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Problem

I'm trying to complete an assignment for my astronomy course, but I'm having trouble understanding how accuracy is determined when converting equatorial coordinates in sexagesimal format to degrees. The problem text is below:

Two stars have the following equatorial coordinates:
$(\alpha,\delta)_1=(12:30:00,+14:30:00)$ and $(\alpha,\delta)_2=(17:30:00,+54:00:00)$.
Use the formula in Appendix D3 to calculate the wide-field angular separation $\theta_{1,2}$ between these two stars in decimal degrees with appropriate accuracy. Compare your answer to the separation in degrees you would expect if the coordinates were on a flat, Euclidean surface rather than the curved celestial sphere surface.

I was able to do the conversion and the calculation with the following results:

Spherical calculation: $\theta_{1,2}=69.52210417º$
Flat Euclidean calculation: $\theta_{flat}=84.76585397º$

However, my professor also wants us to round the answers using proper accuracy or "significant figures." Generally, I know how to use sig figs for normal number systems, but I'm super confused about how it works when starting from a sexagesimal format.

Professor's Example

When my professor was doing an example conversion in class, we had RA: 05h 15min 45.7s and Dec: -08:10:26.4.

For RA, we got the conversion: $$ \left( 5\text{h}*\frac{15º}{1\text{h}}\right) + \left( 15\text{min} * \frac{1\text{h}}{60\text{min}} * \frac{15º}{1\text{h}} \right) + \left( 45.7\text{s} * \frac{1\text{h}}{3600\text{s}} * \frac{15º}{1\text{h}} \right) = 78.9404º $$ Then, for the rounding, he said that 45.7s --> 0.190º (from our conversion of secs-->deg, which gave us 0.190416667º), so our final answer (with accuracy) was RA = 78.940º.

For Dec, we got the conversion: $$ -\left[ 8º + \left( 10' * \frac{1º}{60'} \right) + \left( 26.4'' * \frac{1º}{3600''} \right) \right] = -8.17433º $$ Then, for the rounding, he said that 26.4'' --> 0.00733º (from our conversion of arcsecs-->deg, which gave us 0.007333333º), so our final answer (with accuracy) was Dec = -8.17433º.

Summary

All that being said, I'm super confused about how he got the "proper accuracy" in his example and how that would translate to the problem that I'm trying to complete. I know it has something to do with the smallest measure of precision, but when all the coordinates are 0, then how would you know what precision to use, and what precision should my answer be in?

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3 Answers 3

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The point is if seconds are expressed as a decimal (e.g., $05^{\circ }15^{\prime }45.7^{\prime \prime }$), the decimal part contributes to higher precision, similar to trailing zeros after a decimal point in base-10. When your converting from sexagesimal to decimal, the number of significant figures should be preserved by matching the precision of the original, smallest unit (seconds) to the corresponding decimal place.

Thus, your professor's example is exactly correct.

$45.7^{\prime \prime}$ has 3 sig-figs. Hence when you convert to decimal degrees, you should preserve it into 3 sig-figs too ($0.19041666...$ become $0.190$. zeros after of between sig-figs is considered as sig-figs). Decimal place is kept three digits after decimal point. In the case of $26.4^{\prime \prime}$ which has the same sig-figs with before; if we convert to decimal degrees, even it has 3 sig-figs ($0.0073333...$ become $0.00733$), but it has five digits after decimal point (zeros before sig-fig is not considered as sig-fig). Hence, for declination, it should be trailed into five digits after decimal point (compare with right ascension RA which rounded into three digits after decimal point only because when the smallest unit (which is arcseconds) converted into decimal degrees, it matched between sig-figs and decimal place [three sig-figs and three digits after decimal point]).

Another case: $1.8^{\prime}$ and $5.4^{\prime}$ have 2 sig-figs. They should be: $1.8^{\prime} = 0.030$ or $5.4^{\prime} = 0.090$. Even $1.8^{\prime} = 0.03$ (exactly) and $5.4^{\prime} = 0.09$ (exactly), they have 1 sig-fig only (zero before 3 and 9 respectively is not considered as sig-fig). Thus, matching with sig-figs in smallest unit (arcminutes), you should add zero after the sig-fig. Hence, $0.03$ become $0.030$ and $0.09$ become $0.090$ respectively. When you convert arcminutes into decimal degrees, you should keep three digits after decimal point.

For the problem you asked previously, because it has no decimal point in the smallest unit (which is seconds of hour in RA and arcseconds in declination), zeros at the end of a value (which is 00 in arcsecond or second of hour) indicate precision. Or simply, you can ignore $^h$, $^m$, $^s$, $^{\circ}$, $^{\prime}$, and $^{\prime \prime}$ from the digits and concatenate them all together

Case one:
$14^{\circ} 30^{\prime} 00^{\prime \prime}$
it become 143000. Hence, it has 6 sig-figs.
(three zeros after 3 are sig-figs)
smallest unit / precision: arcsecond

Case two:
$54^{\circ} 00^{\prime} 00^{\prime \prime}$
it become 540000. Hence, it has 6 sig-figs.
(four zeros after 4 are sig-figs)
smallest unit / precision: arcsecond
two zeros in arcminute is also significant because two zeros in arcsecond is already significant

Case three:
$104^{\circ} 03^{\prime} 00^{\prime \prime}$
it become 1040300. Hence, it has 7 sig-figs.
(two zeros after 3 and zero between 1 and 4 are sig-figs)
smallest unit / precision: arcsecond

Case four:
$205^{\circ} 20^{\prime}$
it become 20520. Hence, it has 5 sig-figs.
(one zero after 2 and zero between 2 and 5 are sig-figs)
smallest unit / precision: arcsecond

Case five:
$45^{\circ} 00^{\prime}$
it become 4500. Hence, it has 4 sig-figs.
(two zeros after 5 are sig-figs)
smallest unit / precision: arcminute

Case six:
$07^{\circ} 00^{\prime}$
it become 0700. Hence, it has 3 sig-figs.
(two zeros after 7 are sig-figs, and zero before 7 is not sig-fig)
smallest unit / precision: arcminute

Case seven:
$09^{\circ} 00^{\prime} 00^{\prime \prime}$
it become 090000. Hence, it has 5 sig-figs.
(four zeros after 9 are sig-figs, and zero before 9 is not sig-fig)
smallest unit / precision: arcsecond two zeros in arcminute is also significant because two zeros in arcsecond is already significant

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  • $\begingroup$ Thank you for your response; that makes more sense. How would I consider sig figs when I have all 0s in the seconds and arcseconds, like in my original problem? $\endgroup$ Commented Jan 28 at 22:16
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    $\begingroup$ @Atlas I've been edited my answer. please check it out $\endgroup$ Commented Jan 28 at 22:44
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    $\begingroup$ unless for the zeros before significant number is not considered as significant. either for the case $09^{\circ}00^{\prime}$ or $07^{\circ}00^{\prime}00^{\prime \prime}$, 0 after 9 or 7 are not sig-fig. the rest digits are significant. two zeros indicate precision either the smallest unit is in arcminute/minute of hour of arcsecond/second of hour. $\endgroup$ Commented Jan 28 at 22:59
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    $\begingroup$ if two zeros in arcsecond is significant, thus two zeros in arcminutes or any digits before the digits of arcsecond is also significant. it also confirmed the rule of zero[s] between significant is also significant @Atlas $\endgroup$ Commented Jan 28 at 23:12
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    $\begingroup$ Ah, that makes sense. Thank you for your help! $\endgroup$ Commented Jan 28 at 23:13
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I believe the RA should go to 4 places after the decimal point. Taking the professors conversion:

45.7 s (15 deg / 3600 s) = .19041666...

Try converting .190 to seconds: .190 (3600s / 15 deg) = 45.6 (not 45.7)

Now try .1904:

.1904 (3600/15) = 45.696, which rounds to 45.7 (the accuracy of the instrument or table of values).

For Dec, 26.4 s (1 deg/3600 s) = .0073333 deg.

Then use .0733, which converts to 26.388 and round off.

Using .073 converts to 26.3 (not 26.4).

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    $\begingroup$ I agree that rounding to 3 sig fig as 0.190 is slightly too crude and corrupts the original value on a round trip. Your answer 0.1904 is indeed the closest to the target value expressed as decimal degrees. However, rounding degrees to the nearest multiple of 0.0005 might be considered more nearly in the spirit of expressing the value with appropriate precision (so that 0.1 arcsecond quantisation is preserved). $\endgroup$ Commented 2 days ago
  • $\begingroup$ thus, for precision, we cannot be fixated with sig fig rule which corrupts the original value when we compute, because it will accumulate numerical error. instead, we compute all formula with sufficient precision, then we trill the zero as sig fig should be required @MartinBrown $\endgroup$ Commented yesterday
  • $\begingroup$ even 45.5; 45.6 and 45.7 has the same value if we trail zero based on sig-fig: 0.190; but for 26.4" it has decimal degrees of 0.0733; while 26.3" and 26.5" has decimal degrees of 0.0731 and 0.0736 respectively. @MartinBrown $\endgroup$ Commented yesterday
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What precision should my answer be in?

Pragmatically, since the professor has expressed a method, follow it for the class.

how would you know what precision to use, (?)

I also read some fuzziness in <how he got the "proper accuracy">?

Should you care to dig deeper ...

Consider following angle formats where d is a decimal digit and s is a digit 0-5

ddd.dddddº
dddºsd′sd.d′′
dddhsdmsd.ds

All have 8 digits, but they do not all have 8 decimal digits.

Further, the range for each to represent 1 revolution differs:

360.00000º
360º00′00.0′′
24h00m00.0s

Let us compare the arc (in revolutions) with a step of the smallest non-zero arc value.

Revolution fraction.. Format/value
0.0000002778............ 000.0001º
0.0000000278............ 000.00001º
0.0000000772............ 000º00′00.1′′
0.0000011574............ 00h00m00.1s

The ddhsdmsd.ds form has the largest arc steps and
ddd.dddddº has the smallest.

In general, when converting from one form to another, the arc step of the destination should be just smaller (or as small) as the arc step of the source. This almost matches the prof's guidance of
ddhsdmsd.ds to ddd.dddº (I'd say 1 digit more needed) and
dddºsd′sd.d′′ to ddd.dddddº.

As we can add additional decimal digits of info to the format, the new format's arc step becomes 10 times smaller. Yet up to how many should we add, if any? [More on this later, need to review.]

All 3 below round to the same result. 078.940º converts back to 05h15m45.6s. We have failed to round-trip two of three cases - thus a loss of information.

05h15m45.5s to 078.940º
05h15m45.6s to 078.940º
05h15m45.7s to 078.940º

Had we used another digit in the destination, the round trip would succeeded.

05h15m45.5s to 078.9396º
05h15m45.6s to 078.9400º
05h15m45.7s to 078.9404º

  • Conclusion

To use proper accuracy, use enough digits in the output format with just smaller (or same) destination arc steps than the source.

If we used a destination format with larger arc steps, then for a complete revolution, there will be more source steps than destination steps. Given the Pigeonhole principle), we have lost information.

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