[fireman71]

Hello everyone,

I am wanting someone to check my math and see if I am approaching this problem in the correct manner and obtaining the correct result.

I have a rope travelling from an elevated anchor to a ground level anchor. The rope is set at a 30 degree angle and we will assume the rope has no sag and is taunt.

Somewhere along the rope a 600 pound weight is being suspended.

I need to determine the amount of force applied to each anchor.

I worked this out as follows for the elevated anchor:
B = down rope force vector = Tan 30 times 600 for a value of 346 pounds
C = actually vertical force vector = 600 divided by Cos 30 = 692 pounds
Since the elevated anchor must support both vectors (B & C) the total force applied to it is 1038 pounds.

For the ground level anchorI assumed that it must only support enough force to offset the down rope force so it should only have 346 pounds applied to it.

I am trying to determine the forces applied to the anchors in a rope highline (tyrolean traverse) that is not horizontal when the weight that must be supported is known and the angle of the incline in the tyrolean is also known. I realize that this is only theoretical since the rope will sag some when loaded.

Am I approaching this correctly and getting the correct results or am I missing something?

Thank you,
Ian

[ebaines]

I find it easiest to think in terms of horizontal and vertical components of force at the two anchors and at the load point on the rope.

If we call the force at the elevated anchor E and at the ground anchor G, and realizing that E and G are pulling in opposite directions along the direction of the rope, you can set up 2 equations in two unknowns:

Vertical Force components: E*sin(60) - G*sin(30) - 600 = 0
Horizontal components: E*cos(60) -G*cos(30) = 0

When you solve these you'll find that the value you got for the upper anchor is correct, but the value for the lower anchor is not. Remember, when you're done the total of all forces must equal 0.

[fireman71]

First, thank you ebaines for you quick reply.

Now back to the math!

I understand that for the load to be static all forces must be neutralized (equal to 0).
Using the formulas you provided I came up with the following results:
E*sin(60) - G*(sin(30) -600 = 0 (G here equates to ~597)
E*cos(60) -G*cos(30) = 0 (G here equates to 599)
G = 599 + 597 = ~1196

What force vectors did I miss in my method for computing the ground anchor load?
The only one I see right off is the force of the rope pulling at an upwards angle of 30 degrees.

Thank you again!

[ebaines]

Looking at this problem again, and I think what you are trying to solve is impossible. You say the string is completely taut, correct? If the angle to the ground is a constant 30 degrees, that would require an infinite amount of tension in the string, and hence an infinite amount of force at the two anchors. There must be a change in slope of the string at the point where the weight is attached.

I had an error in the equations I gave you - sorry. Attached is a force diagram, showing that the correct equations to solve are:

Ecos(30) = G cos(30)
Esin(30)-Gsin(30)-600 = 0

The first equation leads to E = G, and the 2nd then says 600 = 0. Impossible!

[fireman71]

Once again thanks!

hmmmmmm, infinite tension is of course an impossability (infinite tension = rope failure = BAD! :)

and I realize that there must be some sag at the point of loading.

As it looks I can not just "ignore" the sag factor as I initially thought.

I guess we are going to have to drop back to the beginning to accomplish the task, so let me state the main problem I am trying to solve.

Given a 1/2" rope that is rigged from an elevated anchor to a ground level anchor. This rope will be used to support a 600 pound load at an undetermined point along the sloped rope so that the load is suspended above a particular point an unknown distance from either end. What is the required strength of the anchors at both the elevated and ground level points to support the load. Then what is the required strength to support the load with a safety factor of 10:1.

25/09/2007--
ACK! ignore the numbers I have listed for the angles in the below diagram. There is no way these angles would be generated with those distances. Sorry/

I perform rope rescue work in my main career and am attempting to obtain a better understanding of the physics involved in these operations for personal knowledge. It also seems that noone in my area seems to know how to do this math and determine an accurate answer to this problem.

An analogy for this situation is as follows:
NFPA 1983 defines a general load as being 600 pounds or less. The same standard also recommends the use of 1/2" diameter rope in these operations. This 600 pound load would be a victim and a rescue attendant at the top of a building that are lowered at an angle across some type of obstacle to a ground level location where the victim can be treated and transported. What strength is required for the anchors at each end of the rope to withstand the involved forces with a system safety factor of 10:1. The acceleration from gravity is offset by a second rope attached to the load that applies an opposite force vector to allow a controlled descent. On another note, would the forces affecting the anchors change as the load is moved along the rope significantly?

Once again... hmmmmm...

Thank you for your invaluable assistance in this matter!

After thinking some more on this matter (I am adding the following with edit instead of just making a new reply) it seems that the amount of sag allowed in the line is going to be vitally important in determining the loads placed upon each anchor since the anchors will have to withstand more force as the amount of sag is decreased in the rope up to the theoretical infinite force if there is no sag in the rope at all. So if we allow a predetermined amount of sag this problem should become solvable.

Known factors:
1 - The elevation distance of the elevated anchor in feet
2 - The lateral separation distance between the elevated and ground anchor in feet
3 - The weight of the applied load
4 - The angle of the loaded rope at both the elevated and ground level anchors

Primary unknown factor:
1 - The force the anchors must withstand to support the load

Supplimentary unknown factors:
1 - The amount of sag (in feet) allowed in the unloaded rope to generate the known (or expected) angles at the anchors when the rope is loaded.
2 - The angles a, a1, a2, b, b1, and b2 that will result when the rope sags

The more I think about this the more I think it might be unsolvable without more information. I cannot figure what the sag will be until the rope is loaded which might then be too late and without the sag distance I cannot compute the angles involved with the sag.

Is there a way to accurately compute this with worst case scenario and determine the maximum amount of force applied to each anchor so that we can state that the maximum amount of force applied to the elevated anchor is X when the load is suspended at position Y (whatever position would generate a realistic worst case loading) and the maximum amount for the ground level anchor is X1 with the load suspended at position Y1 (again worst possible position to generate the maximum load)?

That would let us know what each anchor's maximum loading would be to allow us to multiply by 10 to obtain the needed strength of the anchor with a 10:1 safety factor.

Thanks once more!:)

[ebaines]

I think what would help is if you can tell us the max load that the rope is designed to hold. Then perhaps we can figure out how to constrain this problem so that it doesn't end up with infinite force on the anchors. For example, if you know the rope will fail at, say, 5000 pounds, that gives us something to work with.

[fireman71]

I think what would help is if you can tell us the max load that the rope is designed to hold. Then perhaps we can figure out how to constrain this problem so that it doesn't end up with infinite force on the anchors. For example, if you know the rope will fail at, say, 5000 pounds, that gives us something to work with.

Sure thing.

The maximum permissible load is 600lbs. This means 2 people with all equipment cannot exceed 600 pounds. The rope's minimum breaking strength is at least 9000lbs and the rope is the weakest component in the whole setup.

One of the issues I run into is that multiple ropes can be used to increase the amount of force in the system. IE - 2 ropes could hold 18000lbs with 9000lbs per rope, 3 could hold 27000lbs, etc.

Which is why I am trying to figure out the maximum possible force that the 600 pound load could generate in a worst case scenario. That would give me the number of ropes that I would need to use for maximum safety plus tell me the amount of force that the anchor system must withstand.

This is traditionally done with a rope stretched between 2 points at the same elevation with the following formula:
Tension applied to each anchor = (load weight / 2) / (Cos (angle created when the rope sags / 2)).

This means that when a rope supporting 600 pounds sags enough to form an angle of 120 degrees each anchor will be supporting 600 pounds and there will be 1200 pounds of force exerted at the point where the rope is actually loaded. (2 600 ft.lb vectors intersecting at the load point) This requires 2 ropes to be used which reduces the anchor load per rope to 300 pounds and reduces the load point force vectors to 600 pounds per rope which is the maximum permissible amount.

At least this is the formula I have always been told to use for two points elevated to the same distance.

My question has always been what is the load force generated on each anchor when the anchors are not at the same elevations? All of my former instructors and the books I have read have'nt been able to answer this question and I am repeatedly told to just use the biggest anchor I can find. Yeah, that's a generally good idea but how big is big enough is what I want to mathematically determine if at all possible.

Thanks once more.

[cousinjack]

Working with ropes and lines and formula, you can get bogged down in mathematics. IN an emergency situation you normally do not have time for niceties. Your main objective is to complete a rescue safely and as quickly as possible with the minimum risk to life.
If you have two static points, experience will tell you if the static points will be strong enough to take 600lbs load + 20-30%.

In your case, use a rope/line with the absolute minimum safe working weight of 600lbs as the breaking strain is far in excess of this.

If the rope or line that you are using is bar taught with no sag, then it is probably close to its maximum safe working weight and probably closer to is breaking strain. Unless of course you have a weight tensioner available to indicate the strain on the rope.

My advice is to use the strongest rope that you can get your hands on, use your skill and experience to get it taught, If it does not sag when you put weight on it the chances are that it will snap and you will both plummet to the ground.

If you are working from ship to ship or, ship to shore you have to allow for snatch as the vessels roll back and forth. If you have too much tension on the rope then as the ships roll apart the rope will snap.
To avoid this lash the end of the rope to a heavy duty vehicle tire and then lash the tire to a static point, this will help alleviate the snatch.
Most ships have vehicle tires hung over the sides to stop damage to the sides of the ships.

My heart bleeds for you if you you require an answer for an examination as I have been through it.
Mathematics will not make you a good rescuer, but a thorough understanding and knowledge of ropes knots and rigging, plus experience will make you a good safe rescuer and keep you alive.
My experience: Ex Fireman (3years) Ex Rigger (7 years) Deck hand / Master mariner (27 years.)