Balancing an equatorial

In any equatorially mounted telescope there are three axes about which

moments have to be determined. It is important which order these are

considered. They are as follows:

i) radial equilibrium about the tube axis

ii) longitudinal equilibrium about the declination axis

iii) declination equilibrium about the hour axis

If the telescope tube is not in radial equilibrium it will not be possible

to adjust the system cg to the intersection of the declination and hour

axes, for all possible attitudes of the tube assembly. Radial balance is

the most awkward to achieve, and is crucial. Radial out-of-balance is the

most common cause of the complaint, "My telescope will not balance in all

positions."

To determine the radial momenta about the tube axis, measure the offset of

all the non-symmetric tube furniture (i.e. rackmount, finder, guide 'scope,

accessory mounting brackets &c). Ignore symmetric items, i.e. the primary

and secondary mirrors & cells, the spider, cradle rings &c. Weigh, or

calculate the weight of each item. Adopt an arbitary reference plane,

transverse to the tube's mechanical axis. A useful reference plane might be

in the direction of gravity when the tube is horizontal, pointing due south.

Call this plane X-X'. Prepare a scale diagram depicting the tube profile,

and each weighed item. Measure or estimate the angular distance of each

item, anti-clockwise, looking down the front of the tube from the reference

plane.

The radial momenta of each item will then be:

Moment = radial offset x weight x cos(angular distance from X-X')

Always measure the angular distance from the same reference, using 360°

notation, in the same anti-clockwise direction. Repeat the procedure using a

reference plane rotated 90° clockwise. Call this plane Y-Y'. Sum the momenta

about each plane.

As an example, I performed the calculation for my 10-inch f/10.6 Calver, with

the following results:

Moments about X-X'

rackmount = 12" x 7lbsf x cos270° = 0inlbs

lower counterwights = 8".75 x 56.68lbsf x cos270° = 0inlbs

Cooke 2-inch finder = 12" x 7lbsf x cos330° = 73inlbs

mounting bracket + 400mm lens = 9".75 x 11.8lbsf x cos0° = 115inlbs

mounting bracket + 150mm lens = 9".75 x 10.8lbsf x cos180° = -105inlbs

upper counterweight = 8".75 x 22.42lbsf x cos90° = 0inlbs

4-inch Guide 'scope = 10".25 x 7lbsf x cos135° = -51inlbs

Ottway 2-inch finder = 9".25 x 2.625lbsf x cos210° = -21inlbs

Moments about Y-Y'

rackmount = 12" x 7lbsf x cos0° = 84inlbs

lower counterwights = 8".75 x 56.68lbsf x cos0° = 496inlbs

Cooke 2-inch finder = 12" x 7lbsf x cos60° = 42inlbs

mounting bracket + 400mm lens = 9".75 x 11.8lbsf x cos90° = 0inlbs

mounting bracket + 150mm lens = 9".75 x 10.8lbsf x cos270° = -0inlbs

upper counterweight = 8".75 x 22.42lbsf x cos180° = -196inlbs

4-inch Guide 'scope = 10".25 x 7lbsf x cos225° = -51inlbs

Ottway 2-inch finder = 9".25 x 2.625lbsf x cos300° = 12inlbs

Sum of momenta about X-X' = 11inlbs

Sum of momenta about Y-Y' = 387inlbs

To calculate the weight (W) and orientation (A°) of a counterpoise necessary

to bring the tube into radial equilibrium, we must adopt a suitable radial

offset for the counterpoise (e.g. r = 8".75) and solve the following

simultaneous equation:

i) 8".75 x W x cosA + 11 = 0

ii) 8".75 x W x cos(A+90°) + 387 =0

divide (ii) by (i) (cos(A+90°) = -sinA)

hence 8.75.W.sinA/8.75.W.cosA = tanA = 387/-11 = -35.182

(sinA positive - 2nd quadrant - recall all trig ratios +1st quad; sin +2nd; tan +3rd; cos +4th)

therefore A = 91°.63

subst. A in ii) hence 9.75.Wcos181°.63 = -387

& solving for W
W = 44.25lbsf

To check the calculation, determine the resultant sum of the radial momenta:

about X-X' = 8.75x44.25cos91°.63 + 11 = 0inlbs

about Y-Y' = 8.75x44.25cos181°.63 + 387 = 0inlbs

Where the counterpoise is located along the length of the tube is arbitary.

It is its orientation with respect to X-X' which is important. I decided to

split the counterpoise into two equal weights, placed either side of the

dec. axis.

Once you have determined the radial equilibrium, calculating mometa about

the declination and hour axes is a matter of simply weight x axial distance.

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