Hello
all,
I've been
thinking.
For every plane dial
of whatever orientation in a given location, there is another location on the
planet where a plane horizontal dial has the same orientation with respect to
the sun and the Earth's aixs. Sundials don't care about latitude and longitude
or about a wall's declination or inclination, only about where the sun
is.
So if we can
determine this "position of horizontal equivalence" we can calculate the
parameters of a simple horizontal dial which will show local apparent time at
our original site. We only need the longitude correction.
Here's a way to do
it. Picture a spherical triangle with angles P, O and E. P is the Pole, O is the
original site, E is the equivalent site. The sides of the triangle are p,o and
e, each side opposite its corresponding angle. We know e: it's the original
site's co-latitude; we know O: it's the azimuth of the direction in which the
surface at the original site faces; and we know p: it's the inclination from the
horizontal of the surface at the original site. Picture yourself seizing the
dial plate of your dial at the original location and, all the time preserving
its orientation, rushing off along the great circle in the direction in which it
faces until it becomes horizontal. You will be at the position of horizontal
equivalence.
With two sides and
an included angle we can solve all the other sides and angles of the spherical
triangle. Specifically, o is the co-latitude of the equivalent site; P is the
difference in longitude; and E is also important ;-)
Having built your
horizontal plane dial at location E, corrected for longitude according to
P and nicely rectangular with sides oriented NS/EW, you pick it up and
travel back to the original site, O, all the time preserving its orientation
with respect to the sun and the Earth's axis. When you get back it fits
perfectly on the surface for which you planned it but it's not oriented
vertically. It's at an angle, the angle E.
In fact, all of this
is implied in Waugh. Chapter 5 ends with a section on using a wedge to position
a horizontal dial designed for one latitude at a different latitude. I'm reading
that as, "A horizontal plane dial will work anywhere if you orient it right".
And in chapter 10 on vertical declining dials he instructs us to calculate the
quantity DL, "the difference of longitude" but he doesn't explain it. It's
exactly equal to our angle P, the difference of longitude. He also has us
calculate the "style distance", the angle through which to turn the gnomon from
the vertical. This is angle E.
The advantage of the
spherical trignometry solution is, first, I find it easier to understand. Images
of little men rushing about the planet are easier for me than raw algebra. And,
two, we only have to plug in the walls inclination and we have a universal
solution. Waugh can only recommend repeated observation as a way to construct a
dial that both "reclines and declines". I'm not criticising Waugh; he's the one
who set me off on dialling so I'm pretty protective of him.
A spreadsheet to do
the calculations above is astoundingly simple - six lines to do the basics. I
can supply one, somewhat raw, if anyone's interested. If all this has been
published before then I'm sorry to have put you thorugh it again though I have
to say I haven't found it anywhere.
best regards to all,
Richard
