Hyperbola Peak for given length and Coordinates of Intercepts

Hi All

I'm hoping someone can help me on this:

I have a piece of paper. (200mm in length) I take the two sides of the paper and push them closer to each other until they are 50mm apart. How "high" is the hyperbola's peak?

Basically, if I have two points: A(0, 0) and B(200, 0) with a straight line running between, and I shift point B closer to A so that B(50, 0), yet the length of the curve now still connecting A and B is at 200mm, how do I calculate the "peak" of the hyperbola in coordinates similar to A and B?

Is there a function / formula / equation to work this out? I hope that whoever is reading this understands the question... :)

Thanks in advance, and as always: Any help is greatly appreciated! :thumb:

I forgot to mention: The hyperbola should be perfectly symmetrical along the Y axis at the peak... :thumb:

For one, I doubt the paper is going to form a parabola. But that was probably just an example and doesn't matter to your question :p


Anyway... I'm kinda blank at the moment... But you could try using the formula to calculate arc length of a curve...

I'm not sure if I got it right but I believe it's something like:


length = integral(start to end) ( sqrt ( 1 + (f ' (x))² ))
where f(x) is your function, so f ' (x) is the derivative.


For a parabola like f(x) = - x² + 4 you would get something like:
(assuming to integrate from -2 to 2 which are the parabola's x-intercepts)

length = integral(-2 to 2) ( sqrt ( 1 + (-2x)²)) = int(-2to2)(sqrt(1+4x²))


If you keep the length constant you could work something out perhaps.



EDIT
Or maybe, just maybe, I didn't check this, just a wild thought, but maybe the area between the x-axis and the parabola should be constant? You could calculate that easily with an integral.


EDIT2
See, I said I was blank... I thought you mentioned a parabola but now I see it was a hyperbola :p Still, it doesn't change how you could try it though... I think.

Thanks for the reply Nick.

I will have a "whack" at this, and let you know what I find. :)

Thanks Nick

I found that the formula for calculating an Arc Length is:http://www.vbforums.com/images/ieimages/2007/12/1.gif

Now, to implement this is a bit harder than I thought.

I'm still scratching to see if I can get this into my head, though I don't really see how the formula works.

Also, if I can get this to work, I can "reverse" the formula to say that http://www.vbforums.com/images/ieimages/2007/12/1.gif = 200 (In the example that I first used)

Will check...

Just thinking geometrically, there should be an infinite number of solutions to your question--try thinking about what happens if the paper happens to make 3 peaks, as opposed to 1 peak. The paper can be in that shape and the arc length will still be right, but the maximum height will be different in each case.

The physical constraint you need to apply is the force of gravity on the paper. For that, zaza would probably be good or some other physicist/physics person.

You can make the *assumption* that the paper is shaped like a parabola and get a unique solution to your integral constraint, but it may not match reality exactly.