Tyrolean Simulator

One of the most common questions we receive on our rigging courses is how to calculate the loads on Tyrolean traverses and cableways. Whilst the exact calculations are extremely complex once you account for rope weight, stretch functions, sliding point loads and momentum, a very basic 'point load on simple rope' simulation gives a ballpark figure, and here's a nice little simulator for you to play with!

Frequently-asked Questions on Tyrolean Traverses

What shape is the rope in a Tyrolean?

Without anyone on it, the rope forms a catenary - a hyperbolic function similar to a parabola. The real-world mathematics of catenary systems is extremely complicated but when a rope is loaded the person(s) are usually much heavier than the rope(s) and the system turns more into a simple triangle. You can approximate the maths using this triangle, just be aware the forces are a little higher in reality.

Does the load change at each end if i'm not in the middle of the rope?

Yes. the horizontal force at each end is always identical but the vertical force (and so the angle of the rope) depends on the distances from each end to the central load. It also depends on how the load is being kept in position - is it tied to the rope, or being pulled by another one?

Why always work things out with the load in the middle?

In the centre, the loads at each end are the same, so the maths is easy. Loads can increase beyond these values but unless your load is tied to the rope it will always slide to the centre point anyway.

What sag or angle should I use?

It's entirely relative! You have to make sure that the load at each anchor doesn't exceed your safe limits on equipment (pulleys, clamps, knots, anchors). A smaller sag is better for work as you don't have to haul the load back up the rope each side, but the forces on the system grow very fast as the angle gets small. We'd suggest never allowing the end load to exceed twice the supported load, and that (in the simple model) is an angle of about 15 degrees.

What about stretch in the rope?

This is why the real-world maths needs computers to solve! All the student-level equations assume the rope has weight but doesn't stretch. Nylon climbing ropes have a complex stretch-vs-tension graph, and this must be balanced into the catenary equation. The result is more tension needed for a certain sag than you'd expect on paper.

What if one end is higher than the other?

Obviously the load on each end changes - the horizontal load is the same but the vertical load is higher at the higher end of the traverse (obvious as the rope angle is steeper at that end too). The maths is again possible using a small room of monkeys with iMacs, but the obvious result is to always put your hauling and tensioning gear at the low end of the rope where the tension is smaller!

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