1. Originally Posted by Schneiderlein
The following method has worked well for me (warning: some math talk follows). For a baffled quilt, you take the loft L as the height of your baffle (and then overstuff), so you have a down area per unit length of L * W, where W is the width of each down chamber. I figured with a differentially cut quilt, you want to have the same area per unit length for insulation. So, I computed the parabola with zero crossings at -W/2 and W/2 that would give me an area under the curve of L * W. Then I took the arc length of that parabola as the chamber width for the outside. Since catenary curves are so popular around here, you could go through the same exercise with a catenary curve. It would be interesting to see how much of a difference it makes.

This approximation has worked well for me, but I split the down volume between baffle and differential cut. On a sewn through quilt, the approximation may not be as good (or it might be better), but it's a starting point. YMMV.
Among all two-dimensional closed curves, the circle maximizes the ratio of area enclosed to length of the enclosing curve. In nature, when shapes take form under uniform force they tend towards such maximal configurations. If you blow up a long ballon, the cross-section tends to a circle for this reason.

So on the sew-through thing I've been thinking circular arcs. Given a base
(diameter) b the cross-sectional area is b*pi . Depending on b (the width of a baffle chamber) b*pi may be too small for the area you need for the amount of down you'll stuff into the chamber. For this then I approximate the chamber cross-section as a rectangle with base b and height h, such that the cross-sectional area b*h + b*pi is equal to what I've computed I need for the baffle chamber. That's for small b of course (narrow chambers). For larger ones I figure the base b is a chord on a circle and work from there. Like you, when I have the shape of the curve of the outside fabric, I compute the arc length.

Well that's theory anyway. I'm heading out to the lab right now....

Grizz

Among all two-dimensional closed curves, the circle maximizes the ratio of area enclosed to length of the enclosing curve. In nature, when shapes take form under uniform force they tend towards such maximal configurations. If you blow up a long ballon, the cross-section tends to a circle for this reason.

So on the sew-through thing I've been thinking circular arcs. Given a base
(diameter) b the cross-sectional area is b*pi . Depending on b (the width of a baffle chamber) b*pi may be too small for the area you need for the amount of down you'll stuff into the chamber. For this then I approximate the chamber cross-section as a rectangle with base b and height h, such that the cross-sectional area b*h + b*pi is equal to what I've computed I need for the baffle chamber. That's for small b of course (narrow chambers). For larger ones I figure the base b is a chord on a circle and work from there. Like you, when I have the shape of the curve of the outside fabric, I compute the arc length.

Well that's theory anyway. I'm heading out to the lab right now....

Grizz
Well that just clear everything up for me.

How about you do all the math and I will copy you in the end. I think that will work better. You be the brain and I'll supply the good looks.

Among all two-dimensional closed curves, the circle maximizes the ratio of area enclosed to length of the enclosing curve. In nature, when shapes take form under uniform force they tend towards such maximal configurations. If you blow up a long ballon, the cross-section tends to a circle for this reason.
well this is true. It does not necessarily follow though that the shape of the
minimal length curve over a straight line that achieves a given area is the arc of a circle though....

There is a small difference between what I get for the circular arc and a parabolic arc, with the parabolic one being smaller!

It has the attraction of being an easier calculation to do. Could be my program has a bug in it. I'm not seeing it though.

Working on a mock-up now.

Originally Posted by Dutch
Well that just clear everything up for me.

How about you do all the math and I will copy you in the end. I think that will work better. You be the brain and I'll supply the good looks.
Well I'll do my part. Bound to be easier than me doing your part

Grizz

4. I am not understanding "differentially cut". can you explain that a little.
Uhh... What Grizz said (before they got to the math!)

With the differential cut you will not compress your UQ as long as your shockcords are not fully extended. I think the differential cut is more important for the 1/2 quilts to maximize the entire area for insulation.

Among all two-dimensional closed curves, the circle maximizes the ratio of area enclosed to length of the enclosing curve. In nature, when shapes take form under uniform force they tend towards such maximal configurations. If you blow up a long ballon, the cross-section tends to a circle for this reason.
Well, that's true, but maybe not 100% applicable. I don't think the force is uniform, since there is less down near the edge of the chamber pushing outwards. And then, there is gravity, which will have a different effect on each chamber if you run your chambers length wise (which I believe you intend to do). Plus, the liner is supposed to be snug against the hammock, which means the chamber itself will take on a crescent shape. On a bridge hammock, the inner radius will probably be much larger than on an end-gathered hammock.
So on the sew-through thing I've been thinking circular arcs. Given a base
(diameter) b the cross-sectional area is b*pi . Depending on b (the width of a baffle chamber) b*pi may be too small for the area you need for the amount of down you'll stuff into the chamber. For this then I approximate the chamber cross-section as a rectangle with base b and height h, such that the cross-sectional area b*h + b*pi is equal to what I've computed I need for the baffle chamber. That's for small b of course (narrow chambers). For larger ones I figure the base b is a chord on a circle and work from there. Like you, when I have the shape of the curve of the outside fabric, I compute the arc length.
Grizz
Are you assuming the entire chamber takes on a circular cross-section, or just the bottom half? I think the top of the chamber (the side against the hammock) is already fixed by the hammock shape. You could assume a circular arc with a relatively large diameter to approximate the shape of a hammock. I chose an infinite diameter and used a straight line. For the bottom of the chamber, you could assume a half-circle, but that already fixes the cross-sectional area. Is that why you are using an additional height h?

It has the attraction of being an easier calculation to do. Could be my program has a bug in it. I'm not seeing it though.
I think you may have a problem with the area calculation. Your equation should be A = b * h + 1/8 * b^2 * pi, assuming the chamber is made up of a rectangle b * h and a half circle. You had the formula for the chord length of the entire circle.

6. Originally Posted by Schneiderlein
Well, that's true, but maybe not 100% applicable. I don't think the force is uniform, since there is less down near the edge of the chamber pushing outwards. And then, there is gravity, which will have a different effect on each chamber if you run your chambers length wise (which I believe you intend to do). Plus, the liner is supposed to be snug against the hammock, which means the chamber itself will take on a crescent shape. On a bridge hammock, the inner radius will probably be much larger than on an end-gathered hammock.
you weren't around when I was moving virtual trees around to do a force calculation on a ridgeline, were you? simplifying assumptions...

Are you assuming the entire chamber takes on a circular cross-section, or just the bottom half? I think the top of the chamber (the side against the hammock) is already fixed by the hammock shape. You could assume a circular arc with a relatively large diameter to approximate the shape of a hammock. I chose an infinite diameter and used a straight line. For the bottom of the chamber, you could assume a half-circle, but that already fixes the cross-sectional area. Is that why you are using an additional height h?
[/QUOTE]
just the bottom half.

Anyway, the h business is needed only in a region of parameter values where there is a lot of down going into a narrow chamber, and in that case I shouldn't be thinking of a rectangle...I wouldn't be surprised if the equation I tossed down had a bug.

For the sizes I'm working with right now, the chamber approximation is the area between a circle's arc and a chord of length b. I did find a bug in my calculations, the circular curve does indeed have a shorter length, so there is order in my universe again. That said, the difference is too small to bother with relative to the parabola.

Grizz

you weren't around when I was moving virtual trees around to do a force calculation on a ridgeline, were you? simplifying assumptions...
Did you actually use the method of virtual work to do force computation? It is one of my favorite things I remember from my undergrad education. Now I've got to dig through the archives and look it up.

I'm all for simplifying assumptions, that's why I went with the parabola and called it good.
For the sizes I'm working with right now, the chamber approximation is the area between a circle's arc and a chord of length b. I did find a bug in my calculations, the circular curve does indeed have a shorter length, so there is order in my universe again.
Glad to hear the world makes sense again.
That said, the difference is too small to bother with relative to the parabola.
Grizz
Before I started writing a program to figure everything out, I thought to just make each bottom panel wider by about 2". In the end, the computer said 2.00474". Could have saved the whole exercise, but it was fun.

8. Originally Posted by Schneiderlein
Did you actually use the method of virtual work to do force computation? It is one of my favorite things I remember from my undergrad education. Now I've got to dig through the archives and look it up.
no, wasn't that. As I recall, I was approximating how initial tension put on the ridgeline from tightening up on the suspension (before someone gets in) would be represented. I was assuming Hooke's Law, and (I think) represented that tension by picking up the tree at its position before the tension, and moving an appropriate distance (determined by the force applied, and the spring constant Hookes Law assumes).

I can't find that post anymore. I even remember within a month when that discussion took place, but if I go to get a listing of all my postings, the system yields up only the most recent 500.

wasn't that interesting anyway...

Grizz

(later edit : in the bright light of morning I remembered that Blackbishop had been part of the conversation. He stopped posting a while ago, and so I thought that I might catch a trace of his end of the conversation in the last 500 of his posts. Did. Here's the page with the relevant dialog. Mostly it shows me being thick-headed. )

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