How much would a minimal UHMWPE hammock (i.e. a net woven from ultra-high-molecular-weight polyethylene) weigh?
A standard hammock setup reaches 4–6 meters from anchor to anchor, with a hang angle 30° below the horizontal, so the bottom is 1–1.5 m below the anchor points when you’re sitting at a point in the middle, and the total length of the hammock then is about 4.5–6.7 m.
UHMWPE is springy enough that you don’t need any safety factor for dynamic loads. Suppose the hammock needs to be able to support 200 kg, i.e. 2.0 kN, vertically, or 1.0 kN on each support. This requires the tension in the cord to be 2.0 kN diagonally to have a 1.0 kN vertical component, so the support cord needs to resist 2.0 kN. I don’t have stats on UHMWPE handy at the moment, but suppose its tensile strength is 500 MPa, which is probably in about the right range — that’s the strength of poorly-heat-treated ordinary steel bolts. Then you need 4 mm² of cross-sectional area to support you.
If your hammock were just a cord, that would be all — you need 6.7 m of 4 mm² cord, 2.3 mm in diameter if it’s circular, and you can sit on it. And it would weigh 26 grams at 0.96 g/mℓ.
However, the hammock net spreads out over an area, and you need to not have it break when you’re sitting on just part of it. It probably needs to be able to spread out to at least 700 mm wide, and any, say, 100 mm of the width needs to not break when you sit on it. That means the net needs to be about 7 times stronger: 14.0 kN, and thus 28 mm², which will form a sort of rope about 6.0 mm in diameter when you twist it up. Also, the spacing of the lines in the net needs to be close enough that any particular 100 mm of its width has about the same strength, so you need at least 4 lines in that width — and that seems challenging to fabricate, because it means you need 28 knots and at times 56 lines across the whole width, so let’s stick with 4. Then you need knots along the length at somewhat regular intervals — let’s say, every 50 mm.
This sort of implies that each line only has a strength of 14/56 = 250 N, or 25 kg. This may be a little low — you might be able to break it by hand if you have some kind of tool to keep the line from cutting your hand, like a stick or something — but maybe it’s okay.
If the net part of the hammock is 2 m long, then it contains 2 m · 28 mm² = 56 mℓ or 54 g of line, which is divided among 56 lines of 0.80 mm diameter and 0.5 mm² cross-sectional area. There are about 28 · 2000 mm / 50 mm = 1120 knots in the net, so tying it by hand is probably a few hours of work. The remaining 4.7 m of support are still only 4 mm² and so are only 19 mℓ = 18 g of line, giving a total of 72 grams.
The circumference of each net hole there is 100 mm, which is hopefully small enough to prevent injury.
This doesn’t include anything to connect the support to, such as a tree-hugging loop or a carabiner or a hook.
If you wanted to include an airproof barrier, 100 microns of boPET under the hammock would probably be adequate, although you could make reasonable arguments for 200μm LDPE or HDPE — they would be quieter and less likely to rip. 2 m · 700 mm · 100 μm, disregarding the narrowing at the ends, works out to 140 mℓ, which is almost twice the volume of the net hammock itself; it’s also about than twice the mass, as PET is almost the same density but slightly denser. Sheets and blankets are heavier still. So we can see that the structural support is a minority of the mass of the overall hammock.
Update: UHMWPE fiber normally has, apparently, 2.4 GPa of tensile strength. This means the 4.7 m of 2kN support line need only be 0.83 mm², 1.3 mm in diameter, and the 250 N net lines need only be 0.104 mm² individually (360 μm in diameter), 5.8 mm² collectively — 2.7 mm in diameter. The overall hammock would occupy 2 m · 5.8 mm² + 4.7 m · 0.83 mm² = 15.5 mℓ, 14.9 g.
If it’s possible to actually make this work, it would be pretty stunning: a hammock you can sleep in that’s roughly one tablespoon in volume when rolled up.
I tried a minimal version of this: six strands of 500μm UHMWPE stretched between two metal wall hooks, a total of about 24 m. I was able to sit on it for a while, but then it broke in two places. The snapped cord showed a bit of curling, suggesting either that the shock of breaking deformed it plastically, or that it had deformed plastically ahead of time. It’s possible I might have mistreated the cords while letting them out, running them around the metal hooks — I might have generated too much heat, melting the cord.
I don’t have my scale handy, but this amount of cord ought to occupy 4.7 mℓ and thus weigh about 5 g. If it were a single cord, it should be 1.2 mm in diameter (500 μm √6). This is close to the 1.3 mm I calculated for the 2kN support line, but the fact that it broke without me even jumping around on it makes me think that I should probably double or triple it. Also, though, it’s possible that my weight distribution might have been unequal across the lines.
Knots are consistently weak points, not just in the usual way where they put extra stress on rope fibers, but also because UHMWPE is so slippery that it tends to slide out of knots. My first attempt to reconstitute my backpack buckles using UHMWPE and steel rings failed when I pulled on it, not because the thread broke, but because the sheet bend came untied.
A bit of analysis suggests that stretching 1 m of 1mm² UHMWPE cord to its 2.4 GPa limit requires 2.4 kN; if this elongates it by 2% = 20 mm, implying a 120 GPa Young’s modulus (a bit over half of steel’s) and close to the 66–124 (Spectra) and 115 (Dyneema) given in http://www.mse.mtu.edu/~drjohn/my4150/props.html. Assuming linearity, that’s 24 J of energy stored in 1 mℓ (≈ 1 g) of fiber. The 5 g I broke should then be able to store about 120 J as spring energy before breaking. Unfortunately, that’s my body moving at a bit under 2 m/s.
(This is similar to calculations I did in Spring energy density for the specific energy of different spring materials.)
This gets worse if, as suggested above, the main body of the hammock has a cross-sectional area much larger than that of the support line, while being made of the same material, because the main body of the hammock then won’t stretch much at all when you sit in it; all the stretch will be taken up by the thinner support lines.
I was thinking that sticking it in series with some kind of spring might help to protect it against shocks, but I don’t know what kind of spring. Nylon, from the same page, has 1.36 g/cc, 2.5 GPa Young’s modulus, and 100 MPa tensile yield strength, but supposedly also has like 90% elongation at break rather than the 4% you’d calculate from those figures. (https://www.azom.com/article.aspx?ArticleID=477 explains: 6,6 nylon does indeed have 4.5% strain at yield, but also 60% elongation at break.) If we take the 60% figure, but figure that the engineering stress is limited to 100 MPa over nearly all of the distance, we get the remarkable result that our 1m × 1mm² fiber can elongate to 1.6 m under a force of 100 N, absorbing 60 J of energy in the process, more than twice what the UHMWPE fiber can absorb. This is an improvement, but not good enough to justify it.
Maybe a hyperelastic material like latex rubber would be a better choice. Latex has roughly 1500% elongation at break. Some loops cut from a bicycle inner tube might be most practical, despite their suboptimal properties. The 2012 “Characterization of Natural Rubber Latex Film Containing Various Enhancers” https://core.ac.uk/download/pdf/82660375.pdf got an ultimate tensile strength of 0.34 MPa and 1400% elongation at break, but unfortunately they did not plot the stress-strain curve. Surely the ultimate tensile strength here is simply incorrect; other sources give tensile strengths in the neighborhood of 10–20 MPa, such as https://vtechworks.lib.vt.edu/bitstream/handle/10919/26306/1JTS_ETD.pdf.
If we assume that it’s linear (probably conservative for hyperelasticity) and that in practice we don’t want to exceed 500% elongation, then we can put it under, say, 5 MPa of stress.
If we were to do the same experiment with our hypothetical 1 m × 1 mm² shape, but this time made from rubber, we would stretch it out to 6 m while averaging 2.5 MPa and 2.5 N. This is a rather pathetic 12 J.
Rhett Allain did some simple experiments https://www.wired.com/story/how-much-energy-can-you-store-in-a-rubber-band/ to derive a specific energy of 1.7 kJ/kg in tension for the rubber bands he had lying around the office. That means that absorbing an impact of 250 J would require 150 g of rubber.
So rubber is good for limiting forces given a fixed acceleration, like when you have something mounted on a vibrating chassis, but it isn’t particularly good at absorbing a fixed impact energy. UHMWPE is many times better than rubber at that, and nylon is twice as good as UHMWPE, albeit dissipatively. (And presumably plastics like LDPE (7 MPa, 400%), PET (100 MPa, 300%), polycarbonate (100 MPa, 200%), and polycaprolactone (10 MPa, 300%, though I think I’ve seen more like 1000%) are better still (see https://www.smithersrapra.com/SmithersRapra/media/Sample-Chapters/Physical-Testing-of-Plastics.pdf and https://www.makeitfrom.com/material-properties/Polycaprolactone-PCL), if you’re willing to allow plastic deformation, making the shock absorber consumable.)