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  1. #61
    Senior Member TeeDee's Avatar
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    Quote Originally Posted by blackbishop351 View Post
    TeeDee, either you guys aren't understanding each other or he's got you on this one. I've thought about it some more and drew some diagrams. The forces due to the ridgeline - pre-tension and elasticity - can be ignored by simply measuring the sag angle. Basically, what I think hitchman is saying is that a non-ridgeline hammock with a certain sag angle and load will put the same forces on its suspension as a ridgeline hammock with the same load at the same sag angle. Think about it for a minute.
    Okay - I have thought about this and drawn some diagrams.

    Lets look at diagram 1:



    This is the classical formula. The force on the suspension is as everybody says. No argument

    Now lets add a structural ridgeline:



    Now in this case the force F2 is, as before, some force divided by the sine of the angle alpha. The difference is that the force being divided is no longer the weight of the occupant. The force of the ridgeline has entered the scene and altered the angle from beta to alpha. The force f-one in the diagram is now given by the usual formula - no argument. The force f-2 is given by the same formula using the angler beta. But the force used to compute the force F-2 is no longer the weight of the occupant.

    The wife is ringing the dinner bell and getting anxious - time to go.

    So hitchcraft and BB are right the force on the suspension rope is given by:

    D/(2* sin(alpha))

    Then I guess where we disagree is that hitchcraft seems to have wanted to say that D is equal to the weight of the occupant and I disagree. Don't have time right now to derive what D is though. The wife is calling for dinner.

  2. #62
    Senior Member TeeDee's Avatar
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    Okay - I've been able to think about this over dinner time. What else do you think about??

    hitchcraft and BB351 are right. The force D in my previous post is indeed the weight of the occupant . The effect of the ridge line is to amplify the force of the weight of the occupant on the suspension ropes.

    But the discussion has proven fruitful for me in another regard. I have been slightly curious about the force on the ridge line, but too lazy and busy to do the analysis. It suddenly occurred to me that the computation of that force is now obvious.

    Using the diagram from my previous post:



    I can now see that the force on the ridge line is simply:

    P = (W/2)(cot(alpha) - cot(beta))

    For those mathematically inclined this now gives you a method of estimating the force on your ridge line. If you desire to do so.

    Simply lay in the hammock and estimate the two angles alpha and beta or have someone do it for you. Plug your weight and angle values into the formula and you have the force acting on the ridge line.

    For my HH Safari, I have estimated alpha at approximately 10 deg. and beta at approximately 60 deg. The ridge line amplifies the force considerably. With my weight at 170 lbs, the force on the ridge line computes out to approximately 432 lbs, give or take the error in my estimation of the angles.

    Since I have used 3/32" Yale Crystalyne line for the short ridge line, rated at 1,000 lbs, I am probably over the safety factor that hitchman says to use, but less than 1/2 the rated capacity of the ridge line.

    Hennessy has rated the Safari at 350 lbs. Assuming the same angles are obtained, the force on the ridge line comes out at 891 lbs. That is quite a bit of force on that ridge line.

    That is for my short ridge line. I will have to swap the ridge line and estimate the angles and compute the force for the long ridge line. Since alpha is about the same, but beta is much smaller, the force on the long ridge line will be smaller. From memory, I would estimate the angle at between 45 deg and 50 deg, which would bring the force down to between 397 lbs and 410 lbs. That is way over half the rating of the BPL guy line cord I used for the long ridge line. Now I feel more comfortable that I decided not to use the Mountain Laural Designs guy line cord rated at 200 lbs for the long ridge line.
    Last edited by TeeDee; 06-04-2007 at 19:01.

  3. #63
    Senior Member blackbishop351's Avatar
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    I'm glad we're all on the same page now...I hate arguing, and I hate doubting my own analyses even more
    "Physics is the only true science. All else is stamp collecting." - J. J. Thompson

  4. #64
    Senior Member TeeDee's Avatar
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    Quote Originally Posted by blackbishop351 View Post
    I'm glad we're all on the same page now...I hate arguing, and I hate doubting my own analyses even more
    Ahem - "discussing" please

  5. #65
    Senior Member GrizzlyAdams's Avatar
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    all in an afternoon

    Y'all covered a lot of ground in an afternoon.

    Understanding that my last physics course was, ahem, 30 years ago, I'm not seeing why force applied to the lines before someone gets into the hammock is not part of the picture here.

    Speaking of pictures, consider



    A line is suspended between vertical supports, and a total tension P is applied before tying things off. P manefests itself as two opposing horizontal force vectors with equal magnitudes whose sum is P.

    A downward directed force with magnitude W is applied at the center of the line.


    Because the point mass with weight W isn't moving, there is a force with equivalent magnitude and opposite direction to W, caused by the tension placed on the lines to begin with.

    According to received wisdom, the magnitude of the tension on the diagonal is found by considering W as the vertical projection of the diagonal's magnitude. But by symmetry, P/2 is likewise the magnitude of the force on the line due to the original tensioning. Thus it seems as though the stress F1 on the diagonal must be (W/2)/sin(alpha) + (P/2)/cos(alpha).

    Given W, P, and the distance between posts, ignoring the effects of the rope material, one ought to be able to compute from a force diagram the distance the weight descends, and hence alpha. I don't remember exactly how that goes (given that forces on the diagaonal have opposing directions) and not enough time to work it out from first principles. Supposing I could remember the first principles.

    I had some more scribblings on the ridge line, but that's pointless if I'm wrong in the observations above.

    Grizz the befuddled

  6. #66
    Senior Member blackbishop351's Avatar
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    You're going about it the right way, but I think missing the detail that TeeDee missed earlier - the sag angle alpha will vary depending on the amount of initial tension (for a fixed load, of course). Thus the whole system can be analyzed by considering the hammock/user/ridgeline as a rigid block, because we account for the initial tension by accounting for the sag angle.
    "Physics is the only true science. All else is stamp collecting." - J. J. Thompson

  7. #67
    Senior Member GrizzlyAdams's Avatar
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    solid blocks

    Quote Originally Posted by blackbishop351 View Post
    You're going about it the right way, but I think missing the detail that TeeDee missed earlier - the sag angle alpha will vary depending on the amount of initial tension (for a fixed load, of course).

    No, I don't think I'm missing that at all. It comes into play when you think about the force vector that opposes the gravitational one. We know its magnitude---W. We know that that vector must be a projection of the tension vector, for tension is the only opposing force here. The projection will be something like T sin(alpha). Held constant at W, when T grows then alpha must shrink.

    I'm not facile enough with the force vectors to pin the tension vector down exactly. But I'm convinced with that detail you can compute alpha from the rest of it.

    Thus the whole system can be analyzed by considering the hammock/user/ridgeline as a rigid block, because we account for the initial tension by accounting for the sag angle.
    I'm not arguing that you can't treat hammock/user/ridgeline as a rigid block. It is clear to me that you can. There is nothing fundamentally different about the analysis on the suspension lines if we take the point mass W, turn it into a rigid bar with mass W and length r, and hang the suspension cords from the endpoints (under the initial tension of course, an admittedly gnarly detail).

    My point is that I don't see that you can ignore the inital tension if the object is to compute the overall tension on the suspension lines. Treating the hammock/user/ridgeline as a rigid block doesn't change that. The tension was applied, and exists outside of that block.

    Grizz the disputant
    Last edited by GrizzlyAdams; 06-04-2007 at 22:11. Reason: clarification

  8. #68
    Senior Member blackbishop351's Avatar
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    I really hate not being able to explain things clearly...

    If you change the initial tension on the ridgeline, the sag angle alpha changes to compensate. Thus the total tension in the suspension remains constant to some reasonable approximation.
    "Physics is the only true science. All else is stamp collecting." - J. J. Thompson

  9. #69
    Senior Member GrizzlyAdams's Avatar
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    Quote Originally Posted by blackbishop351 View Post
    I really hate not being able to explain things clearly...

    If you change the initial tension on the ridgeline, the sag angle alpha changes to compensate. Thus the total tension in the suspension remains constant to some reasonable approximation.
    Might be true. Some of y'all believe it's true. Probably is true.

    What you and the others are saying is that if you measure alpha, and measure W, then alpha is smaller than it would really be if the only force involved here is W. Agreed. Then you say that everything we need to know about tension forces are coded by alpha's value, so much so that we don't need really to figure them at all. That's the "then a miracle occurs" step whose validity I'm not sure of. At one level I can imagine it being true; the vertical force (happens to be gravity) is a geometric projection of the force on the suspension rope (happens to be an aggregate force), knowing the projection and the angle, you can compute the aggregate force.

    I'm a first principles kind of guy, and if it's true then it should fall out of the math. I tinkered around with forces and such, I saw that exactly the same argument for computing what the stress on the suspension rope is due to weight ( (W/2)/sin(alpha)) has a symmetric argument when you think not about the vertical gravitational force, but about the horizontal tension force. Seemed there'd be a way to compute the actual force on the suspension rope, by decomposing it into its constituent parts.

    So I figure that if I can dot all the i's and cross all the t's and nail the relationships between the tension force, the weight, and the angle, then I can understand why this trick works, if indeed it does.

    Grizz the pedantic

  10. #70
    Senior Member blackbishop351's Avatar
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    I'd love to continue the discussion, but maybe we shouldn't clutter the open Forums with it...PM me if you want?
    "Physics is the only true science. All else is stamp collecting." - J. J. Thompson

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