[TriLUG] OT: the optimum location for inner track/cylinder on rotating media?

[Steve Litt]

On Wednesday 12 August 2009 09:53:51 Joseph Mack NA3T wrote:
> Some storage media (vinyl records, hard disks, cd/dvds?,
> floppies) rotate at constant angular velocity AND operate at
> a constant data rate. Thus the number of bits
> stored/cylinder is constant, whether the cylinder is near
> the center of the platter or on the outside. If the media is
> unifor (the maximum storage density is the same anywhere on
> the platter), and you're interested in maximising the
> storage, then the lower storage density used on the outer
> cylinders represents wasted area. The maximum bit density is
> determined by the physical length of the inner cylinder.
>
> Here's the storage possible on a platter of radius R in
> which bits are stored at maximum density (assume = 1
> bit/area)
>
> bits = pi*R^2
>
> Let's derive the storage on a platter in which the max
> density is determined by the inner cylinder of radius r.
>
> number of bits on inner cylinder
>       = 2*pi*r
>
> This number of bits is the same on all cylinders, so the
> total storage is
>
>         2*pi*r(R-r)
>
> The maximum storage possible is 0 for r=0, r=R and has a
> maximum at r=R/2. The storage relative to the platter at
> constant density is
>
>         2*pi*(R/2)^2/pi*R^2=1/2
>
> The maximum number of bits that can be stored on a disk
> operating at constant data rate and constant angular
> velocity is half that of the platter operating at maximum
> storage density (I'm surprised that it's that high) and
> occurs when the innermost track is at r=R/2.
>
> If this is correct, then vinyl records and CDs etc should
> all have their inner most track at r=R/2. However inspection
> of these media shows that the inner track is much closer to
> the center of the platter and would indicate that the amount
> of storage used on the platter is sub optimal.
>
> What am I missing?
>
> Thanks Joe

I didn't take the time to really think about it, but one thing you're missing 
is this equation:
2*pi*r(R-r)

That's wrong -- if the inner radius is 0 that implies no bits -- clearly 
wrong.

If tracks per radial inch is constant and K is some constant, then for outer 
radius R and inner radius r, the number of bits recorded is:

K*pi*R^2 - K*pi*r^2 = K*pi*(R^2 - r^2)

But wait, there's more. You're saying that on CDs and vinyl records the outer 
tracks are less dense, and  bits per radian are constant. In that case, if we 
assume every track has C bits, and there are Z tracks per radial centimeter, 
then bits is represented like this:

(R-r)*Z*C

When you did the area calculation, that was valid only if bits are constant 
over tangential lengths, not if they are constant over radial angles.

HTH

SteveT


-- 
Steve Litt
Recession Relief Package
http://www.recession-relief.US
Twitter: http://www.twitter.com/stevelitt