# Thread: Hitchcraft for hammock tensioning

1. ## tension on the rope

BB and JJ--before we create the vertical tension on the rope, I break a sweat and pull on that rope as hard as I can, and fasten it. P lbs is what I can apply on a good day. So there is that much tension on the rope before climbing into the hammock. Now I climb in and create the additional stress whose formula we know. P did not go away. So is the total tension P + ASWFWK ? Does trigonometry come sneaking back here too?

BB--yeah, I can see that the stress on the ridgeline could have some interesting geometry.

Thanks.

Grizz

BB and JJ--before we create the vertical tension on the rope, I break a sweat and pull on that rope as hard as I can, and fasten it. P lbs is what I can apply on a good day. So there is that much tension on the rope before climbing into the hammock. Now I climb in and create the additional stress whose formula we know. P did not go away. So is the total tension P + ASWFWK ? Does trigonometry come sneaking back here too?

BB--yeah, I can see that the stress on the ridgeline could have some interesting geometry.

Thanks.

Grizz
The ridgeline is how that extra tension gets added in...like I said, I'll see about drawing something up later this evening.

3. ## simple rope tension problem

Originally Posted by blackbishop351
The ridgeline is how that extra tension gets added in...like I said, I'll see about drawing something up later this evening.
Appreciated. If you haven't started already, here's a sketch of the simplest problem I can think of that whose solution would illustrate the principle I'm asking about.

A cord is originally horizontal, stretched to be under tension at an equivalence of P lbs (at the center, if position matters).

A weight of W lbs is attached to the center, causing the line to sag
and creating an angle theta between the line and its former horizontal position. What now is the tension on the rope?

Let the mathemagic begin!

Thanks,
Grizz

4. OK, lets say we get this math thing all worked out. But now lets throw in a new twist. What's the failure rating at below freezing temperatures? I'm over fifty and I tend to bounce less when I'm cold!

5. Originally Posted by Just Jeff
The force vector on the rope is the hypoteneus of a triangle.

h = (.5 x user weight) / sin(support angle)
JJ - that is true only for a hammock without a ridgeline or for those with a ridgeline which isn't pulled tight when hung. For those with a ridgeline that people pull tight, that formula still ignores the horizontal force exerted in pulling the ridgeline tight.

6. Originally Posted by blackbishop351
The ridgeline is how that extra tension gets added in...l
Well - yes and no. The ridgeline doesn't create the extra force, the extra force is created when the ridgeline is pulled tight.

Appreciated. If you haven't started already, here's a sketch of the simplest problem I can think of that whose solution would illustrate the principle I'm asking about.

A cord is originally horizontal, stretched to be under tension at an equivalence of P lbs (at the center, if position matters).

A weight of W lbs is attached to the center, causing the line to sag
and creating an angle theta between the line and its former horizontal position. What now is the tension on the rope?

Let the mathemagic begin!

Thanks,
Grizz
Okay, you now have the original force applied in pulling the rope/line/cord tight, plus the weight applied as indicated, plus the elastic force exerted by the fibers in the rope/line/cord used when stretched, both the stretching when the rope/line/cord is pulled tight and the additional stretching when the weight is added. Lacking a lot of experimental data, the last force is totally unknown to me. I would guess though that it can be quite high.

Also, you will note that the vertical force is the applied weight plus the weight of the rope/line/cord. That vertical force is translated into a force along the rope/line/cord which is equal to (ignoring the weight of the rope/line/cord):

W / (2 * sin(A))

where A is the angle the rope/line/cord makes with the horizontal. Now if the initial force was purely horizontal, then the force on the rope initially was:

P / (cos(B))

where B is the angle the rope made with the horizontal before adding any weight. Both of the above ignore any stretching of the rope and hence any of the elastic forces of the rope.

Now the total force, excluding the elastic force, along the rope is the sum:

(W/(2* sin(A))) + (P/cos(B))

If B is less than, say 10 deg., then the second term can be closely approximated by P, ignoring the cos(B) term. for A equal to the much used 30 deg, then the first term reduces to W as stated previously.

What the 30 deg assumption used previously blithely ignores is that for those using a structural ridgeline, it is my experience that A is very much less than 30 deg.It is more common in my experience that A is closer to 10 deg (estimated, not measured). For a value of 10 deg, the first term reduces to:
W * 2.9

to the first decimal place. For 15 deg., we have:

W * 1.9

and for 5 deg., we have:

W * 5.7

The value of A is going to very dependent on P and the elastic properties of the suspension rope and the elastic properties of the ridgeline cord. If the ridgeline and suspension cords have very little stretch, then A is going to be smaller relatively than if both are fairly elastic.

Thus, the force applied along the suspension rope for a hammock with a structural ridgeline can be much greater than the weight of the occupant. And that ignores the elastic forces which may be much greater ( a guess on my part based on my experience with bent steel hooks and without the benefit of data on the physical properties of the ropes used).

8. ## sum of forces

got it, thanks for the run down. So for a line starting at the horizonatal it's

tension = P + ASWFWK + unknown-elastic-force-that-bends-steel-S-hooks

Given how I've been cranking up the tension on the ridgeline of my HH, I'm
(a) mighty glad I've doubled up the Dyeema cord on my rings, and
(b) considering taking a step ladder along on my backpacking trips so I can just attach the lines higher, still keep my posterior off the ground, and reap the benefits of large sine(theta).

Grizz (who, having just crossed the magic threshold of 30 postings exits the state of probationary membership, wonders if anyone will now teach him the secret electronic handshake)

9. Glad we have you mathematically-inclined fellows here to let us know it's OK to use the 200# cleats. I'll have to get some - that would make testing so much easier.

10. man... i didn't know yall's going to be talkin greek on here

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