Goalkeeper Reaction Window

Test whether a goalkeeper can reach a target before the ball arrives.

Worked example — Goalkeeper Reaction Window

Ball flight time	0.720 s
Time needed for target	0.975 s
Reach at ball arrival	0.81 m

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Goalkeeper Reaction Window

This page runs two clocks at once and asks which one finishes first. The ball’s clock is trivial — it travels at a fixed speed, so its flight time is just distance over speed. The goalkeeper’s is not: they cannot move at all for a fixed reaction delay, and when they finally do, they start from a standstill. Almost everything interesting here comes from that standstill.

What each symbol means

t_flight: the ball’s flight time, d/v. The entire budget the keeper has, and nothing they do changes it.
t_react: the reaction delay — sight to first movement. A fixed cost, paid in full before a single step, and it does not shrink when the shot gets easier.
a: the keeper’s sideways acceleration. Not their top speed: over these distances and times they never get near it, so acceleration is the only thing about the athlete that matters.
x_reach: how far sideways they actually get before the ball arrives. The page compares this against the target offset, and that comparison is the verdict.

Where the formula comes from

Start the ball’s clock, because it takes one line. Speed is constant, so t_flight = d/v — a 20 m shot at 80 km/h gives 0.900 s, and that is the whole budget. Note what is not in it: nothing about the keeper. The defender does not get a say in how much time they have.
Now the keeper, who starts from rest. Under constant acceleration distance is x = ½at², so the time to cover a sideways offset x is that inverted: t = √(2x/a). The square root is the shape of the whole problem. Doubling the offset does not double the moving time — it multiplies it by √2, about 1.41. Reaching wide is far less expensive than intuition suggests.
But the reaction delay obeys no such rule. It is a flat tax, paid before any movement, identical for a tap-in and a screamer. Total time needed is therefore t_react + √(2x/a): one term that barely grows and one that does not grow at all. So for small offsets the delay dominates completely, and only for large ones does acceleration start to matter.
The page turns that around, and the inversion is the useful form. Rather than asking how long the keeper needs, ask how far they get in the time that exists: x_reach = ½a(t_flight − t_react)². Look at what is squared — not the flight time, but the leftover after the delay is taken out. That is why the margin collapses so violently. Shorten the flight and reach falls quadratically, from a quantity the fixed delay has already eaten into first.

How to read what you see

Two readings and a verdict. Ball flight time is the budget; time needed for the target is the keeper’s bill; reachable distance converts that bill back into metres so it can be compared against where the shot is aimed. The pitch view marks the reach as a line, which is the quickest way to see the answer — if the marker sits outside the target, the keeper gets there. The three presets are chosen to straddle the boundary, and the middle one is worth sitting on: 0.704 s of flight against 0.716 s needed. Twelve milliseconds decide it, and the reach falls 0.76 m short of an 0.8 m target.

Assumes: A ball at constant speed — no drag slowing it, no dip, no curve, so the flight time is a single division. A keeper who accelerates at a constant rate from a standing start, in the correct direction, having already chosen it. And a reaction delay that is a fixed property of the person rather than something that varies with how readable the shot was.
Breaks when: Take the page’s own "Beaten" preset seriously and it stops being an edge case. A penalty is 11 m; struck at 125 km/h it lands in 0.317 s. Subtract a 0.25 s reaction and the keeper has 0.067 s of movement left, which at 5 m/s² buys a reachable distance of 0.01 m. One centimetre. The model is not being pessimistic — it is stating that a penalty cannot be saved by anyone who waits to see where the ball goes. Which is precisely why real goalkeepers do the one thing this page has no variable for: they commit before the strike, reading the run-up and the plant foot, and convert a reaction problem into a guess. The verdict on screen is the reason the real tactic exists.

Why a non-spinning ball can wobble

A non-spinning or very-low-spin shot does not organize its wake into the stable force direction of a conventional curling shot.

C_L = C_L(t), C_S = C_S(t), ω ≈ 0

Asymmetric vortex shedding can make lift and side-force coefficients vary with time, producing a wandering path rather than a smooth Magnus curve.

Why a spinning ball bends

A spinning ball deflects its wake and experiences a force perpendicular to its velocity and spin axis.

F_M = ½ρAC_Lv² · direction(ω × v)

Reverse the spin vector and the cross product reverses, so the ball bends the other way. The lift coefficient also depends on seams, speed, and spin-to-speed ratio.

How an off-center kick creates spin

Aim the foot's impulse to one side of the ball's center. The forward part launches the ball; the tangential part gives it angular momentum.

ΔL = r × J = Iω

A larger tangential impulse or lever arm creates more spin. Reversing the side of contact reverses the spin direction. Real contact is brief and deformable, so friction, foot path, and contact time all affect the result.

How drag changes the keeper's clock

The reaction calculator uses distance ÷ speed, which deliberately treats ball speed as constant.

m dv/dt = −½ρC_DAv²

A real ball slows. If the input is launch speed, drag generally makes arrival later than the constant-speed estimate; modeling that requires integrating velocity.

Why real goalkeeper reach is different

The blue keeper point represents lateral body displacement, not the fingertip position.

s_ball = s_body + r_reach

A real save also includes arm reach, an initial push step, changing acceleration, and direction-choice time. Those belong in a richer biomechanical model.

Problem solved in full

An 80 km/h shot aimed 1.8 m to the keeper's side 8 steps
A shot is struck from 16 m at 80 km/h, aimed 1.8 m to the keeper's side. The keeper needs 0.20 s to react, then accelerates sideways from rest at 6.0 m/s². Does the keeper get there — and if not, how slow would the shot have to be? Ball speed is constant and the keeper is a point, which is the model on this page and not the whole of football.
1. You cannot divide metres by kilometres per hour. Convert once, at the start, and every later step is in one system.
  
  $v = \frac{80}{3.6} = 22.22 m/s$
2. Constant speed makes the flight a single division. That number is the entire budget — everything the keeper does has to fit inside it.
  
  $t_{f} = \frac{d}{v} = \frac{16}{22.22} = 0.720 s$
3. The delay is subtracted from the front of the flight, not scaled out of it. What survives is the only interval in which the keeper is actually moving.
  
  $Δ t = t_{f} - t_{r} = 0.720 - 0.20 = 0.520 s$
4. From rest under constant acceleration the displacement goes as the square of the moving time. That squaring is why the delay hurts out of proportion: 0.20 s is 28% of the flight, but it removes 48% of the ground the keeper could otherwise have covered.
  
  $s = \frac{1}{2} a Δ t^{2} = \frac{1}{2} (6.0) (0.520)^{2} = 0.81 m$
5. Set that against the 1.8 m the shot demands. The shortfall is bigger than the distance covered — the keeper is closer to standing still than to saving it.
  
  $1.8 - 0.81 = 0.99 m$
6. Run the same algebra backwards to ask what window 1.8 m would need. Solving ½aΔt² = 1.8 for Δt is a square root, so the required window grows slowly as the target moves further out.
  
  $Δ t_{req} = \frac{2 \times 1.8}{6.0} = 0.6 = 0.7746 s$
7. Add the delay back. 0.975 s is the flight time this save needs; 0.720 s is what the striker granted.
  
  $t_{req} = t_{r} + Δ t_{req} = 0.20 + 0.7746 = 0.975 s$
8. A flight time is a distance over a speed, so a requirement on the clock converts directly into a speed limit on the shot.
  
  $v_{crit} = \frac{d}{t _{req}} = \frac{16}{0.9746} = 16.42 m/s = 59.1 km/h$
Answer

The keeper ends 0.99 m short — a gap wider than the 0.81 m actually covered. Effort does not close it. Reach grows as the square of the moving window, so the target is met only once the shot falls to 59.1 km/h: a 26% cut in speed buys a 2.22× gain in reach, and that leverage runs entirely in the striker's favour. The keeper's own lever is much shorter. Cut the reaction delay from 0.20 s to 0.15 s and the reach rises only to 0.97 m, still 0.83 m adrift; even at 0.10 s it reaches 1.15 m and the ball is in. Within this model the clock belongs to whoever kicked the ball — from 16 m to an offset of 1.8 m, every shot above 59.1 km/h arrives first, no matter how fast the keeper reads it.

$\frac{v _{crit}}{v} = \frac{t _{f}}{t _{req}} = 0.739, \frac{1.8}{0.81} = 2.22$

References (1)

The lesson's closing point — why keepers commit before the strike rather than react to it: G. J. P. Savelsbergh, A. M. Williams, J. Van Der Kamp and P. Ward, "Visual search, anticipation and expertise in soccer goalkeepers." Journal of Sports Sciences 20(3), 279–287, 2002.

By the 42Math team · Updated 2026-07-22

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

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Goalkeeper reaction window