C.B. the circumference of the Circle is based on the perimeter of the square. Both are the same, but because the square is a fixed polygon, it's results are always accurate.
Pi, in that instance can be found by using the perimeter as a circumference and divide it by twice the height of the Pyramid.
But that also involve the Pythagorean Theorem, and Tycho Brahe's Right Triangle (which was stolen by Kepler after he poisoned him)
The radius of the circle and 1/2 side length of the square Form the height and base of the Tycho Brahe Right Triangle
Jude Andre Charles the circumference of the Circle is based on the perimeter of the square.
How?
Jude Andre Charles Both are the same, but because the square is a fixed polygon, it's results are always accurate.
How do you know both are the same?
Jude Andre Charles Pi, in that instance can be found by using the perimeter as a circumference and divide it by twice the height of the Pyramid.
But you still don’t know the value of the circumference as to replace it with the perimeter of the square
Jude Andre Charles But that also involve the Pythagorean Theorem, and Tycho Brahe's Right Triangle
Still need to find out the length of the circumference. till then no theorem will work.
4 times 1/2 of the side length of the square divided by the radius is the same as 4 divided by the square root of the Golden Ratio. The result is always 3.144606 if done correctly.
Jude Andre Charles The radius of the circle and 1/2 side length of the square Form the height and base of the Tycho Brahe Right Triangle
If you work that way you’re just supposing beforehand π is 3.1446. It doesn’t work that way.
You’re inserting from the beginning the value you’re looking for. It is not valid.
The of the square to the circle is 1:1.272019649514.
The circumference will eventually always be the same as the perimeter.
Jude Andre Charles 4 times 1/2 of the side length of the square divided by the radius is the same as 4 divided by the square root of the Golden Ratio. The result is always 3.144606 if done correctly.
Still working on straight lines. Where is the connection to the circle.
Jude Andre Charles The of the square to the circle is 1:1.272019649514.
The circumference will eventually always be the same as the perimeter.
You need to prove that, otherwise is only an assumption.
C.B. not quite. One starts with Pi 3.14159265359 first to figure out the obvious pattern.
4/pi = 1.27324 which resembles closely the square root of the Golden Ratio 1.27202
From there, it's really just a question of replacing pi with the result found in 4/sqrt(phi) and test using the perimeter of the square as a basis for the circumference of the Circle and use the circumference (which truly is the same as a square's perimeter) and divide it with 4/sqrt(phi) to find 3.144606, because if we were to take the circumference (which is the same as the perimeter) and divide it by 4/1.27324 (1.27324 being the result found in 4/pi), then the result will eventually be 3.14159265359, however, errors in accuracy will also creep up more often with 3.14159265359 than with 3.14460551103 for the utmost certainty.
You’re approximating the value to the assumed 3.1446. But there is no mathematical demonstration for it. It is not a proof. It is not evidence for the “exact” value of π, which is the important point here.
C.B. tell how I should do it and I will try to demonstrate it to the best of my abilities and with what I have learned so far.
However, I am currently at work, so you might not get an answer before the end of my shift.
But by all means, tell me how I should demonstrate it in order to come to an understanding.
Didn’t you pay attention to my derivation ? It is all there what you need to know.
C.B. because of Hush's incessant posts, I might've skipped it. I'll back up a little and see what is good.
Let’s take a Circle and a Square of same perimeter π and set the equality
4b = π
Now we need to find out the area of both figures:
For the Square is evident
b2
As for the Circle we use 4b as π to calculate the area:
4br2
4b(1/2)2
4b*1/4= b
We take the values b and b2 to construct a right triangle of legs b and b2 inscribed in a semicircle
We take a semicircle of Diameter or Hypotenuse 1, because it is the only semicircle where b and b2 will have their real values π/4 and (π/4)2.
We apply Pythagoras:
b4 + b2 – 1 = 0
And obtain the value of b or π/4:
b = 0.78615
And, following, the value of π or 4b:
Π = 3.1446055
C.B. I see. I will test it out and come back to you with the results.
the result is already there. Try to understand the method.
C.B. is b squared the height of the triangle? I would suppose the base instead as it comes up the shortest. But so far you have exactly the same results as I do.
Before the end of the day, I will reproduce it on Desmos as a comparative assessment.
