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-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-line or six a clock 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-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-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
the ascensional difference 1 houre 1 minute added to 6 houres maketh 7 houres 1 minute the semidiurnal arch or the time of Sun setting which being doubled giveth 14 houres 2 minutes for the Diurnal arch this being taken out of 24 houres leaveth 9 houres 58 minutes for the Nocturnal arch the halfe of which 9 houres 58 minutes is 4 houres 59 minutes the time of Sun rising or 7 houres 1 minute taken out of 12 houres leaveth 4 houres 59 minutes the time of Sun rising But if the declination be Southwards then take this 1 houre 1 minute the ascensional difference out of 6 houres and there will rest 4 houres 59 minutes for the semidiurnal arch which being doubled giveth 9 houres 58 minutes for the length of the day the complement unto 24 houres is 14 houres 2 minutes the length of the night the half of which is 7 hours 1 min. the time of Sunrising CHAP. XIX Having the latitude of the place the declination of the Sun to finde the time of the beginning or ending of twilight THe beginning of twilight is when the Sun approacheth within 18 degrees of the Horizon in the East and the ending of the same at 18 deg depression in the West wherefore to finde the same work thus First by the 16 Chap. get the height of the Sun at the houre of six if the declination be North but if it be South take his depression at six for all is one thing if his quantity of declination be alike the height or depression at six being knowne if the declination be North add the sine of this height to the sine of 18 deg but if the declination be south take the difference betwixt the sine of his depression and the sine of 18 deg so shall the sum of difference be the second proportional number then say As the cosine of the Latitude To this second proportional So is the radius To a fourth proportional sine Then As the cosine of the declination To this fourth sine So is the radius To the sine of the quantitie of time betweene the beginning or ending of twilight and the houre of six Thus in the Diagram of the 6 Chap. having the declination of the Sun ♈ O 11 deg 30 min. North and thereby the Suns height at six RO 9 deg 6 min. we have given us in the Triangle byO first the angle ybO the complement of the latitude 37 deg 30 min. and then the side O y 467 the radius being 1000 this side is made up of the sine of the Suns height at six 158 and the sine of 18 deg 309 or by adding of OR 158 to R y 309 we have O y 467 for the second proportional these being known with the right angle at y we may finde the base bO by the 10 Chapter or 27 Chap. of the second Book For if we place the threed upon the intersection of the contrary sine of 37 deg 30 min. with the right parallel of 467 it will cut the contrary sine of 90 deg at the right sine of 50 deg 7 min. which is the base bO But seeing the Radius of this sine bO is but the cosine of the declination OH 78 deg 30 min. therefore I place the threed again at the intersection of the contrary sine of 78 deg 30 min. with the right sine of this 50 degr 7 min. and it will cut the contrary sine of 90 deg at the right sine of 51 deg 33 min. and such is the distance of time between the houre of six and the beginning or ending of twilight being 3 houres 26 min. Now at all times if you take this distance out of six houres your remainder shall be the beginning of twilight and if you adde this distance found unto 6 hours the sum will be the ending of the same Except it be in those latitudes where the depression of 6 is greater then 18 deg and then you must in that Latitude and South declination whose depression at six is more then 18 deg work by the contrary rule And when the Sun hath North declination if you take the ascensional difference out of the distance of time betweene the houre of six and the beginning or ending of twilight you shall have the distance of time betweene Snn rising or setting and the beginning or ending of twilight And when the declination is South adde the ascensional difference to the distance of time found and you shall have the quantitie of twilight as before As for example let the distance of time betweene the beginning or ending of twilight and the houre of six being found to be 3 houres 26 min. if I take this 3 houres 26 min. out of 6 houres there will remain 2 houres 34 min for the beginning of twilight and if I add this 3 hours 26 min. to six hours it will make 9 houres 26 min. for the ending of twilight And now being the Sun hath North declination if I take the difference of ascensions 1 houre 1 min. as it was found by the last Chapter out of 3 houres 26 min. there will remain 2 houres 25 min. for the quantitie of twilight before or after Sun rising or setting CHAP. XX. Having the distance of the Sun from the next Equinoctial point to finde his right ascension AS the radius To the cosine of the greatest declination So is the Tangent of the distance To the Tangent of the right ascension Thus in the right angled Triangle ♈ X ♉ in the Diagram of the 6 Chap. having the angle X ♈ ♉ the Suns greatest declination 23 deg 30 min. and the base ♈ ♉ the distance of the Sun from the equinoctial point ♈ 30 deg with the right angle at X we may finde the side ♈ X the Suns right ascension by the 33 Chap. of the second Book For if we place the threed at the intersection of the contrary sine of 90 deg with the right tangent of 30 deg it will cut the contrary sine of 66 degrees 30 minutes at the right tangent of 27 degrees 54 minutes and so much is the side ♈ X the right ascension required CHAP. XXI Having the right ascension of the Sun or Star together with the difference of their Ascensions to finde the oblique Ascension and descension IF the declination of the Sun or Star be North substract the ascensional difference from the right Ascension and the residue will be the oblique Ascension but if you adde them together the sum will be the oblique descension But if the declination be South add the Ascensional difference and right ascension together so will the sum thereof be the oblique ascension but if you take the Ascensional difference out of the right ascension the remainder will be the oblique descension As for Example the right ascension of the Sun in the beginning of ♉ being found by the last Chapter to be 27 degrees 54 minutes and the Ascensional
the base 132 inches and his altitude 9 foot we may finde his solid content in feet For if we place the the threed to the intersection of the contrary parallel of 42 55 100 with the right parallel of 132 it will cut the contrary parallel of 3 which is the third part of his altitude at the right parallel of 9 3 10 then keeping the threed in the same position it will cut the contrary parallel of this 9 3 10 at the right parallel of 28â…ž and so many solid feet is in that cone whose bases circumference is 132 inches and altitude 9 foot CHAP. LIX Having the Diameter of the base of a cone given in inches and his altitude in feet to finde the solid content in feet AS 13 54 100 inches To the Diameter in inches So is the third part of his altitude in feet To a fourth number Then As 13 54 100 inches To the diameter in inches So is this fourth number To the solid content in feet Thus in the cone of the 57 Chapter having the Diameter of his base 42 inches and his altitude 9 foot we may finde his solid content in feet For if we place the threed to the intersection of the contrary parallel of 13 54 100 with the right parallel of 42 inches it will cut the contrary parallel of 3 foot at the right parallel of 9 3 10 and lying in this same position it shall cut the contrary parallel of this 9 3 10 at the right parallel of 28â…ž the solid content of the cone as before CHAP. LX. Having the side and semidiameter of a cone to finde his altitude or perpendicular falling from the top thereof perpendicularly upon his base FIrst place the bead to the side of the cone given upon the side of the Quadrat and then open the threed till the bead fall directly upon the right parallel of the semidiameter of the base so shall the bead cut the contrary parallel of the altitude required Thus in the cone of the 57 Chapter having the semidiameter of his base 21 inches and his side 9 foot and somthing better then 2 inches that is 110 inches if you place the bead to 110 inches counted on the side of the quadrat and then open the threed till the bead fall directly upon the right parallel of 21 inches it shall cut the contrary parallel of 108 inches which is just 9 foot for the altitude of the cone which was required CHAP. LXI Having the side of a segment of a cone with the semidiameter of each base or end thereof to finde the altitude of tâ—e whole cone as it were before it was cut off with the altitude of the lesser cone which was cut off FIrst take the lesser semidiameter out of the greater and note the difference then say As this difference To the side of the segment of the cone So is the greater semidiameter To the altitude of the whole cone And the lesse semidiameter To the altitude of the lesser cone which was cut off Thus in the cone of the 57 Chapter although it should be cut off by the plane EF equidistantly to the base BC yet having the side of the segment BEFC of the cone ABC which is BE or CF 6 foot and one inch or 73 38 100 inches with the semidiameter of the greater base 21 inches and the lesser 7 inches we may finde the perpendicular of the whole greater cone as it were before it was cut off and also of the lesser which was cut off For if you take 7 inches out of 21 there will remain 14 inches and then if you place the threed to the intersection of the contrary parallel of 14 inches with the right parallel of 73 38 100 inches it will cut the contrary parallel of 21 inches the semidiameter of the base of the greater cone at the right parallel of 108 inches which is 9 foot the altitude of the whole greater cone as if it had not been cut off The threed lying still in the same position doth likewise cut the contrary parallel of 7 inches which is the semidiameter of the base of the lesser cone which was cut off at the right parallel of 36 inches or 3 foot the altitude of the lesser cone which was required CHAP. LXII How the segment cut off from any cone may be measured THe segment of a cone is no other then a round tapering piece of Timber cut off before it commeth to a point and may be measured by that which before is delivered when therefore at any time you meet with any such solid piece which will be more ordinarily then with any other first get the length of his side and then his semidiameter at each end and so by the last Chapter the altitude of the whole greater cone and also of the lesser which is supposed to be added to the segment to make it a perfect cone this being done by the 59 or 60 Chapters measure the content of the whole greater cone and also of the lesser and taking the content of the lesser out of the greater the remainder without out doubt must be the content of the segment given As for example let the segment BEFC of the cone in the 57 Chapter be given to be measured let the side thereof be 73 38 100 inches and the semidiameter of the greater base BD 21 inches and the semidiameter of the lesser base EK 7 inches and so by the last Chapter the altitude of the whole greater cone AD 9 foot and the altitude of the lesser added cone AK 3 foot and then by the 59 Chapter the content of the whole greater cone will be 28â…ž feet and the lesser added cone 1 foot and somthing better or 1 5 72 foot the which if you take out of 28â…ž feet the content of the whole greater cone there will remain 27 29 36 feet for the content of the segment BEFC which was required CHAP. LXIII To finde the solid content of any pyramis LEt the base of the pyramis be of what manner it will you need do no more but take one third part of his altitude and then suppose it to be a prisma having equal bases like and parallel whose length is equal to the third part of the pyramis you may finde the solid content by one or other of the former Chapters concerning prisma's As for example let the following quadrangular pyramis be given to be measured the side of whose base is 36 inches and the side of the pyramis 76 36 100 inches with the semidiameter of the circle inclosing the base which is AB 25 45 100 inches by which and the side given with the help of the 60 Chapter we may finde the altitude AD to be 72 inches or 6 foot now having the side of the base in inches and the altitude in feet we may finde the content in fect For As 12 inches to the side of the base in inches So is the third part of the altitude in feet
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
the Almicanters the threed being thus placed upon the 11 deg 30 min. in the 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-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