What would the
north south main cables look like if we switched from a
spherical to a parabolic dish.
Computations use the following constraints:
- The curves should be anchored at the rim wall
- The bottom of the curves should match the current
bottom of the spherical reflector : 158.031 feet below the
dish edge at 500'
- Look at the central north/south main cable (spans the 1000
foot diameter).
The equations used:
- parabola:
- y=x^2/(4*f)
- f= distance from focus to vertex of parabola.
- sphere:
- radius of curvature=870feet
- diameter of dish 1000 feet
- catenary:
- y=Acosh(x/A)
- Horizontal component of tension in cable:
- T0=A*lambda
- lambda= cable weight (lbs/foot)
- use 1 1/4 " cable: 3.28 lbs/foot
The plots
show a parabola, sphere, and catenary spanning the dish
(.ps) (.pdf)
- Page 1: parabola focal height vs depth of parabola
below rim wall
- The x axis is the depth of the parabola vertex below the
parabola height at x=500ft (the rim wall)
- The blue vertical line shows the current depth of the
spherical reflector (below the rim wall).
- The green line is the height of the paraxial surface of
the sphere along a radius.
- A parabola touching the rim wall and the bottom of the
dish has a focal height of 395.492 feet.
- Page 2: compare catenary, sphere, and parabola spanning
the dish.
- each curve goes through the bottom of the dish and the
rim wall.
- Top: overplot the 3 curves
- black: catenary (A=816.040)
- red: parabola (focus=395.492 feet
- green: sphere with radius of 870 feet.
- Bottom: plot curve differences: catenary - OtherCurve
- red: Catenary - parabola
- the parabola sits above the catenary
- To form a parabola, you would make the bottom of the
catenary higher than normal (increase tension) and
then pull it down into a parabola
- green:Catenary - sphere
- The sphere sits below the catenary
In the real world you'd probably have the catenary hang
above the bottom of the reflector so you could pull it even at
the center.
Tension in the cable:
- Compute the horizontal tension in the cable assuming:
- 1 1/4" cable, 3.28 lbs/foot
- cable spans 1000 feet
- bottom of the cable is 158.031 feet below the height of
the cable a x=500'
- Assume there is no other load on the cable (no dish, no
tiedown cables)
- T0=A*cableDensity. A= 816.040, density=3.28 lbs/foot
- Cable horizontal tension = 2676 lbs.
- The tension along the cable is 2676/cos(theta) where
theta is the cable angle relative to the horizontal.
processing: x101/210506/parabola.pro