# a parabolic dish

#### 06may21

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:
• 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
•  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

home_~phil