A Pi Question: Why do we use the awkward approximation 22/7 ?

[Agent Smith]

So you will need to take measurement accuracy into account if you want to distinguish between a “good” circle and a “great” circle.

Si and I also say that the closer c/d is to $\pi$ the more circular it is.

Is $\pi$ a measure of a circle's curvature?

 

[DrJohn]

Also, consider the speed of calculation without a calculator compared to using a three or five figure decimal value.
And the final result is often close enough for the majority of non-scientific requirements.

 

[fresh_42]

Is $\pi$ a measure of a circle's curvature?

The usual curvature of a circle depends on its radius. Imagine a circle that is astronomically large. Wouldn't you agree that such a circle is far less curved than one you can draw in the sand? Therefore, $1/r$ is the usual measure for the curvature. The larger the radius the less curved is the circle.

 

[fresh_42]

Cogito it should be $\frac{21 + 1}{6 + 1}$

The Babyloninas thought it should be $\dfrac{3}{\dfrac{57}{60}+\dfrac{36}{3600}}$ and Archimedes thought it should be $\dfrac{223}{71}.$

 

[Frabjous]

Si and I also say that the closer c/d is to $\pi$ the more circular it is.

Is $\pi$ a measure of a circle's curvature?

The curvature of a circle is one over its radius.

 

[Agent Smith]

Gracias to @Frabjous , @fresh_42

I've heard of negative curvatures, positive curvatures, and 0 curvature. How does that work?

 

[fresh_42]

Gracias to @Frabjous , @fresh_42

I've heard of negative curvatures, positive curvatures, and 0 curvature. How does that work?

You can walk through a valley $\cup$, over a hill $\cap$, or on flat land ____ . The details are as so often a bit more complicated.

 

[Mark44]

Cogito it should be $\frac{21 + 1}{6 + 1}$

Again, why is this interesting?

After a brief closing, the thread has been reopened.

 

[PeterDonis]

We use a compass, draw an object such that all of its points are equidistant from a given point.

Which you will never be able to do perfectly. The compass point you're drawing with has a finite width, the compass itself has a finite amount of "play" in its spacing, and you have a finite error in your ability to keep the "center" point of the compass exactly in the same place. So no, nothing you draw this way will be an exact circle. Whether it is a good enough approximation to a circle will depend on what you are going to use it for.

 

[Agent Smith]

You can walk through a valley $\cup$, over a hill $\cap$, or on flat land ____ . The details are as so often a bit more complicated.

Yes but what are the mathematical definitions?
1. Negative curvature = ?
2. 0 curvature = ?
3. Positive curvature = ?

@Frabjous was kind enough to inform us that the curvature of a circle is $\frac{1}{\text{radius}}$. What about simple unclosed loops or ellipses? What's the formula for curvature there?

 

[Agent Smith]

Again, why is this interesting?

After a brief closing, the thread has been reopened.

$\frac{22}{7} = \frac{21 + 1}{7} = 3 + \frac{1}{7} = 3\frac{1}{7}$

 

[PeterDonis]

what are the mathematical definitions?

Two good mathematical definitions for a 2-surface use the sum of the angles of a triangle and the number of lines through a point not on a given line that do not intersect the given line:

1. Negative curvature = ?

Sum of angles in a triangle less than 180 degrees.
More than one line through a point not on a given line that does not intersect the given line.

2. 0 curvature = ?

Sum of angles in a triangle equal to 180 degrees.
Exactly one line through a point not on a given line that does not intersect the given line.

3. Positive curvature = ?

Sum of angles in a triangle greater than 180 degrees.
No lines through a point not on a given line that do not intersect the given line.

Here "line" means "geodesic", and "triangle" means "three-sided figure made up of geodesic segments".

 

[PeterDonis]

@Frabjous was kind enough to inform us that the curvature of a circle is $\frac{1}{\text{radius}}$. What about simple unclosed loops or ellipses? What's the formula for curvature there?

https://en.wikipedia.org/wiki/Curvature#Plane_curves

 

[Agent Smith]

Two good mathematical definitions for a 2-surface use the sum of the angles of a triangle and the number of lines through a point not on a given line that do not intersect the given line:


Sum of angles in a triangle less than 180 degrees.
More than one line through a point not on a given line that does not intersect the given line.


Sum of angles in a triangle equal to 180 degrees.
Exactly one line through a point not on a given line that does not intersect the given line.


Sum of angles in a triangle greater than 180 degrees.
No lines through a point not on a given line that do not intersect the given line.

Here "line" means "geodesic", and "triangle" means "three-sided figure made up of geodesic segments".

Gracias. What about lines? Like a loop in 2D?

 

[Agent Smith]

Which you will never be able to do perfectly. The compass point you're drawing with has a finite width, the compass itself has a finite amount of "play" in its spacing, and you have a finite error in your ability to keep the "center" point of the compass exactly in the same place. So no, nothing you draw this way will be an exact circle. Whether it is a good enough approximation to a circle will depend on what you are going to use it for.

We won't ever be able to draw a perfect circle, ok. But for a set of imperfect circles A = {x, y, z}, if x's circumference to diameter ratio is closest to $\pi$ it is more circular than either y or z, oui?

 

[fresh_42]

Yes but what are the mathematical definitions?
1. Negative curvature = ?
2. 0 curvature = ?
3. Positive curvature = ?

@Frabjous was kind enough to inform us that the curvature of a circle is $\frac{1}{\text{radius}}$. What about simple unclosed loops or ellipses? What's the formula for curvature there?

The usual curvature of a circle depends on its radius. Imagine a circle that is astronomically large. Wouldn't you agree that such a circle is far less curved than one you can draw in the sand? Therefore, 1/r is the usual measure for the curvature. The larger the radius the less curved is the circle.

Terms like slope and curvature are locally defined. Locally means in the neighborhood of points where we speak about them. We use derivatives to describe them, linear approximations. The curvature of a circle is everywhere the same, so $1/\text{radius}$ is sufficient to describe it. Here is a nice picture from Wikipedia where $K$ is the number we call curvature.

To determine $K$ as a number at a certain point of a certain object involves derivatives and the measurement of the change of changes.

We won't ever be able to draw a perfect circle, ok. But for a set of imperfect circles A = {x, y, z}, if x's circumference to diameter ratio is closest to $\pi$ it is more circular than either y or z, oui?

What do you mean by more circular? The best definition of such a statement would be to determine the distance of the curvature of such an object from $1/r.$ I'm afraid that we will not proceed on this question as long as we don't have a mathematical description of $A$ that allows us to calculate an average curvature. Only then we can speak about more or less circular. If we only have our eyesight then we need a microscope to investigate how close your objects in $A$ are to a circle.

There is quite some way to go from $\pi$ to curvature as a mathematical term, and it requires calculus. There are no easy answers that are also satisfactory.

​

 

[Agent Smith]

@fresh_42 arigato gozaimus. I now have a fair idea of what curvature is. For surfaces we could think of cross-sections and then use the formula to compute the curvature of that line formed when we take the cross-section to get an idea of the curvature of the surface itself, no? Was it you who posted $\cap$ and $\cup$ (even the LaTex commands "\cap" and "\cup" seem to support my thesis.

By more circular I mean as close to being a circle as possible, where a circle is defined as points equidistant from a chosen point, the center. An imperfect circle's points may vary in distance from the center. This I believe can be tracked by how close to $\pi$ the circumference/diameter of a given circle-candidate is.

 

[Vanadium 50]

more circular

You know, like this thread.

By more circular I mean as close to being a circle as possible

This reasoning seems kind of...um...circular.

Is a rectangle more circular than a rhombus? Is a square more circular than a sphere? Is a convex n-gon more or less circular than a concave (n+1)-gon?

 

[Agent Smith]

In my world these are good questions. Perhaps I'm talking about how close is the resemblance of a given object to a circle.

 

[A.T.]

In my world these are good questions. Perhaps I'm talking about how close is the resemblance of a given object to a circle.

How do you quantify that "resemblance"?

 

[Agent Smith]

How do you quantify that "resemblance"?

With $\pi$

 

[A.T.]

With $\pi$

I didn't ask with what. I asked how.

Provide a formula to compute the "resemblance of a given object to a circle".

 

[PeterDonis]

for a set of imperfect circles A = {x, y, z}, if x's circumference to diameter ratio is closest to $\pi$ it is more circular than either y or z, oui?

For an imperfect circle, there is no unique "diameter". So how do you compute the ratio of circumference to diameter?

 

[jbriggs444]

For an imperfect circle, there is no unique "diameter". So how do you compute the ratio of circumference to diameter?

There is a class of figures with a unique diameter. Curves of constant width. The circle is one element of this class.

 

[PeterDonis]

The circle is one element of this class.

Sure, but only an exact circle. The post I responded to was talking about curves that aren't exact circles.

 

[jbriggs444]

Sure, but only an exact circle. The post I responded to was talking about curves that aren't exact circles.

Such as an n-gon (for odd $n$) with each side rounded out to be a circular arc about the opposing vertex? Would that be an example of an "inexact circle" with constant width?

 

[PeterDonis]

Such as an n-gon (for odd $n$) with each side rounded out to be a circular arc about the opposing vertex? Would that be an example of an "inexact circle" with constant width?

Please read the subthread between myself and the OP. It makes clear what kind of "inexact circle" we were talking about.

 

[PeterDonis]

an "inexact circle" with constant width?

There is no such thing. Nor have I claimed that there is. Again, please read the subthread.

 

[jbriggs444]

There is no such thing. Nor have I claimed that there is. Again, please read the subthread.

An "imperfect circle" is:

An imperfect circle's points may vary in distance from the center.

The points on the figure that I described vary in distance from the centroid. Which counts, I think, as the "center" of an imperfect circle.

 

[PeterDonis]

An "imperfect circle" is

You didn't go back through the whole subthread. (To be fair, the OP didn't link to previous posts in the subthread in the post you quoted.) See post #45:

Which you will never be able to do perfectly. The compass point you're drawing with has a finite width, the compass itself has a finite amount of "play" in its spacing, and you have a finite error in your ability to keep the "center" point of the compass exactly in the same place. So no, nothing you draw this way will be an exact circle. Whether it is a good enough approximation to a circle will depend on what you are going to use it for.

We won't ever be able to draw a perfect circle, ok. But for a set of imperfect circles A = {x, y, z}, if x's circumference to diameter ratio is closest to $\pi$ it is more circular than either y or z, oui?