Drawing an ellipse from two tangents.

Factual Questions
21

All right, here’s what I’ve come up with, though I’m not sure that it will be of much help:

I’ve been trying to find an ellipse that matches everything exactly, though I notice that flight has been working on finding a close approximation, which might be the best approach after all.

Anyway, the first ellipse I gave was under the assumption that the ellipse would be oriented horizontally/vertically. It matched perfectly, but at least one focus was inside the building.

After the adjustment was made to the measurements, I tried the same approach (again, assuming that the horiz/vert orientation). I’m pretty confident that no such ellipse exists, actually, though I may have made a mistake. The only conic that would match such a situation is a hyperbola, and I don’t think you want that.

So the next step is to drop the assumption that the ellipse is oriented horiz/vert. The general form for a conic is:

Ax[sup]2[/sup] + Bxy + Cy[sup]2[/sup] + Dx + Ey + F = 0

I’m taking the origin to be located at the bottom right corner; this forces F=0, since the ellipse goes through that point.

The slope at any point on the ellipse is given by:

  • (2Ax + By + D) / (Bx + 2Cy + E)

At the origin, we want this slope to be horizontal (zero). This forces D=0.

The ellipse must go through the point (-26.5, 7.5) (the left end of the connection). This gives the linear equation:

702.25A - 198.75B + 56.25C + 7.5E = 0

The slope at that same point must be -.5. This gives the linear equation:

53A - 20.75B + 7.5C + .5E = 0

Now comes the tricky part (for me, anyway). At what angle should the ellipse be rotated? It would be nice to specify such an angle, since this will give us another linear equation in A, B, C, and E. Then we’ll have three equations, four unknowns; solving the system will give us a single parameter, and we can see what happens as we let that parameter vary–hopefully giving us a workable solution (i.e., “usable” foci).

Say the ellipse is rotated by an acute angle t from horizontal or vertical. Then by rotation of axes we get the equation

cot(2t) = (A-C)/B, or, equivalently, our third linear equation:

A - (cot(2t))B - C = 0

(see here if you’re not sure where this comes from (warning: pdf file))

Now, what I did first was to assume that one axis of the ellipse was parallel to the straight line through the points to be connected (those points being the origin (0,0) and (-26.5,7.5))–an angle of about 15.8[sup]o[/sup].

I threw that in to the last linear equation listed above. However, barring a stupid mistake on my part, all conics satisfying those three linear equations were hyperbolas, not ellipses.

I’m getting a bit tired now, but I thought some of this might be useful to someone. With this approach, the key is to find an appropriate angle t such that:

  1. We actually get ellipses and not some other conic, and
  2. The foci are located in a clear spot.

Sorry I couldn’t be of more help.

22

Well, I think you got farther than I did.

You’re right, I need an ellipse since it will be easy to lay out using a string stretched from the two focal points. Thanks again for the help.

23

Well, here’s one other idea: use a spline. It would use a lot of string to mark out, but it would be relatively simple to make.

The dimensions are 7 1/2 by 26 1/2. Take a string and mark out a line along both sides. Then, take another string, and have it connect the midpoints of each side; 3 3/4 on the left and 13 1/4 on the bottom. Then, you connect the midpoint of the top section of the left side to the midpoint of the left section of the bottom side, and the midpoint of the bottom section of the left side to the midpoint of the right section of the bottom side. You keep subdividing like that, and get a spiderweb-like thingy.

Alternatively, you could step along the left side by a given length and then multiply by 3 8/15 to step along the bottom. If you stepped by 6 inches each step, you’d get line segments like:

left side - bottom side
0’ - 26’ 6"
6" - 24’ 9"
1’ - 23’
1’ 6" - 21’ 2 1/2"
2’ - 19’ 5"
2’ 6" - 17’ 8"
3’ - 15’ 11"
3’ 6" - 14’ 1 1/2"
4’ - 12’ 4 1/2"
4’ 6" - 10’ 7"
5’ - 8’ 10"
5’ 6" - 7’ 1"
6’ - 5’ 3 1/2"
6’ 6" - 3’ 6 1/2"
7’ - 1’ 9"
7’ 6" - 0’

24

Hey cornflakes, I think I might have something that’ll work for ya.

I gave up on the approach last night. I was trying to use Maple to solve the linear systems and graph the results for me. I am by no means a Maple expert, and I got tired of wrestling with it along with my own algebra mistakes, so I gave up on that idea.

Then I came back to the hyperbola idea. I had forgotten there is a way to construct a hyperbola using stakes and string, similar to the method used for ellipses. Here’s a Java applet demonstrating it:

http://www.sciences.univ-nantes.fr/physique/perso/gtulloue/conics/drawing/hyper_string.html

The basic idea: Drive two stakes in the ground, one at each focus of the hyperbola. Tie a length of string to each stake (two different strings). Have one person hold the ends of the strings together, taut. Have another person slip a ring over the ends of the strings. Move the ring down the string, keeping everything taut. This will trace out the hyperbola for you.

The hyperbola is:

[(y - k)[sup]2[/sup] / a[sup]2[/sup]] - [(x - 11.5)[sup]2[/sup] / b[sup]2[/sup]] = 1

with:

k = -24.642857

a[sup]2[/sup] = 607.27041

b[sup]2[/sup] = 1001.3214

The origin in this setup is the bottom left corner of the sidewalks. Units are in feet.

I haven’t double checked my algebra; if there’s any mistake it’s probably an easily fixed trivial one. I did graph it on Maple and it looks quite nice.

The center of this hyperbola is (11.5,-24.642857) (i.e., 24.642857 feet south of the bottom right sidewalk corner).

The foci are at (11.5,15.464397) and (11.5,-64.750111), i.e., 15.464397 feet north and 64.750111 feet south of the bottom right corner, respectively.

I don’t know offhand what the string lengths should be to construct it as I described above. If you’re still interested I’ll do that for you.

Hell, you could even turn it into a tourist attraction! “World’s Only Hyperbolic Sidewalk!” :wink:

(Not that I know that to be necessarily true, but that’s never stopped tourism advertisers before).

25

Sorry, I should have clarified one thing in the above. In the string/stake construction, the position of the ring will trace out the hyperbola for you.

26

Hey, the hyperbola might work! There is a building just south of the east-west sidewalk (which I didn’t mention since we were talking about ellipses), but it’s far enough away that I could find a second focal point. A hyperbola could be worked out by trial and error at the site, and my son will have an excuse to have class outside next year!

27

I’m glad to hear this might work!

If you are able to use the foci I mentioned earlier, then the string lengths should be:

27.67095263 feet for the northern focus

76.95666664 feet for the southern focus

These are, of course, the lengths necessary to reach from each focus to the upper left end of the sidewalk.

(Of course, I realize it’s a bit ridiculous for me to be giving you measurements down to the nearest hundred millionth of a foot. I guess, as Spock would say, “I endeavour to be accurate, Captain.”)

28

Not ridiculous at all! Next time that somebody asks “How long is a string”, I’ll have an answer! :smiley: