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# for curve problem

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Hi I have some shape problem PLEASE first of all see this picutre and zoom please . http://www.geocities.com/nareshn2003/curve.jpg In Figure-1 u can see circle-circle intersection . it''s normal Now let define something first in figuer-2 condition A is circle and it''s length(L) and width(W) is same. So guess for condition-A Length,Width=5 value; and result will be circle. Now in condition-B Let length=7 and width=5 is changing so result will be diffrent in picture . In condition -C let length=9 and width=5 so result is more different in picture . And another thing is that i don''t want shape like oval or ellipse i want shape having sharpness at both end of length. I don''t want directly through the arc.. I want x and y cordinate of surface points of that shape..as like in circel we have x=rcos(theta) ,y=rsin(theta) like that i want x and y cordinates.. I had tried with superellispe algoritham but not worked So anyone can suggest me good algoritham or calculations ? devil20

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Assumptions (some might be more or less obvious):
- Axis-labeling: in your figure the horizontal is the x-axis with positive going to the right. Vertical is the y-axis with positive going up.
- Origin of your coordinate system is the center of the figure (the point where mirroring on x- or y-axis is a symmetry).
- "Angles" mean the angle between the x-axis and the connection line between the point they are "located" (see below) in and the point (x,y)=f(alpha) you want to calculate unless mentioned otherwise. They increase counter-clockwise.
- You want (x,y) = f(alpha) where the two circles are fixed (fixed centers C1 and C2=-C1 and fixed radi r1=r2 > 0.5*|C1-C2|) and symmetric in the sense that I put into parentheses. alpha-angle is located in the origin.

Possible solution:

1) Use the symmetry of the figure: If (x,y) = f(alpha) => -(x,y)= f(alpha+180°). Thus, we can reduce the problem to (x,y) = f(alpha), -90°<= alpha <= 90°. Let this be the part of the curve that is created by circle 1.

2) Use an additional symmetry operation: (x,y)(alpha) = (x,-y)(-alpha). => 0°<=alpha<=90°. This reduction on the 1st quadrant of the coordinate system makes thinking about the problem much easier but is porbably not nessecary.

3) (x,y) = C1 + (r*cos(theta), r*sin(theta)) where theta-angle is located in the point C1=(Cx,0). Problem: Find theta(alpha).

4) Take a look at the triangle formed by the points C1, O=Origin and P=(x,y)=f(alpha). You know 2 sides (C1->P, C1->O) and the angle at O: 180°-alpha.
At this point I was pretty sure the problem is almost solved but after 20 minutes of trying to figure out theta I gave up and took a look into my handbook of mathematics:

5) This handbook tells me a single definite solution for this problem exists (what a surprise) under certain conditions that apply here. The angle x between P->O and P->C1 is x = arcsin(|Cx|*sin(180°-alpha)/r1). Hope I didn´t mess things up here.

6) Since the sum of angles in a triangle equals 180° => x+(180°-alpha)+theta = 180° => theta = ... (now I´m getting really lazy)

7) Plug theta(alpha) into equation at 3) => f(alpha).

8) Have fun trying to simplify this equation

This is a step-by-step sketch of my solution. Better ones might exist and there certainly would have been a better way to represent it. But I wrote this post parallel to working on a solution so I couldn´t present it in a more elegant form since I didn´t knew where I´d end up. I really didn´t want to rewrite this all so you´ll have to live with this post...

Btw.: If an implicite equation y=y(x) is sufficient for you: That´s much easier since you can use p*q=h² (dunno the english name for that equation but I hope you know which one I mean) with h=y, p=x, q=r-x and the symmetry operation in 1).

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I''ll try to do that

thanx

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