User:Iratheclimber/Sandbox


 * $$F_{max} = mg + \sqrt{(mg)^2 + 2mghk} = mg + \sqrt{(mg)^2 + 2mgEqf}$$


 * $$ mgh = \frac{1}{2}kx_{max}^2 + mgx_{max}\ ; \      F_{max} = k x_{max} $$

form:


 * $$F_{max} = mg + \sqrt{(mg)^2 + F_0(F_0-2m_0g)\frac{m}{m_0}\frac{f}{f_0}} $$

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 leads to an effective rope length smaller than the available length L and thus increases the impact force. Dry friction is also responsible for the rope drag a climber has to overcome in order to move forward. It can be expressed by an effective mass of the rope that the climber has to pull. This effective mass (which is always larger than the rope mass itself) depends exponentially on the sum of the angles of the direction changes the climber has made.

In rock |rock climbing, rope drag is the friction of the rope plus its weight that the climber feels when pulling a rope through a number of protection points, or over rock prominences. A large number of anchor placements, especially if they form a zig-zag rather than a straight line, can make the rope drag so bad that the climber can hardly move forward. Let’s assume a situation where the rope runs through several carabiners which are not all aligned. By using the equations of Euler-Eytelwein to describe the dry friction between the rope and the protection points, the rope drag can be calculated. It can be expressed by an effective mass of the rope that the climber has to pull, which is always larger than the rope mass itself. This effective mass depends exponentially on the sum of the angles of the direction changes the climber has made. “Early errors” by not using longer runners to reduce the angles at the first protection points are less severe than “later errors” which is in contrast to intuition.