Thread Subject: How to find the point of intersection from two 2d curves?

here is my equations for my curves, they are actually arcs of circles.

x3=([-1:0.01:1]*8)+1;
ypos3=(((8^2)-((x3-1).^2)).^0.5);

x2=([-1:0.01:1]*5)+8.5;
ypos2=((5^2)-((x2-8.5).^2)).^0.5;

I want to find the coordinates where they cross over using a code rather than just looking at my graph.

Looking at the graph...
plot(x2,ypos2,'r')
plot(x3,ypos3,'g')
the coordinates are about (7.35, 4.86)

any ideas what to do?

Since you have the equations, you could use fzero.

f1 = @(x) (((8^2)-((x-1).^2)).^0.5)
f2 = @(x) ((5^2)-((x-8.5).^2)).^0.5;

pt = fzero(@(x) f1(x)-f2(x),7) % The x-coord
coord = [pt,f1(pt)] % The coordinates.

James wrote:
> here is my equations for my curves, they are actually arcs of circles.
>
> x3=([-1:0.01:1]*8)+1;
> ypos3=(((8^2)-((x3-1).^2)).^0.5);
>
> x2=([-1:0.01:1]*5)+8.5;
> ypos2=((5^2)-((x2-8.5).^2)).^0.5;
>
> I want to find the coordinates where they cross over using a code rather than just looking at my graph.
>
> Looking at the graph...
> plot(x2,ypos2,'r')
> plot(x3,ypos3,'g')
> the coordinates are about (7.35, 4.86)
>
> any ideas what to do?

Putting it through the symbolic solver and using 17/2 instead of 8.5, and using
sqrt() instead of raising to the decimal power 0.5, the exact intersection
is at x = 147/20 which is 7.35 . Substituting back into the first formula,
the y coordinateis sqrt(9471)/20, which is about 4.865953144.

Walter Roberson <roberson@hushmail.com> wrote in message
> Putting it through the symbolic solver and using 17/2 instead of 8.5, and using
> sqrt() instead of raising to the decimal power 0.5, the exact intersection
> is at x = 147/20 which is 7.35 . Substituting back into the first formula,
> the y coordinateis sqrt(9471)/20, which is about 4.865953144.


Funny, that is the same thing I got using fzero, as shown above.

Matt Fig wrote:
> Walter Roberson <roberson@hushmail.com> wrote in message
>> Putting it through the symbolic solver and using 17/2 instead of 8.5, and using
>> sqrt() instead of raising to the decimal power 0.5, the exact intersection
>> is at x = 147/20 which is 7.35 . Substituting back into the first formula,
>> the y coordinateis sqrt(9471)/20, which is about 4.865953144.
 
> Funny, that is the same thing I got using fzero, as shown above.

Using fzero, you got an x that was 7.35 to within the search
tolerance; using the symbolic solver showed the x was _exactly_ 7.35 .
No concern about "suppose we'd had quadruple precision arithmetic,
how would the result change" and so on.