SPIRAL LENGTH

By Mark O. Worthington

Posting date August 5, 2004

The Physical Length of the CD, DVD Spiral

An expression can be resolved by converting the polar representation of the Spiral of Archamedes equation to rectangular coordinates.

For a spiral of constant thickness or pitch – Length = Pi * R^2/ T

For the donut geometry of an optical disc we can approximate;

Length = Pi * ( R(outer) ^2 – R(inner) ^2 )

Track Pitch

CD Path Length;

Router = 59.mm

Rinner = 24.8mm

TP = 1.5 – 1.7 um

Length = Pi * (3481 – 615)/ .0015 = 6,002,536 mm

So,

1 mile = 1,604,344 mm , (CD Track Length = 3.7414 Miles)

Pit/Land Lengths are 3..11 clock cycles or (0.8-3.0 ums) with an average length of 2 um.

Modulated marks on a CD = 6,002,536 mm / 2.0um = 3,001,268,000

For DDCD;

For DDCD TP = 1.2

Length = Pi * (3481-615)/ .0012 = 7,503,170 mm

(DDCD Track Length = 4.6767 Miles)

Modulated marks on a DDCD = 7,503,170 mm / 2.0 um = 3,751,585,227

DVD Path Length; *** (Opposite Track Path discs may have a smaller outer diameter)

For DVD-5;

Router = 59 mm

Rinner = 23.6 mm

TP= 0.73 – 0.75 um

Length = Pi * (3481-557) / .00073 = 12,583,585 mm

So,

(DVD-5 Track Length = 7.8434 Miles)

Pits and Lands on a DVD are 3…11 clock cycles (14T for sync) or 0.4 – 1.84 ums. Average Length is approx ~ 0.72um

Modulated marks on a DVD-5 = 12,583,585 mm / 0.72 um = 17, 477,201,390

For DVD– 9/10, 2 X Physical Density of DVD-5 (Parallel Track Path) ***

Length = 25, 167,170 mm

So,

DVD-9/10 Track Length = 15.6868 Miles

Pits and Lands on a DVD-9/10 are ~ 0.44 – 2.0 um. Average Length ~ 0.78

Modulated marks on a DVD-9/10 = 25, 167,170 mm / 0.78 um = 32,265,602,560

For DVD-18 , 2X Physical Density of DVD-9/10 (Parallel Track Path)***

Length = 50,334,340 mm

So,

DVD-18 Track Length = 31.3736 Miles

Modulated marks on a DVD-18 = 50,334,340 mm/ 0.78um = 64,531,205,130

Copyright © Mark Worthington 2004