In micro scale, you treat outer limit and inner limit as the SAME value, that's why you conclude the circle line is the SAME as both outer limit and inner limit whose value would validate Pi.
No, the way it works is you make something which is an upper bound, and something which is a lower bound. Then the real value lies somewhere between.
e.g. with your demonstration of a 4 sided shape, you have a lower bound of 2*sqrt(2), i.e. 2.8284, and an upper bound of 4.
This upper and lower bound does not tell you it is the actual value of pi, it just provides an upper and lower limit.
You then need to go to a smaller angle, and as you do you get a better upper and lower limit, as they converge to the value of pi.
In effect, what you are trying to do is find an n sided shape, in one case, the distance from the centre to the corner is 1 unit, so it fits inside the circle and the perimeter of this shape is a lower limit on the value of 2*pi; in the other case the distance from the centre to the centre of a side is 1 unit, so the circle fits inside it and the perimeter of this shape is an upper limit on the value of 2*pi.
One thing for sure: IT'S BEYOND 1.7453 meters, as THE PI TERRITORY.
Or you can do the math properly.
But before you even try doing that, your calculator is internally using pi.
But doing this, you want to use an angle of 1 degree, this means you are going for a 360 sided polygon. But you take half the angle to calculate the value of tan (the upper bound) and sin (the lower bound), and multiply that by 360 to get the bounds for pi.
Doing this we get 3.1416724047 as an upper bound and 3.1415527794 as a lower bound.
The value of pi, 3.1415926536, fits nicely between these.
So still no problem there.
In your case, you are adding in steps which are ultimately unnecessary
First you are scaling up the size of the circle, which means you then need to divide by it to get back to pi.
And you are using both sides of the sector/chord/tangent, which means the perimeter will bound 2*pi, so you then need to divide by 2.
And in doing so, you show extreme dishonesty by failing to round correctly and stating them to the point they are equal.
The value you provided for tan was 17.453
735581...
Notice how the digit after you rounded is a 7?
The standard rules of rounding, as 7 is greater than or equal to 5, YOU ROUND UP! not down.
And what makes it even more damning is that the standards don't apply when you are finding an upper and lower bound. When doing that, it doesn't matter what the value is you always round down for the lower bound and up for the upper bound.
So doing that honestly, you still get 17.454 for the upper bound and 17.543 for the lower bound.
You have dishonestly rounded to get your upper bound as less than the actual lower bound.
No wonder you have entirely failed.
You then proceed to add in extra BS about "the space between chords" that is entirely irrelevant and does not affect the calculation at all.
Then you basically just complain that it doesn't match your BS number which doesn't work at all.
Where was this difference in chords in your prior BS about 90 degrees? No where, because it is pure BS.
We see that the sine version is a straight line. The tangent version produces an apex further away, and the arc is cutting that corner, meaning it is shorter, but still longer than the straight line.
Suppose the angle = one millionth degrees, then there will be one million tiny segments that are MISSING from pi based calculation.
No, they will be missing from your calculation, because you don't do the math properly.
Circle Area equals MORE than Pi
It equals pi*r^2.
For a unit circle, it is pi. Not more. No matter how much BS you invent, it will not be more.
Because the smaller angle, sin & tan chords will be relatively in the same length, for human sight only reach to +/- milimeter scale while there always SPACE between tan & sin chords.
Which just further shows the 2 values converge.
But no, just like the length of sin and tan converge, the space between them shrinks to nothing.
And do you know why there is the disparity?
Because the circumference is proportional to radius.
So that extra bit gets split up among every chord.
e.g. in the previous example, you had a difference of 0.000665.
If you multiply that by 360, you get 0.239, which is more than you 0.038.
So no problem there.
The curved lines also never be handled by pi based calculation which is stuck on Straight Lines.
Yes it is, by having 1 value which overestimates, and 1 value which underestimates.
That means the curve lies between.
You accept that with your example at 90 degrees. Why pretend it doesn't work now?
in a tiny angle, chords - either for sin and tan - both lengths are are relatively the same, and less than the arc length.
No. For tiny angles, all three lengths are basically the same. But tan is slightly longer than the arc which is slightly longer than the arc.
It doesn't matter how small you go, the tangent will always be longer than the arc length which will always be longer than the sin.
We can see this by doubling up.
The tangent is a long path with a single corner.
The arc is a curved and more direct path which is shorter, it cuts the corner.
And sine is a straight path being the shortest.
Anyone who says otherwise either has no idea what they are talking about, or they are lying.
Challenge for Pi based calculation:
Can you do the same as Phew?
Considering phew is wrong, why would we want to?
Under 1000 meters of radius and 1° of angle, two chords: as long as 17 meters and 45 centimeters and 3 milimeters each, with a space of 3.8 centimeters in which an arc stays, the consequence is, the arc becomes longer than both cords.
Why don't you try this in practice, filming the entire thing without any cuts, and show what you get?
And no, one is 17 m, 45 cm and 3 mm; the other is 17 m, 45 cm and 4 mm. The arc is in between.
AND THERE ARE SIMILARITY OF SIZE BETWEEN CERTAIN AREAS IN THE CIRCLE AND AREAS OUTSIDE THE CIRCLE.
Is there? or are you just lying like you do so often?
e.g. I assume you are claiming the yellow square is the same area as either one or both of the yellow portions inside the circle you have highlighted. But can you prove it?
No. You can just assert it must be true.
e.g. going for the yellow, the yellow square outside is easy.
You have a unit circle. So the length from the centre along the diagonal to the yellow square is 1. So the horizontal distance is 1/sqrt(2)=sqrt(2)/2. So the width of the yellow square is (1-sqrt(2)/2). So the area of the yellow square is (1-sqrt(2)/2)^2 = 1+2/4-2*sqrt(2)/2 = 1+1/2-sqrt(2)=1.5-sqrt(2), which is roughly 0.08578643763.
Inside the circle is more complex. We have the quarter circle (with an area of pi/4), from which we then subtract the square and 2 triangles.
The square is sqrt(2)/2 wide. The rectangles have 1 length of sqrt(2)/2, and one length of (1-sqrt(2)/2).
This gives us a total area we need to subtract of (sqrt(2)/2)^2 + 2*(1/2)*(sqrt(2)/2)*(1-sqrt(2)/2) = (sqrt(2)/2)*((sqrt(2)/2) + 1-sqrt(2)/2)=sqrt(2)/2
So the area in yellow is pi/4-sqrt(2)/2, or roughly 0.0782913822109.
So no, the area is not the same. You falsely claim it is to pretend pi must be 3.171573