GCSE Maths Coursework – Cables

Introduction

For this assignment I have been requested to study a particular design of steel cables used by engineering companies in construction work. The design consist of four circular cross-sectioned strands (radius 1cm) bonded together with plastic and wrapped around with a fine steel thread to assemble a ‘square’ (i.e. 2 strands by 2 strands) like so:

Within my assignment I have investigated thread lengths and amounts of plastic fill for different sized and ‘squares’, which I throughout exposed various useful patterns that could be used in the industry.

Thread Lengths

To commence this branch of my assignment I have worked out the thread length for the modular cable with four strands shown in diagrams 1 and 2:

X = arc touching the thread

Y= the space between two strands

From looking at the above diagram we can perceive that the length of the thread going all around the cable =

This is because if we look carefully we can see that we have four arcs touching the thread (4x) and four sides each with one space between two strands (4y) and together these make the length of thread going around the cable (4x +4y = thread length).

Knowing that the radius is 1 cm we can work out the value of (y):

Below is a close up diagram of the length (y). The two parts of strands shown are both exactly half a strand with an arrow to show the radius (1cm) from the middle of the strand to the edge of the strand. We can see that the length (y) is equivalent to twice this radius i.e.

We can substitute this value into our previous equation for the value of the thread length:

Now we need to know the value of (x) to conclude.

(4 x ¼ πd) x (12 x 2)=

 πd + 24

 (5x5)                 25

From the table we can make out that as we go further along the square number sequence there is a difference of 8 between the thread lengths at each step. We can find the nth term:

This nth term is exceedingly useful; throughout it the company can work out exactly how much thread they need for any ‘square’ size by simply replacing (n) with the number of the number they want in the square sequence.

From the above information we can see that it not an easy job to actually find the place of the number you want in the sequence (n), but we can make it easier using our knowledge.

Diagram 6 shows us the place of some ‘square’ sizes in the sequence

Now, it is much easier for the company to use the nth term, turn over to see how …

E.g. the company wants to create a cable ‘square’ size 6 strands x 6 strands, how much thread do they need?

Let us check this:

Shapes (regular polygons)

This is quite simple if you have understood the method of finding the thread length of the ‘squares’ we have just studied. You only need a little bit more of maths.To commence, I will show you how to work out the thread lengths of different shapes using the information we have gained previously:

Shape 1; triangle

Assuming that the radius is r cm we can work out the value of (y):

Below is a close up diagram of the length (y). The two parts of strands shown are both exactly half a strand with an arrow to show the radius (r cm) from the middle of the strand to the edge of the strand. We can see that the length (y)

Area of cross section = Area of big square – total area of corners

Substitute: 16 – 0.88 = 15.12

This is the area of the cross-section, the volume is the same number multiplied by the length (L) as I said previously:

15.12 x L = 15.12L

If you remember I made an equation before for the amount of plastic:

The plastic fill = volume of cable – total volume of strands

We worked out the total volume of the strands; 12.56L and now we have the volume of the cable as well; 5.12L so let us substitute:

The plastic fill = 15.12L – 12.56L = 2.56L

Having learnt the method of finding the amount of plastic fill in the cable, I have worked out the amount of plastic fill for several other ‘square’ sizes to see if I can spot any patterns. Here is a record of my results:

Yes, there is clearly a pattern; as we go further along the square number sequence there is a difference of roughly 1.7 between the differences between the amounts of plastic fill. We can work out the nth term, but this time we will have to use another formula that is used to work out the nth term of sequences that have a changing difference like ours. Here it is:

This nth term is as useful as the nth term we found previously; now the company will know exactly how much plastic to buy which will prevent any leftovers and save the company’s money. They will only have to replace (L) by the actual length of the cable their making and (n) with the number of the number they want in the square number sequence.

E.g. the company makes a 10 strands x 10 strands cable of length 1cm, how much plastic do they need to buy?

Nth term = 0.85n²L + 1.74nL – 0.03L

10 x10 = 100 which is the 10th number in the square sequence so n = 10

Length (L) = 1

Substitute:

(0.85 x 10² x1)+ (1.74 x 10 x 1) – (0.03 x 1) =

85 + 17.4 – 0.03 =

102.37 cm³