# Catenary Equation [Pending...]

**mycz** 24 May 2022, 11:16

### My question

I understand that a bridges main cable can be modelled using the parent catenary function, y = a cosh(x/a), where cosh(x)=(e^x+e^-x)/2. However, I was confused on how to obtain the parameter 'a'. I've read online that 'a' is a constant to do with horizontal tension? I also understand that 'a', essentially changes the overall wideness of the curve if I'm correct.

Can someone please help me to understood how to solve for a?

### Relevant page

http://euclid.trentu.ca/aejm/V4N1/Chatterjee.V4N1.pdf

### What I've done so far

∫_0^x▒〖√(1+(dy/dt)^2 ) dt=1016.5〗

Secondly, I will consider the equation which describes the height of the towers:

y(x)=206.4

After integrating the first equation and substituting the expression for y from the rationale into the second equations, my two main equations become:

a sinh〖(x/a)=1016.5〗

And

a cosh(x/a)=206.4+a

X

I understand that a bridges main cable can be modelled using the parent catenary function, y = a cosh(x/a), where cosh(x)=(e^x+e^-x)/2. However, I was confused on how to obtain the parameter 'a'. I've read online that 'a' is a constant to do with horizontal tension? I also understand that 'a', essentially changes the overall wideness of the curve if I'm correct.
Can someone please help me to understood how to solve for a?

Relevant page
<a href="http://euclid.trentu.ca/aejm/V4N1/Chatterjee.V4N1.pdf">http://euclid.trentu.ca/aejm/V4N1/Chatterjee.V4N1.pdf</a>
What I've done so far
∫_0^x▒〖√(1+(dy/dt)^2 ) dt=1016.5〗
Secondly, I will consider the equation which describes the height of the towers:
y(x)=206.4
After integrating the first equation and substituting the expression for y from the rationale into the second equations, my two main equations become:
a sinh〖(x/a)=1016.5〗
And
a cosh(x/a)=206.4+a

## Re: Catenary Equation

**mycz** 26 May 2022, 07:06

Nevermind! I ended up being able to solve it

X

Nevermind! I ended up being able to solve it

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