Marvel Posted July 20, 2003 Share Posted July 20, 2003 Does anyone know how to figure the effective gauge of using a pair of wires as one wire, i.e., using a twisted pair of 20 gauge as a single conductor? Using the cross sectional area of 20 gauge wire and doubling it (two wires), I get close to the same cross sectional area as a 17 gauge wire. Does this make sense? Marvel Quote Link to comment Share on other sites More sharing options...
lynnm Posted July 20, 2003 Share Posted July 20, 2003 I assume that according to Ohms Law you are in effect paralleling two resistances and therefore winding up with a resistance that is lower than a single run of AWG20. I suspect based on your previous posts that you know the formula but I'll post it here for other users: For Series Connection: R1+R2=Rt 8 ohms + 8 ohms = 16 ohms For Parallel Connection: R1XR2/R1+R2 = Rt Thus (8x8)/8+8 = 64/16 = 4 That works rather neatly when one is dealing with equivalent values but just for the hello of it let's look at a resistance of 6 ohms and another of 1700 ohms. Series: 6+1700 = 1706 Parallel: 6 X 1700 = 10200/1706 = 5.97 The above calculations are based on DC. in other words 0 cps. The A.C. impedance varies with frequency. Getting back to Marvel's original question I would assume that the DC. resistance would be a little lower than the resistance presented by a single line. That said - twisting the wires would undoubtably introduce a degree of inductance which would impact on the impedance as "seen" by the system. In other words the DC resistance would for all intents and purposes be 1/2 of the resistance of a single run. The AC impedance would definitely be impacted but I don't have sufficient background in electronics or math to say whether that impact would help or hinder the system's performance. My gut feeling is that unless we are talking about exceptionally long runs of cable the difference will amount to: (0 X 0)/ (0 +0) = 0 Quote Link to comment Share on other sites More sharing options...
Malcolm Posted July 20, 2003 Share Posted July 20, 2003 Marvel: You got it right. Lynn: WTF? Quote Link to comment Share on other sites More sharing options...
tpg Posted July 20, 2003 Share Posted July 20, 2003 I understood what Lynn said... just am trying to figure out how it was relevant to paralleling cable... I have not tested, but I am assuming you are suggesting that the formula for paralleling resistances will also work for wire guages... I did some tests with some special CAT5 cable on my system. Results were not pleasing to my ears. I have two lengths of a special CAT5 that has only 5 conductors, four of which are shielded 18-guage solid copper, the last is stranded and tinned bare 16-guage copper. Had the 18-guage strands been stranded, the experiment would have gone better. A tube amp maker I know also makes interconnects and speaker wire using stranded pieces of 18-guage wire. He said that paralleling two 18-guage wires produces a "single" 15-guage wire. Then, adding another (3 parallel) makes 12.5-guage. 4 akes 10-guage (I think)... and so on. With my experiment, I used two paralleled 18-guage solid core wires paralleled for signal and for ground, also frounding the tinned, unshielded wire (which was actually connected to the cable's shield). What was produced was great bass (tight, detailed), crisp and clear lower mids, but in the upper mid/treble, everything fella part... highs were almost inaudible. I then thought, use 2x18-guage solid+ 1x24-guage solid wire for the highs... but it was just too much trouble. Quote Link to comment Share on other sites More sharing options...
lynnm Posted July 20, 2003 Share Posted July 20, 2003 Malcom "Marvel: You got it right. Lynn: WTF?" What is it about my response that you have trouble with ? I admit that I went into more detail than some would see as necessary but is that necessarily a bad thing ? I have seen a lot of evidence in my travels here that indicates that many enthusiasts are a little fuzzy regarding the issues surrounding basic electronics and where possible I have attempted to clarify matters. If that is unacceptable to you and a significant portion of the others on the forum I will STFU and leave technical commentary to those better qualified. Quote Link to comment Share on other sites More sharing options...
Malcolm Posted July 20, 2003 Share Posted July 20, 2003 Just wondering what your response had to do with the question asked. He asked about cross sectional area, not resistance. I used to work with a guy who did what you did all the time. He was very entertaining, actually. However, your explanation was, as usual, quite clear. Keep up the good work! Quote Link to comment Share on other sites More sharing options...
lynnm Posted July 21, 2003 Share Posted July 21, 2003 My assumption was that Marvel was asking a question as to what the equivalent cross -sectional area of a pair of AWG 20 wires would be. I further suspected that he might be wondering whether there might be some theoretical reasons to choose one approach over the other. Given that I could not answer that question in an absolutely specific manner my choices were to respond as I did or simply to keep my thoughts to myself. I happen to believe that speculation and or interprolation are legitimate components within a conversation. There is no line item in the rules of use for this site that requires absolute certainty regarding factual matters or any item that states that speculation is prohibited. Quote Link to comment Share on other sites More sharing options...
Marvel Posted July 21, 2003 Author Share Posted July 21, 2003 It is actually a practical application. Our department is building lecterns for classrooms -- the lecterns have a PC, dvd player and VCR built in. One already built also has an amplifier. We are going to use a multiconductor cable with a MIL-SPEC connector on it so our faculty only have two things to plug in (power and i/o to the components). It won't be a long cable, but we are having to run the speaker outs on the one lectern through the snake. It has 20 gauge wire. I want to use a complete pair for each conductor for the speaker wires. Doubling the wire gives me 17 gauge, not as heavy as I would like, but certainly not as bad as it could be. For the power requirements and speaker volume needed in a classroom, I don't think the 17gauge will cause issues. As a further explanation: Ohms for 1000 ft of copper at 68 degrees F: 20 gauge 0.0008023 ohms 17 gauge 0.001609 Almost double for the gauge, but being in parallel, it may/would drop to 0.0004012 ohms I don't really think this is going to be a problem, as the overall effect on the speaker load won't really be much of an issue. Marvel Quote Link to comment Share on other sites More sharing options...
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