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APPENDIX E:
USING THE QHM IN 'ODD PI' MODE
QHMs are used in 1 pi, 2 pi, 3 pi,
modes making the assumption that
= 1 pi, 2 pi = 2 pi, etc.. While it is obvious that 2 pi = 2 pi, it is not
quite so obvious that 1 pi = 1 pi. What this means is that it may not be true
that the front and back mirrors of a QHM are exactly parallel. For instruments
that are used in odd pi modes, it is worthwhile confirming that the mirrors are
close enough to parallel, or if not, determining the error. Of course it may not
be possible to compare some 1 pi instruments to the 2 pi mode unless they are
taken to a place of high field by natural or artificial means, so Antarctic QHMs
should be tested in Australia.
1 pi
Let
Then
m = an odd integer
e = the angle between the back and front mirrors of the QHM
a = the residual deviation angle, alpha
phi+' = apparent deviation of the magnet in the +m pi position
from the zero position
phi-' = apparent deviation of L~e magnet in the -m pi position
from the zero position
phi' = average of phi+' and phi-'
phi+
actual deviation of the magnet in the +m pi position
from the zero' pas it i on
phiactual deviation of the magnet in the -m pi position
from the zero position
phi
= average of phi+ and phiM = moment of the QHM magnet.
H = horizontal field strength.
QHM fibre torsion constant.
T
B = residual torsion in the fibre.
phi+'
phi+ + e
phi- -e
phi-'
phi+' - phi-' = phi+ - phi- + 2e
phi+' + phi-'
phi+ + phiphi' = phi.
During a 'plus odd pi' rotation,
the QHM head
[m.pi + phi+'], and the QHM magnet rotates through phi+.
developed in the fibre is In.p'i t· ph'i I·' - phi+ = m.pi + e.
rotates through
Hence the twist
During a 'minus odd pi' rotation, the QHM head
[-m.pi - phi-'], and the QHM magnet rotates through -phi-.
developed in the fibre is -m.pi - phi-' + phi- =-m.pi + e.
rotates through
Hence the twist
See Figure 8 for some help in visualising the geometry of the situation.
So the equations
become:
of equilibrium
ill the zero, +n.pi
and -n.pi
positions
M.H.sin(a) = T.B
M.H.sin(phi+ + a) = T.(B + m.pi + e)
M.H.sin(-phi- + a) = ,T.(B - m.pi i' e)
and so:
M.H.[sin(phi+ + a) + sin(-phi- + a)]
=
2.T.8 + 2.T.e
2.M.H.sin(a) + 2.T.e
and
Appendices
E-1
.... (1)