This proves that the formula was correct.
Size 8 Grid Formulas
Upright T :
900 T:
1800 T:
2700 T:
5n-56 = T-total
5n+7 = T-total
5n+56 = T-total
5n-7 = T-total
Predictions
It seems that, for each of the two formulas, the number that must be added or subtracted from 5n is a multiple of 7. For example, 63 and 56. I will test this in a size 7 grid.
Size 7 Grid
Differences
The differences add up to –49, so I think the formula will be: -
5n-49 = T-total
Testing the Formula
5n-49 = T-total
5(16)-49 = T-total
31 = T-total
Also:
1 + 2 + 3 + 9 +16 = 31
5n-49 = T-total
5(27)-49 = T-total
86 = T-total
Also:
12 + 13 + 14 + 20 + 27 = 86
5n-49 = T-total
5(47)-49 = T-total
186= T-total
Also:
32 + 33 + 34 + 40 + 47 = 186
This proves that the formula is correct.
Rotating T through 1800 clockwise
Differences
The differences add up to +49, so I think the formula will be: -
5n+49 = T-total
Testing Formula
5n+49 = T-total
5(2)+49 = T-total
59 = T-total
Also:
2 + 9 + 15 + 16 + 17 = 59
5n+49 = T-total
5(13)+49 = T-total
114 = T-total
Also:
13 + 20 + 26 + 27 + 28 = 114
5n+49 = T-total
5(33)+49 = T-total
214 = T-total
Also:
33 + 40 + 46 + 47 + 48 = 214
This proves that the formula was correct.
Rotating T through 900 clockwise
Differences
The differences add up to +7, so I think the formula will be: -
5n+7 + T-total
Testing the Formula
5n+7 = T-total
5(8)+7 = T-total
47 = T-total
Also:
3 + 8 + 9 + 10 + 17 = 47
5n+7 = T-total
5(19)+7 = T-total
102 = T-total
Also:
14 + 19 + 20 + 21 + 28 = 102
5n+7 = T-total
5(39)+7 = T-total
202 = T-total
Also:
34 + 39 + 40 + 41 + 48 = 202
This proves that the formula was correct.
Rotating T through 2700 clockwise
Differences
The differences add up to –7, so I think the formula will be: -
5n-7 = T-total
Testing Formula
5n-7 = T-total
5(10)-7 = T-total
43 = T-total
Also:
1+ 8 + 9 + 10 + 15 = 43
5n-7 = T-total
5(21)-7 = T-total
98 = T-total
Also:
12 + 19 + 20 + 21 + 26 = 98
5n-7 = T-total
5(41)-7 = T-total
198 = T-total
Also:
32 + 39 + 40 + 41 + 46 = 198
This proves that the formula was correct.
Size 7 Grid Formulas
Upright T :
900 T:
1800 T:
2700 T:
5n-49 = T-total
5n+7 = T-total
5n+49 = T-total
5n-7 = T-total
My Prediction that the numbers in the formula are multiples of 7 seems to be correct. This means that a general formula for any grid size can be created, by replacing the number (i.e. 63, 56, 49) by 7g (seven multiplied by the grid size). Here are the general formulas for each rotation: -
General Formulas
Upright T :
900 T:
1800 T:
2700 T:
5n-7g = T-total
5n+7 = T-total
5n+7g = T-total
5n-7 = T-total
Testing the General Formulas - Size 10 Grid
5n-7g = T-total
5(22) -7(10) = T-total
40 = T-total
Also:
1 + 2 + 3 + 12 +22 = 40
5n-7g = T-total
5(45) -(10) = T-total
155 = T-total
Also:
24 + 25 + 26 + 35 + 45 = 155
5n-7g = T-total
5(59) -(10) = T-total
225= T-total
Also:
38 + 39 + 40 + 49 + 59 = 225
This proves that the general formula is correct.
Rotating T through 1800 clockwise
5n+7g = T-total
5(2)+7(10) = T-total
80 = T-total
Also:
2 + 12 + 21 + 22 + 23 = 80
5n+7g = T-total
5(25)+7(10) = T-total 195 = T-total
Also:
25 + 35 + 44 + 45 + 46 = 195
5n+7g = T-total
5(39)+7(10) = T-total 265 = T-total
Also:
39 + 49 + 58 + 59 + 60 = 265
This proves that the general formula was correct.
Rotating T through 900 clockwise
5n+7 = T-total
5(11)+7 = T-total
62 = T-total
Also:
3 + 11 +12 + 13 + 23 = 62
5n+7 = T-total
5(34)+7 = T-total
177 = T-total
Also:
14 + 34 + 35 + 36 + 46 = 177
5n+7 = T-total
5(48)+7 = T-total
247 = T-total
Also:
40 + 48 + 49 + 50 + 60 = 247
This proves that the general formula was correct.
Rotating T through 2700 clockwise
5n-7 = T-total
5(12)-7 = T-total
53 = T-total
Also:
1 + 10 + 11 + 12 + 19 = 53
5n-7 = T-total
5(18)-7 = T-total
83 = T-total
Also:
7 + 16 + 17 + 18 + 25 = 83
5n-7 = T-total
5(43)-7 = T-total
208 = T-total
Also:
32 + 41 + 42 + 43 + 50 = 208
This proves that the formula was correct.
All this proves that the general formulas I predicted were correct. Here are the correct general formulas for any grid size: -
Upright T:
900 T:
1800 T:
2700 T:
5n-7g = T-total
5n+7 = T-total
5n+7g = T-total
5n-7 = T-total
Further Investigation
The formulas I have found can be made simpler to use by making both signs the same. By using sine and cosine we can do this.
For the angle 00, sine in equal to 0 and cosine in equal to 1
For the angle 900, sine in equal to 1 and cosine in equal to 0
For the angle 1800, sine in equal to 0 and cosine in equal to -1
For the angle 2700, sine in equal to -1 and cosine in equal to 0
Because these values of sine and cosine are negative and positive they can be used to make the signs in the formulas the same. This is because of the rule of signs: like signs give a positive value and unlike signs give a negative value. Here are the revised formulas: -
Upright T:
900 T:
1800 T:
2700 T:
5n-7g(cos 00) = T-total
5n+7(sin 900) = T-total
5n-7g(cos 1800) = T-total
5n+7(sin 2700) = T-total
As you can see the formulas are now easier of follow as the two sets of like formulas have the same signs. Taking the negative values of sine and cosine, then multiplying them by the original formula has achieved this.
These equations can now be used in incorporate vectors. A vector is a translation of the T which if put into the formula in the correct way can allow you to find the T-total which first finding that T’s T-number. I will now investigate vectors on a size 9 grid. Here is the vector I will be using in the formulas: . So in the vector 1 is represented by x and 0 is represented by y.
Here the original T is in red and translation through the vector is in blue.
I will now look at the differences between the original T (red T) and its translation (blue T).
5n-7g(cos 00) = T-total
(5x20)-(7x9) x (1) = T-total
100-(63x1) = T-total
37 = T-total
5n-7g(cos 00) = T-total
(5x21)-(7x9) x (1) = T-total
105-(63x1) = T-total
42 = T-total
Here, the difference between the red T-total and the blue T-total is +5
5n-7g(cos 00) = T-total
(5x51)-(7x9) x (1) = T-total
250-(63x1) = T-total
187 = T-total
5n-7g(cos 00) = T-total
(5x52)-(7x9) x (1) = T-total
255-(63x1) = T-total
192 = T-total
Again, the difference between the red T-total and the blue T-total is +5. This means that I can say x is equal to +5.
I predict that part of the vector formula will be +5x. I will now try and find what y is equal to.
Here the original T is in red and translation through the vector is in blue.
I will now look at the differences between the original T (red T) and its translation (blue T).
5n-7g(cos 00) = T-total
(5x29)-(7x9) x (1) = T-total
145-(63x1) = T-total
82 = T-total
5n-7g(cos 00) = T-total
(5x20)-(7x9) x (1) = T-total
100-(63x1) = T-total
37 = T-total
Here, the difference between the red T-total and the blue T-total is –45
5n-7g(cos 00) = T-total
(5x52)-(7x9) x (1) = T-total
260-(63x1) = T-total
197 = T-total
5n-7g(cos 00) = T-total
(5x43)-(7x9) x (1) = T-total
215-(63x1) = T-total
152 = T-total
Again, the difference between the red T-total and the blue T-total is -45. This means that I can say y is equal to -45. So I think the formula for an upright T translated using a positive vector will look like this: -
5n-7g(cos 00)+5x-45y = T-total
Testing the Formula
Here the original T is in red and translation through the vector is in blue.
Using the new formula the T-total of the blue T should be: -
5n-7g(cos 00)+5x-45y = T-total
(5x47)-(7x9) x (1) + (5x5) – (45x2) = T-total
235-(63x1)+25-90 = T-total
172+25-90 = T-total
107 = T-total
Using the old formula the T-total of the blue T should be: -
5n-7g(cos 00)= T-total
(5x34)-(7x9) x (1) = T-total
170-(63x1) = T-total
107 = T-total
Here the original T is in red and translation through the vector is in blue.
Using the new formula the T-total of the blue T should be: -
5n-7g(cos 00)+5x-45y = T-total
(5x47)-(7x9) x (1) + (5x4) – (45x3) = T-total
235-(63x1)+20-135 = T-total
172+20-135 = T-total
57 = T-total
Using the old formula the T-total of the blue T should be: -
5n-7g(cos 00)= T-total
(5x24)-(7x9) x (1) = T-total
120-(63x1) = T-total
57 = T-total
This proves that the formula was correct.
I will now test the formula using negative vectors.
Here the original T is in red and translation through the vector is in blue.
Using the new formula the T-total of the blue T should be: -
5n-7g(cos 00)+5x-45y = T-total
(5x34)-(7x9) x (1) + (5x-5) – (45x-2) = T-total
170-(63x1)-25+90 = T-total
107-25+90 = T-total
172 = T-total
Using the old formula the T-total of the blue T should be: -
5n-7g(cos 00)= T-total
(5x47)-(7x9) x (1) = T-total
235-(63x1) = T-total
172 = T-total
Here the original T is in red and translation through the vector is in blue.
Using the new formula the T-total of the blue T should be: -
5n-7g(cos 00)+5x-45y = T-total
(5x34)-(7x9) x (1) + (5x-4) – (45x-1) = T-total
170-(63x1)-20+45 = T-total
57-20+45 = T-total
132 = T-total
Using the old formula the T-total of the blue T should be: -
5n-7g(cos 00)= T-total
(5x39)-(7x9) x (1) = T-total
195-(63x1) = T-total
132 = T-total
The same formula appears to work for negative vectors as well as positive vectors. I will now test it using mixed vectors (i.e. one positive, one negative).
Here the original T is in red and translation through the vector is in blue.
Using the new formula the T-total of the blue T should be: -
5n-7g(cos 00)+5x-45y = T-total
(5x20)-(7x9) x (1) + (5x6) – (45x-2) = T-total
100-(63x1)+30+90 = T-total
37+30+90 = T-total
157 = T-total
Using the old formula the T-total of the blue T should be: -
5n-7g(cos 00)= T-total
(5x44)-(7x9) x (1) = T-total
220-(63x1) = T-total
157 = T-total
Here the original T is in red and translation through the vector is in blue.
Using the new formula the T-total of the blue T should be: -
5n-7g(cos 00)+5x-45y = T-total
(5x20)-(7x9) x (1) + (5x6) – (45x-2) = T-total
100-(63x1)+30+90 = T-total
37+30+90 = T-total
157 = T-total
Using the old formula the T-total of the blue T should be: -
5n-7g(cos 00)= T-total
(5x44)-(7x9) x (1) = T-total
220-(63x1) = T-total
157 = T-total
This proves that the formula is correct for all vectors (negative, positive and mixed). I will now test different grid sizes.
Size 8 Gird
Here the original T is in red and translation through the vector is in blue.
5n-7g(cos 00) = T-total
(5x18)-(7x8) x (1) = T-total
90-(56x1) = T-total
34 = T-total
5n-7g(cos 00) = T-total3
(5x19)-(7x8) x (1) = T-total
95-(56x1) = T-total
39 = T-total
As with the size 9 grid, the difference between the translations is +5.
5n-7g(cos 00) = T-total
(5x46)-(7x8) x (1) = T-total
230-(56x1) = T-total
174 = T-total
5n-7g(cos 00) = T-total
(5x47)-(7x8) x (1) = T-total
235-(56x1) = T-total
179 = T-total
Again, the difference between the red T-total and the blue T-total is +5. This means that I can say x is equal to +5.
I predict that y will be the same for the 8 grid as the size 9 grid (-45).
Here the original T is in red and translation through the vector is in blue.
I will now look at the differences between the original T (red T) and its translation (blue T).
5n-7g(cos 00) = T-total
(5x26)-(7x8) x (1) = T-total
130-(56x1) = T-total
74 = T-total
5n-7g(cos 00) = T-total
(5x18)-(7x8) x (1) = T-total
90-(56x1) = T-total
34 = T-total
Here, the difference between the red T-total and the blue T-total is –40
5n-7g(cos 00) = T-total
(5x47)-(7x8) x (1) = T-total
235-(56x1) = T-total
179 = T-total
5n-7g(cos 00) = T-total
(5x39)-(7x8) x (1) = T-total
195-(56x1) = T-total
139 = T-total
Again, the difference between the red T-total and the blue T-total is -40. This means that I can say y is equal to -40. So I think the formula for an upright T translated using a positive vector will look like this: -
5n-7g(cos 00)+5x-40y = T-total
Testing the Formula
Here the original T is in red and translation through the vector is in blue.
Using the new formula the T-total of the blue T should be: -
5n-7g(cos 00)+5x-45y = T-total
(5x21)-(7x8) x (1) + (5x-3) – (40x-3) = T-total
105-(56x1)-15+120 = T-total
49-15+120 = T-total
154 = T-total
Using the old formula the T-total of the blue T is: -
5n-7g(cos 00)= T-total
(5x42)-(7x8) x (1) = T-total
210-(56x1) = T-total
154 = T-total
Here the original T is in red and translation through the vector is in blue.
Using the new formula the T-total of the blue T should be: -
5n-7g(cos 00)+5x-45y = T-total
(5x27)-(7x8) x (1) + (5x4) – (40x1) = T-total
135-(56x1)+20-40 = T-total
79+20-40 = T-total
59 = T-total
Using the old formula the T-total of the blue T is: -
5n-7g(cos 00)= T-total
(5x23)-(7x8) x (1) = T-total
115-(56x1) = T-total
59 = T-total
Here the original T is in red and translation through the vector is in blue.
Using the new formula the T-total of the blue T should be: -
5n-7g(cos 00)+5x-45y = T-total
(5x19)-(7x8) x (1) + (5x4) – (40x-3) = T-total
95-(56x1)+20+120 = T-total
39+20+120 = T-total
179 = T-total
Using the old formula the T-total of the blue T is: -
5n-7g(cos 00)= T-total
(5x47)-(7x8) x (1) = T-total
235-(56x1) = T-total
179 = T-total
This proves that the formula was correct.
Here are the vector formulas I have found so far: -
Size 9 grid: 5n-7g(cos 00)+5x-45y = T-total
Size 8 grid: 5n-7g(cos 00)+5x-40y = T-total
As with the formulas at the beginning there seems to be link between the number of y in each formula. It would appear that they are both multiples of 5. For example, 5x9=45, 5x8=40. I have also noticed that the number that you must multiply 5 by to get the amount of y is the same as the gird size. If this is true throughout all grid sizes, then a general formula can be used. I predict that it will look like this: -
5n-7g(cos 00)+5x-(5g)y = T-total
I will now test this formula using different grid sizes and different vectors.
Testing the Formula – Size 7 Grid
Here the original T is in red and translation through the vector is in blue.
Using the general formula the T-total of the blue T should be: -
5n-7g(cos 00)+5x-(5g)y = T-total
(5x16)-(7x7) x (1) + (5x3) – ([5x7]x-4) = T-total
(5x16)-(7x7) x (1) + (5x3) – (35x-4) = T-total
80-(49x1)+15+140 = T-total
31+15+140 = T-total
186 = T-total
Using the old formula the T-total of the blue T is: -
5n-7g(cos 00)= T-total
(5x47)-(7x7) x (1) = T-total
235-(49x1) = T-total
186 = T-total
Here the original T is in red and translation through the vector is in blue.
Using the general formula the T-total of the blue T should be: -
5n-7g(cos 00)+5x-(5g)y = T-total
(5x30)-(7x7) x (1) + (5x4) – ([5x7]x2) = T-total
(5x30)-(7x7) x (1) + (5x4) – (35x2) = T-total
150-(49x1)+20-70 = T-total
101+20-70 = T-total
51 = T-total
Using the old formula the T-total of the blue T is: -
5n-7g(cos 00)= T-total
(5x20)-(7x7) x (1) = T-total
100-(49x1) = T-total
51 = T-total
Here the original T is in red and translation through the vector is in blue.
Using the general formula the T-total of the blue T should be: -
5n-7g(cos 00)+5x-(5g)y = T-total
(5x27)-(7x7) x (1) + (5x-3) – ([5x7]x-2) = T-total
(5x27)-(7x7) x (1) + (5x-3) – (35x-2) = T-total
135-(49x1)-15+70 = T-total
86-15+70 = T-total
141 = T-total
Using the old formula the T-total of the blue T is: -
5n-7g(cos 00)= T-total
(5x38)-(7x7) x (1) = T-total
190-(49x1) = T-total
141 = T-total
This is sufficient evidence that the general formula is correct for a size 7, 8 and 9 grid. I will now test one example from a size 4, 5, 6 and 10 size grid to be such the formula is universal.
Size 4 Grid
Here the original T is in red and translation through the vector is in blue.
Using the general formula the T-total of the blue T should be: -
5n-7g(cos 00)+5x-(5g)y = T-total
(5x10)-(7x4) x (1) + (5x1) – ([5x4]x-2) = T-total
(5x10)-(7x4) x (1) + (5x1) – (20x-2) = T-total
50-(28x1)+5+40 = T-total
22+5+40 = T-total
67 = T-total
Using the old formula the T-total of the blue T is: -
5n-7g(cos 00)= T-total
(5x19)-(7x4) x (1) = T-total
95-(28x1) = T-total
67 = T-total
Size 5 Grid
Here the original T is in red and translation through the vector is in blue.
Using the general formula the T-total of the blue T should be: -
5n-7g(cos 00)+5x-(5g)y = T-total
(5x29)-(7x5) x (1) + (5x-2) – ([5x5]x3) = T-total
(5x29)-(7x5) x (1) + (5x-2) – (25x3) = T-total
145-(35x1)-10-75 = T-total
110-10-75 = T-total
25 = T-total
Using the old formula the T-total of the blue T is: -
5n-7g(cos 00)= T-total
(5x12)-(7x5) x (1) = T-total
60-(35x1) = T-total
25 = T-total
Size 6 Grid
Here the original T is in red and translation through the vector is in blue.
Using the general formula the T-total of the blue T should be: -
5n-7g(cos 00)+5x-(5g)y = T-total
(5x23)-(7x6) x (1) + (5x-3) – ([5x6]x-2) = T-total
(5x23)-(7x6) x (1) + (5x-3) – (30x-2) = T-total
115-(42x1)-15+60 = T-total
73-15+60 = T-total
118 = T-total
Using the old formula the T-total of the blue T is: -
5n-7g(cos 00)= T-total
(5x32)-(7x6) x (1) = T-total
160-(42x1) = T-total
118 = T-total
Size 10 Grid
Here the original T is in red and translation through the vector is in blue.
Using the general formula the T-total of the blue T should be: -
5n-7g(cos 00)+5x-(5g)y = T-total
(5x22)-(7x10) x (1) + (5x6) – ([5x10]x-3) = T-total
(5x22)-(7x10) x (1) + (5x6) – (50x-3) = T-total
110-(70x1)+30+150 = T-total
40+30+150 = T-total
220 = T-total
Using the old formula the T-total of the blue T is: -
5n-7g(cos 00)= T-total
(5x58)-(7x10) x (1) = T-total
290-(70x1) = T-total
220 = T-total
This proves that the general formula, 5n-7g(cos 00)+5x-(5g)y = T-total, is correct for and upright T on all grid sizes. This formula can be used to calculate the T-total of a T, which you know the vector from its T-number to another number of the grid. I will now investigate the different rotations of the T shape. As with the formulas at the beginning, the upright T and 1800 T’s formulas where the same except for the sign. I think that this will be the same for the general vector formula. I predict that it will look like this: -
5n-7g(cos 1800)+5x-(5g)y = T-total
Rotating T through 1800 clockwise
Testing the Formula
Here the original T is in red and translation through the vector is in blue.
Using the general formula the T-total of the blue T should be: -
5n-7g(cos 1800)+5x-(5g)y = T-total
(5x21)-(7x9) x (-1) + (5x1) – ([5x9]x1) = T-total
(5x21)-(7x9) x (-1) + (5x1) – (45x1) = T-total
105-(63x-1)+5-45 = T-total
(105+63)+5-45 = T-total
168+5-45 = T-total
128 = T-total
Using the old formula the T-total of the blue T is: -
5n-7g(cos 1800)= T-total
(5x13)-(7x9) x (-1) = T-total
65-(63x-1) = T-total
65+63 = T-total
128= T-total
Here the original T is in red and translation through the vector is in blue.
Using the general formula the T-total of the blue T should be: -
5n-7g(cos 1800)+5x-(5g)y = T-total
(5x47)-(7x9) x (-1) + (5x6) – ([5x9]x5) = T-total
(5x47)-(7x9) x (-1) + (5x6) – (45x5) = T-total
235-(63x-1)+30-225 = T-total
(235+63)+30-225 = T-total
298+30-225 = T-total
103 = T-total
Using the old formula the T-total of the blue T is: -
5n-7g(cos 1800)= T-total
(5x8)-(7x9) x (-1) = T-total
40-(63x-1) = T-total
40+63 = T-total
103 = T-total
Here the original T is in red and translation through the vector is in blue.
Using the general formula the T-total of the blue T should be: -
5n-7g(cos 1800)+5x-(5g)y = T-total
(5x26)-(7x9) x (-1) + (5x-5) – ([5x9]x-1) = T-total
(5x26)-(7x9) x (-1) + (5x-5) – (45x-1) = T-total
130-(63x-1)-25+45 = T-total
(130+63)-25+45 = T-total
298-25+45 = T-total
213 = T-total
Using the old formula the T-total of the blue T is: -
5n-7g(cos 1800)= T-total
(5x30)-(7x9) x (-1) = T-total
150-(63x-1) = T-total
150+63 = T-total
213 = T-total
This proves that the formula is correct. There is no need to test the formula on differing grid sizes as I have proved that the upright T general vector formula is work for all grid sizes, and the two are the same in that respect. I will now test the general vector formulas as for a T shape rotated through 900 and 2700 clockwise.
Rotating T through 900 clockwise
I will try the formula as above but with the correct rotation included: -
5n-7g(sin 900)+5x-(5g)y = T-total
Here the original T is in red and translation through the vector is in blue.
Using the general formula the T-total of the blue T should be: -
5n-7g(sin 900)+5x-(5g)y = T-total
(5x29)-(7x9) x (1) + (5x1) – ([5x9]x1) = T-total
(5x29)-(7x9) x (1) + (5x1) – (45x1) = T-total
145-(63x1)+5-45 = T-total
(145-63)+5-45 = T-total
82+5-45 = T-total
42 = T-total
Using the old formula the T-total of the blue T is: -
5n+7 = T-total
(5x21)+7 = T-total
105+7 = T-total
112 = T-total
Here the formula is wrong. The T-total=112, not 42. There is a difference of 70 between the two answers.
Here the original T is in red and translation through the vector is in blue.
Using the general formula the T-total of the blue T should be: -
5n-7g(sin 900)+5x-(5g)y = T-total
(5x43)-(7x9) x (1) + (5x-2) – ([5x9]x1) = T-total
(5x43)-(7x9) x (1) + (5x-2) – (45x1) = T-total
215-(63x1)-10-45 = T-total
(215-63)-10-45 = T-total
152-10-45 = T-total
97 = T-total
Using the old formula the T-total of the blue T is: -
5n+7 = T-total
(5x32)+7 = T-total
105+7 = T-total
167 = T-total
Again, the formula was wrong, but by the same amount (70). To put the formula right I need to add and additional 70 at the end. I will now test the following formula: -
5n-7g(sin 900)+5x-(5g)y +70= T-total
Testing the Formula
Here the original T is in red and translation through the vector is in blue.
Using the general formula the T-total of the blue T should be: -
5n-7g(sin 900)+5x-(5g)y +70 = T-total
(5x24)-(7x9) x (1) + (5x-3) – ([5x9]x-2) +70 = T-total
(5x24)-(7x9) x (1) + (5x-3) – (45x-2) +70 = T-total
120-(63x1)-15+90 +70 = T-total
(120-63)-15+90 +70 = T-total
57-15+90 +70 = T-total
202 = T-total
Using the old formula the T-total of the blue T is: -
5n+7 = T-total
(5x39)+7 = T-total
195+7 = T-total
202 = T-total
The formula is correct, here.
Here the original T is in red and translation through the vector is in blue.
Using the general formula the T-total of the blue T should be: -
5n-7g(sin 900)+5x-(5g)y +70 = T-total
(5x53)-(7x9) x (1) + (5x-7) – ([5x9]x4) +70 = T-total
(5x53)-(7x9) x (1) + (5x-7) – (45x4) +70 = T-total
265-(63x1)-35-180 +70 = T-total
(265-63)-35-180 +70 = T-total
202-35-180 +70 = T-total
57 = T-total
Using the old formula the T-total of the blue T is: -
5n+7 = T-total
(5x10)+7 = T-total
50+7 = T-total
57 = T-total
This proves that the general vector formula for T rotated through 900 is correct. I predict that the general vector formula for a T rotated through 2700 will be the same as the 900 T, except the extra 70 needed will be negative. This is follows the pattern of the first formulas I found: 900T-total=5n-7, 2700T-total=5n+7. I will now test the formula, which I think will look like this: -
5n-7g(sin 900)+5x-(5g)y -70= T-total
Rotating T through 2700 clockwise
Here the original T is in red and translation through the vector is in blue.
Using the general formula the T-total of the blue T should be: -
5n-7g(sin 2700)+5x-(5g)y –70 = T-total
(5x29)-(7x9) x (-1) + (5x6) – ([5x9]x1) –70 = T-total
(5x29)-(7x9) x (-1) + (5x6) – (45x1) –70 = T-total
145-(63x-1)+30-45 –70 = T-total
(145+63)+30-45 –70 = T-total
82+30-45 –70 = T-total
123 = T-total
Using the old formula the T-total of the blue T is: -
5n-7 = T-total
(5x26)-7 = T-total
130-7 = T-total
123 = T-total
This proves the formula is correct.
Here the original T is in red and translation through the vector is in blue.
Using the general formula the T-total of the blue T should be: -
5n-7g(sin 2700)+5x-(5g)y –70 = T-total
(5x43)-(7x9) x (-1) + (5x-4) – ([5x9]x3) –70 = T-total
(5x43)-(7x9) x (-1) + (5x-4) – (45x3) –70 = T-total
215-(63x-1)-20-135 –70 = T-total
(215+63)-20-135 –70 = T-total
278-20-135 –70 = T-total
53 = T-total
Using the old formula the T-total of the blue T is: -
5n-7 = T-total
(5x12)-7 = T-total
60-7 = T-total
53 = T-total
This proves my general vector formula is correct. There is, again, no need to test the 900 T and the 1800 T general vector formulas, as they both follow the patterns of the earlier 900 and 1800 formulas, which work for any grid size Here are all the general vector formulas: -
Upright T: 5n-7g(cos 00)+5x-(5g)y = T-total
900 T: 5n-7ga(sin 900)+5x-(5g)y +70= T-total
1800 T: 5n-7g(cos 1800)+5x-(5g)y = T-total
2700 T: 5n-7g(sin 900)+5x-(5g)y +70= T-total
These formulas do not work if the T goes outside the grid. This is because the shape is no longer a T shape. Using these formulas it is possible to work out any T-total of any rotation of the T shape, on any grid size, from the T-number of an upright T and a vector to the desired T-number.
