I finally looked at my discharge setup, into which I had put some research (re: our recent discussion). I used an 800K 25W resistor. The reason I used a wirewound resistor is that a small carbon resistor can arc inside and fry, leaving you with an open circuit, so when you think you're discharging, nothing is being drained. V=IR, right? Assuming a tube voltage of 10,000V: V/R=I, 10000/800000 = .0125 So that's .0125A, or 12.5mA. But at 10000V: P=IV, or P=(.0125)(10000)=125W That's a whole lot of watts. But as we discharge that voltage very quickly, over a wirewound resistor, that's not too bad. I used a new screwdriver with a heavy handle and a 7" metal shank terminating in a narrow flat tip. I used a 16 gauge lamp (I used both wires) cord with heavy insulation, wrapping the wire around the screwdriver for a good mechanical connection soldering the wire to the metal shank (I chose a screwdriver with a frosted finish for better adherence) and then tied the cord to the handle with a cable tie, and then used multiple layers of heavy heatshrink over the solder joint to serve as a strain relief. I put the resistor inline, with multiple levels of heat shrink over the ends, wrapping the upper levels of heatshrink onto the resistor body to prevent lead fatigue and failure. I then used two final layers of heatshrink on top of the entire resistor portion of the cable, both to insulate the resistor and provide additional strain relief. Silicone-insulated high-voltage test lead wire would be a better choice, but I did not have any at the time. I would always measure the resistance of the discharge lead before and after use, so as to be certain the cable did not fail. And remember that the portion of the wire between the screwdriver and the resistor is at the anode cap voltage potential during discharge, so avoid contact with it and inspect it for damage. Treat it like an expensive test tool, and if the resistor or wiring becomes bent or damaged, throw it away and make a new one. -Chris