To a fourth number And this fourth number To the solid content in feet Wherefore if you apply the threed to the intersection of the contrary parallel of 12 with the right parallel of 2 foot the third part of the altitude it will cut the contrary parallel of 36 inches the side of the base at the right parallel of 6 the threed lying still in the same position will cut the contrary parallel of this 6 at the right parallel of 18 and such is the solid content of this pyramis in feet which was required Now if you would measure any segment of a pyramis being cut off observe the former rules given for the measuring of the segment of a cone and you shall soon have your desire For having the side of the segment of the pyramis with the semidiameters of the circumscribing circles at each base you may by the 61 Chapter finde the altitude of the whole greater pyramis as if it had not been cut off and also of the lesser pyramis which is supposed to be added to the segment to make it a perfect pyramis this being done you may by the rules delivered first get the content of the whole greater pyramis and then of the lesser added pyramis and taking the lesser out of the greater the remainder will be the content of the given segment as was required The end of the first Book The Second Book Shewing the most plentifull easie and speedy use of the Vniversal Quadrat in the Resolution of the whole Doctrine Trigonometrical as well plain as sphericall and that two severall wayes upon the Instrument with surpassing facility and with the least intricacie that may be CHAP. I. To finde the chord of any arch the radius being given not exceeding the side of the quadrat THe radius of a circle is the semidiameter of the same circle or the totall sine or sine of 90 degrees so AE is the semidiameter or radius upon which the divided arch EG was dramn and AB is the radius upon which the arch BFC was drawne And a Chord is a right line subtending an arch so BC is the chord of the arch BFC for the finding of which chord for any arch required place the radius given upon the side of the quadrat from the center and note the point where the same endeth and there place the bead then opening the threed to the arch given counted in the quadrant the distance between the point noted in the side of the quadrat and the head shall be the chord required As for example let AB be the radius of a circle placed upon the side of the quadrat and let it be required to finde a chord of 40 degrees First I place the bead at the point B where the given radius ended and opening the thred to 40 degrees in the quadrant I take the distance betwixt the point B and the bead which is the chord of 40 degrees viz. the chord of the arch BFC drawn upon the given radius AB as was required and thus may you finde the chord of any arch as well as upon the sector CHAP. II. To finde the right sine of any arch given the radius being put 1000. HAving shewed the use of these parallel lines as they stand in their own proper signification it resteth now to shew the use of them as they signifie right fines and tangents and first to finde the right fine of any arch seeke the degree given upon the quadrant counted from any side of the quadrant but here let it be from the beginning of the degrees at E so shall the right parallel running through the degree given in the quadrant give the number of the fine required but if you would have the fine in a right line take the length of the contrary parallel from the side of the quadrat AE to the degree given in the quadrat and that shall be your desired sine whereof the whole side is radius As for example let it be required to finde the right sine of 53 degrees 8 minutes I look 53 degrees 8 minutes in the quadrant counted from the begining of the degree at E unto O through which point O runeth the right parallel of 800 this 800 is the right sine of 53 degrees 8 minutes 100 being radius But if you would haue this sine in a right line take the length of the contrary paaallel from the point O unto N for the right sine of 53 degrees 8 minutes which was required CHAP. III. To finde the arch of any sine given the radius being put 1000. IF your sine be given in numbers seeke the right parallel of the number given for where that parallel cutteth the quadrant there is the arch required counted from the beginning of the degrees at E as before But if your sine be given in a right line then from the side of your quadrat AE extend your given sine upon any of the contrary parallels and that right parallel that shall cut through the extream point thereof shall cut the quadrant in the required arch As for example let 800 be a sine given unto which it is required to finde the arch belonging thereto I seeke 800 in the right parallels and finding it I follow it to the quadrant and there I see it cut the arch of 53 degrees 8 minutes counted from E which 53 degrees 8 minutes is the arch belonging to the sine 800 the radius being 1000. But if the sine be given in a right line take it betwixt your compasses and setting one foot in the side of the quadrat AE extend the other upon any of the contrary parallels as here I set one foot in the point N and extend the other unto O through which point the right parallel of 800 runneth and cutteth the quadrant in 53 degrees 8 minutes counted from E as before CHAP. IV. Any radius not exceeding the side of the quadrat being given to finde the right sine of any arch or angle thereunto belonging FIrst take the given radius and place it upon the contrary parallel of 100 which is the total sine and where it shall end there place the threed which being so placed see where it cutteth the contrary parallel that passeth through the given arch counted in the quadrant from G for the segment so cut which lieth betweene the threed and the side AE is the fine required Let the line A be the radius given and let it be required to finde the sine of 36 degrees 53 minutes First I take the line A and place it upon the contrary parallel of 100 as from E to D upon which point D I place the threed this done I count the given arch viz. 36 degrees 33 minutes from the end of the quadrant at G unto O and see what contrary parallel there I finde which is 60 intersecting the threed at M so shall the distance NM be the line B the sine of 36 degrees 53
I seek in the Ecliptick and following the parallel that passeth thereby to the Horizon I finde it there to cut 19 deg 7 min. counted from the Center and such is the Amplitude required the Sphere being set to the latitude of 52 deg 30 min. CHAP. XXX Having the latitude of the place with the declination of any Star to finde his Amplitude FIrst the Sphere to the latitude being set if the star be therein placed follow the parallel of Declination whereon he standeth to the Horizon and if he be not in the Instrument get his Declination by the seventh Chap. or otherwise by some Table for that purpose which found seek it in the Axletree-line among the parallels of declination follow the parallel thereof to the Horizon and if it cutteth the Horizon it there sheweth the Amplitude but if it cutteth it not then doth not that star set or go down under our Horizon As for example let it be required to finde the Amplitude of Arcturus I look in my Instrument for Arcturus and I finde him placed upon the parallel of 21 d. 10 min. of North Declination as it were just in the Horizon at 36 deg 23 min. from the Center and such is his Amplitude And if you would have the Amplitude of the Lions Heart look in your Instrument and you shall finde him placed upon the parallel of 13 deg 42 min. of North Declination which if you follow to the Horizon you shall finde it there cut 22 deg 54 min. and such is the Amplitude of the Lions Heart in the latitude of 52 deg 30 min. But if you would seek the Amplitude of the Bright Star called the Harp you shall finde him placed upon the parallel of 38 deg 30 min. North Declination but this parallel cutteth not the Horizon therefore he goeth not under our Horizon at all in the aforesaid latitude of 52 deg 30 min. but when he is in his lowest directly in the North he is just one degree of height CHAP. XXXII Having the Declination and Amplitude of the Sun to finde the latitude of the place COunt the Declination upon the axletree-Axletree-line among the parallels of Declination and the Amplitude upon the Horizon from the Center the same way with the declination then moving the Sphere to and fro until the Amplitude and the parallel of Declination meet both in one point the South end of the Horizon shall cut the Meridian at the complement of the latitude required As for example let the Suns Declination be 11 deg 30 min. and his amplitude 19 deg 7 min. if you move the Sphere untill the 19th deg 7 min. one the Horizon fall directly on the parallel of 11 deg 30 min. the South end of the Horizon will cut the limbe at 37 deg 30. min. and such is the height of the Equinoctial which being taken out of 90 deg leaveth 52 deg 30 min. for the latitude of the place which was required CHAP. XXXII The latitude of the place and the amplitude of the Sun being given to finde his Declination FIrst set the Sphere to the latitude proposed then count the amplitude upon the Horizon and where it shall end shall meet you a parallel which being followed to the Axletree-line will there shew you the Declination required As in the latitude of 52 deg 30 min. the Amplitude being 19 deg 7 min. if you first place the Sphere to the latitude and then count 19 deg 7 min. upon the Horizon there shall meet you a parallell which being followed to the Axle-tree line shall there give you 11 deg 30 min. the Declination required CHAP. XXXIII Having the latitude of the place and the Declination of the Sun to finde his height in the vertical Circle FIrst set the Sphere to the latitude given then count the Declination given among the Almicanters from both ends of the Horizon towards the Zenith and thereto place the threed parallel to the Horizon and where the threed so placed cutteth the axletree-Axletree-line or six a clock hour-hour-line there is the altitude required As in the latitude of 52 deg 30 min. the Declination of the Sun being 11 deg 30 min. if you count this 11. deg 30 min. from both ends of the Horizon among the Almicanters and place the threed thereto it shall lie parallel to the Horizon and cut the Axle-tree-line at 14 deg 33 min. and such is the Suns height when he is directly in the East or West points which was required Or if you place the Horizon to 52 deg 30 min. in the limbe counted from the Equator and count the Declination upon the Axletree-line viz. 11 deg 30 min. the parallel thereof shall cut the Horizon at 14 deg 33 min. for the height of the Sun in the vertical Circle the same as before CHAP. XXXIV Having the latitude of the place and the Declination of the Sun to finde the time of his comming to the East or West Points FIrst set the Planisphere to the latitude given then place the threed upon the Zenith and the Nadir so shall it cut the Horizon at right angles in the Center and the parallel of Declination at the time from six counted among the Meridians As for example let the latitude given be 52 deg 30 min. and the Suns Declination 11 deg 30 min. First place the Horizon to 37 deg 30 min. the complement of the latitude below the South end of the Equator then place the threed to the Zenith and Nadir that is to 52 deg 30 min. above the South end of the Equator and as much below the North end thereof the threed being sixed shall cut the Horizon at right angles in the Center and also the parallel of 11 deg 30 min at the Meridian of 9 deg or in time at 36 minutes from the hour of six and such is the distance of time betwixt the hour of six and the Suns coming to the East or West points which was required CHAP. XXXV Having the latitude of the place and the Declination of the Sun to finde his height at the hour of six FIrst set the Horizon to the complement of the latititude below the South end of the Equator or the Equator to the complement of the latitude above the South end of the Horizon which is all one then seek the Declination upon the axletree-Axletree-line and place the threed thereunto parallel to the Horizon by help of the Almicanters at both ends thereof so shall the threed shew among those Almicanters the height of the Sun required Thus in the latitude of 52 deg 30 min. the Suns declination being 11 deg 30 min. I first set the Equator 37 deg 30 min. above the Horizon and then place the threed to 11 deg 30 min. counted in the axletree-Axletree-line and moving the ends thereof to and fro until the threed lie parallel to the Horizon which will be when both ends of the threed cut like degrees from the Horizon upon
it falleth as was shewed in the 28 Chapter therefore if we take either halfe the perpendicular and the whole base or the whole perpendicular and halfe the base whereon it falleth the one for the breadth and the other for the depth of a squared solid we may measure it in all respects according to the 44 Chapter and that to any length required As for example let the Triangle BCD be the base of a prisma or peece of Timber to be measured now taking 10 the whole perpendicular AB for the breadth and 8 the halfe base CD for the depth of a squared prisma or peece of Timber we may finde the solid content thereof by the 44 Chapter and that according to any length required CHAP. XLVII How that prisma or peece of Timber whose base or end is a Rhombus or Diamond form may be measured A Rhombus or Diamond is a figure of 4 equall sides but no right angles such as is the figure ABCD. Now seeing every Rhombus is equall in their superficiall content to that right angled parallelogram or long square whose length is one of the sides and breadth equal to the parallel distance of the same sides If we take one of the sides for the breadth and the length of the perpendicular falling thereon from the opposite angle for the depth of a squared solid we may finde the solid content thereof by the 44 Chapter As for example let the Rhombus ABCD be the base or end of some peece of timber to be measured now taking one of the sides which is 10 for the breadth of a squared solid and the perpendicular which is something better then 8â…— for the depth of the same solid we may finde the solid content thereof for any length required by the 44 Chapter CHAP. XLVIII How that prisma or peece of Timber whose base or end is a Rhomboides or Diamond like may be measured A Rhomboides or Diamond like is a figure whose opposite sides and opposite angles are only equall and it hath no right angles Such as is the figure EFGH Now seeing that the Rhomboides is also equall in his superficiall content to that right angled parallelogram whose length is one of the sides and breadth equall to the parallel distance of the same sides If we take one of the sides for the breadth of a squared solid and the length of the perpendicular falling thereon from the opposite angle for the depth of the same solid we may finde the solid content thereof by the 45 Chapter Let this Rhomboides EFGH be the base of a peece of Timber to be measured Now if you take 16 the side GH or EF for the breadth of a squared solid and 10 the length of the perpendicular EL for the depth of the same solid you may finde the solid content thereof by the 45 Chapter for any length required CHAP. XLIX How that prisma or peece of Timber whose base or end is a trapeziam may be measured A Trapeziam is any irregular fouresided figure of what fashion soever as this figure ABCD is a trapeziam and may be cast into two Triangles by drawing the diagonall line AC and so each Triangle measured as is before shewed which being done adde the contents of them both together you shall have the content of the whole trapeziam Or when you have drawn the diagonall line and from the two opposite angles let down the perpendiculars thereon joyne the two perpendiculars together and take the sum thereof for the breadth of a squared solid and take one half the diagonall line for the depth of the same solid with which breadth and depth proceed by the 44 Chapter according to your length given and you shall thereby finde the solid content required Let the trapeziam ABCD be the base or end of a peece of Timber now if you take the two perpendiculars BF and DE and joyne them together they will make 10 for the breadth of a squared solid and then take one halfe the diagonall AC which will be 8 for the depth of the same solid you may by the 44 Chapter finde the solid content thereof according to any length given CHAP. L. How that prisma or piece of Timber whose base or end hath many sides as 5 6 7 8 9 10 or more so they be equall may be measured MAny sided figures are those which have more sides then foure and are generally called polygons as this figure A having equall sides and Angles is called a polygon Every regular polygon being equall in the superficial content to that long square whose length is equall to one halfe the perimiter and breadth to a perpendicular drawn from the Center to the middle of any side thereof If we adde all the sides together and take halfe the summe for the breadth of a squared solid and then take the perpendicular for the depth of the same solid we may by the 44 Chapter finde the solid content thereof Let the figure A be the base or end of a peece of Timber to be measured consisting of 5 equall sides and each side containing 12 inches which being added together into one summe maketh 60 the halfe whereof is 30 for the breadth of your peece then take the length of the perpendicular falling from the Center A to the midst of one of the sides which is 8 inches for the depth of the same peece with which breadth and depth you may proceed to finde the solid content by the 44 Chapter acording to any length given A note to finde the centers of those equiangle figures If the sides be even the Center is found by drawing lines from one angle to his opposite angle but if they be odde it is found by drawing lines from the middle of a side to his opposite angle so shall the cutting or intersection of those straight lines shew the place of the center CHAP. LI. Having the circumference and Diameter of a Cylinder given in inches and the length in feet to finde the content in feet As 12 inches to the Diameter in inches So is the length in feet to a fourth number Then As 12 inches to one quarter of the circumference So is this fourth number to the solid content in feet As for example let the figure A be a cylinder or some round peece of Timber to be measured whose circumference is 44 inches and Diameter 14 inches and let the length thereof be 10 foot now for to finde the solid content First place the threed to the intersection of the contrary parallel of 12 with the right parallel of 14 and it will cut the contrary parallel of 10 at the right parallel of 11â…” now where this right parallel of 11â…” cutteth the contrary parallel of 12 there place the threed again so shall it cut the contrary parallel of 11 at the right parallel of 10 7 10 and so many solid feet is contained in this cylinder or round peece of Timber CHAP. LII Having the
the Almicanters the threed being thus placed upon the 11 deg 30 min. in the axletree-Axletree-line and parallel to the Horizon will cut at each end thereof 9 deg 6 min. among the Almicanters and such is the height of the Sun at the hour of six Or if you place the Equator 52 deg 30 min. above the South end of the Horizon and then count 11 deg 30 min. upon the Horizon where you shall finde meet you a parallel which being followed to the Axletree-line shall there give you 9 deg 6 min. for the altitude of the Sun at six as before CHAP. XXXVI Having the latitude of the place and the Declination of the Sun to finde his Azimuth at the hour of six FIrst set the Sphere according to the latitude given and count the Declination upon the horizon and look what parallel there meeteth you the same follow to the axtree-Axtree-line where it wil shew you the Azimuth desired As in the latitude of 52 deg 30 min. the declination of the Sun being 11 deg 30 min. First I place the Equator at 37 deg 30 min. above the South end of the horizon and count the Declination 11 deg 30 min upon the horizon where I finde to meet me the parallel of 7 deg 4 min. and such is the Azimuth from the East or West points Northwards CHAP. XXXVII Having the latitude of the place and the Declination of the Sun to finde the Ascensional difference and consequently the time of Sun-rising and setting with the Diurnal and Nocturnal arches FIrst set the Planisphere to the latitude given and reckoning the declination upon the Axletree-line follow low the parallel thereof to the Horizon where you shall finde meet you the meridian of the ascensional difference counted from the houre of six and this same meridian giveth you the time of Sun setting counted from the South part of the limbe or general meridian and likewise the time of his rising counted from the North part of the limbe and now if you still note this intersection of the Horizon and Suns parallel and count the houres thereupon between the South part of the limbe and the Horizon you shall have the Suns semidiurnal arch and from the Horizon to the North part of the limbe his seminocturnal arch both which being doubled giveth you both the diurnal and nocturnal arches As for example let the latitude given be 52 degrees 30 minutes and the declination 11 degrees 30 minutes first I set the Equator 37 degrees 30 minutes above the Horizon then counting 11 degrees 30 minutes upon the Axtree-line I follow the parallel thereof to the Horizon where I finde meet me the meridian of 15 deg 22 min. and such is the difference of Ascensionals required Now upon the same parallel of declination I seek how many houres I can finde betweeene the South part of the limbe and the Horizon which is 7 houres 1 minute and so much is the semidiurnal arch and the time of Sun setting which being doubled maketh 14 houres 2 minutes the diurnal arch so likewise counting the houres betweene the North part of the limbe and the horizon I finde them 4 houres 59 minutes and such is the seminocturnal arch and the time of sunrising which being doubled giveth 9 houres 58 min. for the Nocturnal arch the semidiurual and seminocturnal arches of any Star whose declination is known may be found in the self same manner CHAP. XXXVIII Having the latitude of the place and the declination of the Sun to finde the time of the beginning and ending of twilight FIrst set the sphere to the latitude given then count the declination upon the axtree-line either Northward or Southward according as it is and then place the threed parallel to the Horizon to the 18 deg of depression below the same and note where it cutteth the parallel of declination for those houres to the Southward of the threed giveth the time of twilight ending and the other to the Northward the time of its beginning As for example let the latitude given be 52 degrees 30 minutes and the declination of the Sun 11 deg 30 min. first I set the Equator 37 deg 30 min. above the horizon and counting 11 deg 30 min. upon the axletree-line Northward because his declination is towards that pole I place the threed parallel to the horizon 18 deg below the same and I finde it cut the said parallel of 11 deg 30 min. at 2 houres 34 minutes from the North part of the limbe and such is the time of the begining of twilight so doth it likewise cut the same parallel at 9 houres 26 minutes from the South part of the meridian and such is the time of the ending of twilight which was required CHAP. XXXIX The latitude of the place the altitude and declination of the Sun being given to finde the houre of the day FIrst place the planisphere to the latitude given and seek his declination on the axletree-line among the parallels of Declination then place the threed parallel to the horizon to the degree of the Suns altitude counted from the horizon among the Almicanters and where the threed so placed shall cut the parallel of Declination there passeth by the Meridian which giveth you the hour of the day As for example in the latitude of 52 deg 30 min. the Declination of the Sun being 11 deg 30 min. and his altitude 25 deg 56 min. I set the Equator 37 deg 30 min. above the horizon and count 11 deg 30 min. upon the Axletree-line then placing the threed to 25 deg 56 min. among the Almicanters I finde it cut the parallel of 11 deg 30 min. at four hours 8 minutes from the Meridian and so much was the hour of the day if the observation was made in the afternoon but if the observation was made before noon then was it 7 a clock and 52 min. in the forenoon Again let the latitude and declination be the same as before and the altitude given 42 deg 26 min. now if I place the threed to 42 deg 26 min. among the Almicanters it will cut the parallel of 11 deg 30 min. at just two houres from the Meridian so shall it be either ten a clock in the morning or two afternoon according to the observation CHAP. XL. The latitude of the place the Suns altitude and Declination being given to finde his Azimuth FIrst set the Sphere to the latitude given and then count your altitude among the parallels of Declination and your Declination among the Almicanters so shall your Meridians become Azimuths and the work will be the same as in the former Chapter Only here by the way I would have you note that as your parallels and Almicanters changed offices so must your North and South part of the general Meridian change names so that when your threed cutteth any parallel at any Azimuth if you count that Azimuth from the South part of the limbe it shall
90 equall degrees and subdivide each degree into as many parts as quantity will give leave and let them be numbered with 10 20 30 c. unto 90 both ways as there you see and if upon the center A with any convenient distance you describe another Quadrant and divide it into six equal parts as in my other Quadrat set down and draw slope lines to each 15th deg in the Quadrant BC and number them backward then these slope lines may be accounted as hours in the art of dialling and will be sufficient to perform the whole Art of Dialling as I have at large taught in my Book lately published called the Compleat Diallist This done by laying a Ruler upon the center A and each degree in the Quadrant BC you may divide the two opposite sides from the center each into 45 unequal degrees called Tangents let these be numbered by 10 20 30 c. unto 90 degrees from B at the left hand to 45 in the opposite angle and so on to 90 at C the right hand of the Quadrant and back again as in the Quadrant and as you see in the figure draw also the diagonal line AD. To this Instrument as to all other of this kinde in their use is added Sights with a threed bead and plummet according to the usual manner The fore-side of this Quadrat is of it selfe being of a portable size and exactly drawn an absolute Instrument for the speedy resolution of both kinds of Trigonometria as well spherical as plain as shall be made manifest in the ensuing Treatise by applying it in several kindes But seeing the backside of this Universal Instrument standeth yet blank we will make use thereof in fixing a moveable Planisphere of Brasse thereon on which Planisphere we may project the Sphere in plano with streight lines and so by help of a fixed Horizontal line and fixed Almicanters we shall make this side also become Universal to performe most Spherical conclusions and that most demonstratively and generally for all Latitudes The description whereof is after this manner CHAP. II. The description of the back-side of the Universrl Quadrat FIrst upon the back-side draw foure straight lines neer to the edges and parallel thereto all cutting one the other at right angles as here in the figure you may behold upon which lines you shall insert the degrees of Altitude as you shall be shewed hereafter within these degrees draw four other parallel lines all cutting one the other at right angles unto which you shall draw every tenth degree and set his number thereto as here you see This being done inscribe a circle as large as you can within this Square within which circle inscribe two other Concentrical Circles representing the Ecliptique the dividing whereof shall hereafter appear Then upon the same Radius with this inwardmost circle describe another Circle upon some thin smooth plate this Circle divide into equal 24 parts or hours and let every hour be divided into halfs and quarters or rather minutes if quantity will give leave and let them be numbered towards the right hand as here you see That done draw another Concentrical Circle upon the margine of the houre Circle representing the plane of the general Meridian which divide into four equal parts in 00 90 00 90 and draw the two Diameters 00 00 90 90 crossing one the other at right Angles in the Center of the Planisphere so shall the Diameter 00. 00 Represent the Equator and the Diameter 90. 90. the Circle of the houre of Six and it is also the Axis of the World wherein 90 at the houre of 18 stands for the North pole and 90 at the houre of 6 stands for the South pole then let each Quarter of the Meridian be divided into 90 degrees and numbered from the Equator towards the Poles The Meridian being thus divided lay a Ruler upon each degree thereof counted from the Equator and thereby draw parallel lines to the Equator these parallels shall represent the Declination of the Sun Moon and Stars and the Latitude of Towns Cities and Countries and shall divide the Axis according to the line of Sines If again we divide the Equator and each of his Parallels in the like sort as the Axes of the World is divided and then carefully through each degree draw a line so as it makes no angles the lines so drawn shall be Elliptical and represent the houre Circles the Meridian Circle the houre of 12 at Noon and midnight and that 15 degrees next unto it drawn through 75 degrees from the Center the hours of 11 1 and that which is drawn through 60 degrees from the Center the houres of 10 and 2 and so of the rest let these houre Circles be prickt to know them from the other Meridians Then in the Meridian number 23 deg 30 minutes the greatest Declination of the Sun from the Equator to ♋ Northward and to ♑ Southward and draw the line ♋ ♑ for the Ecliptique then numbering 23 degrees 30 minutes from the North-pole North-ward and from the South pole South-ward draw the Axes of the Ecliptique which let be divided in like sort as the Axis of the World is divided and through each degree or each fifth degree draw-lines parallel to the Ecliptique these lines so drawn shall represent the Latitude of the planets and fixed stars and for distinction sake prick every fifth or tenth as before now if we divide the Ecliptick and his parallels as the Equator and his parallels are divided and then carefully draw a line through each 30th degree so as it makes no Angles the lines so drawn shall represent the signes the first 30 degrees from the Center towards ♋ shall stand for the signe ♈ the 30 degrees next following for ♉ and so of the rest as you may see by the figure The Sphere being thus projected in plain let us resort to the back of our Quadrat againe and draw a line through the Center cuiting two of the sides just in the middle so shall this Diameter make right Angles with the sides and represent the Horizontal line for any Region then take the just length of the Semidiameter of your Plainisphere and set it above and below the Horizon at both ends thereof in the side of the Quadrat as here you see and let this line be divided in like sort as the Equator and his Axis is divided and numbered with 10 20 30 c. unto 90 from the Horizon both wayes these are the degrees of Altitude All this being done let us come to the dividing of the Ecliptique which incompasseth the Sphere which may be done out of a Table of Right Ascensions for placeing your Sphere upon the Center of the Quadration the back side putting some pin through both their Centers so that they may be kept both at a stay for slipping one besides the other then turning the Sphere about untill the beginning of the houres come to the
Declination and Magnitude of some of the principal fixed Stars Rectified for this present year 1652. The Names of the Stars Right Ascension Declination Magnitude H. M. D. M. Â The Pole Star 00 32 87 19 N 2 Andromeda's girdle 00 52 33 48 N 2 The Rams horn 01 34 17 27 N 3 The Rams head 01 48 21 44 N 3 The head of Mednsa 02 46 39 32 N 3 Aldebaran the Bulls eye 04 16 15 48 N 1 Hircus the Goat 04 51 45 36 N 1 Orions left foot 04 58 08 40 S 1 Orions left shoulder 05 06 05 59 N 2 The first in Orions girdle 05 14 00 38 S 2 Orions right shoulder 05 36 07 17 N 2 Sirius the great Dog 06 31 16 13 S 1 Procyon the lesser Dog 07 21 06 07 N 2 The heart of Hydra 09 11 07 09 S 1 Cor Leonis the Lions heart 09 49 13 42 N 1 The Lions neck 09 50 21 49 N 2 The Lions back 10 54 22 33 N 2 The formost star in â–¡ of the great Bear 10 40 58 23 N 2 The Lions tail 11 30 16 34 N 1 The Root of the great Bears tail alio tail 12 38 57 51 N 2 The Virgins spike 13 08 09 16 S 1 The middlemost star in the great Bears 13 10 56 45 N 2 The last of the said tail 13 26 51 05 N 2 Arcturus 13 59 21 10 N 1 The bright star of the Crown 15 24 27 43 N 3 Antares the Scorpions heart 16 08 25 34 S 1 The bright star of the Harp 18 25 38 30 N 1 The Eagles heart 19 33 07 58 N 2 The bright star of the Vultur 19 34 08 01 N 1 Fomahant 22 38 31 25 S 1 Pegasus shoulder 22 48 13 14 N 2 Pegasus leg 22 52 26 09 N 2 CHAP. IV. To shew the Generall Use of the universall Quadrat IN the Use of this Quadrat let us for distinction sake call those Parallels which make Right Angles with the line whereon the sights stand Right Parallels and the other which Run parallel to the same line Contrary Parallels the one signifying Right shadow the other Contrary shadow the Generall Use whereof consists in the solution of the Golden Rule where three lines or numbers being given of a known denomination a fourth proportionall is to be found and this solution is diverse in regard of the signification of these parallel lines and also of the entrance into the worke The signification of these Parallels is sometimes simple and sometimes compound sometimes they signifie right lines or numbers only such as they are simple in themselves and sometimes they signifie naturall sines which is halfe the Chord of the double Arch and sometimes naturall Tangents which are perpendiculars at the end of the Diameter sometimes the work may be begun with one kinde of signification and ended with another it may be begun with the lines as they are in themselves and ended as they are sines it may begin as they are sines and end as they are Tangents The solution in regard of the entrance into the worke may be either right or contrary I call that right entrance or entrance on the Right Parallels when the two lines or numbers of the first denomination are applied or found in the Right parallels and the third line or number and that which is sought for on the contrary parallels and on the contrary I call that contrary entrance when the two lines or numbers of the first denomination are applied or found on the parallels of contrary shadow and the third line or number and that which is to be found out in the parallels of right shadow As for example let there be given three numbers 50. 60 and 70. unto which I am to finde a fourth proportionall Let the Question be this if 50 pounds gain 60 shillings what shall 70 pounds gaine here are numbers of two denominations one of pounds the other of shillings and the first with which I am to enter must be that of 50. pounds If then I would enter my Question upon a parallel of right shadow see where the right Parallel of my first number which is 50 cutteth the contrary parallel of the second which is 60 and upon the point of intersection there lay the threed the threed lying in this position look along upon the right Parallel of the third number which is 70 and see what contrary parallel cutteth it directly under the threed for that is the fourth proportionall number required which here is found to be 84 wherefore I conclude that if 50 pounds gaine 60 shillings 70 pounds will gaine 84 shillings for as 50 is to 60 so is 70 to 84 the like will apeare if you enter your worke upon the contrary parallels for it mattereth not on which kind of Parallels you enter so be it you enter your first and third number on the same kinde of Parallels so shall the third and fourth be of the other kinde Thus much for the Generall Use of this fore-side of the universall Quadrat which being considered and well understood there is nothing hard in that which followeth CHAP. V. To divide a line given into any number of equall parts LEt the line given be AB and let it be required to divide the same into 5 equall parts Take with your compasses the given line AB and lay it down upon any parallel whose number may be equally divided into five parts without any remainder as lay it upon the fifth parallel and at the point where the line endeth there lay the threed the threed lying in this position and in regard that 10 is â…• part of 50 therefore the segment of the tenth parallel which is cut by the threed shall be the distance which shall divide the given line into 5 equall parts As for example take the given line AB and lay it upon the fiftieth contrary parallel as from A to B seeing the given line endeth in B lay the threed upon the same point as here you see in the following figure then in regard that 10 is â…• of 50 therefore I take the segment cut by the threed upon the tenth parallel viz. CD and with it I divide the given line AB into 5 equall parts as here you see The fore-side of the Quadrat CHAP. VI. To increase a line in a given proportion TAke the line given with a paire of Compasses and lay it one the parallel of the number given and upon the point where the given lines endeth there lay the threed which being kept in this position the segment of the parallel of the number required intercepted by the threed shall give the line required Let A be a line given to be increased in the proportion of 3 to 5. First I take the given line A with my compasses and in regard that 30 and 50 is the same with 3 and 5 if from them you give o therefore I take the given line A and lay it upon the thirtieth parallel
as from F to G in the last Chapter and upon I the point where the line endeth there I lay the threed which lying in this position cutteth the parallel of 50 in the point B so shall AB be the length of the line B which was required CHAP. VII To diminish a line in a given proportion TO diminish a line in a given proportion is no other then that which was shewed in the last Chapter but only whereas there you entred with the lesser line which was to be increased upon the lesser parallel number and so the line required was found upon the greater parallel number so here you must enter with the greater line which is to be diminished upon the greater parallel number and that which is required shall be found upon the lesser parallel number Let A be a line given to be diminished in the proportion of 5 to 3 I take the given line B and lay it upon the fiftieth parallel and at the point where the line endeth I lay the threed as at the point B in the figure of the fifth Chapter the threed lying in this position I see it cut the thirtieth parallel in the point G so shall the distance FG give me the line A which was required and is ⅗ of the given line B. Here note that you may use other numbers to worke your proportion upon besides these for you may multiply or divide the proportionall numbers given either by 2. 3 or 4 c. and so work by their numbers as for 3 and 5 we may worke by 6 and 10 or else by 9 and 15 or by 12 and 20 or 15 and 25 or 18 and 30 or 24 and 40 or 30 and 50 as here we have done CHAP. VIII To finde a proportion betwixt two or more Right lines TAke the greater line betwixt your Compasses and extend it upon that parallel whose number you intend to make the denominator of the fraction at the end of which extent place the threed the threed resting in this position take the lesser line betwixt your Compasses and keeping one foot in the side of your Quadrat carry it parallel unto tthe greater line untill the other foot touch the threed which you may very well do by help of the parallels so shall the number of that parallel whereon the Compasses shall stay be the Numerator to the former Denominator Let the lines given be AB and CD first I take the line CD and place it upon the fiftieth parallel as from A to B upon which point B I lay the threed which being kept in this position I take the lesser line AB betwixt my Compasses and keeping one foot in the side of the Quadrat EH I carry them parallel to the greater line untill the other foot toucheth the threed at G so shall the fixed foot rest in the parallel of 30 which is the Numerator to the former Denominator wherefore I conclude that the line AB is 30 50 or ⅗ of the line CD or the line AB is in such proportion to the line CD as 30 to 50 which was required But if the line CD be greater then these parallels can contain take ½ thereof and enter with the same as before you did with the whole line so likewise take ½ the lesser line and carry it parallel as before you did the whole line and you shall produce the same proportion as before CHAP. IX To lay down sodainly 2 3 or more Right lines in any proportion required FIrst lay your threed by chance upon the Quadrat so as it may cut the parallels of all the numbers given so shall the threed cut off from each parallel the proportional lines required which may be taken with your Compasses and laid down presently As for Example let it be required to lay down foure streight lines which shall bear such proportion one to the other as foure given numbers which let be these 60 50 30 10. then opening the threed by chance so as it may cut the parallels of all these given numbers so shall the threed cut from each parallel the proportionall line required which you may take from your Quadrat and lay them down as here you see viz. from the parallel of 60 the line KL from the parallel of 50 the line AB from the parallel of 30 the line FG and from the parallel 10 the line CD CHAP. X. Having two lines given to finde a third proportionall line to them FIrst place both the lines given on the contrary side of the Quadrat and marke how many parallels of contrary shadow they do extend unto then take the second line againe and place it upon the right side of the Quadrat and note the extension amongst the parallels of right shadow now see where the contrary parallel of the first line cutteth the right parallel of the second line and there lay the threed the threed lying in this position see what right parallel is cut by the contrary parallel of the second line directly under the threed for that is the third proportionall line required Let the two lines given be EO and EF and it is required to finde a third line which shall be in such proportion to EF as EF is to EO first place the two lines upon the contrary side of the Quadrat and you shall finde the line EO to extend unto 20 and EF unto 30 now extend EF on the right parallels also which will be likewise unto 30 therefore where the contrary parallel of EO 20 cutteth the right parallel of EF 30 there lay the threed as at P the threed lying still in this position I see where the contrary parallel of EF 30 cutteth the threed which is at G upon the right parallel of 45 therefore FG shall be the third proportionall line required and shall be 45 of such divisions as EO is 20 and EF 30. To performe the same more speedily with the numbers for as 20 is to 30 so is 30 to a third number Therefore at the intersection of the twentieth contrary parallel with the thirtieth right parallel there lay the threed which being kept in this position see where the thirtieth contrary parallel cutteth the threed which will be at G upon the 45 parallel of right shadow which 45. is the third proportionall number as before CHAP. XI Having three lines given to finde a fourth proportionall to them HEre the first line and the third are to be placed on the contrary parallels and the second on the right parallels therefore at the intersection of the contrary parallel of the first line with the right parallel of the second there lay the threed the threed lying still in this position whese the contrary parallel of the third line cutteth the threed there is the right parallel of the fourth proportionall line Let the three lines given be EO OP and EK and let it be required to finde a fourth line which shall have such proportion to EK as
minutes as was required CHAP. V. The right sine of any arch being given to finde the radius FIrst take the given sine and place it from the side AE of your quadrat along upon that contrary parallel which cutteth the degrees of the sine given counted in the quadrant from G and where the sine endeth there place the threed so shall it shew upon the contrary parallel of 100 the radius required Let the line B in the foregoing Chapter be a given sine of 36 degrees 53 minutes First place it from N to M upon that parallel which runneth through the degrees of 36 and 53 minutes as at O then upon the point M I place the threed so doth it cut the contrary parallel of 100 in the point D making DE the radius which is the line A in the foregoing Chapter as was required CHAP. VI. The radius being given with a streight line resembling a sine to finde the quantitie of that unknown sine FIrst place the given radius upon the contrary parallel of 100 and thereto apply the threed then take the right sine given and setting one foot of your compasses in the side of the quadrat carry it parallel to the former till the other foot cut the threed and there stay for the parallel where the compasses so resteth shall cut the quadrant in the degrees answering to that unknown sine given As if the line A in the fourth Chapter were the radius given and B the streight line resembling a sine First I place the line A from E to D upon the contrary parallel of 100 and apply the threed thereto the threed lying in this position I take the line B and fixing one foot in the side of the quadrat AE I carry it parallel to the former until the other foot touch the threed so will one foot rest in the point N and the other in M upon the parallel of 60 which parallel cutteth the quadrat in the point O at 36 degrees 53 minutes counted from G this 36 degrees 53 minutes is the arch of which the given line B is the sine the line A being radius CHAP. VII The use of these parallels as they signifienaturall sines I Told you in the fourth Chapter of the foregoing Book that these lines did somtimes signifie themselves alone somtimes sines and somtimes tangents and now for distinction sake when I call them parallels I would have you to understand them as they are in their own proper signification and when I call them Sines I would have them understood as the Sines of those arches through which they run and when I call them Tangents I would have them understood as the Tangent of those degrees against which they butt in the sides of the quadrant And further for distincton whereas in their own signification I call them right and contrary parallels so now I call them right and contrary sines and right and contrary tangents These lines as they signifie sines hath like use in finding a fourth proportional sine as the ordinary Canon of Natural sines and the manner of finding it is alwayes such as in this example As the sine of 90 degrees unto the sine of 30 degree So is the sine of 23 degrees 30 minutes unto a fourth sine Wherefore place the threed at the intersection of the contrary Sine of 90 degrees with the right Sine of 30 degrees and it will cut the contrary Sine of 23 degres 30 minutes at the right Sine of 11 degrees 30 minutes and such is the fourth proportional Sine required Or you may place the threed at the intersection of the contrary Sine of 90 degrees with the right Sine of 23 degrees 30 minutes and it shall cut the contrary Sine of 30 degrees at the right sine of 11 degrees 30 minutes as before And thus may all the rest of the sinical propositions be wrought both wayes CHAP. VIII The use of these parallels as they signifie both Sines and Tangents HEre the work is in a manner the same as before as will appear by this example As the Sine of 90 degrees to the Sine of 51 degr 30 min. So is the Tangent of 30 degrees to a fourth Tangent Thus placeing the threed upon the intersection of the contrary Sine of 90 degrees with the right Sine of 51 degrees 30 minutes it will cut the contrary Tangent of 30 degrees at the right Tangent of 24 degrees 20 minutes Or if you place the threed at the intersection of the contrary Sine of 90 degrees with the right Tangent of 30 degrees it will cut the contrary Sine of 51 degrees 30 minutes at the right Tangent of 24 degrees 20 minutes which is the fourth proportional Tangent required And this manner of work will hold until the Tangents be greater then 45 degrees the side of the quadrat But when the given tangent or tangents exceed 45 degrees the side of the quadrat if the two first given numbers be Sines and the third a Tangent set the second Sine in the place of the first and the first Sine in the place of the second and in steed of the given Tangent take his complement unto a quadrant so shall the fourth proportional be the complement of the Tangent required As for example in this proportion As the Sine of 90 degrees To the Sine of 20 degrees So the Tangent of 70 degrees To a fourth Tangent And because the Tangent of 70 degrees is greater then 45 degrees the side of the quadrat therefore I turn the proportion thus As the Sine of 20 degrees To the Sine of 90 degrees So the Co-tangent of 70 degrees To the Co-tangent of the sum required Thus placing the threed upon the intersection of the contrary sine of 20 degrees with the right Tangent of 20 degrees the complement of 70 degrees it should cut the contrary Sine of 90 degrees at the right Tangent of the complement of the arch required but the threed intersecteth not the said contrary Sine of 90 degrees upon the quadrat but besides it which never happeneth but when the one Tangent is greater then 45 degrees and the other lesse Wherefore you are to note that at any such time though the threed cutteth not the contrary Sine of 90 degrees upon the quadrat yet is the threed by the former position placed at the true angle of the Tangent required which is 46 degrees 47 minutes the complement whereof 43 degrees 13 minutes is the arch required And if the first given numbers be Tangents and the third a Sine take the complements of the two Tangents and exchange their places as in the examples following But if one be more and the others lesse as in this proportion As the Tangent of 70 degrees To the Tangent of 43 degrees 13 minutes So the Sine of 90 degrees To a fourth Sine Herein regard the Tangent of 70 degrees is more