ON THE AGREEMENT OF THE LAST TWO ECLIPSES OF THE SUN AND MOON WITH MY TABLES, FOR FINDING THE ACTUAL TIMES OF THE HALF-MOON AND NEW MOON

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1 Sur l'accord des deu dernieres eclipses du soleil et de la lune avec mes tables, pour trouver les vrais momens des pleni-lunes et novi-lunes Synopsis of Leonard Euler s E4 Paper Andrew Fabian and Hieu D. Nguyen March 6, 008 ON THE AGREEMENT OF THE LAST TWO ECLIPSES OF THE SUN AND MOON WITH MY TABLES, FOR FINDING THE ACTUAL TIMES OF THE HALF-MOON AND NEW MOON BY MR. EULER In this 750 paper Leonard Euler comments on his predictions from a previous Memoir [E7] regarding the times and durations of two eclipses that occurred in 748: the solar eclipse on July 5 and the lunar eclipse between August 8 and 9. In the first section Euler discusses his accurate prediction for the solar eclipse. In particular his calculations for the beginning and end times of the solar eclipse are in close agreement with observation, but not for the duration of the annulus. He corrects his error for this duration by using a more accurate measurement for the latitude of Berlin recently obtained by Maupertuis. In the second section Euler discusses his somewhat inaccurate prediction for the lunar eclipse. He is surprised by this, especially since he had updated his astronomical tables for many other lunar eclipses, and devotes the remainder of the paper section III-XV towards reworking his calculations to obtain values that are in much closer agreement with observation. I. In this section Euler discusses the accuracy of his prediction for the solar eclipse on July 5, 748 based on observation. He points out that his time predictions for the beginning and end of the eclipse were accurate, but not the duration of the annulus, in comparison with the observed times, as seen in the table below: Solar Eclipse July 5, 748 Euler s Prediction Observation Difference Beginning Time 0 h, 7, 45 < 0 h, 8-5 Ending Time h, 4, 0 h, 4, Duration Annulus 5, 0, Euler then eplains the source of his error for the duration of the annulus as stemming from the latitude of Berlin, which he places too far north by 4, 30. He argues that if the more recent observed calculation for the latitude of Berlin were used instead 5, 3, 30, then he would have obtained a much more accurate prediction.

2 II. In this section Euler mentions that his predictions for the lunar eclipse on December 8-9, 748 are not in agreement with observation. His Astronomical Almanac had predicted that it begin at h, 0, 4 :00 PM on December 8 and end at h, 4, 4 :4 AM on December 9, yet it was observed to begin at h, 5 :05 PM and ended at h, 8 :8 AM. He is surprised by this since he had earlier corrected his astronomical tables for many earlier lunar eclipses. Euler therefore devotes the rest of the paper in reworking his calculations to see if there were any errors in his original calculations. In the end he obtains a more accurate prediction in hindsight but does not point out any of these errors if any and so it is unclear where his calculations were improved upon. III - VIII. In these sections Euler provides tables of astronomical values relating to the positions and movements of the Sun and Moon that will be used in his calculations in later sections. IX. Finding a relation for UL: To the right is a more complete figure of the problem. The known values are denoted by a, n, m, and ω. Euler proceeds to find relations for the distances UL and z and the elapsed time in terms of these known values. To obtain a relation for UL at the moment of opposition, Euler begins by applying the Law of Cosines for spherical trigonometry, which states that in a spherical triangle with sides a, b, c and an angle C opposite to side c, we have cos c cos acos b sin asin bcos C. This is applied to the spherical triangle UL in the figure, which gives the relation cosul cos sin a cos a.. The Pythagorean Identity states that cos sin, thus cos a sin a. Hence, equation. can be rewritten as cosul sin a cos.. But the Double Angle Formula for cosine states that cos sin which finally gives cos. Hence, equation. becomes sin sin sin UL a, X. Finding a relation for z in terms of the variable : sin, thus sin UL sin a sin..3 Now that Euler has solved for the simple case at the moment of opposition, he now proceeds to solve for z, the distance between the centers hours after the opposition. Euler begins by again applying the Law of Cosines to the spherical triangle ul, giving cos z cos sin a msin a n cos a mcos a n..4 Euler then states the Product-to-Sum Formulas:

3 sin b sin c cos b c cos b c, cos b cos c cos b c cos b c. Making these substitutions into.4 and simplifying gives cos z cos cos cos cosa cos cosa. Net, Euler groups like terms to obtain cos z cos cos cosa cos, but because of the Half Angle Formulas this reduces to cos z cos n m cos cos a n m sin..5 From his previous reasoning cos z sin z and cos sin, thus.5 becomes sin z cos sin cos sin cos a.6 Euler then states the Sum-Difference Formula for cos a : cos a cosacos sinasin, which applied to.6 gives sin z cos sin cos cos asin cos sin asin sin Euler then uses the approimations cos n m, and n m, cos sin, since the angles are relatively small, to simplify.7: 4sin z sin sin cos asin cos asin sin a sin,..7.8 but the Double Angle Formula for sine states that sin sin cos, thus equations.8 becomes 4sin z n m sin sin n m cosasin n m cosasin 4 sin a cos asin. Grouping like terms gives 4sin sin sin z cosa n m cosasin 4 n m sin acos asin, but from previously sin cos and cosa sin a, thus yielding the first result of this section: 3

4 4sin z 4sin asin cos..9 cos asin 4 sin acos asin Another Double Angle Formula for cosine states that cos cos sin, thus.9 becomes sin z 4sin asin n m cos n m cosasin 4 sin asin 4 sin cos sin a a. 4sin a 4 sin acos a cos asin Euler then perceives that sin a cos a sin, hence finally yielding the second result of the section: 4sin z sin a cos a sin cos sin asin.0 XI. Finding and plugging it into the relation for z: Euler begins by minimizing the first result of part X by differentiating with respect to and equating to zero, which gives cos cos sin sin cos sin a a a 0. Then solving this equation for yields sin a sin. cos cos a sin. Substituting solution. for into.0, the second result of part X, gives cos sin a sin.. cos cos sin z a a sin sin sin Observe that this substitution is not as simple as Euler presents it. Appendi A is a stepby-step verification of this substitution. Formula. is somewhat complicated. To obtain a more simple formula for sin z, he considers an approimation for. Observe that the terms in. that are multiplied by sin are etremely small, thus can be epanded as a geometric power series since it can be epressed in the form. The condition that sin r relatively small guarantees that the ratio r is less than. To see this, first rewrite the numerator and denominator for in. as such: is or equivalently, sina sin cos, cos cos sin a cos 4

5 Denote by sina tan. cosa tan sin a tan and r cos a tan so that r. Since r, the formula r r is a good approimation for. Thus, r r sin a tan cos a tan..3 Euler substitutes this new approimate value for into.9 the first result from part X to obtain cos a tan sin z sin asin..4 This reduction is even lengthier than the first! For those that are interested in the details of this substitution, see Appendi B. XII. Working Backwards: Now Euler supposes that z is known, so the only unknown value is now the time. He begins by rewriting relation.3 in terms of sin z and an angle φ defined implicitly as follows: sin asin sin z,.5 cos where sin UL cos. sin z Substituting.5 into.9 and simplifying gives 4sin asin tan cos cos asin.6 sin asin But the last two terms containing sin on the right hand side of.6 are negligible and so 4sin a sin tan cos. Thus, solving for yields the first-order approimation sin a tan tan. 5

6 This value of is not eact because of the neglected terms; hence there is an error y defined by sina tan tan y..7 To find a formula for y and thus obtain a second-order approimation for, Euler substitutes.7 back into.6, rearranged as 4sin asin tan sin asin cos, cos asin and simplifying gives 4sin asin tan 4 sina sin asin tan tan sin asin y 4sin a cos tan tan 4 sin acos asin tan tan 4 sina cos tan tan y 6 4 sina cos asin tan tan y cos y cos asin y But the last two terms containing y on the right-hand side above are negligible in comparison to y assumed to be relatively small, and the remaining equation can be grouped as 4sin asin tan 4sin acos tan tan 4 sina cos tan tan y 4 sin acos asin tan tan 4 sin a sin a sin tan tan a y sin sin.8 4 sina cos asin tan tan y Again, the last two terms containing y on the right-hand side above are relatively small in comparison to first term containing y, thus.8 reduces to 4sin asin tan 4sin acos tan tan 4 sina cos tan tan y 4 sin acos asin tan tan 4 sin a sin a sin tan tan.9

7 Making some trigonometric substitutions,.9 in turn reduces to 0 4 y sin asin cos tan 4 sin acos asin tan tan 4 sin asin asin tan tan Now solve this equation for y to get 3 y sin a tan sin acos a tan 3 tan...0 Lastly, substitute this value for y into.7 and simplify to get the desired value of : sin a tan tan cos a tan cos a tan tan. XIII. In this section Euler determines what state the umbra and Moon have to be in when the Eclipse is starting and ending. Using z, the distance between the centers of the umbra and the Moon, he reasons that z is the sum of the radii of the umbra and the Moon when the Eclipse is beginning and ending. This is because with this condition, the disks are just touching at the edges. The immersion and emersion would be the moments where z is the difference of the radii. Euler also notes that the condition sin sin sin z a is required to obtain a meaningful result. If this condition is not met, theuch accuracy would be lost in the computations due to the relative insignificances of some of the terms in the formula. XIV. Euler previously derived sin sin sin UL a where UL is the z value at the moment of opposition. Euler uses his Astronomical Tables to find the values of a and ω, which are 0 s, 8, 48, 46 and 0 s, 5, 6, 33 respectively. Euler then uses base-0 logarithms to simplify his work. Either using the log tables or computing directly, UL = 48, 9, thus at the moment of opposition z = 48, 9. XV. Euler provides the values for m and n, the hourly movements of the Sun and the Moon respectively, to be m = 5 and n = 77. Euler can now solve for at the sina tan moment of closest proimity since where this formula is the optimization of the position versus time formula, and where the error caused by approimation is negligible. Using the more accurate numbers from the Opera Omnia gives = 4, 7, or the moment when the disks centers are closest is the time of opposition + = August 8 d, h, 0, in 748 true time. XVI. In this section the formula cos a tan sin z sin asin is used to find z at. This formula can also be epressed as 7

8 sin z sin UL sin asin cos a tan, or approimately as z UL sin asin cos a tan.. Thus, doubling. gives z UL sin a sin cos a tan.3 where UL has already been found. Calculating sin asin cos tan a gives 4, thus z = 48, 9 4 = 48, 5 at the moment when the centers are closest. This allows the setup of the following ratio: radius of the Moon : 6 :: z of the greatest obscuration : magnitude of the Eclipse Euler measured the magnitude of Eclipses using digits, in which digits represent the diameter of the disk being covered, in this case the Moon. This ratio is solved by: 004 : 6 :: 839 : 5.04 thus the magnitude of the Eclipse is 5.04 digits, or the Eclipse covers 5.04 / = 0.48 of the diameter of the Moon. XVII. Euler now wants to find the beginning and ending times for the Eclipse. At these times he knows that z = 6, 4, or the sum of the radii of the umbra and the Moon. sin asin Euler must now solve cos so he can use his formula from section XII. sin z All of the variables on the right-hand side are known, so finding φ is easy. φ = 39,, 8. Net, Euler calculates using the formula.: sin a tan tan cos a tan cos tan a tan, and when the leading value is distributed, there are three terms to calculate. Euler finds their values to be:.04 h, 0.073, and respectively. XVIII. The first and last terms both have tan in them, and to find the start time, this value is negative, while finding the end time makes this value positive. Thus: I. = = II. = =.80 To find the starting and ending times, adjust the time of opposition by these values to get: Time of the opposition, Aug. 8 d, h, 4, 39, 0, 49 +,, 6 Beginning of the Eclipse, 3, 50 8

9 End of the Eclipse 3, 6, 55 XIX. In conclusion, Euler compiles all of the values he has determined about the Eclipse and is ecited that they agree so well with observations: The beginning h, 3, 50 observed at h, 5 The largest obscuration, 0, The opposition in orbit, 4, 39 The end of the Eclipse 3, 6, 55 observed at 3 h, 8 The magnitude of the Eclipse 5.04 digits and the duration of the Eclipse h, 3, 05 observation h, 3 Appendi C contains a short Mathematica program that compares Euler s approimations with the original equations to see how much error was introduced into the calculations in these last few sections. In section XV, Euler uses an approimation of to find the moment of greatest obscuration. His approimation results in the time = August 8 d, h, 0, in 748 true time. This approimation results in an error of only second. In section XVI, Euler uses his formula for finding z at to get z = 48, 5 at the moment when the centers are closest. The error here is negligible, about 0.5 sitieths of an arcsecond. The rest of the paper is devoted to finding the starting and ending times of the Eclipse. Euler s calculations here give the start to be h, 3, 50, where the eact equation gives, 3, 43, and he gives the end to be 3, 6, 55, where the eact equation gives 3, 7, 3. Even though this approimation causes greater error than the others, it is still a good approimation since predicting the start and end of an eclipse only have to be accurate to the minutes. References [] A. Fabian, Translation of Leonard Euler s E4 Paper: Sur l'accord des deu dernieres eclipses du soleil et de la lune avec mes tables, pour trouver les vrais momens des pleni-lunes et novi-lunes On the agreement of the last two eclipses of the Sun and Moon with my tables, for finding the actual times of the half-moon and new moon, Memoires de l'academie des sciences de Berlin pp Opera Omnia: Series, Volume 30, pp Available online at the Euler Archive: Andrew Fabian Department of Mathematics Rowan University Glassboro, NJ 0808 fabian78@students.rowan.edu Hieu D. Nguyen Department of Mathematics Rowan University Glassboro, NJ 0808 nguyen@rowan.edu 9

10 Appendi A The following is a step-by-step justification of. in section XI. We begin with equation.0, 4sin z sin a cos a sin cos, sin asin and substitute formula. for into each of the three terms on the right hand side. The first term can be rewritten as follows: sin a cos a sin sin a sin sin a cos a sin cos cos a sin sin acos a sin sin a sin cos cos a sin sin a cos a sin sin a sin cos cos a sin cos a sin sin a sin cos cos a sin cos a sin 4sin asin cos cos a sin Net, create a common denominator and use the Double Angle Formula for cosine to simplify the last line above as sin a cos a sin cos sin a sin 4sin sin a cos cos a sin The second and third terms in.0 can be simplified as follows: cos sin asin 4 4 sin a cos a sin cos cos a sin cos sin asin 4 4 sin a cos a sin asin cos sin.4.5 cos sin asin cos cos a sin Now sum up all three reduced terms in.4 and.5, then combine like terms to get 0

11 cos sin a sin 4sin 4sin sin z a cos cos a sin cos a sin cos sin asin Again, create a common denominator for the last factor on the right-hand side above to get cos sin a sin 4sin z 4sin asin cos cos a sin cos sin asin cos a sin cos sin asin Net, use the Double Angle Formula for cosine to simplify the last factor: cos sin a sin 4sin z 4sin asin cos cos a sin cos cos a sin cos sin asin Notice that this epression is of the form, which reduces to. Thus, cos sin a sin 4sin 4sin sin, cos cos sin z a a which simplifies to the promised result: sin z sin asin cos cos n n m sin a sin m cosa sin.

12 Appendi B The following is a step-by-step justification of.4 in section XI. r sin a tan cos a tan We begin with equation.9: 4sin z cos cos asin 4 sin acos asin 4sin asin Substituting.3 into the first term on the right-hand side of.6 gives cos cos asin 4sin asin cos a tan tan cos a cos 4 cos asin cos 4sin asin cos a tan 4 4 cos a tan cos a cos a tan The second term on the right-hand side of.6 reduces to 4 sin a cos asin sin a cos a tan cos a tan 4 sin a cos asin 4sin asin cos a tan cos a tan Combining both terms gives 4sin z 4sin asin [ cos a tan cos a tan 4 4 cos acos a tan cos a tan cos a tan ] To simplify the notation, we denote by 4sin asin We then simplify n n m m cos a tan.6

13 4sin z 4 4 [ cos a tan ] cos a cos a tan 6 6 [ cos a tan cos a cos a tan ] [ cos a cos a tan cos a cos a tan cos acos a tan ] 4 4 4sin asin [ cos a tan cos a cos a tan cos a cos a tan cos a cos a tan ] Euler s Approimation Euler recognizes the following factorization: cos a tan 4 cos a tan cos a tan He then drops the last two terms from the previous result since they contain higher-order powers of tan, thus negligible, to get 4sin z 4 4 4sin asin [ cos a tan cos a cos a tan ] The last term above is also negligible, and we are led to assume that Euler replaces it by a different negligible term, namely 4 cos a tan we see no other way to justify for this step: 4sin z sin asin [ cos a tan 4 cos a tan ] 4sin asin cos atan We thus arrive at our desired relation: cos a tan sin z sin asin. 3

14 Appendi C This Mathematica program checks Euler's approimation in E4 a 0.538; n ; m ; w ; eq 4 Sin a ^ Sin w ^ n m ^ Cos w ^ n m ^ Sin a ^ Sin w ^ n m ^ Cos w ^ n m ^ Cos a Sin w ^ n m ^ Cos a ^ Tan w ^ eqappro 4 Sin a ^ Sin w ^ n m ^ ^ z : eq 80 Π zsign : Floor z 30 zdegwr : z zsign 30 zdeg : Floor zdegwr zmwr : zdegwr zdeg 60 zmin : Floor zmwr zswr : zmwr zmin 60 zsec : Floor zswr zsswr : zswr zsec 60 Print "EXACT: ", zsign, " signs ", zdeg, " degrees ", zmin, " minutes ", zsec, " seconds ", zsswr, " sitieths of a second" EXACT: z : eqappro 80 Π zsign : Floor z 30 zdegwr : z zsign 30 zdeg : Floor zdegwr zmwr : zdegwr zdeg 60 zmin : Floor zmwr zswr : zmwr zmin 60 zsec : Floor zswr zsswr : zswr zsec 60 Print "APPROX: ", zsign, " signs ", zdeg, " degrees ", zmin, " minutes ", zsec, " seconds ", zsswr, " sitieths of a second" 0 signs 0 degrees 48 minutes 4 seconds sitieths of a second APPROX: 0 signs 0 degrees 48 minutes 4 seconds 5.80 sitieths of a second

15 E4 approimation.nb n m Sin a Sin w ^ time n m ^ Cos w ^ n m ^ Cos a Sin w ^ tappro n m Sin a Tan w ^ n m ^ Cos a Tan w ^ n m ^ n m ^ tappro n m Sin a Tan w ^ n m ^ : Abs time h : Floor mwr : h 60 m : Floor mwr swr : mwr m 60 If time 0, Print "EXACT: ", h, " hours ", m, " minutes ", swr, " seconds ", Print "EXACT: ", h, " hours ", m, " minutes ", swr, " seconds " : Abs tappro h : Floor mwr : h 60 m : Floor mwr swr : mwr m 60 If tappro 0, Print "APPROX: ", h, " hours ", m, " minutes ", swr, " seconds ", Print "APPROX: ", h, " hours ", m, " minutes ", swr, " seconds " 3 : Abs tappro 3h : Floor 3 3mwr : 3 3h 60 3m : Floor 3mwr 3swr : 3mwr 3m 60 If tappro 0, Print "APPROX: ", 3h, " hours ", 3m, " minutes ", 3swr, " seconds ", Print "APPROX: ", 3h, " hours ", 3m, " minutes ", 3swr, " seconds " EXACT: APPROX: 0 hours 4 minutes seconds 0 hours 4 minutes seconds APPROX: 0 hours 4 minutes seconds A : n m ^ Cos w ^ n m ^ Cos a Sin w ^ B : n m Sin a Sin w ^ F : 4 Sin a ^ Sin w ^ Tan φ ^ B B^ 4 A F start : A startn : start. a startn : startn. n startn3 : startn. m

16 E4 approimation.nb 3 startn4 : startn3. w startn5 : startn4. φ stappro : Sin a Tan w n m n m ^ Tan φ Cos a Tan w Cos a Tan w ^ Tan φ n m n m n m ^ san : stappro. a san : san. n san3 : san. m san4 : san3. w san5 : san4. φ B B^ 4 A F end : A endn : end. a endn : endn. n endn3 : endn. m endn4 : endn3. w endn5 : endn4. φ endappro : Sin a Tan w n m n m ^ Tan φ Cos a Tan w Cos a Tan w ^ Tan φ n m n m n m ^ ean : endappro. a ean : ean. n ean3 : ean. m ean4 : ean3. w ean5 : ean4. φ st : startn5 sth : Floor st stmwr : st sth 60 stm : Floor stmwr stswr : stmwr stm 60 Print "START EXACT: ", sth, " hours ", stm, " minutes ", stswr, " seconds before opposition" sta : san5 stah : Floor sta stamwr : sta stah 60 stam : Floor stamwr staswr : stamwr stam 60 Print "START APPROX: ", stah, " hours ", stam, " minutes ", staswr, " seconds before opposition" doneh : Floor endn5 donemwr : endn5 doneh 60 donem : Floor donemwr doneswr : donemwr donem 60 Print "END EXACT: ", doneh, " hours ", donem, " minutes ", doneswr, " seconds after opposition"

17 4 E4 approimation.nb dah : Floor ean5 damwr : ean5 dah 60 dam : Floor damwr daswr : damwr dam 60 Print "END APPROX: ", dah, " hours ", dam, " minutes ", daswr, " seconds after opposition" START EXACT: hours 0 minutes seconds before opposition START APPROX: hours 0 minutes seconds before opposition END EXACT: END APPROX: hours minutes 4.3 seconds after opposition hours minutes seconds after opposition

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