- #36
Again, why is this interesting?Agent Smith said:Cogito it should be ##\frac{21 + 1}{6 + 1}##
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 said:We use a compass, draw an object such that all of its points are equidistant from a given point.
Yes but what are the mathematical definitions?fresh_42 said: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.
##\frac{22}{7} = \frac{21 + 1}{7} = 3 + \frac{1}{7} = 3\frac{1}{7}##Mark44 said:Again, why is this interesting?
After a brief closing, the thread has been reopened.
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:Agent Smith said:what are the mathematical definitions?
Sum of angles in a triangle less than 180 degrees.Agent Smith said:1. Negative curvature = ?
Sum of angles in a triangle equal to 180 degrees.Agent Smith said:2. 0 curvature = ?
Sum of angles in a triangle greater than 180 degrees.Agent Smith said:3. Positive curvature = ?
https://en.wikipedia.org/wiki/Curvature#Plane_curvesAgent Smith said:@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?
Gracias. What about lines? Like a loop in 2D?PeterDonis said: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".
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?PeterDonis said: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 said: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?
fresh_42 said: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.
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.Agent Smith said: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?
You know, like this thread.fresh_42 said:more circular
This reasoning seems kind of...um...circular.Agent Smith said:By more circular I mean as close to being a circle as possible
How do you quantify that "resemblance"?Agent Smith said:In my world these are good questions. Perhaps I'm talking about how close is the resemblance of a given object to a circle.
With ##\pi##A.T. said:How do you quantify that "resemblance"?
I didn't ask with what. I asked how.Agent Smith said:With ##\pi##
For an imperfect circle, there is no unique "diameter". So how do you compute the ratio of circumference to diameter?Agent Smith said: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?
There is a class of figures with a unique diameter. Curves of constant width. The circle is one element of this class.PeterDonis said:For an imperfect circle, there is no unique "diameter". So how do you compute the ratio of circumference to diameter?
Sure, but only an exact circle. The post I responded to was talking about curves that aren't exact circles.jbriggs444 said:The circle is one element of this class.
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 said:Sure, but only an exact circle. The post I responded to was talking about curves that aren't exact circles.
Please read the subthread between myself and the OP. It makes clear what kind of "inexact circle" we were talking about.jbriggs444 said: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?
There is no such thing. Nor have I claimed that there is. Again, please read the subthread.jbriggs444 said:an "inexact circle" with constant width?
An "imperfect circle" is:PeterDonis said:There is no such thing. Nor have I claimed that there is. Again, please read the subthread.
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.Agent Smith said:An imperfect circle's points may vary in distance from the center.
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:jbriggs444 said:An "imperfect circle" is
Agent Smith said: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?PeterDonis said: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.
If we broaden our scope beyond the particular case that started the subthread (trying to draw a circle with a compass), yes, I would say the figure you describe could qualify as an "imperfect circle"--just not one with constant width.jbriggs444 said: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.
Why not constant width?PeterDonis said:If we broaden our scope beyond the particular case that started the subthread (trying to draw a circle with a compass), yes, I would say the figure you describe could qualify as an "imperfect circle"--just not one with constant width.
Because, as you note, the distance of the points on the curve from the centroid varies.jbriggs444 said:Why not constant width?
Ah, I see the disconnect: you are using "width" instead of "diameter". They're not the same thing. "Constant diameter" is the property of the exact circle that "imperfect circles" do not share, which is what I was referring to in my earlier post about that. That is not the same property as "constant width".jbriggs444 said:There is a class of figures with a unique diameter. Curves of constant width.
I would have wanted to ask the legislators how they planned to deal with all the wheels in the state that weren't hexagonal...Nik_2213 said:Dishonourable mention for those infamous US politicians who, in 1897. nearly enacted Pi as three (integer 3) to ease calculation...
Machining motorcycle parts.Vanadium 50 said:I'm trying to imagine a case where you need to know π in advance when building something and where 1/25 of a percent isn't good enough.
Nescio, measure the perimeter and divide it by the "width". The closer the ratio approaches ##\pi## the more circular it is. Archimedes & Zu Chongzhi used polygons to compute ##\pi##. A circle is an infinite-sided polygon?A.T. said:I didn't ask with what. I asked how.
Provide a formula to compute the "resemblance of a given object to a circle".
And that's just for surfaces. Try dimensions 3 or higher.fresh_42 said: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.