Dome za balance when ch at stow (usng td tensions)
Az swings and the tiedown tension were
used to find the dome za that balanced the carriage house at stow
(8.835 degrees). When the dome balances the ch, doing an azimuth swing
should not change the tensions at any of the tiedowns.
The procedure used was:
The plots show the
results of the fits (.ps) (.pdf):
- bring the telescope into focus using tietrk then turn off
the tiedowns (so they don't move during the test).
- move the carriage house to stow position (8.835 degrees).
- swing the azimuth from 90 <--> 450 degrees at +/- .38
degs/sec. Do this with the dome at 7,7.5,8,8.5 degrees.
- For each azimuth swing fit y=c0 + c1*cos(az) + c2*sin(az).
- Compute the amplitude of the 1a term: sqrt(c1^2 + c2^2)
- evaluate the fit at the azimuth of each tiedown.
- Top: the 1 az amplitude for the azwings at za=7,7.5,8,8.5.
- black:td12, red:td4,green:td8
- It goes to 0 close to za=8
- Bottom: evaluated each 1az fit at the tiedown azimuth.
- the tension goes from positive (dome side to light) to negative
(dome side to heavy) at the balance point.
- A linear fit to all the points vs za gives a balance of
za=7.968 degress (optic axis).
- With the ch at stow, the azimuth arm balances with
dome za=7.968 degrees.
- Saying that the azimuth is balanced does not mean that all 3
tiedowns have the same tensions, just that the tensions don't change as
you do an az swing.
- Tiedowns will have an unequal tension since the weight
distribution on the triangle is not uniform (there is more weight in
- Using the expected weights, distances for the various things gave
a balance of 7.5 degrees:
- chCounterWeight=45 kips
- chCounterWeightdistance:145 feet
- Radius for dome,ch : 430 feet.
- dome enter of mass 1.9 degrees downhill from the encoder value.
- What is missing is the residual unbalance from the az arm
stairway, vertex shelter, etc.. see computing