Using this T, I am now able to work out an algebraic T Total for a regular T. It would be n-21+n-20+n-19+n-10+n which simplifies to 5n – 70. I can check that this is correct by putting the T number of any T into this equation and then work it out. I am going use T number 22, one which I have already worked out and I know the T total should be 40. 5*22=110 and 110-70=40 which proves that this formula is correct. Another way to find out whether this formula is correct if to plot a graph of all the points and find the gradient of the line. I can also try and work out a T Total value using the graph. Below is the graph I have drawn plotting T number against T total.
As you can see from the graph, I worked out the gradient by dividing the height of my little triangle, which was 30, by its base, which was 5. From this calculation I knew that the gradient was 6. Spotting from the graph I found the intercept of the line was –70 and using the equation y=mx+c I found that the equation of the line was 5X-70. The equation I came up with when I worked out a formula for T Total was 5n-70 but because the T Number is n I should have use the formula: y=mn+c which means that I got the same equation proving my equation is correct. On the graph below I have show how to work out the T Total for a T whose number is 43.
The two methods I have used have helped me to successfully find a pattern between the T Total and the T number.
2nd T- Total – (5 square T rotated 90º about n)
I am going to try and establish a relationship between the T number and the T total and try to come out with a formula which will help me find the T Total from the T Number using different methods for a Normal 5 squared T which has been rotated about 90º about n.
From the previous experiment, I discovered that I could not establish a pattern by drawing the table that I did earlier so this time I am going to go straight to the second method I used to find a formula and confirm it with the third method I used which was the graph.
I can now work out an algebraic T total for this T. The formula would be n+n+10+n+19+n+20+n+21, which amounts to 5n+70 as opposed to the 5n-70 that I obtained from the previous experiment. Putting the T number of any T should now give me the right T Total if I put it into the equation. I will use 22 as the n and because I already know that a T total for a T where N=22 is 22+32+41+42+43=180. Putting it into my formula – 5n+70 then 5*22=110 and 110+70 =180 so this proves my formula works. Just to confirm it I am going to draw the graph of y=5n+70 and see if when I look the T number 22, I get 180.
