Name this curve
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Name this curve
Here's a strange curve I stumbled upon.
Draw a unit circle centered at (1,1), take the lowerleft quarter:
Now, from any point on the arc draw a line of length 1 through the origin:
Here's how the locus of all the endpoints of such lines looks:
Here's the curve itself:
The question is: does it have a name? What can be said about it? Is it useful for... something?
Draw a unit circle centered at (1,1), take the lowerleft quarter:
Now, from any point on the arc draw a line of length 1 through the origin:
Here's how the locus of all the endpoints of such lines looks:
Here's the curve itself:
The question is: does it have a name? What can be said about it? Is it useful for... something?
I'm not good, I'm not nice, I'm just right.
Re: Name this curve
I'm not totally sure that this is it, but it definitely looks like a rose curve.
http://en.wikipedia.org/wiki/Quadrifolium
http://en.wikipedia.org/wiki/Quadrifolium
Re: Name this curve
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{12\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.
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.
Re: Name this curve
I shall name it Richard.
Re: Name this curve
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{12\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]
For the first time ever! flatland in a mindboggling 3D!
import antigravity
import antigravity
Re: Name this curve
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]
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]
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.
http://en.wikipedia.org/wiki/DSV_Alvin#Sinking wrote:Researchers found a cheese sandwich which exhibited no visible signs of decomposition, and was in fact eaten.
Re: Name this curve
Yeah, I used Mathematica, and my equations looked like this:
First, the circle:
x=x, y[x] = [imath]1\sqrt{(x2)x}[/imath]. The quartercircle 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 coordinates 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...
gorcee, were you ever asked for directions by two men in a hotair balloon?
First, the circle:
x=x, y[x] = [imath]1\sqrt{(x2)x}[/imath]. The quartercircle 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 coordinates 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...
gorcee, were you ever asked for directions by two men in a hotair balloon?
I'm not good, I'm not nice, I'm just right.
Re: Name this curve
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"
http://en.wikipedia.org/wiki/DSV_Alvin#Sinking wrote:Researchers found a cheese sandwich which exhibited no visible signs of decomposition, and was in fact eaten.
Re: Name this curve
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...
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...
For the first time ever! flatland in a mindboggling 3D!
import antigravity
import antigravity
Re: Name this curve
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):
I'm not good, I'm not nice, I'm just right.
Re: Name this curve
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?
what program do you use to do the animation and stuff in?
For the first time ever! flatland in a mindboggling 3D!
import antigravity
import antigravity
Re: Name this curve
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.
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.
Spoiler:
Re: Name this curve
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.
cba, for the animation I'm using Mathematica (v. 7), it can export animated GIFs.
I'm not good, I'm not nice, I'm just right.
Re: Name this curve
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
I was trying to do it all the wrong way lol
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
I was trying to do it all the wrong way lol
For the first time ever! flatland in a mindboggling 3D!
import antigravity
import antigravity
 Yakk
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 Posts: 11129
 Joined: Sat Jan 27, 2007 7:27 pm UTC
 Location: E pur si muove
Re: Name this curve
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 lowerleft quarterunitradiuscircle 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.
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 lowerleft quarterunitradiuscircle 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.
One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision  BR
Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.
Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.
Re: Name this curve
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
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.
Spoiler:
Re: Name this curve
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 coordinates [imath](\sqrt{x(C)^2 + y(C)^2}  1; angle[C])[/imath] (this is a polar coordinate, 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
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 coordinates [imath](\sqrt{x(C)^2 + y(C)^2}  1; angle[C])[/imath] (this is a polar coordinate, 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
For the first time ever! flatland in a mindboggling 3D!
import antigravity
import antigravity
 Yakk
 Poster with most posts but no title.
 Posts: 11129
 Joined: Sat Jan 27, 2007 7:27 pm UTC
 Location: E pur si muove
Re: Name this curve
And the general version:
[imath](\sqrt{x(Cp)^2 + y(Cp)^2} + m; angle[Cp])+p[/imath]
does the transformation through the point p with magnitude m (where a magnitude of 1 corresponds to the original post).
[imath](\sqrt{x(Cp)^2 + y(Cp)^2} + m; angle[Cp])+p[/imath]
does the transformation through the point p with magnitude m (where a magnitude of 1 corresponds to the original post).
One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision  BR
Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.
Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.
Re: Name this curve
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 closedform, nonparametric, equation for it first?
Now here's a new question: how would I calculate the curve's length and area?
Would it help to get a closedform, nonparametric, equation for it first?
I'm not good, I'm not nice, I'm just right.
 Xenomortis
 Not actually a special flower.
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Re: Name this curve
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.
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.
Re: Name this curve
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:
a=AE_{1}=AE_{2}
Translating the circle such that the origin is in it or on it makes some interesting shapes too.
As for naming it... uhm... Sirkle!
a=AE_{1}=AE_{2}
Translating the circle such that the origin is in it or on it makes some interesting shapes too.
As for naming it... uhm... Sirkle!

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Re: Name this curve
GO FREE SIRKLE! GO FREE!
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Re: Name this curve
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).
 gmalivuk
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Re: Name this curve
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 geometryrelated 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.
I did come across a few geometryrelated 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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