Selected quad for the lemma: end_n

Word A	Word B	Word C	Word D	Occurrence	Frequency Band	MI	MI Band	Prominent
end_n	lie_v	side_n	south_n	1,453	5	9.2242	5	false

Text snippets containing the quad

ID	Title	Author	Corrected Date of Publication (TCP Date of Publication)	STC	Words	Pages
A93912	The description and use of the universall quadrat. By which is performed, with great expedition, the whole doctrine of triangles, both plain and sphericall, two severall wayes with ease and exactness. Also the resolution of such propositions as are most usefull in astronomie, navigation, and dialling. By which is also performed the proportioning of lines and superficies: the measuring of all manner of land, board, glasse; timber, stone. &c. / By Thomas Stirrup, Philomathemat.	Stirrup, Thomas.	1655 (1655)	Wing S5687; Thomason E823_8; ESTC R207671	96,908	228

first placing the bead to one of the sides then opening the threed untill the bead fall on the Right parallel of one halfe of the base so shall it shew the contrary parallel of the perpendicular CHAP. XX. How speedily to finde the perpendicular of a Right Angled Triangle the sides being given as also to finde the segments of the base cut by the perpendicular A Right angled Triangle is that whose sides comprehending the Right angle falleth square or perpendicularly one upon the other thereby making the Angle contained betwixt them a Right Angle or an Angle of 90 degrees To finde the perpendicular of such a Triangle lay the threed upon the side of the Quadrat rectifie the bead to the lesser side of the Triangle given then open the threed to the intersection of the contrary parallel of the greater side with the Right Parallel of the lesser side so shall the bead shew the contrary parallel of the perpendicular required Let the Triangle ABC be a right angled Triangle given whose perpendicular is required First laying the threed upon the side of the Quadrat rectifie the bead to 45 the lesser side BC of the given Triangle ABC then opening the threed to the point of intersection betwixt the contrary parallel of 60 the greater side AB and the Right Parallel of 45 the lesser side BC the threed lying upon this point you shall see it cut the contrary parallel of 36 which is the length of the perpendicular BD which was required Or the bead being fixed as before if you open the threed to the point of intersection betwixt the contrary parallel of the lesser side BC 45 and the right parallel of the greater side AB 60 the bead shall fall upon the right parallel of 36 which is the length of the perpendicular BD as before or having two beads upon your threed if you place the one to 45 the lesser side and the other to 60 the greater side and opening the threed as before unto the point of intersection betwixt the contrary parallel of 60 the greater side and the right parallel of 45 the lesser side the threed lying in this position the bead which signifieth the greater side shall shew the right parallel of 36 and the other bead shall shew the contrary parallel of the same which is the perpendicular BD as before And farther the threed and beads lying still in this position the bead which signifieth the greater side 60 shall cut the contrary parallel of 48 which shall be the greater segment of the base divided by the perpendicular viz. AD and the bead which signifieth the lesser side 45 shall cut the right parallel of 27 which is the lesser segment of the base viz. CD CHAP. XXI To finde the perpendicular of any Triangle the three sides being given FIrst for this purpose lay the threed upon the side of the Quadrat and rectifie the bead to the first side then take the second side with your Compasses and setting one foot in the end of the third side open the three untill the moving foot of your Compasses and the bead rest directly both at a point so shall that parallel which passeth by the point where the Compasse foot and the bead meeteth parallel to the side wherein the other foot of the Compasses standeth be the perpendicular required As for Example let it be required to finde the perpendicular of the Triangle in the last Chapter First lay the threed to the side of the quadrat and rectifie the bead to the greater side AB which is 60 then take 45 betwixt your Compasses which is the lesser side BC and setting one foot in the side of the Quadrat in 75 which is the base AC with the other open the threed untill the point of the Compasses and the bead meet both in one point so shall the parallel distance betwixt the bead and that side of the Quadrat wherein the foot of the Compasses standeth be 36 the perpendicular required CHAP. XXII To finde a proportion betweene two or more like superficies FIrst place the threed upon the side of the Quadrat then rectifie the beads one to the one given side and the other to the other given side then at the point of intersection betwixt the greater sides contrary parallel with the right parallel of the lesser side there lay the threed the threed lying in this position the bead which signifieth the greater side shall cut the contrary parallel of the greater terms required and the other bead which signifieth the lesser side given shall cut the right parallel of the lesser terme required which two termes or numbers shall beare such proportion the one to the other as the two given sides which was required Let A and B be two sides of like superficies as the sides of two squares or the Diameters of two Circles the side A being 45 and the side B 60 First therefore lay the threed upon the side of the Quadrat and rectifie the beads one to 45 and the other to 60 then at the point where the contrary parallel of 60 intersecteth the right parallel of 45 there lay the threed the threed lying in this position the bead for 60 shall cut the contrary parallel of 48 and the bead for 45 shall cut the right parallel of 27 these two numbers viz. 48 and 27 shall be in such proportion one to the other as the two given superficies wherefore the proportion of the superficies whose side is B to that whose side is A is as 48 to 27 which is in lesser termes as 16 to 9. CHAP. XXIII Two Geometricall Squares being given to adde them together into one Square and also to adde two circles together into one circle FIrst from the center of your Quadrat lay or count the two sides of the given squares one upon the one side and the other upon the other side of the Quadrat then setting one foot of your Compasses in the extreame point of one of the given sides extend the other foot to the extreame point of the other given side so shall you have betwixt your Compasses a third side which shall make a square equall to both the given squares Note what is saide here of the sides of squares the like may be understood of the Diameters of Circles Let the two lines A and B be the sides of two given squares and let it be required to adde them together into one square or to make a third square equall to them both seeing the side A is 45 and the side B 60 therefore I set one foot of my Compasses in 60 upon one side of my Quadrat and extend the other unto 45 upon the other side of my Quadrat so shall I have betwixt my Compasses the line C which being applyed to the side of the Quadrat giveth 75 upon which line C if I make a square it will be equall to both the squares of A and B which was required And what is

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 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