Richard Clark, KB7QHC wrote:
"Simple trig would reveal that with an 88 foot caternary with a 5 foot
sag would develop roughly an angle of 6;5 degrees below the horizontal."

I think we are about on the same page. Ed Laport says:
"If W is the equivalent total weight at the center, then the tension of
the triatic is:
T=W/2sinA

Where T and W are in identical units."

A is the deflection angle below an imaginary horizontal line.

Best regards, Richard Harrison, KB5WZI