Conference rusure::math

          <<< Note 1398.0 by HPSRAD::BUSCH "Dave Busch, MRO1-2/S10" >>>
                              -< Catenary curve. >-

The form of a perfectly flexible rope, suspended from two points, and only
affected by gravity, is a section of the hyperbolic cosine curve (apart from
constants to make things fit).

The equation for the hyperbolic cosine is

	y = .5 * (exp(x) + exp(-x))

        The base note mentioned a suspension bridge.  That's like
        hanging extra weights from the rope.  What happens then?
        
        Dan

if you add weights, I believe the curve becomes a parabola, provided that

	the weights are large compared to the weight of the rope

	the weights are uniform along the horizontal axis

if you remove these restrictions, then the curve becomes very complex.

Hi,

	I'm currently working on a similar problem. I am trying to work out the
    length of a rope involved in the 'half-catenary' from a boat to the sea bed.
    I have an equation (below) that, for a given weight of cable, depth of water
    and cable tension, will print the 'half-catenary' length and the horizontal
    distance from the boat to the sea-bed touchdown.

    However, my results don't tally with those in the little handbook I took the
    equation from (but they are close). Can anyone tell if the equation is
    correct ? Also, how might one determine enough information to draw a graph
    of the curve ?

Thanks,
Ross

Below is a BASIC program.

1   input "weight (lbs per ft) of cable ";w
    input "cable tension (kips) on cable";f
    f = f*1000
    input "water depth (ft) below sheeve";d

    y = (f/w)
    c = y-d

    s = sqr(d*(2*c+d))
    print "Half-catenary length = ";s
    x = c*log((s+y)*(1/c))
    print "Horizontal x co-ord = ";x

    Re .4:
    
    I would guess that there is no closed-form formula for the length of a
    catenary, so any formula is going to be only an approximation.
    
    The basic form of a catenary curve is y = cosh x.  The cosh function
    is:
                  x    -x
    	cosh x = e  + e
    	         --------
    		    2
    
    You can modify that by stretching and translating in the x and y
    directions, so for your situation, you might have y = a (cosh(bx) - 1). 
    This will give y = 0 when x = 0, so the cable is on the seabed at x=0. 
    Then a and b determine your vertical and horizontal scales.  I'd have
    to do some figuring to relate a and b to cable tension, but if you just
    want to draw the curve, you can draw y = cosh(x)-1; it will touch the
    origin and the shape will be the same regardless of scale.
    
    
    				-- edp

        If y = cosh x, then dy/dx = sinh x and 1 + (dy/dx)^2 is
        cosh^2 x.  So sqrt(1 + (dy/dx)^2) is cosh x.  Isn't that
        what you integrate to get arc length?
        
        Dan

Hi,

	by 'playing' with some additional calculations, I came up with the sample
    of code below. Using the cosh function from reply .5, I get a curve when it's
    plotted. Anyone care to speculate it it's correct as I'm by no means a maths
    wizard ?

    Thanks for the help so far.

Ross



1   input "weight (lbs per ft) of cable ";w
    input "cable tension (kips) on cable";k
    f = k*1000
    input "water depth (ft) below sheeve";d

    y = (f/w)
    c = y-d

    s = sqr(d*(2*c+d))
    x = c*log((s+y)*(1/c))
    print "catenary length = ";s
    print "cable touchdown = ";x

    v = sqr((2*f*d*w)-(d*w)^2)
    print "vertical force  = ";v; w*s

    f = w*((s^2+d^2)/(2*d))
    k = f / 1000
    print "tension (kips) to hold cable = ";k

    l = x/40
    m = l*20
    e = 2.71828

    for j = 0 to x step 20

	b = j/c
	h = c * (( e^b + e^-b )/2) - c
	print x-j,h
	
    next j

    Re .6:
    
    Yes, but how about integrating sqrt(1 + (ab sinh(bx))^2) to get the
    length when the x and y scales differ?  Consider that we can integrate
    y = cos t and x = sin t to get the arc length of a circle, but not y =
    b cos t and x = a sin t to get the arc length of an ellipse.
    
    
    Re .7:
    
    I can't comment on whether tension is calculated accurately, but your
    basic formula for the catenary looks okay.  However, your BASIC
    probably has a function such as EXP(x) which is more accurate than e^x.
    I am also not sure about calculating "b=j/c" and using that in
    (e^b+e^-b) -- when j equals x, is h approximately equal to d?
    
    
    				-- edp

    It's possible to rectify the catenary in terms of elliptic functions
    if I recall... (in the general case)

    I'll double check the formulas tonight if there is any
    interest.  But I don't remember them off the top of my head
    though I probably could rederive them.

    - Jim