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#1
Posted to sci.electronics.repair




Coax cable calculation
Here's a question that stumped me back in the old slide rule days:
A coaxial cable, having an inner diameter of 0.0254mm (0.001") and using an insulator with a dielectric constant of 2.56, is yo have a characteristic impedance of 2,000 Ohms. What must be the outer conductor diameter? You guys can locate the appropriate method of calculation.  Trevor Wilson www.rageaudio.com.au 
#2
Posted to sci.electronics.repair




Coax cable calculation
On Tuesday, April 10, 2018 at 2:42:50 AM UTC4, Trevor Wilson wrote:
Here's a question that stumped me back in the old slide rule days: A coaxial cable, having an inner diameter of 0.0254mm (0.001") and using an insulator with a dielectric constant of 2.56, is yo have a characteristic impedance of 2,000 Ohms. What must be the outer conductor diameter? You guys can locate the appropriate method of calculation.  Trevor Wilson www.rageaudio.com.au It's pretty hard to get impedances above that of free space. (377 ohms) An air gap would be better. 
#3
Posted to sci.electronics.repair




Coax cable calculation
On Tue, 10 Apr 2018 16:42:43 +1000, Trevor Wilson
wrote: Here's a question that stumped me back in the old slide rule days: A coaxial cable, having an inner diameter of 0.0254mm (0.001") and using an insulator with a dielectric constant of 2.56, is yo have a characteristic impedance of 2,000 Ohms. What must be the outer conductor diameter? https://www.rfcables.org/coaxcalculator.html  Select inches.  Plug in any number for outside dia (D) trial an error.  Plug in 0.001 for inner diameter (d).  Plug in 2.56 for permittivity (er) (same as dielectric constant in this example).  Punch "calculate" I used 1,000,000 inches for outside diameter and still obtained only 776 ohms. 1,000,000,000 inches yielded 1035 ohms. I don't think it's going to make it to 2,000 ohms unless the coax cable is absurdly huge.  Jeff Liebermann 150 Felker St #D http://www.LearnByDestroying.com Santa Cruz CA 95060 http://802.11junk.com Skype: JeffLiebermann AE6KS 8313362558 
#4
Posted to sci.electronics.repair




Coax cable calculation
On Tue, 10 Apr 2018 08:11:37 0700, Jeff Liebermann wrote:
On Tue, 10 Apr 2018 16:42:43 +1000, Trevor Wilson wrote: Here's a question that stumped me back in the old slide rule days: A coaxial cable, having an inner diameter of 0.0254mm (0.001") and using an insulator with a dielectric constant of 2.56, is yo have a characteristic impedance of 2,000 Ohms. What must be the outer conductor diameter? https://www.rfcables.org/coaxcalculator.html  Select inches.  Plug in any number for outside dia (D) trial an error.  Plug in 0.001 for inner diameter (d).  Plug in 2.56 for permittivity (er) (same as dielectric constant in this example).  Punch "calculate" I used 1,000,000 inches for outside diameter and still obtained only 776 ohms. 1,000,000,000 inches yielded 1035 ohms. I don't think it's going to make it to 2,000 ohms unless the coax cable is absurdly huge. Almost sounds like an OLD ("slide rule") homework problem.... Jonesy 
#5
Posted to sci.electronics.repair




Coax cable calculation
On 11/04/2018 8:10 AM, Allodoxaphobia wrote:
On Tue, 10 Apr 2018 08:11:37 0700, Jeff Liebermann wrote: On Tue, 10 Apr 2018 16:42:43 +1000, Trevor Wilson wrote: Here's a question that stumped me back in the old slide rule days: A coaxial cable, having an inner diameter of 0.0254mm (0.001") and using an insulator with a dielectric constant of 2.56, is yo have a characteristic impedance of 2,000 Ohms. What must be the outer conductor diameter? https://www.rfcables.org/coaxcalculator.html  Select inches.  Plug in any number for outside dia (D) trial an error.  Plug in 0.001 for inner diameter (d).  Plug in 2.56 for permittivity (er) (same as dielectric constant in this example).  Punch "calculate" I used 1,000,000 inches for outside diameter and still obtained only 776 ohms. 1,000,000,000 inches yielded 1035 ohms. I don't think it's going to make it to 2,000 ohms unless the coax cable is absurdly huge. Almost sounds like an OLD ("slide rule") homework problem.... **Correct. It was. I thought someone would have come up with an answer before now. It's not an overly difficult calculation.  Trevor Wilson www.rageaudio.com.au 
#6
Posted to sci.electronics.repair




Coax cable calculation
In article ,
Trevor Wilson wrote: Here's a question that stumped me back in the old slide rule days: A coaxial cable, having an inner diameter of 0.0254mm (0.001") and using an insulator with a dielectric constant of 2.56, is yo have a characteristic impedance of 2,000 Ohms. What must be the outer conductor diameter? Almost sounds like an OLD ("slide rule") homework problem.... **Correct. It was. I thought someone would have come up with an answer before now. It's not an overly difficult calculation. Well, just for grins... using the formula Z0 = 138 log10 ((Dd/Dc)*(1/sqrt(eR))) it seems to transform to Dd = 10^(Z0/138) * sqrt(eR) * Dc with Z0 = 2000, eR = 2.56, and Dc = .001, we'd get approximately Dd = 10^14.5 * 1.6 * .001 Dd = 1.6 * 10^11.5 Dd = 5.1 * 10^11 or about half a trillion inches. That's if I didn't screw up something using my slide rule (yes, I have a nice 10" Hemmi here at my desk) or slip a decimal point somewhere along the way. This seems a bit impractical for a coaxial cable. The transmissionline cutoff frequency would be ridiculously low, and construction of a matching section would be, well, interesting, to say the least. And, I rather doubt that the classic formula is actually applicable for impedances (and diameters) this large. A solid cylinder of dielectric material this large would probably insist on collapsing under its own gravitational selfattraction, leaving you with a neutron star or a black hole. 
#7
Posted to sci.electronics.repair




Coax cable calculation
On 11/04/2018 9:42 AM, Dave Platt wrote:
In article , Trevor Wilson wrote: Here's a question that stumped me back in the old slide rule days: A coaxial cable, having an inner diameter of 0.0254mm (0.001") and using an insulator with a dielectric constant of 2.56, is yo have a characteristic impedance of 2,000 Ohms. What must be the outer conductor diameter? Almost sounds like an OLD ("slide rule") homework problem.... **Correct. It was. I thought someone would have come up with an answer before now. It's not an overly difficult calculation. Well, just for grins... using the formula Z0 = 138 log10 ((Dd/Dc)*(1/sqrt(eR))) it seems to transform to Dd = 10^(Z0/138) * sqrt(eR) * Dc with Z0 = 2000, eR = 2.56, and Dc = .001, we'd get approximately Dd = 10^14.5 * 1.6 * .001 Dd = 1.6 * 10^11.5 Dd = 5.1 * 10^11 or about half a trillion inches. That's if I didn't screw up something using my slide rule (yes, I have a nice 10" Hemmi here at my desk) or slip a decimal point somewhere along the way. This seems a bit impractical for a coaxial cable. The transmissionline cutoff frequency would be ridiculously low, and construction of a matching section would be, well, interesting, to say the least. And, I rather doubt that the classic formula is actually applicable for impedances (and diameters) this large. A solid cylinder of dielectric material this large would probably insist on collapsing under its own gravitational selfattraction, leaving you with a neutron star or a black hole. **According to my calculations (performed ca. 1973) the answer is around 1.59 X 10^20 inches. My instructor ticked my answer as the correct one. Or, for a more useful figure, 427 light years. That's one big coax. A few hundred Metres would likely require more mass in it's construction than exists in our galaxy (maybe the entire universe). I haven't tried to work that out.  Trevor Wilson www.rageaudio.com.au 
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