If cosh(x)= y  is the height of the arc,
then the ratio of the height at the center 
to the width/2 should give us the extent of 
the cosh curve we plan to us.

           y = (2H/W)x 

  The hyperbolic cosine function looks like a 'U' shaped
curve whose lowest point passes through x=0, y=1. So to
find out what portion of this curve will need to be
'magnified' to make the proper ratio for a bridge design,
we must first subtract 1 from the middle point to set
the curve on the origin. 

      (y-1)= (2H/W)x 

  For example, if the width is to be 12" from the
center to the side, and 3" inches high, maximum, then
the equation, y=cosh(x) must find the value where

    (y-1)= (2H/W)x 

    Substituting 3 for the height, and 24 for the 
width, we get:

   ( y -1 ) =x/4 or (y-1) =(.25)x

    We want to find out what range of the cosh function 
will satisfy the  ratio of height to width for our bridge.
Once we find out the value of x which satisfies this ratio,
we plug in values between zero and the maximum x, and multiply 
the result by a magnification factor.

    Ok, first set up a table. 

 
   x    y    ratio of (y-1)/x 
  ===   ====      ============
   0.1  1.005         0.05
   0.2  1.02          0.1
   0.3  1.045         0.15
   0.4  1.08          0.202
   0.49 1.12247       0.2499  (* this is pretty close to 0.25)  
   0.5  1.127         0.2552
   0.6  1.185         0.309
   

   So, from the above, we see that the range of x is really 
only from zero to 0.49 for an arc whose height will be 
roughly 1/8 its width. ( remember, we're flipping the 
curve upside down)


   To plot the curve, simply pick as many interim points 
between 0 and 0.49 and evenly space them. For example, you
might like to just say, one per inch, in 12 inches would 
need a point at say (0.04,0.08,0.12,0.16,0.20,0.24,0.28,0.32,
0.36,0.40,0.44,0.48) 

   And what magnification factor,M, will we need to get this to
yield the proper heights at each inch? Well, the ratio of 12 to
0.49 ought to do it -- the answer...  24.489 

   12/(0.49) = 24.489 = M  

   So, setting up another table, we get the values for the heights.
We also need to subtract one from y to normalize our curve.

   x    y=cosh(x)    M(y-1)
  ===  =========     ====

   0     1             0           (CENTER OF BRIDGE)
   0.04  1.008         0.0195
   0.08  1.0032        0.078
   0.12  1.0072        0.176
   0.16  1.0128        0.314
      .......
   0.48  1.1174        2.875
   0.49  1.12247       2.999       ( FOOT OF BRIDGE ARC)

   Ok, now, to plot this, you just need to plot the points and flip it
over, then mirror them for the other side.