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  1. #21
    krshome's Avatar
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    A pointy cap thanks

    i'm definitely not a pointy cap, all I know is ask the pointy cap. BTY the bridge cut to Grizz's measurements is on point. I used his 85" length, 6.5 sidecut, 55" head and 46" foot for the fabric body. Different from Grizz's is the 40" head spreaders 30" foot Spreader. This setup is perfect for my 6'2" 220# body and 1.9 ripstop works great. finish of with Dynaglide suspension. Haven't had any problems yet this is my 6th bridge and I think this is by far my favorite. Thank you for all you help and hard work Grizz man.

  2. #22
    Senior Member DemostiX's Avatar
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    Archives on this topic show past sensitivity to users / fabricator's stability in the hammock and to tippiness with different curves. That would be for bridge hammocks, not for tarps. With tarps, the user looks at the edge and surface and adjusts the amount and direction of additional tension, conditional on the loads and origins of the ones she has previously placed. Use the wrong regular curves and it may still be hard to not pitch a taut tarp even with three or four tie-outs on each side.

    I have not read the original OZ paper, but from the metaphor for the design, I'd expect that with stretchy fabric which stretched vertically with load, a measure of the appropriateness of the cut for the occupant would be that the bed is horizontal when loaded.

    Is that understanding correct?

    If the Griz is out foraging on wild salmon and berries, I'd trouble him and take issue with the conclusion from the graphs that there isn't much difference in some of the calculated curves. I know the eye / brain fails to measure from pairs of curves the Y distances, substituting the irrelevant proximity on the X axis. That's just a perceptual error most of us make with curves of the same period and similar shape, no matter how familiar we are with graphs. I don't know what curve or model is "right", or how much success depend on that selection, but some of the differences you have calculated and displayed in the past seem larger than cutting and hem-making tolerance.
    Last edited by DemostiX; 09-07-2012 at 13:32. Reason: wrong inference about where flatness is

  3. #23
    DivaB's Avatar
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    After reading that........I give up. Thanks for causing me to forever loose my pointy cap as my brain just POPPED! I was so close too, so close
    Last edited by DivaB; 09-06-2012 at 20:06.

  4. #24
    GrizzlyAdams's Avatar
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    Quote Originally Posted by DemostiX View Post
    If the Griz is out foraging on wild salmon and berries, I'd trouble him and take issue with the conclusion from the graphs that there isn't much difference in some of the calculated curves. I know the eye / brain fails to measure from pairs of curves the Y distances, substituting the irrelevant proximity on the X axis. That's just a perceptual error most of us make with curves of the same period and similar shape, no matter how familiar we are with graphs. I don't know what curve or model is "right", or how much success depend on that selection, but some of the differences you have calculated and displayed in the past seem larger than cutting and hem-making tolerance.
    "Trip, trap, trip, trap! " went the bridge.

    "Who's that tripping over my bridge? roared the troll"
    So. For those inclined, buzz over to this site and pick your handy-dandy catenary curve spreadsheet. Modify cell F7 to be 80 inches, a fairly typical bridge hammock length. Modify cell F8 that deflects the curve to value 135. This gives maximum deflection, at the center of 6", such as you might cut on a bridge hammock. Column C gives the deflection of the curve from the line y=0, as a function of x. Starts at 0 at x=40, drops to -6 at x=0, rises back to 0 all in rows 14 to 44. For comparison, in some un-occupied column, say E, from row 14 to 44 put in the corresponding equation for a parabola over that range with depth 6". For this put alpha = 4*6/(80*80) in cell E13 (that's the coefficient in y = alpha*x^2), and the formula
    =$E$13*B14*B14-($F$9)

    in the cell for the parabola's x=40, noting here that B14 is the x position for that cell, and F9 is the maximum depth. Copy that formula all the way down the column to get a corresponding deflection curve for the parabola.

    The catenary curve and parabola curve values are given in inches and 16ths of inches.

    They are exactly the same.

    Exactly. The. Same.

    "It's I! The big Billy Grizz Gruff ," said the billy grizz, who had an ugly hoarse voice of his own.

    "Now I 'm coming to gobble you up," roared the troll.

    Well, come along! I've got two spears,
    And I'll poke your eyeballs out at your ears;
    I've got besides two curling-stones,
    And I'll crush you to bits, body and bones.

    That was what the big billy grizz said. And then he flew at the troll, and poked his eyes out with his horns, and crushed him to bits, body and bones, and tossed him out into the cascade, and after that he went up to the hillside.
    Last edited by GrizzlyAdams; 09-06-2012 at 23:15.
    Grizz
    (alias ProfessorHammock on youtube)

  5. #25
    Senior Member DemostiX's Avatar
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    One pair of curves I had in mind and was referring to is yours, from 5 years ago, in http://www.hammockforums.net/forum/showpost.php?p=22382 .

    If I am correct in reading the distances after scaling the y-scale, some of the differences approach a half inch if the depth of the curve is 6 inches.

    BTW, the location in the Cloud of this alternative calculator will be of interest to some here,

    http://www.angrysparrow.com/hammock_...eGenerator.xls.

    The proof given in the Shailish Shirali article you cited is that under the realized condition that the suspended mass of the uniform road bed (or body) vastly swamps the negligible mass of the cable (or fabric), the curve is a parabola not a catenary curve. So, the matter is not the difference in curves, but that in practice the curve is never catenary and just should be called what it is, a parabola.


    On method, wrote Tee Dee in that same thread:

    They are parabolic - think suspended-deck suspension bridge with the hammock occupant the road bed. The threads of the fabric act as the vertical cables on a suspended-deck suspension bridge. You can go ahead and use the cat cuts, but I don't think it will support you as well as the parabolic arcs.

    The parabolic arcs are lots easier to make than the catenary arcs. You don't need to make all those measurements and connect the dots.

    Just get a good flexible rod, fiberglass or carbon fiber or AL tent poles work very well. I used fiberglass chimney brush handles, 2 4' sections coupled together to make an 8' rod. You can get them at Lowes for about $4 per section.

    Mark out the end points, fix the rod at the end points and then flex the rod to the desired shape. Use a marker of some sort, a Sharpie works good, to trace the rod on the material and cut and you are done.

    A great parabolic arc.

    Note that the head end of the arc is steeper than the foot end to accommodate the extra weight there. You have to adjust the center point where you pull the rod to get that.

  6. #26
    DivaB's Avatar
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    Blah! There are just times when cheating is easier Lets pretend my friend is 5'11" and 210lbs.....and my friend would like to try a bridge....can someone just draw my friend a darn picture of what the ideal length and width would be, and where the two dots should be to make my friends parabola curve?....say my friend only wants a 6" depth because she is a bit broad

    This is where I got thrown off a bit, "Note that the head end of the arc is steeper than the foot end to accommodate the extra weight there. You have to adjust the center point where you pull the rod to get that".
    Last edited by DivaB; 09-07-2012 at 16:04.

  7. #27
    Senior Member stevebo's Avatar
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    Quote Originally Posted by Ramblinrev View Post
    I have never used a formula. I mark the end points of the arc, take a fiberglass tent pole (they gotta be good for something) and bend it to fit the depth I want at the center and trace the resulting curve. Has worked for me.
    I'm more of a "hands on" kind of guy, than a mathematical person------the tent pole trick works really well! ( You may have to hold it in place while a second person marks the arc) As a friend at work once said "always lazy, never stupid!" (really funny when said with a west Texas drawl!)
    FYI: If you want to know what type a certain bear is, sneak up behind it and kick it. Then,
    run like crazy and climb up a tree. If the bear climbs the tree and eats you, it's a black
    bear. If the bear just pushes the tree over and eats you, it's a grizzly bear : )


    Do not walk behind me, for I may not lead. Do not walk ahead of me, for I may not follow. Do not walk beside me, either, just leave me alone.
    --unknown

  8. #28
    GrizzlyAdams's Avatar
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    DemostiX : apologies, I assumed in response to your earlier posting that you knew what you were talking about. I see now instead an opportunity for the professor to educate!

    Quote Originally Posted by DemostiX View Post
    One pair of curves I had in mind and was referring to is yours, from 5 years ago, in http://www.hammockforums.net/forum/showpost.php?p=22382 .

    If I am correct in reading the distances after scaling the y-scale, some of the differences approach a half inch if the depth of the curve is 6 inches.
    But you see, the curve being compared against a parabola is not at all a catenary curve, and therefore not at all what I was taking about. You won't find the word 'catenary' in that posting anywhere. The curve is one derived from application of conservation laws to load transfer under the assumption of my rather non-uniform weight. Perhaps one might call it a grizznary curve.

    The catenary curve is a plot of points (x,f(x)) where f(x) is a function of one variable with very specific form. A function that is not equivalent to one in that form does not, rigorously, yield a catenary curve. Exactly the same thing is true of functions g(x) said to be parabolic. If g(x) is not equivalent to a function satisfying the form, then rigorously it is not a parabolic curve.

    The most visible difference between a catenary curve and a parabolic one that matches it in certain characteristics is how fast the function changes as you get farther away from the point (0,0). The cat curve rises faster, is steeper. The behavior of the parabolic curve is best understood by considering the slope of a line that is tangent to the curve. At (0,0) that slope is 0; as you move further along the curve, the tangent's slope gets steeper and steeper. You may remember the slope of the tangent as being known as the first derivative, from freshman calculus. What characterizes the parabola is the functional form of this slope function---it is a scalar times x, i.e., it is proportional to x. Which means that the further away from (0,0) you get the steeper the slope of that tangent. Still with me? There's an application here.

    Quote Originally Posted by DemostiX View Post
    On method, wrote Tee Dee in that same thread:

    They are parabolic - ...
    The parabolic arcs are lots easier to make than the catenary arcs. You don't need to make all those measurements and connect the dots.
    ...
    Just get a good flexible rod, fiberglass or carbon fiber or AL tent poles work very well. ...
    Mark out the end points, fix the rod at the end points and then flex the rod to the desired shape. Use a marker of some sort, a Sharpie works good, to trace the rod on the material and cut and you are done.

    A great parabolic arc.
    Observe from TeeDee's pictures, your thought experiments, or someone else's experience how the slope of the rod changes at the endpoints of the curve...for a few inches of the rod it does not. More of the bending takes place between the endpoints and the center. If you allow me to get all quanty about it, the 2nd derivative of the curve near those endpoints is quite a bit smaller---close to zero---than it is in spots nearer the center. Has to do with the stiffness of the pole. If I were to guess I would guess the bend is sharpest midway between center endpoint, halfway between the points where the ends of the pole are constrained. A parabola's 2nd derivative is constant, independent of position.

    All of which is to say, that rigorously, the curve you get from a bent rod is not a parabola. TeeDee's use of the term is qualitatively descriptive, whereas I'm speaking of quantitative description, and have been throughout the discussion.

    Later in that same thread I opine that the curve really does matter, but I was speaking from ignorance then, my hanging experience at the time being limited to my garage where I was not getting suspension angles correct. I have repented of that error, and repudiate the statement.

    TeeDee's early work (and for all I know he still does this) on the suspension curve is based on template for the naval bridge hammock from the Aussie site, that creates a curve that is not symmetric and has sharper slope over the torso. I tried that, I tried the poles, I tried tailoring a curve to my estimated weight distribution. I can't tell the difference, but I have often been accused of being an insensitive lout. So I do parabolas because it's easy to bang out scripts or spreadsheets to mark a curve with desired length and depth, transfer it to a strip of heavy paper, and pull out for use when I need to. But I say more power to the pole people, go to it the hanging a string on the wall people, and marvel at the confidence of the eyeball and winging it people. For the catenary cut people I say their character is stronger, their spreadsheet-foo is mighty, and their curves are indistinguishable in any practical way from a parabola.

    The point of my earlier comparison from a spreadsheet of a quantitatively catenary curve and a quantitatively parabolic curve that match points in the center and at length endpoints with values consistent with bridge hammocks is that whatever difference there is (and there surely is) it is smaller than the granularity of the function value reported, one-sixteenth's of an inch. I don't think I can draw cut and sew anything approaching one-sixteenth of an inch in accuracy; an eighth or a quarter of an inch is more like it.

    hang your own hang, cut your own suspension curve
    Grizz
    (alias ProfessorHammock on youtube)

  9. #29
    Senior Member stevebo's Avatar
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    Grizz, so are you saying that using a pole to cut a curve, and using your chart, you get a different arc, but in the end it really doesn't matter-performs the same? Your bridge-foo is very, very strong!
    Last edited by stevebo; 09-08-2012 at 03:37.
    FYI: If you want to know what type a certain bear is, sneak up behind it and kick it. Then,
    run like crazy and climb up a tree. If the bear climbs the tree and eats you, it's a black
    bear. If the bear just pushes the tree over and eats you, it's a grizzly bear : )


    Do not walk behind me, for I may not lead. Do not walk ahead of me, for I may not follow. Do not walk beside me, either, just leave me alone.
    --unknown

  10. #30
    GrizzlyAdams's Avatar
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    Quote Originally Posted by stevebo View Post
    Grizz, so are you saying that using a pole to cut a curve, and using your chart, you get a different arc, but in the end it really doesn't matter-performs the same? Your bridge-foo is very, very strong!
    I haven't checked to see just how different the curves are, to the eye or measured. But yes, my opinion is that in the end it doesn't much matter. At least it has not made a perceptible (in terms of "feel" or "lay") to me in the various DIY bridges I've made. The depth of the curve does. The precise shape of the curve between deepest point and the ends hasn't. YMMV.
    Last edited by GrizzlyAdams; 09-08-2012 at 10:19.
    Grizz
    (alias ProfessorHammock on youtube)

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