i have a question i was hoping to gather some opinions on.

do you think an extra 5 lbs. and 1 inch in diameter in rotor size (i.e; 2 piece 332 vs 355mm) would be at all noticable as unsprung weight? the main reason i ask is that there is only one extra inch moving outward from the hub- from 13" to 14" and it seems that such a short distance wouldn't generate too much extra effort. i am basing this on a stock 17" wheel size and 12" stock rotors.
i can see, and i have actually felt, how upsizing one or two sizes with a heavy rim can really affect braking as unsprung weight. but i guess i wasnt too sure how a similiar amount of weight would be percieved when not moving so far from center.

thanks for any insights

...what is not negligble is the additional effort required to rotate the larger disc.

The concept is called "Rotational Moment of Intertia". As you increase the diameter of an object that you are rotating, there is a natural point at which the object will rotate (center of mass I believe?). If that center of mass is further out from the center, more effort is required to rotate the object.

A brake disc is a perfect example. The mass of the weight will be centered in the disc face, not the hub (since there is no material there). The disc face is typically 54mm in annulus, and at a 332mm disc, puts the center of mass at something like 305mm. The same 54mm annulus on a 355mm disc puts the center of mass at something like 328mm.

I'm taking a lot of mathematical liberties for simplicity sake, but while you're likely notice no difference in unsprung weight (do to very little difference in weight change), you may notice the difference it takes to rotate the larger object.

Wheels are another great example of this. Think about that next time you're racing the guy with the same car rollin' on 22's and you're on stock 17's.

centre. What you talked about is the mass averaged radius of the object, which is only necessary if you insist upon not doing integration when calculating rotational moment of inertia <I>I</I>. :)

This is for MISTERJONES: bottom line, torque needed to accelerate an object is directly proportional to <I>I</I>, and <I>I=mR</I>^2, where <I>m</I> is mass and <I>R</I> is radius from rotational centre; note the quadratic radius dependence. That being said, compared to the massive wheel and tire, something like a 0.5" radius increase in the rotor is relatively small when it comes to the total <I>I</I>, and thus *may* not be noticeable.

but its obviously going to take less effort for the 332 disk to be started, stopped, and supported by the suspension.