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Catenary Equation [Pending...]

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

Nevermind! I ended up being able to solve it

X

Nevermind! I ended up being able to solve it

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