Dear Willy,
That is an interesting sundial...
> I made the calculation for the hour lines and datelines
> for a sundial ... on a concave cylindrical wall...
When I first heard about this project, I imagined that the
nodus would be hidden in shadow for most of the day by the
wings of the wall.
Now I see the finished sundial, all is clear. The top of
the wall appears to be your horizon line. Is this right?
> I am in search of other sundials on a concave cylindrical
> wall.
I do not know of any other examples but, in my experience,
there is no such thing as a flat wall. Every stone wall
has undulations, some of which are concave and some are
convex! On a big dial you have to allow for this in the
calculations...
I am interested to know whether you assumed that your
surface was a perfect mathematical cylinder? Did you
survey the wall carefully to see where it goes in and
out?
With a big wall, which is supposed to be flat, I use the
following procedure:
1. Note the latitude.
2. Survey the wall using, say, a 500mm grid.
3. Determine the best-fit *vertical* plane.
4. Determine the declination of the best-fit
vertical plane.
5. For each intersection point on the 500mm grid,
determine how far it is behind or in front of
the best-fit vertical plane.
6. Determine the perpendicular distance of the
centre of the nodus from the best-fit vertical
plane (the ortho-style distance).
7. For each feature (e.g. constant-declination
line) calculate the positions of a number of
points assuming the ortho-style distance is
constant.
8. For each calculated point, estimate the offset
from the best-fit vertical plane and add or
subtract this from the ortho-style distance.
Then recalcualte the point. Repeat as necessary.
Of course, what I am describing is an iterative
procedure and I certainly claim no originality
for it! Essentially, I am assuming that the
wall consists of lots of parallel planes, each
with its own ortho-style distance.
I developed this procedure for a wall where the
undulations were of the order of +/- 10mm over an
area 10m x 4m. Some of the deviation was accounted
for just by the wall leaning over (9mm in 10m).
In fact, the procedure works perfectly wall if the
wall is any shape at all. Since a cylinder is easy
to describe mathematically it works well for that.
Even your wall has a best-fit vertical plane! It
is just that the ortho-style distance varies rather
a lot from the mean!
What I am interested to know, is:
1. `How close to a true cylinder is your wall?'
2. `How did you allow for the inevitable
undulations?
All the best
Frank King
Cambridge, U.K.
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