1. Problem statement: You have two horizontal spans (one S=80 ft and one S=40 ft) with ropes hanging from their midpoints and you asked how to solve for the rope shape, the sag, or the area under the rope given the span and either the rope arc length L or the area A. 2. Model and formulas: A hanging uniform rope (idealized) takes the catenary shape, whose vertex is at the lowest point and whose equation (vertex at y=0) is $$y = a\cosh\left(\frac{x}{a}\right)-a$$ where $a$ is the catenary parameter (controls steepness). Half the span is $x_0=\frac{S}{2}$. The rope arc length across the whole span satisfies $$L = 2a\sinh\left(\frac{S}{2a}\right) $$ The vertical sag (drop from lowest point to an endpoint) is $$\text{sag}=a\left(\cosh\left(\frac{S}{2a}\right)-1\right) $$ The area between the catenary and the horizontal line through the vertex over the full span is $$A = 2a^2\sinh\left(\frac{S}{2a}\right)-aS $$ 3. Important rules and remarks: You need two independent known quantities to get concrete numbers, typically the span $S$ plus either the arc length $L$ or the area $A$. 4. How to solve for $a$ when $S$ and $L$ are given: rearrange $$f(a)=2a\sinh\left(\frac{S}{2a}\right)-L=0$$ This equation cannot be solved in closed form for $a$; solve it numerically (Newton or a root-finder). The derivative needed for Newton's method is $$f'(a)=2\sinh\left(\frac{S}{2a}\right)-\frac{S}{a}\cosh\left(\frac{S}{2a}\right) $$ so Newton iteration is $$a_{n+1}=a_n-\frac{f(a_n)}{f'(a_n)} $$ 5. When solving from area $A$: start from $$A=2a^2\sinh\left(\frac{S}{2a}\right)-aS$$ Bring like terms together: $$A+aS=2a^2\sinh\left(\frac{S}{2a}\right)$$ Divide both sides by $2a$: $$\frac{A+aS}{2a}=\frac{2a^2\sinh\left(\frac{S}{2a}\right)}{2a}$$ Show cancellation on the right-hand side: $$\frac{2a^2\sinh\left(\frac{S}{2a}\right)}{2\cancel{a}}=a\sinh\left(\frac{S}{2a}\right)$$ Hence $$\frac{A+aS}{2a}=a\sinh\left(\frac{S}{2a}\right)$$ and again you must solve this equation numerically for $a$. 6. Numerical example (first span S=80 ft with given area A=2513.27 ft^2): set $S=80$ and solve $$2513.27=2a^2\sinh\left(\frac{80}{2a}\right)-80a$$ Using a numerical root-finder (starting guess around 13), one finds $$a\approx 13.29\ \text{ft}$$ Check the sag using $\text{sag}=a(\cosh(\tfrac{S}{2a})-1)$ gives approximately $$\text{sag}\approx 121.9\ \text{ft}$$ which is the vertical distance from the catenary vertex to the endpoints for these parameters. 7. How you proceed for your two spans: treat each span separately as a problem with its own $S$ and known quantity (either $L$ or $A$), then use the formulas above and a numerical solver to find $a$, then compute sag or verify area/length. Final answer (first problem solved): For the 80 ft span with area 2513.27 ft^2 you get $a\approx 13.29$ ft and sag $\approx 121.9$ ft.