What equation describes the curve of the perfect hammock?

February 2, 2006 2:05 PM

You have a hammock of length l and width w, and a person of height h. At what height should the hammock be hung? What is the optimum distance between the hanging points?
posted by obiwanwasabi to grab bag (13 comments total)

Lo, I am not a mathy type, but I come hither to deliver a profound statement nonetheless: After a year or so, the hammock will stretch like a bitch. Hold it up as high as you can imagine hanging it, then move it up six inches to a foot.
posted by booksandlibretti at 2:09 PM on February 2, 2006

Just attach lengths of chain to each end of the hammock, and then attach the chain to a hook on each of your hanging points. This allows you to raise and lower the hammock by hanging different links of the chain on the stationary hooks.
posted by odinsdream at 2:13 PM on February 2, 2006

Hang it as low as you need to in order to be able to get into it, and no lower. Giving you a hard number would depend on some constants that you didn't (and probably can't reasonably) provide.
posted by cortex at 2:14 PM on February 2, 2006

You could assume the hammock is a one-dimensional curve for simplicity's sake, of length l, since the width is equal along the length and doesn't play too much into the behavior of the hammock as it hangs from opposite points on the tree.

Drawing a catenary curve for your hammock's dimensions will give you a good start for calculating an optimum distance between two trees given

Taking the derivative along this curve will give you length of that catenary curve. At a minimum, the person's height should equal this curve's length.

Given you've calculated the length of the hammock you'd need to hold the person, you can use the catenary curve function to figure out the distance between the two trees.

There's no easy answer to a precise formula for a hammock-off-the-ground calculation, since you'll need to somehow take into account:

• the elasticity of the hammock material
• the weight of the person

But at least you'll know how long a hammock you'll need!
posted by Rothko at 2:22 PM on February 2, 2006

You can't do this mathematically in any reasonable sense -- it depend on how bendy the person is (a frozen corpse will fill a hammock differently from a live person who bends in the middle), the stretchiness of the hammock material, etc. As mentioned above, chains are your friends.

Taking the derivative along this curve will give you length of that catenary curve.

This, like most of the rest of this answer, is nonsensical. Please ignore.
posted by gleuschk at 2:41 PM on February 2, 2006

If you want to get the length of the curve, you'll need to be able to be able to differentiate the function representing the curve.
posted by Rothko at 3:29 PM on February 2, 2006

Half of h at the centre should work. When no one was in it, the centre (mid span?) of my hammock was about waist high to me (5' 4"). At that height it was pulled down easily enough to get my butt (and the rest of me) into it. Work from there and you should be able to figure out how high the ends should be fastened (mine was fastened at about 5 feet high, it depends on how long your hammock is).
posted by deborah at 3:46 PM on February 2, 2006

What is the optimum distance between the hanging points?

You need to give us a metric for "optimum". Do you mean...

...firmness? Set the distance between hanging points close to l.
...convenience? Set the distance between hanging points equal to the distance between the trees in your backyard.
...comfort? You'll have to determine this experimentally.

Taking the derivative along this curve will give you length of that catenary curve.

Like gleuschk said, this is incorrect.
posted by event at 4:26 PM on February 2, 2006

/shrugs
posted by Rothko at 5:02 PM on February 2, 2006

You're such a drama queen, Rothko :)

Like he seems to do often in threads I comment in, odinsdream has the correct answer:

Just attach lengths of chain to each end of the hammock, and then attach the chain to a hook on each of your hanging points. This allows you to raise and lower the hammock by hanging different links of the chain on the stationary hooks.

My family does this too with our (outdoor) hammock, and it's quite useful. And yes, no matter what, they do stretch like a bitch after a while, so I'd suggest placing the hooks *way* up and using longer chains if you have to.
posted by cyrusdogstar at 5:28 PM on February 2, 2006

Oh, also, don't hang one under a Chinese chestnut tree if you can help it. Nice tree to look at, but having those spiny mofos drop on your face on a lazy afternoon tends to ruin your nap.
posted by cyrusdogstar at 5:29 PM on February 2, 2006

Taking the derivative along this curve will give you length of that catenary curve. At a minimum, the person's height should equal this curve's length.

Unless I'm misunderstanding you, this won't work.

You need to determine the arc length of the function f(x), which is the integral from point a to point b of sqrt[1 + f'(x)²], and for us f'(x) is the formula of a catenary given in Rothko's link.

Your arc length is a given, l. But I don't think the math will help you, since l is fixed, and that once you get on the hammock, it no longer takes the shape of a catenary (which only exists if the weight of the hanging body is the only force involved).

You can vary the distance of the anchor points, which changes the shape of the catenary, but I'm aware of no nifty way to relate the way that shape will change to your comfort in the hammock (and I'd be pretty surprised if such theory existed ;-p).

In short: it's time for some empiricism!
posted by teece at 5:35 PM on February 2, 2006

Rothko: your answer is nothing but posturing, and serves only to further reinforce the idea that calculus is black voodoo. To "take the derivative along a curve" has no meaning, and even if it did, wouldn't address the question. Reread it: the length of the hammock is known. No matter how close together you put its ends, the length of the catenary curve connecting them is l.
posted by gleuschk at 5:56 PM on February 2, 2006