How many clips do I require in any run of armoured cable?
One of the classic exam-type questions is testing knowledge on the calculation for how many clips or straps are required for a specific run of cable, such as armored cable.
This is one of those questions that is a bit of a contradiction to what goes on at the job site. In the real world we would order a number of boxes of clips based on simple division of the length of the run of cable. Typically, we would use the formula of the length of the run divided by 1.5 m. This calculation is very practical and we’d end up having a clip every 1.5 m. We would also install a clip comfortably close to the boxes on each end for the insulation to follow Code. As usual, there will be some clips left over that we would use for the next job.
Having said that, when this very question is found on an exam then we electricians have to change our approach to the way we answer the question. So, let’s take a look at a classic example.
How many clips are required for a 14 m run of armored cable between a submain switch and a distribution panel?
a) 1
b) 9
c) 10
d) 11
As usual, let’s start with consulting the Code. Since the armored cable Rule 12-618 tells us to follow the non-metallic sheathed cable Rule 12-510, we know that both cables, as far as clipping distances are concerned, are identical.
The Rule tells us clips are required:
- within 300mm of every box or fitting
- spaces of maximum 1.5 m along the installation run
This is interesting because the Code uses millimeters and meters for measurements, so let’s use only one – meters.
At first glance, the Rule advises us to install a clip within 0.3 m of the box or enclosure at each end- that’s two clips and then divide the remainder of the length of the run by 1.5 m, but is that true every time? in every run? in every situation? The answer is NO, and that’s the trick in this question.
As in understanding any motor control sequence we work left to right and top to bottom – logical and in a straight line. That means:
- place the first clip within 0.3 meters of the first enclosure on the left
- divide the remaining length of the run by 1.5 m
- if leftover is 0.3 m or more, add one extra clip
In our case of this question: (1rst clip at 0.3 m) + (14 -.3 = 13.7 / 1.5 = 9.13) the leftover was 0.13m, which is less than 0.3 m and therefore that last clip was already accounted for, and installed within the 0.3 m of the box. So, therefore the extra clip was not necessary which brings our answer choice to c).
It’s a great trick question when it’s not frustrating!