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This is from a letter sent around 1859:

notation

I think the curly braces are just to be treated like parentheses, so that part is fine (if that's not the case, do tell), but I struggle to understand what that dot represents. From context I would assume that it's actually a decimal point, but when it's mixed with the degree symbol in that manner inside the braces it almost looks more like a multiplication symbol (but that doesn't really make sense in the other expressions, like "1 dot 00", since in that case that term would just be 0). So, how exactly should I be reading this? Does "585° dot 26" actually mean 585.26°?

Outis Nemo
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  • (i) Who wrote the letter? (ii) That is presumably a transcription of the original hand-written letter. What did the letter itself look like? – TonyK Jun 25 '23 at 12:05
  • @TonyK: The original letter was written by astronomer and mathematician Johann Rudolf Wolf; and yes, you're right that it's a transcription, actually a translation, published in Monthly Notices of the Royal Astronomical Society in 1859. – Outis Nemo Jun 25 '23 at 12:09
  • Here is a link to the translation. Is there a link to a copy of the original letter? – Тyma Gaidash Jun 25 '23 at 12:26

3 Answers3

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The translation that Tyma Gaidash links to in the comments explains what these terms mean: they are the orbital positions of Venus, Earth, Jupiter, and Saturn at time $t$ years. So the value for a given planet is $360^\circ/P_\text{orb}$, where $P_\text{orb}$ is its orbital period in earth years.

The Royal Astronomical Society of Canada's Calgary Centre has this web page which gives the orbital periods of the planets in years to a very high precision (strangely, it gives the orbital period of Earth as $1.0000007$ years, which perhaps somebody can explain in the comments):

$$\begin{array}{c|c|c|} & P_\text{orb} & 360^\circ/P_\text{orb} \\ \hline \text{Venus} & 0.61517237 & 585.2018^\circ \\ \hline \text{Jupiter} & 11.8663142 & 30.3380^\circ \\ \hline \text{Saturn} & 29.47305083 & 12.2145^\circ \\ \hline \end{array}$$

This clearly shows that "585° dot 26" should be interpreted as $585.26^\circ$, and similarly for Jupiter and Saturn. The third column matches your quoted figures to better than one part in $2000$; if we interpreted the last two digits as arcminutes, the match would be no better than one part in $80$.

Updated to add: The Calgary Centre's Larry McNish was kind enough to reply to my query about that $1.0000007$:

It was the result of applying Kepler's third law to the published JPL values for the planetary orbits as they were back in 2009.

I've attached the spreadsheet (which had a JPL value of the semi-major axis for the Earth's orbit as 1.00000018 a.u.)

Either because I used too many digits of precision in the answer or because it was measured as sidereal years.

365.256622 / 1.0000007 = 365.25636632054357561949706635205 which is very close to the number of days in a sidereal year https://en.wikipedia.org/wiki/Earth%27s_orbit 365.256363004 days[13]

TonyK
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  • Thanks; I guess I should've bothered to actually check the exact orbital periods myself, since I already knew where it was from and what the terms represented, I was just very confused by that notation. As for why the orbital period of Earth is not exactly 1 year, it's because the astronomical definition of a year without further qualifiers refers to a tropical year, i.e. the time for Sol's longitude to increase 360°, which is slightly shorter than the orbital period due to general precession. I'm not sure how they got that exact number, though, as it should be ~1.00004 years. – Outis Nemo Jun 26 '23 at 18:19
  • @OutisNemo: I looked up the figures for the Earth's perihelion precession, and it seemed about a factor of ten too large to agree with $1.0000007$. So I'm still puzzled. But I e-mailed them, and if I get a reply, I'll update here. – TonyK Jun 26 '23 at 18:49
  • It's not the perihelion precession that yields the tropical year, but the general precession (i.e. equatorial precession and ecliptic precession). They could be using the anomalistic year (i.e. the year based on the perihelion precession) instead of the sidereal year for the orbital period, but that would be odd given that it's not the astronomical definition of the orbital period at all, and still you'd only get ~1.00005 instead of ~1.00004, two orders of magnitude off in either case. Not sure what you did to only get a single order of magnitude off. – Outis Nemo Jun 27 '23 at 06:06
  • @OutisNemo: this page gives the earth's observed perihelion precession as $11.45$ arcseconds per year; the quoted figure of $1.0000007$ corresponds to $0.0000007\times 360\times 60\times 60=0.9072$ arcseconds per year. (Admittedly, I don't know what I'm doing here.) – TonyK Jun 27 '23 at 10:53
  • As I mentioned, the perihelion precession is unlikely to be relevant here; that's primarily relevant for what's called an anomalistic year, i.e. the period from perihelion to perihelion. What you do is divide the sidereal year (aka. orbital period) by the tropical year (which is what is referred to as a year when the word "year" is used without qualifiers). This yields ~1.00004. That is however not ~1.0000007, so I also wonder how exactly they arrive at that number. Maybe they're using outdated estimates for those periods. If they answer you, let me know. – Outis Nemo Jun 28 '23 at 11:21
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    @OutisNemo: They answered, as I reported yesterday $-$ see my edit. But it didn't really help me. – TonyK Jun 28 '23 at 14:25
  • I see, looks like they're comparing automatically generated values for the orbital period (sidereal year) based on the ephemeris data and comparing this against a set value for the orbital period, yielding a small mismatch due to the difference in precision only. That explains it, although in that case it doesn't have anything to do with the difference between the orbital period and the tropical year; that's a bit strange given how a year without qualifiers should refer to a tropical year, but in this context they're evidently using it to refer to a sidereal year (orbital period). – Outis Nemo Jun 29 '23 at 15:16
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It's a smaller form of the unit( ° ) known as minutes, 1°= `60 or 60 minutes, there's also another unit known as seconds, 60 seconds (`` 60 ) = 1 minute.

Just to clarify, it isn't related to the unit of time.

Creamy Dice
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    But it's not a back-tick, it's a dot, just like the decimal points in 50$\cdot$31 etc. – TonyK Jun 25 '23 at 12:13
  • @TonyK it's clearly at the top of the character though so it doesn't make sense as a period or center dot. I think ' is the only interpretation that make sense in context. – CyclotomicField Jun 25 '23 at 12:14
  • That doesn't make sense at all in the context of the expression; and it's clearly not the symbol used for arc-minutes either. – Outis Nemo Jun 25 '23 at 12:16
  • @CyclotomicField: I am not an expert on decimal point placement in the mid-19th century. Are you? In any case, it's the only reasonable explanation. – TonyK Jun 25 '23 at 16:05
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This is a decimal place as the $3.73$ excludes arcminutes.

CyclotomicField
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    That doesn't make sense at all in the context of the expression; and it's clearly not the symbol used for arc-minutes either. – Outis Nemo Jun 25 '23 at 12:16
  • @OutisNemo you're right, the .76 doesn't make sense. It's got to be a decimal place. – CyclotomicField Jun 25 '23 at 12:18
  • Yes, that much I've figured; but then the problem is what e.g. "585° dot 26" means, as that seems like a strange way to notate "585.26°", but that's my best guess so far. I'm just looking for someone who perhaps has seen it notated this way before and can confirm. – Outis Nemo Jun 25 '23 at 12:24
  • @OutisNemo The 360° is odd when it's also 0° so perhaps it literally means 585 but it's possibly a typo. The repeated 5 makes me wonder. – CyclotomicField Jun 25 '23 at 12:32
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    @CyclotomicField: the expression is $360^\circ t$, not $360^\circ$. This is not the same as $0^\circ t$! It is the (approximate) position of the Earth in its orbit at time $t$ years, as the linked paper says. – TonyK Jun 25 '23 at 16:10
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    I don't see any $3.76$. Did you mean $3.73$? But that doesn't include the degree sign, so it's not relevant to the question of whether these figures denote arcminutes or hundredths. – TonyK Jun 25 '23 at 16:14