# How Bridges Work 9, Exercise 3: how much corrosion can a cable stand?
# A main cable has 11,618 wires. The working stress is 581 MPa when the wires
# are sound, and the wire breaking strength is assumed to be 1500 MPa. If a
# fraction of the strength is lost, the factor of safety is
# (1 - loss) x 1500 / working stress. (a) Print the factor for losses of 0, 5, 10,
# 20, 30 and 40 percent. (b) Find the loss at which the factor falls to 2.0 and to
# 1.5, and the equivalent number of wires fully broken. (c) Repeat (b) for
# working stresses of 450 and 700 MPa. Which design has the most margin?
WIRES, BREAK = 11618, 1500

def factor(loss, working):
    return (1 - loss) * BREAK / working

for loss in (0, 0.05, 0.10, 0.20, 0.30, 0.40):
    print(loss, round(factor(loss, 581), 2))

for working in (450, 581, 700):
    for target in (2.0, 1.5):
        lost = 1 - target * working / BREAK
        print(working, target, round(lost * 100, 1), round(lost * WIRES))

# Output when run:
# 0 2.58
# 0.05 2.45
# 0.1 2.32
# 0.2 2.07
# 0.3 1.81
# 0.4 1.55
# 450 2.0 40.0 4647
# 450 1.5 55.0 6390
# 581 2.0 22.5 2618
# 581 1.5 41.9 4868
# 700 2.0 6.7 775
# 700 1.5 30.0 3485
#
# Part (a) lines: loss fraction, factor at 581 MPa working stress. Part (b)/(c)
# lines: working stress (MPa), target factor, loss at that factor (percent), the
# equivalent number of wires. At 581 MPa the factor is 2.58 when sound and
# 2.32 at 10% loss, still above 2. A factor of 2.0 is reached at a 22.5% loss,
# equal to about 2,618 wires. A lower working stress gives a larger margin:
# at 450 MPa a loss of 40% (4,647 wires) is needed to reach 2.0, whereas at
# 700 MPa a loss of 6.7% (775 wires) is enough. Margins of this kind explain why
# an estimated 8% to 10% loss of strength needed action but was not an
# emergency, and why monitoring and dehumidifying the cables made sense.
# The 1500 MPa breaking strength is an assumption.