fixed points in survex

[John Halleck]

On Thu, 6 Feb 2003, Lev Bishop wrote:

> On Thu, 6 Feb 2003, John Halleck wrote:
 
> > On Thu, 6 Feb 2003, Lev Bishop wrote:

> [...]

> > > 1) Tidal effects on the earth. The two sources of earth tides are the sun,
> > > causing effects of 17cm amplitude, and the moon, with 36cm (the next most
> > > influential body is venus, whose effects are less than 0.05% of the
> > > above). However, as long as our measurements cover an area of limited
> > > dimensions then the effects will be small. For example, over a 100km
> > > distance, the tides should come to less than a cm.
> > 
> >    Where does this figure come from?  It is higher than what I've been told.
> >    (Unless you mean "MUCH less than a cm".)
> 
> Rough calculation as follows: take combined tidal effect to be sinusoidal
> and of 50cm amplitude (eg a spring tide when sun & moon line up
> 36cm+17cm=53cm). assume we're on the equator and feel the full +-53cm

  But where are THOSE figures come from?  The claim for the NGS HARN
  point around the corner from me is that the earth tide here is in mm's.
  It is possible I've been told incorrectly, but I don't know where
  your figures come from.

> displacement over the course of a day.  100km on the equator implies an
> angle of roughly 100/6400 radians subtended at the centre of the earth. we
> must double this to get the phase angle for the tide (because there are 2
> tides per day, one on the side closest to the sun/moon one on the opposite
> side). assuming we're midway between tides at the current moment in time
> then the difference in vertical position is 50cm * sin(2*100/6400) or
> using small angle approx 50cm*2*100/6400=1.5cm, actually a little more
> than a cm (for this worst-case scenario). However, this calculation is for
> vertical positions referenced to the ellipsoid - presumably the Geoid will
> distort almost the same amount as the earth tides (perhaps the earth tides
> will slightly lag the Geoid since its the effective gravity (gravity plus
> centrifugal forces) that drives the earth tides in the first place). So
> for Geoid-referenced heights the differences should indeed be MUCH less
> than a cm. I guess what you've been told was for land surveying - 
> measuring using levels and so on it is the geoid height that matters. I 
> was thinking in terms of high-accuracy DGPS in which case its the 
> ellipsoid height that matters, although thinking about it a little more 
> carefully I suppose that the software which does the postprocessing for 
> such types of phase-ranging dual-frequency long-baseline DGPS probably has 
> the option to make tidal corrections for you - its something that you 
> wouldn't be able to ignore in such cases (unless you made observations for 
> several days on the same point and let time-averaging take care of it). 
 
> [...]

> >   If I remember aright, the earth curvature drops 1foot/mile, and
> >   the refraction drops line of sight 1/2 ft/mile, for a net APPARENT
> >   curviture of about 0.6 feet/mile.
> 
> Actually I had forgotten refraction - thanks for bringing that up. The
> earth curvature does drop about 0.7feet in 1mile, but it depends on the

  I was working from memory.  (and only giving the curvature in a mile,
  not a figure per mile.)  And I do have the units wrong.

> [...]

> What I'm really thinking about is a little different, though. When you're
> calculating the directions of legs in cartesian space you can't really
> assume (as survex does) that up is the +z direction, north is the +y
> direction, east is the +x direction. To take an extreme example, lets say

  Yes.

  But then you can't assume that the back sight is 180 degrees out,
  either which is assumed by many programs.

  [For an extreme example, sit 100 feet south of the pole, Go 100 feet north
  turn right, and go 100 feet forward.  The foresight from the original point
  to the new point is NE, and the backsight is NW.]

  You need to know the latitude...  the difference (if I remember correctly)
  is a few seconds of arc here.

> In that coordinate system, a point 1/4 of the way east around the equator,
> you would still have north in the +y direction, but up would now be +x and
> east would be -z. This is a direct result of how we define directions on
> the globe.Refraction shouldn't play much part so long as the legs are
> short, since the refraction curvature will depend on the square of the leg
> length.
> 
> > > NOTE: because the direction "north" is actually a direction which is 
> > > perpendicular to the direction "down" rather than just the direction 
> > 
> >   I'm not sure I agree with the argument of this paragraph.
> >   But private email is probably more suited for a discussion of this.
> >
> > > parallel to the earth's axis (conventional or instantaneous) any errors in 
> > > measuring either the gravity direction or the axis direction will be 
> > > multiplied by tan(lattitude) when calculating "north". Ie a 1' deflection 

> As an example, lets imagine we're at a point 1m south of the north pole.  
> We only have to walk 3.14m east (or west) for the direction "north" to
> change by almost 180 degrees - remember walking east or west just takes us
> around a circle of radius ~1m centred on the pole. Or to show the role of
> the deflection of the vertical: Lets assume again we're very close to the
> north pole and we're attempting to determine the north direction by
> astronomical observations and we've already worked out which point in the
> sky is the north celestial pole and lets pretend there's a star there. If
> you want to walk north you want to "follow the star" but since the star is
> almost vertically above you its very hard to know which direction is north
> - anything which interferes with your idea of "vertical" will strongly
> affect your idea of north. If that still doesn't make sense then feel free
> to send me an email.

  It doesn't.  What you argue is a matter of measuring north, not truely
  a matter of what direction north is.  It is in the direction of that
  pole one meter away from you.

  There seems to be some confusion with actual angles vs. posible to measure
  angles in your discussion.

> [...]