Astronomical Applications Department, U.S. Naval Observatory wedges Page 9
This time the independent variable of interest is z', so we would like to express the radius
in
terms of the perturbed "height," z', and use that in the z' component of (22). From the
' compo
nent of (22), we have
(23)
q
Â
=
q
+
2a
+
h

q cos h
tan x
q c
Solving for
, we find
(24)
q
=
q
Â

2a
+
h

q
Â
cos h
tan x
q
Â
c
+
O
c
2
Hence, dropping the primed notation, the z component in the perturbed coordinates becomes
(25)
z
=
2a
+
h

q cos h
tan x
+
1
2
4a
2

1
+
cos
2
h
tan
2
x
q
2
+
h
+
4a tan x
tan
2
x
h
c
+
O
c
2
Invert this to get
as a function of z. We find
(26)
q
=
h
+
(2a

z) tan x
cos h
+
1
2
2h
+
(4a

z) tan x
cos h
z

h
+
(2a

z) tan x
2
cos
3
h
tan x
c
+
O
c
2
Convert to Cartesian coordinates to obtain
(27)
x
=
h
+
(2a

z) tan x
+
1
2
2h
+
(4a

z) tan x
z

x
2
+
y
2
x
2
h
+
(2a

z) tan x
2
tan x
$ c
+
O
c
2
Solve (27) for x. We find
(28)
x
=
h
+
(2a

z) tan x
+
Ds
where
s is given below. Hence the perturbed surface height in the case of a transverse tempera

ture gradient is
(29)
s
=
h
+
(2a

z) tan x
+
Ds
where
(30)
Ds
=
z h
+
1
2
z (4a

z)

h
2

y
2
tan x

1
2
(2a

z)
2h
+
(2a

z) tan x
tan
2
x $ c
+
O
c
2
3.2. Resulting Change in the CMA Basic Angle; Constraint on Gradient Magnitude.
Since the two wedge mirrors are tilted in opposite directions (Figure 1), a positive temperature
gradient across one mirror is a negative gradient across the other, in terms of the effect on the tilt
angle. Therefore the change to the basic angle will again be twice the change in tilt of one of the
mirrors:
Basic Angle Temperature Gradient Sensitivity
FTMUSNO9501
9