The question cameronjreed asked peaked my interest and so I got the spreadsheet and had a look.
First of all, you guys that use this have no idea of the sophistication that is lurking under the covers. IvLeph clearly put a lot of effort into both the calculations and the presentations part. Nice job!
But round-off error is something I know about, and there is more than round-off error going on here. I expected (as did cameronjreed) that the function would be "symmetric" (ain't that right CJR?), by which is meant that if you flip the graph around the center-line axis (x=72/2 in the example case), then nothing changes. CJR was looking at the initial points. If the function was symmetric, then the value 4 inches from the left (1.25") would be the same as the value 4 inches from the right (1.75"). Now a difference of 0.5" relative to 1.25 or 1.75
is really pretty big, even when you're rounding to 0.125".
EDIT : the spreadsheet doesn't make the error that I thought it did, described below. The non-symmetry of the curve is still much larger than I would expect due to round-off, but I've not an answer for that now. Maybe I'll try this same calculation in a different language and see what emerges there.
So I dug a little deeper to see just exactly what function is being plotted here (parabolas I know and love, catenary curves are new to me). Well, the formula is
y = H - alpha*( exp(x) + exp(-x) - 1 )
where exp(x) is the natural number e raised to power x. The thing is, this function is not symmetric over the interval [0,72], e.g., if c = the center point of the x domain (e.g., 72/2 = 36) then if you compute the function value s units to the right of c it is
H - alpha*( exp(c+s) + exp( -c -s ) - 1 )
and if you compute the function value s units to the left of c it is
H - alpha*( exp(c-s) + exp( -c + s ) - 1 )
and these values aren't the same. Round-off or not.
But the catenary curve is symmetric about its axis, and so something is amiss here. What is amiss is that c needs to be 0.
So I'm thinking Ivleph that your loop that computes the values out needs to run from x= -L/2 to x=L/2 rather than x=1 to L.
Then the center point of the curve is relabeled to be 0 rather than L/2, and all is well in the universe. Rather, you can run x over the same interval (because you use that in the output column) but instead you should compute
H - alpha*cosh( x - L/2 )
Not sure what you're solving for with the Newton method, so can't really tell if this offset affects it to. Guessing not, as it seems the iterative solution is working out the necessary relationships between some of the parameters in the model.
Just as I experienced with cutting a curve for the bridge hammock, it seems that getting the details "exactly right" don't matter too much for functionality of the tarp, and the variance introduced by sewing is of the same scale as the deviations from symmetry. But thought you'd want to know.... (along with the rest of the Internet universe )