Calculate shock load or impact force from a falling weight, its drop height and stopping distance, or from impact speed and stopping time.
- All Automotive Calculators
- Average Impact Force Calculator
- Impact G-Force Calculator
- Average Force Calculator
- Force From Height Calculator
- Gravitational Force Calculator
Shock Force Formula
The distance and elastic models estimate peak contact force under explicit ideal assumptions. The duration method instead gives average net stopping force. Inputs are converted to SI using conventional standard gravity g = 9.80665 m/s²; real stopping histories and equipment performance are not established.
Dropped Object Formula
- F = model peak contact force, in newtons (N)
- W = weight force of the object, in newtons (N)
- h = free-fall distance, in meters (m)
- d = actual final stopping travel after contact, in meters (m)
- k = 1 for constant contact resistance, or 2 for force rising linearly from zero over the entered travel.
Rope / Cable Formula
- F = model peak contact force, in newtons (N)
- W = weight force hanging on the rope, in newtons (N)
- h = free-fall distance before the rope catches, in meters (m)
- EA = rope or cable axial stiffness, in newtons (N)
- L = rope length in play, in meters (m)
The contact work is W(h + d), including gravity’s work during stopping. Dividing by d gives travel-averaged contact force W(1 + h/d). Constant resistance has that same peak; an ideal linearly increasing contact force has twice the travel average because its force-versus-distance area is triangular. These are assumed profiles, not verified properties of padding or ground. The square-root expression W[1 + √(1 + 2h/δ)] uses static spring deflection δ = W/K, not an independently entered actual stopping travel d. See the displacement work relation in OpenStax Work.
The rope model assumes an initially unstrained, massless linear elastic member with fixed anchor, K = EA/L and no damping. At the first maximum extension x, 0.5Kx² = W(h + x), peak force F = Kx and x = FL/EA. At fixed W and h, increasing EA/L increases this ideal peak. Actual ropes may be nonlinear, dissipative, rate-dependent or preloaded; no real rope property or rating is supplied by a material name. At h = 0, sudden release into the unstrained elastic member gives peak 2W; static equilibrium gives W. For constant resistance, h = 0 is only a formal zero-net-force bound, not finite deceleration travel from rest. The elastic energy basis is described in OpenStax Potential Energy.
Stable distance and height inverses are d = kWh/(Flimit − kW) and h = d(Flimit − kW)/(kW). Minimum distance requires h > 0 and Flimit > kW; at equality a positive fall would require infinite travel. Height allows exact Flimit = kW with h = 0, and rejects a lower limit. These are model constraints only; h + d is not a complete person/equipment clearance requirement.
For a downward vertical impact stopping from speed v to rest in duration t, impulse gives average upward NET force mv/t; average upward CONTACT force is mv/t + W. Duration alone identifies neither peak nor actual stopping travel. A triangular NET-force pulse would have peak 2mv/t, and constant NET deceleration would give travel vt/2; these are separate conditional illustrations, not predicted force history. See OpenStax Impulse and Collisions.
Unvalidated Numerical Illustrations
The numbers below are retained only as unvalidated arithmetic illustrations, not typical or measured surface, rope, cable or absorber properties. Enter independently supported actual travel and stiffness. No numerical helper establishes a product rating or regulatory limit.
| Numerical illustration | Unvalidated travel range | Within the same ideal model |
|---|---|---|
| Travel example A | 0.5 to 2 cm | Smaller d gives higher model force |
| Travel example B | 2 to 10 cm | Force depends on W, h and profile |
| Travel example C | 20 to 50 cm | Larger d lowers the model force |
| Travel example D | 50 cm or more | No absorber rating is established |
| Numerical EA illustration | EA value used | At fixed W, h and L |
|---|---|---|
| EA example A | 25,000 N | Lower EA gives lower ideal peak |
| EA example B | 60,000 N | Increasing EA increases ideal peak |
| EA example C | 120,000 N | Increasing EA increases ideal peak |
| EA example D | 8,000,000 N | Highest listed EA gives highest ideal peak |
Shock Force Examples
Example 1: Dropped Object
An 80 kg object falls 1 m and has actual stopping travel 0.05 m. Assume force rises linearly from zero over that travel, k = 2.
The ideal linear-rise peak contact force is 32.950344 kN. The travel-average contact force, also the constant-resistance result with k = 1, is 16.475172 kN. Contact work is 784.532 × (1 + 0.05) = 823.7586 J. This is a numerical model, not a measured force or equipment rating.
Example 2: Rope / Cable
An 80 kg load falls 2 m before engaging an ideal unstrained elastic member of length 10 m. Use EA = 25,000 N as an unvalidated numerical illustration, not a property of any climbing rope.
The ideal peak contact tension is about 3.693279 kN, with extension x = FL/EA = 1.477311693 m. The balance 0.5(EA/L)x² = W(2 + x) gives about 2728.062297 J on both sides. No product stiffness or rating is verified.
FAQ
Why is shock force higher than the object’s normal weight?
Weight is the steady force from gravity. Shock force is the force needed to stop motion after a fall. When an object stops quickly, its kinetic energy must be absorbed over a short distance, so the peak force can be many times higher than its normal weight.
Does a longer stopping distance always reduce shock force?
Within a fixed force-versus-travel profile and positive fall height, increasing actual stopping travel reduces its modeled peak. Real pads, ropes and absorbers can change profile, stiffness, damping and geometry; the calculation does not infer their travel or performance from a material name.
Is this result safe to use for rigging or fall protection design?
This calculator does not approve rigging, lifting, climbing or fall-arrest design. Real systems include preload, damping, plasticity, knots, slip, friction, moving anchors, load geometry and person/harness response. A force alone cannot establish compliance with the system criteria in OSHA 1926.502 or with other applicable rules. No universal EN/ANSI limit, regulatory preset or equipment recommendation is supplied. Optional comparison is only against a user-entered number with the same force basis: peak contact for distance/rope, average net for impulse. An already adjusted reference is used directly; a divisor is applied only when explicitly selected for an unadjusted reference, with no automatic rigging factor or second reduction. Qualified system-specific assessment and testing remain separate.
