THE Seaman's Companion BEING A Plain Guide to the Understanding OF ARITHMETICK GEOMETRY TRIGONOMETRY NAVIGATION and ASTRONOMY Applied Chiefly to NAVIGATION AND Furnished with a Table of Meridional Parts to every third Minute With excellent and easie ways of keeping a Reckoning at Sea never in Print before ALSO A Catalogue of the Longitude and Latitude of the principal Places in the World With other useful things The Third Edition corrected and amended By MATTHEW NORWOOD Mariner LONDON â—â—â—â—â—ed by Anne Godbid and John Playford for William Fisher at the Postern-Gate near Tower-Hill Robert Boulter at the Turks-Head and Ralph Smith at the Bible in Cornhill Thomas Passinger at the Three Bibles on London-Bridge and Richard Northcot next St. Peter's Alley in Cornhill and at the Anchor and Mariner on Fishstreet-Hill TO THE READER THE famous and ever to be admired Art of Navigation having been so learnedly handled and written of not only in all other Languages but also in our Mother-Tongue by so many learned and able men both of former and of our present Age that it may seem impossible to write any thing more thereof that hath not already been done by others yet in my experience which I have seen being at Sea in several Vessels where divers young Mariners have been I have heard this general Complaint among them that though the things that are chiefly useful for them in their Art may be found in several Books here and there dispersed yet they could wish that there were such a Book contrived that might be soley useful for them entire by it self which then would be more convenient for them and might be purchased at a more reasonable rate than otherwise they could be by buying of so many sorts of Books which they must be constrained to do if ever they intend to be able Proficients in that most noble Profession of Navigation at which they chiefly aimed This Complaint of theirs was one chief Motive which induced me to collect and compose the subsequent Treatise which I have endeavoured to handle in such a methodical manner as it ought to be read and practised by the young Seaman For First There is a Treatise of ARITHMETICK containing all the Rules thereof which are necessary for the Seaman to know and practise all or most of the Questions thereof being made applicable in one kind or other to Nautical Affairs Secondly There is a Treatise of GEOMETRY containing the first Grounds and Principles thereof with the making and dividing of the Mariners Scale and Compass with the projecting of the Sphere in Plano and the resolving of many Questions in Astronomy which are useful in Navigation thereby Thirdly you have a Treatise of the practick part of NAVIGATION wherein is shewed after a new experienced way used by the Author how to keep a Reckoning at Sea the making and use of the Plain Sea-Chart the Doctrine of plain Triangles made applicable to such Questions in Navigation as concern Course Distance Rumb Difference of Longitude and Departure likewise a Table of Meridional parts to every third Minute and the application and use thereof exemplified in Questions of Sailing by Mercator's Chart with Tables of Longitude and Latitude of Places of right Ascension and Seasons of certain Fixed Stars with Rules to keep a Reckoning and to find the Latitude by the Meridian Altitude of the Sun or Stars Fourthly There is a short Treatise of ASTRONOMY wherein you have the Doctrine of Spherical Triangles applied to Questions in Astronomy and Navigation This is the brief Sum and Substance of the following Treatise which I commend to the Practice of all young Sea-men desiring their kind acceptance of these my first Labours which if I shall find to be kindly entertained by them it will encourage me to lanch farther into the more nice and critical part of this most noble Science In the mean time I commend this to them wishing good success in all their honest and laudable undertakings and in the interim bid them Farewel MATTHEW NORWOOD THE CONTENTS Of ARITHMETICK NUmeration 1 Addition 2 Subtraction 5 Multiplication 7 Division 10 Reduction 16 The Golden Rule or Rule of Three 19 The Rule of Interest and Interest upon Interest 24 The Rule of Fellowship 25 Of GEOMETRY Geometrical Definitions 28 How to raise a Perpendicular 30 How to divide a line into two equal parts 31 How to raise a Perpendicular on the end of a line ibid. From a Point given to let fall a Perpendicular 32 How to draw Parallel lines 33 How to make a Geometrical Square 34 How to make an Oblong or Square whose length and breadth is given 34 How to make a Diamond-figure whose Side and Angles shall be limited 35 To make a Rhomboiades whose Sides and Angles are given 36 To find the Center of a Circle 37 To find the Center of that Circle which shall pass through any three given points which are not situate in a streight line 37 How to divide a Circle into 2 4 8 16 32 equal parts 40 The Projection of the Mariners Compass 41 The Projection of the Plain Scale 42 The Projection of the Sphere in Plano 45 The Names and Characters of the Signs with the Months they belong to 48 How to project the Sphere 49 The meaning of certain Terms of Art 51 How to find the Suns Meridian Altitude 55 How to find the Amplitude of rising or setting ibid. How to find the Azimuth at six of the Clock 57 To find the Altitude at six 58 To find the Altitude the Sun being due East or West 59 How to find the Ascensional Difference ibid. To find the time of the Suns rising or setting 60 To find what hour it is the Sun being due East or West 61 To find the time of Day breaking and Twilight ending 62 To find the Continuance of Twilight ibid. To find the length of the longest day in any Latitude 63 To find the Suns Place and right Ascension 68 To find the Suns declination c. 69 Terms of Art used in Navigation explained 70 Prâ—positions of Sailing by the Plain Scale 72 Questions in Navigation resolved from page 73 to page 92 Of a Travis 92 The manner of keeping a Reckoning at Sea 97 Concerning the Variation of the Compass 102 The Use of a Plain Sea-Chart 108 Of Oblique TRIANGLES THe application whereof are in ten Questions perfectly explained all which Questions are applied to Questions in Sailing and are wrought both Geometrically by the Plain Scale and also by the Tables of Sines Tangents and Logarithms from page 113 to page 146 A Table of Meridional parts to every third minute 146 A Declaration of the Table 158 The Use of the Table of Meridional parts exemplified in five Questions appertaining to Navigation from page 160 to page 166 Of the Longitude and Latitude of Places A Table of them 167 The Use of them 171 How to keep a Reckoning of the Longitude and Latitude a Ship makes at Sea 172 The Names Declinations and Seasons
of their intersection draw your line to the end of your given line and it shall be a Perpendicular you see it is the same the other was PROP. IV. To let fall a Perpendicular from a Point to a given Line LEt the given line be D A the point from whence the Perpendicular is to be let fall be at C. From the point C draw a white line to the given line by guess as C A divide it into two equal parts which is done at B then continuing ½ the line C A which is A B or C B in your Compasses and your Compasses fixed one foot at B describe the Arch C D and where it cuts the given line there will your Perpendicular fall from the given point for C D is Perpendicular to the given line D A. PROP. V. To draw a Line parallel to a Line given LEt the given line be A S It is required to draw a line so that the two lines may run at both ends one by the other and never meet which is parallel one to the other Open your Compasses to that extent as you would have the two lines asunder and go towards one end of the given line as at S and describe the Arch u and with the same distance come towards the other end as at A and describe another Arch which is N and by the top of these two Arches draw the line R O which is parallel to A S. PROP. VI. To draw a Line parallel to a given Line from any Point assigned LEt the given line be S L the point assigned be A take the distance from S to A and carry it towards the other end of the given line as at L describe the Arch n then take the distance from L to S and fixing one foot of your Compasses in the given point A cross the Arch n with the arch o and by the place of their intersection and the point assigned draw a line which shall be parallel to S L. PROP. VII To make a Square of a Line given LEt the Line given be A equal to which draw the side of the Square B E and from one end of it raise a Perpendicular and by it set off the length of A as here from E the Perpendicular was raised and the length of A set upon it which is the side of the Square E D continue the same distance in your Compasses and go to D and describe the Arch 8 carry the same distance to B and cross the Arch 8 with f and from the intersection of those two Arches draw the sides C D and C B which makes the Square BEDC this is a true square PROP. VIII To make a Square whose Length and Bredth is given THese sorts of Squares are called Geometrical Squares when but two sides are equal namely the two longest sides or the two shortest sides The Angles are all equal namely right Angles Suppose the Length of the Square be A the bredth B I desire to make it first draw a line equal to A for the length of it as S V then from any end of that Line raise a Perpendicular as here from V and set off the line B for the bredth of it upon it which falls in the Perpendicular line at L then take the length of A and describe the Arch n fixing your Compasses one foot in L then take B the bredth and cross that Arch by another fixing your Compasses one foot in S and draw L K and K S from the place of their intersection as you did in the other Thus the two opposite sides in this Square are equal and the Angles in both all equal for they are right Angles PROP. IX To make a Diamond Figure of a Line and an Angle given A Diamond Figure is a Figure of four equal sides but the Angles are two of them acute and two of them obtuse the acute Angles are equal and the obtuse Angles are equal one to another Let A B be the given line C B the measure of the given Angle A being the angular point first take the line AB and draw a line of its length for one side of the figure namely 8 0 then take the Semidiameter of the Arch C B which is C A and fixing one foot of your Compasses in 8 describe the Arch S 0 and take the Arch B C and set it off from O to S then draw the line S 8 equal to 8 O this done keep the length of O 8 in your Compasses and from S and O describe the Arches n and t and draw the sides O R and S R as you did in the other figures and thus S R is equal to 8 O or R O is equal to 8 S and the opposite Angles also equal I forbear to shew the reasons of their being equal because it hath been handled by others and indeed it is so plain that with a little consideration you may know it PROP. X. To make a Rhomboiades of two given sides and an Angle included A Rhomboiades is a figure whose opposite sides and opposite Angles are equal as a Geometrical Square is but in this they differ a Rhomboiades hath never a right Angle but two obtuse and two acute whereas the other hath all right Angles it differs from the Diamond figure also for in one the sides are all equal and in this but two equal sides I need not shew the working of it because it differs not from a Diamond figure only in taking the two sides apart to describe the Arches at L I suppose you may conceive how it is made by seeing this which is here made the given sides and Angle is s O N I have not set down the Arch to measure the Angle at O. I suppose from what hath been said you will conceive how that is PROP. XI To find the Center of a Circle Draw a line from side to side of the Circle at a venture as A C and divide that line into two equal parts by a Perpendicular as was shewed before that Perpendicular line draw through the Circle from side to side as is u S and it shall be the Diameter of the Circle the half of which is the Semidiameter or very Center ⊙ It is possible to find the Center of a Triangle after the same manner Suppose the Triangle whose Center you would find were A C n divide any side into two equal parts by a Perpendicular and it will go through the Center of the Triangle as the side A C is divided into two equal parts by the Perpendicular B S then I say B S goeth through the Center of this Triangle but to find whereabouts in this Perpendicular the Center of the Triangle is I know by no other means but by removing your Compasses in this line from place to place till you find it which is here found to be at u. But this is but a botchingly way and with a little more labour you may find it at once therefore mind
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
Fellowship THis Rule is the Rule of Three done several times in one Sum or Question for here are several Stocks and several men that own them now if the principal or all their Stocks gain so much how much shall each man gain according to the Stock he put in I 'll say if the whole Stock gained so much what shall the first mans Stock gain and so for the rest Quest 1. Six men make a Stock the first puts in 30 l. the second 40. l. the third 52 l. the fourth 58 l. the fifth 60 l. and the sixth 78 l. If the whole Stock together gains 200 l. what shall each man have so that there may be no wrong First find the whole Stock by adding every mans Stock together which is 318 l. Then say If the whole Stock 318 l. gains 200 l. what shall the first mans Stock gain  l. First 30 Second 40 Third 52 Fourth 58 Fifth 60 Sixth 78  318 For the Second If the whole Stock 318 l. gains 200 l. what shall 40 l. gain which is the second Stock The like is to be understood of the rest I forbear to work them as being out of my intentions I leave their operation to your genius who never learned them Quest 2. There where five men made a Stock the first put in 200 l. the second 59 l. the third 180 l. the fourth 78 l. the fifth 240 l. they lost at the return of the Ship 120 l. I demand each mans Loss proportionable to his Venture First find the whole Stock by adding every mans Venture together this done say for the first mans Loss if the whole Stock lost 120 l. what shall the first mans Stock 200 l. lose For the second If the whole Stock lose 120 l. what shall the second man lose which put in 59 l And so for the rest The way to prove one of these Questions is less trouble than the Rule of Three for after you have done add all your Facits together and it will make the same that was lost in the whole Stock or gained in the whole Stock if the Question be for Gain and this stands to good reason for every mans Loss or Gain must together be equal to the whole or else they do not contribute to the Loss or enjoy the Gain that they have lost or gained There is more intricacy in this Rule than I have here cited but the Foundation-work lies on the Rule of Three and the intricacy that I speak of will be understood with some consideration I had thoughts to have shewed it but it is shewed as well as can be I think by Mr. Record and Mr. Hodder besides it would take up much paper and I should digress from my intentions to Navigation What I have shewed since I treated of the Rule of Three is only a shew how all Rules depend upon this Golden Rule Blame me not for being so large in Arithmetick but both that and all faults else season with the Salt of a charitable Construction remembring that Navigation is imperfect without it I end GEOMETRY THat which I shall handle in Geometry will be only that part of it which is used in Navigation He that will treat of it at large had best to put it in a Treatise by it self and he that will learn to be a good Geometrician let him apply himself to Euclid's Works There is no Art but hath a dependency on it and Navigation depends much on it which enforceth me to treat upon some principles of it Geometrical Definitions A Point or Prick is this and is void of length breadth or thickness A Line is length without bredth or thickness and is properly called the nearest distance if a streight line between two places but if not streight but circular it is termed an Arch. A Triangle is when three lines meet making three angular points now there be two sorts right lined and spherical spherical being all the sides Arches of great Circles There be three sorts of Angles namely obtuse acute and right angles If divers Circles be described having all one Center they be called Concentricks but if they have divers Centers they be called Excentricks The Circumference of any Circle consists of 360 degrees every degree being 60 minutes The Diameter of a Circle is a line drawn from one side of the Circle to the other through the Center As A C. The Semidiameter of a Circle is half the Diameter or the distance from the Circumference to the Center PROP. I. To raise a Perpendicular from the middle of a Line given THe line given is A B divide it into two equal parts at F then open your Compasses to any convenient distance above half the length of the given line and fix one foot in B and with the other describe the Arch C with the same distance fix one foot of your Compasses in A and cross the other Arch by the Arch D mind where they intersect and from that place which is ⊙ draw a line to the middle of your given line A B namely to F then is F ⊙ a true perpendicular from the middle of the line A B for it is so directly from it that it leans no way PROP. II. To divide a Line into two equal Parts by a Perpendicular THe given line is A B fix one foot of your Compasses in A and setting them at any convenient distance above half the given line describe the Arches D and f carry your Compasses to B with the same distance and describe the Arches C and e so as they may cross the other Arches as they do in ⊙ mind it and by those intersections lay your scale and draw the line ⊙ ⊙ which will cut the given line in the middest and be a perpendicular to it which was required PROP. III. To raise a Perpendicular to the end of a Line given HEre your given line is A B fix one foot of your Compasses in that end of your given Line which you would raise your Perpendicular from which is B and extend your Compasses to any convenient distance as in the example from B to R and describe the Arch R c D continue the same distance in your Compasses and fix one foot in R extend the other upwards to c then making c the Center describe the Arch 1 2 3 and beginning at R set off that distance three times upon that Arch as 1 2 3 from the place where your Compasses fall the third time which is 3 draw a line to the place where you began your work which is B and it is a Perpendicular to your given line Another way After you have described the Arch R c D with the same distance which described it extend your Compasses from R to c from c with the same distance describe the Arch u which is part of the Arch R 1 2 3 then extend the same distance again which falls at D and cross the Arch before described as N and from the place
the Course For the Course As the Distance run A B 108 miles comp arith 7,966576 Is to Radius   So is the Difference of Latitude A D 90 miles 1,954242 To the Course or Rumb Sine comp B 33 d. 33 m. 9,920818 For the Departure As Radius  To the Distance run A B 108 miles 2,0334237 So is Sine the Course A 33 deg 33 min. 9,9424616 To D B the Diff of Longi here found to be 60 miles 1,7758853 You may ask the reason why the Course is not 33 deg 45 min. as it was first given to be I answer because of the part of an unite that the Difference of Latitude was set down more than it should be for if you observe I said it was almost 90 miles when I found it before Now the error is not worth minding in Sailing for the Difference that the Fraction causeth is but 12. min. of a degree of a point of the Compass which is no more than the 56 th part of a point If you mind I have worked the Course thus found and it doth not alter the Difference of Longitude or Departure which should be 60 miles if I had worked to the least Fraction for the Logarithm here is nearest to the Logarithm of 60 in the Tables Indeed I worked on purpose thus because those that I have taught when they could not find it come out right as the other was could not tell where the fault lay now this will direct them to know that some part of a unite missing in the finding of the Sides may make some minutes Difference in the Course But now here in the following Examples we will work to a Fraction namely to the 10 th or 100 th part of a unite in the sides to shew how it is done though in Sailing if you work to half an unite it is near enough or for any ordinary uses and that the Tables give and most commonly to less Departure from the Meridian and Distance run given to find the Course and Difference of Latitude QUESTION V. Admit the Departure from the Meridian were the same that the Difference of Latitude was in the first Question 60 miles the Distance run also 108 miles I demand the Course and Difference of Latitude let the Latitude you set from be 50 deg HEre the Course from the Meridian will be the Complement of the Course there and the Difference of Latitude will be the Difference of Longitude or Departure from the Meridian there But instead of taking the Logarithm answering to the number here take the Logarithm answering to ten times the number and so you shall always have your sides come out to the tenth part of a unite I say it will come out in tenths for you must divide by 10 to bring it into miles after it is done Now instead of multiplying your given sides by 10 do but set a Cipher more to the number and it is done as now for 60 set down 600 which is ten times 60 for 108 set down 1080 which is ten times 108 and thus for any number that will be comprehended in the Radius of the Tables Proportion for the Course As Distance run given B A 1080 tenths comp arith 6,9665762 Is to Radius  So is the Departure from the Meridian 600 tenths D B 2,7781512 To the Sine of the Course A 33 deg 45 min. 9,7447274 For the Difference of Latitude As Radius  To the Distance run 1080 tenths B A 3,0334327 So is the Sine comp of A 33 deg 45 m. 9,9198464 To the Diff. of Latitude in tenths D A 898 tenths 2,9533708 Divide these tens by 10 and it gives the miles contained in the side D A the Difference of Latitude which is 89 miles 8 tenths of a mile whereas before it was 90 miles whis is 2 tenths of a mile more Now when you have the sides thus found in tenths or Centisms work for the Angles and you will find them to come out roundly alike But if you desire more exactness work then in Centisms provided that the numbers in Centisms do not out-run the Tables A Centism is the hundredth part of an unite or of any thing and if I would put the number 60 into Centisms I will set two Ciphers behind it and it is the same as before I multiplied 60 by 100 for it is 6000 the like is to be understood of any other number I would work to the nearest Fraction for the Angles here but I conceive it to be no way beneficial and therefore I 'll refer it to the work of the Sphere where it is of â—uch more use this finds the Angles to a minute which is near enough It is like that some will be so curious that this way of finding things to the tenth or hundredth part of unit will not suffice them but they would have the real number answering to the Logarithm to the 1000 th part of a unite because I find it of no real consideration or use in Navigation I omit that here and desire such to look in Book 1. Page 11. of my Fathers Trigonometry and there it is plainly shewed By the Plain Scale This Question differs nothing from the other in its operation only as you work in the other from a North and South line here you work from an East and West line The work is this First draw an East and West line and from the west end of it set off your Departure from the Meridian which is from D to B 60 miles from D raise the Perpendicular D A which is a North and South line and draw it at length then take the Distance run 108 miles and fix one foot of your Compasses in B with the other cut the Meridian line D A which it doth in A. Draw the side A B then is B the place your Ship is at after her Sailing A is the place of her setting out A D is the Difference of Latitude and N O s is an Arch of 60 deg from the Angle D A B as a Center the measure of N O is the Course and is performed as before in the last Example and also of the Difference of Latitude as in other Examples The Difference of Latitude is D B 90 miles The Course is North 33 deg 45 min. Easterly D A B or N O. The Plain Scale of equal parts will not resolve to the least Fraction as the Tables will now to a tenth part of a mile I confess a good Diagonal Scale will produce things to a small Fraction If you would know what Latitude you are in divide 90 by 60 and the Quotient will be degrees namely 1 deg the Remainder minutes namely 30 min. Now because your Latitude was Northerly and you have gone to the Northwards you have increased it 1 deg 30 min. so that you must add the Difference of Latitude to 50 deg and it makes 51 deg 30 min. if you have gone to the Southwards you must
purpose to shew the truth of the general Rule The same way other Questions of this nature are wrought For the Distance that the Southermost Ship sailed A O. As A O B Sine 67 deg 30 min. comp arith 0,034384 Is to A B 58 miles  2,763428 So is Sine A B O 45 deg 00 min.  9,849485 To A O 44 4 10 miles  2,647297 If you desire to find the Difference of Latitude between the two places B and A you have the Distance A B and the Course A B n given to find it or the Longitude If you desire to find the Latitude the Ships are in when they meet let fall a Perpendicular from O upon the South line which is B N and you have the distance O B given 58 miles and the Angle O B R 67 deg 30 min. to find it or the Departure between the Ships at their meeting and the Head-land O r. The like is to be understood of any other Question of this nature Two Sides and their contained Angle being given to find the third Side and the other Angles QUESTION III. Two Ships set from one Port and make an Angle of 58 deg one sailed N b E and the other sailed N W 1 deg 45 min. Westerly the Eastermost Ship sailed 70 miles and came to her Port The Westermost Ship sailed 89 miles and came to her Port I demand the Distance of these Ports asunder and how they bear one from the other FIrst draw the North and South line A S. Let A B represent the Port you set from Then from A as a Center describe the Arch of 60 deg F S D From S to the Eastwards set off one point for N b E the Eastermost Ships Course draw the line A D at length and set the first Ships distance 70 miles upon it which is A B then from A draw a N W line 4 deg 45 min. Westerly and set off the Westermost Ships distance run upon that which is 89 miles and it reacheth from A to C draw the line C B from the ends of the other two sides which is the Distance between the Ports for the ends of each Ships run must be the Ports then is A C B the Angle of the bearing of the Eastermost Port from the Westermost or A B C the bearing of the Westermost Port from the Eastermost C A is a S E line 1 deg 45 min. Easterly from C see how many points are contained between the C B and C A by the way that is shewed in the Question before or because B A is N b E line from B you may find the Point of the Compass that B C runs upon from B you need find it but one of these ways and the opposite point must be the other as I have shewed before then take the length of the side C B which is the distance of the Ports asunder I find their distance and bearing to be as followeth The Distance between the two Ports is C B 79 miles The Bearing of them is East 5 deg 35 min. Northerly or West 5 deg 35 min. Southerly By the Tables Here is given the Angle at A 58 deg 00 min. the side A B 70 miles the side A C 89 miles which is two sides and their contained Angle and here is required A C B or A B C and the side B C. First For the Angles For all Questions of this nature this Rule is beneficial and will work your Question Note That in all plain Triangles the Sum of two Sides are in such proportion to their Difference as the Tangent of the half Sum of their opposite Angles is to the Tangent of an Angle which Angle shall be the Difference between the half Sum and the Angles so that if it be added to the half Sum it shall make the greater Angle if it be subtracted it shall make the lesser I have said something of this Rule before which makes me demonstrate it no farther here Proportion As the Sum of A B and A C 159 miles comp arith 6,79860 Is to their Difference 19 miles  2,27875 So is the Tan. of the half Sum of B and C 61 d. om 10,25624 To Tangent of an Angle 12 10 9,33359 This added to the half Sum makes the Angle at B 73 d. 10 m. subtracted it makes the Angle at C 48 deg 50 min. which is 4 points and 3 deg 50 min. then say I if C A be a S E Course 1 deg 45 min. Easterly from C 't is certain that C B must be 4 points and 38 deg 50 min. from it which is East 5 deg 35 min. Northerly the bearing of the Eastermost Port from the Westermost To find the half Sum of the Angles Note that the three Angles are 180 d. therefore if one of them be known subtract that from 180 d. and the Remainder must be the other two Now for their Distance asunder C B. As Sine A C B 48 deg 50 min. comp arith 0,123321 Is to B A 70 mil.  2,845098 So is Sine B A C 58 deg co min.  9,928420 To the Distance of the Ports asunder C B 78 9 10 2,896839 The three Sides of a Triangle given to find the Angles QUESTION IV. Two Ships set from one Place one sails N b E 70 miles the other sails between the North and the West 89 miles and at the end of their sailing they are distant asunder 78 miles 9 10 of a mile I demand what Course the Westermost Ship sailed and how the two Ships bear one from the other FIrst draw the North and South line A e make A the place of the Ships setting out set off a N b E Course upon the Arch of 60 deg I e which is e D because the Eastermost Ship sails N b E likewise set off her Distance which she sailed upon that Course which is A B 70 miles then take the other Ships Distance which is 89 miles and fixing one foot of your Compasses in A with the other describe the Arch at C Lastly take the Distance of the two Ships asunder 78 miles 9 10 and fixing one foot of your Compasses in B cross that Arch and from the place of intersection draw the sides of the Triangle C A and C B then measure the Angles from an Arch of 60 deg as hath been shewed before and to know them consider that A B is a N b E line so that A F will be upon a Course so many points or degrees and minutes from a N b E as there are points or degrees and minutes in the Arch D F. Secondly for the bearing of the Ships which is the Angle at B consider that B A is a S b W line from B and then the degrees and minutes or points contained in the Arch O R is so many degrees and minutes or points from the S b W. I have resolved the Demands which are as followeth The Course that the
which is A C and cuts the the Westermost Head-land Lastly draw out a South line from A which is A D and cuts the Island Lastly A s and R s are the Courses that lead from the two Stations to the Island and therefore s is the Island for there they cut Draw a line from every one of them as from s to O and from O to I from I to s and take the distance of them asunder then for their bearing one from another draw a North and South line from each place and by an Arch of 60 deg measure it as hath been shewed By the Tables First I consider that I have the Distance between the two Stations R A which is a side of the triangles A R I A R O and A R s and in every of these Triangles the Question gives me all the Angles so that I can find my Distance from my Station to the Head-lands after the form of this Example which I shall do from each Station to the Eastermost For A I. As Sine R I A 38 deg 45 min. comp arith 0,20347 To A R 8 miles  1,90309 So is Sine A R I 28 deg 45 min.  9,68213 To A I 6 miles 2 10  1,78869 For R I. As R I A Sine 38 deg 45 min. comp arith 0,20347 To A R 8 miles  1,90309 So is R A I Sine its comp to 180 d. 67 d. 30 m. 9,96561 To R I 11 miles 8 10  2,07217 Here I have found the Distance from each Station to the Eastermost Head-land next find the distance from the Westermost Head-land to the first Station A O. For A O. As Sine R O A 6 points or 67 d. 30 m. co ar 0,034384 Is to R A 8 miles  1,903090 So is O R A 67 deg 30 min. Sine  9,965615 To A O 8 miles  1,903090 Thus you might find R O R s and A s for you have the same things given namely R A and the Angles Here we have found the sides A I and A O and because A I is upon a N N E Course and A O upon a N W Course the Angles contained between O A I must be 6 points or 67 deg 30 min. so that we have two sides and their contained Angle to find the Angles of the bearing of these Head-lands A I O or A O I and the Distance of them asunder O I this is wrought as the third Question is Thus as this is found so you may find the bearing of the rest of the places one from the other if there were twenty of them for you have as much given in the other Triangles but I leave it to your own practice also to make such Experiments or to frame such Questions Two Sides and a contained Angle being given to find the third Side and the Course that each Side runs upon provided that no Course be named only the half of the Compass that you sail in or the quarter which is most commonly and the Difference of Latitude that is made between the extent of the two Sides QUESTION VI. Two Ships make an Angle of 50 deg the one sails between the South and the West 40 leagues the other sails between the South and the East 50 leagues and at the end of their sailing the Eastermost Ship is more Southerly than the Westermost so that they differ their Latitude 15 leagues I demand each Ships Course and their Distance asunder THe place the Ship set from I make to be B take off 60 deg from your Scale and describe an Arch making B your Center as the Arch T R then I take 50 deg from my Scale and set off somewhere upon that Arch namely from T to R and from B upon the Course that leads to T I set off 40 leagues my Westermost Ships Distance and so conclude my Westermost Ship is at C then from B upon the Course that leads to R I set off my my Eastermost Ships Distance run which is 50 leagues from B to A. Lastly draw A C which is the Distance of the Ships at the end of their sailing Then because the Question saith the Eastermost Ship was 15 leagues more Southerly in Latitude than the Westermost I will take 15 leagues from the Scale of equal parts and fix one foot of my Compasses in A and with the other describe the Arch I S N and draw the line from C by the upper edge of the Circle at length which must be an East and West line because A is just 15 leagues to the Southwards of the nearest part of it then let fall a Perpendicular from B upon C N and extend it as far as the Arch of 60 deg which you first described T R and it cuts in O this must be a South line from B namely B S then measure upon the Arch of 60 deg from O to R for the Eastermost Ships Course and from O to T for the Westermost Ships Course as you do in any other Question I find them here as followeth The Eastermost Ship sailed S b E 4. d. 40 m. Easterly The Westermost Ship sailed S W 11 deg Southerly The Distance asunder is C A 39 leagues By Proportion Here are the two sides B A 50 leagues and B C 40 with the contained Angle C B A given to find the Angles A C B and B A C therefore for the Angles As B A and B C 90 leagues comp arith 7,045757 Is to their Difference 10 leagues  2,000000 So is Tan. half the Angles B A C and B C A 65 d. 0 m. 10,331327 To Tang. their Difference 13 deg 24 min. 9,377084 Which added to the half sum 65 deg is 78 deg 24 min. the Angle A C B subtracted from 65 deg it makes 51 deg 36 min. the Angle C A B. Then for the Ships Distance asunder C A. As Sine A C B 78 deg 24 min. comp arith 0,008962 Is to B A 50 leagues  2,698970 So is Sine C B A 50 deg 0 min.  9,884254 To C A 39 deg 1 10 leagues  2,592186 Here we have found the required Angles and the required side but we know not what point of the Compass either of the sides that bounds the Triangles runs upon but this which we have found makes way to it for now we know the side A C to be 39 1 10 leagues and if you let fall a Perpendicular from A upon the East and West line C N it cuts it in S then A S is known to be the Semidiameter of the Arch I N S u 15 leagues so that in the Triangle A S C right angled at S we have A S 15 leagues given and A C 39 1 10 leagues to find A C S or C A S we will find A C S. As A C 39 1 10 leagues 7,407823 Is to Radius or A S C 90 deg  So is A S 15 leagues 2,176091 To Sine A C S 22 deg