When I was a kid, about third grade or so, we did an art project that involved drawing X and Y axes, marking an equal number of points on each, and then drawing lines from the greatest of X to the least of Y. The next point down X went one point farther out Y, and so forth, until you connected the least of X to the greatest of Y: X | Y 9 | 1 8 | 2 7 | 3 6 | 4 5 | 5 4 | 6 3 | 7 2 | 8 1 | 9 The result was an aesthetically pleasing arc that followed the outmost intersections. No, I never learned what practical application these things have. But I've seen them here and there over the years, mostly used in some artistic context. However, I'm convinced there is something more to the things than simply being pretty. What are they called? No, really, I wouldn't even know how to Google these things. And if there is a practical application to them beyond simply looking pretty, I'm happy to receive the lesson. Thanks.
You put 9 twice in the Y's, I assume the first is an 8. And I think i remember something like this too... only they include 0?
Tess Aren't those coordinates? They graph a sloped line. Maybe I'm misinterpreting, seeing as how you mention an arc: Are you plotting points on a graph, out on an X axis then up on a Y?
Wonderful stuff! I remember seeing them made with string and nails. Please Register or Log in to view the hidden image! Try googling "String Art" and "Bezier curves"
Here's the pattern Tiassa was describing: Please Register or Log in to view the hidden image! String art at Math Cats
Cool stuff, indeed You have it exactly, Pete. Thank ye! Is there really no technical name for the outcome? Or equation that describes the process? I mean, I can't just start throwing words out because, not being a mathematician, I would only embarrass myself. It's not that I'm trying to be stubborn about it. Rather, brief surveys leave me unable to connect Bezier or other parametric curves to string art in any functional context. I mean, first impressions for us non-mathletes is often, "Wow, cool stuff." But, you know, there must be something more technical about it. Update/Correction: D'oh! At 3:30 this morning, or whatever time it was, I completely missed the sentence in the Wikipedia entry on Bézier curves about string art.
Let's see if we can't derive the equation of the curve you'd get if you smoothed over the edge of the thing Pete's animation shows (I'm typing this on the fly, I haven't put pen to paper so I might go down a dead end...). By the OP we connect 9 to 1, 8 to 2 etc. What's the equation of a particular line? We join N to 10-N. In terms of (x,y) coordinates (not the X,Y of the OP) a line joins (0,N) to (10-N,0). The equation of a line from \((x_{1},y_{1})\) to \((x_{2},y_{2})\) is \(y-y_{1} = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}(x-x_{1})\), so \(y = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}(x-x_{1}) + y_{1} \equiv mx+c\). Putting in the numbers we get \(y = \frac{0-N}{(10-N)-0}(x-0) + N = N - \frac{N}{10-N}x\) so \((y-10)(10-N) = -Nx\). Note that we've got \(y_{1}+x_{2}=10\) for the tangents, as stipulated by the original post's definition of them. Now if this forms part of a curve in the way Pete's picture shows it is the tangent to the curve at some point. It might be a bit confusing but x parametrises the tangent line but N parameterises the point on the line we're trying to find. Unfortunately the curve cannot be written as y = f(x) because it is multivalued (plot a slew of the tangents using Mathematica or the like and you'll see the curve being formed). As such you have to do everything with vectors and now I don't feel like going down that road...
\(s \in \left[0,1\right] \\ X(s) = 10 s \\ Y(s) = 10 - 10 s \\ L(s,x,y) = Y(s)x + X(s)y -X(s)Y(s) = 10 s y + 10 (1-s) x + 100 (s^2 -s)\) So each line is given by \(L(s, x, y) = 0\) for some fixed value of s and has intercepts at (X(s),0) and (0, Y(s)). So we find an intersection by solving: \(L(s,x,y) = 0 \\ L(s+\Delta s,x,y) - L(s,x,y) = 0 \\ \\ 10 s y + 10 (1-s) x + 100 (s^2 -s) = 0 \\ 10 \Delta s y - 10 \Delta s x + 100 ( (2 s - 1) \Delta s + ( \Delta s ) ^2 ) = 0 \\ \\ y = 10 (s -1) (s + \Delta s -1) \\ x = 10 (s \Delta s + s^2)\) Check: \(L \left( s, 10 (s \Delta s + s^2, 10 (s -1) (s + \Delta s -1)) \right) = 100 s (s - 1) (s + \Delta s -1) + 100 (1-s) (s \Delta s + s^2) + 100 (s^2 -s) = 100 (s - 1) \left( s^2 + s \Delta s - s - s^2 - s\Delta s + s \right) = 0 \\ L(s+\Delta s,x,y) = L \left( s + \Delta s, 10 (s \Delta s + s^2, 10 (s -1) (s + \Delta s -1)) \right) = 100 (s + \Delta s) (s - 1) (s + \Delta s -1) + 100 (1-s - \Delta s) (s \Delta s + s^2) + 100 (s + \Delta s) (s + \Delta s - 1) \\ = 100 \left( s^3 + 2 s^2 \Delta s - 2 s^2 + s (\Delta s)^2-3 s \Delta s +s- (\Delta s)^2+\Delta s) \right) + 100 \left( -s^3-2 s^2 \Delta s+s^2-s (\Delta s)^2+s \Delta s \right) + 100 \left( s^2+2 s \Delta s-s+(\Delta s)^2-\Delta s \right) = 0\) In the limit of infinite lines, the smooth curve is made up entirely of intersections. \( \left(\tilde{x}(s), \; \tilde{y}(s)\right) = \left. \lim_{\Delta s \to0} (x,y) \right| _{s} \quad = \quad \left( 10 s^2 , \; 10 s^2 - 20 s + 10 \right)\) Say \(P = \left( X(s), \; 0 \right) = \left( 10 s , \; 0 \right) ; \; Q = \left( 0 , \; Y(s) \right) = \left( 0 , \; 10 (1-s) \right) \) Then \( s P + (1-s) Q = \left( 10 s^2 , \; 10 s^2 - 20 s + 10 \right) = \left(\tilde{x}(s), \; \tilde{y}(s)\right) \) /// This is the same as the graph of \(y = \left( \sqrt{x}-\sqrt{10} \right)^2\) or \(x^2 -2 xy + y^2 - 20 x - 20 y + 100 = 0\) and so geometrically it is a parabola.
From \(L(s,x,y) = 0\) we get \(y = Y(s)-\frac{Y(s)}{X(s)}x = 10 - 10 s - \frac{10 - 10s}{10 s} x = 10 - 10 s + (1 - \frac{1}{s}) x \) and can read off the slope of the line as \(1 - \frac{1}{s}\). From \(x = 10 s^2\) and \(y = 10 + x - \sqrt{40 x}\) we can compute the slope of the tangent at any point as \(1 - \sqrt{\frac{10}{x}} = 1 - \frac{1}{s}\) And I've already demonstrated that the point \(\left( 10 s^2, \, 10 (1-s)^2\right)\) lies both on the parabola and on the "string art" line \(L(s,x,y) = 0\). Thus the string art is a collection of tangents to a limiting parabola. // Since the parabola is given by: \(\begin{pmatrix} x & y & 1\end{pmatrix} \begin{pmatrix} \cos \frac{\pi}{4} & \sin \frac{\pi}{4} & 0 \\ -\sin \frac{\pi}{4} & \cos \frac{\pi}{4} & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \frac{\sqrt{2}}{20} & 0 & 0 \\ 0 & 0 & -\frac{1}{2} \\ 0 & -\frac{1}{2} & \frac{5\sqrt{2}}{2} \end{pmatrix} \begin{pmatrix} \cos \frac{\pi}{4} & -\sin \frac{\pi}{4} & 0 \\ \sin \frac{\pi}{4} & \cos \frac{\pi}{4} & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ 1 \end{pmatrix} = 0\) it is geometrically congruent to the parabola \(y = \frac{\sqrt{2}}{20} x^2 + \frac{5\sqrt{2}}{2}\) with focus at \(\left( 0, \, 5 \sqrt{2} \right)\) and directrix at y=0. So it follows that for our original parabola, the directrix is \(y=-x\) and the focus is at \(\left( 5, 5 \right)\). For an arbitrary point on the parabola, the distance to the focus is \(\sqrt{ \left( 10 s^2 - 5 \right)^2 + \left( 10 (1 - s)^2 - 5 \right)^2 } = 5 \sqrt{2} ( 2 s^2 - 2 s + 1)\) The distance to the directrix is \(\frac{x + y}{\sqrt{2}} = 5 \sqrt(2) ( s^2 + s^2 - 2s + 1)\) and that they are identical is one of the important geometric properties of the parabola.
Look at http://en.wikipedia.org/wiki/Envelope_(mathematics) Example 2 is the case for the OP. It is (as rpenner has said) a parabola.
