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Frank Reed opened this thread on 30 May 2024 with “Suppose your sundial is wrong. Its gnomon or "shadow stick" is inclined at an incorrect angle, not equal to your latitude (for most common sundial designs). How bad would that be? What error would be introduced in the time reading from the sundial?”

Adrian F. used an undisclosed iteration method in addressing that question in his post of 11 June 2024. Geoff Hitchcox, in his post of 13 June 2024, provided a link to a paper by Fred Sawyer that derived an equation to calculate the error:

    Fred Sawyer in NASS Compendium Vol 1 No 4,  December 1994:
       T2 = atan( sin(T) / (cos(T) * cos(M) + tan(Dec) * sin(M) ) ), 
       where T is the true solar hour angle, M is the difference between the installed
       and design latitudes (here = 41.36° - 48.64° = -7.28°), Dec is solar declination,
       and T2 is the resulting hour angle.

       (For more on this, see [A] and [B] at the end of this post.)

Details for what follows are given in the attachment to this post.

I used Sawyer’s method to calculate errors for the two April 2024 photos that Frank posted, along with two other photos with date and time information that I found online; one from 14 September 2015 (shortly after the sphere had been refurbished), and the other from 10 July 2018.

I found local apparent time, L.A.T., values for the four photos and compared them to my estimates of the actual dial readings. These dial readings ranged from about four minutes less than, to about twelve minutes greater than, L.A.T.

I added corrections calculated by Sawyer’s equation to the dial readings and compared the corrected dial readings to L.A.T. The four corrected dial readings were between four and five  minutes fast with respect to L.A.T. These reasonably consistent errors, averaging 4.5 minutes fast, indicate there is an error in addition to the gnomon angle error in the construction or orientation of the Mystic Seaport armillary dial.

I added a fourth correction of -4.5 minutes (labeled “Add” in the attachment) to correct for this. With this, along with the corrections for longitude offset, equation of time, and gnomon angle error, the results for the four photos were within a half-minute of EST. As seen in the attachment, all this was done using times of hours and tenths of minutes. The attachment also includes results using times of hours and whole minutes. This method, which I personally feel is more appropriate given the 30‑minute marking interval on the dial equatorial ring, gave results that were within one minute of EST.

The question is what might be the source of an error averaging about 4.5 minutes fast over the range of hour angles and solar declinations that existed when this set of photographs was taken? One possibility is that if the set of time marks on the equatorial ring were rotated 1.125° clockwise, when viewed from the south, it would increase the displayed times by 4.5 minutes. Physically, what would 1.125° look like?

On 25 July I sent this query to the Mystic Seaport Museum: “At your museum there’s an armillary sphere sundial near the Treworgy Planetarium. Can someone tell me what the diameter of the sphere is?” Less than 45 minutes later I received this reply: “Thank you for the email. We just had a member of the planetarium team measure. It is approximately 42 inches in diameter.”  (A prompt thank you was sent to the museum folks for their help.)

Neglecting thicknesses of the steel structure, a 42 inch diameter translates to 5.5 inches between hour marks, and ~0.41 inch for 1.125°. I can’t tell from the available photos if an error of this magnitude exists on the Mystic sphere.

Chuck V.

[A]          

Sawyer wrote in his paper: "From the perspective of spherical trigonometry, this same result can be obtained by noting that our problem requires solving the triangle with vertices at the north pole, the sun's position, and a point on the meridian M degrees from the pole."

I used M = -7.28°, Latitude, L = 90 + M = 82.72° and Longitude, Lo = -71.964°, with dates and times (UT) for the four photos discussed above to verify his statement using an Excel embodiment of a Basic language sun data and sight reduction prgram I wrote in 1991:

Date                  14Sep15                10Jul18              26Apr24             28Apr24

Time               14:38:42 UT           14:39:00 UT       18:09:10 UT       20:35:30 UT

L °                     82.72                     82.72                 82.72                 82.72

Lo °                   -71.964                  -71.964              -71.964              -71.964

T (LHA°)            328.80                   326.49               20.91                 57.57

T (LHA - 360°)    -31.20                    -33.51                - - - - -                - - - - -

Z°                      148.37                   144.50               158.27               120.60

T2 (Z - 180) °      -31.63                    -35.51                - - - - -                - - - - -

T2 (180 - Z) °      - - - - -                    - - - - -                21.73                 59.40

[B]

I recently found a document from 18 Jan 2017 in which I’d derived this equation for azimuth angle, Z, by projecting the sight line of a cardboard ‘Astro Compass’ model I’d made onto a horizontal surface:

Z = atan (sin(T) / (cos(T) * sin(L)  - tan(Dec) * cos(L)))   [1]

Z is the same as T2 in Sawyer’s equation. Replacing it, and replacing L in [1] with
L = 90 + M as used in [A] above:

T2 = atan (sin(T) / (cos(T) * sin(90+M) - tan(Dec) * cos(90+M)))   [2]

which can be rewritten as:

T2 = atan (sin(T) / (cos(T) * cos(-M) - tan(Dec) * sin(-M)))   [3]

Both [2] and [3] are essentially Sawyer’s equation, and give the same results as Sawyer’s equation.

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