Finally, I will find the formula for the total number of joints in a cube. As I have taken n to be the number of rods per side, this must be turned to the number of joints per side. This is simply (n + 1). Then, if I cube this figure, it actually gives me the correct answer for the total joints in a cube of a certain size. So the formula is simply:
t = (n + 1)³
I have had to add 1 to n, as I have taken n to be the number of RODS per edge of the cube. I am trying to find the total JOINTS. As there is one less rod per edge than joints, adding 1 to n changes it to the number of joints per edge. This must be cubed to find the total joints. As squaring it would find a face of the cube, and there are the equivalent of (n + 1) faces worth of joints in a cube, so (n+1)² must be multiplied by (n + 1), to get the total joints, or simply put as being cubed.
I now have all my formulae for the cube problem. I must now prove that my formula work. To do this, I must put all of my formulae for each joint type in a large sum. This will start off as shown below:
t = 8 + 12 x (n - 1) + 6 x (n - 1)² + (n - 1)³
I must remove all the brackets. The first bracket would change from 12 x (n - 1) to 12n - 12. The second bracket would have to be expanded from 6 x (n – 1)². Firstly I will expand the brackets. I do this by converting (n – 1)² to (n – 1) x (n – 1). I can then remove the brackets by multiplying everything in one bracket by everything in the second bracket. This comes up with the result (n – 1)(n – 1) = n² - 2n + 1. 6 must now multiply all figures in this sum, this end up being 6n² - 12n + 12. The final bracket is then done as above, with the exception of multiplying everything by 6, and then multiplying the sum by (n – 1). So, we have the sum
(n² - 2n + 6)(n – 1). We expand out the brackets by multiplying all figures in one bracket by the figures in the other bracket to finally give n³ - 3n² + 3n – 1. We have now expanded out all the brackets to give us a final sum of:
t = 8 + 12n – 12 + 6n² - 12n + 6 + n³ - 3n² + 3n – 1
Some of these figures cancel each other out. These are highlighted below:
t = 8 + 12n – 12 + 6n² - 12n + 6 + n³ - 3n² + 3n – 1
We can instantly remove these from the sum, now simplify the sum. This eventually comes to:
t = n³ + 3n² + 3n + 1
Now, to finally prove our formula works, we must expand out the sum for t that we found before. If these two sums are the same, my formulae all work, and my cube problem will be finished. The current sum we have for the total is (n + 1)³. If we expand out the brackets, we must first get (n + 1)(n + 1). We expand the brackets as before to get the answer n² + 2n + 1. We must then expand the final (n + 1). We get the problem:
(n² + 2n + 1)(n + 1) = n³ + 2n² + n + n² + 2n + 1
This answer can be simplified to:
n³ + 3n² + 3n + 1
This result for t expanded, is the same as the answer for each of the types of joint added together. This is the result I expected, and am pleased with the result as it shows I have found the correct formulae for the entire cube.
Now that I have proven my formulae work, I will test them on the next size cube, this will be a 5 x 5 x 5 cube. Here are my results.
c = 8
e = 12 x (5 – 1) = 12 x 4 = 48
y = 6 x (5 – 1)² = 6 x 4² = 96
X = (5 – 1)³ = 4³ = 64
t = (5 + 1)³ = 6³ = 216
8 + 48 + 96 + 64 = 216
These are the results I found. A picture of a 5 x 5 x 5 cube has been drawn, and is on isometric paper. I manually counted the number of each joint in this cube; they came to the same totals as shown above.
Extension (1)
I have completed my original problem, and will now extend it. I will find the same type of joints, except I will vary the size of one of the lengths. This will create a square-faced cuboid.
As in the previous investigation, I will be looking for the 3-Joints, 4-Joints, 5-Joints and 6-Joints. To start off I will draw the first four square-faced cuboids. These are a 1 x 1 x 2 cuboid, a 2 x 2 x 3 cuboid, a 3 x 3 x 4 cuboid and a 4 x 4 x 5 cuboid. These are drawn on isometric paper.
I manually counted the number of each joint type in each cuboid and have put them in a similar table as to the one in the previous investigation:
I now have my results, so I will now find and explain the formulae for each joint type. Firstly I will do the 3-Joint. After looking at my results I notice that, as before, all the answers are 8. Therefore the formula is again:
c = 8
Like a cube, a cuboid can only have eight corners, and as we know 3-Joints only occur on corners, then there will ALWAYS be eight 3-Joints on any sized cuboid. This is the reason for the above formula.
I will now find and explain the formula for the 4-Joints. The formula for the 4-Joints in the cube was 12(n – 1). This was because there are 12 edges on a cube. As we know, 4-Joints only occur on edges, and as the corners are taken up with 3-Joints, we must convert n and m to joints by adding 1, then minus 2 to make up for the corner pieces. This means we will have (n - 1) and (m – 1) in our formula. As there are also 12 edges on a cuboid, then the numbers used for multiplying by will add up to 12 also. Eight of the edges are equal to n, and 4 of the edges are equal to m. So, 8 will be in the formula by the n, and 4 will be by m. Our formula ends up being:
e = 8(n – 1)4(m – 1)
As I’ve previously explained, I have taken n and m to be the number of RODS on a side, not joints, so they must be converted to joints, as I did with the cube. This is simply done by adding 1 and subtracting 2, or just bye subtracting 1. We also know that eight edges on a square-faced cuboid are equal to n, while four edges are equal to m. As there are twelve edges on a cuboid, and as 8 + 4 = 12, then we are correct with these numbers.
Now I will find the formula for the 5-Joints in a square-faced cuboid. As in the cube’s formula for y, we must multiply each different axis by each other. As there are only two different lengths, there will be a (n – 1)² and a ((m-1)(n-1)). In the 5-Joint formula for the cube, it has a number 6, which represents the number of sides on a cube. As a cuboid also has 6 sides, but of different sides, then there will be two numbers in the formula, which will add up to 6. I predict that as only two sides have n as both axis, and four sides have the length of m, then the formula will be:
y = 2(n – 1)²4((m – 1)(n – 1))
As in most of the previous formula, we have had to convert n and/or m from rods to joints, this is simply done by subtracting 1 from either n and/or m. This explains why there is an (n – 1) and an (m – 1) in the formula. The first (n – 1) must be squared as this is one of the edges of one of the square faces of the cuboid, and as both axis are the same, rather than writing (n – 1)(n – 1), it is easier to write (n – 1)². (m – 1) must be multiplied by (n – 1) as this will find all of the 5-Joints on a rectangular face. As there are two square faces and four rectangular faces, then there must be 2(n – 1)² for the ends, and 4((n – 1)(m – 1)) for the sides. This is the explanation for the formula for the 5-Joints (y).
I will now find the formula for the total number of 6-Joints in a square-faced cuboid. We notice that the number of 6-Joints in a cuboid, is the same size as the total joints in the cuboid one size down, then we know all we have to do is minus 1 off both n, n and m and multiply all of these by each other. If we do this, we get the formula (n – 1)(n – 1)(m – 1). In better algebraic form, this can be written as:
X = (n – 1)²(m – 1)
As I have previously explained, the number of 6-Joints in any square-faced cuboid or cube, is the equivalent to the total number of joints to the structure, which is the next size down. To find the total number of joints, all we have to do is multiply the number of joints in each length by each other ((n +1)²(m + 1)). We just have to do this, but by subtracting 2 off each edge of joints to account for the corners, this gives us the above formula.
Finally I will find out the formula for the total number of joints in a square-faced cuboid. This formula for the cube was simply (n+1)³. This was because we had to turn the n from the number of rods per edge into the number of joints per edge; adding on 1 did this. We then had to multiply n by n by n; cubing it easily did this. If we take these rules to the cuboid we get:
t = (n + 1)²(m + 1)
n + 1 is squared as there are two edges which use n as their length, only one edge uses m. I have explained how I came to the above formula in the above paragraph.
As I now have all the formula for the square-faced cuboid problem, I will prove that they all work. This is done as I have done it in the cube problem. I will simply add together all the formulae for the different joints, and see if the result comes to the same answer as the total joint formula after having been expanded. If I take all my joints formulae, and put them in a sum, this is what I get:
t = 8 + 8 x (n – 1) + 4 x (m – 1) + 2 x (n – 1)² + 4 x ((m – 1) x (n – 1)) + (n – 1)² x (m – 1)
I must now remove all the brackets from the sum. I will explain how the more complex ones were done as I do them:
8 x (n – 1) = 8n – 8
4 x (m – 1) = 4m – 4
2 x (n – 1)² = 2n² - 4n + 2
4 x ((m – 1) x (n – 1)) = 4 x (nm – n – m + 1) = 4nm – 4n – 4m + 4
(n – 1)² x (m - 1) = (n² - 2n + 1)(m – 1) = n²m – 2nm + m – n² + 2n - 1
Now I have successfully removed all the brackets, I can put a final sum of:
t = 8 + 8n – 8 + 4m – 4 + 2n² - 4n + 2 + 4nm – 4n – 4m + 4 + n²m – 2nm + m – n² + 2n - 1
As in the cube, some of these numbers cancel out each other. These cancellations are shown below:
t = 8 + 8n – 8 + 4m – 4 + 2n² - 4n + 2 + 4nm – 4n – 4m + 4 + n²m – 2nm + m – n² + 2n - 1
We are then left with the formula t = n² + 1 + 2nm + n²m + m + 2n. In a better algebraic order it is written as:
t = n²m + n² + 2nm + 2n + m + 1
Now, to test that our formulae work, I will expand the formula for the total number of joints, which is (n + 1)²(m + 1). I will do this as shown below.
t = (n + 1)² x (m + 1)
= (n² + 2n + 1) x (m + 1)
= n²m + 2nm + m + n² + 2n + 1
= n²m + n² + 2nm + 2n + m + 1
After having expanded the formula for the total joints in a square-faced cuboid, I notice that both the formula for all the different joint types added together and the formula for the total joints expanded are the same. This proves that my formulae work. Now that I have actually proven my formulae work, I will test them on the next size square-faced cuboid, this will be a 5 x 5 x 6 cuboid. I will take n to be 5, and m to be 6.
Extension (2)
I will now extend my investigation further. I have had one variable and two variables. I will now have three variables. My cuboid will always have three different length edges (e.g. 4 x 5 x 6 etc.). As in the previous investigations, I will start by drawing the first four simplest cuboids, and manually count the amount of the four different joint types and the total joints. These will be a 1 x 2 x 3 cuboid, a 2 x 3 x 4 cuboid, a 3 x 4 x 5 cuboid and a 4 x 5 x 6 cuboid. Pictures of these are shown on the isometric paper included.
I shall now attempt to find and explain the formulae for each joint type in this cuboid type. Firstly I will do the 3-Joints. As before, for all four cuboids, there are eight 3-Joints. This means that the formulae is the same as before:
c = 8
This formula is as above for the same reason as in the previous two investigations. As there can only be 8 corners on any cuboid or cube, and as 3-Joints only occur on corners, then the number of 3-Joints will always be 8.
I shall now find and explain the formulae for the 4-Joints. As in the previous two extensions, the numbers used to multiply anything in brackets add up to 12.This allows me to predict that, as there is an equal number edges for each value (n, m and p), that everything in brackets will be multiplied by 4. I also notice that in each formulae for the 4-Joints, the different values always have 1 subtracted off them, so I predict that the formulae is:
e = 4(n – 1)4(m – 1)4(p – 1)
I will now explain the above formula. As we know, 4-Joints only occur on edges, as previously explained, and as there are 12 edges on a cuboid, then the numbers used for multiplying by, will add up to 12. As there are an equal number of edges which are the length of each value, then 12 must be divided by 3 (for the three different values n, m and p) and put in front of each edge length, which gives us 4 in front of each. As I have taken n, m and p to be the number of rods per edge, we must convert these to the number of joints per edge. As I have explained in the previous investigations, subtracting 1 off each value does this. This explains why each value is followed by “- 1”.
Now, I will find the formulae for the 5-Joints. After looking at the 5-Joint formulae for the previous structures, I notice that each variable must be multiplied by all other available variables, after having being converted to the number of joints on each edge (minus 1). These figures must also be multiplied by a number, which when added to the other two numbers used for multiplying equals 6. Therefore I predict the formulae is:
y = 2((n – 1)(m – 1)) + 2((n – 1)(p – 1)) + 2((m – 1)(p – 1))
I will now explain the formula. All three values are taken in rods per edge. I must always convert these to the number of joints per edge, simply by subtracting 1 off each. All values must be multiplied by a different value, as this finds out the number of 5-Joints on a face (e.g. (n – 1) x (m – 1)). As there are always 2 of each sized side, then each side must be multiplied by 2. This rule applies to all three parts of the formulae.
I shall now find the formulae for the number of 6-Joints in this type of cuboid. I notice in the last two formulae for this, 1 must be subtracted off each value, and then the new values must be multiplied together. Therefore I predict that the formulae for this is:
X = (n – 1)(m – 1)(p – 1)
As I have described in the last two investigations, 6-Joints only occur in the middle of a cube/cuboid, and are the equivalent to the total joints in the cube/cuboid of the next size below. The total number of joints is simply each value, plus 1 (to convert it into joints per edge) multiplied by each other (this gives, in this cuboid, (n + 1)(m + 1)(p + 1)). We simply have to subtract two off each value in the formulae for the total joints; this accounts for the corner joints being added on. Then we take our new figures and multiply them by each other.
Finally, I will find the total joints in this type of cuboid. As I have looked at the previous formulae for the total joints in a cube/cuboid, I notice that all we have to do is convert each of our three values into the number of JOINTS on an edge, and multiply the resulting three figures by each other. In this case, this gives:
t = (m + 1)(n + 1)(p + 1)
Each value must have 1 added onto it to convert it from the number of rods per edge, to the number of joints per edge. We then simply multiply our three new figures by each other. This is all that makes the total joints formula.
I will now prove that all my formulae work. As I have done in the previous investigations, I will add up the formulae for the four different types of joint, and expand the formula for the total joints. If the answers are the same, my formulae work.
t = 8+4x(n-1)+4x(m-1)+4x(p-1)+2x((m-1)x(n-1))+2x((p-1)x(n-1))+
2x((p-1)x(m-1))+(n+1)x(m+1)x(p+1)
After having expanded the brackets out of this particular formula, and then simplifying it, it gave me the resulting formula:
t = nmp+pm+pn+nm+p+m+n+1
I will now expand the formula I have for the total joints in this cuboid:
t = (n+1)x(m+1)x(p+1)
= (nm+m+n+1)x(p+1)
= nmp+pm+pn+p+nm+m+n+1
As the formulae have ended up as the same, this proves my formulae work.
Finding out the total Rods Formula
In this part of my investigation, I will find a formula, which will use n, m and p to find the total number of rods in a cube or cuboid. As I have noticed in my previous investigations, the formulae for the 3 variable cuboid work on all the shapes, so I will find it’s formulae, and these will then be able to work on any other structure. Firstly, I will find the amount of rods on the largest face of the cuboid. m will represent the longest edge of the cuboid, n the shortest edge and p the other edge. An example of one of these is shown on the included isometric paper.
On the largest edge of the cuboid, I notice, that if I multiply m with (p plus1), this will give me all the vertical rods on that face. If I swap m and p, this gives me the formula px(m+1). This formula gives me the total number of horizontal rods on the largest face. As there are always one more of these sized ‘faces’ in any cuboid than n, then our two previous formulae must be added together, and multiplied by (n+1). This gives us the formulae:
(m(p+1)+p(m+1))(n+1)
This formula totals the number of vertical and horizontal formulae in any cuboid or cube. Now we must find all the rods, which are in-line with n. To the side of the largest face, the number of rods going backwards is always one more than the number of vertical rods in-line with m. Therefore in our formula somewhere is (m+1). At the top of the largest face, there is also 1 more rod, than the number of horizontal rods in-line with p. This proves that in the end of our formula there will be (p+1). If I multiply (m+1) with (p+1), I get the number of rods in-line with n, between two of these large ‘faces’. So, if I multiply my result by n, this will give me the total number of rods in-line with n (n(m+1)(p+1)). So final formula therefore is:
TR = (m(p+1)+p(m+1))(n+1)+n(m+1)(p+1)
As I stated before, this formula will work on a square-faced cuboid, and a cube. If it needs to be used on a square-faced cuboid, all you need to do is substitute p with n. If you need to use it on a cube, you just need to replace both p and m with n. The reason that 1 is added onto the first p, is because you always have one more vertical rod going across than p. This reason applies to all the values which need to have 1 added onto them. I have explained the formula in the above paragraphs.
