The catenary is the shape that a free hanging flexible cable or wire assumes. The following gives equations for the case where the wire is of uniform mass per unit length, and the supporting points are at the same height. Cases where the ends are not at the same height, or where there are point loads (i.e. dipole antennas supported at ends with a feedline hanging in the middle), can also be calculated, but these equations won't do it. (They are "left as an exercise for the reader"...<grin>)

The picture below defines some terminology and the reference points.

We assume that the origin is at the center of the span.

Total span = L

Sag in the cable = h

So, the coordinates of the endpoints are (+/- L/2,h).

The weight per unit length = w

Total length of wire/cable = S

Length along the cable from the origin = s

Fh is the horizontal force component everywhere, and is equal to half the tension at the center.

The horizontal force, in terms of total cable length and sag is

Fh = w / (8*h) * (S^2 - 4*h^2)

The y coordinate (height) of any point in terms of the horizontal force

y= Fh / w * (cosh(w * x / Fh) - 1 )

(Change suggested by Stephen Argles, 24 Nov 2003)

The span, given horizontal force, weight, and length of cable

L = (2 * Fh / w ) * arcsinh(S * w / (2 * Fh))

The total cable length, given span and horizontal force (useful for computing how long a span can be supported)

S = (2 * Fh / w ) * sinh( w * L / (2 * Fh))

The arc length from origin (center):

s = Fh / w * sinh(w * x / Fh)

It's also useful to know the peak tension in the cable, which occurs at the end points. The Vertical force at support is

Fv = w * S / 2,

*i.e.* the total weight of the cable divided by two. And, the Horizontal
force, computed above, is Fh. So the tension is simply the combination of the
two:

T^2 = Fh^2 + (w * S / 2)^2

The minimum tension is, of course, Fh, at the center point where the cable
doesn't support any of it's own weight. If you need the tension in between,
you just need to compute the vertical force at a given point, which is equal
to the weight of the cable from that point to the center (*i.e.* s*w).

Some equations assuming that the cable forms a parabola (generally a good approximation if the sag is not too great) taken from the Reference Data for Radio Engineers book:

Two supports, 1 and 2, at heights h1 and h2, respectively

The low point is L1/2 from the lower support and L2/2 from the higher support.

L0 is the distance between the supports in a straight line

L is the distance between the supports in a horizontal distance

S is the total cable length

w is the weight per unit length

T is the tension in a straight line parallel to the line between supports

Fh is the horizontal component of the tension

alpha is the angle above the horizontal from support 1 to support 2

h = h0 = w*L0^2*cos(alpha)/(8*T)

h1 = w*L1^2/(8*Fh)

h2 = w*L2^2/(8*Fh)

L1/2 = L/2 - ( (h2-h1)*Fh*cos(alpha)/(w*L))

L2/2 = L/2 + ((h2-h1)*Fh*cos(alpha)/(w*L))

S = L + 4/3 *( h1^2/L1 + h2^2/L2)

Assume a 40 meter dipole, supported in three points, center and the two ends, made of AWG#12 copper wire. The actual "leg length" is only 10 meters nominally, and we want to allow no more than 10 cm (4") of sag. Will the wire fail?

Looking up copper wire in the tables: AWG 12 is 2.05 mm in diameter, and weighs 29.4 kg/km, with an area of 3.309 sq mm. The breaking strength of the wire is 337 pounds for hard drawn (65.7 kips), Medium hard drawn 51kips, 261.6lb, soft 38.5kips, 197.5lb. Copper clad steel is much stronger at 711 or 770 pounds for 40% and 30% copper respectively.

Calculating: Fh = 36 N; The span will be 9.997 meters; The tension is 37.03N, for a stress of 10.9 MPa. This is well below the failure loads for copper wire, even soft annealed.

An excel spreadsheet with the equations for the equal support height case and adding in wind forces can be found here.

Wind loads are added to the gravity load; worst case.. normally, they'd act at right angles, so you could take the square root of the sum of the squares. The load is calculated by assuming a Cd of 1.2 (typical for round cylinders at these Reynolds numbers)

http://www.du.edu/~jcalvert/math/catenary.htm

radio/math/catenary.htm - 24 Nov 2003 - Jim
Lux

(changed to update formula for y coordinate, based on email from Stephen Argles)

(math home page) (radio home
page) (Jim's home
page)