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Sliding glass doors are typically of double pane construction, with multiple 
sheets of glass between
the inside and the outside.  Then we will need to consider the parallelism of each 
sheet of glass to
the others.  Not all double pane windows are air filled, some are gas filled to 
increase the insulation
value.  We would need to consider the index of refraction for that gas, should it 
exist, compared to
the nominal value for the atmosphere.

The stiffness of the glass also comes into play, independent of the age of the 
glass.  As the extent
of the sheet of glass gets larger, the propensity of the glass to bend and 
displace under self and wind
loading will be increased.  How significant this is will be a function of the 
thickness of the glass, the
angles to vertical it is installed and the like.  Just because it is installed in a 
house, you cannot assume
that the walls are perfectly vertical.  Non-verticality will cause sag and 
deflection, which will vary with
location within that pane of glass.  Is the frame that the glass is installed in 
perfectly flat and planar?
If not, then that glass will have bends in it, to comply with the greater 
stiffness of the frame.

Assuming perfect parallelism with a singular sheet of glass (or shade), when we place 
that glass in the
optical path, it will cause a shift (displacement) of the beam as a function of the 
thickness of the glass.
First the beam will shift in angle as it enters the glass, then when the beam 
encounters the opposite and
parallel surface, the original angle of the beam will be recovered.  The parallel shift 
of the beam is a
function of the thickness of the glass.  To the observer, will that result in a slight 
angular error in observation?
I believe the answer to that to be yes, albeit small.

How significant is all of this?  I do believe that the dip short to the opposite 
lakeshore will be a
far larger effect than any prismatic effect.

One quick way to tell would be to adjust the sextant whilst sitting indoors and 
now that the sextant
is adjusted for the objects, step outside those same panes of glass.  Without 
adjustment, perform the
same observation, just this time do it outdoors.  Assuming that the lake really 
isn't a pond, then the
dip short correction outdoors will be very close the dip short correction indoors. 
 That is, all things being now
equal, do you observe the same altitude?

And one small correction to Scott: there really aren't dumb questions, except the 
ones you don't ask!

Best Regards
Brad



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