Happy New Year, Frank,
How about a T-shaped dial consisting of a vertical east dial backed by a
vertical west dial and sharing a solid sloping roof whose edges act as the
gnomons. The hour lines close to noon could be marked as a horizontal dial
on the "floor" of the dial.
I think that this
I used to subscribe to the CALNDR-L mailing list but I gave it up because
I did not have time to consider the numerous fascinating postings. It seems
to be still available at:
http://myweb.ecu.edu/mccartyr/calndr-l.html
Geoff
2018-04-13 14:43 GMT+01:00 graham stapleton via sundial <
Dear Frank,
Can I help the POTUS? No.
Can I answer your question? Maybe but I am not used to working in imperial
units.
Consider a horizontal cross section of the gnomon at any height above the
dial plate. The shape of the section would be an ellipse with minor axis
equal to 0.5" aligned
> mathematical language of beauty. Was AUD $2. $2 to get you to purchase the
> 1st book. And then you may want to purchase the series of books. Of course
> they will not $2.
>
> Regards,
>
> Roderick Wall.
>
> - Reply message -----
> From: "Frank King" <f...
Frank,
I think the most elegant proof that the diagonal to side ratio in a
pentagon equals phi is as shown in the attachment.
Geoff
On 23 June 2017 at 08:08, Frank King wrote:
> Dear All,
>
> Referring to the Golden Ratio and Sundials, Donald
> Snyder wrote:
>
> I see
Just a guess, but might San 1790 be a misinterpretation of F.an 1790 (an
abbreviation of Fecit anno 1790). This would then match Mrs Gatty's
description pretty well.
Geoff
On 20 June 2017 at 10:53, Patrick Powers
wrote:
> Hello all,
>
> It’s always interesting to
Michael,
I seem to recall that sec^2(x)=1+tan^2(x)
Therefore sec^2(lat).sec^2(dec)=(1+tan^2(lat)).(1+tan^2(dec))
=1+tan^2(lat)+tan^2(dec)+tan^2(lat).tan^2(dec)
=(1+tan dec tan lat)^2 + (tan dec - tan lat)^2
I guess that this relationship, which is just a variant of sin^2+cos^2=1,
should have
Doug,
1. You could do what the Egyptians and the Greeks did 1500 years
earlier. Use a mural quadrant in the meridian to establish the greatest and
least altitude of the noon sun ( from this you could calculate the orbital
inclination and your latitude) and mark the mid altitude. You
Thank you, Frank, for that comprehensive analysis of the problem. However,
I wonder if the errors might be masked by the 32 arc minute solar penumbra.
Best wishes,
Geoff
On 19 January 2017 at 16:33, Frank King wrote:
> Dear John,
>
> I wondered when someone would spot that
Happy New Year to one and all,
The University Clock has gained 0.5 seconds against UTC in the past week
but is now 2.0 seconds ahead of UTC because of the leap second. If Frank
removed coins in accordance with the LWG for 0.5 seconds, then the clock's
rate would be rectified but the clock would
I like David's solution and I am in awe of his stone-cutting capability.
However, to return to your problem where the ellipse has been returned
without any markings, why not draw an ellipse of the same specified
dimensions on paper with the axes already marked on it? This could then be
matched to
Frank,
Thanks for the puzzle. I think that these instructions might work if you
ventured to the antarctic circle during the southern winter and then
trecked to a position such that your latitude is greater than 90-Dec (so
that the midnight sun is visible) but less than 90. The shadow of the stake
Thanks for the mental exercise, Frank. I think that Willy might squeeze
another 15 seconds of illumination at the end of December owing to the
rapid change in the EoT. A useful graph showing the rate of change of EoT
is provided by the USNO at:
http://aa.usno.navy.mil/faq/docs/eqtime.php
Geoff
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