<blockquote>
  <p><strong>How would you interpret this?</strong> "(1780 + 80) * 3 on base 808, and add 2"</p>
</blockquote>

<p>As before, there is a confusing avoidance of specifying the unit under consideration, but the pattern is still that the last number is in minutes of arc, the middle in degrees, and the first in $30^\circ$ for the Zodiac divisions, so:
$$\begin{eqnarray*}
\frac{(1780+80)\times 3}{808} &amp;=&amp; 6+\frac{27}{30} + \frac{10}{30\times 60} + \frac{41}{30\times 60^2}+\ldots \\
&amp;=&amp; (6;\;27\!:\,10)\text{.}\end{eqnarray*}$$
The $+2$ makes this $(6;\;27\!:\,12)$, although this makes the second addend in minutes of arc while the first one is not. As before, the calculation is not rounded.</p>

<blockquote>
  <p>Eade has <code>(8; 11 : 7) - (6; 27 : 12) = (1; 3 : 55)</code>, but that doesn't work.</p>
  
  <p>That specific substr. gives me <code>(1; 13 : 55)</code>, which could be a typo in the paper, but that value doesn't carry things forward either.</p>
</blockquote>

<p>I think that you are correct that this is a typographical error, but the corrected value probably <em>is</em> carried forward--the error is just repeated again in Step 16. Note that $(1;\;13\!:\,55) = 43^\circ 55'$, which makes the corrected Step 16
$$296\sin(1;\;13\!:\,55) = 205.309 = (0;\;3\!:25)\text{,}$$
following the convention that the numerical value is re-interpreted as minutes of arc. On the other hand, if we follow what Step 16 actually says, we get a very different value:
$$296\sin(1;\;3\!:\,55) = 165.164 = (0;\;2\!:45)\text{.}$$</p>

<p>Granted, the value actually carried forward to Step 17 is $(0;\;3\!:24)$ instead, but the former is definitely much closer. This further discrepancy could also be an undocumented geographical correction (the procedure does, after all, have things like $+3$ and $+2$ at times), but I'm more inclined to guess that this is due to imprecision in the calculation of sine, as the paper indicates that a table was used.</p>

CC BY-SA 3.0
2015-10-08T15:33:12.357Z
2015-10-08T20:27:11.567Z
2015-10-08T20:27:11.567Z
Stan Liou
<p>This is a follow-up to <a href="https://astronomy.stackexchange.com/q/11753/9153">How to interpret this old degree notation?</a></p>

<p>It is about an old calendar system which generates a luni-solar calendar with a procedural method, using a handful of calculations and some constants derived from astronomical observations.</p>

<p>This method originates from ancient India. Today it is used in Thailand to determine public holidays and observance days (<em>uposatha</em> days) for Buddhist monks, which fall on New- and Full Moon days.</p>

<p>Actually it's a neat system that had allowed village "astronomers / astrologers" construct a calendar with a method they could memorize by rote learning, without astronomical tools to make precise measurements. It is dreadfully underdocumented, and I've been collating the method and it's practices in <a href="https://github.com/profound-labs/calculating-the-uposatha-moondays" rel="nofollow noreferrer">Calculating the Uposatha Moondays</a>.</p>

<p>Most useful have been the actual formulas in <a href="http://www.siamese-heritage.org/jsspdf/1991/JSS_088_0r_Eade_RulesForInterpolationInThaiCalendar.pdf" rel="nofollow noreferrer">Rules for Interpolation in the Thai Calendar</a> by J.C. Eade, it is just some basic arithmetic, but following his notation is a puzzle in itself.</p>

<p>Effectively I'm trying to implement these formulas (see image below), in a <a href="https://github.com/splendidmoons/suriya-go/blob/master/suriya.go" rel="nofollow noreferrer">golang package</a>.</p>

<p>With the earlier help, I arrived at <strong>step 13 (Mean Moon)</strong>. Hooray :)</p>

<p>Step <strong>14.</strong> and <strong>15.</strong> is a puzzle again.</p>

<h2>Step 14.</h2>

<p><strong>How would you interpret this?</strong> "(1780 + 80) * 3 on base 808, and add 2"</p>

<h2>Step 15.</h2>

<p>Eade has <code>(8; 11 : 7) - (6; 27 : 12) = (1; 3 : 55)</code>, but that doesn't work.</p>

<p>That specific substr. gives me <code>(1; 13 : 55)</code>, which could be a typo in the paper, but that value doesn't carry things forward either.</p>

<p>Things straighten up at step <strong>16., 17., 18.</strong>, which would produce the paper's results, to get there I faked the values for step 14. and 15, to work at least in the example case given in the paper.</p>

<p>In <a href="https://github.com/splendidmoons/suriya-go/blob/master/suriya.go" rel="nofollow noreferrer">suriya.go</a>
this business is happening in:</p>

<pre><code>func (suDay *SuriyaDay) Init(ce_year int, lunar_year_day int)
</code></pre>

<p><a href="https://i.sstatic.net/bMGRW.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/bMGRW.png" alt="JC Eade, Rules for Interpretation"></a></p>

<notation>
From Mean Moon to True Moon in an old procedural calendar
Gambhiro
# From Mean Moon to True Moon in an old procedural calendar

(Asked by: username_0 on 2015-10-08)

This is a follow-up to [How to interpret this old degree notation?](https://astronomy.stackexchange.com/q/11753/9153)

It is about an old calendar system which generates a luni-solar calendar with a procedural method, using a handful of calculations and some constants derived from astronomical observations.

This method originates from ancient India. Today it is used in Thailand to determine public holidays and observance days (*uposatha* days) for Buddhist monks, which fall on New- and Full Moon days.

Actually it's a neat system that had allowed village "astronomers / astrologers" construct a calendar with a method they could memorize by rote learning, without astronomical tools to make precise measurements. It is dreadfully underdocumented, and I've been collating the method and it's practices in [Calculating the Uposatha Moondays](https://github.com/profound-labs/calculating-the-uposatha-moondays).

Most useful have been the actual formulas in [Rules for Interpolation in the Thai Calendar](http://www.siamese-heritage.org/jsspdf/1991/JSS_088_0r_Eade_RulesForInterpolationInThaiCalendar.pdf) by J.C. Eade, it is just some basic arithmetic, but following his notation is a puzzle in itself.

Effectively I'm trying to implement these formulas (see image below), in a [golang package](https://github.com/splendidmoons/suriya-go/blob/master/suriya.go).

With the earlier help, I arrived at **step 13 (Mean Moon)**. Hooray :)

Step **14.** and **15.** is a puzzle again.

## Step 14.

**How would you interpret this?** "(1780 + 80) \* 3 on base 808, and add 2"

## Step 15.

Eade has `(8; 11 : 7) - (6; 27 : 12) = (1; 3 : 55)`, but that doesn't work.

That specific substr. gives me `(1; 13 : 55)`, which could be a typo in the paper, but that value doesn't carry things forward either.

Things straighten up at step **16., 17., 18.**, which would produce the paper's results, to get there I faked the values for step 14. and 15, to work at least in the example case given in the paper.

In [suriya.go](https://github.com/splendidmoons/suriya-go/blob/master/suriya.go)
this business is happening in:

```
func (suDay *SuriyaDay) Init(ce_year int, lunar_year_day int)

```

[![JC Eade, Rules for Interpretation](https://i.sstatic.net/bMGRW.png)](https://i.sstatic.net/bMGRW.png)

**Tags:** notation


---

## 1 answers

### Answer by username_1 on 2015-10-08. Score: 1 | Accepted answer

> **How would you interpret this?** "(1780 + 80) \* 3 on base 808, and add 2"

As before, there is a confusing avoidance of specifying the unit under consideration, but the pattern is still that the last number is in minutes of arc, the middle in degrees, and the first in $30^\circ$ for the Zodiac divisions, so:
$$\begin{eqnarray\*}
\frac{(1780+80)\times 3}{808} &=& 6+\frac{27}{30} + \frac{10}{30\times 60} + \frac{41}{30\times 60^2}+\ldots \\
&=& (6;\;27\!:\,10)\text{.}\end{eqnarray\*}$$
The $+2$ makes this $(6;\;27\!:\,12)$, although this makes the second addend in minutes of arc while the first one is not. As before, the calculation is not rounded.

> Eade has `(8; 11 : 7) - (6; 27 : 12) = (1; 3 : 55)`, but that doesn't work.
>
> That specific substr. gives me `(1; 13 : 55)`, which could be a typo in the paper, but that value doesn't carry things forward either.

I think that you are correct that this is a typographical error, but the corrected value probably *is* carried forward--the error is just repeated again in Step 16. Note that $(1;\;13\!:\,55) = 43^\circ 55'$, which makes the corrected Step 16
$$296\sin(1;\;13\!:\,55) = 205.309 = (0;\;3\!:25)\text{,}$$
following the convention that the numerical value is re-interpreted as minutes of arc. On the other hand, if we follow what Step 16 actually says, we get a very different value:
$$296\sin(1;\;3\!:\,55) = 165.164 = (0;\;2\!:45)\text{.}$$

Granted, the value actually carried forward to Step 17 is $(0;\;3\!:24)$ instead, but the former is definitely much closer. This further discrepancy could also be an undocumented geographical correction (the procedure does, after all, have things like $+3$ and $+2$ at times), but I'm more inclined to guess that this is due to imprecision in the calculation of sine, as the paper indicates that a table was used.


---

