# How Bridges Work 7, Exercise 1: sizing the cables of a small suspension bridge
# A bridge has a main span of 300 m and two cables, each carrying 60 kN/m
# (deck and cable weight together). For sag-to-span ratios of 1/8, 1/10 and
# 1/12 find the horizontal pull, the vertical pull at each tower, the largest
# tension, the approximate cable length, and the steel area needed if the
# working stress is 600 MPa. Which sag gives the smallest cable area, and what is
# the cost of choosing it?
import math

def cable(w, span, sag):
    h = w * span**2 / (8 * sag)
    v = w * span / 2
    t = math.hypot(h, v)
    length = span * (1 + 8 / 3 * (sag / span) ** 2)
    return h, v, t, length

for ratio in (8, 10, 12):
    h, v, t, length = cable(60, 300, 300 / ratio)
    area = t * 1000 / 600
    print(f"1/{ratio}", round(h), round(v), round(t), round(length, 1), round(area))

# Output when run:
# 1/8 18000 9000 20125 312.5 33541
# 1/10 22500 9000 24233 308.0 40389
# 1/12 27000 9000 28460 305.6 47434
#
# Columns: sag ratio, horizontal pull H (kN), vertical pull V (kN), largest
# tension (kN), cable length (m), steel area needed (mm2). V is the same in
# every case (60 x 300 / 2 = 9,000 kN) because it depends only on the load. H
# and the tension grow as the cable is made flatter: from 1/8 to 1/12 the
# tension rises from 20,125 to 28,460 kN and the area from 33,541 to 47,434
# mm2, about 41% more steel. The deeper sag (37.5 m at 1/8) therefore saves
# cable steel, but it needs taller towers to hold the cable up, and a longer
# cable (312.5 m against 305.6 m). Tower cost against cable cost is the real
# trade-off; this is the same relation as the arch thrust of How Bridges Work 3.