Name this curve

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Aldarion
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Name this curve

Postby Aldarion » Mon Nov 01, 2010 11:49 am UTC

Here's a strange curve I stumbled upon.

Draw a unit circle centered at (1,1), take the lower-left quarter:

Image

Now, from any point on the arc draw a line of length 1 through the origin:

Image

Here's how the locus of all the endpoints of such lines looks:

Image

Here's the curve itself:

Image

The question is: does it have a name? What can be said about it? Is it useful for... something?
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xepher
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Re: Name this curve

Postby xepher » Mon Nov 01, 2010 1:21 pm UTC

I'm not totally sure that this is it, but it definitely looks like a rose curve.
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Tirian
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Re: Name this curve

Postby Tirian » Mon Nov 01, 2010 2:22 pm UTC

From what I can recall, the polar equation for a unit circle centered at (1,1) is [imath]r=f(\Theta)=\sqrt{2}\cos(\Theta-\frac{\pi}{4})+\sqrt{1-2\sin^2(\Theta-\frac{\pi}{4})}[/imath]. So you'd need to find the range of [imath]\Theta[/imath] that delivered just the lower right quadrant of that circle and then just calculate [imath]g(\Theta)= f(\Theta)-1[/imath] over that range to get what you want. Perhaps it would be less ugly as a parametric equation.

I just looked over the small bestiary of polar curves in Schaum's Mathematical Handbook and this did not seem to appear. I imagined it's been studied and named by someone along the line, but whatever it is would not appear to be in the Hall of Fame. It's pretty, though.

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Re: Name this curve

Postby gorcee » Mon Nov 01, 2010 2:33 pm UTC

I shall name it Richard.

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Re: Name this curve

Postby cba » Mon Nov 01, 2010 6:38 pm UTC

Tirian wrote:From what I can recall, the polar equation for a unit circle centered at (1,1) is [imath]r=f(\Theta)=\sqrt{2}\cos(\Theta-\frac{\pi}{4})+\sqrt{1-2\sin^2(\Theta-\frac{\pi}{4})}[/imath]. So you'd need to find the range of [imath]\Theta[/imath] that delivered just the lower right quadrant of that circle and then just calculate [imath]g(\Theta)= f(\Theta)-1[/imath] over that range to get what you want. Perhaps it would be less ugly as a parametric equation.


a far nicer equation for a circle centered at (1,1) in polars is: [imath]r(\theta) = sin(\theta) + cos(\theta) - \sqrt{sin(2\theta)}[/imath]
this is the bottom left corner of the circle over the range [imath]0 \le \theta \le \frac{\pi}{2}[/imath]
but it might be easier to get the equation using complex numbers... ie. [imath]|z-(1+i)| = 1[/imath]
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Re: Name this curve

Postby Sagekilla » Mon Nov 01, 2010 7:11 pm UTC

I managed to derive the equation that plots that curve you're talking about. Talk about an ugly expression..

Just out of curiosity, are you using Mathematica? Cause I am and my graph looks identical to yours.

I thought it was a bit easier working it out as a parametric equation.
You get a very nice expression for the quarter circle, with [imath]0 \le t \le \frac{\pi}{2}[/imath]
[math]X_c(t) = 1 - cos(t)[/math]
[math]Y_c(t) = 1 - sin(t)[/math]

(Lots of manipulating equations happened here)


Final result looked like:
[math]x(t) = m(t)X_c(t)[/math]
[math]y(t) = m(t)Y_c(t)[/math]

Where:
[math]m(t) = 1 - \frac{1}{\sqrt{3 - 2 cos(t) - 2 sin(t)}}[/math]
Last edited by Sagekilla on Wed Nov 03, 2010 3:25 am UTC, edited 1 time in total.
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Aldarion
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Re: Name this curve

Postby Aldarion » Tue Nov 02, 2010 9:27 am UTC

Yeah, I used Mathematica, and my equations looked like this:

First, the circle:
x=x, y[x] = [imath]1-\sqrt{-(x-2)x}[/imath]. The quarter-circle is for [imath]0\leq x \leq1[/imath].

Then, let d[x] be the distance between a point on the circle (x,y[x]) and the origin. [imath]d[x]=\sqrt{x^2+y[x]^2}[/imath].

And now, the co-ordinates of the curve are (p[x],q[x]), [imath]0\leq x \leq1[/imath], when p[x_] := -Sin[ArcTan[y[x]/x]]*(1 - d[x]) and q[x_] := -Cos[ArcTan[y[x]/x]]*(1 - d[x]) .

Not as elegant as yours, Sagekilla, but perhaps more basic...

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Re: Name this curve

Postby Sagekilla » Wed Nov 03, 2010 3:24 am UTC

Aldarion wrote:Not as elegant as yours, Sagekilla, but perhaps more basic...


Well the real elegance I think is that you can take multiple approaches to one problem and end up at one common result ;)
The way you did it was actually where I started, but before I finished I thought it might be a bit easier if I worked it out parametrically.

My equations above are only nice because I found a way to simplify all my results.
The first solution, along with all the intermediary stuff, was actually quite messy.

I actually ended up calculating the arc length at one point instead of just saying "Duh, use euclidean distance"
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Re: Name this curve

Postby cba » Wed Nov 03, 2010 1:20 pm UTC

what about extending this curve to be formed from a whole circle?
just change it so that the line of length 1 always points towards the origin...
I would try it myself, but I dont have the program you used to make those graphs...
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Aldarion
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Re: Name this curve

Postby Aldarion » Wed Nov 03, 2010 8:37 pm UTC

cba wrote:what about extending this curve to be formed from a whole circle?
just change it so that the line of length 1 always points towards the origin...
I would try it myself, but I dont have the program you used to make those graphs...


Oh, I'm going to do that! Actually, I want to generalize it in two ways: first, to have it work for any kind of curve, and second - for any point, not just the origin. Of course, I've already generalized it for any length of the line, not just 1. The results are before you (the length cycles from -1 to 2):

Image
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Re: Name this curve

Postby cba » Wed Nov 03, 2010 9:55 pm UTC

the part of the curve in the first quadrant is an interesting shape...
what program do you use to do the animation and stuff in?
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Re: Name this curve

Postby Esquilax » Thu Nov 04, 2010 12:31 am UTC

I made a nice version of this curve in Geogebra, unfortunately I don't seem to be allowed to attach my actual geogebra source, but it's pretty neat.
I plotted the curve around the whole circle.
Also, since it seemed related, I plotted the curve which is the locus of points on the line through the origin and a point on the circle, a distance 1 from the point on the circle, but away from the origin rather than towards it. I also added the parameter L on the slider, which when animated gives the graphic Aldarion posted.
Attachments
geoCurve.png
curve
Spoiler:
Image

Aldarion
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Re: Name this curve

Postby Aldarion » Thu Nov 04, 2010 11:26 am UTC

Very nice, Esquilax! I'd better get working on my improvements now.

cba, for the animation I'm using Mathematica (v. 7), it can export animated GIFs.
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Re: Name this curve

Postby cba » Thu Nov 04, 2010 3:17 pm UTC

Esquilax, any chance you could like, type up or somehow attach the paramaters/file for geogebra?
I downloaded it last night to have a play around with and was trying to get the same result, but I just can't get my head around how the locus tool works (or in my case doesnt seem to work at all lol)
normally just playing around I can figure it out, but the examples they give in the help file arent that great for figuring out how to model this situation...

edit: actually, after more playing around I managed to figure it out :P
I was trying to do it all the wrong way lol
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Re: Name this curve

Postby Yakk » Thu Nov 04, 2010 4:43 pm UTC

So you have shape S:[0,1]->RxR. You fix a point p not in the image of S, and a signed magnitude m.

For each t in [0,1], let u(t) := (S(t)-p) / || S(t)-p ||.

Then we define a new curve C:[0,1]->RxR := S(t)+n u(t)

The original question has S being the lower-left quarter-unit-radius-circle centered at (1,1), n is -1, and p is (0,0).

That should let you do it with any shape, and any 'size', and any point.

Oh, and the natural next step would be to replace p with p(t) (where the above corresponds to a constant p(t) function), and the restriction of p is not in image of S with p(t) != S(t) for any t.

Also of note -- stick p at infinity along a particular direction, and the above corresponds to a translation by m.
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Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

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Re: Name this curve

Postby Esquilax » Thu Nov 04, 2010 6:22 pm UTC

cba wrote:Esquilax, any chance you could like, type up or somehow attach the paramaters/file for geogebra?
I downloaded it last night to have a play around with and was trying to get the same result, but I just can't get my head around how the locus tool works (or in my case doesnt seem to work at all lol)
normally just playing around I can figure it out, but the examples they give in the help file arent that great for figuring out how to model this situation...

edit: actually, after more playing around I managed to figure it out :P
I was trying to do it all the wrong way lol


Even though you say you've figured it out, I'll explain just in case anybody else wants to try.
In the diagram I uploaded, I actually have a hidden circle centered at A of radius 1 (L actually, varying based on the slider), and a hidden line which contains both A and O. B and F are the intersections of that circle and that line on either side of O, and the curves are the loci of B and F, respectively, as A moves along the circle. To make a locus you just select the locus tool, click the dependent point you want to use, and then click the point that your previous point depends on, which can range freely along a curve.

I actually did play around with a lot more generalizations, for example it's very easy to move the point O = (0,0) around to any point, and you can use any curve you can manage to plot to have A travel along, rather than simply using a circle.
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Re: Name this curve

Postby cba » Thu Nov 04, 2010 6:54 pm UTC

I think I actually found an easier way of doing it. A tiny bit more mathematical though...
I set a circle centered at (1,1) with radius 1, then added another point restricted to lie on the circle.
I then made one more point with co-ordinates [imath](\sqrt{x(C)^2 + y(C)^2} - 1; angle[C])[/imath] (this is a polar co-ordinate, hence the ';' instead of a ',') where C is the point I put on the circle...
I then just used the locus tool and clicked on C and the other point and voila! lol
the -1 after the square root sign can then just be changed to a variable, say m, which is linked to a slider :)
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Re: Name this curve

Postby Yakk » Tue Mar 26, 2013 9:17 am UTC

And the general version:
[imath](\sqrt{x(C-p)^2 + y(C-p)^2} + m; angle[C-p])+p[/imath]
does the transformation through the point p with magnitude m (where a magnitude of -1 corresponds to the original post).
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Aldarion
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Re: Name this curve

Postby Aldarion » Tue Mar 26, 2013 9:23 am UTC

I was going through my old documents, and came across the Mathematica notebook with my work on the curve.

Now here's a new question: how would I calculate the curve's length and area?
Would it help to get a closed-form, non-parametric, equation for it first?
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Xenomortis
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Re: Name this curve

Postby Xenomortis » Tue Mar 26, 2013 9:29 am UTC

The curve is generated by a line from the origin sweeping out an arc.
How the line's length varies shouldn't be too difficult and the angle it makes crossing the origin is simple enough.
Just some rambling thoughts on how I might try to calculate the area.
Image

Demki
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Re: Name this curve

Postby Demki » Tue Mar 26, 2013 11:50 am UTC

I've made it in geogebra with varying length. I can't upload the file here, if anyone wants I'll upload it to dropbox or something. I exported a gif of the distance going from 0 to 3, and it looks neat:
Image
a=|AE1|=|AE2|
Translating the circle such that the origin is in it or on it makes some interesting shapes too.

As for naming it... uhm... Sirkle!

Dark Avorian
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Re: Name this curve

Postby Dark Avorian » Tue Mar 26, 2013 2:24 pm UTC

GO FREE SIRKLE! GO FREE!
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nadando
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Re: Name this curve

Postby nadando » Tue Mar 26, 2013 7:08 pm UTC

It's a pretty straightforward numerical integration (assuming I didn't screw something up). I get an area of -0.198140234735592 (which is not, by the way, equal to 1 - 1 / Gauss's constant) and a curve length of 1.639346234237188 (with 0 <= t <= 1).

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Re: Name this curve

Postby gmalivuk » Thu Mar 28, 2013 1:17 am UTC

A bit of searching around failed to bring up any name for this, though I was unable to find an article I'm sure I saw once about the perhaps counterintuitive fact that the locus on the other side of the origin from the circle will never itself be a circle. (A naive argument from symmetry might seem to suggest that, if you have two wheels set up, with a rod between, say, the rightmost point on each, and then you spin the wheels in opposite directions, the rod ought to slide through a stable pivot point at the origin.)

I did come across a few geometry-related books that point out the fact that this locus can be treated as the path of a particular point on the rod of an oscillating cylinder steam engine, but none of them gave it a name.
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