Catenary Equation of a hose.
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Hi,
Given are the elevations of a hose connected to a building on one side and to a ship on the other. The below attached sketch shows the POC-Points of Connection as 19.09 ft on building and 49.9 ft on the ship. The distance from the building to the mid-span of hose is r2= 5 ft and r1 from ship is r1=6.5 ft. The hose is not to go below the elevation of the deck at y0=7 ft. I would like someone to calculate the necessary length of hose via catenary formula showing all steps with comments.
Regards,
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Solution Summary
The 9 pages solution discusses the integration of a catenary equation and using the constraints of the problem to determine the different coefficients that describe the shape of the hose.
in additon the solution shows how to utilize the Secant method to find a numerical solution to an equation.
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The solution is attached below in two files. the files are identical in content, only differ in format. The first is in MS Word format, while the other is in Adobe pdf format. Therefore you can choose the format that is most suitable to you.
The hose equation is given by:
However, since the interval is between -5 and 6.5 it is useful to use the parity property of the hyperbolic cosine, namely to write:
The line segment is:
So the total length of the cable is:
In our case:
Using:
We obtain:
And:
The ...
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