First things first, please excuse my maths, its been over a decade since I did trig and the like at school and, real world, I haven’t much cause to use them every day…

Second the problem –

I need to find the centre points [or rather distance between them] we’ll call them A & B of an ellipse.

I have a plan which shows me a section of the ellipse but the only further info I can get from the architect is that “he made it using the standard ellipse tool on CAD” he doesn’t know what I’m talking about when I ask for the distance AB let alone why I need it to form an extrusion that follows the edge of his ellipse.

The plan shows me two dimensions, in Cartesian terms, x and (y/2).

Question one) I’m making the assumption that the section of the shape shown is half of it, cut along a the axis x (so values of x range from –x max to +x max, and values of y are >0). Do I need to make this assumption – I’m of the belief if I don’t my solution will work out circular as that’s the simplest solution to the equations (i.e. AB=0)

Question two) so… using the above assumptions and substituting some conveniently round numbers when needed I reckon the following should be correct:

X = 11, (y/2) = 7 :. y=14

For calculation purposes I need to have the points (Xmin,0) (Xmax,0) (0,Ymin) and (0,Ymax), for simplicity C, D, E and F respectively

A is located at (0,<0) B at (0,>0)

Given y > x then A & B lie on axis x=0 :. y = AB+ AE + BF

AE=BF so…

y=AB+2AE=14

——-

Given the shape is an ellipse: AB+AC+BC=2(AB)+2(AE) or AC+BC=AB+2(AE)

Using Pythagoras and that given C lies on the axis y=0 and that A and B are equidistant from the same (hence the half an ellipse assumption):

AC = ((AB/2)2+(x/2)2)0.5 = BC

So substituting in to the above:

2*((AB/2)2+(x/2)2)0.5 = AB+2(AE) or 2*((AB2/4)+(x2/4))0.5 = AB+2(AE)

given y = AB+2AE we have 2*((AB2/4)+(x2/4))0.5 = y
so square out to simplify:
4*((AB2/4)+(x2/4)) = y2
reduce
(4AB2)/4+(4×2/4) = y2
reduce
AB2+x2 = y2 or AB2 = y2-x2

Insert values for y and x to get AB2 = 142 – 112 Solve AB = (142-112)0.5 = 8.66 (ish)

Now that I have AB and I know that for values at y= +or- 7 x = 0 and I can plot my ellipse such that A is found at (0,-4.33) and B at (0,4.33)

Is this right?

Is there a simpler more elegant solution or way of getting there that doesn’t involve me having to inflict physical violence on the architect?

Thanks

http://www.mathopenref.com/ellipsefoci.html

does this help

centres would be, if i understand you correct, 8.5 from the axis.

also

did you add a diagram to your initial post?

Hi rendo,

No image I’m afraid as I’d have to upload it to else where then link and I haven’t a suitable service to upload to – I could sign up to Flickr or some such I guess.

The link however does help, the first section is what I’m after and confirms my working above thank you.

The architect is being a bit dumb as you can’t practically set out a true ellipse

Get the architect to draw it again using 2 or more part circles with centres placed on the preceding circles radius. This can be set out properly, will be close enough and everyone will be happy

This is what I had to do after setting out something with an ellipse and asking someone to build it….