Dismiss
InnovationQ will be updated on Sunday, Oct. 22, from 10am ET - noon. You may experience brief service interruptions during that time.
Browse Prior Art Database

Stiction Force Calculation Method

IP.com Disclosure Number: IPCOM000107302D
Original Publication Date: 1992-Feb-01
Included in the Prior Art Database: 2005-Mar-21
Document File: 4 page(s) / 109K

IBM

Related People

Aoyagi, A: AUTHOR [+4]

Abstract

A program is disclosed where stiction of HDA which is assembled with 2 or more heads is estimated with the average and standard deviation of stiction force of the one head to some disk surfaces.

This text was extracted from an ASCII text file.
This is the abbreviated version, containing approximately 52% of the total text.

Stiction Force Calculation Method

A program is disclosed where stiction of HDA which is
assembled with 2 or more heads is estimated with the average and
standard deviation of stiction force of the one head to some disk
surfaces.

As usual, the component level data of stiction is obtained in
the early phase of development.  This simulation can estimate the
stiction distribution of 2 or more heads and disks assembled in HDA
using stiction data in component level.

One HDA assembled with one disk and two heads is considered
here.  Fig. 1(a) shows a 2-head and 1-disk model. Stiction forces
which occur on the interfaces of Head #1 and Head #2 are referred to
as F1 and F2 (F1 >= F2), respectively.  When a motor starts up, the
force applied to each head is expressed as follows.
F = ( T / R ) / 2                                   (1)
where T and R are motor torque and head rest radius, respectively.

In the case of F1 <= 2 * F2, when the force applied to
head #2 is more than the stiction force which occurs on the interface
of head #2 and disk surface, and head #2 moves to the disk surface
relatively, here the force F by motor torque is the same as the
stiction force F2 of head #2. Then the force F by motor torque which
the stiction force F1 is less than 2 * F2, the force applied to head
#1 becomes larger than the stiction force F1, so head #1 can move to
the disk surface relatively.  Thus, the force which is necessary to
2 * F2 (2)

In case of F1 > 2 * F2, when the force F by motor
torque is the same as the stiction force F2 of head #2, head #2 can
move to the disk surface and then double force is applied to head #1.
However, head #1 cannot move to the disk surface, as stiction force
F1 is more than stiction force 2 * F2.  In this case, head #1 can
move to the disk surface, if the force F by motor torque reaches to
the stiction force F1 to head #1.

Thus, as mentioned above, the force that is necessary to break
two heads to each disk interface i...