This is most likely a bridge engineering question:
A monofilament line that is 2000 meters
long and weighs 7 lbs. (weight is based on an available shark fishing
line which has a 130 lb. test rating) is suspended horizontally by each
end. Assuming the monofilament is immune to stretching (unless you
know how to calculate that too based on available products), what
strength rating should the monofilament have to support itself in as
nearly a horizontal line as possible (for example: "130 lb. test")?
How much force will need to be applied to each end? How much
deflection will occur at the midpoint?
I'm looking for the lightest/strongest solution - it only needs to
support itself under favorable weather conditions.

---

Request for Question Clarification by krobert-ga on 30 Dec 2002 16:36 PST

This is not as simple a question as it seems. There will be
stretching.  If we assume that deflection is linear, the total
deflection will be related to the integral of the weight across the
length of the line. More than likely (read: I haven't run any
numbers), the deflection will be nonlinear. That is, the deflection
itself will be related to the total deflection (2 points if you
inderstood that!).  This is a problem for a finite element analysis.
The shape of the curve that the line takes will be a catenary, which
is nongeometric (I think that's the right term). Unfortunately, unless
another researcher is fluent in a simple FEA package that he can use
to whip out an answer, this question is not going to get answered for
$5.00.

I hope that this information helps you out rowpole-ga.

Best,

krobert-ga

---

Clarification of Question by rowpole-ga on 31 Dec 2002 07:56 PST

I understand that calculating for stretch adds a great deal to
complexity to the problem - which is why I'd be satisfied with an
answer that ignores that variable. My own online research into bridge
engineering uncovered terms such as "catenary" but none of the sites I
found offered any simplified formulas in the area of my question. This
is not a $5 question for a non-engineer, but for someone fluent in
finite element analysis, I'd assume it would be a fairly common
calculation.

---

Request for Question Clarification by haversian-ga on 31 Dec 2002 08:13 PST

How accurate an answer are you looking for?

The problem can be simplified by making the line neither stretch nor
have mass, but place a 7lb weight in the center of the span and
calculate what force at each end would be required to raise the mass
to some arbitrary height.  The real line would show deflection
strictly less than that value.

---

Request for Question Clarification by krobert-ga on 31 Dec 2002 09:07 PST

haversian-ga and rowpole-ga,

haversian's argument is true except that now your span does not equal
2000 meters, but some value less than that.  If the line doesn't
stretch, your span must decrease or the length of the line must
increase.

Yes, the real line would not show as much deflection because in the
real line, that 7 lbs is distributed approximately evenly over the
length of the line.

rowpole, in response to part of your original question, "what 
strength rating should the monofilament have to support itself in as 
nearly a horizontal line as possible", you would need to specify the
allowable deflection in order to calculate the value of strength.
Conversely, you could calculate the deflection based on a specified
strength rating, but I would think that the mentioned strength rating
would be inadequate.

In retrospect I think that in the real world any of these arguments
are deceiving in respect to a real line out in the open.  I would
guess that at such a length, wind would be pushing this line around
much more that any deflection due to it's own weight.

krobert-ga

---

Clarification of Question by rowpole-ga on 31 Dec 2002 16:52 PST

Wow - thanks for the fast responses! Ok, let's see if I can simplify
the question.  The length of the line may increase - only the distance
covered (2000m) is fixed. Lets set the maximum allowable deflection at
5 meters. Lets set the strength of the line at "130lb test" - meaning
that a short length of the line will suspend a dry weight of 130lbs.
The total weight of the line is still 7lbs. Let's assume wind does not
exist - and that we're at sea level. With those parameters, will the
line fail (answer with some certainty, please)?
Thanks!

Hello rowpole

      I reckon you want a simple, even if aproximate answer.

      So,-- in situation described, the tensile strength of the
      fiber would be exceeded about ten times. Calculated as follows:

      tension in the fiber is about:  7 lb * (1000m / 5m) = 1400 lb


 This follows from the triangle H------C
                                       |
                                       |
                                       W
  which is half of your rope. C is center point bearing weight 7 lb
                               Distance HC = 1000m

 and rope is streached from H to W, aproximating the catenary by
 the straight line HW.

  This aproximation is good only for small angles VHW (=alfa),
 as we are putting sin alfa=tang alfa = alfa (in radians).

  The force in HW (tension) is  more exactly 7 lb / sin(alfa).


  For picture of catenary look here:

 Mudd Math Fun Facts: Suspended Rope Trick
  ... the catenary y = cosh x and suspending a rope in front of it.. ...
   www.math.hmc.edu/funfacts/ffiles/10001.3.shtml

This problem (suspended rope) is a typical example in textbooks on
the calculus of Variation. See e.g.:

 [PDF]Lecture 3: Calculus of Variations
 phyastweb.la.asu.edu/phy501-shumway/2001/notes/lec3.pdf

 The question as originally formulated was hard because of the term
 'nearly horizontal'. As apparent from formula above, tension goes to
 infinity as alfa goes to zero. The question: what deviation from
 horizontal is needed to suspend 7 lb with rope which can support 130 lb
 seems closer to what you are after, based on your clarifications.
 Answer follows from the formula and is left as homework :-)

 Search terms:

 Calculus of Variations,
 Catenary
 Hyperbolic trigonometry (for cosh)

 Neat problem. Thanks.

 Hedgie

Thanks Hedgie! Great work!

A catenary curve 2000m wide with a 5m sag has slope -0.010 at the
ends, so the tension is 100 times the weight each end needs to
support, which is half the total weight of the line.  Therefore the
tension in a 2000m fishing line weighing 7 lb and sagging 5m is 350
lb, almost three times the breaking strength.  A tension of 130 lb
occurs when the end slope is -0.027, which occurs when there is 13.5m
sag in the line.

My main point being, hedgie should have used 3.5 lb rather than 7,
because only half the weight of the line is supported by each end.