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  1. #71
    Senior Member angrysparrow's Avatar
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    Quote Originally Posted by blackbishop351 View Post
    I'd love to continue the discussion, but maybe we shouldn't clutter the open Forums with it...PM me if you want?
    Please don't do that. This is clearly a hammock-related topic, and I quite enjoy following along.
    I think that when the lies are all told and forgot the truth will be there yet. It dont move about from place to place and it dont change from time to time. You cant corrupt it any more than you can salt salt. - Cormac McCarthy

  2. #72
    Senior Member NCPatrick's Avatar
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    Agreed. Even though I have no idea what you guys are talking about, please proceed.


    "Civilization is the limitless multiplication of unnecessary necessities."
    - Mark Twain
    I go to nature to be soothed and healed, and to have my senses put in order.
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  3. #73
    Senior Member blackbishop351's Avatar
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    OK OK OK...

    Let x be the length of the suspension from the ridgeline attachment point to the tree. Let T be the initial tension applied to the system (along the ridgeline/suspension).

    Since the materials are elastic (ALL materials are, to some approximation), we can model their static properties using the harmonic oscillator equation F = -kx . This is kind of back-of-the-napkin (as I'm sure an engineer would tell you) but should suffice for our macro-scale purposes here. Here, F is the restoring force due to displacement, so whatever tension T applied to the system must oppose this in order to stretch the materials. Also, the spring constant k is an inherent property of the material, which would have to be determined experimentally to achieve a thorough analysis. Fortunately, we don't need to do that for a broad-strokes analysis like this.

    In order for the system to be static after the initial tension T is applied, we must then have T = kx for some stretch amount x . This then gives x = T/k for a given tension.

    Upon adding a load to the hammock, we know that the tension in the materials will change by an amount delT = W/sin(alpha), where W is the user's weight and alpha is the sag angle yet to be determined. Combining this with the previous result, we have T + delT = k(x + delx), where delx is the resulting change in length of the suspension.

    Because of the geometry involved - the stretching changes the length, which allows a right triangle with apex angle alpha - we also have cos(alpha) = x/(x + delx), which can be substituted to give Tsin(alpha) = Ttan(alpha) - W. This is transcendental in alpha, but still gives a functionality between sag angle alpha, user weight W, and initial tension T.

    What this boils down to is something that's common sense - if you add a weight to a tensioned line, the line will sag. Increase the weight (holding the tension constant) and the line sags more. Increase the tension (with the weight constant) and the line sags less. We can see now that the sag is actually due to a change in length of the (necessarily) elastic material.

    I hope that helps.
    "Physics is the only true science. All else is stamp collecting." - J. J. Thompson

  4. #74
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    It looks like Jeff and Troll need to add greek letters to this site. The lack of them is really hurting the look of your equations. Us engineers don't have those issues you physics book readers do.
    Is that too much to ask? Girls with frikkin' lasers on their heads?
    The hanger formly known as "hammock engineer".

  5. #75
    Senior Member blackbishop351's Avatar
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    I know, right? We need TeX compatibility...
    "Physics is the only true science. All else is stamp collecting." - J. J. Thompson

  6. #76
    GrizzlyAdams's Avatar
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    then a miracle occurs

    Quote Originally Posted by blackbishop351 View Post
    OK OK OK...

    ...
    In order for the system to be static after the initial tension T is applied, we must then have T = kx for some stretch amount x . This then gives x = T/k for a given tension.
    I think for consistency with what you say later, really this is

    T = k*delx

    Upon adding a load to the hammock, we know that the tension in the materials will change by an amount delT = W/sin(alpha), where W is the user's weight and alpha is the sag angle yet to be determined. Combining this with the previous result, we have T + delT = k(x + delx), where delx is the resulting change in length of the suspension.

    Because of the geometry involved - the stretching changes the length, which allows a right triangle with apex angle alpha - we also have cos(alpha) = x/(x + delx), which can be substituted to give Tsin(alpha) = Ttan(alpha) - W. This is transcendental in alpha, but still gives a functionality between sag angle alpha, user weight W, and initial tension T.
    Let's go slow here. x is the length before any kind of tension. Putting tension on the
    line stretches it by a length d, which gets wrapped around the tree or pulled past the
    grip point of my rings or your buckles. I'm going to approximate this by assuming instead that
    I dig up the tree and plant it a distance d further away, making a length x+d to reflect the
    original tension T. Since T = k*d, we can use d = T/k.
    Now I hang the hammock and see an angle alpha. The length of the line
    from tree to hammock is

    ( x+(T/k) )/cos(alpha)

    This means that the line has stretched ( x+(T/k) )/cos(alpha) - x )
    so that the restoring force (complete tension) vector is along the
    line, pointed at the tree, with magnitude

    k*((x+(T/k))/cos(alpha) - x )

    We know that the horizontal projection of this force exactly counter-balances the
    downward force W due to hanging the hammock/user. That gives us

    W = sin(alpha) * k * ((x+(T/k))/cos(alpha) - x )

    So in principle we can express alpha in terms of W,T, k. But, interestingly,
    also the original length x is showing up. We've not seen that before in our
    back-of-the-napkin analyses.

    But then a miracle does indeed occur . We re-arrange the equation above to

    W/sin(alpha) = k * ((x+(T/k))/cos(alpha) - x )

    The left-hand-side is the equation you guys like. The right-hand-side
    is the magnitude of the tension force on the suspension rope, which is what I was wanting to compute.

    I am contented now, thanks for the patient explanations.

    Grizz

  7. #77
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    Quote Originally Posted by blackbishop351 View Post
    I know, right? We need TeX compatibility...
    Your just jealous I write my whole thesis without a single equation in it.
    Is that too much to ask? Girls with frikkin' lasers on their heads?
    The hanger formly known as "hammock engineer".

  8. #78
    Senior Member sk8rs_dad's Avatar
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    Any idea where can I order some massless rope and maybe a frictionless pulley too?

  9. #79
    Senior Member NCPatrick's Avatar
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    I'm going to approximate this by assuming instead that
    I dig up the tree and plant it a distance d further away
    Now, now... if everyone solved theoretical problems like this, there would be no theoretical trees left.

  10. #80
    GrizzlyAdams's Avatar
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    leave no trace

    In retrospect, we don't need to dig that tree up.

    One thing that makes the analysis go is that the tension is a vector pointed along the suspension line with magnitude T, which is what the engineering guys knew intuitively but I had to be told. The other thing that makes it go is that the counter-force to gravity on the hammock/user is sin(alpha)*T, so W = sin(alpha)*T, no matter what T is. So we can leave the tree alone, T will be a little bit different, but the same principle applies.

    leaving no trace,
    Grizz
    Last edited by GrizzlyAdams; 06-05-2007 at 13:36. Reason: I need an on-line grammar checker

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