Author Topic: Zipline Trajectories  (Read 1807 times)

DDK

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Zipline Trajectories
« on: December 22, 2020, 06:16:24 PM »
The shape of a cable strung within a span and hanging under the influence of its own weight is known as a Catenary.  Attaching a load at any point on the cable alters its shape.  Additionally, the cable will elastically stretch due to the fact that it is under tension.   If the cable is infinitely resistant to stretching, so called, "Rigid", it will not stretch. 

Background information can be found in topic "Rope Sag - Catenary" @ Link: https://igkt.net/sm/index.php?topic=6887.0

If the catenary for the zipline rider (the "Load") is calculated at each point along the zipline cable, the trajectory, that is, the path that the rider takes, for the zipline can be determined.

Attachment #1 is a graph showing an example of a zipline trajectory for a 100 kg load and two of the catenaries.  This 30-meter zipline starts at a height of 3 meters and finishes at a height of 2 meters for a drop of 1 meter.  The lowest point in the trajectory has a height of about half a meter.  Given that the height is known at each point in the trajectory, the change in height from any starting point determines the potential energy, kinetic energy and speed at every point.

In attachment #2, the trajectories for a Rigid cable and Elastic cable for a 100 kg load are shown.  The Rigid (Analytical) curve is for a cable that does not stretch and has a constant total length of 30.2 meters.  This curve is elliptical, see page 33 of Link: http://www.ropelab.com.au/files/physics.pdf (suggested to me by agent_smith) or even ABoK #2592.  The Rigid (Numerical) data are calculations using an extremely large cable strand elastic modulus (essentially no stretch).  The Elastic (Numerical) data are using a cable strand elastic modulus of 47,500 (newtons/millimeters squared).  The Elastic cable has approximately 0.30 meters more sag than the Rigid cable.

Edit: The Rigid cable trajectories will not look exactly symmetrical or elliptical.  This is due to the difference in the scaling of the Height and X-Position axes in the graphs. 
« Last Edit: December 22, 2020, 06:26:50 PM by DDK »

KC

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Re: Zipline Trajectories
« Reply #1 on: December 23, 2020, 09:34:56 AM »
Nice.
As we take this to motion tho, as a trajectory.
The inerita of the weight throwing across with squaring factor of velocity would seem in motion less downward on rope as speed increased?
.
Like, if did incremental calculated 'sits' on ice to show the downward force of going across in micro-slices, STATIC time slices and show how breaks
>>vs. throwing hard across ice with run across thinnest ice w/o falling thru?
Back into rope model; if high speed across rope, the across factor could be great enough for the load to run past the position of an overhead termination w/right slack in line under the load, more fully revealing the sidewards force of the motion over the static?
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DDK

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Re: Zipline Trajectories
« Reply #2 on: December 23, 2020, 06:12:34 PM »
Thank you.

A very good point that my current (static) calculations do not include an important dynamical consideration.  That is to say, the motion of the load and in partcular the change of direction of that motion requires an additional force.  This is known as centripetal force which the cable must apply to the moving load to change its direction in addition to applying a force to counteract gravity acting on the load.  This will induce additional tension in the cable which does not affect the trajectory for the "Rigid" cable since it does not stretch (like a roller coaster) but will affect the trajectory for the "Elastic" cable which will have additional stretch and sag.

This phenomena is part of common experience as one swings a ball attached to the end of a string in a vertical plane.  At just the right speed and at the highest point of the circular swing, this centripetal force may be exactly supplied by the weight, gravitational force, of the ball and the tension in the string will be zero.  Continuing the swing at the same speed to the bottom of the swing, one will feel a tension in the string equal to twice the weight of the ball.  At the bottom, the string must supply both the force to counteract gravity as well as the force to continue changing the direction of the ball's motion. For more background, see Link: https://en.wikipedia.org/wiki/Circular_motion.

The magnitude of the centripetal force required at a particular point is equal to mv2/R where m is the mass, v is the magnitude of the velocity, a.k.a. speed, and R is the radius of curvature of the trajectory at that point.  At the lowest point, the Elastic zipline trajectory in the OP has a radius of curvature of about 70 meters.  At speeds of approximately 7 meters/second and a 100 kg mass, this results in a required centripetal force of 70 newtons or the equivalent of an additional 7 kg.  At double that speed, the centripetal force will quadruple to 280 newtons (an additional 28 kg).  Certainly worth adding to the next set of calculations.

The thin ice failure would seem to be more of a time dependent mechanical deformation and crack propagation effect.   In this case, as the velocity increases, the time the load is on the thin ice spot and able to do damage decreases.  The less time, the less damage.



 

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