Speed Line Force?

i have not read these posts, just some observation this past season though that i thought was amazing.

steep angled speedline with a no control line (let down line) = less pull on the anchor attachment than I would think.

much less than if we were lowering from the same tree.

steep speedline with no control line = safer rigging than traditional lowering.
 
Did anything posted give you useful information? If so, what?

The speedline equations are on the money. Were they helpful, or could you apply them to your work? They're most useful for learning about tension in a speedline before a person rigs with one.

Joe
 
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Did anything posted give you useful information? If so, what?

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Indeed. The links were awesome. I especially liked the slack line link because we can find out the force quickly.

To be honest, I'm installing a zip-line for a customer and his children.
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Since the zip-line wouldn't be angled as much as a controled speedline in a tree operation, I knew I'd be messing with different higher forces. I needed tools to figure that out.

However, it has been awakening for me when I use speed-lines in future tree operations.
 
The method of calculating the forces is the same whether it si a zipline, slackline, Y-hang, of tightrope. It is just simple trigonometry. The slacklineexpress calculator asks fot the amount of sag. If you know the sag or the angles, the problem is solved. The real problem is predicting the sag or angles. To get the sag and angles for a given span and load, you need to know the stretch characteristics of the rope. This information is hard to get.

The wolfram site uses a catenary. You can ignore the catenary if the weight of the rope is much less than the weight of the load.

If you photograph the line to get the forces, take the photo perpendicular to the zipline with the lens axis horizontal. Do not worry about wasting space at the bottom of the picture.

Many books on rigging or climbing show a Y-hang and show how the tension in the rope can be large as the included angle approaches 180 degrees. Much of this is an exaggeration that is impossible to achieve with a real rope. IF there is 2 percent stretch in the rope, the tension will be 2.5 times the load. High-modulus ropes, such as Kevlar or Spectra, stretch more than 2 percent at break. To get a 175 degree included angle, the stretch must be less than 0.01 percent.

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Edit: for load with 2 percent stretch
 
Sorry to bring this up again, however I have a new question.

Anyone know where I could calculate force on the speedline, after tensioning to certain number of pounds?
 
"Joe",

I'm not sure what your inferring there, so I won't bother addressing your timeline accusation until you make that a bit clearer. I hope that you will consider in your clarification, that the TCIA publication was some time after the job.

What I will point out is that opposing pendulums is NOT for speed lines, and the formula posted by "Joe Hash" on the old forum in 2000, would NOT work accurately for speed lines.
Please note some of xmans observations.

I will also point out that the day Joe Hash made that post, it included the same intentional arithmetic error we put into the formula.
Each time we see the same error replicated, we laugh and know how it got there, and who it was given to.
 
Jamin,

Working with speed line formulas is difficult because the load is very dynamic. Tyroleans are a lot easier, because the load is generally slower moving, but they are not the same.

When you pretension a system, you're adding quite a few variables. In practical terms, there is no simple formula that will do what you want.

Pulley friction, Deflection rate, Stretch, Line weight/meter, Angle of line, Friction coefficient of load, Length of sling and Anchor stretch are some of the variables that would need to be taken into account.

Putting a dyno on would be the easiest way, and working with as much deflection as possible (less pretension).
 
Here is my scenerio:

I set up a zip line (for a customer). The span is about 120'. It is such a low angle that is more like a tyrolean. I would guess the angle around 5°. The customer didn't like the sag (which I wanted more for safety). He also wanted more speed. I had anchors and zipline spliced with brass thimbles. All made of 7/16" Amsteel.

I set up the system with the idea of making it adjustable with a turn buckle. However, it wasn't fast enough/tight enough for the customer.

Since the whole process didn't seem like it was going to end, I suggested using the GRCS to take up slack (and eliminate the turnbuckle idea). So, I did.

With the turn buckle, I may have had 6'-7' of sag (under load). After tensioning with the GRCS I think there is 4'-5' of sag (under load).

And yes, they are aware of the risks and I had them sign a waiver.
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5° over 120', with 4'-5' sag under load.

From the numbers I've jotted down, your high point is about 10.5' above your low point.
I wouldn't have thought that is anywhere near enough if speed is an issue.

With 4'-5' sag, that means you are losing roughly 50% of your potential energy in sag.
I would've thought your only option is to raise your high point. Is that an possible?
 
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"Joe",

I'm not sure what your inferring there, so I won't bother addressing your timeline accusation until you make that a bit clearer. I hope that you will consider in your clarification, that the TCIA publication was some time after the job.

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Angus, I'm not inferring anything other than I put a version of those trig equations on the web before your version was put in the TCIA publication(which were "published" incorrectly). There is no accusation being made about anything.

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What I will point out is that opposing pendulums is NOT for speed lines, and the formula posted by "Joe Hash" on the old forum in 2000, would NOT work accurately for speed lines.

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Those equations, just like the ones you wrote, can give a reasonable account of the tensions in a line when a weight is at a point along that line with 2 anchor points at different heights like when using speedlines. The equations represent differing tensions at the rigging points when the load is static. It's useful information when we don't have anything better.


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I will also point out that the day Joe Hash made that post, it included the same intentional arithmetic error we put into the formula.
Each time we see the same error replicated, we laugh and know how it got there, and who it was given to.

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You're more observant than I, I could use better information if you're willing to give it.

I used those equations to get an idea of what I could expect to experience when using a speedline. So, now, I set anchors on the lower side of the speedline and take even smaller pieces when I don't have a lower anchor point available. I could have learned this through experience, but the math made it much more clear for me.

Joe
 
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Angus,

I really appreciate you taking time to put these numbers to work for us. It's always good to get more clarity in the science behind our work!
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Then, what was used to come to the conclusions?

The working equations start the process of a better understanding in rigging. Not what was left out.

Joe
 
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