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On 2019-07-12 9:46, Fred Hebard wrote:
>
> But if the azimuth is 0 in the direction of tilt, then when rotated to an 
azimuth of 90 degrees and raised to 90 degrees altitude, the azimuth will be 
off by the amount of tilt.  Those were, more or less, the conditions for me.  
I could find Venus when it was low, altitude 10-20 degrees, pointing near the 
direction of tilt, but raising the telescope to 45 and rotating 90 degrees in 
azimuth made it hard to find.

That sounds like a problem similar to the Laplace correction of the
surveyor. It accounts for the fact that the direction of gravity is not
perpendicular to the ellipsoid (the mathematically perfect surface that
is the basis for geodetic coordinates). This "deflection of the
vertical" is typically a few arc seconds, and is caused by the irregular
mass distribution in the Earth.

In Fred's case the tilt of the telescope mount corresponds to deflection
of the vertical. Geometrically, the problem is the same as the Laplace
correction.

For a simple example, suppose you need to set a stake in the ground
precisely north from a theodolite. Its scope has been pointed carefully
to Polaris, the azimuth axis locked in position, and the azimuth circle
zeroed. (We'll pretend Polaris coincides exactly with north.) However,
the instrument is inclined a little to the east. Therefore, the line of
sight diverges to the left as you tilt down, so your assistant will
drive the stake west of the correct point.

The Laplace correction can be computed if the amount of dislevelment is
known. The math in this document is quite an eyeful, but we can skip the
derivation and accept equation 10 as correct:

http://www.mygeodesy.id.au/documents/Laplace.pdf

In surveying the deflection of the vertical is split into two
perpendicular components ξ (xi) and η (eta). These angles are positive
if the instrument is tilted respectively to the north and east of the
true vertical.

Variables αA and αG are respectively the azimuth indicated on the tilted
instrument, and the true azimuth.

The quantity φA is the altitude of the azimuth reference that you use to
align the instrument to north.

The quantity zA is the zenith distance (90 - altitude) of the target body.

The result of equation 10 is δα, which you subtract from the instrument
indication to obtain true azimuth.

Let's try equation 10 on my example of setting a stake true north of the
instrument. Suppose latitude is 40°, hence Polaris altitude (φA) is
likewise 40°. Assume the instrument is tilted one degree east, hence ξ =
0 and η = 1° = .0175 radian. Since the stake is on the horizon, its
zenith distance zA is 90. Plug those values into the equation:

δα = .0175 * tan 40 + (0 * sin 0 - .0175 * cos 0) * cot 90

Since cot 90 = tan 0 = 0, the second term is zero, so the equation
simplifies to

δα = .0175 * tan 40

δα = .0146 radian = .84°

Thus, the instrument says 0 (aka 360) when pointed at the stake, but
with the Laplace correction, true azimuth is 360 - .84 = 359.16.

We can reverse the problem and use the stake as the known azimuth. Sight
the stake and set the azimuth circle to read 359.16°. Of course the
theodolite reading doesn't change when you aim up at Polaris, but it
must be corrected for an instrument tilt of 1° to the east.

The azimuth reference is at altitude 0, so φA = 0. As before, the tilt
components are ξ = 0 and η = 1° = .0175 radian. Polaris azimuth per the
instrument (αA) is 359.16, and its zenith distance zA is 50°. Plug those
values into equation 10:

δα = .0175 * tan 0 + (0 * sin 359.16 - .0175 * cos 359.16) * cot 50°

δα = -.0175 * cos 359.16 * cot 50°

δα = -.0146 radian = -.84°

Subtract that from the instrument reading to obtain corrected Polaris
azimuth: 359.16 - (-.84°) = 0.00.

There's also a tilt correction for the zenith distance in that same
article, equation 12.

   

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