# How Bridges Work 7, Exercise 3: catenary or parabola?
# A cable hanging under its own weight forms a catenary, y = a (cosh(x / a) - 1).
# With a deck load spread along the span, it forms a parabola. For a 300 m span and
# sag-to-span ratios of 1/20, 1/10 and 1/5, find the catenary parameter a that
# gives the stated sag (use bisection), then the difference in metres between the
# catenary and the same-sag parabola a quarter of the way across. Which ratio
# makes the choice of curve matter most?
import math

def catenary_sag(a, half):
    return a * (math.cosh(half / a) - 1)

def solve_a(span, sag):
    lo, hi = 1.0, 1e7
    for _ in range(200):
        mid = (lo + hi) / 2
        if catenary_sag(mid, span / 2) > sag:
            lo = mid
        else:
            hi = mid
    return (lo + hi) / 2

span = 300
for ratio in (20, 10, 5):
    sag = span / ratio
    a = solve_a(span, sag)
    x = span / 4
    y_cat = catenary_sag(a, span / 2) - catenary_sag(a, span / 2 - x)
    y_par = sag * (1 - (1 - x / (span / 2)) ** 2)
    print(f"1/{ratio}", round(a), round(y_cat, 3), round(y_par, 3), round(y_cat - y_par, 3))

# Output when run:
# 1/20 752 11.259 11.25 0.009
# 1/10 380 22.573 22.5 0.073
# 1/5 197 45.532 45.0 0.532
#
# Columns: sag ratio, catenary parameter a (m), height of the catenary at a quarter
# span (m), height of the parabola (m), difference (m). At 1/20 the two curves
# differ by 9 mm, at 1/10 by 73 mm and at 1/5 by 0.53 m. For flat cables the
# difference is negligible, which is why engineers use the simpler parabola for a
# cable carrying a deck, while the catenary matters for a cable under its own
# weight and for steep, deeply sagging cables.