The Holographic Universe and Pi = 4 in Kinematics!

[gravityblock]

Let's try it with a "time variable" then. Draw a big circle and a square around it. You walk around the square and I'll walk around the circle, at the same speed. Who will walk completely around, first?

(And I note that you did not provide a single credible reference or support for your position. Nor did you provide an example of problem-solving using your value.)

You conveniently left out the first four steps.  We'll start walking at the same speed in the fifth tile or the fifth step in the illustration below, and we'll finish at the same time.

In addition to this, references and support for my position has been provided.  Your disagreement with those references doesn't make them not credible!  You have only asserted those references aren't credible without providing one scientific argument against any of those references posted in this thread.

Gravock

[MarkE]

If there's an increase in X, then there will be a proportional decrease in Y which maintains the same distances and perimeter.  Also, you're trying to derive 3.14 as Pi in this example by using radians where a full circle equals 2 * 3.14 or Tau.  This is no different than me saying 2 * 4  = 8 to represent the eight points on the circumference that lie on arcs that are multiples of 4/4 or 1.

There's a reason why the taxicab geometry correctly represents the true value of Pi being four in a real circle with a time variable.

Gravock

For any finite path approximation by turning the corners instead of traveling from vertex to vertex that are closest to the circumference you artificially increase your travel distance.  For every approach towards the circumference you make a matching turn away from it.  If you make enough and small enough squares subtracting the area they consume from the inset square will give you an approximation of the circle's area.  The more and smaller squares you use, the better the area approximation.  Because you keep turning away from the circumference your estimate of the perimeter length never improves.  If the method took the diagonal paths then the path approximation would improve with more and smaller squares and eventually approach Pi*D.

[MarkE]

You conveniently left out the first four steps.  We'll start walking at the same speed in the fifth tile or the fifth step in the illustration below, and we'll finish at the same time.

In addition to this, I have provided references and support for my position.  Your disagreement with those references doesn't make them not credible!

Gravock

TinselKoala wins every time following the circular path.  You keep turning away from the perimeter and have to go back over and over again increasing your travel distance to 4/Pi Tinsel Koala's.

[gravityblock]

For any finite path approximation by turning the corners instead of traveling from vertex to vertex that are closest to the circumference you artificially increase your travel distance.  For every approach towards the circumference you make a matching turn away from it.  If you make enough and small enough squares subtracting the area they consume from the inset square will give you an approximation of the circle's area.  The more and smaller squares you use, the better the area approximation.  Because you keep turning away from the circumference your estimate of the perimeter length never improves.  If the method took the diagonal paths then the path approximation would improve with more and smaller squares and eventually approach Pi*D.

TinselKoala wins every time following the circular path.  You keep turning away from the perimeter and have to go back over and over again increasing your travel distance to 4/Pi Tinsel Koala's.

TK does not win, for the square will be as uniform as the circle itself at the planck length.

Gravock

[MarkE]

TK does not win, for the square will be as uniform as the circle itself at the planck length.

Gravock

No, making the diversionary steps forced by the squares method you have specified smaller increases the number of diversionary steps.  By your own assertion, the length of such a path remains stuck at 4*D.  The path length of the circumference is Pi*D which has been approximated to eight digits as 3.1415953.