Impact Force Calculator

Using the impact force calculator

Choose what to calculate: the impact force, the body's mass, its velocity, the collision distance or the collision duration. The form shows the inputs that mode needs, each with a unit menu (force in N to GN or lbf, mass from mg to US tons, speed from in/s to miles/s, distance in mm to yards, time in ms to minutes).

The motion has to be described by either the collision distance or the collision duration. A divider on the form reminds you which pair of options applies in the current mode. If you fill in both, the calculator uses the first one it lists: the distance when solving for force, mass or velocity, and the force when solving for distance or duration.

• Impact force: returns the average force and the peak force. The unit steps from N to kN, MN and GN as the average grows past 1,000 N, 1,000,000 N and 1,000,000,000 N.
• Mass: shown in kg and lb, in g below 1 kg, or in tonnes and (US) tons from 1,000 kg.
• Velocity: shown in m/s, km/s, miles/s and ft/s.
• Collision distance: shown in mm, cm or m depending on its size, and in inches.
• Collision duration: shown in milliseconds, seconds, or both.

Note the difference between a tonne (metric, 1,000 kg) and a ton (US short ton, 2,000 lb or about 907.18 kg) when choosing the mass unit. A pound-force (lbf) is counted as 4.448222 N.

Impact force formulas

A body that is stopped loses all its kinetic energy, ½mv², to the work done by the stopping force over the stopping distance. Setting the two equal gives the average force:

F(avg) = m × v² ÷ (2 × d)m in kg, v in m/s, d in metres: the work–energy approach

If the time Δt of the collision is known instead, use the change in momentum (the impulse–momentum theorem):

F(avg) = m × v ÷ Δt

F(peak) ≈ 2 × F(avg)the usual rule of thumb for a force that builds up and fades away

The two methods agree when the body slows down at a steady rate, because then d = v × Δt ÷ 2. This relationship is also how the calculator turns a duration into a distance (or back) when you solve for one of them.

Rearranging gives the other modes: m = 2 × F × d ÷ v² (or F × Δt ÷ v), v = √(2 × F × d ÷ m) (or F × Δt ÷ m), d = m × v² ÷ (2 × F) and Δt = m × v ÷ F.

Worked examples

A fall onto a cushion. An 80 kg person lands at 10 m/s and is brought to rest over 0.5 m. F = 80 × 10² ÷ (2 × 0.5) = 8,000 N (8 kN), with a peak of about 16 kN. The stop lasts 2 × 0.5 ÷ 10 = 0.1 s, and entering that duration in place of the distance gives the same 8 kN: 80 × 10 ÷ 0.1.

A car hitting a barrier. A 1,500 kg car at 50 km/h (13.89 m/s) crumples by 0.6 m. F = 1,500 × 13.89² ÷ (2 × 0.6) ≈ 241 kN, with a peak near 482 kN. That corresponds to an average deceleration of about 16 g.

A dropped object. A 2 kg tool falls 1.5 m and its tip stops in 5 mm. The speed at impact is √(2 × 9.81 × 1.5) ≈ 5.42 m/s, so F = 2 × 5.42² ÷ (2 × 0.005) ≈ 5,886 N, and the peak is about 11.8 kN. A shortcut gives the same number: F = m × g × h ÷ d = 2 × 9.81 × 1.5 ÷ 0.005.

How much room is needed? To keep the average force on a 70 kg person landing at 4 m/s below 5 kN, solve for distance: d = 70 × 4² ÷ (2 × 5,000) = 0.112 m, with the stop lasting 70 × 4 ÷ 5,000 = 0.056 s.

What drives impact force

• Stopping distance. The force is inversely proportional to it. Crumple zones, padding, bent knees and airbags all work by lengthening the distance (and the time) over which the energy is absorbed.
• Speed. With a fixed stopping distance the force grows with the square of speed, so doubling the speed quadruples the force. With a fixed stopping time it only doubles.
• Mass. The force is directly proportional to mass, as long as the speed and the stopping distance are the same.

The energy that has to be absorbed is the kinetic energy; you can check it with the kinetic energy calculator. For speeds derived from a drop or a rate of change, the acceleration calculator and speed calculator are useful companions. A compressed spring stores energy in a different way, covered by the elastic potential energy calculator.

Assumptions and limits

• The body is treated as a rigid point mass that decelerates smoothly over the stated distance. Real collisions are messy: forces spike when stiff parts meet, then drop.
• The peak value is a simple estimate of twice the average. A force that is more sharply peaked than a triangle can exceed this considerably.
• All of the kinetic energy is assumed to be absorbed over the stopping distance. If the object rebounds, the impulse is larger than m × v.
• Gravity, friction and rotation during the impact are ignored.

Impact estimates are for learning and rough planning only. They are not a safe design limit for helmets, vehicles, structures or people. See the disclaimer.