High Jump Fosbury Flop

Question. Why can a jumper whose centre of mass (COM) rises only about 2.4 m clear a 2.35 m bar? Because the Fosbury flop lets the COM pass under the bar while the body goes over it. This page models the three things that decide the result: the take-off, the flight and the arch.

1. Take-off: a leg spring. During the 0.10-0.14 s ground contact the plant leg acts like a spring of rest length L₀ = 0.55 H and stiffness k joining the foot to the COM (the SLIP model of Alexander 1990 and Seyfarth 2002). With r the foot-to-COM distance and F = k (L₀ − r) along the leg:

The COM arrives at the plant angle θ₀ with the approach speed v_a and a small downward speed; the leaning-back leg turns horizontal momentum into vertical speed. The factor ε makes the extension phase slightly stiffer, which stands for the muscular work done during the push. Integration is classical RK4 (step 0.1 ms); take-off is when r returns to L₀ while still lengthening. Arms and the free leg lift the COM by a further h_x at take-off.

2. Run direction versus bar. The jumper approaches along a curve and takes off at an angle ψ to the bar, so only the component v sinψ carries the COM across the bar plane while v cosψ carries it along the bar. The take-off foot is placed a distance d_T from the bar plane.

3. Flight. After take-off the COM is a projectile. Apex height and time are

so the take-off vertical speed fixes the height gained: a jumper leaving the ground at 4.6 m/s gains 1.08 m. The best take-off distance puts the apex above the bar plane (graph 4); too close and the COM is still rising at the bar, too far and it is already falling.

4. The arch. Model the body in the air as a thin circular arc of length ℓ = 0.9 H and half-angle α. Its COM lies on the concave side, below the highest point of the arc by

For a flat body (α = 0) d = 0 and the COM must pass over the bar; at α = 68 deg the COM travels about 16 cm below the bar. The bar is cleared when the COM height over the bar plus d is at least the bar height; the highest bar this jump clears is y_bar + d.

Three independent checks run in the page. (1) With ε = 0 the spring is conservative, so kinetic + gravitational + spring energy must stay constant through the whole contact. (2) A separate numerical integration of the flight must reach the closed-form apex height and time. (3) The closed-form arch offset must equal the numerical centroid of 20,000 points on the arc.

Honest simplifications. A point mass on a massless spring (no segment motion, no foot rotation, no ankle or knee joints); the free-limb lift h_x is a fixed bonus, not computed; the arch is a circular arc and the body is assumed to pass the bar at the moment of maximum arch; the 3D mannequin movement during flight is a kinematic illustration of the phases, not a simulation of angular momentum; no wind, no bar vibration, no tilt of the take-off surface. Numbers are plausible for trained jumpers but not a coaching prescription. Standard heights, mat and track proportions follow common competition layouts without any event branding.

Pick a preset and watch the full jump in slow motion, or press Pause and change a slider; the graphs and the verdict update at once. Try: (1) Too close to the bar and Too far from the bar against the elite flop: the apex is no longer over the bar. (2) Straddle against the flop with the same take-off: the arch alone lifts the clearable height by about 10 cm. (3) Soft leg, shallow plant: the vertical speed collapses. (4) Raise the bar until the verdict flips.