In the problem Cables provided by K Sengupta, the solution required the knowledge that a hanging flexible cable takes the form of a catenary, the general equation of which is y=a*cosh(x/a), combined with calculating the overall length of the catenary between the two fixed points of the cable.
This time we have a cable strung between two towers on a flat level plane. The low point of the cable is 10m above the plane. Later, the cable heats up and expands to a length of 80m. The towers also expand and gain 0.2m in height – they are now exactly 50m high. The low point of the cable is still 10m above the plane. How far apart are the towers?
I think we need to assume the coefficient of expansion is the same for the towers as that for the chain.
The towers went from 49.8 meters to 50 meters. The cable is now 80 meters and so was 80*498/500 = 79.68 meters long.
The dip now needs to be .2 meters more now than before in order to maintain the same distance from the ground.
The program solves for a in the length and dip calculations (designated aa as the solution) based upon an assumed distance apart and each of the two lengths of the cable, and reports the difference in dip.
The distance trial values were determined by previous runs with a wider range and larger jumps in value.
clc,clearvars
syms a
for dist =52.38:.0001:52.39
dip=[];
for L=[79.68 80]
% L=2*a*sinh((dist/2)/a);
aa=vpasolve(L==2*a*sinh((dist/2)/a),a,14);
y1=aa*cosh((dist/2)/aa);
y0=aa*cosh(0);
dip(end+1)=y1-y0;
end
fprintf('%8.5f %8.5f\n',dist,dip(2)-dip(1))
end
52.38730 0.1999996
52.38740 0.1999998
52.38750 0.2000001
52.38760 0.2000003
The towers seem to be about 52.3855 meters apart.
The dip went from about 27.02238 to 27.22238.
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Posted by Charlie
on 2024-03-27 14:20:36 |
solution
