# How Bridges Work 3, Exercise 3: an abutment that stays put
# An arch pushes on an abutment with a vertical force V = 300 kN and a
# horizontal thrust H per metre run of bridge, with H acting 4 m above the
# base of a masonry block of height 4 m. Masonry density 2200 kg/m3,
# friction coefficient 0.6 at the base. For widths from 1.0 m in steps of
# 0.1 m, find the narrowest block with a sliding factor of at least 1.5 and
# an overturning factor of at least 2.0 (moments about the outer toe, with
# the vertical forces taken through the middle of the block), for
# (a) H = 150 kN (a rise of 6 m on a 12 m span carrying 50 kN/m) and
# (b) H = 300 kN (a flatter arch, rise 3 m).
G, RHO, MU = 9.81, 2200, 0.6

def factors(b, height, v, h_thrust, h_arm):
    wt = RHO * G * b * height / 1000
    sliding = MU * (wt + v) / h_thrust
    overturn = (wt + v) * b / 2 / (h_thrust * h_arm)
    return sliding, overturn

def narrowest(h_thrust):
    b = 1.0
    while True:
        s, o = factors(b, 4, 300, h_thrust, 4)
        if s >= 1.5 and o >= 2.0:
            return round(b, 1), round(s, 2), round(o, 2)
        b = round(b + 0.1, 1)

print(narrowest(150))
print(narrowest(300))

# Output when run:
# (3.9, 2.55, 2.07)
# (6.0, 1.64, 2.04)
#
# Columns: narrowest width (m), sliding factor, overturning factor.
# The overturning check, not sliding, decides the width in both cases.
# Doubling the thrust (the flatter arch) raises the minimum width from
# 3.9 m to 6.0 m, which is why flat arches need massive abutments or ties.