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epact

In The Calendar, Table Iv., Look For April, And The Epact 28 Is Found Opposite The Second Day.

10In The Calendar, Table Iv., Look For April, And The Epact 28 Is Found Opposite The Second Day.

10Epact Is A Word Of Greek Origin, Employed In The Calendar To Signify The Moon'S Age At The Beginning Of The Year.

00Another Addition Of Eleven Gives Thirty Three For The Epact Of The Fourth Year; But In Consequence Of The Insertion Of The Intercalary Month In Each Third Year Of The Lunar Cycle, This Epact Is Reduced To Three.

00In Like Manner The Epacts Of All The Following Years Of The Cycle Are Obtained By Successively Adding Eleven To The Epact Of The Former Year, And Rejecting Thirty As Often As The Sum Exceeds That Number.

00They Are Therefore Connected With The Golden Numbers By The Formula (I), In Which N Is Any Whole Number; Andor A Whole Lunar Cycle (Supposing The First Epact To Be 11), They Are As Follows: 11, 22, 3, 14, 25, 6, 17, 28, 9, 20, I, 12, 23, 4, 15, 26, 7, 18, 29.

00But The Order Is Interrupted At The End Of The Cycle; For The Epact Of The Following Year, Found In The Same Manner, Would Be 29 11=40 Or 10, Whereas It Ought Again To Be 1S To Correspond With The Moon'S Age And The Golden Number 1.

00The Reason Of This Is, That The Intercalary Month, Inserted At The End Of The Cycle, Contains Only Twenty Nine Days Instead Of Thirty; Whence, After 11 Has Been Added To The Epact Of The Year Corresponding To The Golden Number 19, We Must Reject Twenty Nine Instead Of Thirty, In Order To Have The Epact Of The Succeeding Year; Or, Which Comes To The Same Thing, We Must Add Twelve To The Epact Of The Last Year Of The Cycle, And Then Reject Thirty As Before.

00In Consequence Of The Solar And Lunar Equations, It Is Evident That The Epact Or Moon'S Age At The Beginning Of The Year, Must, In The Course Of Centuries, Have All Different Values From One To Thirty Inclusive, Corresponding To The Days In A Full Lunar Month.

00The Third Column Corresponding To The Golden Number 3, Has For Its First Epact 12 = 23; And In The Same Manner All The Nineteen Columns Of The Table Are Formed.

00The Epact Of The Following Year Is Therefore Twenty Nine.

00The 2Nd Of January Is Therefore The Day Of The New Moon, Which Is Indicated By The Epact Twenty Nine.

00In Like Manner, If The New Moon Fell On The 4Th Of December, The Epact Of The Following Year Would Be Twenty Eight, Which, To Indicate The Day Of Next New Moon, Must Correspond To The 3Rd Of January.

00When The Epact Of The Year Is Known, The Days On Which The New Moons Occur Throughout The Whole Year Are Shown By Table Iv., Which Is Called The Gregorian Calendar Of Epacts.

00For Example, The Golden Number Of The Year 1832 Is (= 9 19 R And The Epact, As Found In Table Iii., Is Twenty Eight.

00This Epact Occurs At The 3Rd Of January, The 2Nd Of February, The 3Rd Of March, The Znd Of April, The 1St Of May, &C., And These Days Are Consequently The Days Of The Ecclesiastical New Moons In 1832.

00From This It Appears That If The Golden Number Of The Year Exceeds Ii, The Epact 25, In Six Months Of The Year, Must Correspond To The Same Day In The Calendar As 26; But If The Golden Number Does Not Exceed Ii, That Epact Must Correspond To The Same Day As 24.

00In Using The Calendar, If The Epact Of The Year Is 25, And The Golden Number Not Above Ii, Take 25; But If The Golden Number Exceeds Ii, Take 25'.

00The Epact 19' (Also Distinguished By An Accent Or Different Character) Is Placed Table Iii.

00It Is, However, Only Used In Those Years In Which The Epact 19 Concurs With The Golden Number 19.

00Hence, If In That Year The Epact Should Be 19, A New Moon Would Fall On The 2Nd Of December, And The Lunation Would Terminate On The 30Th, So That The Next New Moon Would Arrive On The 3 Ist.

00The Epact Of The Year, Therefore, Or 19, Must Stand Beside That Day, Whereas, According To The Regular Order, The Epact Corresponding To The 31St Of December Is 20; And This Is The Reason For The Distinction.

00Under 9, And In The Line C, We Find The Epact 28.

00Under 17, In Line B, The Epact Is 25'.

00In The Calendar This Epact First Occurs Before The 2Nd Of December At The 26Th Of November.

00Hence We Derive The Following Rule For Finding Easter Sunday From The Tables: 1St, Find The Golden Number, And, From Table Iii., The Epact Of The Proposed Year.

002Nd, Find In The Calendar (Table Iv.) The First Day After The 7Th Of March Which Corresponds To The Epact Of The Year; This Will Be The First Day Of The Paschal Moon.

00Sometimes a misunderstanding has arisen from not observing that this regulation is to be construed according to the tabular full moon as determined from the epact, and not by the true full moon, which, in general, occurs one or two days earlier.

00But the fourteenth of this moon falls at the latest on the 18th of April, or 29 days after the 20th of March; for by reason of the double epact that occurs at the 4th and 5th of April, this lunation has only 29 days.

00Now, if (1840+1) = 17, and the epact (Table III.

00line C) is 26.2nd, 19 r After the 7th of March the epact 26 first occurs in Table III.

00In Order To Investigate A Formula For The Epact, Let Us Make E=The True Epact Of The Given Year; J =The Julian Epact, That Is To Say, The Number The Epact Would Have Been If The Julian Year Had Been Still In Use And The Lunar Cycle Had Been Exact;, S =The Correction Depending On The Solar Year; M =The Correction Depending On The Lunar Cycle; Then The Equation Of The Epact Will Be E=J S M; So That E Will Be Known When The Numbers J, S, And M Are Determined.

00The Epact J Depends On The Golden Number N, And Must Be Determined From The Fact That In 1582, The First Year Of The Reformed Calendar, N Was 6, And J 26.

00For The Following Years, Then, The Golden Numbers And Epacts Are As Follows: 1583, N= 7, J=26 Ii 30= 7; 1584, N= 8, J= 7 11 =18; 1585, N= 9, J = 18 Ii =29; 1586, N = To, J = 29 I I 30 10; And, Therefore, In General J = J (N Io(N I)) 30 R On Account Of The Solar Equation S, The Epact J Must Be Diminished By Unity Every Centesimal Year, Excepting Always The Fourth.

00Hence The Correction Of The Epact, W Or The Number Of Days To Be Intercalated After X Centuries Reckoned From The Commencement Of One Of The Periods Of Twenty Five (X I Centuries, Is ? ?

00Let (C7) W =A, Then For All Years After 1800 The Value Of M Will Beiven By The Formula (C 18 A), G' Y W Therefore, 3 Counting From The Beginning Of The Calendar In 1582, C 15 A ' M = 3 W By The Substitution Of These Values Of J, S And M, The Equation Of The Epact Becomes E C 16 (C 16 (C 15 A) 3 O R () 4 W 3 W.

00The Above Formula For The Epact Is Given By Delambre (Hist.

00Having Determined The Epact Of The Year, It Only Remains To Find Easter Sunday From The Conditions Already Laid Down.

00The Value Of L Is Always Given By The Formula For The Dominical Letter, And P And 1 Are Easily Deduced From The Epact, As Will Appear From The Following Considerations.

00When P =I The Full Moon Is On The 21St Of March, And The New Moon On The 8Th (21 13 =8), Therefore The Moon'S Age On The 1St Of March (Which Is The Same As On The 1St Of January) Is Twentythree Days; The Epact Of The Year Is Consequently Twenty Three.

00When P=2 The New Moon Falls On The Ninth, And The Epact Is Consequently Twenty Two; And, In General, When P Becomes I X, E Becomes 23 X, Therefore P E = I X 23 X =24, And P =24 E.

00But It Is Evident That When 1 Is Increased By Unity, That Is To Say, When The Full Moon Falls A Day Later, The Epact Of The Year Is Diminished By Unity; Therefore, In General, When 1=4 X, E=23 X, Whence L E =27 And 1=27 E.

00By Means Of The Formulae Which We Have Now Given For The Dominical Letter, The Golden Number And The Epact, Easter Sunday May Be Computed For Any Year After The Reformation, Without The Assistance Of Any Tables Whatever.

00The Last Period Of Twenty Five N=17 For The Epact We Have (N Io(N 1) _ 117 160 (1 771 30 R 30 R 30) R =?

00The Gregorian epact being the age of the moon of Tebet at the beginning of the Gregorian year, it represents the day of Tebet which corresponds to January I; and thus the approximate date of Tisri I, the commencement of the Hebrew year, may be otherwise deduced by subtracting the epact from Sept.

00epact training package A series of two-day, computer-based interactive training courses were developed, organized and run throughout 1997/98.

00Epact Is A Word Of Greek Origin, Employed In The Calendar To Signify The Moon'S Age At The Beginning Of The Year.

00Another Addition Of Eleven Gives Thirty Three For The Epact Of The Fourth Year; But In Consequence Of The Insertion Of The Intercalary Month In Each Third Year Of The Lunar Cycle, This Epact Is Reduced To Three.

00In Like Manner The Epacts Of All The Following Years Of The Cycle Are Obtained By Successively Adding Eleven To The Epact Of The Former Year, And Rejecting Thirty As Often As The Sum Exceeds That Number.

00They Are Therefore Connected With The Golden Numbers By The Formula (I), In Which N Is Any Whole Number; Andor A Whole Lunar Cycle (Supposing The First Epact To Be 11), They Are As Follows: 11, 22, 3, 14, 25, 6, 17, 28, 9, 20, I, 12, 23, 4, 15, 26, 7, 18, 29.

00But The Order Is Interrupted At The End Of The Cycle; For The Epact Of The Following Year, Found In The Same Manner, Would Be 29 11=40 Or 10, Whereas It Ought Again To Be 1S To Correspond With The Moon'S Age And The Golden Number 1.

00The Reason Of This Is, That The Intercalary Month, Inserted At The End Of The Cycle, Contains Only Twenty Nine Days Instead Of Thirty; Whence, After 11 Has Been Added To The Epact Of The Year Corresponding To The Golden Number 19, We Must Reject Twenty Nine Instead Of Thirty, In Order To Have The Epact Of The Succeeding Year; Or, Which Comes To The Same Thing, We Must Add Twelve To The Epact Of The Last Year Of The Cycle, And Then Reject Thirty As Before.

00In Consequence Of The Solar And Lunar Equations, It Is Evident That The Epact Or Moon'S Age At The Beginning Of The Year, Must, In The Course Of Centuries, Have All Different Values From One To Thirty Inclusive, Corresponding To The Days In A Full Lunar Month.

00The Third Column Corresponding To The Golden Number 3, Has For Its First Epact 12 = 23; And In The Same Manner All The Nineteen Columns Of The Table Are Formed.

00The Epact Of The Following Year Is Therefore Twenty Nine.

00The 2Nd Of January Is Therefore The Day Of The New Moon, Which Is Indicated By The Epact Twenty Nine.

00In Like Manner, If The New Moon Fell On The 4Th Of December, The Epact Of The Following Year Would Be Twenty Eight, Which, To Indicate The Day Of Next New Moon, Must Correspond To The 3Rd Of January.

00When The Epact Of The Year Is Known, The Days On Which The New Moons Occur Throughout The Whole Year Are Shown By Table Iv., Which Is Called The Gregorian Calendar Of Epacts.

00For Example, The Golden Number Of The Year 1832 Is (= 9 19 R And The Epact, As Found In Table Iii., Is Twenty Eight.

00This Epact Occurs At The 3Rd Of January, The 2Nd Of February, The 3Rd Of March, The Znd Of April, The 1St Of May, &C., And These Days Are Consequently The Days Of The Ecclesiastical New Moons In 1832.

00From This It Appears That If The Golden Number Of The Year Exceeds Ii, The Epact 25, In Six Months Of The Year, Must Correspond To The Same Day In The Calendar As 26; But If The Golden Number Does Not Exceed Ii, That Epact Must Correspond To The Same Day As 24.

00In Using The Calendar, If The Epact Of The Year Is 25, And The Golden Number Not Above Ii, Take 25; But If The Golden Number Exceeds Ii, Take 25'.

00The Epact 19' (Also Distinguished By An Accent Or Different Character) Is Placed Table Iii.

00It Is, However, Only Used In Those Years In Which The Epact 19 Concurs With The Golden Number 19.

00Hence, If In That Year The Epact Should Be 19, A New Moon Would Fall On The 2Nd Of December, And The Lunation Would Terminate On The 30Th, So That The Next New Moon Would Arrive On The 3 Ist.

00The Epact Of The Year, Therefore, Or 19, Must Stand Beside That Day, Whereas, According To The Regular Order, The Epact Corresponding To The 31St Of December Is 20; And This Is The Reason For The Distinction.

00Under 9, And In The Line C, We Find The Epact 28.

00Under 17, In Line B, The Epact Is 25'.

00In The Calendar This Epact First Occurs Before The 2Nd Of December At The 26Th Of November.

00Hence We Derive The Following Rule For Finding Easter Sunday From The Tables: 1St, Find The Golden Number, And, From Table Iii., The Epact Of The Proposed Year.

002Nd, Find In The Calendar (Table Iv.) The First Day After The 7Th Of March Which Corresponds To The Epact Of The Year; This Will Be The First Day Of The Paschal Moon.

00Sometimes a misunderstanding has arisen from not observing that this regulation is to be construed according to the tabular full moon as determined from the epact, and not by the true full moon, which, in general, occurs one or two days earlier.

00But the fourteenth of this moon falls at the latest on the 18th of April, or 29 days after the 20th of March; for by reason of the double epact that occurs at the 4th and 5th of April, this lunation has only 29 days.

00Now, if (1840+1) = 17, and the epact (Table III.

00line C) is 26.2nd, 19 r After the 7th of March the epact 26 first occurs in Table III.

00In Order To Investigate A Formula For The Epact, Let Us Make E=The True Epact Of The Given Year; J =The Julian Epact, That Is To Say, The Number The Epact Would Have Been If The Julian Year Had Been Still In Use And The Lunar Cycle Had Been Exact;, S =The Correction Depending On The Solar Year; M =The Correction Depending On The Lunar Cycle; Then The Equation Of The Epact Will Be E=J S M; So That E Will Be Known When The Numbers J, S, And M Are Determined.

00The Epact J Depends On The Golden Number N, And Must Be Determined From The Fact That In 1582, The First Year Of The Reformed Calendar, N Was 6, And J 26.

00For The Following Years, Then, The Golden Numbers And Epacts Are As Follows: 1583, N= 7, J=26 Ii 30= 7; 1584, N= 8, J= 7 11 =18; 1585, N= 9, J = 18 Ii =29; 1586, N = To, J = 29 I I 30 10; And, Therefore, In General J = J (N Io(N I)) 30 R On Account Of The Solar Equation S, The Epact J Must Be Diminished By Unity Every Centesimal Year, Excepting Always The Fourth.

00Hence The Correction Of The Epact, W Or The Number Of Days To Be Intercalated After X Centuries Reckoned From The Commencement Of One Of The Periods Of Twenty Five (X I Centuries, Is ? ?

00Let (C7) W =A, Then For All Years After 1800 The Value Of M Will Beiven By The Formula (C 18 A), G' Y W Therefore, 3 Counting From The Beginning Of The Calendar In 1582, C 15 A ' M = 3 W By The Substitution Of These Values Of J, S And M, The Equation Of The Epact Becomes E C 16 (C 16 (C 15 A) 3 O R () 4 W 3 W.

00The Above Formula For The Epact Is Given By Delambre (Hist.

00Having Determined The Epact Of The Year, It Only Remains To Find Easter Sunday From The Conditions Already Laid Down.

00The Value Of L Is Always Given By The Formula For The Dominical Letter, And P And 1 Are Easily Deduced From The Epact, As Will Appear From The Following Considerations.

00When P =I The Full Moon Is On The 21St Of March, And The New Moon On The 8Th (21 13 =8), Therefore The Moon'S Age On The 1St Of March (Which Is The Same As On The 1St Of January) Is Twentythree Days; The Epact Of The Year Is Consequently Twenty Three.

00When P=2 The New Moon Falls On The Ninth, And The Epact Is Consequently Twenty Two; And, In General, When P Becomes I X, E Becomes 23 X, Therefore P E = I X 23 X =24, And P =24 E.

00But It Is Evident That When 1 Is Increased By Unity, That Is To Say, When The Full Moon Falls A Day Later, The Epact Of The Year Is Diminished By Unity; Therefore, In General, When 1=4 X, E=23 X, Whence L E =27 And 1=27 E.

00By Means Of The Formulae Which We Have Now Given For The Dominical Letter, The Golden Number And The Epact, Easter Sunday May Be Computed For Any Year After The Reformation, Without The Assistance Of Any Tables Whatever.

00The Last Period Of Twenty Five N=17 For The Epact We Have (N Io(N 1) _ 117 160 (1 771 30 R 30 R 30) R =?

00The Gregorian epact being the age of the moon of Tebet at the beginning of the Gregorian year, it represents the day of Tebet which corresponds to January I; and thus the approximate date of Tisri I, the commencement of the Hebrew year, may be otherwise deduced by subtracting the epact from Sept.

00

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