upon the point 1 in the South line T L describe a little Ark at a also out of the South line take T 1 and with it on the West Line at 1 describe an Ark at a crossing the former so the cross at the point a shewes the place where the ship is at the first Course 2. Again take T 2 out of the West line and setting one foot of the Compasses upon 2 in the South line describe a small piece of an Ark near b also take T 2 out of the South line and setting one foot at 2 in the West line describe an Ark crossing the former at b which point is the place where the Ship was at the end of the second Course 3. Take T 3 out of the South Line and setting one foot of the Compasses on 3 in the West Line describe a small Ark near c Again take T 3 out of the West Line and upon 3 in the South Line draw an Ark crossing the former at c and c is the Point where the Ship is according to this dead reckoning and the former Point a and b need not have been found for there is no question proposed concerning them To finde the Course and Distance from c to N. 1. For the Distance the extent c N measured on the Scale of Leagues sheweth it to be 114 Leagues and a half 2. For the Course to find it without drawing lines on the Chart lay a ruler over N and c which we must suppose to cross somâ— Meridian or parallel in the Chart here it crosseth the South line at e and the Scale of Leagues at f then place the Radius or 90d of the line of greater Sines which before was pricked from T to R from e to g and the nearest distance from g to the edge of the ruler measured on the Sines sheweth the Course from the Ship to St. Nicholas Island to lye 43d 17′ to the Westward of the South then if you look for 43d 17′ in the greater Chord you shall finde in the Rumbes against it that this Course is 3 points and above 3 quarters more to the Westward of the Meridian that is almost Southwest and if e g had been placed from f toward A the nearest distance to the edge of the ruler would have shewed the complement of the Course required 3. To finde what Course the Ship must steer to bring her self about 23 Leagues or 22 Leagues 8 tenths East from St. Nicholas Island draw a Line from N to L and prick down 22 ⌊ 8 out of the Scale of Leagues from N to d and this may be performed by the edge of a ruler without drawing any such Line or otherwise it may be found by the intersection or crossing of two Arks as the former Traverse Points were found Now laying a ruler over d and c it crosseth the South Line at h then place T R from h to k and the nearest distance from k to the edge of the ruler measured on the greater Sines sheweth the Course to be 33d 45′ to the Westward of the South that is three points to wit South-west and by South 4. The extent c d measured on the Scale of Leagues sheweth the distance to be 100 Leagues Lastly we suppose the Ship to sail this Course and distance till she come into the parrallel or Latitude of St. Nicholas Island at d and then she sailes almost 23 Leagues West and arrives at the Island being her desired Port. If it were required to finde the Ships Course and distance from the Point c to Tenariff Lay a ruler over the Points T and c the nearest distance from R to the edge of the ruler measured on the greater line of Sines sheweth that the Course from Tenariff to the Ship is 30d 47′ to the Westward of the South which is almost two points three quarters and contrarily Tenariff bears from the Ship as much to the Eastward of the North and the extent c T measured on the Leagues sheweth the distance of Tenariff from the Ship to be 158 Leagues This I think sufficient to explain the use of the Plain Chart upon which in the laying down of any two places in their Rumbe and distance there is framed a right angled Plain Triangle one side whereof T L is the difference of Latitude the other side L N the difference of Longitude and the third side N T the distance And after the same manner a Traverse is to be platted when the Chart is made true as to the Latitudes of places and as neare the truth as to the Rumbe and distance as it can waving the longitude and as the whole Navigation of a Voyage is performed on this Chart without drawing any lines thereon to deface it so after the same manner and well nigh with as much ease may it be performed on the true Sea Chart commonly called Mercators as shall afterwards be shewed But before I part from this Discourse I think it necessary to shew how the former Scale of Leagues may very well be spared and still account the distance run in Leagues and tenths provided the degrees of Latitude on the Sides of the Chart be divided each of them into 10 equal parts accounting each degree for 10 Leagues Take 40 Leagues and setting one foot in 80 leagues draw the Ark G which may be done by any other numbeâ—s in the same Proportion and draw a line from H â—ust touching the said Ark and the Scale of leagues A B may be spared Suppose I would take out 60 leagues take the nearest distance from six in the latitude side of the Chart to the line H G and it is the measure of 60 leagues and so much is the distance at her first Traverse from T to a. But supposing the distance T a in the Chart were unknown and I would measure it take the said extent and prick it twice in the side of the Chart from H and it will reach to 60 and so many leagues is the distance required and so any other extent turned twice over shewes the distance in leagues accounting every degrees 10 leagues and being turned but once over shewes the distance in degrees and Decimal parts How in estimating the Ships Course and Distance to allow for known Currents This subject is handled by Mr. Norwood in his Sea-mans Practice at the end and by Mr. Phillips in his Advancement of Navigation page 54 to 64. As also how to how find them out by comparing the reckoning outwards between two places with that homewards wherefore I shall be very brief about it and perform that with Scale and Compasses which is by them done with Tables 1. If you stem a Current if it be swifter then the Ships way you fall a stern but if it be slower you get on head so much as is the difference between the way of the Ship and the race of the Current Example If a Ship sail 7 Miles North in an hour by estimation against a Current that sets South
4 Miles in an hour then the ship makes way three Miles or a League in an hour on head but if the Ships way were 4 Miles in an hour by estimation North against a Current that sets 7 Miles in an hour South the Ship would fall 3 Miles or a league a stern in an hour But suppose a Ship to cross a Current that sets S S E about 3 miles an hour and first the Ship in 4 hours sayls 8 leagues East and by South by the Compass then in 8 hours more she sayls 12 leagues East South-east by the Compass Now it is demanded what Course and Distance the Ship hath made good from the first place where this recknoning began In resolving this we shall account every ten leagues of the former Scale of leagues in the Plat to be but one Draw two lines at right Angles at the Center A thus First draw the line A F then with 60 degrees of the Chords describe the prickt quadrant F I L and therein set off one point from the East to I and draw a line to A this is the line of the Ships first Course wherein prick off 8 leagues from A to B. Prick off the course of the Current being 6 points to the Southwards of the East from F to E and draw the line L A then because the Current in 4 hours sets 4 leagues forward in its own race draw the line B C parallel to A L by the former directions and prick down 4 four leagues from B to C and the line A C shews what course and distance the Ship hath made good the first Watch or four hours Then for the second Course draw C H parallel to the line A F and with the Radius upon C as a Center draw the Arch H D wherein prick two points for the Ships second Course from the East from H to D and draw D C wherein prick down C D the Ships distance run in that Course and draw D G parallel to A L as you did B C then because the Current sets 8 leagues in 8 hours prick down 8 leagues from D to G and joyn A G so the extent A G being measured on the Scale of leagues sheweth that the Ships direct distance from the first place is about twenty eight leagues and a half then measure the Arch F M on the Rumbes and you will finde it to be a little above 3¼ points from the East so that the Ship hath made her way good South-east and by East and above a quarter of a point more Southwardly and is now at the point G whereas if there had been no Current she had been but at N in the same line with D G and distant from it equal to B C. And what we have done here with drawing many Lines after the old fashion of keeping a reckoning on the Plain Chart may be done by help of the Traverse-quadrant by Intersection and points onely without drawing any other Lines at all but the two Lines making right angles at A in which are plotted the Ships two Courses and distances as also the Currents Course and distances proper to each of the Ships Courses and the Variation and Separation proper to each Course hath the figures 1 2 3 4 on the South and East-line truly set down whereby may be found the 4 points B C D G after the same manner as the reckoning was platted between Tenariff and St. Nicholas Island before and in stead of laying down the Current at the end of each Course and distance it may be done at once for many courses and distances if you bring them first into one right line from the first place and cast up into one Sum how much ought to be allowed to each Course and distance during the time of its continuance and from the Traverse Point of the Ships place so found set off the allowance for the Current in a Rumbe line that runs the same way with the Current How to Rectifie the Account when the dead Latitude differs from the Observed Latitude By the Dead Latitude is meant the Latitude in which the Ship is by the Dead Reckoning or Estimation and this is always given as in the former Chart between Tenariff and St. Nicholas Island by the points 1 2 3 in the South-line if they be measured in the other side of the Chart amongst the degrees of Latitude The whole Practise of the art of Navigation in keeping a due reckoning consists chiefly of three Members or Branches 1. An experienced judgement in estimating the Ships way in her Course upon every shift of Wind allowing for Leeward-way and Currents 2. In duly estimating the Course or Point of the Compass on which the Ship hath made her way good allowing for Currents and the Variation of the Compass 3. In the frequent and due Observing the Latitude The reckoning arising out of the two former Branches is called the Dead Reckoning and of these three Branches there ought to be such an harmony and consent that any two being given a third Conclusion may be thence raised with truth As namely from the Course and distance to find the Latitude of the Ships place Or by the Course and difference of Latitude to finde the distance or by the difference of Latitude and distance to finde the Course But in the midst of so many uncertainties that daily occur in the practise of Navigation a joynt Consent in these three particulars is hardly to be expected and when an error ariseth the sole remedy to be trusted to is the observation of the Latitude or the known Soundings when a Ship is near land and how to rectifie the Reckoning by the observed latitude we shall now shew Those that will not yeild unto truth in this particular that a bout 24 of our common English Sea leagues are to be allowed to vary a degree of latitude under the Meridian do put themselves into a double incapacity First in sailing directly North or South under the Meridian where there is no Current finding their Reckoning to fall short of the observed latitude they take it to be an error in their judgement in concluding the Ships way by Estimation or guess to be too little And secondly if there be a Current that helps set them forward that there is a near agreement between the observed and the dead latitude they conclude there is no such Current Or lastly if they stem the Current they conclude it to be much swifter then in truth it is and thus one error commonly begets another but supposing a conformity to the truth we shall prescribe four Precepts for correcting a simple or single Course Prec I. First therfore if a ship sail under the Meridian if the difference of latitude be less by estimation then it is by observation the Ships place or Variation onely is to be corrected and enlarged under the Meridian and the error i◠to be imputed either to the judgement in guessing at the distance run in making
more to wit 180d 54. But supposing two places to be in the same Latitude and to have but 58d 33 Centesms difference of Longitude the example will be the same with one of those before put between the Berbadoes and St. Helens and the distance found by the middle Arch or Latitude is 62d 00 and the same by the middle space 61d 38 centesms but should be in truth 64d 7. Also in the former Example between the Berbadoes and the Lizard the true distance is 58d 54 and by the middle Latitude or Arch is 58d 7 but by the middle space it is 57d 66. Also in the Example between the Bermudas and the Lizard the true distance was found to be 44d 31 Centesms by the middle Arch or Latitude it is 45d 13 Centesms and by the middle space it is 44d 97 Centesms Where places are in the same Latitude or Parallel the Compasses must be set down in the Meridian-line at the Latitude given and the half extent applyed both upwards and downwards as before and if the distance be large the measure thereof will be much more erronious then when the Rumbe lyes nearer the Meridian Some Examples of a parallel Distance Let there be two places in the Latitude of 35d.   True Distance Distance by the Meridian-line Difference of Longitude 180d 147d 45 123d 93. 18 14 74 13 98. Another Example of two places in the Latitude of 50d.   True Distance Distance by the Meridian-line Difference of Longitude 180d 115d 8 111d 87. 18 11 58 11 59. A third Example of two places in the Latitude of 70d.   True Distance Distance by the Meridian-line Difference of Longitude 180d 61d 57 76d 41. 18 6 15 6 18. From which Examples we may observe that a large Distance cannot be so certainly measured in the Meridian-line as a small one whereof Mr. Wrâ—ght was very sensible and therefore prescribes Rules for the measuring of a small part of the Distance at a time and argues for the truth thereof but where the whole extent between two places is not above ten of the degrees of Longitude I see nothing to the contrary but that it may well enough be measured in the Meridian-line and so for a great distance we may measure a tenth or a twentieth part of the whole and by multiplying the known part finde the whole for the ready performing whereof another Scale of equal parts whereof the degrees are twice as large as those in the Scale of Longitude will be of much conveniency and ease Example Suppose it were required to measure the distance between the Points f and d in the former Chart take the same distance and measuring it in the inches finde how much it is to wit 1 Inch 61 Centesmâ— then take the same number out of the Scale of Longitude and setting one foot at the middle Latitude to wit 19d 18 Centesms the other will reach Northwards to 20d 69 Centesms and Southwards to 17d 64 Centesms the difference of which two Arks is 3d 05 the distance sought which allowing 20 leagues to a degree is 61 leagues as before and this is more easily done then to take the half of any extent and by the same reason you may finde the middle space between both Latitudes So also when you would measure the tenth part of a great distance measure the whole extent in the Scale of Longitudes and take the twentieth part found by the pen or memory out of the said Scale and set it at the middle Latitude or middle Space turning the other foot in the Meridian-line both upwards and downwards and the degrees so intercepted will be the tenth part of the whole distance Now the taking of the twentieth or fortieth part of an Extent is easily done by help of these two Scales of equal parts Suppose I would finde the twentieth part of 3 inches or degrees in the greater Scale I say 3 of the small parts in the lesser Scale is the length required and so the twentieth part of 3 inches 5 tenths is three and a half of the smaller parts in the lesser Scale and the half of that is the fortieth part of the whole I need not insist further upon these ways of measuring seeing I have before delivered others which as they are more ready in the practise so also they are built upon better foundations To keep a Reckoning on the true Chart. Here I shall insist upon a new Method never before published which will render this Chart very easie and acceptable to Seamen and having made our Example that before laid down being the same with that in the Plain Chart we shall here also retain the same Traverses The first Operation is to finde the Latitude The first Course the Ship sayls is South South-west 60 leagues from Tenariff to protract this Traverse I shall make use of another Traverse-quadrant bigger then that which was used before which may be made upon state Take 60 leagues out of the Scale of Longitudes T W and enter it in the Traverse-quadrant on the second point from C to A the nearest distance from A to C W prick in the Scale of Inches from W to A + and it shews me now that the Ship is in the Latitude of 25d 23 Centesms in the Meridian-line T S set the figure 1 to this Latitude Secondly to finde the difference of Longitude Take the extent T 1 in the Meridian-line and enter it so in the Traverse-quadrant on the second Rumbe that one foot resting thereon as at a the other turned about may but just touch C W then is the nearest distance from a to C S the difference of Longitude required to wit 1d 24 Centesms which prick in the Scale of Longitude from T towards W and set the figure 1 at it Thirdly to plot the Traverse Point Set one foot of the said extent at 1 in the South line of the Chart and with the other draw a small Ark at a then take the extent T 1 out of the South line and setting one foot at 1 in the West line with the other cross the former Ark at a and there is the point where the Ship is at the end of this first Traverse Demonstration The Proportion for finding the difference of Latitude we have before handled the Proportion for finding the difference of Longitude is As the Radius Is to the Meridional parts between any two Latitudes ∷ So is the Tangent of the Rumbe To the difference of Longitude ∷ The extent T 1 being taken out of the South line is the parts of the Meridian-line between the Latitude of 28d and the Latitude of the Ships place namely 25d 23 Centesms which being entred as before in the Traverse-quadrant at a becomes the Radius to the Tangent of that Rumbe and so the Tangent of the said Rumbe to that Radius being the nearest distance from a to C S becomes the difference of Longitude required the Proportion for finding it being duly observed
thy Works or try the Rumbs N ever desist but let 's have more of thine H ere 's but a Tangent but let 's have a Sine O r bosom full of thy industrious toyl I t will inform the weak enrich our Soyl. Your loving Friend Sylvanus Morgan The CONTENTS of the First Book In the Proportional Part. GEometrical Definitions Page 1 2 To raise Perpendiculars 2 3 4 To draw a line parallel to another Line 5 To bring three points into a Circle 6 To finde a right line equal to the Arch of a Circle 9 10 Chords Sines Tangents Secants Versed Sines c. defined 11 12 13 The Scale in the Frontispiece described 14 Plain Triangles both right and obliqued Angled resolved by Protraction from p 15 to 25 Proportions in Sines resolved by a Line of Chords p 21 to 25 Proportions in Tangents alone so resolved p. 25 27 28 29 30 Proportions in Sines and Tangents resolved by a Line of Chords p. 26 31 32 33 Particular Schemes fitted from Proportions to the Cases of Oblique Angled Sphoerical Triangles To finde the Azimuth p. 34 35 As also the Amplitude p. 36 The Azimuth Compass in the Frontispiece described p. 38 The Variation found by the Azimuth Compass p. 39 To finde the Hour of the Day p. 40 As also the Azimuth and Angle of Position p. 41 To finde the Suns Altitudes on all Hours p. 43 46 Also the Distances of places in the Arch of a great Circle p. 44 To finde the Suns Altitudes on all Azimuths p. 48 The Latitude Declination and Azimuth given to finde the Hour p. 50 to 54 ☞ To finde the Amplitude with the manner of measuring a Sine to a lesser Radius p. 55 To get the Suns Altitude by the shadow of a Thread or Gnomon p. 56 The Contents of the Treatise of Navigation OF the Imperfections and Uncertainties of Navigation p. 1 to 5 To measure a Course and Distance on the Plain Chart. p. 7 8 9 Of the quantity of a degree and of the form of the Log-board p. 9 10 A Reckoning kept in Leagues how reduced by the Pen to degrees and Centesmes p. 11 12 Of a Traverse-Quadrant p. 13 A Traverse platted on the Plain Chart without drawing Lines thereon p. 14 to 18 A Scheme with Directions to finde what Course and Way the Ship hath made through a Current p. 18 to 21 Divers Rules for Correcting of the Dead Reckoning from p. 21 to 33 Of the errors of the Plain Chart. p. 33 And how such Charts may be amended p. 34 To finde the Rumbe between two places p. 35 Proportions having one tearm the middle Latitude how far to be trusted to p. 35 to 38 To finde the Rumbe between two places by a Line of Chords onely p. 39 to 42 The Meridian-line of Mercators Chart supplied generally by a line of Chords p. 42 to 47 The Meridian-line divided from the Limbe of a Quadrant with the use thereof in finding the Rumbe p. 48 to 51 The error committed by keeping of a Reckonâ—ng on the Plain Chart removed p. 52 to 54 Of the nature of the Rumbe on the Globe p. 55 to 57 Mereators Chart Demonstrated from Proportion p. 58 to 60 Objections against it answered p. 60 to 63 To finde the Rumbe between two places in the Chart. p. 64 Distances of places how measured on that Chart. p. 65 to 71 Another Traverse-Quadrant fitted for that Chart with a Traverse platted thereby without drawing lines on the Chart. p. 71 to 78 To measure a Course and Distance in that Chart without the use of Compasses p. 79 Of Sailing by the Arch of a great Circle p. 81 To finde the Latitudes of the great Arch by the Stereographick Projection p. 82 to 83 Of a Tangent Projection from the Pole for finding the Latitudes of the great Arch p. 84 to 88 With a new Method of Calculation raised from it p. 89 90 And how to measure the Distance in the Arch and the Angles of Position p. 91 Another Tangent Projection from the Equinoctial for finding the Latitudes of the Arch. p. 93 to 100 And how to finde the Vertical Angles and Arkes Latitudes Geometrically p. 100 to 102 To draw a Curved-line in Mercators Chart resembling the Arch with an example for finding the Courses and Distances in following the Arch. p. 102 to 104 The Dead Reckoning cast up by Arithmetick p. 106 to 108 A brief Table of Natural Sines Tangents and Secants for each point of the Compass and the quarters p. 107 The difference of Longitude in a Dead Reckoning found by the Pen. p. 109 That a Table of Natural Sines supplyes the want of all other Tables p. 110 Many new easie Rules and Proportions to raise a Table of Natural Sines p. 111 to 113 And how by having some in store to Calculate any other Sine in the Quadrant at command p. 114 Of the contrivance of Logarithmical Tables of Numbers Sines and Tangents and how the want of Natural Tables and of a Table of the Meridian-line are supplied from them p. 117 The Sides of a Plain Triangle being given to Calculate the Angles without the help of Tables two several ways p. 118 119 An Instance thereof in Calculating a Course and Distance p. 119 CHAP. I. Containing Geometrical Definitions A Point is an imaginary Prick void of all length breadth or depth A Line is a supposed Length without breadth or depth the ends or limits whereof are Points An Angle derived from the word Angulus in Latine which signifieth a Corner is the inclination or bowing of two lines one to another and the one touching the other and not being directly joyned together If the Lines which contain the Angle be right Lines then is it called a Right lined Angle A right Angle when a right Line standing upon a right Line maketh the Angles on either side equal each of these Angles are called Right Angles and the Line erected is called a Perpendicular Line unto the other An obtuse Angle is that which is greater then a right Angle An acute Angle is that which is less then a right Angle when tvvo Angles are both acute or obtuse they are of the same kinde othervvise are said to be of different affection An Angle is commonly denoted by the middlemost of the three Letters set to the sides including the said Angle The quantity of an Angle is measured by the arch of a Circle described upon the point of Concurrence or Intersection where the two sides inclosing the said Angle meet By the complement of an Arch or Angle is meant the remainder of that Arch taken from 90d unless it be expressed the complement thereof to a Semicircle of 180d. A Circle is a plain Figure contained under one Line which is called the Circumference thereof by some the Perimeter Periphery or Limbe a portion or part thereof is called a Segment The Center thereof is a Point in the very midst thereof from which Point all right lines drawn
to the Circumference are equal if divers Circles are described upon one and the same Center they are said to be Concentrick if upon divers Centers they are in respect of each other said to be Excentrick The Diameter of a Circle is a right Line drawn through the Center of any Circle in such sort that it may divide the whole Circle into two equal parts The Semidiameter of a Circle commonly called the Radius thereof is just one half of the Diameter and is contained betwixt the Center and one side of the Circle A Semicircle is the one medeity or half of a whole Circle described upon the Diameter thereof And a Quadrant is just the one fourth of a whole Circle All Circles are supposed to be divided into 360 equal parts called Degrees consequently a Semicircle contains 180d and a Quadrant 90d and so much is the quantity of a right Angle A Minute is the sixtieth part of a degree being understood of measure but in time a Minute is the sixtieth part of an hour or the fourth part of a degree 15 degrees answering to an hour and 4 Minutes to a degree A Minute is marked thus 1′ a second is the sixtieth part of a Minute marked thus 1″ Where two lines or arches cross each other the point of meeting is called their Intersection or their common Intersection CHAP. II. Containing some Geometrical Rudiments Prop. 1. To raise a Perpendicular upon the end of a given Line Let it be required to raise a Perpendicular upon the end of the Line C D upon C as a Center describe the arch of a Circle as D G prick the Extent of the Compasses from D to G and upon G as a Center with the said Extent describe the ark D E G in which prick down the Extent of the Compasses twice first from D to E and then from E to F then joyn the points F D with a right line and it shall be the Perpendicular required Otherwise in case room be wanting Upon G with the Extent of the Compasses unvaried describe a small portion of an Ark near F then a Ruler laid over the Points C and G will cut the said Ark at the Point F from whence a line drawn to D shall be the Perpendicular required In this latter case the Extent C F becomes the Secant of 60d to the Radius C D which Secant is always equal to the double of the Radius As to the ground of the former Way if the three Points C G F were joyned in a right line it would be the Diameter of a Semicircle an Angle in the circumference whereof made by lines drawn from the extreamities of the Diameter as doth the Angle C D F is a right Angle by 31 Prop. 3 Book of Euclid Otherwise To raise a Perpendicular upon the end of a given Line Set one foot of the Compasses in the Point A and opening the other to any competent distance let it fall in any convenient point at pleasure as at D then retaining that foot in D without altering the Compasses make a mark in the line A C as at E Now if you lay a Ruler from D to E and by the edge of it from D set off the Extent of the Compasses it will find the point B from whence draw the line A B and it shall be the Perpendicular required Thus upon a Dyal we may raise a Perpendicular from any point or part of a Line without drawing any razes to deface the Plain Otherwise upon the Point D having swept the Point E with the same Extent draw the touch of an Arch on the other side at B then laying the Ruler over D and E as before it will intersect the former Arch at the Point B from whence a line drawn to A shall be the Perpendicular required The converse of this Proposition would be from a Point assigned to let fall a Perpendicular on a line underneath To be done by drawing a line from the Point B to any part of the given line A C as imagine a streight line to pass through B D E and to finde the middle of the said line at D where setting one foot of the Compasses with the Extent D E draw the touch of an Arch on the other side the Point E and it will meet with the said line A C at A from whence a line drawn to B shall be the Perpendicular required Prop. 2. To raise a Perpendicular from the midst of a Line In the Scheme annexed let A B be the line given and let it be required to raise a Perpendicular in the Point C. First set one foot of the Compasses in the point C and open the other to any distance at pleasure and mark the given line therewith on both sides from C at the point A and B then setting one foot of the Compasses in the point A open the other to any competent distance beyond C and draw a little Arch above the line at D then with the same distance set one foot in B and with the other cross the Arch D with the arch E then from the Point of Intersection or crossing draw a streight line to C and it shall be the Perpendicular required The converse will be to let a Perpendicular fall from a Point upon a given Line Let the Point given be the Intersection of the two Arks D E setting down one foot of the Compasses there with any Extent draw two Arks upon the line underneath as at B and A divide the distance between them into halfs as at C and from the given Point to the Point C draw a line and it shall be the Perpendicular required and if this be thought troublesom upon the Points A and B with any sufficient Extent you may make an Intersection underneath and lay a Ruler to that and the upper Intersection and thereby finde the Point C. Prop. 3. To draw one Line parallel to another Line at any distance required In this Figure let the Line A B be given and let it be required to draw the Line C D parallel thereto according to the distance of A C. First open the Compasses to the distance required then setting one foot in A or further in with the other draw the touch of an Arch at C then retaining the same extent of the Compasses set down the Compasses at B and with the former extent draw the touch of an Arch at D then laying a Ruler to the outwardmost edges of these two Arks draw the right line C D which will be the Parallel required This Proposition will be of frequent use in Dyalling now the drawing of such pieces of Arks as may deface the Plain may be shunned for having opened the Compasses to the assigned Parallel distance set down one foot opposite to one end of the line proposed A B so as the other but just turned about may touch the said Line and it will finde one Point Again finde another Point in like manner opposite to the
it were needful It is not necessary to press Examples if what before is written be well understood especially in this case where all directions are slippery Thus in imitation of Maetius a Hollander though a Latine Author we have prescribed several rules for the correction of a single Course which Mr. Phillips in his Geometrical Sea-man makes but one rule retaining always the same Course and correcting the distance run therein by drawing a parallel through the observed Latitude and so for many Courses they are first brought all into one line and the distance corrected by the same rule But concerning it we must give a double Caution First that no three places can be laid down true in their Courses and Distances from each other on the Plain Chart as shall afterwards be handled however the error in small distances will be inconsiderable And secondly admitting they could the said general Direction is unsound but the nearer the truth the nearer the Courses are to the Meridian and when all the Courses do either increase or diminish the Latitude but very erronious when some Courses increase and others lessen the Latitude in all which Cases it is most safe to allot to the Variation or dead Difference of Latitude of every Course its proportional share of the whole error between the Dead and Observed Latitude and then to correct each course by the former directions First therefore in the following Chart let us suppose a Ship to sayl from A in the Latitude of 28d South South-west almost 65 leagues to B this Course is set off in the Arch E F and by the Dead Reckoning she should now be in the Latitude of 25d. Again from B she sayls South-west and by West 72 leagues to C which Course being three points from the West is set off in the Arch G H and now by the dead Reckoning she should be in the Latitude of 23 degrees whereas by a good observation she is found to be in the Latitude of 23d 30′ wherefore to correct this Reckoning draw the line C A which is the compound Course arising from the two former Courses and through the parallel of observed Latitude draw L K parallel to A W so is the point K the corrected point of the Ships place according to Mr. Phillips and agreeing with the truth as we have fitted the Example But now as to the other way of correcting a compound Course it is to be done by this Proportion First finde the Variation or Difference of Latitude proper to each Course then it holds As the sum of all the Variations or Differences of Latitude Is to the whole error between the Dead and Observed Latitude ∷ So is each particular Difference of Latitude To its proportional share of the whole error ∷ Then if the differences of Latitude fall all the same way if the the estimated difference of Latitude be too much you must abate out of each dead Latitude its proportional error so in this case the said error is prickt from E to N. But when the estimated difference of Latitude is too little the proportional error must be added to each difference of Latitude then prick the second dead difference of Latitude being equal to E S from N to O and place the said extent from A the Center to Y and take the nearest distance to M A as before and prick it from O to L being the second error this is needful when there are more Courses then two but for the last Course not at all necessary neither is it for this then through the point N draw the line N F parallel to A W so is F the corrected Point of the Ships place at the first Course then draw F K parallel to B C and where the parallel of Latitude cuts it as at K is the corrected Point of the Ships place at the second Course being the same we found it before the other way But in stead of the second Course and distance which was 72 Leagues South-west and by west let us now suppose the Ship sayls the same distance from the point B North-west and by west which Course being as much on the other side the west make G I equal to G H and draw the Course B I therein pricking off the former distance to D so is D the point of the Ships dead reckoning in the Latitude of 27d and now supposing the observed Latitude to be 27d 30′ the error and difference of Latitude are as much now as they were before wherefore draw D A the compound rumbe draw Q T parallel to A W where it cuts the compound rumbe as at P by M◠Phillips his reckoning is the corrected point of the Ships place at the end of the second Course whereas in truth it should happen at T and so P bears from A in this example 76d 43′ from the Meridian and is distant from it 43 Leagues and a half whereas the Ships true Course from A to T is 83d 32′ from the Meridian and the distance almost 89 leagues which is very considerable Now for as much as the Sum of the differences of Latitude A E and E f in this latter example is equal to A S in the former example also the error f Q here is equal to S L there therefore the proportional part of each error will be the same as before Then if some Courses decrease the Latitude Southwardly and others increase it North-wardly if the dead Latitude be too little as in this example consider that to place the Ship more North-wardly so as to allot to each difference of Latitude its proper error that the South-wardly differences of Latitude must be decreased or lessened and the North-wardly increased wherefore the proportion of the error is placed from E to N and the Point F found as before In like manner if the dead Latitude were too much to bring the Ship more South-wardly the Southern differences of Latitude must be increased and the Northern decreased now the point T is found by drawing a line from F the corrected point of the first Course parallel to B D and so the line F V being equal to B D is the Ships second Course and distance from the corrected point F then in regard part of the error in the Latitude is supposed to be committed as well in the latter as in the former Course which error being too little the distance F V is to be enlarged and where the parallel of observed Latitude cuts it as at T is the corrected point of the Ships place at the end of the second Course And though what we have here performed be done by the drawing of many Lines yet by help of the Traverse-quadrant it may may be inserted into the Chart without drawing any Lines therein at all for in each the Course and corrected Difference of Latitude is given and that two things are sufficient to dispatch the work we have shewed before Those that are prompt in Plain Triangles may
Scheme the Vertical Angles being the Angles at the Pole on each side the Perpendicular are measured from the Point B and the Chords of those Angles from the Point A are the Cosines of those Angles and so the extent A L is the Cosine of the lesser Vertical Angle which prick from A in the line A S the point found we may call the Vertical point then prick down the Sines of every fifth degree from 90 as of 85d 80d c in the line I B from I towards B and laying a ruler over the Vertical Point and all those Sine Points divide the Diameter into as many points then prick down the Sine of the Latitude to wit 50d in the Semicircle A B from A upwards towards B and a ruler over the said Latitude point and all the former points found in the Diameter will cut the under Semicircle B A in as many points or Arkes more which being counted or measured from B downwards towards A are the respective Latitudes of the great Arch sought Nota if a line of sines equal to the diameter be graduated on the floap edge of a ruler the sines in the line B I may be easily pricked down with the pen by the edge thereof without Compasses and the Arkes in the under Semicircle B A may be readily measured by view by laying the beginning of that line of Sines to the point B and moving the edge of the ruler to each respective Ark before found in the said Semicircle Here note that a Semicircle being divided into 90 equal parts the distances of each degree from one end of the Diameter are Sines of those Arkes the whole Diameter being their Radius wherefore the use of Sines is as Geometrical as the use of Chords In the same Scheme the Reader may first finde the Perpendicular and then the Distances on each side of it by Proportions before set down Now for the Latitudes of the great Ark. Having some of these ways found the Latitudes of the Arch make a Mercators Chart for your Voyage laying down the two places suppose the Lizard at L in the following Chart in the Latitude of 50 degrees and Trinity Harbour at T in the Latitude of 36 degrees the difference of Longitude to wit L M being 68 degrees and a half then because the Perpendicular falls between both places count off the lesser Vertical Angle 14 degrees 63 Centesms from L to P and there raise the Perpendicular P R. This following Chart we have made as large as the page would admit and having a Meridian-line fitted to the degrees of Longitude in that Chart by the edge thereof on the Perpendicular P R you may prick down the Latitudes of the Arch before found and 5 degrees difference of Longitude on each side of the Perpendicular at a time and thereby graduate the respective Points or Crosses through which the curved prickt line is drawn Example Now to the apprehension the right Line L T will seem nearer then to sayl along by the prickt Arch also it may seem improbable that a Ship bound from the Lizard to Trinity Harbour in Virginia being a place nearer the Equinoctial by 14 degrees of Latitude should yet run into a more Northwardly Latitude then the Lizard To which I answer That Mercators Chart being no Projection of the Sphere doth not represent things to the fancy as they are in the Sphere but as they are accommodated to that Chart by which Trinity Harbor in that Chart bears from the Lizard 74d 17′ from the Meridian which is above 6½ points to the Westwards of the South to wit West and by South ¼ of a point and one degree 39 minutes more Southwardly and the Distance in the Rumbe is 1034 leagues whereas the Distance in the Arch if it could be precisely followed is 1003 leagues and following it from point to point through every 5 degrees difference of Longitude from the Perpendicular it is 1020 leagues almost being less then the Distance in the Rumbe by 14 leagues as appears by Calculation or the Chart it self The respective Courses and Distances to be sayled   Course  Distance   Deg. Min.  Leagues Parts From L to 1 80 25 to the Westwards of the North 60 06 1 2 84 14½ 63 86 2 3 88 00½ 61 02 3 4 88 00½ to the Westwards of the South 61 02 4 5 84 14½ 63 86 5 6 80 14½ 64 88 6 7 76 28 66 66 7 8 72 51 69 18 8 9 68 56 72 32 9 10 65 30 69 20 10 11 61 32 81 40 11 12 59 00 87 76 12 13 55 02 93 86 13 T 52 20 77 88    Whole Distance 1019 96. This Example serves fully to explain the Sailing by the great Arch though it might not be safe to follow it by reason of haling too near the Coast of Ireland OF Arithmetical Navigation ALl performances by the Pen require the help of some Tables by which the Question proposed may be speedily resolved if such Tables be at hand and if no other Table be at hand but that of Natural Sines it will serve to do the whole work and how such a Table may be made or the Sine of any Ark at command we shall handle For conveniency and dispatch there follows a Brief Table of Natural Sines Tangents and Secants to every one of the eight Points of the Compass and their Quarters from the Meridian and such a kinde of Table but false printed is in the Works of Maetius the use whereof he explains in some of the following Propositions Given Sought  1. The Longitudes and Latitudes of two places Rumbe which is required at the beginning of a Voyage Distance 2. The Rumb and Distance given Difference of Latitude whereby Dead Reckoning kept  Departure from the Meridian 3. Difference of Latude and Rumbe Departure from the Meridian Whereby the Dead Reckoning is corrected  Distance 4. Difference of Latude and Distance Rumbe Departure  To give the difference of Latitude and Departure whereby in some Cases to correct the Dead Reckoning is the same with the first Proposition and that we shall handle last 5. Rumbe Difference of Latitude Departure Distance 6. Departure Rumbe Distance Difference of Latitude These two last Propositions are of small use Some Examples of the Use of the said Table 1. In keeping the Dead Reckoning As the Radius Is to the Distance run ∷ So is the Cosine of the Rumbe To the Difference of Latitude The Table of Sines is numbred with the Points of the Compass on the left side in the Variation Column from the bottom upward and shew the difference of Latitude to 1000 leagues sayling upon any Rumbe taking the figures that stand against the Point of the Compass as far as the Comma and the other two figures beyond it may be taken in if more preciseness be required so if a Ship sayl 1000 leagues South South West being the second Point from the
Proportion of the same nature and after the same manner as we found the Azimuth at six before by the Analemma was the said Proportion protracted yet here it is to be suggested that in the Analemma there are three Proportions in Sines wrought instead of the one in Sines and Tangents before expressed 1. As namely to finde the Suns Altitude at Six As the Radius is to the Sine of the Latitude So is the Sine of the Declination To the Sine of the Suns height at six 2. To finde his Azimuth in that parallel of Altitude As the Radius is to the Cosine of the Latitude So is the Sine of the Declination to the Sine of the Azimuth in the said Parallel 3. To reduce it to the common Radius As the Cosine of the Altitude at six Is to the Radius So is the Sine of the Azimuth in that parallel To the Sine thereof in the common Radius The two latter Proportions in Sines may be brought into one as I have shewed in a Treatise the Sector on a quadrant Pag. 111 114. and that will be As the Cosine of the Altitude at six Is to the Cosine of the Latitude So is the Sine of the Declination To the Sine of the Azimuth sought And thus in effect the Analemma performs that single Proportion intermingled with Tangents after a more laborious manner in Sines or if you will the Altitude at six being found it holds As the Cotangent of the said Altitude Is to the Radius So is the Cotangent of the Latitude to the Sine of the Azimuth sought and this Proportion lies visible in the Analemma By these Directions derived from the Analemma together with the Proportions for each Case all the 16 Cases of right angled Sphaerical Triangles may be resolved some whereof seem to require the drawing of an Ellipsis as namely if the Suns place and right Ascension were given to find his greatest Declination which notwithstanding according to these Directions is easily shunned CHAP. VIII Shewing how to come by the Suns Altitude or Height without Instrument UPon any Flat or Plain that is level or parallel to the Horizon erect or set up a Wire without inclining or leaning to either side admit in the Point C and when you would finde the Suns Altitude or height make a mark at that instant in the very end or extreamity of its shadow suppose at G the shadow be-being the Line C G. Then upon the same Flat draw the quadrant of a Circle C A F with 60d of your Chords and make C E equal to the height of the Wire and through the point E draw the line E D parallel to C A and therein prick down the length of the shadow from E to D a Ruler laid from the Center to D cuts the quadrant at B and the Arch B A measured on the Chords sheweth the height required in this Example 20d 25′ in like manner if the length of the shadow were E K the height would be N A 38d 16′ Otherwise This may be performed otherways by drawing the quadrant C A F upon any plain board whatsoever then stick in a Pin at the Center C and hold the board so towards the Sun that the shadow thereof may fall upon the line C A then imagine C G to represent or supply the use of a Thread and Plummet hanging upon the Pin in the Center at liberty and mark where it cuts the Arch of the quadrant F A suppose at H measure the Arch F H on the line of Chords and it shews the height requiâ—ed By the next Chapter we shall finde the Suns Azimuth belonging to the Altitude 20d 25′ according to the Latitude and Declination there given to be 31d 19′ from the Meridian admit to the Westward of the South then doth the shadow happen so much to the Eastward of the Noâ—th wherefore if 31d 19′ be set off in the quadrant from the line of shadow from H to N a line drawn into the Center as N C shall be a true Meridian Line or line of North and South CHAP. IX Shewing the Resolution of such Propositions wherein the Suns Altitude iâ— either given or sought THe Latitude of the place Declination of the Sun and his Altitude given to finde the hour of the day and the Azimuth of the Sun Example Declination 13d South Altitude 14d 40′ With 60d of the Chords draw the Circle S Z N E the Center whereof is at C and draw the Diameter S C N and Perpendicular therto Z C prick off the Latitude 51d 3′ from N to P and from Z to AE and draw the Axis B P and the Equator AE C Q prick off the Declination from AE to D and from Q to E being 13d from the Chords and draw the parallel of Declination D B E then from S to A and from N to O out of the Chords prick off 14d 40′ the Altitude and draw the parallel of Altitude A O so is B ⊙ the hour from six towards Noon in the parallel of Declination and M ⊙ the Azimuth of the Sun from the East or West Southwards To measure the Hour Set the extent B D from C to F and upon F as a Center with the extent B ⊙ draw the Arch G a Ruler laid from C just touching that Arch findes the point H the Arch N H measured on the Chords sheweth 45d and so much is the hour from six to wit in time three hours either nine in the forenoon or three in the afternoon To measure the Azimuth In like manner set M A from C to R and upon R with the extent M ⊙ draw the Arch I a Ruler laid thereto from C cuts the Limbe at L the Arch L S measured on the Chords is 44d 35′ and so much is the Sun to the Southwards of the East or West To finde the Angle of Position Place the complement of the Altitude Z A from P to g then place the extent A g from g to f a Ruler laid over A and f cuts the parallel of the Suns Declination at t and D t is the versed Sine of the Angle of Position To measure it Thorough the point t draw k t y parallel to the Axis B P and draw D C then the extent C y measured on the Sines sheweth the complement of it to wit 62d 57′ therefore the said Angle is 27d 3′ Otherwise With the extent B D setting one foot in the Center at C with the other cross the former parallel line k t at k a Ruler laid from C to k cuts the Limbe at X and the Arch AE X is 27d 3′ the measure thereof as before the Complement whereof is the Arch V E and might be measured from the South Pole V if the Equinoctial AE Q were not drawn Three sides given to finde an Angle Another Example the Declination being North Latitude 51d 32′ Declination 13d North Altitude 37d 18′ In the following Scheme Upon the Center C with
former directions Having found the angle C A B it is equal to the side Z P in the first Triangle which is the complement of the Latitude required And the complement of the angle A C B in the second Triangle being taken to 180d is equal to the side ⊙ P in the first Triangle the Suns distance from the elevated Pole required Example in this Scheme The Latitude S T 30d whose complement is Z T 60 equal to the side of the Triangle A B. Z Y is the Polar distance 69d 46′ the complement whereof is N Y. L G is the Sine of 6d 36′ in that parallel the excess of the Azimuth above a quadrant from the north C F is equal to C G the extent G Z reaches from F to A and the arch Z A being 35d 59′ is equal to C B in the second Triangle or to the angle Z ⊙ P in the first Triangle being there the angle of Position To finde the Hour Make N P equal to S T drawing C P and set off 69d 46′ from P to D and E and draw the parallel D E. Then draw the parallel of Altitude H A and B ⊙ is the Sine of the hour from six C R is equal to B D upon R with B ⊙ describe K a Ruler from the Center just touching it findes M and the arch M O being 38d 28′ is the angle C A P in the second Triangle equal to the side Z P in the first Triangle which is the complement of the Latitude there required To finde the Angle of Position Place Z H from P to f and make f g equal to f H a Ruler from H to g cuts D E at I upon R with B I draw Q a Ruler from the Center touching it findes V and the ark O V being 66d 29′ is the measure of the angle A C B in the second Triangle which is equal to the side ⊙ P in the first Triangle the Suns distance from the Pole required and thus this question hath exercised most of the Rules before delivered and thus solely upon the consideration of the Projection of the Sphaere I have shewed how all the twenty eight common Cases may be resolved Orthographically CHAP. X. Shewing how to project all the Cases of right angled Sphaerical Triangles ANd because Projections of the Sphaere do not depend upon Proportions for the resolving of any Case propounded but on the contrary Proportions are derived from Projections I shall therefore further enlarge in shewing how the right angled Cases are resolved without the knowledge of the Proportions In Projecting the right angled Cases Orthographically the Triangle V A ⊙ right angled at A shall be the Example used in which the arch S AE is equall to the Complement of the Latitude at London 38d 28′ equall to the Angle A V ⊙ The Hipotenusal V ⊙ is the hour from Six The legg A ⊙ the Altitude or Suns height at the time proposed The legg V A the Suns Azimuth from the Vertical The angle A ⊙ V the angle between the Suns Azimuth Circle and the Equinoctial Case 1 2 3. Given a Leg and its adjacent Angle to finde the rest Data A V 31d. Angle A V ⊙ 38d 28′ Having drawn the Quadrant S Z and S V and Z V at right angles in the Center place the given angle 38d 28′ from S to AE and draw AE V draw AE L parallel to Z V then prick the Sine of the given Leg from V to A and F the nearest distance from A to AE V prick from L to B. 1. To finde the other Side or Leg. A ruler over V and B cuts the Limbe at D and the arch S D being 22d 16′ is the measure of the side A ⊙ 2. To finde the Hipotenusal Through D draw D ⊙ G parallel to S V and the extent V ⊙ measured on the Sines is 37d 30′ the Hipotenusal sought 3. To finde the angle V ⊙ A. Through F draw F E parallel to V S and make V C equal to V ⊙ a Ruler from the Center over C cuâ—s the Limbe at L and the arch S L being 57d 55′ is the measure of the angle sought Otherwise Make V I equal to V C then take the distance between C and Z and the said extent shall reach from I to L in the Limbe as before after the same manner the point D might have been found Also the extent V F may be doubled or tripled upwards and at the end thereof a Perpendicular raised accordingly multiply or increase the extent V ⊙ and one foot resting in the Center cross the said Perpendicular and a ruler laid over the point of Intersection or crossing and the Center V shall pass through the Limbe at L as before Otherwise Place V ⊙ from V to Q and upon Q with V A describe the ark H a ruler from the Center just touching the outward extreamity of it cuts the Limbe at L as before the arch S L being the angle sought Or to keep the said arch more remote from the Center prick V ⊙ twice from V to P then double the extent V A and therewith upon P describe the ark K a rule from the Center just touching the outward extreamity of that ark cuts the Limbe at L as before These Cases being the most difficult as requiring an Ellipsis to be drawn which we have shunned and may be otherwise avoided according to the Example in Page 51 I thought fit first to handle The Triangle resolved is not represented unless there be an Elliptical a k drawn from ⊙ to A. The finding of the Point D was carried on from a proportion of this kinde As the Radius AE L to the Tangent of the ark S AE being the measure of the given angle So is the Sine of the given Leg L B proportioned out to that Radius to the Tangent of the ark S D being the measure of the Leg sought Case 4 5 6. Given a Side and its opposite angle to finde the rest Data the Side A V 31d. The Angle V ⊙ A 57d 55′ Place the given Angle at the Center of the quadrant wherefore prick it to wit 57d 55′ from S to L and draw L V. Then place the given Legg from S to D to wâ—t 31d and draw D ⊙ G parallel to S V. To measure the Arks sought 1 The Hipotenusal The Extent V ⊙ applyed to the Sines is 37d 30′ being the measure thereof 2. The other Leg. Place D G from V to R and upon R with the extent G ⊙ draw the ark H a ruler from the Center touching the extreamity of it cuts the Limbe at E and the arch S E being 22d 16′ is the measure of the other Leg. 3. The other Angle Draw E F parallel to S V and make V C equal to V ⊙ a ruler from the Center over C cuts the Limbe at AE and the arch S AE being 38d 28′ is the measure of the other oblique angle Lastly the Triangle
be drawn through the three points will fall altogether within the outward Circle and may very well be described and it will cut the Meridian S N in two points the one towards the South the other towards the North a ruler from the East point O over each of the former points will give the Suns south and north Meridian Altitudes in the Limbe And if the ark between these Meridian Altitudes be divided into halfs the point in the middle so found is the Latitude point L the ark between which and N shews the Latitude required after the same manner might the point L be found in all other Cases which notwithstanding to avoid the drawing of a whole vast Circle we shunned though when the Sun is near the Winter Tropick the drawing of part thereof cannot be shunned without the help of a Steel-bowe or some such like remedy whereby to describe part of a Circle thorow three points given which is the great and onely inconvenience of the Stereographick Projection Clavius handles this Proposition after another manner but not so convenient and makes no further use of it But here we shall also apply it to the making of all Dyals that have Centers except the Equinoctial In a Treatise of Geometrical Dyalling from page 72 to 82 I have shewed a general method both by Calculation and with Scale and Compasses how from three shadows to make all Dyals with Centers from whence those that have leisure may calculate the Arks found in the former Scheme whereto I now further adde that the former Proposition here insisted on performeth the same the three shadows C D C E and C F here are the same in length with those there and make the same angles at the Center here as those did at the foot of the perpendicular Stile there and supposing the former Plain to be a Declining Leaning Plain in some other Latitude the Line S C N is the Substilar Line and the Arch Z AE is the Stiles height the same as we found it in that Treatise Suppose this be a Plain that looks southward not much leaning from the Zenith if the former shadows happened in the Summer half year in north Declinations the Scheme will describe a Winter parallel of as much south declination also if the Sun had south declination the Scheme would describe a parallel of as much north declination but this is onely true when the elevated Pole is elevated above the northern or other face of the Plain to illustrate this In the Scheme above let S C N represent the Horizon of a certain place the Latitude whereof is Z AE or N L the Equator is represented by AE Q and D E represents a parallel of North Declination let Z C represent a Vertical or upright South Plain in that Latitude but if the said Vertical Plain become an Horizontal Plain the height of the North Pole will be Z L and the height of the South Pole H K I say then in respect of the first Horizon the Sun being in the parallel of North Declination between D and A his Altitudes on any hour from Noon being the Angles between the Wall and the Sun above the Vertical Plain Z A in the first Horizon are in respect of the second Horizon if the North Pole be elevated his depressions under the same upon the like hour from midnight which are equal to his Winter Altitudes above the second Horizon Z C H the Sun having as much South Declination being between B and G in the parallel F G or which is all one retaining the Suns Declination the same they are his Winter Altitudes upon the like hour from Noon when the South Pole K is elevated above the Horizon H C Z which is the reason why the Scheme describes a parallel of South Declination when his Declination was as much North. To place the Meridian Line This as I have shewed in the afore-mentioned Treatise may be projected by the eye on the Plain by help of a Thread and Plummet and if the Plains Re clination In clination be given together with the Substiles distance from the Plains perpendicular both which may be got without dependence on the Sun the Meridians place may be Calculated or at least the Inclination of Meridians as was there suggested either will serve But for the placing of the same and for finding the Latitude of the place and the Plains Declination though not required in the making of the Dyal the Converse of the Dyalling Scheme there used will perform it We shall take the Example there used Let the Substiles distance from the Plains perpendicular be 32d 16′ The Stiles height 41 30. The Plains Inclination be 15. Upon V as a Center describe the Circle S E N W and draw the Diameters S V N and W V E making right angles in the Center then assuming V N to represent the Plains perpendicular set off the Substiles distance from it the same way it happened from N to Y to wit 32d 16′ and from Y prick off the Stiles height to K from the point K let fall a perpendicular on the Substile at I and draw I C parallel to N V from which point let fall I Q produced perpendicular to the Plains perpendicular V N and make Q T equal to I K and upon V as a Center with the extent V T describe an ark in the other quadrant of the same Semicircle as at B then from the same end of the Horizontal Line prick the Inclination from E to R upwards and draw V R but if the Plain recline it must be pricked downward towards N then with the extent I K draw a Line parallel to R V and it will pass through the former ark at B which here we found by entring one foot of that extent in that ark so that the other turned about would but just touch V R Having discovered the point B through it draw B O parallel to E C then take the nearest distance from C to R V and place it in the Line O B from P on that side thereof which is farthest from R V here we placed it from P to M then from the Center draw a line passing through the point M as doth V X and it shall be the Meridian Line required Through the point O draw V D and the arch N D shews the Declination of the Plain to be 40d and the extent V O is the Cosine of the Latitude Or through the point B draw G B L parallel to V N and the arch N L is the measure of the Latitude of the place to wit 51d 32′ Now having placed the Substile Stile and Meridian the Hour-lines are easily inscribed either in a Circle or Paralellogram as in that Treatise is largely shewed For the Readers recreation and practice of what is here delivered and for tryal how well this kinde of Dyalling agrees with other kindes he may make a Dyal true to any known Latitude and Scituation and
other end of the Line at B and through these two Points draw a right Line and it shall be the Parallel required This way though it be not so Geometrical as the former yet in other respects may be much more convenient and certain enough Prop. 4. To draw the Arch of a Circle through any three Points not lying in a streight Line In the Figure adjoyning let A B C be the three Points given and let it be required to draw a Circle that may pass through them all Set one foot in the middle Point at B and open the Compasses to above one half of the distance of the furthest point therefrom or to any other competent extent and therewith draw the obscure Arch D E F H with the same extent setting one foot in the point C draw the Arch F H Again with the same extent setting one foot in the point A draw the Arch D E then laying a Ruler to the Intersections of these Arches draw the lines D G and G H which will cross each other at the point G and there is the Center of the Circle sought where setting one foot of the Compasses and extending the other to any of the three points describe the Arch of a Circle which shall pass through the three points required Prop. 5. Two lines being given inclining each to other so as they seem to make an acute angle if they were produced To finde their angular point of concurrence or meeting without producing or continuing the said Lines And if they be multiplied the other way from D to H and from C to E then the lines E H and F G being produced find the same point I without continuing either of the lines first given and with much more certainty An Oblong a Rectangle a right angled Parralellogram or a Long Square are all words of one and the same signification and signifie a flat Figure having onely length and breadth the four Angles whereof are right Angles the opposite sides whereof are equal In Proportions the product of two tearms or numbers are called their Rectangle or Oblong because if the sides of a flat be divided into as many parts as there are units in each multiplyer lines ruled over those parts will make as many small squares as there are units in the Product and the whole Figure it self will have the shape of a long Square A Rhombus or Diamond is a Figure with four equal sides but no right Angle But a Rhomboides or Diamond-like Figure is such a Figure whose opposite Sides and opposite Angles onely are equal either of these Figures are commonly called Oblique Angled Parralellograms Thus either of the Figures A B F E or B C E D are Rhomboides or Oblique Angled Parralellograms This foregoing Figure is much used in Dyalling thereby to set off the Hour-lines Admit the Sides A B and B E were given and it were required on both sides of B E to make two oblique Parralellograms whose opposite sides should be equal to the lines given this may be done either by drawing a line through the point E parallel to A B C and then make F E E D and B C each equal to A B and through those points draw the sides of the Parralellogram or continue A B and make B C equal thereto and with the extent B E upon A and C draw the cross of an Ark at F and D Again upon E with the extent A B draw oâ—her Arks crossing the former at F and those crosses or intersections limit the extreamities of the sides of the Parralellogram A line drawn within a four-sided Figure from one corner to another is called a Diagonal-line A Parralellipipedon is a solid Figure contained under six four-sided figures whereof those which are opposite are parallel and is well represented by two or many Dice set one upon another or by the Case of a Clock-weight To finde a right Line equal to the circumference of a Circle given Let the given Circle be B D C divide the upper Semicircle B D C into halfs at D and the lower Semicircle into three equal parts and draw the lines D E D F which cut the Diameter at G and H and make G I equal to G H then is the length D I a little more then the length of the quadrant B D neither doth the excess amount unto one part of the Diameter B C if it were divided into five thousand and four times the extent D I will be a little more then the whole circumference of the Circle To finde a right Line equal to any Arch of a given Circle Let C D be an arch of a given Circle less then a Quadrant whereto it is required to finde a right line equal Divide the Arch C D into halfs at E and make the right line F G equal to the Chord C D and make F H equal to twice C E and place one third part of the distance between G and H from H to I and the whole line F I will be nigh equal in length to the Arch C D but so near the truth that if the line F I were divided into 1200 equal parts one of those parts added thereto would make it too great albeit the Ark C D were equal to a Quadrant but in lesser Arches the difference will be less and if the given Arch be less then 60 degrees or one sixth part of the whole Circle the line found will not want one six thousandth part of its true length But when the given Arch is greater then a quadrant it may be found at twice thrice or four times by former Directions These two Propositions are taken out of Hugenius de magnitudine Circuli Page 20 21. In Dyalling to shun drawing of Lines on a Plain it may be of frequent use to prick off an Angle by Sines or Tangents in stead of Chords it will therefore be necessary to define these kinde of Lines 2. The right Sine of an Arch is half the Chord of twice that Arch thus G F being the half of G L is the Sine of the Arch G A half of the Arch G A L whence it follows that the right Sine of an Arch less then a quadrant is also the right Sine of that Arks residue from a Semicircle because as was shewed above the Chord of an Ark is the same both to an Ark lesser then a quadrant and to its complement to a Semicircle What an Arch wants of a quadrant is called the Complement thereof thus the Arch D G is the Complement of the Arch A G and H G is the Sine of the Arch D G or which is all one it is the Cosine of the Arch A G and the Line H G being equal to C F it follows that the right Sine of the Complement of an Arch is equal to that part of the Diameter which lieth between that Arch and the Center From the former Scheme also follows another Definition of a right Sine as