# How Bridges Work 10, Exercise 1: where a truss stops working
# Using truss_steel and cable_steel from the chapter (deck load 144 kN/m, truss
# depth span/10, allowable 165 MPa for the truss and 600 MPa for the cable),
# (a) find the longest span, in steps of 10 m, for which truss_steel still settles
# on an answer, (b) find the shortest span, in steps of 10 m, at which the truss
# needs more than 20 times the steel of the cable, and (c) print the
# truss-to-cable ratio at 50, 100, 200 and 300 m. What does the ratio's growth
# say about how steel use scales with span for the two systems?
import math

RHO, G = 7850, 9.81                      # steel
W_DECK = 144                             # kN/m: deck, surfacing and traffic (illustrative)

def truss_steel(span, depth_ratio=10, sigma_mpa=165, extra=1.5):
    """Steel (t per metre) for a truss of depth span/depth_ratio, found by repeating
    the calculation until its own weight stops changing. None if it never settles."""
    w = W_DECK
    for _ in range(500):
        moment = w * span**2 / 8                      # kN m
        force = moment / (span / depth_ratio)         # chord force, kN
        area = extra * 2 * force / (sigma_mpa * 1e3)  # m2 (kN / kPa), both chords plus bracing
        w_new = W_DECK + area * RHO * G / 1000
        if abs(w_new - w) < 1e-6:
            return area * RHO / 1000
        w = w_new
        if w > 1e6:
            return None
    return None

def cable_steel(span, sag_ratio=10, sigma_mpa=600):
    """Main-cable steel (t per metre), same method. Towers and anchorages are not counted."""
    w = W_DECK
    for _ in range(500):
        h = w * span**2 / (8 * span / sag_ratio)
        v = w * span / 2
        area = math.hypot(h, v) / (sigma_mpa * 1e3)
        w_new = W_DECK + area * RHO * G / 1000
        if abs(w_new - w) < 1e-6:
            return area * RHO / 1000
        w = w_new
        if w > 1e6:
            return None
    return None


last = None
for span in range(10, 1210, 10):
    if truss_steel(span) is None:
        break
    last = span
print("longest span with a solution:", last)

for span in range(10, last + 10, 10):
    if truss_steel(span) > 20 * cable_steel(span):
        print("first span with ratio over 20:", span)
        break

for span in (50, 100, 200, 300):
    print(span, round(truss_steel(span) / cable_steel(span), 1))

# Output when run:
# longest span with a solution: 550
# first span with ratio over 20: 300
# 50 11.0
# 100 12.1
# 200 15.0
# 300 20.2
#
# The truss's steel per metre grows faster than the span (roughly with its
# square, plus the weight of the steel itself), so beyond 550 m the model
# never settles: each extra tonne of steel adds more weight than it
# carries. At 50 m the truss uses 11 times the cable's steel, at 100 m 12.1 times,
# at 200 m 15 times and at 300 m 20.2 times, the first span (300 m) over 20.
# The cable's main-cable steel grows roughly in step with the span. The
# comparison is only a trend. It counts only the cable, leaving out the
# towers, anchorages, hangers and the deck that the cable carries, and it
# uses a constant truss depth of span/10, whereas the Forth Bridge (521 m)
# gets past 500 m by tapering its depth, as in How Bridges Work 6.