To find the equations for different grid widths I used the algebra method shown above.
8 width:
n + (n – 8) + (n – 16) + (n – 15) + (n – 17)
= 5n – (8 + 16 + 15 + 17)
= 5n – 56
9 width: 5n – 63
10 width:
n + (n – 10) + (n – 20) + (n – 19) + (n – 21)
= 5n – (10 + 20 + 19 + 21)
= 5n – 70
11 width:
n + (n – 11) + (n – 22) + (n – 21) + (n – 23)
= 5n – (11 + 22 + 21 + 23)
= 5n – 77
12 width:
n + (n – 12) + (n – 24) + (n – 23) + (n – 25)
= 5n – (12 + 24 + 23 + 25)
= 5n – 84
This can be put into a table to solve the equation for a T-shape for any width grid.
w = width
Again we can prove this using algebra:
n + (n – w) + (n - 2w) + (n – 2w + 1) + (n – 2w – 1)
= 5n – (7w + 1 – 1)
= 5n – 7w
Now I have the equation for finding the T-total of this T shape on any grid I can move on to the next and final task. This was to investigate the relationship between the T-total when using a transformation or combination of transformations.
There would be no point in studying both rotation AND reflection because a diagonal reflection is the same as a 90º rotation and horizontal the same as 180º.
For this same reason there is no point in studying reflection AND translation as vertical reflection is a translation right or left.
I have also decided not to do enlargements a they pose a difficulty when trying to deduce a T-number e.g.
As you can see there is now two different possibilities for the T-number and that is with only a scale factor of 2. It would also not be possible to do anything with a scale factor that is not an integer.
I would be possible to take the average between the two possible T-numbers but when you get into decimals it becomes complicated to spot patterns.
Because of this I have decided to study rotation and translation. The first step was to find the equation for finding the T-total of a rotation any where on any grid width.
n + (n - 1) + (n – 2) + (n – 2 - w) + (n – 2 + w)
= 5n – 7
The w’s cancel each other out. There is still 5 numbers though hence the 5n.
n + (n + w) + (n + 2w) + (n + 2w – 1) + (n + 2w + 1)
= 5n + 7w
All the numbers are bigger than the T-number so you must ad something after multiplying it by 5.
n + (n + 1) + (n + 2) + (n + 2 + w) + (n + 2 – w)
= 5n + 7
Again the w’s cancel each other out and as the majority of numbers are valued above the T-number multiplying the T-number by five will not be enough. That is why you add 7.
The next step is to move on to a combination of transformations, which means a rotation THEN a translation.
To work systematically using vectors you have to do (1) then (2)
(1) (2)
etc.
For convenience reasons I am going to start in the top left corner and move across 1 and down one each time.
This diagram shows a rotation of 90º then a translation with a vector of (1)
(1)
The T-number has gone from 10 to 20 we need to find the relation between this and the vector.
If I do a rotation of 90º and then a translation with vector (2)
(2)
The T-number goes from 10 to 30.
Now we already know that for every 1 square you move to the right you add on 1 and for every row you move down you add one grid width. This means that we can use the vector to find the new T-number and simply insert this into the normal equation for a 90º rotation.
We can use algebra if we substitute the top vector for x and the bottom for y.
The top vector represents how many squares you move right or left. As the grids always read like a book if you move right you get a positive vector, move left and you get a negative vector. The vector shows how many you have to add on or take away from the original T-number to find the new one. This means we have the first part of the equation (n + x).
It is not complete yet, however because the number changes if you move up or down. This change is relative to the grid width. If the shape moves up, y is positive, yet you have to minus a grid width. Move down, y is negative, and you add one. This gives you (n + x – yw).
In full the equation for finding the T-total for a rotation of 90º and THEN a translation of any vector is:
5(n + x – yw) + 7 = t
To prove this I will take a 90º rotation ad then a random translation.
The translation has a vector of (4)
(-2)
This means that x = 4, y = -2 and n = 10. The grid width (w) is 9.
5(n + x – yw) + 7
= 5(10 + 4 – (-2 x 9) + 7
= 5(10 + 4 + 18) + 7
= (5 x 32) + 7
= 160 + 7
= 167
This can be proved by doing the long calculation for the T-total of the new rotation and translation of the shape.
32 + 33 + 34 + 25 + 43 = 167
In conclusion you have to know what to add on to the 5n or 5(n + x – yw) to get the T-total. This is different for every rotation and there is no way of knowing. You can remember it quite easily though. 7w for 0º and 180º. 7 for 90º and 270º. Whether to add or subtract his is simple to deduce. If you times the T-number by 5 and all the other numbers are less than the T-number you will get an answer too high s you will have to minus something. And visa versa.
Whilst doing this investigation I tried to work systematically and clearly to prove my points thoroughly. I am satisfied with my answers and would like to use the extra knowledge I have gained in years to come.