I've seen the answer to this before in passing, but I want to get a fresh YES, NO, or KINDA BUT-.

I got a ACT flywheel for my S5 turbo motor, and I know I need to pop the appropriate counterweight on from a flexplate. Everything I've read so far says the weight works for both turbo and atmo(N/A) of same series; S4 and S5s don't intermix because of different rotor masses.

My question though is why doesn't the turbo or atmo class matter? Especially the S5, where there is a significant difference in compression--which translates to different rotor shapes, which translates to different rotational mass. Wouldn't putting an S5 atmo counterweight on a turbo be slightly off?

 

Come on, somebody should know this one for sure. The S5 NA's are a dime a dozen, but trying to get something from an actual factual automatic turbo isn't as easy.

 

Because the S5 Turbo and NA Rotors Weigh the same.

S4 Turbo and NA Rotors Weigh.

S4 Rotors are Heavier than S5 Rotors.

Different Shape does not mean different mass. A pound of water weighs the same whether it is in a round glass or square.

 

True only if spinning around the center of mass. As you are well aware, the rotors move in a Spirograph orbit so they exert a centrifugal force away from the orbital center. I'm running under the assumption that a higher compression rotor has more of its mass further out from center, in order to leave less airspace under full compression thus increasing the compression ratio. Spin a tennis ball on a string, then spin it with a longer string at the speed needed to complete one rev in the same amount of time--it'll 'pull' harder.

But, I don't know for sure Mazda's exact method of increasing the compression since I've never stacked all four FC rotor versions on top of each other. So, that's the reason for my asking.