I often need to draw large ovals on fabric. I always use the method of
attaching a string to two points and then swinging the marker around
to make the oval.

To make the oval wider I move the foci down from the center line. To
make the oval taller I move the foci closer together.

However, I just experiment until I get the oval I want. I'd like to
just use equations to determine, given a desired width and height for
the oval, the horizontal distance between the foci and the vertical
distance from the center line.

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Clarification of Question by mxnmatch-ga on 25 Oct 2006 22:28 PDT

Ok, I guess I had two problems then. That information is exactly what
I needed to do ellipses.

But, when I say move the foci down from the center line I'm talking
about drawing a certain type of curve. In particular, I recently
needed to create what you might call "orange slices" which, when sewn
together, make a sphere.

Those slices aren't elapses. I draw them using the same method except
I move the foci off the axis. For instance, let's say that I want to
create a 1 foot diameter sphere. By experimentation I found that if
the axis goes from (-1.5,0) to (1.5,0) (that's 3 feet long) then I can
draw one side of the slice by attaching a string at (-1.25,-1) then
running it to (-1.5,0) then to the other foci at (1.25,-1). I can then
draw the curve starting at (-1.5,0) then going around to (1.5,0) just
like I would with an ellipse.

To draw the other side I just fold the cloth along the y axis at 0 and
cut it along the line I just drew.

Take six of those, sew them together, and I get a sphere with a 1 foot diameter.

I'd like to know what the equation is that would tell me, given a
desired sphere diameter, what the locations of the foci and the length
of the axis are. Even better would be if I could choose how many
slices make up the sphere.

If I could find out how to draw the slices to make an egg shape then
that would be fantastic. But, that's probably getting outside the
initial scope of the question.

I am not quite sure what you mean by "mov[ing] the foci down from the
center line."  Moving the foci up or down from the center line will
just move the ellipse up or down.  BTW an "oval" can mean lots of
shapes; the shape we're talking about here is more specifically an
ellipse.

The "center line" (by which I mean the line drawn between the two
foci) can be either vertical or horizontal, and this will give
different shapes for your ellipse.  If the foci are in line
horizontally, the ellipse will be wide; if the foci are in line
vertically, the ellipse will be tall.  If the foci are coincident
(i.e. they are in the same place) then you will draw a perfect circle,
i.e. an ellipse that is neither tall nor wide.

This discussion will assume a horizontal center line, i.e. the two
foci are horizontally in line with each other.  This will give results
that you can use to create wide ellipses, and if you need a tall
ellipse then you can just turn the fabric sideways (which is the same
as switching "width" for "height" in the following discussion).

Let's say you want an ellipse that will be T tall and W wide.  We need
to figure out L, the length of the string, and D, the distance between
the foci.

The length of the string is simply the same as the desired width of
the ellipse. L=W.  Easy enough.  The distance between the foci is more
complicated:

D = 2*SQRT((L/2)^2+(H/2)^2)

where (just to be clear) * is multiply, SQRT is square root, / is
divide, and ^2 is square.  So take the length, divide it by two, and
square it.  Call this "A".  Do the same for the height, and call it
"B".  Add A+B.  Take the square root of this sum, and then multiply
that by two.  This is the distance between the two foci.  Again
remember that this whole discussion is based on the foci being
horizontally next to one another.

Perhaps that is correct, but the length of the string is:
1) double the distance from one focus to the most distant point on the
ellipse beyond (in line with) the other focus;
2) the distance between the foci plus the distance from each focus to
the point of the ellipse's maximum width, which is the point that a
perpendicular from the midpoint of the line between the foci would
intersect the ellipse.

Since I can't solve that this morning, here is the formula:
http://www.analyzemath.com/EllipseProblems/EllipseProblems.html

"The length of the major axis is 2a, and the length of the minor axis
is 2b. The two foci (foci is the plural of focus) are at (± c , 0) or
at (0 , ± c),
where c^2 = a^2 - b^2."

Just plug in your a and b amounts and solve for c, which will be the
distance of each focus from the center point of your ellipse, that is
the distance between the foci will be 2c.

myoarin is correct.

badgerw is close to correct, with different terminology, except for a
sign error and inconsistent notation.

using badgerw's terminology, L=W is correct.  badgerw uses both H and
T for the height of the ellipse; let's use H.  the plus sign in
badgerw's equation for D should actually be a minus sign.  also, it
can be simplified down to:

D = sqrt (W^2 - T^2)

you can check this by using a circle as a special case.  W=H=DIAM, L=DIAM, and D=0.

-cab

um... the tiny little insignificant error in my post above was
intentional, just to see if anyone was paying attention.  yeah, that's
it, intentional.  also, badgerw confused me with his T vs. H thing. 
so really, if you think about it, my error is badgerw's fault.

because otherwise, the slightly superior tone of my comment would have
made me look like an ass.

anyway, to be clear, the "T" in my equation should be "H".

now, where is that "Edit Comment Before Anyone Sees It" button?

-cab (temporarily humble)

With unusual modesty, let me say that I just found a good website.  :-)  Myo

Oops!  Yeah.  I accidentally switched the sign (should be a -, not a
+) and used H instead of T.

As for the clarified question: It took me a while to figure out what
you meant, but now I get it.  You are talking about (basically) the
panels on a beach ball.  Each of these panels is roughly the shape of
a flattened football, with two pointy tips and two uniform curves.  So
your question is you're making a beach ball with 6 panels and X
diameter, how do you cut the panels?  That's quite a bit harder, and I
doubt I can answer that without some kind of software.

Ok, now we're going to say that R is the radius of the desired sphere
(X/2).  I can tell you that the long dimension of the panels (from
pointy tip to pointy tip) will be pi*R in length (you came close by
saying that your 3-foot shape would make a 1-foot diameter beach ball)
and the short dimension of the panels (the widest part of the curves)
would be one-third of that, or pi/3*R.  To make a sphere with a
different number of panels, change the denominator.  6 panels = 3; 8
panels = 4; 10 panels = 5; etc.

Of course, this is ignoring the extra that you will need to leave in
order to sew the panels together!

HI,
I am sure there is a mathematical formula for your "orange slices",
but if someone can't find it, you could make a pattern from a "beach
ball", either finding one with those panels and make several
measurements of the "latitudes" (you would just need several between
the "equator" and the pole, since the other half would be identical);
OR find a large globe, and work from that.

Once you have the dimensions and your pattern, you can draft it to the
necessary scale by multiplying the dimensions.

Cheers, Myoarin