I did notice the replication of the posts. I deleted them, but I
don't know why google repeated them so.

To make it easier on me to understand, I have updated the image with
numbers, so now it is a matter of matching numbers (I hope). I think
this is always the difficulties with tech support on the phone or
online because it is really hard to visualize with words sometimes.

http://davidd.250free.com/resistor3.gif

I tried to read through what you had written and got a little
confused, so this might make it easier.

OK. Assuming that the wire to the left of your battery pack is "-" and
the
one to the right is "+", then it connects pretty much as you have it
lying
there. The short leg of the LED goes straight to battery "-". The battery
"+" wire goes to the right-hand tag of the three that are together on the
pot, as we are looking at it on the picture. The centre and left tags
should
be joined together. You then need a fixed resistor of say 22 ohms
connected
between the joined-together pot tags, and the long lead on the LED. This
resistor is important, as it will limit the maximum current that the LED
can
draw with the pot turned right up. Without it, the LED will burn out.

Ah! OK. Now I can see what the two additional terminals on the pot are. They
are a switch, which presumably goes "click" when you go fully anticlockwise,
yes ?

Yes, it clicks off/on

Assuming yes, then the correct hookup will be right hand side yellow to mid
body righthand tag, as shown. New wire from left hand mid body tag to
righthand tag of the block of three tags, ie the tag where you are currently
showing a red wire on the right. Lefthand red wire where it is currently
shown ie to the centre and left tags that are joined together. Lefthand
yellow wire now irrelevant. Bottom red or yellow wire as currently
indicated.

Now, when the pot is clicked fully anticlockwise, the feed from the battery
to the righthand pot tag, will be broken by the switch, so no current will
flow through the LED, and it will extinguish. Once the switch has clicked
back on, as you rotate clockwise, you will be returned to the 'red' circuit
that worked ok to dim the LED. Do you follow that OK ?

As far as not having any additional series resistor, if that's the way it
was done in the original unit that you canibalised it from, then that's
fine. I would however suggest that you do not use any other battery type
than the originals, as it probably gets away with current limiting by the
internal resistance of this type of cell. Whilst the LED may have its own
internal current limiting, from the photo, it looks like a pretty
bog-standard type, which if you do allow it to draw too much current, will
destroy itself.

Incidentally, your post appeared multiple times on my news server, separated
by a couple of minutes each, so I'm not too sure what happened there.

Arfa