Force calculation for a unusual hang

[james90755]

Try the ridgepole idea.

I thought about the ridgepole idea, but I don't like a pole above my head... and in my application the max length of pole is 56" in order to pack it in the Jeep. Having a two piece ridgepole would be a risky proposition IMO. Other factors include time/complexity to set up...

I have a special screw stake coming that might work... Then it becomes a sink two stakes, biner in the hammock, tighten whoopie, done.

[necaver]

XexorZ has the right idea in his videos. He's basically using homemade cable stakes like the ones trappers use to anchor their traps. They can be a hard to remove, but they are cheap to make if you do lose one. The ridgepole idea would transfer some of the force from the stakes to the inverted V poles, which could give you some added security.

[james90755]

XexorZ has the right idea in his videos. He's basically using homemade cable stakes like the ones trappers use to anchor their traps. They can be a hard to remove, but they are cheap to make if you do lose one. The ridgepole idea would transfer some of the force from the stakes to the inverted V poles, which could give you some added security.

On the cable stakes idea, I like the stability of the anchor... I don't like the work to retrieve it and the chance of losing it.

[OutandBack]

In loose dirt, sand, or gravel a deadman type anchor would probably work best. I used them for wenching when 4x4'ing. Just size it smaller for this purpose.
hth

[GrizzlyAdams]

I set off to try and do the math, but cannot figure it out.
I searched hf and can't extrapolate from other threads.
....

So the question is: Any idea what force is exerted on the spike from the rope??

With a 250 lb load, I think that translates into 125 lb
horizontal force vectors pointing in from the eyebolts.
That force gets translated down the cord to the ground, as a function
of the steepness of the angle theta between the cord and the ground,
at the stake. Getting all quanty, that force is 125 lbs / cos(theta).
Now cos(theta) = h / L where h is the height of the eyebolt
above the ground and L is the length of the cord to the ground. So with
your 97 inch cord and a height around 48", the force trying to pull out the stake is 250 lbs. Which you knew but couldn't quantify.

What to do?

Spreader bar. Two five foot long sectoins of fencerail. Eliminate the foces, or greatly control them at least.

Bingo! If you can, get something rigid that keeps the two eyebolts from collapsing inward when the hammock is loaded.

Without calculating the force, you can use the auger type anchors used for staking out dogs. Walmart for about $6 each. I have used them in similar circumstances and you cannot pull them out without unscrewing them from the ground. These are the type anchors utilities use to anchor guy wires for poles.

Bingo again! With a caveat. Been there and done that and have had the anchors pull up out of the ground when it is really soggy.

Another trick (involves more work and more stakes) is to rig up a sequence of stakes, see diagram. There's a name for this, but I've forgotten...the idea is that as the first stake is pulled forward by the main force, it is checked in part by the connection to the second stake, and so on.
Screen shot 2011-02-10 at 5.16.35 PM.png

[oldgringo]

I think if you incline the bipods so that the angle of incidence between the hammock suspension and the bipods is reduced, the bipods will take more of the load, reducing the load on your anchor points. Grizz, what say you?

[Youngblood]

With a 250 lb load, I think that translates into 125 lb horizontal force vectors pointing in from the eyebolts.

Grizz, will it not translate into 125 lb vertical force at the eyebolts and we will not know the horizontal force without knowing the angle of the hammock suspension rope/webbing?

Probably the minimum force would be with a hammock suspension rope/webbing at a 30 degree angle, which would make the force on the hammock suspension rope 250 lbs (125 / sin (30)) with a horizontal component of 216.5 lbs (125 / tan (30)). If I'm figuring it right, it would be 216.5 lbs or more if the hammock rope/webbing suspension angle was less, instead of the 125 lbs you used in your example. If they are using a hammock with a structural ridgeline it could get up to 250 lbs pretty easily.

[matto]

I think if you incline the bipods so that the angle of incidence between the hammock suspension and the bipods is reduced, the bipods will take more of the load, reducing the load on your anchor points. Grizz, what say you?

I think that's a good thought. Or why not use tripods for better stabilty?

Also, for the stakes, making the line longer (i.e. staking down farther away from the hammock) would be an easy way to reduce the amount of "pull out" force. The stake is going to hold better if there's less upwards and more sideways force acting on it.

[PuckerFactor]

Spanish Windlass is the term, I believe, Grizz.
+1 for a small deadman.

PF

[oldgringo]

Spanish Windlass is the term, I believe, Grizz.
+1 for a small deadman.

PF

Hmm...I was taught that a Spanish Windlass is a spike is inserted between two parallel lines and rotated so as to twist the two lines around each other, thus shortening their effective lengths.