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Astronomical Applications Department, U.S. Naval Observatory - wedges (Page 5)

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Astronomical Applications Department, U.S. Naval Observatory - wedges
(5)
q
=
q
-
h
+
q
cos h tan x
q
c
+
O
c
2
Use (5) for
in the z' component of (2), expanding on
, to get
(6)
s
=
s
+
1
2
s
2
-
q
2
c
=
h
+
q
cos h tan x
-
1
+
cos
2
h tan
2
x
q
┬ 2
-
h
2
c
+
O
c
2
Switching back to Cartesian coordinates and dropping the primes, we have the equation for the
perturbed glass thickness:
(7)
s
=
h
+
x tan x
+
Ds
where
(8)
Ds
= -
1
2

x
2
cos
2
x
+
y
2
-
h
2

c
+
O
c
2
Notice that
is the position along the mirror surface, which is physically of length
x/ cos x
2a/ cos x.
A positive gradient along the z axis produces a warp in both the x and y directions. Relative to
x=0, the far end of the mirror at x=2a sags along x in the shape of a parabola. Along the y
dimension (x=0), we also have a downward-sagging parabolic warp.
2.2. A Numerical Example.
The warps introduced by a positive temperature gradient of 10 mK/m with
are
a
=
2 $ 10
-
8
K
-
1
illustrated in Figures 4 and 5. In Figure 4, the surface perturbation units are nanometers. The
warp across the short dimension (Figure 5) is sho
wn in picometers.
2.3. Resulting Change in the CMA Basic Angle; Constraint on Gradient Magnitude.
The longitudinal excursion of the end of the mirror
(x=2a) relative to the beginning (x=0) is
(9)
Ds
c, y
=
h
=
0, x
=
2 a
=
-
2 a
2
cos
2
x c
+
O
c
2
This corresponds to an angle
(10)
dy
2
=
Ds
2 a
= -
a
cos
2
x c
+
O
c
2
The change in the basic angle
is approximately twice this amount. For a tolerance
such that
, the constraint on the longitudinal temperature gradient is then
dy [ t
(11)
b [
cos
2
x
2 a a t
+
O(t
2
)
For the specific case
,
,
, and
, we have
x
=
45 deg t
=
25 lrad a
=
30 cm
a
=
2 $ 10
-
8
K
-
1
Basic Angle Temperature Gradient Sensitivity
FTM-USNO-95-01
5

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