And the actual “T”-total I was trying to find. 72 – 37 = 35.
So I tried to add this number instead, to the previous “T”-total. 30 +35 = 65> But it still did not give me the number I wanted (72). However there was only a difference 7. So I added 7 to the formula and came up with the formula “T”-total=pt + 35 + 7 for the first rotation in the 5 by 5 grid.
“T”-total = pt + 35 + 7
=30 + 35 + 7
= 72 this was correct, so I tried to use this formula on the 6 by 6 grid.
“T”-total = pt + 35 + 7
=33 + 35 + 7
=75 which is not equal to 82 the next “T”-total in the sequence. However I found that there was a difference of 7 between 82 and 75. I then looked at the formula I had got from my earlier investigation, “translating “T”-shapes horizontally” and found that the number you add to the previous “T”-total to get the next “T”-total in the sequence e.g. in the 5 by 5 grid 35 is actually equal to 7g where g is the grid size.
7 * 5 = 35 and in the 6 by 6 grid 7 * 6 =42. I then tried the formula
“T”-total = pt + ( 7g + 7 ) to calculate the next “T”-total in the first rotations of the 6 by 6 and 7 by 7 grids.
“T”-total = pt + ( 7g + 7 )
= 33 + ( 42 + 7 )
= 33 + 49
= 82
“T”-total = pt + ( 7g + 7 )
= 36 + ( 7g + 7 )
= 36 + ( 49 + 7 )
= 36 + 56
=92.
This formula worked for both grids so I concluded that this was the correct formula for the first rotation.
Now I needed to find the formula for the second rotation. From the information I had from the tables I had drawn I realised that the difference between the “T”-totals of the second rotation was less than the difference between the “T”-totals of the first rotation. So I subtracted the difference between the “T”-totals of the second rotation in the 6 by 6 grid (35)from the difference between the “T”-totals of the first rotation of the 6 by 6 grid (49) (refer to diagram 1B). 49 – 35 = 14. After I found that there was a difference of 14 I knew that there was there was a difference of 7 between 7g which for the 7 by 7 grid is 49 and 35.This was because I had added 7 to 42 to find the formula for the first rotation. So if you imagine 35, 42 and 49 on a number line. Then there is a difference of 14 between 35 and 49 and 42 being the mid-point between the two numbers. Then there is a difference of 7 between 42 and 49 and a difference of 7 between 42 and 35(refer to diagram 2). So I tried to put this number in my formula for the second rotation.
“T”-total = pt + ( 7g –7 )
= 82 + ( 42 – 7 )
= 82 + 35
= 117 this was correct so I tried to use the same formula on the 7 by 7.
“T”-total = pt + ( 7g – 7 )
= 92 + ( 49 – 7 )
= 92 + 42
= 134 this was also correct so I concluded that this was the formula for the second rotation.
When I was looking for the formula for the third rotation I realised that the difference between the “T”-totals of the third rotation was the same as the difference of the first rotation except that instead of adding you subtract. E.g. in the 7 by 7 grid the difference between the “T”-totals of the first rotation is 56 and the difference between the “T”-totals of the third rotation is –56 (refer to diagram 3). So instead of adding (7g + 7) which in the 7 by 7 grid is equal to 56, to the previous “T”-total as I had done in the first rotation to get the next “T”-total for the first rotation, I subtracted (7g + 7) from the previous “T”-total to get the next “T”-total in the sequence for the third rotation. This is because if the next “T”-total in the sequence is –56 from the previous “T”-total in the sequence then if I subtract ( 7g + 7 ) which is equal to 56 in 7 by 7 grid from the previous “T”-total in the sequence then I should get the same answer. So I tried this formula.
“T”-total = pt – ( 7g + 7 )
=134 – ( 49 + 7 )
= 134 –56
=78 this is correct so I tried to use the same formula on the 8 by 8.
“T”-total = pt – ( 7g + 7 )
= 151 – ( 56 + 7 )
= 151 – 63
= 88 this was correct so I concluded that this was the correct formula for the third rotation.
When I was looking for the formula for the fourth rotation I realised that just like the first and third rotation. The difference between the “T”-totals of the second and fourth rotations where the same except that instead of adding 42 you subtract 42 in the 7 by 7 grid ( refer to diagram 3). So as I had done to find the formula for the third rotation I subtracted ( 7g – 7) from the previous “T”-total to get the next “T”-total in the sequence. So I tried the formula
“T”-total = pt - ( 7g – 7 )
= 78 – ( 49 – 7 )
= 78 – 42
=36 this was correct so I tried this formula on the 8 by 8 grid.
“T”-total = pt – ( 7g – 7 )
=88 – ( 56 – 7 )
=88 - 49
= 39
Now I have found the formulae for all the rotations where pt = previous “T”-total in the and g = grid size
“T”-total = pt + ( 7g + 7 ) for the first rotation.
“T”-total = pt + ( 7g – 7 ) for the second rotation.
“T”-total = pt – ( 7g + 7 ) for the third rotation rotation.
“T”- total = pt – ( 7g – 7 ) for the fourth rotation.
( refer to diagram 4 )