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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.

  • $\begingroup$ 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? $\endgroup$ Aug 6 at 15:22
  • $\begingroup$ 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. $\endgroup$
    – JohnHoltz
    Aug 6 at 15:32
  • $\begingroup$ @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. $\endgroup$ Aug 6 at 16:19
  • $\begingroup$ 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? $\endgroup$
    – PM 2Ring
    Aug 6 at 17:03
  • 1
    $\begingroup$ how did he arrive to 18deg 51 minutes $\endgroup$ Aug 6 at 17:42

1 Answer 1


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 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 ...

  • $\begingroup$ From the OP: how did he arrive to 18deg 51 minutes? $\endgroup$ Aug 7 at 14:32
  • 2
    $\begingroup$ @GregMiller see the 3rd paragraph? $\endgroup$
    – Glorfindel
    Aug 7 at 14:34
  • $\begingroup$ 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. $\endgroup$
    – PM 2Ring
    Aug 7 at 18:23

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