My next step would be to see if the formula works on a larger grid sizes, as I would like to find out if this is a general rule for any shape grid however large or however smaller. The same algebraic method will help me distinguish this.
I would like to start of with a grid larger than 4 x 4 but not too large as starting of with a very large grid may confuse the results.
I have decided a 7 x 7 grid will be appropriate for me to investigate my theory further.
7x7 grid
My t-number is – 16
My t-total is – 31
Using the same rule I will substitute these values into the formula.
5t – 7x (grid number)
5 x 16 = 80
7 x 7 = 49
80 – 49 = 31
Now I will try the formula on a different part of the 7 x 7 grid this time I have chosen:
T-number – 20
T- Total – 51
Using the same rule I will substitute these values into the formula
5t – 7x (grid number)
5 x 20 = 100
7 x 7 = 49
100 – 49 = 51
I will try the same formula once more on an 11 x 11 grid to make sure it works
On even larger grids.
T-total = 1+2+3+13+24 = 43
T-number = 24
The formula is: 5t – 7x (grid number)
So by substituting my own values into the formula I get:
5 x 24 =120
7 x 11= 77 120 – 77 = 43
The formula has been proven to work on the 11 x 11 grid.
Now I will investigate if It still works when put in different places on the grid.
My t-number will be: 28
The t-total will be: 63
The formula: 5t – 7x (grid number
So when I substitute the values I get:
5 x 28 = 140
7 x 11 = 77 140 – 77 = 63
The formula has been proven to work how ever big or small the grid and also in different positions upon the different size grids.
PART 3
Investigating relationships between the t-total and the t-number, and the grid size. This will also include translations.
I am first going to try and rotate the t-number 180 degrees so it is upside down.
I will do this as we already have a formula for the t-total (5t-63). I have used a 9x9 grid as this is the grid which the formula works on.
The t-shape I am going to use has a t number of:
T- number: 2
The old formula stated that we had to minus 63 away from 5 x the t-number but as this is upside down (rotated 180 degree) I am going to try and add the 63.
So 5 x t number makes 10 and + 63 = 73
To see if the formula has worked I will add all the numbers in the t-shape:
2+11+19+20+21 =73 THIS NOW PROVES THAT THE MINUS SIGN
HAS WORKED
We have now found a formula for both the upright and upside down T- shape.
The next step is to move the shape on its side (90 degrees from the upright position) I move the t-number 90 degrees to the left.
This time we also keep the same kind of formula and also change the minus number.
We work out the formula by finding out the differences between the numbers in the t-shape:
12-1 =11
12-10= 2
12-19= -7
12-11 = 1
TOTAL = 7
To check if the formula is 100 percent correct I substitute it into the expression:
(5 x 12) – 7 = t- total
(t- number = 12)
T-total = 1 +10 +19 +11 +12 = 53
If we rotated the t-shape 180 degrees, the same will happen, as what happened when the t-shape was turned 180 degrees from it is first original position.
The formula for this t-shape must be:
5t-number +7 = t – total as the one above is the opposite.
I will now test this formula:
5 x 70 + 7 = 357
T-number = 70
T-total = 70+71+72+63+81 = 357
This formula works for all grid sizes and anywhere on the grid.
CONCLUSION
During this project we have learned how to find data and summaries it into smaller parts.
We have learnt to find and make rules and have put them into practice. This project has taught us how to make algebraic rules which can be put into use.
Here are the various rules we found out:
PART 1
T-total is 5t – 63
(This only works for 9x9 grid anywhere)
PART 2
5tn – 7x(grid number)
(This works for all grid sizes anywhere.
PART 3
Normal way up – 5t – 7x
90 degrees to the left – 5t – 7 x
90 degrees to the right- 5t + 7 x
180 degrees from upright : 5t + 7 x
