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

Coax cable calculation
 

On Tuesday, April 10, 2018 at 2:42:50 AM UTC-4, 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.

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/coax-calculator.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 831-336-2558

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/coax-calculator.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

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/coax-calculator.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

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
transmission-line 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 self-attraction,
leaving you with a neutron star or a black hole.

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
transmission-line 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 self-attraction,
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