riseth at 6 of the Clock and so there is no Difference of Ascension for he is then in the Aequinoctial which cuts the Horizon in the two opposite points of East and West In this Question the Difference of Ascension is O P and is a distance upon a line that goeth not through the Center therefore take half the length of the parallel I D and proceed as you did in the third Question and after you have found it in degrees and minutes convert it into hours and minutes of time and set it down I find it to be here 16 deg 18 min. which is 1 hour 5′ 3 15. By the Difference of Ascension thus found you may find the length of the day or night the hour of Suns Rising or the hour of uns Setting QUESTION VII For the time of the Suns Rising or Setting in this Example I Consider as much as the Sun riseth before 6 so much he sets after 6 here he riseth before 6 of the Clock 1 hour 5′ subtract that from 6 hours and you have the hour of the Suns Rising 4 h. 55′ add it to 6 hours and you have the time the Sun sets 7 h. 5′ double that and it is the length of the whole day which is here 14 deg 10 min. Subtract the length of the day from 24 hours and it leaves the length of the night 9 h. 50′ I omitted the Fractions But if the Sun riseth after 6 of the Clock and you have a desire to find these things as he doth when he is South Declination add the Difference of Ascension to 6 hours and it gives the time of the Suns Rising subtract it from 6 and it gives the time of Suns Setting and that doubled is the length of the whole day Again that subtracted from 24 hours is the length of the night QUESTION VIII To find the hour of the Suns being due East or West THe Sun is due East in this Example or any other when he goeth by the East point of the Horizon or West when he goeth by the West point In this Example the Suns parallel of Declination cuts the East and West Azimuth in S which is later in the morning than 6 of the Clock by the distance S P therefore see what S P is by The third way of Measuring and convert it into time and add it to 6 hours it shall give the hour of the Suns being due East I find it in this Example to be 6 h. 46′ Subtract this from 12 hours and it gives the time of the Suns being due West that day for as many hours and minutes as the Sun is due East before 12 of the Clock so many hours and minutes must it be due West after 12 According to this Example the Sun is due West at 5 of the Clock 14 min. If the Sun hath South Declination he passeth the point of East before he riseth and is set as long before he comes to the point of West provided the Latitude be Northerly as this is but if the Latitude and Declination be both one way the Sun is always up before he cometh to the point of East and the work is as I have shewed QUESTION IX To find the time of Day breaking and Twilight ending IT is an antient Observation and concluded Opinion That the Sun makes some shew of Day when he is 17 degrees under the Horizon therefore take 17 deg from your Scale of Chords and set it from both ends of the Horizon downwards and draw the line T r then fix your Compasses in r the place where the Suns parallel of Declination intersects that line and extend them to 6 of the Clock set it off by The third way of Measuring and convert it into time Subtract that from 6 of the Clock and it gives the time of Day breaking I find in this Example Day breaks at 3 of the Clock 6 min. â—8 15. Add it to 6 hours and it gives the time of Twilight ending 8 of the Clock 3 53 14 13. It may happen so sometimes that it may be past 6 of the Clock before the Day breaks in such a case you must add in the morning for break of Day and subtract for Night from 6 hours which is the contrary These things your own Reason will give you after you are used to it which makes me forbear to give any more reasons of it QUESTION X. To find the Continuance of Twilight THe Continuance of Twilight is the time between the Day breaking and Sun rising which is r o take it off by the Third way of Measuring and convert it into time I find it to be 1 hour 48 min. QUESTION XI To find the Length of the longest Day in that Latitude VVHen the Sun is nearest the Zenith in any Latitude that day must be the longest now in places near the Aequinoctial as between the Tropicks there is but little difference all the year long but in places nearer the Poles there is more The Sun is nearest the Zenith in this Latitude when he is in the Tropick of Cancer so that then must be the longest Day Imagine the Tripick of Cancer to be the parallel of the Suns Declination as indeed it is that day take the distance between Y and R which is between Suns Rising and 6 of the Clock that day and set it off by the Third way of Measuring the number of degrees and minutes convert into time and add it to 6 hours which makes the length of the Forenoon and that doubled is the length of the whole day as in Quest 7. which is equal to the longest Night in that Latitude I find it here to be 16 hours 10 minutes Subtract the length of this longest day from 24 h. 00 min. And it leaves the length of the shortest night 7 h. 50 min. Equal to which is the length of the shortest day 7 h. 50 min. But you may measure the length of the shortest day and subtract that from 24 hours and it will be the length of the longest night which is equal to the longest day Now the shortest day is when the Sun is in the Tropick of Capricorn for then it is latest before he riseth Now to measure it take the distance from 6 of the Clock which is T and the Suns Rising which is U and set it off by the Third way of Measuring convert the degrees into time and subtract that time from 6 hours it leaves the length of the Forenoon which doubled is the length of the shortest day as in Quest 7. I have spoke to that purpose I need say no more that 7 th Question is light sufficient I might do an Example of what is before done in a South Latitude but Reason gives that the same which the South Pole or Southern parts of the Heavens are deprest in a Northern Latitude the same will the Northern parts of the Heavens be deprest in a Latitude as far Southerly so that there will be no difference
a great Circle for by great Circles is the Sphere measured QUESTION I. To find the Meridian Altitude of the Sun IN this or any other Sphere M G E is that part of the Heavens that is visible the other half invisible to us for it is parted from our sight by the Earth and Sea and the furthest part of it which seems to mix as it were with the Heavens we call the Horizon which is the great Circle M E for it is a great Circle though here we are forced to represent it by a streight Line M is the South point of the Horizon Now the Suns Meridian Altitude is his distance between that point and the place he cuts the Meridian that day which is M I fix your Compasses in M and extend the other foot to I and apply it to your Scale of Chords and as many degrees as you find it there so many degrees is the Sun high when he is upon the Meridian that day which is the thing required Note that at O the Sun riseth at P it is 6 of the Clock and P I is equal to ♈ K when it is extended to a great Circle and both the Sine of 90 deg which is extended to a Chord must be the Chord of as many degrees which is 6 hours in time The time between 6 a Clock and 12 which proves that M I is the Meridian Altitude of the Sun and this measuring any Distance from the Meridian is called The first Way of Measuring QUESTION II. To find the Suns Amplitude of Rising and Setting FIx one foot of your Compasses in the place of the Suns Rising which is O and extend the other foot to the Center of the Sphere which is termed the East point of the Horizon and this Distance apply to your Line of Sines if you have any but if you have no line of Sines extend it to the Chord of the same Arch thus Fix your Compasses with that distance one foot in the Meridian so that the other may just sweep a line that goeth through the Center of the Sphere then say I the Arch between that foot of your Compasses that stands in the Circle and the place where this line you sweep cuts the Circle is the Chord of the thing required and will be the same number of degrees upon the line of Chords as O R would be upon the line of Sines and after this manner is the Chord of any distance taken from a line that goeth through the Center of the Sphere found and this is called The second Way of Measuring I find that the Sun riseth 20 deg 53 min. to the Northwards of the East for she hath North Declination and the Latitude is Northerly or sets so much to the Northwards of the West QUESTION III. To find the Suns Azimuth at six of the Clock THe Suns Azimuth at 6 of the Clock is the nearest distance between the Sun at 6 of the Clock and the East and West Azimuth which is P z Now if you mind it P z is taken off from a Circle which is not so great as the Horizon and yet is parallel to it as the line n w is parallel to the line M E and as many degrees as the Sun is from the nearest part of the East and West Azimuth in the little Circle so many he is from it if an Azimuth where drawn in the great Circle for P z is as many degrees in the little Circle as n ♈ is in the great one Now because your Scale is made by the great Circle therefore extend the distance taken from the lesser Circle to the measure of the greater which is done thus Take half the length of the pricked line which is z w and fixing one foot of your Compasses in the Center of the Sphere describe an Arch from some line that goeth through the Center of the Sphere as the Arch M l n then set the distance z P as a Sine upon that Arch for it is a Sine upon that Arch as well as o ♈ was a Sine upon the Meridian I find the Sine of it is the Sine of the Arch l M lay your Scale from the Center of the Sphere by l and draw the line l q then shall q M be the same quantity of degrees upon the great Circle as l M is upon the little one therefore take M q and apply it to the Scale of Chords and it answers your desire And thus is the Suns Azimuth at 6 of the Clock P z found to be 8 deg 36 min. to the Northwards of the East and this is called The third way of Measuring to measure any distance from a line that doth not go through the Center which must represent a small Circle And thus you may find the Suns Azimuth at any time of the day QUESTION IV. To find the Suns height at six of the Clock THe Sun is at P at 6 of the Clock fix one foot of your Compasses in P and extend the other to sweep the Horizon which is the same as though you let fall the Perpendicular P n set it off by The second way of Measuring as was shewed in Quest 2. and apply it to your line of Chords and the reason is because P n represents a part of a great Circle and so is to be understood to be of the nature of those lines that go through the Center of the Sphere for all Azimuths pass through the Zenith and Nadir which are two opposite points I find the Sun is 10 deg 7 min high at 6 of the Clock QUESTION V. To find the Suns height being due East or West VVHere the Suns parallel of Declination cuts the East and West Azimuth is the place the Sun is in when he is due East in the morning for you see he is then over the point of East in the Horizon which is ♈ or the Center therefore take the distance between that place and the Center which is S ♈ and apply it to the line of Sines but if you have no line of Sines extend it to a Chord after the manner of the second Question which I call the The second way of Measuring for it is the distance of a line which goeth through the Center I find the Suns height being due East or West is 17 deg 25 m. Note that the same height that the Sun is being over the East point so high he is being over the West point in the afternoon QUESTION VI. To find the Difference of Ascension THe Difference of Ascension is the portion of time that is between the Suns Rising and six of the Clock If the days be longer than the nights the Sun riseth before 6 but if shorter after 6 but whether it be before or after 6 that he riseth so many hours and minutes as it is from 6 so much is the half day or night longer or shorter than 6 hours from whence it is evident that if the Sun riseth due East he
but in any case take this in general that if the Difference of Ascension be before 6 of the Clock subtract it if after 6 add it to 6 hours and you have the time of Sun Rising This stands to good reason forasmuch as the Difference of Ascension is the portion of time that the Sun riseth before or after 6 of the Clock In this Example you see the Difference of Ascension is before 6 of the Clock 1 hour 5′ 3 15 we will omit the part of a part of a minute I would know the time of the Suns Rising Example Subtract the Difference of Ascension 1 h. 5 m. 3 15 From 6 hours 5 60 The remainder is the time of Sun Rising 4 54 12 15 or â…˜ Thus I conclude the Sun riseth at 4 a Clock 54 min. â…˜ If you have a desire to find the time of Sun setting subtract the time of Sun rising from 12 hours and you have it for as many hours and minutes as the Sun riseth before 12 so many hours and minutes he sets after 12. Example Here the Sun riseth at 4 of the Clock 54 m. â…˜ which we express thus 4 h. 54 m. â…˜ This subtracted from 11 60 Leaves the time of Sun setting 7 05 â…• Which is 5 min. â…• past 7 of the Clock in the afternoon This doubled is the length of the whole day which is 14 hours 10 min â…– This subtracted from 24 hours is the length of the night 9 hours 49 min. 8 5. To find the length of the longest Day in that Latitude before proposed VVHen the days are at the longest in any North Latitude the Sun is in the Tropick of Cancer in a Southern Latitude in the Tropick of Capricorn This is a Northern Latitude therefore make the Tropick of Cancer the Parallel of the Suns Declination as here O R is the Parallel of the Suns Declination then must I 6 be the Difference of Ascension which is = to F ♈ In the right Angled Triangle I F ♈ right angled at F you have I ♈ F the Complement of the Latitude 40 deg given and I F the Suns Declination 23 deg 30 min. to find F ♈ find it as you was shewed to find the Difference of Ascension before convert it into time and find the length of the day as was shewed before This note that Sine s I F L is an Arch of the Meridian cutting the Horizon in that place of it where the Sun riseth Sine ♈ F + Radius is = to Tang. I F + Tang. comp I ♈ F. Tang. I F Suns Declination 23 deg 30 min. 9,638301 Tang. com I ♈ F Poles Elevation 40 deg 0 min. 10,076186 Sine F ♈ Difference of Ascension 31 d. 12 m. 39 sec 9,714487 But you may ask all this while how this Fraction is found it is thus found Admit I would find the sine of the Arch answering to 9714487 I look in the Sines and find the nearest less than it to be  9,714352 I take the next greater than it and find it to be 9,714560 I subtract the lesser from the greater the remainder is 208 Then from the figures that came forth 9,714487 I subtract the nearest in the Tables less 9,714352 And the Remainder is 135 Then say As the Difference between the nearest less and the nearest greater 280 comp arith 7,681936 Is to the Difference between the same and that less 135 2,130333 So is 60 min. 1,778151 To 39 sec 1,590420 The like is to be understood of any Fraction else if you have occasion for the artificial Sine or Tangent of an Arch that hath a Fraction to it Say As 60 seconds Is to the seconds in the Fraction which is here 39 So is the Difference between the nearest lesser and the nearest greater 208 To the Fraction 135 And this added to the lesser makes the artificial Sine of the Arch required 9,714352 9,714487 The like is to be understood of a Tangent  To find the hour of the Suns being due East or West Latitude 50 deg 0 min. Declination 13 deg 15 min. Northerly I demand the time of the Suns being due East or West And for the time of the Suns being due West you are to subtract it from 6 hours in this case and the remainder is your desire The reason is because as many hours as the Sun is due East after 6 of the Clock in the morning so long time is he due West before 6 of the Clock in the Afternoon I have left the working of this to your own practice only I have set down the Resolution of it The Sun is due East 45 min. 9 15 past 6 of the Clock The Sun is due West 45 min. 9 15 before 6 of the Clock Which is at 5 of the Clock 14 min. 6 15. Note That if the Suns Declination be Southerly then will he be East before 6 so that whereas here you add there you subtract from 6 hours to find the hour of the Suns being due East or add to 6 hours for the time of its being due West but your own Reason if you look well on the Scheme will guide you to know this and also to know that it is useless in such cases for then he is not above the Horizon Latitude 50 deg 0 min. Declination 13 deg 15 min. I demand the Continuance of Twilight AS I have noted before the Sun is accounted 17 deg under the Horizon when the day breaks therefore in the following Scheme let 17 I be an Arch parallel to the Horizon 17 deg under it Let D B C A be an Azimuth cutting the Aequinoctial and the line of 17 deg in the place of their intersection which is at C then in the Triangle r B C right angled at B you have given B C 17 deg the Angle B r C the Complement of the Poles Elevation 40 deg to find C r the continuance of Twilight for C r and B u are equal I leave it to your own Practice I find the continuance of Twilight to be 27 deg 4 min. which converted into time is 1 h. 48 min. 4 13 nearest So I conclude that between the Day breaking and Sun rising it is 1 h. 48 min. 4 15 which is u B Now if you add this to the Difference of Ascension before found B 6 which was 1 h. 5 min. 3 15 of a minute I omit the smaller Fraction you have the time between break of day and 6 of the Clock which may be termed u B 6 2 h. 53 min. 7 15. And because the day breaks so much before 6 of the Clock if you subtract it from 6 hours you have the hour and minute of day breaking which is at 3 of the Clock 6 min. â—8 15. Add it to 6 of the Clock and you have the time of Twilight ending which is at 8 h. 53 min. 7 15. To find the Suns Place and Right Ascension provided the Latitude and Declination be given Latitude 50
in the work at all And indeed my desire to be brief makes me omit things that I think may be understood without treating of them I 'll only touch upon an Example in a Sphere which hath South Declination Latitude Northerly 50 deg 00 min. Declination 13 deg 15 min. Southerly Also if you subtract this Amplitude V ♈ from ♈ H 90 deg the remainder is the Suns Azimuth from the South V H or if you add V ♈ to ♈ A 90 deg it is V A the Suns Azimuth of rising from the North forasmuch as V A is as much above 90 deg as V H wants of it The same is to be understood in any Sphere that the Amplitude and Azimuth of the Sun are Complements one of an other to 90 deg in that Quarter When the Suns Declination is Southerly in any Northern Latitude the days are not so long as the nights so that the Sun cannot be up at 6 of the Clock which is at R therefore the Suns Azimuth at 6 of the Clock is of no use Neither is he up when he passeth the East point of the Horizon which is at P so that you cannot take his height but you may measure how far he is under the Horizon at either of those times or you may find his true Azimuth at 6 which is of no value to us that are Seamen because we cannot have a Magnetical Azimuth they are measured as before The Difference of Ascension is here after 6 of the Clock as much as it was before 6 in the last Example And the reason is because the Latitude is the same and the Declination of the Sun is just as far Southerly as before it was Northerly so that the Sun will now be just as long before he riseth after 6 as before he rose sooner it is V R take it off by the third way of Measuring as you did before and set it down I find it to be 1 hour 5 min. 3 15 it serves for the same uses that it did in the other Example For the length of the day subtract the Difference of Ascension V R from 6 hours which is R f and the remainder is f V the length of the Forenoon which doubled is the length of the whole day that subtracted from 24 hours is the length of the night For the hour of the Suns being due East it is P R which is from the place where the Sun cuts the East and West Azimuth to 6 of the Clock and set it off by the third way of Measuring convert your degrees and minutes into time and subtract that time from 6 hours and it gives the due time of the Morning that the Sun cuts the East point of the Compass the reason why you subtract is because the Sun is due East so long before 6 of the Clock Subtract the hour of the Suns being due East from 12 hours and it will give the true time of the Suns being due West For the time of day breaking draw the line 17 ⊙ parallel to the Horizon and 17 deg under it as was shewed before and measure from P to R that is from the place where the Suns Parallel of Declination cuts that Circle of 17 deg under the Horizon to 6 of the Clock convert it into time and subtract it from 6 of the Clock and it leaves the time of Day breaking namely the time in the morning that the Sun is in P you subtract because the day breaks so long before 6 of the Clock If it where so that the Sun were in the Tropick of Capricorn you must take the Distance S q. You see the Sun is past the hour of 6 before it is break of day now in this case you must add whereas before you subtracted For the continuance of Twilight it is measured as before from the place where the Parallel of the Suns Declination cuts the Circle of 17 deg to the place where it cuts the Horizon convert it into time and set it down In this Example it is P V. What I have not mentioned here I have given sufficient instructions about in the other Example before this Latitude 51 deg 30 min. Northerly Declination 11 deg 10 min. Northerly I demand the Suns place and right Ascension BEfore we can find the Suns place in the Ecliptick you are to consider that the Ecliptick is divided into 12 equal parts by the 12 Signs six of these Signs divide that half of the Ecliptick which is to the Northwards of the Aequinoctial and are called Northern Signs and the other 6 divide the half that is to the Southwaads of the Aequinoctial and are called Southern Signs as I have set them down in this Book before with their Names and Months to their Characters and have set the Northern Signs by themselves and the Southern Signs by themselves This plain Superficies can shew but one part of the Globous Body but you must know it is round which makes the Ecliptick to be divided on both sides from ♈ to ♋ is 90 deg from ♋ to ♎ is 90 deg from ♎ to ♑ is 90 deg and from ♑ to ♈ is 90 deg every one of these quarters containing 3 Signs The Ecliptick being thus divided into Signs we may find the Suns place and right Ascension THe Suns place in the Ecliptick is the nearest Distance between the next Aequinoctial point and the Sun in the Ecliptick by Aequinoctial points is meant ♈ and ♎ the two points of intersection that the Aequinoctial and the Ecliptick make Now in this case the Sun must be nearest the Equinoctial point at ♈ because the Month is belonging to a Sign nearer ♈ than ♎ Therefore fix your Compasses in the point of ♈ and extend the other foot to the Sun in the Ecliptick which is I and apply it to the line of Sines or if you have no line of Sines convert it to a Chord by the second way of Measuring and as many degrees as it is so far is the Sun distant from the nearest Aequinoctial point if it exceed 30 deg the Sun must be in that degree of ♉ that is above 30 deg if it exceed 60 deg the Sun must be in that degree of ♊ above 60 but here I find it in less than 30 namely 29 deg 3 min. therefore I conclude the Sun is in 20 deg 3 min. of ♈ But suppose it were time of year that the Sun were returning from the Tropick towards the Aequinoctial and were the same Declination then would the Suns place be the same from entring into ♎ that it now is in ♈ namely in 57 min. of ♠which wants 29 deg 3 min. of entring into ♎ Thus much for the Suns place The Suns right Ascension is measured from the place where the Suns parallel of Declination cuts the Ecliptick to 6 of the Clock which is I 6 set it off by the third way of Measuring to extend it to a Great Circle for the Sine of I
the Difference of Latitude in the whole run 13 leagues 3 10 if you had West Longitude you must have subtracted thus likewise and that which was greatest the remainder belongs to OF A RECKONING FRom what hath been already shewed you understand that if the Latitude of two places be known and the Difference of Longitude between them you may find the bearing of them one from the other their distance asunder or any thing you desire and this is all so plain that if I set from a place and am bound to a place whose Latitude is known and the difference of Longitude also known between them and keep a reckoning of the departure both East and West that I make setting them down in two distinct Columns it is but subtracting one from the other and the remainder will be the East or West Longitude that I have made that of the two which was most also if I know what Latitude I am in and what Latitude the place lies in it is but subtracting the lesser from the greater and the remainder will be the difference of Latitude between your ship and that place provided the Latitudes be both one way Also subtract the Difference of Longitude that you have made provided that it be that way which shortens your Longitude between the place you set from and are bound to from the whole difference of Longitude between the two places and the remainder will be what is still between you and the place you are bound to but if your departure which you have made hath increased the difference of Longitude between the place you set from and are bound to add the remainder after your subtracting the Columns one from the other to the whole difference of Longitude between the places the like for Latitude Example in the Reckoning following Suppose the Island of Ditiatha lies so that it hath been found to have 950 Leagues departure West from the Meridian of the Lizard according to the Courses that be ordinarily sailed by Plano the Latitude of it is 16 deg 16 min. North Latitude the Lizard lies in the Latitude of 50 deg 00 min. Now upon some Occasion that fell in the term of our Voyage namely the 29th of January on Tuesday the Captain demands of me what Distance Ditiatha was from me by Plano how it bears what Leagues of Departure I have to it and the like I look in my Reckoning and I find in the West Column 434 Leagues I look in the East Column and see 60 Leagues I subtract 60 my whole East Column from 434 the whole West Column and the Remainder is 374 Leagues which signifieth the Ship hath departed from the Meridian of the Lizard 374 Leagues West for the West Column was greatest Subtract this West departure 374 Leagues from 950 the whole Departure and I find there is still wanting 576 Leagues This 576 Leagues is the Departure that is from the Meridian that my Ship is in and Ditiatha Subtract also the Latitudes one from the other namely the Latitude your Ship is in and the Latitude of Ditiatha and the Remainder is the Difference of Latitude in degrees and minutes between the Ship and the Place you are bound to which in this Example I find to be 6 degrees 24 minutes Take the Latitude out of the Column of Latitude for that day in the Reckoning and you will find it so if you subtract 16 deg 16 min. from it Thus you have the Difference of Latitude between you that day and your Port and the Departure that the Port hath from the Meridian that you are in that day given you to find the things demanded I will not work it for it hath beeen shewed sufficiently already I have here following set down a Reckoning and afterwards how I work it and lastly the reason that makes me use this way of keeping my whole Reckoning in the last line so that I am not troubled to add it up when I give account of it January the 3d Anno Dom. 1655 we departed from the Lizard lying in the Latitude of 50 d. 00 m. being bound for Ditiatha which lies in the Latitude of 16 d. 16 m. it having West Departure from the Meridian of the Lizard 950 Leagues according to the courses which we then steered at 4 of the clock the Lizard bore from us N N E about 6 leagues distance by estimation EVery day at Sea you set down at noon that 24 hours work Now I advise you to set it down as it is here In the first Column on the left hand I set the day of the Month in the next the day of the Week in the third the Latitude I am in by dead Reckoning or by Observation only I make a distinction between them by the letter E which stands for Estimation and is set down only when I do not observe that so you may know where to begin to correct In the 4 th Column I set down the Easting in the fifth column the Westing The manner that I set my Easting and Westing down is thus Every day there is the Ships course minded and an observation whereby you have the difference of Latitude and course to find the departure or else if you cannot observe you have the distance run guessed at and the course to help your D. M. Week days Latitude Dep. E. in leag Dep. W. in leag deg min. 4 Friday 49 00 000 021 5 Saturday 47 55 000 059 6 Sunday 46 20 000 085 7 Munday 45 00 000 088 8 Tuesday 43 E 54 000 108 9 Wednesday 42 4 000 147 10 Thursday 41 E 19 000 262 11 Friday 41 10 000 264 12 Saturday 40 46 000 274 13 Sunday 40 45 000 178 14 Munday 39 E 49 000 179 15 Tuesday 37 E 24 000 200 16 Wednesday 35 59 000 229 17 Thursday 34 42 000 249 18 Friday 33 15 000 278 19 Saturday 31 15 000 322 20 Sunday 29 5 000 364 21 Munday 28 E 20 000 400 22 Tuesday 27 5 012 400 23 Wednesday 27 30 022 400 24 Thursday 27 22 031 400 25 Friday 26 E 49 049 400 26 Saturday 25 36 060 400 27 Sunday 24 31 060 406 28 Munday 23 28 060 406 29 Tuesday 22 E 40 060 434 30 Wednesday 22 08 060 434 31 Thursday 21 53 060 444 February 1 Friday 21 40 060 450 2 Saturday 20 47 060 456 3 Sunday 19 22 060 460 4 Munday 19 53 060 472 5 Tuesday 18 23 060 485 D. M. Week days Latitude Dep. E. in leag Dep. W. in leag deg min. 6 Wednesday 17 19 060 518 7 Thursday 16 47 060 556 8 Friday 16 54 060 606 9 Saturday 16 29 060 643 10 Sunday 16 19 060 678 11 Munday 16 21 060 726 12 Tuesday 16 28 060 784 13 Wednesday 16 35 060 840 14 Thursday 16 27 060 889 15 Friday 16 24 060 929 16 Saturday 16 2 060 973 17 Sunday 16 44 060 1003 guess is the Log-line every 2 hours your course and distance is set
down upon the Log-board every 2 hours whereby you may see it and work it every 24 hours to find the departure and difference of Latitude Now whatsoever I find the departure to be I add it from the first day all along thus the departure when I set the Lizard which was N N E 6 Leagues off is 2 Leagues West I keep mind of that till the next day at noon and I find what departure I have made from the Meridian that I was in the day before at noon by the things I have given me and here I found it was 19 Leagues I add it to my departure from the Meridian West yesterday 2 Leagues and it makes my whole departure West 21 Leagues Again the next day at noon I work my Ships Travis as hath been shewed and I find that she hath departed from the Meridian she was in yesterday 38 Leagues West which I add to what I had before which was 21 and it makes 59 Leagues and thus I go forwards adding my last days departure in its true Column to all the rest which is in one sum and by this means the last line is the whole Reckoning 21 38 59 In the East Column I set Ciphers to fill it up till I have some East departure to put in it and as that increaseth so I add it as the other every day Also I carry the West departure along in its full number of Leagues in the same line without increase or decrease till such time as I have more to increase it and then I increase that and carry the whole Easting with it in the same line and thus you have need to look upon nothing but the last line to resolve you any thing that you desire either of your course made good since you set out your distance upon a straight line that you have failed your Ports bearing from you what departure is yet between you and your Port your distance to it upon a straight line for there you have your whole Westing and your whole Easting and the Latitude you are in as also the Latitude of the place you are bound to whereby as I have formerly shewed you may find them If there be Longitude but one way you will see nothing but Ciphers in the other Column this way you see is done without a Plat. I hold a Plat to be necessary only to shew what dangers lie in your way that so you may shape a course clear by it and I should use a Plat for nothing else except it be for Coasting where it is really useful I commonly set down the Easting or Westing between the place I set and the place I am bound to at the beginning of my reckoning as also the Latitude of each place as I have done here You may ask why I do not set down the course that I made good every day indeed that is not unnecessary but the reason why I omit it is First because I find it of little use in my Reckoning and secondly I find it can be better expressed in a Journal which is kept with my Reckoning and indeed there I set it with the Winds and the reasons for steering upon such Courses but I leave every one to their own Judgment for that as also for Distances and Winds for Difference and Variation It would have been necessary here to have set down some Tables to have worked Triangles by for as I have said sometimes in 24 hours you have a Travis of 4 or 5 several courses and to work them this way may seem tedious I commend you to my Fathers Practice where there is as good Tables as can be to every degree of the Compass its use is easie and works to the tenth part of a Mile or League CONCERNING the VARIATION OF THE COMPASS THere is always two things given to find the Variation of the Compass that is the true Amplitude of the Suns rising and the Magnetical Amplitude of the Suns rising or setting the true Amplitude of the Suns rising is as I have said before in another place the true and absolute quantity of degrees that the Sun riseth from the East either Northwards or Southwards or sets from the West and is found as I have before shewed in the use of the Sphere The Magnetical Amplitude of the Sun is what the Sun riseth from the East or sets from the West by the Compass Now because the first gives the Truth how far the Sun riseth from the East or sets from the West therefore whatsoever difference there is between them so much is the variation As now Suppose I find in a certain Latitude such a day of the Month by the Sphere that the Sun riseth to the Northwards of the East 17 degrees that is the true Amplitude but I observe the Sun at her Rising with an Azimuth Compass which is made for that purpose and find that she riseth but East 10 degrees Northerly then I conclude the Variation is 7 deg 0 min. or the Compass is false so much so that whereas if I direct my Course East 10 degrees Northerly by the Compass I do not go on that Course but I go East and by North 5 deg 45 min. Northerly which is just 7 d. from my expectation or East 17 deg Northerly the true Amplitude Now that is a gross error and in long runs may deceive a man much and perhaps be a means to lose a Ship when one little thinks of it and therefore it ought to be looked to An Azimuth Compass is no other in effect but a Compass fitted for the exact taking of the Sun at her Rising or Setting or upon other certain times of the day as you may have occasion In like manner you may find the Variation of the Compass at other times by taking the Suns Azimuth at any time of the day Example Suppose I were in the Latitude of 33 deg 20 min. Northerly and upon the 8 th day of November the Suns Declination is Southerly 19 deg 20 min. I demand the Suns Azimuth at 8 of the Clock The Suns Azimuth at her Rising as I have shewed is the Complement of the Suns Amplitude but after the Sun is up it may be also the Distance of the Sun from the East and West Azimuth Now in this Proposition you desire to know how many Degrees the Sun is from the East and West Azimuth which is that Part of the Heavens that is distant from the Sun parallel to the Horizon over the East Point of it or if it had been at 4 of the Clock in the Afternoon it would have been required from the West Point For the resolving this Question project a Sphere as hath been taught for the Latitude proposed with the Parallel of the Suns Declination drawn as it is given Divide this Parallel of Declination into the hours of the day from 6 to 12 or which serves from 12 to 6 in the Afternoon the way to divide it is thus From 6 of the
Clock set off 15 degrees by the third way of measuring which will be 7 then because 30 degrees is two hours in time next set off 30 degrees by the third way of Measuring from 6 and it makes 8 a Clock then take 45 deg which is three hours and it makes 9 a Clock then 60 deg for 10 a Clock then 75 deg for 11 a Clock and then 90 deg which will just fall in the place where the Sun comes to the Meridian for 12 a Clock Thus you divide the Parallel of the Suns Declination into hours as you may see in the Sphere here following it being thus divided the Suns Azimuth at 8 of the Clock is 8 s taken off by the third way of Measuring I find it to be 33 deg 0 min. by the Scale so it is certain that the Sun should be just 33 deg 0 min. to the Southwards of the East at 8 of the Clock then I will see how high he should be at that time and I find him to be 13 deg 50 min. equal to s ♈ take off s ♈ by the second way of Measuring I will observe the Sun till I find him at that height and then I know it is 8 of the Clock see by your Azimuth Compass whether he be 33 deg 0 min. from the East or no and if not what it differs so much is the Variation of the Compass I have wrought this Example no nearer than the plain Scale works it which possibly may be 15 min. or half a degree of the Compass out of the way but that is no considerable error in a Course which is the thing we here use it for The same may be done at any time of the day as Latitude 32 deg 20 min. North Declination 19 deg 20 min. South Suppose it were cloudy but at some time the Sun breaks out so that I have an opportunity to take his height with my Quadrant and also his magnetical Azimuth with my Azimuth Compass Now I find his height to be r t equal to N P and the Azimuth at that time is N t it being drawn parallel to the Horizon this N t set off by the third way of Measuring and so much as it comes to so much is the true Azimuth of the Sun from the East towards the South Take your magnetical Azimuth at the same time and as much as they differ so much is the Variation And thus much for the Variation of the Compass When a man observes the Sun with a Quadrant which is our usual Instrument he takes the upper edge which is to the Northwards in this Latitude Suppose then I set from the Lizard and am bound to Barbadoes I make observations of the Sun as often as I can all the way till I bring the Sun to the Zenith after I pass the Zenith the edge of the Sun that was highest is lowest Now he that doth not consider this loseth the bredth or Diameter of the Sun so that your observations may differ from your expectation 30 miles which is enough to miss an Island therefore I advise you always to allow the Suns Diameter which some count 30 min. to every observation you make in places where the Sun is to the Northwards of you that is make your Meridian Altitude 30 min. less than it is which is the Altitude of the lower edge which was your upper edge before you crossed the Zenith where the Sun was or if you work your observations by the Complement of his height make that 30 min. more I confess the best way would be at all times to subtract the Semidiameter of the Sun which is 15 min. from your Meridian Altitude But because the Latitude of Places are not set down with this consideration it is more safe to do as before unless you know otherwise by your own experience This only by the by to give notice Example for the laying open of this Error Suppose I observe the Sun at Barbadoes the 11th day of December and I find that the Meridian Altitude of the Suns upper edge is 53 deg 18 min. to which I add the Suns Declination for that day which is Southerly 23 deg 30 min. and it gives the height of the Aequinoctial above the Horizon of the Complement of the Latitude to 90 degrees which is 76 deg 48 min. which subtracted from 90 degrees leaves the Latitude or Aequinoctial's Distance from the Zenith 13 deg 12 min. Upon the 11th of June I observe the Sun again in the same place and I find the Meridian Altitude of the upper edge of the Sun is 80 deg 12 min. Note that this upper edge is opposite to the edge before for the Suns Declination is 23 deg 30 min. Northerly which makes that the Sun gives a South Horizon I add the Suns Declination to the Meridian Altitude and it gives 103 deg 42 min. the Distance between the Aequinoctial and the North Horizon from which subtract 90 deg and the remainder is the Aequinoctial's distance from the Zenith or the Latitude which is 13 deg 42 min. Now this Latitude should agree with the other for the place stands still but for want of this subtracting the Semidiameter of the Sun it differs 30 min. and when men meet with things thus at Sea for want of minding of this they run 30 min. more Southerly than they should do which I suppose may be the greatest reason of their missing this Island and others that lie near the Aequinoctial but for those places that lie far from the bounds of the Sun it is not to be minded Yet it were good methinks if all Places had been laid down with allowance for the Suns Semidiameter but because they are not you ought to be careful when at any time you cross the Zenith of the Sun I have not been my self in a time of year to cross the Suns Zenith and therefore I only set this down by way of Caution to those that may The USE of a PLAIN SEA-CHART MOst Plats have all the Points of the Compass drawn out from several Centers through the Plat and because they are common and the way to work by them so ordinarily known I have thought it necessary to draw the manner of a Plat and to shew its use that hath only Meridians and Parallels drawn in it which is the Plat following Suppose then that I have a Plat that hath the Meridians and Parallels drawn to every fourth degree as this following is You see this Plat is divided into equal Squares throughout by the Meridians and Parallels so drawn also one of these Squares is divided into eight Points of the Compass which is the Square E C D O the manner of dividing it we will shew anon Now if A where the Lizard and B were the Island Madera I demand their distance asunder Let B A represent the edge of a Scale lying between them take the Distance A B and apply it to the Meridian and see