# How Bridges Work 9, Exercise 2: how much scour can a pile group take?
# A friction pile of diameter 1.0 m carries 1000 kN. The soil gives a skin
# friction of 60 kPa over the embedded length, and a flood lowers the river bed
# by the scour depth. For pile lengths of 10, 12, 15 and 18 m find (a) the
# safety factor with no scour, (b) the scour depth at which the factor drops to 1.5,
# and (c) the scour depth at which it drops to 1.0 (failure). Then repeat (b)
# and (c) for 12 m piles carrying 1500 kN.
import math

def factor(length, scour, load, dia=1.0, skin=60):
    return skin * math.pi * dia * max(0.0, length - scour) / load

def scour_for(length, load, target, dia=1.0, skin=60):
    return length - target * load / (skin * math.pi * dia)

for length in (10, 12, 15, 18):
    print(length, round(factor(length, 0, 1000), 2),
          round(scour_for(length, 1000, 1.5), 2), round(scour_for(length, 1000, 1.0), 2))

print("1500 kN:", round(scour_for(12, 1500, 1.5), 2), round(scour_for(12, 1500, 1.0), 2))

# Output when run:
# 10 1.88 2.04 4.69
# 12 2.26 4.04 6.69
# 15 2.83 7.04 9.69
# 18 3.39 10.04 12.69
# 1500 kN: 0.06 4.04
#
# Columns: pile length (m), factor with no scour, scour depth for a factor of
# 1.5 (m), scour depth for a factor of 1.0 (m). A longer pile has a bigger margin
# (a 10 m pile fails at 4.69 m of scour, an 18 m pile at 12.69 m). The last
# line shows how the load matters: the same 12 m pile carrying 1500 kN is down
# to a factor of 1.5 after only 0.06 m of scour, and fails at 4.04 m, the same
# depth as a 1000 kN pile reaches a factor of 1.5. Scour removes the soil that
# the foundation relies on, so the margin drains away as the bed drops.