On Sun, 07 Dec 2008 19:22:55 -0500, Joseph Gwinn
wrote:

In article ,
(dan) wrote:

What's that Lassie? You say that Joseph Gwinn fell down the old
rec.crafts.metalworking mine and will die if we don't mount a rescue
by Sun, 07 Dec 2008 13:03:03 -0500:

The classic approach is two photocells (or one split photocell) feeding
a differential amplifier. If the line is halfway between the cells,
output is zero. It the line strays one way the output is negative, and
the other way positive. To do full X-Y, there will be four cells in a
square, and two differential amplifiers.

This was enough for the gantry servo to steer by.

Do you mean that's what rotated the pickup head? And the position of
the pickup head controlled the X and Y for the gantry?

Or was the head stationary? With a stationary head, what would
determine witch direction to go on the line?

The head would be stationary, and all adjustment would be electronic.

Let's first deal with one axis at a time.

There are two rectangular cells side by side. If the line to be
followed is parallel to the boundary between cells, then the
differential amplifier output will be signed and proportional to the
offset of the line from the border. If the line is instead
perpendicular to the border, the output of both cells will be reduced,
but their difference will remain zero.

Now, add a second pair of cells, so we have four cells total laid next
to one another like tiles in a square pattern. Let us number the tiles
by row and column:

S11 S12

S21 S22

We have two differential amplifiers. The X amplifier has plus inputs
from S12 and S22, and minus inputs from S11 and S21. The Y amplifier
has plus inputs from S11 and S12, and minus inputs from S21 and S22.

In math:

X output = +(S12 + S22) -(S11 + S21)

Y output = +(S11 + S12) -(S21 + S22)

It's clear what will happen if the follow line is vertical or
horizontal, so let's consider the case of a diagonal line crossing the
centers of S21 and S12. If perfectly centered, X and Y outputs are both
zero. If the line drifts towards S22, what happens? S12 and S21 will
remain about the same, while S22 will grow and S11 will shrink. This
will cause X output to become more positive, while the Y output becomes
more negative.

One can go through this exercise for any line orientation, and get the
same answer, so the X and Y outputs provide a sufficient steering signal.

Joe Gwinn

That gets displacement error signals for X and Y, but provides no
motion in the direction of the line, only correction for offsets
perpendicular to it.

I think there must be some means for maintaining set speed in a
direction parallel to the line. This could be done with a form of
difference engine, or it might use more straighforward computation of
Vx^2 + Vy^2 and Vy/Vx. The difference engine might require a start
at known speed and known direction. Could have been analog or
digital. Microprocessors have been around since the mid-'70's, CNC
since the late '50s, and electronic, mechanical and even pneumatic
ways of doing such computations have been known since World War II.