Calculating the diameter of regular polygons

  • #1
LightningInAJar
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TL;DR Summary
How do I calculate the diameter of regular polygons with 1 as side length?
I measured the maximum diameters of 15 regular polygons with side lengths of one. Is there a way to calculate the maximum diameter for regular polygons of any number of sides without using pi or any nonterminating numbers?

03-Sided = 1
04-Sided = 1.4142
05-Sided = 1.6180
06-Sided = 2
07-Sided = 2.2470
08-Sided = 2.6131
09-Sided = 2.8794
10-Sided = 3.2361
11-Sided = 3.5133
12-Sided = 3.8637
13-Sided = 4.1484
14-Sided = 4.4940
15-Sided = 4.7834
 

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  • #2
Yes, you could derive a formula.

Consider the regular polygon inscribed in a circle. Lines drawn from the circle's center to each vertice of the polygon are radii. An 8-sided polygon makes 8 wedge-shaped isosceles triangles, so you can determine the angle between the radii as ##2\pi/8##.

Knowing the length of the n-gon side, you can determine the radius using the sin() and noting that you can drop a perpendicular line splitting the isosceles triangle into two right triangles so the angle is ##1/2 (2\pi/n)##

and the ##sin(\pi/n) = (1/2 side)/radius## since you know the side length solve for the radius.

Or you could use this handy online calculator:

https://calcresource.com/geom-ngon.html
 
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  • #3
LightningInAJar said:
without using pi or any nonterminating numbers?
Are you asking whether all the diameters of regular polygons with side length 1 are algebraic numbers?

Or, more accurately, asking for a constructive proof of that claim?

Off hand, it seems unlikely. I would expect that most of the diameters will be transcendental.
 
  • #4
LightningInAJar said:
TL;DR Summary: How do I calculate the diameter of regular polygons with 1 as side length?
Diameter of regular polygons.jpg
 

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  • #5
jbriggs444 said:
Are you asking whether all the diameters of regular polygons with side length 1 are algebraic numbers?

Or, more accurately, asking for a constructive proof of that claim?

Off hand, it seems unlikely. I would expect that most of the diameters will be transcendental.
Yeah, basically something that can be measured exactly that can be fully written out. I figured a many sided polygon is the closest we get to a circle and the only thing a computer can even represent.
 
  • #6
Also I am referring to a maximum diameter. If the polygon has an even number of vertices it can be split right in half, but if it is an odd number it will have one more side on one side of the diameter line.
 
  • #7
LightningInAJar said:
Yeah, basically something that can be measured exactly that can be fully written out. I figured a many sided polygon is the closest we get to a circle and the only thing a computer can even represent.
Nothing can be measured exactly. What can be fully written out depends on your encoding system. One can write ##\pi## in one symbol.

I have no idea what you have in mind with respect to computer representations. They are not limited to polygons any more than you and I are limited to polygons. 64 bit floating point is not all there is.
 
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  • #8
jbriggs444 said:
Nothing can be measured exactly. What can be fully written out depends on your encoding system. One can write ##\pi## in one symbol.

I have no idea what you have in mind with respect to computer representations. They are not limited to polygons any more than you and I are limited to polygons. 64 bit floating point is not all there is.
I was just wondering if one could create a regular polygon with sides that are a planck length long. I wonder how many sides it would need to have to create a shape that is large enough to be seen. I think .073mm is the smallest thing a person can see, and I'm not sure if curved lines are meaningful at the planck length? So basically it would be a circle for all intent and purposes?

Yeah pi is not terminating, but I was hoping with only straight lines one could limit the number of digits?
 
  • #9
LightningInAJar said:
I was just wondering if one could create a regular polygon with sides that are a planck length long.
Made out of what? Certainly not atoms.
 
  • #10
LightningInAJar said:
I was just wondering if one could create a regular polygon with sides that are a planck length long.
That is not at all a feasible way to accurately measure the ratio between the circumference of a circle and its diameter.

We already know the ratio for Euclidean space. We can calculate the deviation for a circle of a particular scale in a real world region with a known, measured or calculated spatial curvature.

Working the other way, this is essentially the definition of Gaussian curvature:

https://www.newworldencyclopedia.org/entry/Curvature said:
An intrinsic definition of the Gaussian curvature at a point P is the following: imagine an ant which is tied to P with a short thread of length r. He runs around P while the thread is completely stretched and measures the length C(r) of one complete trip around P. If the surface were flat, he would find C(r) = 2πr. On curved surfaces, the formula for C(r) will be different, and the Gaussian curvature K at the point P can be computed as$$K = \lim_{r \to 0} (2 \pi r - C(r)) \cdot \frac{3}{\pi r^3}$$

We can also calculate the deviation of the circumference to diameter ratio from ##\pi## in Euclidean space for a regular polygon with ##n## sides. But that's just boring and obvious.
 
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  • #11
LightningInAJar said:
Yeah pi is not terminating, but I was hoping with only straight lines one could limit the number of digits?
The decimal representation of ##\pi## is non-terminating. But that is also true for the decimal representation of ##\frac{1}{3}##.

There is a base ##\pi## representation of ##\pi## with two digits: ##10_{\pi} = \pi##.

The ratio of the circumference of a unit square (##4##) to its diameter (##\sqrt{2}##) is ##2\sqrt{2}##. The decimal representation of that does not terminate. It is an irrational number approximately equal to ##2.82842712474619##. So it seems that polygons with straight line edges are not a prescription for approximations to ##\pi## whose decimal expansions terminate.

One does get lucky with six sides and a ratio of exactly ##\frac{6}{2} = 3##.
 
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  • #12
jbriggs444 said:
There is a base ##\pi## representation of ##\pi## with two digits: ##10_{\pi} = \pi##.
A bit off-topic, but what are the digits that are used in base ##\pi##? In all other bases the digits are nonnegative integers 0, ..., base - 1. For bases greater than 10, letters of the alphabet are used to represent the integers from 10 on up to the base - 1; e.g., A, B, C, D, E, and F for the integers 10, 11, 12, 13, 14, and 15 in base 16.
 
  • #13
Why not 0,1,2 and 3?
 
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  • #14
Mark44 said:
A bit off-topic, but what are the digits that are used in base ##\pi##? In all other bases the digits are nonnegative integers 0, ..., base - 1. For bases greater than 10, letters of the alphabet are used to represent the integers from 10 on up to the base - 1; e.g., A, B, C, D, E, and F for the integers 10, 11, 12, 13, 14, and 15 in base 16.
I assume 0, 1, 2 and 3. This will result in many almost all numbers having multiple representations.

For instance, in addition to the ##10_\pi## representation, ##\pi## also has a representation that begins with ##3.01102..._\pi = 3 + 0/\pi + 1/\pi^2 + 1/\pi^3 + 0/\pi^4 + 2/\pi^5 + ...##

Yet another representation begins with ##2.31220..._\pi = 2 + 3/\pi + 1/\pi^2 + 2/\pi^3 + 2/\pi^4 + 0/\pi^5+ ...## (if two quick calculations serve).

This is why I was careful to say "a" representation rather than "the" representation. It is probably not worthwhile to chase after a rule for which representation should be canonical.
 
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  • #15
Vanadium 50 said:
Made out of what? Certainly not atoms.
Not made out of atoms. But who knows. Maybe we'll find smaller stuff someday. Lol.
 
  • #16
LightningInAJar said:
Lol.
I assume that means you were just mocking the people who were trying to help you. Your choice, I guess.
 
  • #17
Vanadium 50 said:
I assume that means you were just mocking the people who were trying to help you. Your choice, I guess.
I'm not mocking anyone. I was asking how it's calculated. I didn't claim it's applicable in the real world. Just a thought experiment.
 
  • #18
LightningInAJar said:
I'm not mocking anyone. I was asking how it's calculated. I didn't claim it's applicable in the real world. Just a thought experiment.
If you want to calculate pi then you can do so without any reference at all to the real world. There are various algorithms that can give you its decimal digits. For hexadecimal, there is one that is excellent.

The Planck length has nothing to do with it.
 
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  • #19
If you want to talk about abstractions, that's mathematics.
If you want to talk about physical objects, that's science.
If you want to talk about objects that don't exist, that's just plain silly.
 
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  • #20
jbriggs444 said:
If you want to calculate pi then you can do so without any reference at all to the real world. There are various algorithms that can give you its decimal digits. For hexadecimal, there is one that is excellent.

The Planck length has nothing to do with it.
I was just hoping pi could be avoided with objects without curves.
 

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