Equation for the impact force and its interpretation

When modeling the rope as an undamped harmonic oscillator (HO) the impact force F_max in the rope is given by:

${\displaystyle F_{max}=mg+{\sqrt {(mg)^{2}+2mghk}},}$

where mg is the climber's weight, h is the fall height and k is the spring constant of the portion of the rope that is in play.

We will see below that when varying the height of the fall while keeping the fall factor fixed, the quantity hk stays constant.

There are two factors of two involved in the interpretation of this equation. First, the maximum force on the top piece of protection is roughly 2F_max, since the gear acts as a simple pulley. Second, it may seem strange that even when f=0, we have F_max=2mg (so that the maximum force on the top piece is approximately 4mg). This is because a factor-zero fall is still a fall onto a slack rope. The average value of the tension over a full cycle of harmonic oscillation will be mg, so that the tension will cycle between 0 and 2mg.

Derivation of the equation

Conservation of energy at rope's maximum elongation x_max gives

${\displaystyle mgh={\frac {1}{2}}kx_{max}^{2}-mgx_{max}\$ ;\ F_{max}=kx_{max}.}

The maximum force on the climber is F_max-mg. It is convenient to express things in terms of the elastic modulus E = k L/q which is a property of the material that the rope is constructed from. Here L is the rope's length and q its cross-sectional area. Solution of the quadratic gives

${\displaystyle F_{max}=mg+{\sqrt {(mg)^{2}+2mgEqf}}.}$

Other than fixed properties of the system, this form of the equation shows that the impact force depends only on the fall factor.

Using the HO model to obtain the impact force of real climbing ropes as a function of fall height h and climber's weight mg, one must know the experimental value for E of a given rope. However, rope manufacturers give only the rope’s impact force F₀ and its static and dynamic elongations that are measured under standard UIAA fall conditions: A fall height h₀ of 2 × 2.3 m with an available rope length L₀ = 2.6m leads to a fall factor f₀ = h₀/L₀ = 1.77 and a fall velocity v₀ = (2gh₀)^1/2 = 9.5 m/s at the end of falling the distance h₀. The mass m₀ used in the fall is 80 kg. Using these values to eliminate the unknown quantity E leads to an expression of the impact force as a function of arbitrary fall heights h, arbitrary fall factors f, and arbitrary gravity g of the form:

${\displaystyle F_{max}=mg+{\sqrt {(mg)^{2}+F_{0}(F_{0}-2m_{0}g_{0}){\frac {m}{m_{0}}}{\frac {g}{g_{0}}}{\frac {f}{f_{0}}}}}}$

Note that keeping g₀ from the derivation of "Eq" based on UIAA test into the above F_max formula assures that the transformation will continue to be valid for different gravity fields, as over a slope making less than 90 degrees with the horizontal. This simple undamped harmonic oscillator model of a rope, however, does not correctly describe the entire fall process of real ropes. Accurate measurements on the behaviour of a climbing rope during the entire fall can be explained if the undamped harmonic oscillator is complemented by a non-linear term up to the maximum impact force, and then, near the maximum force in the rope, internal friction in the rope is added that ensures the rapid relaxation of the rope to its rest position.^[2]

Effect of friction

When the rope is clipped into several carabiners between the climber and the belayer, an additional type of friction occurs, the so-called dry friction between the rope and particularly the last clipped carabiner. "Dry" friction (i.e., a frictional force that is velocity-independent) leads to an effective rope length smaller than the available length L and thus increases the impact force.^[3]