How do you terminate 750 Kcmil fine stranded cable with 1080 strands using NEC listed lugs made for 61 strands?

It seems to me that the solution is to use the lug which fits the cable. I don't think the nominal size of the lug has any bearing. Lots of cables have been trimmed to fit smaller connections and I image 99.9% of them are still working satisfactorily.

I would not trust any formula to accurately determine cable diameter since there are variables that are not known, such as the tightness and angle of the twist.

Don Young

The formula I am looking for is to find the diameter of a circle that inscribes a known number of smaller circles each with the same diameter. It appears that with an increase in the number of smaller circles the greater the ratio is of the circumscribed circle diameter squared to the sum of the smaller circles diameters squared. In other words, for a given circular mil area (cma), the diameter of a circumscribed circle in thousandths of an inch squared (cmil area) grows as the number of smaller circles increases. The sum of the smaller circles circular mil areas being equal to the given cma.

The correct lugs for these cables are available from Burndy and other lug manufacturers. They are used all the time in telecommunications. In the Burndy catalog 'Code' wire is the low strand such as you describe. There are separate sections for the flex. You will also have to know the details of the lug attachment hole(s) as to bolt size and spacing if there are two holes (usual for 750 kcm).

Major flexible cable manufacturers such as Cobra Cable stock these lugs.

Graybar is a major electrical distributor covering most of the USA that carries some of these lugs.

Once you find a lug part number, try eBay.

Bill Kaszeta Photovoltaic Resources Int'l Tempe Arizona USA snipped-for-privacy@pvri-removethis.biz

That is what the engineer is proposing - changing the lugs. But this demonstrates that when ever products outside the normal NEC are used, problems arise. But on the formula. It appears for one strand the circumscribed circle diameter ratio to the strand diameter would be 1:1. For two strands the ratio would be 2:1. For three to infinity strands it appears the ratio would go from 2:1 to 1;1. The problem is what would the ratio be for three strands, four strands, five strands.....etc? There must be a mthematical relationship so the ratios can be calculated.

According to equation 4-13 on page 4-20 of "Standard Handbook for Electrical Engineers, 11th Edition" edited by Fink and Beatty , the diameter of the circumscribing circle D, given n layers over the core wire, each wire of diameter d, is D = d(2n+k), where k = 1 for a single wire core and 2.155 for a

3-wire core. I haven't derived this myself (should be possible with a little high-school algebra) and I hope I haven't made a typo.

Let's check another table in the Handbook: It says a 19-strand #1 AWG conductor is made up of 19 wires each 0.0664 inches diameter, in 2 layers over the 1-strand core), so k=1, n=2, d=0.0664, and the formula says D=5d, which is 0.332 inches; and the table says 0.332 inches, so it's all consistent. The tricky part is that n isn't the number of wires, but the number of layers of wire over the core - what's the name for the sequence 1,3,7,19...?

There is also a table on the same page which shows that a stranded cable is between 15% and 20% bigger in diameter than a solid wire with the same total copper cross section. The worst case is 12 strands, everything 19 strands and up is around 15 %.

I can highly recommend this book, I use it regularly, though often just for the copper wire tables.

Bill

Thank you, I will check this out. I used to have an old copy of the EE handbook; maybe it is still in a box.

I think the densest you can pack circles is in a hexagon packing. So each strand actually occupies an area of the hexagon that the circular strand fits in.

If the diameter of a strand is X, then the area of a hexagon that encircles such a strand is 1.5*X. So if you have 1080 strands of diameter X, the total area of perfectly packed strands would be about 1620*X. A circle that has that much area would be 45.4*X. (D= sqrt( 1620*X / (pi/4))

BUT, this doesn't account for skewing of the strands due to the twist. I suspect that wouldn't add very much though unless you have a short twist.

daestrom

A straight cross section would make for a slight oval shape per strand that is in twist around the core. And the angle would be greater for the same twist pitch for more outward conductors. Think of a conductor as wide as the twist pitch to realize this. Of course the twist pitch would be much greater than the conductor diameter. That is an interesting geometry problem to consider.

I wrote a program some time back that figured out numbers of hexagonal lattice points inside a circle for all steps in circle width. I'll have to go find it and see if I can update it for this kind of geometry. Note that this program was to figure out lattices kept inside a circle even though the arrangement was a hexagonal structure. One thing I remember I figured out from it is that the largest hexagon that would fit inside a circle and would break out of that circle with any points added was one of a side of 7 points. Larger than that and you can start adding points at the middle of the sides and still be inside the circle. But this was done with points, not finite width circles or with slight ovals (it's purpose was for hexagonal rather than quadrature lattice data encoding confined well to a circle as opposed to the usual encodings that stay square).