Originally Posted by
bluesam3
The tension in the ridgeline is W(cot(b) - sin(a)/2), where W is your weight, b is the suspension angle, and a is the repeated angle of the isosceles triangle with two edges half the length of your hammock, and the third the length of the ridgeline (to find this, resolve the forces at your body vertically, and the forces at the point where the ridgeline meets the hammock suspension horizontally). Generally, b should be around 30 degrees, and your ridgeline should be around 83% of the length of your hammock, so a is around arccos(.83), which is 33.9 degrees. That makes the tension in the ridgeline W(cot(30)-cot(33.9)/2), which is, conveniently, pretty damned close to 1 (0.988, specifically). Thus, the tension in the ridgeline is pretty much equal to your weight.
Now, consider hanging something on your ridgeline. The effect depends on where you hang it, so we'll take the case that gives the biggest dip, and put it in the centre. Then the angle between the dipped ridgeline and the horizontal is arcsin(X/(2W)), where X is the weight of the thing that you're hanging [NB: I'm ignoring the extra tension on the ridgeline due to the thing that you're hanging, because it makes it easy, and should be small in comparison to the tension from your bodyweight]. The distance that the thing drops, then, is 0.2LX/W, to as much accuracy as anybody could reasonably care about. Let's say you hang 10% of your bodyweight off your ridgeline. Then the total deflection is 0.02L. Assuming an 11' hammock, that's about 2 inches.
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