The totals are the same so my formula seems to work.
Translation
I will now try and figure out the relationship between translation and the T Total. I will translate a T by x units to the right and y units down. I will represent this by the vector. I will use S to represent grid size and use T to represent Old T Total. I will try and find a formula for the new T Total from the old T Total.
(T+7S)/5-7S-1+x+yS+( T+7S)/5-7S +x+yS+( T+7S)/5-7S +1+x+yS+( T+7S)/5-S+x+yS+( T+7S)/5+x+yS=
5(T+7S)/5-7S+5x+5(yS)=
(T+7S)-7S+5x+5(yS)=
T+5x+5(yS)= New T Total
I used (T+7S)/5 because this turns the T Total into the T number of the T. I add x because that is the amount the T moves horizontally. I added yS because y is the amount the T moved vertically and if it moves vertically then for every space it will move down, the grid size is added, which is S. I made an example to check whether this works.
I will use translation of, a grid size of 9 and a T total of 37.
The resulting T total is 22+23+24+32+41= 142.
Using my equation for a translated t total, New T total= T+5x+5(yS), for a T total of 37, a grid size of 9 and translation of then the equation will be T=37+15+90= 142 which the same as the T total on the grid, so my formula works.
Another way of doing this is by representing the number in each square by a letter.
Again I will use to represent the vector movement, S for the grid size T for the new T Total.
a+x+yS+ b+x+yS+ c+x+yS+ d+x+yS+ e+x+yS=
a+b+c+d+e+5x+5yS= New T Total
a+b+c+d+e is equal to the old T Total which I will call T so the formula is
T+5x+5yS= New T Total
This gives the same equation as in the previous result so I know works as I showed it in the previous result.
Rotation
I will now try and work out the relationship between the T number and the T total when it is rotated about the original T number.
n= T number
S= Grid size
T= New T total
I will try a 90º clockwise rotation
n+n+1+n+2+n+2-S+n+2+s= 5n+7
T=5n+7
This is the T for a 180º clockwise rotation
n+n+S+n+2S-1+n+2S+n+2S+1=7n+7S
T= 7n+7S
This is the T for a 270º clockwise rotation
n+n-1+n-2+n-2-S+n-2+S= 7n-7
I can show that these formulas work by drawing a table.
The formula for the 90º clockwise rotation is T= 5n+7 so the T number is 20 so the T total would be (5×20) +7= 107. From the table I can see that the T total is 25+26+27+17+37= 107 so my formula works.
The formula for the 180º clockwise rotation is T= 7n+7S so the T number is 20 so the T total would be (7×20) + (7×10)= 210. From the table I can see that the T total is 25+35+45+46+44= 210 so my formula works.
The formula for the 270º rotation clockwise is T= 5n-7 so the T number is 20 so the T total would be (5×20) -7= 93. From the table I can see that the T total is 25+24+23+13+33= 107 so my formula works.
I will now try and figure out the relationship between the T number and the T total in a 90º clockwise rotation about a point on the table.
T number= n
New T total= T
Grid Size= S
Horizontal distance from the T number
To the point of rotation= x
Vertical distance from the T number
To the point of rotation = y
= Distance from the point of rotation
= Rotation Point
= The T number
To figure it out I would have to
I can prove this by drawing a grid
I will now try a rotation like the last one but 180º clockwise
n+2yS+2xs+n+2yS+2x+S+ n+2yS+2x+2S+ n+2yS+2x+2S-1+ n+2yS+2x+2S+1=
5n+10yS+10x+7S=T
T=5n+10(yS+x) +7S
To show that it works, I made an example.
I will now make a T for a rotation of 270º clockwise in the same format as the last one
n+yS+x+xS-y+n+yS+x+xS-y-1+n+yS+x+xS-y-2+n+yS+x+xS-y-2-S+ n+yS+x+xS-y-2+S=
5n+5yS+5x+5xS-5y-7= T
T= 5(n+yS+x+xS-y)-7
To show that it works, I made an example.
Limits on T
