<p>The author is doing the following: <span class="math-container">$\log_{10} \tan(14^o 26') = -0.58943$</span> and then adding 10 to get to 9.41057. I didn't know about that practice, but it's akin to shifting the decimal point. Anyway, it is compensated for by the 'inverse sine log' ('sine ,,' two lines below).</p>
<p>Next line, <span class="math-container">$\log_{10} \tan(51^o 28') = 0.09888$</span> so that fits. (The operations are just in a different order than listed; first the tangent, then the logarithm.)</p>
<p>From the subsequent line, 9.50945, you're expected to take that value minus 10 (-0.49067), then the inverse of the base-10 logarithm (<span class="math-container">$10^x$</span>), so 0.32309, then the arcsine, which gives 18°51'.</p>
<p>The entire calculation seems to be a variation of the <a href="https://en.wikipedia.org/wiki/Sunrise_equation" rel="nofollow noreferrer">sunrise equation</a> reproduced below; the logarithms change multiplication into addition, and the sine and cosine are just a shift/flip (which happens in the next lines).</p>
<p><span class="math-container">$$\cos \omega _{\circ }=-\tan \phi \times \tan \delta$$</span></p>
<p>A proof that they're equivalent is left as an exercise for the reader.</p>
<p>If it seems weird to you, don't worry. I feel exactly the same. Please keep in mind that people used lookup tables rather than calculators for that kind of calculations at that time, so it shouldn't be such a surprise back then as it is now ...</p>

CC BY-SA 4.0
2022-08-07T14:32:16.363Z
From the OP: how did he arrive to 18deg 51 minutes?
Greg Miller
CC BY-SA 4.0
2022-08-07T14:34:02.990Z
@GregMiller see the 3rd paragraph?
Glorfindel
CC BY-SA 4.0
2022-08-07T18:23:39.160Z
Those of us who went to high school before the era of cheap calculators (~1975) spent a significant amount of time doing calculations using log tables and slide rules.
PM 2Ring
CC BY-SA 4.0
2022-08-06T17:17:37.280Z
2022-08-08T06:28:54.347Z
2022-08-08T06:28:54.347Z
Glorfindel
<p><a href="https://i.sstatic.net/vbTvN.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/vbTvN.png" alt="enter image description here" /></a></p>
<p>How did the author arrive to these numbers? How did he take the <span class="math-container">$\tan(\log(14 \deg 26 \text{ minutes}))$</span>? I tried many times but I am not reaching to his numbers.</p>

CC BY-SA 4.0
2022-08-06T15:22:57.780Z
Please do use links to link the sources you have questions about. Preferentially don't use screenshots to talk about texts and equations, copy&paste or type them. MathJax displays them better than the source you screenshoted. Images do not work with screen readers. Also that's an example, so this is likely explained above?
planetmaker
CC BY-SA 4.0
2022-08-06T15:32:05.090Z
The question is about math and interpreting equations from an (old?) book. Perhaps Math S.E. would be better. For example, is it really the tangent of a log? Is that log base 10? How can the sine of any number be 9.509? The math makes no sense.
JohnHoltz
CC BY-SA 4.0
2022-08-06T16:19:59.847Z
@planetmaker, due to the obvious non-standard notation, I think a screenshot is the best way to show this.  Any attempt to use standard math notation would certainly miss important notation.
Greg Miller
CC BY-SA 4.0
2022-08-06T17:03:38.230Z
Those are common base 10 logarithms. You do the trig function, then take the logarithm, eg $\log(\tan(14°26'))$. Also, he's adding $10$ to the log if the result is negative. But why do you want to use logarithms for these calculations? That method was standard many years ago, before electronic calculators. Are you just trying to check his calculations?
PM 2Ring
CC BY-SA 4.0
2022-08-06T17:13:53.403Z
Thanks......The book is from 1920 and it is explaing planetary hour table
hamzeh musmar
CC BY-SA 4.0
2022-08-06T17:23:39.037Z
But how would you do it in the modern method you are referring to?
hamzeh musmar
CC BY-SA 4.0
2022-08-06T17:42:03.983Z
how did he arrive to 18deg 51 minutes
hamzeh musmar
<sunrise>
Mean Sunrise Time
hamzeh musmar
# Mean Sunrise Time

(Asked by: username_0 on 2022-08-06)

[![enter image description here](https://i.sstatic.net/vbTvN.png)](https://i.sstatic.net/vbTvN.png)

How did the author arrive to these numbers? How did he take the $\tan(\log(14 \deg 26 \text{ minutes}))$? I tried many times but I am not reaching to his numbers.

**Tags:** sunrise
> **Comments:**
> **username_1**: Please do use links to link the sources you have questions about. Preferentially don't use screenshots to talk about texts and equations, copy&paste or type them. MathJax displays them better than the source you screenshoted. Images do not work with screen readers. Also that's an example, so this is likely explained above?  
> **username_2**: The question is about math and interpreting equations from an (old?) book. Perhaps Math S.E. would be better. For example, is it really the tangent of a log? Is that log base 10? How can the sine of any number be 9.509? The math makes no sense.  
> **username_3**: @planetmaker, due to the obvious non-standard notation, I think a screenshot is the best way to show this.  Any attempt to use standard math notation would certainly miss important notation.  
> **username_4**: Those are common base 10 logarithms. You do the trig function, then take the logarithm, eg $\log(\tan(14°26'))$. Also, he's adding $10$ to the log if the result is negative. But why do you want to use logarithms for these calculations? That method was standard many years ago, before electronic calculators. Are you just trying to check his calculations?  
> **username_0**: Thanks......The book is from 1920 and it is explaing planetary hour table  
> **username_0**: But how would you do it in the modern method you are referring to?  
> **username_0**: how did he arrive to 18deg 51 minutes  

---

## 1 answers

### Answer by username_5 on 2022-08-06. Score: 8 | Accepted answer

The author is doing the following: $\log\_{10} \tan(14^o 26') = -0.58943$ and then adding 10 to get to 9.41057. I didn't know about that practice, but it's akin to shifting the decimal point. Anyway, it is compensated for by the 'inverse sine log' ('sine ,,' two lines below).

Next line, $\log\_{10} \tan(51^o 28') = 0.09888$ so that fits. (The operations are just in a different order than listed; first the tangent, then the logarithm.)

From the subsequent line, 9.50945, you're expected to take that value minus 10 (-0.49067), then the inverse of the base-10 logarithm ($10^x$), so 0.32309, then the arcsine, which gives 18°51'.

The entire calculation seems to be a variation of the [sunrise equation](https://en.wikipedia.org/wiki/Sunrise_equation) reproduced below; the logarithms change multiplication into addition, and the sine and cosine are just a shift/flip (which happens in the next lines).

$$\cos \omega \_{\circ }=-\tan \phi \times \tan \delta$$

A proof that they're equivalent is left as an exercise for the reader.

If it seems weird to you, don't worry. I feel exactly the same. Please keep in mind that people used lookup tables rather than calculators for that kind of calculations at that time, so it shouldn't be such a surprise back then as it is now ...

> **Comments:**
> **username_3**: From the OP: how did he arrive to 18deg 51 minutes?  
> **username_5**: @GregMiller see the 3rd paragraph?  
> **username_4**: Those of us who went to high school before the era of cheap calculators (~1975) spent a significant amount of time doing calculations using log tables and slide rules.  

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