm not sure but I suspect by playing with the number you'll find out pretty quickley if this calulation is linear or not. I bet you'll see a relationship between blower rpm and psi is linear.

Remember c = pi (r x 2) or c = r x d ' r 3.1415926

 

I flunked Advanced Algebra at Miami Senior High........... Having said that, yes, 2 + 4 = 6 in this case. Is it 'really' 5.851 lbs or 6.324 lbs--don't know, but it's 'close' to 6 lbs.


Several guys run an upper and a lower and show the expected boost level on their aftermarket(ie: accurate) boost gauge. Get your chip tuner up-to-speed on your pulleys and boost question and the two of you will be successful.

Dan