Figuring out the wedge position on the rotary floor
12apr04
Lynn baker and mike nolan hung a target
under
the alfa location and surveyed its position in dome center line
coordinates:
lynn's memo:
Mike and I did a survey to start locating the ALFA system on
Fri.
We
mounted a single target thru two exisitng holes in the
floor. The
target
was horizontally centered between the two holes in the plane of
the floor
and extended 22 3/16" below the top surface of the floor.
The floor
is
3/4" thick and this is included in the 22 3/16" measurement.
We checked
that the extension was perpendicular to the bottom plane of the
floor
within perhaps a 1/4" or so. We then measured the location
of this
single
target using the usual technique of surveying a reference set on
the
secondary and back fitting a coordinate system in the survey
software.
The reference set used is dome centerline coordinates and units of
inches.
We moved the turret position until the y coordinate of the
target
was
close to zero. The turret angle was 26.5
degrees. The
coordinates of the
target are:
ALFA4
246.5803
0.0677 393.7084
For reference, the focal point of the optics in the same
system is:
Nominal focus: 248.145
.0 389.822
The target was not too far from the nominal focus.
Making the
approximation that the floor is in the xy plane of the dome
C/L
system
with the z axis upward and perpendicular to the floor. (Not
too bad
for
small motions). Moving along the floor 1.5" radially
inward
from the
original bolt holes, (away from the secondary) places a
point above
the
focal point. The focal point would be 18.3 inches
below the
top surface
of the floor at this new point.
The reference is now the two holes that straddle the center of the
pie region. Using:

8.17 degrees for the tilt angle of the horn

18.3 inches focus below the top of the floor (measured by lynn
and mike)

1.5 inches radially in to go from reference to focus

24.4094 inches alfa focus to where alfa attaches to the wedge

4.5 inches as the vertical height of the wedge at it's center

1.063 inches radially inward as german corte's offset (in the
tilted focal
plane).
I get that

the center of the wedge should be 5.82 inches radially
inward from
the holes on the floor.

We need a shim of 1.22 inches between alfa and the
wedge.
The plot
shows
how i arrived at the numbers.
note 13sep18
There is an error in the plot (found by felix..)
 B=18.3*sin(8.17)=2.6".. should be
 B=18.3*tan(8.17)=2.63" .. so there is an error of
.03"
processing: x101/mbeam/wedge.pro
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