# Formula, footage of cable **URL:** \ **Category:** Factual Questions\ **Created:** [December 13, 2002, 6:34pm UTC](https://boards.straightdope.com/t/formula-footage-of-cable/142563 "2002-12-13T18:34:10Z")\ **Posts on this page:** 5\ **Page:** 1
**Author:** ![Dan\_Tana](https://avatars.discourse-cdn.com/v4/letter/d/4af34b/32.png) [@Dan\_Tana](https://boards.straightdope.com/u/Dan_Tana)\ **Post date:** [December 13, 2002, 6:34pm UTC](https://boards.straightdope.com/t/formula-footage-of-cable/142563/1 "2002-12-13T18:34:10Z")
Does anyone know of a formula that can be used to determine the length of cable that is on a spool? Of course the Spool dimensions(Radius, width etc…) and diameter of the cable will be a “known”. Thanks in advance for any help. ---
**Author:** ![GaryM](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/garym/32/241_2.png) [@GaryM](https://boards.straightdope.com/u/GaryM)\ **Post date:** [December 13, 2002, 7:52pm UTC](https://boards.straightdope.com/t/formula-footage-of-cable/142563/2 "2002-12-13T19:52:54Z")
There are electronic devices that you can hook up and determine the “electrical length” of the cable. This number, multiplied by the cables velocity factor gives the physical length. There is one built into my antenna analyzer. Alternativly, if you know the cable weight per foot, and the weight of the empty spool, all you need is a pencil and paper. ---
**Author:** ![sailor](https://avatars.discourse-cdn.com/v4/letter/s/a587f6/32.png) [@sailor](https://boards.straightdope.com/u/sailor)\ **Post date:** [December 13, 2002, 9:57pm UTC](https://boards.straightdope.com/t/formula-footage-of-cable/142563/3 "2002-12-13T21:57:45Z")
That’s fairly simple to calculate if you are talking of _theoretical_ length. In other words, if every turn is side by side and every layer neatly wrapped. But this only happens with thick wires. Now that I think of it we had a similar thread not long ago which turned into quite a deep discussion but I forget exactly what started it. I remember discussing the approximation of considering a spiral or a circle etc. Anyway, give us a bit more info on what you are trying to do. ---
**Author:** ![sailor](https://avatars.discourse-cdn.com/v4/letter/s/a587f6/32.png) [@sailor](https://boards.straightdope.com/u/sailor)\ **Post date:** [December 13, 2002, 10:11pm UTC](https://boards.straightdope.com/t/formula-footage-of-cable/142563/4 "2002-12-13T22:11:04Z")
A quick solution: let’s call the wire diameter “d”, the outside spool radius “R”, the inside spool radius “r” and the spool width “w”, all in the same units. If you have a long spiral starting from the center and forming a solid circle of wire, the area “A” equals pi()\*R^2. Now remove the central part to allow for the spool core and you have a total area A’ = A - a = pi() \* (R^2 - r^2) That is the area of one layer of spiral and the length of wire it contains equals A’/d Now the number of spirals or layers is w/d so the solution is that the total length of wire is L = A’/ d \* w / d = w \* A’/ d^2 = pi() \* w \* (R^2-r^2) / d^2 Unless I have made some mistake along the way, which I may have. And I hope this time I do not get dragged into a long argument like last time. ---
**Author:** ![octothorpe](https://avatars.discourse-cdn.com/v4/letter/o/b77776/32.png) [@octothorpe](https://boards.straightdope.com/u/octothorpe)\ **Post date:** [December 14, 2002, 12:38am UTC](https://boards.straightdope.com/t/formula-footage-of-cable/142563/5 "2002-12-14T00:38:28Z")
You didn’t state what type of cable you have (i.e. wire rope, insulated electrical cable, etc.), but in the case of wire rope the ‘reel capacity’ chart [here](http://www.kulkonihouston.com/Catalog/html/pg011/pg011.htm) should get you close. For other cables substitute (0.2618/d[sup]2[/sup]) for K where d=diameter of cable. I believe this formula was originally published by the **Roebling Wire Rope Company** , but pretty much every wire rope supplier i’ve seen uses this formula. As **sailor** stated, the tighter and more neatly wrapped, the more accurate the answer. This should give you fair estimate. #