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enter image description here I like to draw an ellipse via 2 fixed points and a rope between the fixed points (2 focuses). I wanted to extend the idea. Point A,B,C,D are fixed points and Point E can move freely. Point E,B,C have small pulleys without friction and also their perimeters are very small. (Take zero for theoretical calculation)

If we fix a rope on Point A then it goes Point E and then Point B then C then E and finally fix again on Point D as shown figure above. If we move E while the rope stretched, we can draw a curve similar to ellipse.

We can express the equation of the closed curve as shown graph above :

$$\sqrt{(x+a)^2+y^2}+\sqrt{(x+b)^2+y^2}+\sqrt{(x-a)^2+y^2}+\sqrt{(x-b)^2+y^2}=l-2b$$

Where $l$ is lenght of the rope.

This is a symmetric curve over x and over y lines like ellipse.

My questions:

  • Is there any special name of this kind of curves?
  • What is the area formula of such closed curve? Is the formula similar like circle and ellipse starts with $\pi$ such as $\pi.f(a,b,l)$ ?

I tried polar coordinate transform but I could not find the area.

We can do many combinations with different number of fixed points. There is no limit of such closed curves :

Another example is:

3 fixed points (one fixed point is with a small pulley) We can get this curve if we select Point B and C in same point $P(x_1,y_1)$ in figure above.

$$\sqrt{(x+a)^2+y^2}+2\sqrt{(x-x_1)^2+(y-y_1)^2}+\sqrt{(x-a)^2+y^2}=l$$


First of all, I focused on the simplest case (2 fixed end point, 1 fixed point with pulley on origin)

We can get this curve if we select $b=0$ in figure above.

$$\sqrt{(x+a)^2+y^2}+2\sqrt{x^2+y^2}+\sqrt{(x-a)^2+y^2}=l$$ $x=r\cos \alpha$

$y=r\sin \alpha$

$$\sqrt{r^2+a^2-2ax}+2r+\sqrt{r^2+a^2+2ax}=l$$

$$2r^2+2a^2+2\sqrt{r^2+a^2-2ax}\sqrt{r^2+a^2+2ax}=(l-2r)^2$$

$$2\sqrt{r^2+a^2-2ax}\sqrt{r^2+a^2+2ax}=2r^2-4rl+l^2-2a^2$$

$$4(r^2+a^2-2ax)(r^2+a^2+2ax)=(2r^2-4rl+l^2-2a^2)^2$$

$$4(r^2+a^2)^2-16a^2x^2=(2r^2-4rl+l^2-2a^2)^2$$

$$4(r^2+a^2)^2-16a^2r^2\cos^2 \alpha =(2r^2-4rl+l^2-2a^2)^2$$

$$4(r^2+a^2)^2 -(2r^2-4rl+l^2-2a^2)^2=16a^2r^2\cos^2 \alpha$$

$$(4r^2-4rl+l^2)(4a^2+4rl-l^2)=16a^2r^2\cos^2 \alpha$$

If we expand the terms, We will have a polynomial with degree $3$.

$r^3+(m+n\cos^2 \alpha)r^2+tr+k=0$

And my aim is to find the area of the closed curve.

$$A = 4 \int_{0}^{\pi/2} \frac{r^2}{2} d \alpha $$

I am stuck in this point because I do not see the solution easy after here. Please help me if you see how to solve the integral.

Thanks a lot for helps

Note:We can easily find the points on X axis of the curve ,$ (+\frac{l}{4},0) ; (-\frac{l}{4},0)$

And the point on Y axis of the curve , $(0,+\frac{l}{4}-\frac{a^2}{l}) ; (0,-\frac{l}{4}+\frac{a^2}{l})$

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    $\begingroup$ This is called a multifocal ellipse (also known as $n$-ellipse, $k$-ellipse, ...) $\endgroup$ – achille hui Aug 2 '16 at 14:13
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    $\begingroup$ Believe it or not, great Maxwell stumbled upon the same idea as you - see here $\endgroup$ – Yuriy S Aug 3 '16 at 8:39
  • $\begingroup$ @You'reInMyEye i saw the book in the multifocal ellipse link above in references that achille hui gave. I am happy to know that I have common interest with Maxwell master. Have you found any reference or clue what the area of the closed curve is? I cannot see many page of that book. Thanks $\endgroup$ – Mathlover Aug 3 '16 at 11:02
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    $\begingroup$ @Mathlover, no, for the general case I don't know. I would advise searching for some recent books or articles (Wikipedia page has several links). It's unlikely Maxwell studied that in detail. I'm mainly fascinated by these curves because they can be used to approximate general convex shapes, and don't require much data to be stored (n-ellipse needs 2n+1 numbers to define it). This might be useful in computer vision and other applications $\endgroup$ – Yuriy S Aug 3 '16 at 11:06
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    $\begingroup$ Desmos is a good tool for plotting such expressions. $\endgroup$ – Lucian Aug 7 '16 at 10:54
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As mentioned in the comments, the locus of points whose sum-distance to n given points is constant is called an n-ellipse.

The image shows a 4-ellipse; without loss of generality, we can scale the figure so that B and C are at $(\pm1,0)$ (and A and D at $(\pm a,0)$), then ignore the part of the rope stretched across BC so that $l=\mathsf{AE+BE+CE+DE}$. The locus of E is the locus of intersection of two ellipses centred on the origin:

  • One has foci B and C and axes $k$ and $\sqrt{k^2-1}$
  • The other has foci A and D and axes $m$ and $\sqrt{m^2-a^2}$, with $2(k+m)=l$ $$\frac{x^2}{k^2}+\frac{y^2}{k^2-1}=1; \frac{x^2}{m^2}+\frac{y^2}{m^2-a^2}=1$$

After some algebraic manipulation we find $$x^2=\frac{m^2k^2(m^2-a^2-k^2+1)}{m^2-a^2k^2}$$ $$y^2=\frac{(k^2-m^2)(m^2-a^2)(k^2-1)}{m^2-a^2k^2}$$ Since manipulating all this would be too tedious, take $a=2$ and $l=8$ as an example. The expressions reduce to $$x^2=\frac{(4-k)^2k^2(13-8k)}{16-8k-3k^2}$$ $$y^2=\frac{(8k-16)(12-8k+k^2)(k^2-1)}{16-8k-3k^2}$$ Let $X=x^2$ and $Y=y^2$. The resultant of $X(k)$ and $Y(k)$, the curve's implicit equation, is quintic in X and Y: $$-6720X^5 + (-42688Y + 524160)X^4 + (-104128Y^2 + 2625248Y - 13608000)X^3 + (-123456Y^3 + 4595473Y^2 - 50494240Y + 139856640)X^2 + (-71680Y^4 + 3405920Y^3 - 55511328Y^2 + 349444224Y - 591252480)X + (-16384Y^5 + 911616Y^4 - 19293696Y^3 + 188524800Y^2 - 792115200Y + 870912000)=0$$ And here is a pretty plot of this curve, with the red dots at $(\pm1,0)$ and $(\pm2,0)$: Plot of the 4-ellipse; looks like an eye

Now for the main part of my answer: it is not possible to find the area bounded by this curve analytically, whether by parametric equation or resultant, because the equations for x and y alone contain a square root within them that prevents the use of integration techniques for rational functions, and because the quintic expression in X and Y is not solvable. What I would do to find its area numerically is the following:

  • For many positive values of x, find the corresponding y-value (which will also be positive)
  • Use the points sampled to estimate the area of the curve in the first quadrant
  • Since the curve is symmetric about the x- and y-axes, multiply the estimate by 4 to get the total area

Doing so with the above curve yields a result of 7.614914 (to 6 decimal places).

The computations for the 3-ellipse are similar, but involve the intersection of an ellipse with a circle.

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  • $\begingroup$ It is very nice approach to use the idea of 2 ellipses intersection. It helps to simplify many pre-steps.. Your result is very interesting. It shows that n-ellipses has strong relation with algebric equation solubility too. However, I still have a sense that we may find a way to have a closed form of area formula after some integral transform methods in limits $0, \pi/2$. Thanks a lot for great effort and contribution $\endgroup$ – Mathlover Aug 10 '16 at 10:06
  • $\begingroup$ You do not use polar transforms to find the area of algebraic curves like this. There is no relation between $\pi$ and curves of cubic and higher orders, and generally no way to find an analytic expression for the area contained within an algebraic curve (the degree-6 quadrifolium with area $\pi/2$ is an exception, but it has an integrable polar form). Some things just can't be done in mathematics – and you can prove those impossibilities. Indeed, in practical work we use numerical methods all the time with algebraic curves. $\endgroup$ – Parcly Taxel Aug 10 '16 at 10:39
  • $\begingroup$ @ParclyTaxel: You gave a particular example of $a,l$. Checking with Mathematica, I find that for the special case $l=2(a\pm1)$, the curve's implicit equation is just a quartic in $X,Y$. Would your conclusions regarding the area bounded by this special case be different for general $a,l$? $\endgroup$ – Tito Piezas III Aug 26 '16 at 17:06
  • $\begingroup$ @TitoPiezasIII Hm... if $l=2(a+1)$ then the curve degenerates to a point; its area is then zero. If $l=2(a-1)$ then the curve can't even exist. (You may also be able to help with my hendecagon question – it's been sitting around for a while, and I was looking at your answers trying to find a solution.) $\endgroup$ – Parcly Taxel Aug 26 '16 at 17:42
  • $\begingroup$ @ParclyTaxel: I've done work on transforming quintics from one form to another. A quick look at that post shows there "might" be a way, but I need you to clarify some details more. $\endgroup$ – Tito Piezas III Aug 26 '16 at 17:53

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