M. Blundevile his exercises containing sixe treatises, the titles wherof are set down in the next printed page: which treatises are verie necessarie to be read and learned of all yoong gentlemen that haue not bene exercised in such disciplines, and yet are desirous to haue knowledge as well in cosmographie, astronomie, and geographie, as also in the arte of navigation ... To the furtherance of which arte of navigation, the said M. Blundevile speciallie wrote the said treatises and of meere good will doth dedicate the same to all the young gentlemen of this realme.

THE DESCRIPTION and vse of the Tables of Tan∣gents and Secants.

EVclid in the second proposition of his third booke defineth the line Tangent in this sort. A right line (saith hee) is said to touch a Circle when it toucheth it so, as being drawne out in length, it would neuer cut the said Circle.

The line Secant is not by him any wher defined, but what these two lines are, you shal better vnderstand by this figure Demonstratiue here fol∣lowing, then by any definition that can be made thereof: for a de∣finition ought to bee plaine and briefe, and not long, intricate or doubtfull, which will be hardly performed in shewing the nature of these two lines by way of definition, and therefore marke well this figure following.

[illustration]

In this figure you sée first a Circle drawne vpon the Cen∣tre C. from which Centre is ex∣tended to the circumference of the Circle a right line, called the Semidiameter, marked with the letters A. C. then there is another right line which toucheth the said Circle, and also the outermost end of the said Semidiameter making therewith a right Angle in the point A. and is called the line Tangent, then there is a third line which procéeding from the Centre C. doth cut the circumference of the Circle in the point B. and also méeteth with the line Tangent in the point D. and therefore is called the line Secant, betwixt which two lines, I meane the Tangent and Secant, is intercepted or included a certaine portion or arch of the foresaide Cir∣cle

lesse then a Quadrant marked with the letters A. B. of which Arke the line A. D. is the Tangent, and the line C. D. is the Secant thereof, which must néeds méete with the Tangent in the point D. because that the two Angles C. A. D. and D. C. A. are lesser then two right Angles, for the one is right, and the other sharpe, by reason that the Arke is lesse then a Quadrant. And some doe call the line Tangent the line Ascript, because it is as∣cribed to the Circle, and they call the line Secant the Hipothe∣nuse, because it subtēdeth the right Angle A. & they cal the Semi∣diameter or total Sine, y^• base of the rectangle Triangle C. A. D which is called a rectangle Triangle because it containeth one right Angle marked with the letter A. and note y^t whensoeuer any manner of Angle is propounded by thrée letters, that the middle letter doth alwaies signifie the Angle propounded, be it right, sharpe, or blunt. Now if you would know to what end the foresaid two lines were inuented, & wherto they serue, you haue to vnder∣stand that they chiefly serue in calculating the quantitie of Angles and their sides, as well in right lined Triangles, as in sphericall Triangles, for the sides of Triangles are either right or crooked, and if they haue right sides, such Triangles are either right ang∣led Triangles, or oblique angled Triangles, and you haue to note that the quantity of euery Angle is to be measured by the arch of a Circle subtending that Angle, for the point of euery Angle is imagined to be the Centre of a whole Circle, which you may sup∣pose to be so great or little as you will, for euery Circle (be it great or little) is diuided into 360. degrées, and looke how many degrées and minutes the arch subtending that Angle containeth, so much is the quantitie of that Angle, the practise whereof is very well set down by Clauius in his cōmentaries vpon Theodosius, which I mind (God willing) hereafter to translate into our mother tongue: In the meane time my intention here, is onely to shew you by one example or two, the vse of the tables made for the fore∣said lines Tangent and Secant.

The vse of the said tables according to Clauius, is thus.

IN séeking out the Tangent or Secant of any Arke giuen, or of the complement of any Arke by either of these tables, you haue to

obserue the selfe same order which you did before in finding out the right Sine of any Arke giuen, or of the complement of any Arke, Per tabulas sinuum. As for example, if you would find out the Tangent of an Arke containing 50. degrées, 24′· then resorte to the table of Tangents in the front, whereof looke first for the Arke 50. degrées, and then in the first collum on the left hand of the said table, for 24′· right against which on the right hand vnder the Arke 50. you shall find in the cōmon Angle the Tangent to be 12/087/ 923. the totall Sine whereof is 10/000/000. but if you would find out the Secant of the foresaid Arke 50. and 24′· then you must resort to the table of Secants, and hauing found out the Arke 50. in the front of the said tables, and the 24′· on the left hand as be∣fore, you shall find in the common Angle the Secant to be 15688144. And if you would haue the Tangent of the complement of the said Arke which is 39. degrées, and 36′· you shall find the 39. de∣grées of the cōplement in the foote of the table of Tangents right vnder the Arke 50. & the 36′·. in the outermost collum on the right hand of the said table, with which complement you must enter the table of Tangents, séeking for 39. degrées in the front of the table, and 36′· in the first collum on the left hand of the said table, right against which in the common Angle you shall find 8272720. to be the Tangent of 39. degrées, 36′· which is the complement of 50 degrées, 24′·

And you must work in like manner with the table of Secants: As for example, if you would find the Secant of 72. degrées, 36′· first then enter the table of Secants, looking for 72. degrees a∣boue in the front of the table, and 36′· in the first collum on the left hand of the said table, and in the common Angle you shall find 33/ 440/240. which is the Secant of 72. degrees, 36′· But if you would haue the Secant of the complement of the said arch 72. de∣grees, 36′· then looking in the foote of the table right vnder 72. degrees, you shall find 17. degrees, and in the outermost collum on the right hand, iust against 36′· you shall find 24′· so as you see that 17. degrees 24′· is the complement of 72. degrees, 36′· with which complement you must enter the table of Secants, looking for 17. degrees, aboue in the front of the table, and for 24′· in the first collum on the left hand of the said tables, & in the common An∣gle you shall find 10/479/542. to bee the Secant of the arch 17.

degrees, 24′· which is the complement of the arch 72. de∣grees, 36′·

The vse of which tables in Astronomi∣call matters, I haue here set downe as followeth.

1 To find out the declination of the Sunne, the place thereof being knowne.

MVltiply the Secant of the complement of the greatest declination by the totall sine, and diuide the product by the sine of the sunnes distance from one of the equinocti∣all points, the quotient is the Secant of an arch, whose complement is the declination of the sunne: for example, suppose that the sunne be entering into ♉ to finde the decli∣nation therof, first I multiply 25112030 the Secant of 66. degrées 32′· (which is the complement of 23. degrées 38′ the greatest declination) by 10000000. the product is 251120300000000. which being diuided by 5000000. (the sine of 30. degrées the sunnes distance from the equinoctial point) the quotient is 50224060. for which number I séeke in the table of Secants, the arch answering vnto it, is 78. degrées 32′· the complement whereof is 11. degrées 29′· which is the declination of the sunne.

2 Knowing the declination of the sunne how to finde his distance from the equinoctiall point, and so consequently his place in the Zodiacke.

MVltiply the Secant of the complement of the greatest decli∣nation by the totall sine, and deuide the product by the Se∣cant of the cōplement of the declination giuen, the quotient is the distance of the sun from the equinoctiall point. As for example, the declination of the sunne is supposed to be 11. degrées 29′· then to

finde his distance from the equinoctiall, I multiply 25112030. the Secant of 66. degrées 32′· (which is the complement of the greatest declination) by the totall sine the product is 251120300000000. which I diuide by 50224350. the Secant of 78. degrées 31′ the complement of 11. degrées 29′· the supposed de∣clination, the quotient is 5000000. the sine whereof is 30. de∣degrées 0′ which is the distance of the sunne from the equinoctial: thē for his place you must take the same according to y^e seasō of the yeare: for if it be in Aprill, then the sunne is entering into Tau∣rus, but if it be in August it is entering into Virgo, and being in October, it is entering into Scorpio, and being in February it is in the beginning of Pisces.

3 To finde out the right ascention of the sunne.

MVltiply the Tangent of the distance of the sunne from the e∣quinoctiall point which is nearest vnto it, by the sine of the complement of the greatest declination, and diuide the pro∣duct by the totall sine, the quotient is the Tangent of the right as∣cention of the sunne, for which if you seeke in the table of Tan∣gents, the arch answering vnto it is your desire: For example the sunne being in the first of Taurus, to knowe the right ascention thereof, I multiply 5773502. the Tangent of 30. degrées (for that 30. degrées is the distance of the sunne from the equinoctiall point) by 9172920. the sine of 66. degrées 32′· which is the com∣plemēt of 23 degrées 28′· the greatest declination, the product is 52959871965840. which being deuided by 10000000. the to∣tall sine, the quotient is 5295987. which is the Tangent of 27. degrées, 54′· which is the right ascention of the sunne béeing en∣tered into Taurus.

4 How to finde out the declination of the sunne know∣ing onely the right ascention thereof.

MVltiply the Tangent of the complement of the greatest de∣clination by the total sine, and deuide the product by the sine of the right ascention giuen, the quotient is the Tangent of the complement of the sunnes declination: for example the right

ascention of the sunne being 27. degrées 54′· I would knowe the declination thereof, multiplying 23035062. the Tangent of 66. degrées 32′· the complement of 23. degrées 28′ by 10000000. the product is 230350620000000. which being deuided by 4679298. the sine of the giuen ascention the quotient is 49224856. the arch of which Tangent is 78. degrées 32′· which being subducted out of 90. the remainder is 11. degrées 28′· and so much is the declination of the sunne.

5 How to finde the place of the sunne knowing onely the right ascention thereof.

SVbduct the right ascention giuen out of 90. if it bee lesse then 90. but if the same be more then 90. subtract the ascention gi∣uen out of 180. & being greater then 180. subduct the same out of 270. or being greater then 270. subduct the same from 360. and multiply the Tangent of the remainder by the sine of the comple∣ment of the greatest declination, and deuide the product by the to∣tall sine, the quotient is the Tangent of the complement of the suns distance from one of the Equinoctiall points: which distance being knowne, the place of the sunne can not be vnknowne. For example, supposing the right ascention of the sunne to be 27. de∣grées 54′· the complement thereof is 62. degrées 6′· the Tangent whereof is 18886715. which being multiplyed by 9172920. the product is 173245985757800. which being deuided by 10000000. the quotient is 17324598. for which I looke in the table of Tangents, and I finde the arch thereof to be 60. the comple∣ment whereof is 30. which is the distance of the sunne from the equinoctiall point, that is, from Aries, for that the right ascen∣tion is lesse then 90. I say then that the sunne is in the beginning of Taurus. But if the right ascention had béene more then 90. and lesse then 180. the place of the sunne had béene betwixt Can∣cer and Libra 30. degrées from Libra, so should it haue béene in the first point of Virgo, but if the right ascention had béene more then 180. the place of the sunne should be betwixt Libra and Ca∣pricorne, that is in the beginning of Scorpio, but being more then 270. the place of the sunne should be betwixt Capricorne and Aries, that in the beginning of Pisces.

6 To find out the ascentionall difference of the sunne or any starre in the firmament, knowing the declination thereof, and also the latitude of your region.

MVltiply the Tangent of the declination of the Sunne or star by the Tangent of the latitude of the place, and deuide the product by the totall sine: the sine of the quotient is the sine of the ascentionall difference, the arch wherof is the desired ascentionall difference. For example let the declination of the sunne, starre, or other point in the firmament be 10. degrées 3′· and suppose the la∣titude to be 52. degrées, the Tangent of 10. degrées 3′· is 1772268. the tangent of 52. degrées is 12799416. which being mul∣tiplyed together, the product will be 22674995395488. which being deuided by 10000000. the totall sine, the quotient is 2267499. for which number I looke amongst the sines, the arch an∣swering thereunto is 13. degrées 6′· which is the ascentionall difference desired.

7 To find out the oblique ascention of the sunne.

KNowing the place of the sunne, finde out the right ascention of the same by the thirde proposition, and finde also the ascentionall difference of the same point, then if the declination of the sunne be North, subduct the ascentionall difference out of the right ascention, the remainder is the oblique ascention. For example, suppose the sunne to be in the beginning of Taurus, now to finde the oblique ascention thereof in the latitude 52. degrées, first by the third proposition I finde out the right ascention of the beginning of Taurus, which I finde to be 27. degrées 54′· then by the sixth proposition I finde the ascentionall difference to bee 15. degrées 4′· which being subducted frō 27. degrées 54′· (for that the declination is North) the right ascention the remainder is 12. degrées 50′· and so much is the oblique ascention for the latitude of 52. degrées. But if the sunne be in any of the Southerne sines, and that the declination bée South, then the ascentionall diffe∣rence is to be added vnto the right ascention before giuen.

8 To finde out the oblique descention of the Sunne at any time.

FIrst finde out the right ascention of the sunne by the third pro∣position, the same shall be the right descention thereof, then finde the ascentionall difference by the sixt proposition, and if the sunne be in any of the Northern signes, adde the ascentional diffe∣rence vnto the right ascention, the summe shall be the oblique de∣scention of the sunne: But if it be in any of the Southerne signes subduct the ascentionall difference out of the right ascention, the remainder is the desired descention: as for example the sunne be∣ing in the beginning of Taurus, the right ascention thereof by the third proposition is 27. degrées 54′· the ascentional difference thereof by the sixth proposition is 15. degrées 4′· for as much then as Taurus is a Northerne signe, I adde the ascentionall diffe∣rence vnto the right ascention, the summe is 42. degrées 58′· and so much is the oblique descention of the sunne being in the begin∣ning of Taurus.

9 To finde out the length of the day or night.

HAuing found out the ascentionall difference by the sixth pro∣position, adde the same vnto 90. if the sunne be in any of the Northerne signes, but if it bee in any of the Southerne signes, subduct the ascentionall difference out of 90. then diuide y^e summe of the addition or the remainder of the subtraction by 15. the quo∣tient wil shew the halfe length of the day in houres and minutes, which being doubled you shall haue the whole length of the arti∣ficiall day: For example the sunne being in the beginning of Tau∣rus, the ascentionall difference thereof is 15. degrées 4′· which for that the sunne is in a Northerne signe, I adde vnto 90. degrées, the summe is 105. degrées 4′ which being diuided by 15. the quo∣tient is 7. houres o′· the halfe length of the day which being dou∣bled will be 14. houres the whole length of the artificiall day in the latitude of 52. degrées.

10 To finde the houre of the Sunne his rising or setting in any latitude assigned.

FIrst find out the halfe length of the artificial day by the 9. pro∣position, and subtract the same from 12. houres, the remainder

will shew the houre of the sunnes rising, for example the sunne be∣ing in the beginning of Taurus to know the houre of his rising in the latitude of 52. degrées, by the ninth propositiō, the half length of the artificiall day I find to be 7. houres, which I subtract from 12. houres the remainder is 5. which sheweth that the sunne riseth at 5. of the clocke in the morning, but the halfe length of the day it selfe is the houre of the sunnes setting.

11 To find out the length of the planetary houres and to find what Planet raigneth at any houre of the day.

FIrst find out the length of the artificiall day by the ninth pro∣position, and diuide the same by 12. the quotient is the length of one planetary houre: or thus, hauing an houre of the artificiall day giuen, looke what houre the same is from the sunne rising, and multiply the same by 12. diuide the product by the length of the artificiall day, the quotient is the number of the Planetarye houre. For example the sunne being in the beginning of Taurus, and our latitude being 52. degrées, the length of the artificiall day by the ninth proposition is 14. houres: then doe I diuide 14. by 12. the quotient is 1 ⅙. that is 1. houre 10′· the length of one pla∣netary houre. But if an houre of the artificial day be giuen as y^t I would know what planetary houre it is at 9. of the clock the sunne being in the first of Taurus in the latitude of 52. degrées, hauing found that the sunne riseth at fiue of the clocke by the tenth pro∣position, I sée that 4. houres of the artificiall day are gone at nine of the clocke, I therefore multiply 12. by 4. the product is 48. which I diuide by 14. the length of the artificiall day, the quotient is 3 3/7. which is the planetary houre at the time set downe: like∣wise shall you finde the planetary houre of the night, finding the length thereof, and then worke with it as was shewed before for the day.

Thē to know the Planet which raigneth at the appointed time, you must consider the Planet whereof the day taketh his name, for that Planet ruleth the first houre of that day, the next Planet the second houre, and so foorth. But for your ease I haue set down a table whereby to finde the Planet which raigneth at any time.

Houres	1	2	3	4	5	6	7
of the day.	8	9	10	11	12
Houres	10	11	12			1	2
of the night.	3	4	5	6	7	8	9
Sunday.	☉	♀	☿	☽	♄	♃	♂
Munday.	☽	♄	♃	♂	☉	♀	☿
Tuesday.	♂	☉	♀	♂	☽	♄	♃
Wednesday	☿	☽	♄	♃	♂	☉	♀
Thurseday.	♃	♂	☉	♀	♂	☽	♄
Friday.	♀	♂	☽	♄	♃	♂	☉
Saturday.	♄	♃	♂	☉	♀	☿	☽

The vse whereof is this, hauing founde the number of the Planetary houre, looke for the same in the head of the Table, whe∣ther it be in the day or night, and right vnder it iust against the name of the day, you shall haue the Planet which rayneth at that time. For example the sunne being in the beginning of Taurus the 11. of April, being Thursday at nine of the clocke in the morning, I find the number of the Planetary houre at that time to be 3 3/7. thē looking for thrée amongst the houres of the day, I descend in that collum vntill I come to be iust against thursday, where I sée the Character of Sol to be, I conclude then that the sunne raigneth at that time.

12 To finde the arch of the Equinoctial, comprehended betwixt the Meridian and any Circle of position according vnto Campanus and Gazula.

MVltiply the Sine of the complement of your latitude by the Tangent of the distance of the giuen Circle of position from the Zenith, & diuide the product by the totall Sine, the quotient is the tangent of the arch of the Equinoctiall, which is compre∣hended betwixt the giuen Circle of position, and the Meridian. For example in the latitude of 52. I would knowe what part of the Equinoctiall is comprehended betwixt the Meridian and that Circle of position, which is 30. degrées from the Zenith: the la∣titude being 52. degrées, the complement thereof is 38. degrées,

the Sine whereof 6156615. and the tangent of 30. degrees, (for that 30. is the distance betwixt the giuen Circle of position and the Zenith) is 5773502. which being multiplyed by 6156615. the product is 35545229015730. which being diuided by 10000000. the totall Sine, the quotient is 3554522. for which I séeke amongst the tangents, and I finde the Arke answering thereunto, to be 19. degrées 34′ the Arch of the Equinoctiall, be∣twixt the Meridian and the giuen Circle of position.

13 Knowing the Latitude of your Region, and also the eleuation of the Pole aboue any Circle of position, howe to finde the inclination of the saide Circle of position vnto the Meridian, and so consequently the Arch of the Equinoctiall, which is betwixt the said Circle of position & the Meridian.

MVltiply the Secant of the complement of the eleuation of the Pole aboue the Circle of position by the Sine of your Lati∣tude, and diuide the product by the totall Sine, the quotient is the Secant of the complement of the inclination of the Circle of position vnto the Meridian, and that is the distance betwixt the Circle of Position and the Zenith, by helpe wherof you shall find the Arch of the Equinoctiall, betwixt the Circle of Position and the Meridian, as in the former proposition. As for example, sup∣pose the eleuation of the Pole aboue a Circle of position, in our Latitude of 52. to be 23. degrées 12′· Now to find out the Incli∣nation of that Circle of Position vnto the Meridian, first I mul∣tiply 25384445. the Secant of 66. degrées 48′· (for that is the complement of 23. degrées 12′· the eleuation of the Pole aboue the Circle of position) by 7880108. the Sine of 52. the Latitude of our Region, the product is 200032168120060. which being diuided by 10000000. the quotient is 20003216. for which I looke in the Table of Secants, and the Arch therof is 60. degrées 0′· the complement whereof is 30. degrées 0′· which is the In∣clination of the Circle of position vnto the Meridian, or the di∣stance of the Zenith from the sayde Circle, then to find the Arch of the Equinoctiall betwixt the sayd Circle of Position and the Miridian, repeate the worke of the former proposition.

14 To finde out the Eleuation of the Pole aboue any as∣signed Circle of Position in any giuen Latitude.

KNowing the Inclination of the assigned Circle of Position vnto the Meridian, multiply the Secant of the complement thereof by the totall Sine, and diuide the product by the Sine of your Latitude, the quotient is the Secant of the complement of the Eleuation of the Pole aboue the giuen Circle of Position: as for example, suppose the inclination of a Circle of position to be 30. degrées 0′· Now to finde the eleuation of the Pole aboue the same for the Latitude of 52. degrées 0′· First take the comple∣ment of 30. degrées 0′· which is 60. degrées 0′· the Secant wher∣of is 20000000. which being multiplyed by the totall Sine, the product is 200000000000000. whcih being diuided by 7880108. the Sine of 52. the assigned Latitude the quotient is 25380362. for which I séeke in the Table of Secants, and the Arch answering thereunto I finde to be 66. degrées 48′· the comple∣ment whereof is, 23. degrées 12′· and so much is the eleuation of the Pole aboue the assigned Circle of Position in your Latitude.

And thus much shall be sufficient to haue béene spoken for the vse of the lines Tangent and Secant at this present, of whose ample and infinite vse, you shall haue further taste in our Direc∣torie tables and Horologies, whereof a beginning is made, and shal be ended so soone as may be possible: in the meane time I shal desire the Reader fauourably to accept of these, vntill better lea∣sure and more fit opportunitie will be offered. And now followe the Tables themselues.