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Ed Huntress Ed Huntress is offline
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Default hard black plastic?


"Don Foreman" wrote in message
...
On Fri, 24 Oct 2008 11:47:37 -0400, "Ed Huntress"
wrote:


wrote in message
...
On Oct 24, 10:30 am, Don Foreman
wrote:
On Fri, 24 Oct 2008 08:21:30 -0500, "Karl Townsend"

wrote:
I need to make some new idler wheels for my wire EDM. The material is a
very
hard black plastic. Maybe the same stuff you see on a car electrical
distributor. It needs to be non-conductive and be able to have a light
press
fit for a ball bearing.

Can somebody give me a material name and source to order?

Karl

The most economical way to produce things like this is to mold them.
However, that isn't necessarily the only way, nor the best way if you
don't happen to have the molds.

I'd make an undersized aluminum wheel, press that into an insulating
bushing made of Delryn, Noryl, phenolic or whatever, and then possibly
press that assy into a metal "tire" that would resist wear. The
resulting wheel would be non-conductive from rim to center and would
be quite robust.


What voltage?

On the order of several hundred volts, at very high impedance. If there's
any conductivity, the sparks won't initiate and the machine won't cut.


Ed, this assertion doesn't nearly meet your usual high standard of
editorial rigor.


I'm not following you. What I described is the way it works. You have a high
voltage at high impedance, and that voltage ionizes the channel. When it's
ionized and current starts to flow (the "spark"), the voltage drops to a
very low value, to reflect the low ohmic resistance of the ionized channel.

In an RC relaxation circuit, like most very old EDMs used, the open-circuit
voltage is high. In a somewhat newer electronic circuit, the low- and
high-impedance circuits actually are separate. A Sodick EDM of about 1980
vintage actually has three circuits.

If there is leakage in the high-impedance circuit, you won't have enough
voltage (it will drop in the high-impedance circuit because of the parallel
resistance of the leaky element of the circuit) to ionize the channel. Thus,
no spark will be able to initiate.

Is this editorially sufficient? g

--
Ed Huntress