- #1

- 152

- 0

My question is, where can i learn all the basics about hypebolic functions enough so to learn how to fit an equation?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter um0123
- Start date

- #1

- 152

- 0

My question is, where can i learn all the basics about hypebolic functions enough so to learn how to fit an equation?

- #2

- 59

- 0

http://en.wikipedia.org/wiki/Catenary

http://en.wikipedia.org/wiki/Hyperbolic_function

- #3

- 152

- 0

[tex]\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1[/tex]

- #4

- 152

- 0

- #5

- 59

- 0

Assuming that the two points are at the same potential in an uniform gravitational field, the hanging chain's shape can be shown to be of the form y = a cosh bx, so, you just have to gather enough data points to solve for a and b in this equation. You could also derive an explicit formula for a and b given the length of the chain and the distance between the two points, but this requires calculus.

- #6

- 152

- 0

Okay, so how exactly could i form an equation for hyperbolic cosine using data that i collect?

Secondly, i dont understand how the a value over 2 in the front of the equation and the a value under the x and -x can be the same.

In this equation dont you need those two variables to be different in order to stretch the graph how you want? Why are the values the same?

Secondly, i dont understand how the a value over 2 in the front of the equation and the a value under the x and -x can be the same.

In this equation dont you need those two variables to be different in order to stretch the graph how you want? Why are the values the same?

Last edited:

- #7

- 59

- 0

[tex]y_1 = a cosh bx_1[/tex]

[tex]y_2 = a cosh bx_2[/tex]

These can easily be combined into one equation which can be solved numerically. I can't think of any algebraic way to solve it.

If you experiment, you'll see that the shape of the chain always remains the same, only the scale and height changes.

If x is half the distance between the points, there are infinitely many solutions (y,a), corresponding to various lengths of the chain, to the equation [tex]y=a cosh\frac x a[/tex], since this is one equation in two variables, y and a.

Share: