# How Bridges Work 3, Exercise 2: the thrust of an arch
# A parabolic arch of span 20 m carries a uniform load of 40 kN/m (per
# metre width). (a) Find the vertical reaction and the horizontal thrust
# at each springing for rises of 2, 4, 5 and 10 m, and the resultant force.
# (b) Check that the beam moment wL^2/8 equals thrust x rise in each case.
# (c) Find the smallest rise that keeps the thrust at or below 400 kN.
import math

def thrust(w, span, rise):
    return w * span**2 / (8 * rise)

w, span = 40, 20
v = w * span / 2
print("V =", v)
for rise in (2, 4, 5, 10):
    h = thrust(w, span, rise)
    print(rise, h, round(math.hypot(h, v), 1), w * span**2 / 8 == h * rise)

min_rise = w * span**2 / (8 * 400)
print("minimum rise:", min_rise, "m, rise/span =", min_rise / span)

# Output when run:
# V = 400.0
# 2 1000.0 1077.0 True
# 4 500.0 640.3 True
# 5 400.0 565.7 True
# 10 200.0 447.2 True
# minimum rise: 5.0 m, rise/span = 0.25
#
# The vertical reaction (400 kN) does not depend on the rise; the thrust
# does, in inverse proportion: halving the rise doubles the thrust. The
# check in (b) shows why: the arch turns the beam's bending moment (wL^2/8,
# 500 kN m here) into a pair of forces, the thrust H and its lever arm, the
# rise. A rise of a quarter of the span keeps the thrust at 400 kN.