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  1. #1
    Senior Member DemostiX's Avatar
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    Dynaglide weight

    Would someone here with an accurate scale weigh a good length of this, so there is no question about rounding error influencing the measurement in a major way?

    Say 30 feet (or more) on a weight-weenies gram scale accurate to a gram. (Accurate to a gram, not reading out in gram units.)

    No inference, please, and no reference to manufacturer's / seller's spec sheet, (where .1 lb per hundred feet could be as much as .149 lb rounded down. That's 49% heavier.)

    I think this is important, because Dynaglide has a break strength which brings many here to use it, and more to be attracted by some claims such as the one from Samson on its weight per length.

    I'm not asking its diameter. With hollow braid rope, that is truly a nominal dimension. Just an accurate (and hopefully unbiased) measurement of the weight per unit length.

    ------
    If personal anecdotes matter. I have some 8-strand Spectra sold to me as 1/8", but which is no heavier, by weight / length than Amsteel Blue 7/64". The Spectra is uncoated, (NOS) new old stock , and maybe as strong per weight as Amsteel. But, just as the difference between 7/64" and 1/8" (nominals) is just 1/64", the breaking strength difference between them is 800 lb, consistent with the larger line packing 50% more yarn into a length than the smaller line. And I think this Spectra has a much lower strength than the seller still insists on believing.
    Last edited by DemostiX; 07-27-2011 at 19:30. Reason: correction

  2. #2
    Senior Member rip waverly's Avatar
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    someone here put this together...not sure if it helps

    http://www.hammockforums.net/forum/a...9&d=1311746299
    "Jeff-Becking"

    DOWNTOWN BROWN!!!!

  3. #3
    Senior Member DemostiX's Avatar
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    Rip, that's just what I was referring to and responding to. We both saw it.
    Unless I find out otherwise from good measurement, I suspect the 1.6 oz per 100 ft is a calculated, inferred value, from Samson's spec sheet stating it weighs .1 lb per 100 ft.

    That's not a .10 lb /100 ft claim; that's a .1 lb / 100 ft. claim. So the only thing I'm certain of about the 1.6 oz (=.10 x 16 oz) claim is that it is not a bad inference from .05lb per 100ft rounded up, so that Dynaglide really weighs just .8 oz ( ie 23 grams) per 100 ft. That would be so ridiculous nobody would believe it. (Or would he?)

    Anyone with a good scale and a quantity of Dynaglide to weigh accurately? Maybe one of the vendors here?
    Last edited by DemostiX; 07-27-2011 at 19:55. Reason: correction

  4. #4
    Senior Member GrizzlyAdams's Avatar
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    Quote Originally Posted by DemostiX View Post
    Rip, that's just what I was referring to and responding to. We both saw it.
    Unless I find out otherwise from good measurement, I suspect the 1.6 oz per 100 ft is a calculated, inferred value, from Samson's spec sheet stating it weighs .1 lb per 100 ft.

    That's not a .10 lb /100 ft claim; that's a .1 lb / 100 ft. claim. So the only thing I'm certain of about the 1.6 oz (=.10 x 16 oz) claim is that it is not a bad inference from .05lb per 100ft rounded up, and that Dynaglide really weighs just .8 oz (14 grams) per 100 ft.

    Anyone with a good scale and a quantity of Dynaglide to weigh accurately?
    I will be able to do this, later this p.m.
    Grizz
    (alias ProfessorHammock on youtube)

  5. #5
    Senior Member GrizzlyAdams's Avatar
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    well now, this was interesting.

    First thing, test the accuracy of the scale. It is reported to be accurate to 0.5 grams. I put 3 US quarters on it, it reports 17g. Standard weight of a quarter is 5.67 which is surely an approximation for 5 2/3, times 3 equals 17, check.

    I measured off a length of dynaglide I have, 4 feet at a time on a 48" long measuring stick, counted off almost exactly 28 such lengths, call it 112 ft.

    Put it on my scale...wavers between 82.5 and 83 grams
    dynaglide.jpg

    So do the math and you get 74.1 grams / 100 ft, which is 2.61 oz / 100 ft.
    Known on the street as 0.163125 lbs / 100 ft.

    So the reported 0.1 lbs / 100 ft is an approximation with a large relative error.

    Good intuition there DemostiX!
    Grizz
    (alias ProfessorHammock on youtube)

  6. #6
    Senior Member DemostiX's Avatar
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    Quote Originally Posted by GrizzlyAdams View Post
    well now, this was interesting.

    <snip>

    So do the math and you get 74.1 grams / 100 ft, which is 2.61 oz / 100 ft.
    Known on the street as 0.163125 lbs / 100 ft.

    So the reported 0.1 lbs / 100 ft is an approximation with a large relative error.

    Good intuition there DemostiX!
    Grizz:
    Thanks for the credit, but there are too many years in and of schooling, and many more of practice as a demographer and statistician to credit intuition.

    And, anyway I was partly wrong, (but happy to have everyone here with Dynaglide honestly and capably address the question.) It is a disappointment to learn that rounding error alone (ie. .149999 lb rounded to the nearest .1 lb) does not explain the real weight. Or to distinguish probability from likelihood, the real weight does not explain or account for Samson's product listing. Others' credulity for this long time is something I'll take up later.

    For the record, you didn't have to employ video tech to shoot a photo of a digital scale reading for me to trust that you'd made and correctly reported a best estimate.

    Part of what I'm thinking about is how far and in what direction "if there aren't pics, it didn't happen" is taking us. Not liking it, though, I'd urge that no numbers be shown that are not significant, and by that I mean "significant" in an everyday sense of the word. We read from left to right, and what we read most recently, the digits to the right of the decimal, the least significant part, are our most recent memories. Most of that is clutter.

    Especially, thank you for staying in character, doing what Professor Hammock (and Grizz) do.

  7. #7
    Senior Member DemostiX's Avatar
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    Quote Originally Posted by GrizzlyAdams View Post
    well now, this was interesting.

    First thing, test the accuracy of the scale. It is reported to be accurate to 0.5 grams. I put 3 US quarters on it, it reports 17g. Standard weight of a quarter is 5.67 which is surely an approximation for 5 2/3, times 3 equals 17, check.

    <snip>
    I wonder where 3 for 17 (grams) came from?!!

    A lot of us get lost in digits. Me too.

    For calibration and mnemonic reasons combined, maybe the best calibration unit for central North Americans is a stack of US 5 cent coins, the nickel: For a long time the standard weight has been 5.00 grams.

    Nickel = 5.00 grams.
    Last edited by DemostiX; 07-28-2011 at 16:43. Reason: grammar

  8. #8
    Senior Member more's Avatar
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    Quote Originally Posted by DemostiX View Post
    I wonder where 3 for 17 (grams) came from?!!
    You might be right and we all need to start using nickels, but I think that what he was saying is that the three quarters he measured have a total mass of 17 grams.

  9. #9
    Senior Member GrizzlyAdams's Avatar
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    Quote Originally Posted by more View Post
    You might be right and we all need to start using nickels, but I think that what he was saying is that the three quarters he measured have a total mass of 17 grams.
    I think that was understood.

    What happened was that I weighed the cord and saw that the reading was a lot heavier than I expected. Now I use this scale a lot for shipping, and my measurements are always confirmed to the resolution of an ounce by the post office, but I wanted to check the scale somehow.

    Looked around the workshop, didn't see anything with a known weight, but then thought of coins. I didn't know what their weight ought to be, but I figured I could find out easily enough. I had three quarters in my pocket ....
    Grizz
    (alias ProfessorHammock on youtube)

  10. #10
    Senior Member DemostiX's Avatar
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    Quote Originally Posted by DemostiX View Post
    I wonder where 3 for 17 (grams) came from?!!

    For calibration and mnemonic reasons combined, maybe the best calibration unit for central North Americans is a stack of US 5 cent coins, the nickel: For a long time the standard weight has been 5.00 grams.

    Nickel = 5.00 grams.
    I'll answer my own question, and so offer another useful memory device for central North Americans, the one around nickels being close to foreign:

    Five modern quarters, such as the ones Grizz used to check the bias of his scale, weigh so close to 1.00 ounces as to be good enough for all weight-weenies. Folks here with iPad-equivalents in their hammocks at home might confirm that this is why the new weight standard for U$ quarters was set as it was.

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