The science
Rigid rod on a hinge. Before it breaks the shaft is a rod of length L and mass m rotating about its base. With θ measured from the vertical and I the moment of inertia about the hinge (I = mL²/3 for a uniform rod):
I θ'' = m g d sinθ θ'' = (3g / 2L) sinθ (uniform)
ω² = 2 m g d (cosθ₀ − cosθ) / I
t(θ) = ∫ dφ / √(2ω₀² (cosθ₀ − cosφ)), ω₀² = m g d / I
The integral is of elliptic type; it diverges logarithmically as the initial lean θ₀ → 0, which is why the hinge lean you set controls how long the fall lasts.
Why the tip "outruns" gravity. A point at distance r from the hinge has tangential acceleration rα = (3g r / 2L) sinθ, which exceeds the free-fall component g sinθ for r > 2L/3. The vertical acceleration of the very tip is g(3/2 + 3cosθ − 9/2 cos²θ), which passes g at cosθ = (3+3√2)/9, θ ≈ 36.4°.
Hinge reactions. For the uniform rod the vertical reaction is N = mg(3cosθ/2 − 1/2)² (it just touches zero at cosθ = 1/3, θ ≈ 70.5°) and the horizontal one is H = ¾ mg sinθ (3cosθ − 2), which changes sign at cosθ = 2/3, θ ≈ 48.2°. A real hinge that cannot supply that force slides or kicks back; here the hinge is treated as a perfect pivot and the peak |H|/mg and min N/mg are reported.
Bending moment. Cutting the shaft at height s, the part above must be accelerated by the transverse load q(ρ) = μ(ρ)(ρα − g sinθ) (+ wind). Integrating, M(s) = ∫ₛᴸ q(ρ)(ρ − s) dρ, and the axial force C(s) = ∫ₛᴸ μ(g cosθ − ρω²) dρ. For a uniform rod this gives
M(s) = (μ g L² sinθ / 4) · u (1 − u)², u = s/L
M_max = μ g L² sinθ / 27 at s = L/3
So the maximum bending moment sits one third of the way up from the base (two thirds down from the top). The simulation evaluates the same quantities numerically for any taper and cap mass. The shaft cracks where the extreme-fibre tension σ = M/Z − C/A first exceeds the masonry tensile strength (hollow-section modulus Z = π(rₒ⁴ − rᵢ⁴)/(4rₒ)); the break is snapped to the nearest ring joint.
After the break. The lower part keeps rotating about the hinge as a shorter rod; the upper fragment becomes a free rigid body (ballistic centre of mass, constant spin) until an end meets the ground, where an impulse with restitution 0.1 and Coulomb friction 0.5 is applied. The rings then detach into debris blocks (ground contact points with restitution and friction; debris–debris contact is not modelled, so blocks may overlap).
Hinge and notch. The notch removes a fraction n of the diameter on the fall side. The remaining section carries the weight eccentrically; the stress at its extreme fibre is σ = W/A + W·e·c/I and the hinge gives way when σ exceeds 3 MPa (typical aged brickwork in compression). The slider shows the shallowest notch that does so. The notch-closing angle is β = atan(h/(nD)) with h = 0.18 D; the closing impact itself is not modelled.
Wind. Horizontal drag ½ρCdw|w| per projected width (Cd = 0.7, uniform wind) adds a torque on the standing/hinged parts only; the free fragment ignores it.
Simplifications. Planar motion in the plane of the fall; Euler-Bernoulli bending of a rigid rod (the shaft's own flexibility is neglected, which is reasonable because the break occurs well before the rod stretches); hollow section with wall thickness 10 % of the local diameter; uniform masonry density 1800 kg/m³; no joint-by-joint cracking, no kick-back, no soil model. The "tensile strength" is an effective section strength: aged unreinforced masonry joints are usually far weaker (0.1 to 0.5 MPa), which is why very tall old chimneys often crack almost at once.
References. Classical analyses of the falling chimney (Madelung's early rigid-rod treatment; Varieschi and Kamiya, "Toy models for the falling chimney", Am. J. Phys. 71, 2003) and standard beam theory. This page is an educational model, not an engineering design tool, and is not affiliated with any demolition company.