Twisted-shielded wire lengths 2

[nbourg8]

This is my first post and if someone can answer this, I'll be long-time user of the forum

Situation:
Everyone knows as you twist a pair of wires together, their total lengths will shorten. Also the number of twists and the diameter of the wire are crucial in how much it will shorten.

Problem:
I am searching for an equation relating final, shortened, length to the starting length, number of twists, and guage of the wire.

Givens:
Gauge, starting length of wires, and number of twists.
If it helps:
Gauge 16 AWG
200 m long wire
1 twist per 2 or 4 inches.

THANKS!

 

[VxA]

My guess would be that the new length would be equal to the original length (200m) multiplied by the ratio of the twist length (2" or 4" -- whatever you're going to use) and the sum of the twist length and 2*pi*Diameter of the conductor (#16AWG).

NewL = OldL * [TwistL / (TwistL + 2*pi*D)]

For example, if you have 200m (200,000mm) to start with, and your wire has an outside diameter of 1mm and your going to have one complete twist for every 2" (50.8mm), then the new length would be:

New length = 200,000*[50.8 / (50.8 + 2*3.14159*1)]mm
= 177,986mm = 178m

I guess, you'd still need a design check on that calc, but that's the way I see it anyhow.

Thanks,

VxA

 

[nbourg8]

Okay, I love you guys, I posted 11 minutes ago and already got an accurate answer. Thanks a bunch VxA.

Note: This doesn't mean I have checked it yet, the post is still active incase anyone has any further input on the subject

 

[VxA]

Hey no probs,

The formula I gave assumes you have tight wisting (i.e., the wires are always in contact).

Thanks,

VxA

 

[nbourg8]

Curiosity killed the cat, but I'm willing to risk a cats life so I ask. How did you know to use that method to solve the problem? I have been searching math forums and such for helix complex math stuff and you just skipped over all that.

 

[VxA]

I just figured for every complete twist, one cable will have to circle the other cable one complete time.

Say we have two wires A and B and they are both the same size diameter. Now, you can picture the center of wire A wire making a complete circle around wire B for each twist. The radius of that circle will be equal to the sum of the radii for each wire. Seeign as the wires have the same diameter (and hence the same radius) then the radius of the 'circle' is 2 times the radius of either of the wires. Therefore the diamter is 4 tiems the radius of one of the wires (or twice the diameter). So for each twist, you take away the circumference of a circle with a diameter = to 4 times the radius of one conductor. The total number of twists made will be equal to the final length divided by the twist length.

So Final Length = Old Length - 2*pi*D*[Final Length/Twist Length]

solve the equation and you get the forumla I posted.

Again though, it's possible my logic is wrong. So don't take it as gospel.

Thanks,

VxA

 

[nbourg8]

Makes sense, I have found something similar to this problem but its a different approach, which I believe to be more correct than yours. Please verify this though.

http://mathforum.org/library/drmath/view/53155.html

 

[cswilson]

VxA:

Admirable analysis, but I think you missed one thing. I don't think you should just subtract the "added distance" -- the 2*Pi*D*(FinalLength/TwistLength) term -- from the Old Length. Rather, I think you should treat it as the short side of a right triangle whose hypotenuse is OldLength. Think of unwrapping the twist and laying the resulting straightened wire on a diagonal.

The equation would then be:

FinalLength = sqrt(OldLength^2 - AddedDistance^2)

Curt Wilson
Delta Tau Data Systems

 

[nbourg8]

However, you then have two FinalLength terms on both sides of the equation.

 

[VxA]

If that's the case, I think the formula would reduce to:

NewL = OldL * [TwistL / (TwistL + 4*pi^2*Dwire^2)]
using the example of 200m of cable, 1mm wire diameter, an a twist length of 50mm, this yields a length of 198.43m.

Looking at it closer, I think this is probably the correct approach (the one taken by cswilson and the "dr. math"). Besides, I wouldn't want to argue with someone called "Dr. Math"

It seesm like very little loss in length for so many twists, but I guess that is the glory of the hypotenuse.

 

[nbourg8]

Sorry for not catching on to this, fresh eyes always help.

I still am having trouble incorporating more than 1 twist into Dr. Math's method of doing this problem. Can anyone explain this further please.

 

[nbourg8]

Well I have come up with a spreadsheet showing the relationships of all this in feet and in meters, including how much percent of the line was taken away.

http://www.yousendit.com/transfer.php?action=download&ufid=4A0E40A84DBF5E1D

"Interesting" how 1 twist per inch does not affect it much. I had to hand twist a pair of 16 gauge at 2 inches per twist and it was very difficult. Never really noticed their lengths tho.

 

[nbourg8]

Okay so this was accurate, I just had not considered any thickness of the sleeve the cable sits in. This made a huge difference when accounted for