The mariners magazine, or, Sturmy's mathematical and practical arts containing the description and use of the scale of scales, it being a mathematical ruler, that resolves most mathematical conclusions, and likewise the making and use of the crostaff, quadrant, and the quadrat, nocturnals, and other most useful instruments for all artists and navigators : the art of navigation, resolved geometrically, instrumentally, and by calculation, and by that late excellent invention of logarithms, in the three principal kinds of sailing : with new tables of the longitude and latitude of the most eminent places ... : together with a discourse of the practick part of navigation ..., a new way of surveying land ..., the art of gauging all sorts of vessels ..., the art of dialling by a gnomical scale ... : whereunto is annexed, an abridgment of the penalties and forfeitures, by acts of parliaments appointed, relating to the customs and navigation : also a compendium of fortification, both geometrically and instrumentally / by Capt. Samuel Sturmy.

STƲRMY'S MATHEMATICAL and PRACTICAL ARTS The Sixth Book. Wherein is contained a Definition of the Circles of the Sphere, with the manner how to Resolve all the most necessary Propositions thereunto belonging, by a Line of Chords and Signs, or by Chords and Tangents; as also by Calculation by Tables. (Book 6)

CHAP. I.

IN the former Books are contained such Pro∣blems Geometrically, as are most necessary for every professed ingenious Artist to un∣derstand and Practice; Now to the end the Practitioner may Elevate his Thoughts to the contemplation of those Glorious Bo∣dies, the Sun, Moon, and Stars; I shall here in this place give a brief Survey of the first Rudiments of Astronomy, for the help of young Practitioners and Mariners, for whom chiefly I take these pains; I shall give a brief and succinct Explanation of the se∣veral Circles of the Sphere, better than we could in the foregoing RHYTHMES to be understood, and then shew how to re∣solve the most usual and common Problems thereof; and after the Art of Dialling Geometrically and Instrumentally, and by Cal∣culation, as promised.

1. This is to be understood, that Astronomers do imagine 10 principal Points, and 10 Circles to be in the hollow inside of the first Moveable Sphere, which are commonly drawn upon any Globe or Sphere, besides divers other Circles which are not delineated, but only apprehended in the Fancy.

The Points are the two Poles of the World, the two Poles of the Zodiack, the two Equinoctial Points, the two Solstitial Points, and the Zenith and Nadir.

2. The Poles of the World are two Points, which are directly opposite to one another, about which the whole frame of Heaven moveth from the East into the West, whereof that which is seen on the North-side the Equinoctial, is called the Arctick-Pole.

3. And the other directly opposite to the former is called the Antrctick, or South-Pole, and can be seen only on the South-side of the Equator; a right Line imagined to be drawn from the one Pole to the other, is called the Axis or Axle-tree of the World.

4. All other Lines drawn through the Centre of the Sphere, and limited on each side of the surface of the Sphere, is a Diameter, but not an Axis, unless the Sphere move round about it.

5. The Poles of the Zodiack are two Points directly opposite to each other, distant from the North and South Pole 23 degr. 31 min. and are Diametrically opposite, on which the Heavens move, from the West into the East.

6. The Equinoctial Points are in the beginning of Aries and Libra, to which Points the Sun cometh the 11 of March, and 13 of September, and makes the Days and Nights of equal Length in all places in the World.

7. The Point of the Summer Solstice, is the beginning of Cancer, which the Sun cometh unto about the 12 of June, and longest day, to the Inhabitants on the North part of the World, and the shortest day to the Inhabitants of the South.

8. And the Point of the Winter Solstice, is the beginning of Capricorn, to which the Sun cometh the 11 of December, and maketh the shortest Days of the Year to the Inhabitants of the North Hemisphere, and the longest to the Inhabitants on the South-side the Equinoctial.

9. The Vertical Point, or Zenith, is the Point directly over-head, and is the Centre or Pole of the Horizon, 90 degr. every way from the Horizon.

10. The Nadir is the opposite Point under our Feet.

Circles of the Sphere.

The Ten Circles are as followeth; The Equinoctial, or likewise called the Equator, which is the chief Circle in the Sphere, dividing the Heavens in the middle between the two Poles; the two Points of Aries and Libra, cut this Circle in opposite Points, and make the Days and Nights of equal length over all the World.

2. The Meridian is a great Circle passing through the Poles of the World, and the Zenith of the Place; the Sun when he comes to this Meridian, it is Noon; the number of Meridians is as many as can be imagined Vertical Points from the West to the East, whereof the Cosmographers have described, 180.

3. The Horizon is distinguished by the names of Rational, or Sensible; the first is a great Circle every where Equidistant from the Zenith, and divides the superior or upper Hemisphere, from the lower, and by chance are distinguished by the names of Right, Oblique, and Parallel-Horizon.

A Right-Horizon have the Inhabitants under the Equator, who have the Horizon passing through the Poles of the World, and cuts the Equator at Right-Angles.

An Oblique-Horizon is such an one as cuts the Equinoctial Oblique, or aslope, or hath any degr. of Latitude from the Equator.

A Parallel-Horizon is one that hath the Poles for the Zenith and Nadir, and the Equinoctial for the Horizon.

The Sensible-Horizon is a Circle that divideth the part of the Heavens, which we see, from the part we see not, called a Finitor.

4. The Zodiack is a great Circle, that divides the Equator into two equal parts; the Points of Intersection are Aries and Libra, the one half declining toward the North, the other to the South 23 degr. 31 min. his ordinary breadth is 12 degrees; but later Writers make it 14 or 16 degrees by reason of the Wandring of Mars and Venus.

In the middle thereof is a Line called the Ecliptick, from which the Latitude of the Planets are numbred both Northward and Southward; the Circumference of this Circle containeth 360 degr. which is divided into 12 equal Parts called Signs, every one representing some living Creature, either in Shape or Property, as you read in the De∣nominations; and also every Sign containeth 30 degr. and every degree containeth 60 min. and every min. 60 seconds, and every second 60 thirds.

The Names and Characters of the 12 Signs.

♈	♉	♊	♋	♌	♍
Aries.	Taurus.	Gemini.	Cancer.	Leo.	Virgo.
♎	♏	♐	♑	♒	♓
Libra.	Scorpio.	Sagitarius.	Capricornus.	Aquarius.	Pisces.

5. The Six uppermost are the Northern, and Six undermost the Southern Signs.

Of the Colures.

6. These are two great Circles, or two Meridians passing through the Poles of the World, crossing one the other at right Angles, and dividing the Equinoctial and the Zodiack into four equal parts, making thereby the four Seasons of the Year.

7. The Solstitial Colure is as before, a great Circle drawn through the Poles of the World, the Poles of the Zodiack, and the Solstitial Points of Cancer and Capricorn, shewing the beginning of Summer and Winter.

8. The Equinoctial Colure, is a Circle passing by the Poles of the World through both the Equinoctial Points of Aries and Libra, shewing the beginning of the Spring and Autumn, when Days and Nights are equal.

9. The T••opick of Cancer is a lesser Circle of the Sphere, equally distant from the Equinoctial to the Northward 23 degr. 31 min. 30 seconds, wherein when the Sun is, he is entring Cancer, and making his greatest Northern Declination.

10. The Tropick of Capricorn is also a lesser Circle, equally distant from the Equi∣noctial Southward 23 deg. 31 min. 30 seconds, to which when the Sun cometh, which is the 10th of December, maketh his greatest Southern Declination.

11. Of the two Pole Circles.

These are two lesser Circles, distant so much from the Poles of the World, as the Tropick of Cancer and Capricornus is from the Equinoctial 23 degr. 30 min. which are the Pole Points of the Zodiack, which moving round the Poles of the World, describe by their motion the said two Circles; that about the North-Pole is the Arctick Circle, and that about the South the Antarctick Circle.

12. The first Six are called great Circles, and the other Four lesser Circles; by the Centre of a Circle is meant a Point or Prick in the middle of a Circle, from whence all Lines drawn to the Circumference are equal, and are known by the names of Radius.

13. That is said to be a great Circle, which hath the same Centre as the Sphere, and Divides it into two equal halfs, called Hemispheres; and that is a lesser Circle, which hath a different Centre from the Centre of the Sphere, and Divides the Sphere into two unequal Portions or Segments.

14. Of other Circles imagined but not described in a material Sphere or Globe.

Such are the Azimuths, Almicanters, Parallels of Latitude and Declination.

Azimuth or Vertical Circles pass through the Zenith, and Intersect the Horizon with right Angles; wherein the distance of the Sun or Stars from any part of the Meri∣dian are accounted, which are called Azimuth, and the East and West is called the Prime Vertical Azimuth.

15. The Sun or any Star having Elevation or Depression above or below the Horizon, are then properly said to have Azimuths; but if they be in the Horizon, either rising or setting, the Arch of the Horizon between the Centre of the Sun or Star, and the true Points of East and West, is called Amplitude.

16. Circles of Altitude called Almicanters, are Circles Parallel to the Horizon, and Intersect the Vertical Circles with right Angles, which are greatest in the Horizon, and meet together in the zenith of the place, in which Circles the Altitude of the Sun, Moon, or Stars above the Horizon are accounted, which is the Arch of an Azimuth, contained betwixt the Almicanters, which passeth through the Centre of the Sun, Moon, or Stars, and the Horizon.

17. Parallels of Declinations are lesser Circles, all Parallel to the Equinoctial, and may be imagined to pass through every degree and part of the Meridian, and are de¦scribed upon the Poles of the World; in like manner, the Declination of the Sun or-Star is measured by the Arch of the Meridian between the Sun and Star, and the Equi∣noctial.

18. Parallels of Latitude in the Heavens, are lesser Circles described upon the Poles of the zodiack or Ecliptick, and serve to define the Latitude of a Star, which is the Arch of a Circle contained betwixt the Centre of any Star or Planet, and the Ecliptick Line, making right Angles therewith and counted toward the North or South Poles of the Ecliptick, the Sun never passeth from under the Ecliptick-Line, is said there∣fore to have no Latitude.

19. Longitude of the Sun or Stars is measured by the Arch of the Ecliptick, com∣prehended between the Point of Aries, and a supposed great Circle passing from the Poles of the Ecliptick and the Sun or Stars Centre, and accounted according to the order and succession of the Signs.

20. Longitude on the Earth, is an Arch of the Equinoctial contained between any assigned Meridian where it begins, and the Meridian of any other place, but accounted Eastward from the first place, as the Right-Ascention; but in my Tables it is counted East and West from the Meridian of the Lands-end terminating at 180 degrees.

21. Right-Ascention is an Arch of the Equinoctial accounted from the beginning of Aries, which cometh to the Meridian with the Sun, Moon, or Stars, or any portion of the Ecliptick; and by it there are Tables made for the Sun, Moon, and Stars to know the time of their coming to the Meridian, as by the help of the hour of the Star, the true time of the Night.

22. Oblique Ascention, is an Arch of the Equinoctial between the beginning of Aries and that part of the Equinoctial, that riseth with the Centre of the Sun or Star, and any portion of the Ecliptick in any Oblique-Sphere.

23. Ascentional-difference is the Arch of Difference between the Right-Ascention, and the Oblique-Ascention, and thereby is measured the time of the Sun or Stars before, and after Six.

24. Elevation of the Pole is the Height thereof above the Horizon, which is equal to the distance between the zenith, and the Equinoctial, whose Complement is equal to the distance of the zenith from the North or South Pole, or the Elevation of the Equator above the Horizon; these Circles I have inserted, to the end they may be perfectly known; for without them, the Reader cannot well understand the following Problems of the Sphere that are depending thereon.

CHAP. II. The Projection of the Sphere in Plano, represented by the Analemma, and the Points and Circles before described.

THe Sphere may be Projected in Plano in straight Lines, as in the Ana∣lemma, if the Semi-diameters of the Circles given, be Divided in such sort as the Line of Signs in the Fundamental Diagram of the Scale.

This Scheme is fitted for the Latitude of Bristol 51 degr. 28 min. and represents the Points and Circles of the Sphere before described.

Take with your Compasses the Chord of 60 degr. and upon the Centre C describe the Circle HZON (2.) Draw the Diameter HCO which represents the Horizon; and at Right-Angles thereunto, cross it with another Diameter ZCN.

Then with the Latitude of the place, prick off 51 degr. 28 min. from O to N, and from H to S; and of the same Line of Chords, take the Complement of the Latitude 38 degr. 32 min. and prick off from HAE, and from O to Q, and draw NSC and AECQ.

Then take the Suns Declination 23 degr. 31 min. and prick off from AE to G and T, and with the like Chord do the same from N to Y and g, for the Polar Circle; and the like do from Q to D and P, and from S to X and ♄; and through these Points draw Parallels to the Equator Y g, and TSD, and G h: P, and X ♄.

And through the Centre draw the Ecliptick-Line TGP; and draw RS Parallel to the Horizon HCO, which is the Parallel of Altitude of the Hour of Six; and at any other distance, draw Parallels of Altitude E I f.

(1.) Thus are the Points before defined, represented in this Diagram; N is the North-Pole-Arctick, S the South-Pole-Antarctick; g the North, X the South-Pole of the Ecliptick; C both the Equinoctial-Points of Aries and Libra.

T The Point of the Summer-Solstice— P the Point of the Winter-Solstice, Z the Zenith over our Heads, N the Nadir-points under our Feet.

(2.) The greater Circles are HCO the Horizon, ZCN the Axis thereof, or Prime Vertical Azimuth of East and West; HZON the great Meridian, and also the Colure of the Summer and Winter-Solstice,— AECQ the Equinoctial; T C P the Ecliptick; SCN the Axis of the World, the Hour-Circle of 6; and lso it repre∣sents the Colure of the Equinoxes.

(3.) The lesser Circles are there represented, T D the Tropick of Cancer; GP the Tropick of Capricorn; Y g the Arctick Circle, about the Pole North; X ♄ the Antarctick-Circle, or South-Pole.

(4.) Other Circles not described upon Globes, are there represented; E f repre∣sents a Parallel of Altitude called an Almicanter; the Prickt Arches Z ☉, and S I be∣ing Ellipses represent the Azimuths, or Vertical-Circles.

[illustration] geometrical diagram

The Projection of the Sphere in Plano, by straight Lines or Signs.

The Prickt Arches from the Poles, represent the Meridian or Hour-Circles, which are also Ellipses; the Drawing thereof will be troublesom, and for that reason is not mentioned; and how to shun them in the resolution of any Proposition of the Sphere, by Chords shall be shewn in the several Questions following; But the Sphere may be Projected in Plano by Circular-Lines, as in the general Astrolobe of Gemma Trisius, by help of the Tangent, and ½ Tangents in the Fundamental Diagram of the Scale, and by the Directions in the 4 Book 12 Chapter beforegoing, and will resolve the same things; the directions shall be one and the same, in both, in Letters, and represents the same.

[illustration] geometrical diagram

The Convex Sphere, by Circular Lines.

Any Line drawn Parallel to the Equinoctial AEQ, as pq, TD. Yg: doth repre∣sent Parallels of Declination.

And any Line drawn Parallel to the Ecliptick TP, represents a Parallel of Latitude of the Stars and Planets in the Heavens.

(5.) Divers Arches relating to the motion of the Sun, and seen upon the Globes, and found by Calculation, are in the Convex-Sphere, represented in Right-Lines, and in the Concave-Sphere by Circular-Lines.

(1.) The Suns Amplitude, or Coast of Rising and Setting, from the East and West in the Analemma, is represented by CL in North Signs, and by CF in South Signs.

(2.) His Ascentional-difference, or time of Rising and Setting from Six in Summer by SL, in Winter by FH.

(3.) His Altitude at Six in Summer by RC, and his Depression at Six in Winter, by Cb.

(4.) His Azimuth at the hour of Six in Summer, by RS, or CI, equal to hb in Winter.

(5.) His Vertical-Altitude, or Altitude of East and West, by MC in Summer, and his Depression therein in Winter by CN.

(6.) His hour from Six being East and West, in Summer by MS, in Winter by Nh.

(7.) His Azimuth from the East and West upon any Altitude, is represented in the Parallel of Altitude by the Convex-Sphere, where it Intersects the Parallel of Declina∣tion by I ☉; but in the Concave-Sphere may be measured on the Horizon HO, as CV, or CI, measured on the Line of half Tangents.

(8.) The Hour of the Day from Six, to any Altitude, is always represented in the said Point of Intersection, in the Parallel of Declination, hereby q ☉, or in the Concave-Sphere by S ☉; and all these Arches thus represented in Right-Lines, are the Signs of those Arches to the Radius of that Parallel in which they happen, being ac∣counted in the midst of that Parallel.

How to measure the Quantities of those respective Arches by a Line of Chords and Signs, and by Half-Tangents; and consequently thereby to resolve the most useful Cases of Spherical Triangles; as also by Calculation, is what I intend shall be the sub∣ject of the Pages, viz. and the Art of Dialling by a Gnomical Scale.

The former Sphere or Scheme doth represent the Triangles commonly used in Cal∣culation.

Thus the Right-Angled Triangle CK ☉, Right-Angled at K; supposing the Sun at ☉, is made of CK, his Right Ascention, ☉ K his Declination; — KC ☉ the Angle of the Ecliptick, and the Equinoctial being the Suns greatest Declination 23 deg. 31′ C ☉ K, the Angle of the Suns Meridian and Ecliptick.

In the Right-Angled-Triangle LON, Right-Angled at O; supposing the Sun at LON is the Elevation of the Pole, NL the Complement of the Suns Declination, LO the Suns Azimuth from the North.

LNO the Hour from Midnight, or Complement of the Ascentional-Difference, NLO the Angle of Position, that is, of the Suns Meridian with the Horizon; and of the like parts, or their Complements, is made the Triangle CML.

In the Right-Angled-Triangle CIS, Right-Angled at I; supposing the Sun at S, there is given CS his Declination, IS his Altitude at the hour of Six, CI the Suns Azimuth from the East and West at the hour of Six, ICS the Angle of the Poles Elevation, CSI the Angle of the Suns Position.

In the Right-Angled-Triangle COM, suppose the Sun at M; dM the Suns De∣clination, Cd his hour from Six, CM the Altitude being East or West, dCM the Latitude, dMC the Angle of the Suns Position.

In the Oblique-Angled Triangle Z ☉ N, if the Sun be at ☉. ZN is the Com∣plement of the Latitude, and N ☉ the Complement of the Suns Declination, or di∣stance from the Pole. Z ☉ the Complement of the Suns Altitude, or height; ZN ☉ the Angle of the Hour from Noon; NZ ☉ the Suns Azimuth from the North-part of the Meridian; Z ☉ N the Angle of the Suns Position.

And thus we have shewed how the former Diagram or Analemma represents the Spherical Triangles used in Calculation; whereby, of the Six parts in each Triangle, if any three are given, the rest may be found by Calculation from the Proportions, and that either by Addition and Substraction, by the Artificial Signs and Tangents; and what is resolved by either of these sorts of Tables, we will resolve with the first Tables, and with Scale and Compasses, that you may see the near agreement betwixt them.

CHAP. III. How to Calculate the Sun's true place.

THis Proposition is propounded, in the first place, because many others depend upon it; According to the Hypothesis of Ticho, it is to be sug∣gested, and there is ascribed to the Sun a Triple Motion; first, a Mo∣tion upon his own Centre, whereby he finisheth one Revolution in 26 Days.

2. A daily motion from the East into the West, whereby he causeth the Day and Night.

3. An opposite Motion from the West into the East, called the Annual Motion, whereby he runs once round in a Year through the whole Ecliptick, moving near a degree in a Day; and thereby causeth the several Seasons of the Year.

The two later Motions are fancied out unto us, by a Man turning a Crane-Wheel, or Grind-stone 365 times round, while a Worm struggling against, and contrary to that Motion, creeps once round the contrary way, but Obliquely and a slope; that is, from the further side of the Wheel, towards the hithermost; and by this Motion the Sun is supposed to describe the Ecliptick-Line, and continually to insist in this Course; the other Planets, except the Moon, moving round him, or following after him, like Birds flying in the Air, being subject to his Motions, and many of their own besides; many of which Motions are removed by the Copernican supposition of the Earths Mo∣tion, which is a subject of much controversy among the Learned; However, be it ei∣ther the one, or the other, the Propositions hereafter Resolved, vary not, by reason thereof: And so the Sun being supposed not to vary from under the Ecliptick in re∣spect of Latitude; the Proposition and Quaere, in effect is what Longitude he hath therein, from the nearest Equinoctial-Point, which may be found within a Degree for his Course in each day, from his Entrance into each Sign, from that day of the Month.

The Day of the Month the Sun enters into each Sign.

• ♈ Aries, March 10.
• ♉ Taurus, April 9.
• ♊ Gemini, May 11.
• ♋ Cancer, June 12.
• ♌ Leo, July 13.
• ♍ Virgo, August 13.
• ♎ Libra, September 13.
• ♏ Scorpio, October 13.
• ♐ Sagitarius, November 11.
• ♑ Capricornus, Decem. 11.
• ♒ Aquarius, January 9.
• ♓ Pisces, February 8.

1. If the number of the Day of the Month given, exceed the number of that Day, in which the Sun enters into any Sign; Substract the lesser Number from the greater, and the Remainer is the Degree of the Sign, that the Sun possesseth.

PROBL. I.

Example.

ON the 12 of May I would find the Suns true place by the former Rules; The Sun enters Gemini May 11; which Substract from 12, the Remainer is 1, which shews the Sun to be in 1 deg. of Gemini the third Sign; that is, 61 degr. from the next Equinoctial-Point.

2. Example.

Let it be required to know the Suns Place the 4th of November; on the 11 day of that Month the Sun enters Sagitarius, and the 13th day of September he enters Libra: betwixt the 13 of September, and the 4 of November is 52 days, and consequently 52 degr. from the Equinoctial-Point Libra; then 30 taken from 52, there remains 22 degr. the Suns place in Scorpio, which is the thing required.

But here is a nearer Rule yet than this, to find the Suns place exactly, and that is by Mr. Vincent Wing's Hypothesis, and Tables in Astronomia Britanica, how to Calculate his true place from Earth, the Rule is,

First, enter the Table of the Middle-Motion of the Sun, and write out the Epocha next going before the time given, under which set the Motion distinctly belonging to the Years, Months, and Days, and Hours, and Minutes, if any be; (only in the Bissextile or Leap-Years,) after February a Day is to be added to the number of Days given; then adding them all together, the sum will be the Middle-Motion of the Sun for the time given.

As for Example.

Suppose the time given be the 12 of May at Noon 1667, at which time the Suns place is required.

Time given.	Longitude ☉	Apog. ☉
	S	D	M	S	S	D	M	S
The Epocha 1661 years In∣cluding 6 May. Days 12.	9	20	24	49	3	6	45	5
11	29	33	07	0	0	6	10
3	28	16	39				20
0	11	49	40				2
The Suns Mean-motion, or Longitude.	2	00	04	15	3	6	51	37

2. Substract the Apog. of the Sun from his Mean-Longitude, and the Remain will be his Mean Anomaly.

Example.	S	D	M	S
The Suns Mean-Longitude is	2	00	04	15
The Apogeum Substracted	3	06	51	37
The Suns Anomaly.	10	23	12	38

With the Suns Anomaly enter the Table of his Equation with the Sign on the Head and the deg. descending on the left hand, if the Number thereof be under 6 Signs; but if it be more than 6 Signs, enter with the Sign in the bottom, and the degr. ascending on the right hand, and in the common-Angle you have the Equation answering thereunto; only you must, if need require, remember to take the Proportional part.

Example.

	S	D	M	S
In the Table of Equation answering to 23 degr. is	0	01	11	51
The Suns Mean-Longitude, add	2	00	04	15
The Suns true Longitude.	2	01	16	06

Therefore the true place of the Sun is in 1 degr. 16 min. 6 seconds of Gemini.

Another Example.

In the Year 1583 March 14 at Noon, in the Meridian of Ʋraneburg in Denmark, thrice Noble Ticho-Brahe, most excellently observed the Suns true place in 3 deg. 17 min. 40 seconds of ♈. The time at London was 1583 March 13 day 23 h. 8 m.

[illustration] geometrical diagram

The Convex-Sphere, which resolves all the most useful Problems in Astronomy, by the Direction of 13 Problems following.

[illustration] geometrical diagram

The Concave-Sphere, which resolves 13 Problems, viz. and by them may be resolved, most of the useful Problems in Astronomy.

A Table of the Suns Mean Motion.

The Apoche, or Radius.

Year.	Longit. ☉	Apog. ☉
S	D	M	S	S	D	M	S
Ch. 1	9	07	59	51	2	08	20	03
1581	9	19	48	55	3	05	22	55
1601	9	19	57	54	3	50	43	28
1621	9	20	06	52	3	06	04	00
1641	9	20	15	51	3	06	24	33
1661	9	20	24	49	3	06	45	05
1681	9	20	33	48	3	07	05	38

☉ Mean Motion in Years above 20.

Year.	Longit. ☉	Apog. ☉
S	D	M	S	S	D	M	S
20	0	00	08	59	0	00	20	33
40	0	00	17	57	0	00	41	05
60	0	00	26	56	0	11	01	38
80	0	00	35	54	0	11	22	10
100	0	00	44	53	0	11	42	43
200	0	02	29	45	0	03	25	26
300	0	02	14	38	0	05	08	08
400	0	02	59	31	0	06	50	51
500	0	03	44	23	0	08	33	34
600	0	04	29	16	0	10	16	17
700	0	05	14	09	0	11	59	00
800	0	05	59	01	0	13	41	42
900	0	06	43	54	0	15	24	25
1000	0	07	28	47	0	17	07	08
2000	0	14	57	34	1	04	14	15
3000	0	22	26	21	1	21	21	22
4000	0	29	55	08	2	08	28	30
5000	1	07	33	55	2	25	35	37
6000	1	14	52	42	3	12	42	44

☉ Mean Motion in years under 20

Years.	Longit. ☉	Apog. ☉
S	D	M	S	M	S
1	11	29	45	40	1	02
2	11	29	31	20	2	04
3	11	29	16	59	3	05
L 4	00	00	01	48	4	07
5	11	29	47	28	5	08
6	11	29	33	07	6	10
7	11	29	10	47	7	12
L 8	00	00	03	35	8	13
9	11	29	49	15	9	14
10	11	29	34	55	10	16
11	11	29	20	35	11	18
L 12	00	00	05	23	12	20
13	11	29	51	03	13	21
14	11	29	36	43	14	23
15	11	29	22	23	15	25
L 16	00	00	07	11	16	26
17	11	29	52	51	17	28
18	11	29	38	31	18	29
19	11	29	24	10	19	31
L 20	00	00	08	59	20	33

☉ Mean Motion in Months.

	Longit. ☉	Apog. ☉
S	D	M	S	M	S
Janu.	00	00	00	00	00	00
Febr.	01	00	33	18	00	05
Marc.	01	28	09	11	00	10
April	02	28	42	30	00	15
May	03	28	16	39	00	20
June	04	28	49	58	00	25
July	05	28	24	07	00	31
Aug.	06	28	57	25	00	36
Sept.	07	29	30	44	00	42
Octo.	08	29	04	54	00	47
Nov.	09	29	38	12	00	53
Dec.	10	29	12	22	00	59

The Calculation. Apog. ☉

	Long. ☉
S	D	M	S	S	D	M	S
The ☉ Apocha.
☞ 1581.	9	19	48	55	3	5	22	55
Years added 2	11	29	31	20			2	4
March.	1	28	9	11				10
Days 13		12	48	48				2
Hours 23			56	40				0
Minutes 8				20				0
The Suns Mean Motion.	0	1	15	14	3	5	25	11
Apogeum Substract.	3	5	25	11
The Anomaly of ☉	8	25	50	03
The Equator added to 115′ 14′		2	2	51
The Suns true place, with Observation.	♈	3	18	5	Agreeing.

(3) Example the time given the 10 of April 1665 at Noon; and admit by the former Rules we have found the Suns Mean Motion 29 degr. 0 min. 30′ his Apogeum 3 s. 6 d. 49 m. 29 s. his Anomally 9 s. 22 d. 11′ 1″; first find a Proportional part, the Equat. answering to 22 s. 1 d. 52′ 22″

The Equator answering to 23 d.	1 51 29
their difference	59

Then I say, if 1 deg. or 60 min. give 53 seconds, what shall 11 of the Anomaly give? by the Rule of proportion, 〈 math 〉〈 math 〉

A Table of the Suns Mean Motion.

	Longit. ☉	Apog.	H	Long. ☉	M	Lon. ☉.
S	D	M	S		S	D	M	S	M	S
1	0	0	59	08	0	0	1	0	2	24	31	1	17
2	0	01	58	17	0	0	2	0	4	56	32	1	19
3	0	02	57	25	0	0	3	0	7	24	33	1	21
4	0	03	56	33	0	1	4	0	9	51	34	1	24
5	0	04	55	42	0	1	5	0	12	19	35	1	26
6	0	05	54	50	0	1	6	0	14	47	36	1	29
7	0	06	53	58	0	1	7	0	17	15	37	1	31
8	0	07	53	07	0	1	8	0	19	43	38	1	34
9	0	08	52	15	0	1	9	0	22	11	39	1	36
10	0	09	51	23	0	2	10	0	24	38	40	1	39
11	0	10	50	31	0	2	11	0	27	6	41	1	41
12	0	11	49	40	0	2	12	0	29	34	42	1	43
13	0	12	48	49	0	2	13	0	32	2	43	1	46
14	0	13	47	57	0	2	14	0	34	30	44	1	48
15	0	14	47	05	0	2	15	0	36	58	45	1	51
16	0	15	46	13	0	3	16	0	39	25	46	1	53
17	0	16	45	22	0	3	17	0	41	53	47	1	56
18	0	17	44	30	0	3	18	0	44	21	48	1	58
19	0	18	43	38	0	3	19	0	46	40	49	2	01
20	0	19	42	47	0	3	20	0	49	17	50	2	03
21	0	20	41	55	0	3	21	0	51	45	51	2	06
22	0	21	41	03	0	4	22	0	54	13	52	2	08
23	0	22	41	12	0	4	23	0	56	40	53	2	11
24	0	23	40	20	0	4	24	0	59	8	54	2	13
25	0	24	39	28	0	4	25	1	01	34	55	2	18
26	0	25	38	37	0	4	26	1	04	04	56	2	18
27	0	26	37	45	0	4	27	1	06	32	57	2	20
28	0	27	36	53	0	5	28	1	09	00	58	2	23
29	0	28	35	02	0	5	29	1	11	27	59	2	25
30	0	29	35	10	0	5	30	1	13	55	60	2	28
31	0	00	34	18	0	5	M	m	se.	th.	sec	sec	th.

A Table of the Suns Equation.

	Sig. 0 AE Sub.	Sig. 1 AE Sub.	Sig. 2 AE Sub.	Sig. 3 AE Sub.	Sig. 4 AE Sub.	Sig. 5 AE Sub.
	D	M	S	D	M	S	D	M	S	D	M	S	D	M	S	D	M	S
0	0	0	0	0	59	32	1	44	28	2	2	54	1	48	23	1	3	26	30
1	0	2	5	1	1	21	1	45	34	2	2	56	1	47	20	1	1	32	29
2	0	4	9	1	3	10	1	46	38	2	2	56	1	46	15	0	59	37	28
3	0	6	12	1	4	57	1	47	41	2	2	55	1	45	9	0	57	40	27
4	0	8	16	1	6	42	1	48	40	2	2	52	1	44	1	0	55	42	26
5	0	10	19	1	8	26	1	49	38	2	2	47	1	42	50	0	53	43	25
6	0	12	22	1	10	9	1	50	34	2	2	39	1	41	37	0	51	43	24
7	0	14	25	1	11	51	1	51	29	2	2	29	1	40	22	0	49	42	23
8	0	16	28	1	13	32	1	52	22	2	2	17	1	39	4	0	45	39	22
9	0	18	30	1	15	11	1	53	13	2	2	2	1	37	45	0	45	35	21
10	0	20	32	1	16	49	1	54	1	2	1	46	1	36	24	0	43	31	20
11	0	22	34	1	18	26	1	54	43	2	1	29	1	35	3	0	41	26	19
12	0	24	37	1	20	02	1	55	31	2	1	7	1	32	30	0	39	20	18
13	0	26	39	1	21	36	1	56	14	2	0	44	1	32	2	0	37	14	17
14	0	28	41	1	23	9	1	56	55	2	0	18	1	30	44	0	••5	••	16
15	0	30	42	1	24	41	1	57	34	1	59	49	1	29	13	0	3••	••8	15
16	0	32	43	1	26	11	1	58	10	1	59	19	1	27	41	0	3	50	14
17	0	34	44	1	27	40	1	58	44	1	58	47	1	26	7	0	2••	••1	13
18	0	36	43	1	29	8	1	59	16	1	58	12	1	24	32	0	26	••1	12
19	0	38	41	1	30	34	1	59	46	1	57	35	1	22	56	0	24	••	11
20	0	40	38	1	31	58	2	0	14	1	56	56	1	21	18	0	22	1••	10
21	0	32	35	1	33	20	2	0	40	1	56	14	1	19	38	0	19	5••
22	0	34	81	1	34	41	2	1	04	1	55	30	1	17	56	0	17	4••
23	0	46	27	1	36	0	2	1	26	1	54	44	1	16	12	0	15	36
24	0	48	22	1	37	17	2	1	46	1	53	56	1	14	26	0	13	33
25	0	50	16	1	38	33	2	2	3	1	53	5	1	12	40	0	11	10
26	0	52	09	1	39	48	2	2	18	1	52	12	1	10	52	0	8	57
27	0	54	1	1	41	1	2	2	30	1	51	18	1	9	2	0	6	43
28	0	55	52	1	42	12	2	2	40	1	50	21	1	7	11	0	4	29
29	0	57	42	1	43	21	2	2	48	1	49	23	1	5	18	0	2	15
30	0	59	42	1	44	28	2	2	54	1	48	23	1	3	36	0	0	0
	Add Sig. 11	Add Sig. 10	Add Sig. 9	Add Sig. 8	Add Sig. 7	Add Sig. 6

Place this Table between Pag. 106 and 107.

Multiply 53 by 11, the Product is 583, which Divide by 60, the Quotient will be 9″ 45/60; and because the Equation decreases I Substract it from the Equa. answering 22 degr. which is 1 d. 52″ - 12″ for the true Equation desired, which according to the Title, being added to the Suns Mean-Longitude, giveth the true place of the Sun re∣quired.

Example.

	S	D	′	″
The Suns Mean-Longitude.	0	29	00	30
The Equation added.		1	52	12
The Suns true Longitude.	1	00	52	42

D. ″ ″

Therefore the Suns true place is in 0 52:42 of Taurus; these Examples are suffi∣cient for Direction, to find the Suns true place at any time.

PROBL. II. The Suns Distance from the next Equinoctial-point; and his greatest De∣clination being given, to find the Declination of any Point required.

VVIth your Compasses take the Chord of 60 degr. upon the Centre C, describe the Circle HZON, and draw the Diameter HCO, which represents the Horizon, and at Right-Angles, or perpendicular thereunto, draw ZCN, the Vertical Azimuth of East and West, and take the Latitude of the Place, as in this Example, is 51 d. 28 m. and prick it from O to N, and from H to S, and draw the Axis or Meridian of the Hour of Six NCS; then prick from Z to AE, and from N to Q 51 degr. 28 min, and draw the Equinoctial-Line AECQ, then the Suns place being given, take 23 deg. 31 m. and prick from AE to ♋, and from Q to P, and draw the prickt Line ♋ CP; then take the Suns Distance from the next Equin.-Point, which in this Example shall be 61 deg. 18 m. out of the Line of Signs, and prick it from C to ☉, and through ☉ ♁ draw a Parallel-Line to the Equinoctial, as TD, and it shall be a Parallel of Decli∣nation, and where it cuts the outward Meridian, as at T; apply the Distance AET to the Line of Chords, and you have the Declination 20 degr. 30 min. which was required; Or you may take the nearest Distance from ☉ to the Equator, and apply it to the Line of Signs, and that will give you the Declination 20 degr. 30 min. as before; and if through ☉ ♁ you draw a Line Parallel to the Horizon HO, as ef, it is a Parallel of the Suns Altitude, and so have you the Sphere Orthographically in Right-Lines in the Convex-Sphere; and if you follow the directions of the use of Tangents, and half Tan∣gents in the 12 Chap. of the fourth Book of the Description of the Globe in Plano, you have the Sphere projected in Plain and Circular Lines, and fitted for the use of divers Questions; the Direction in both Spheres by the Letters signify the same thing; but observe what you are directed by Signs in the Convex-Sphere, is likewise to be done by ½ Tangents in the Concave-Sphere.

By the Tables in the Right-Angled Triangle CK O; we have given, first the Hy∣pothenase C ☉ 61 degrees 18′; secondly, the Angle KC ☉ 23 degr. 31′, hence to find K ☉, the Rule is, as the Radius is in proportion 10

to the Sign of the Suns greatest Decl. 23 d. 31′ KC ☉	960099
So is the Sign of the Suns distance from the next Equinoctial-Point 61 degrees 18 min. C ☉	994307
to the Sign of his Declination required 20 degrees, 30′ K ♁	954406

Or extend the Compasses in the Line of Artificial Signs from 90 degr. to 23 degr. 30 min. the same extent will give the distance from the Suns Place, to his Declination.

[illustration] geometrical diagram

The Sun being either in 1 deg. 18 min. of Gemini, or 29 deg. 42 min. of Capricorn, or 1 degr. 18′ of Sagitarius, or 28 degr. 42 min. of Cancer; that is, 61 degr. 18 min. from the next Equinoctial-Point; the Declination will be found to be 20 degrees 30 minutes.

PROBL. III. Having the Suns greatest Declination, and his Distance from the next Equi∣noctial-Point; to find his Right-Ascention.

IN the Foregoing Scheme, having drawn the Parallel of the Suns Declination TD, passing through the Place at ✶ the extent S ♁, is the Sign of the Suns Right-Ascen∣tion from the next nearest Equinoctial-Point, to the Radius of the Parallel TD; and therefore place the extent ST from C to X, and upon X as a Centre with the extent S ♁, describe the Arch at k, a Ruler laid from the Centre just touching the extremity of that Arch, finds the Point N in the Limb of the Meridian or Quadrant, and the Arch ON, applyed to the Line of Chords, is 59 degr. 09 min. and so much is the Suns Right-Ascension in the first quarter of the Ecliptick.

In the Triangle CK ♁ we have given as before, (1.) the Angle of the Suns grea∣test Declination KC ♁ 23 degr. 31 min. (2.) the Longitude of the Sun from the next Equinoctial-Point Aries C ♁ 61 degr. 18 min. hence to find the Suns Rght-As∣scention, the Rule is,

As the Radius	10
to the Tangent of the distance 61 degr. 18 min. C ♁	1026162
So is the Co-Sign of the Suns greatest Decl. 23 deg. 31′ KC ♁	996234
to the Tangent of the Right-Ascention CK 59 deg. 9 m.	1022396

Or in the Concave-Sphere; if you draw the Meridian from N through ♁ to S, whose Centre will be found upon the Equator, it will cut the Equinoctial in K; mea∣sure the distance GK on the Line of half-Tangents, and you have 59 d. 09′, as before.

Or extend the Compasses from 90 d. to 66 d. 29, the same distance will reach from 61 deg. 18 m. to 59 deg. 9 min. which is the Suns Right-Ascention in 61 deg. 18 ♊.

But this you are to observe, that if the Right-Ascention sought, be in the second Qua∣drant ♋ ♌ ♍, then you are to take the Complement of the Arch found to 180 deg. if in the third Quadrant ♎ ♏ ♐, adde 180 deg. to the Arch found; but in the last Quadrant, Substract the Arch found from the whole Circle 360 degr. and you shall have the Right-Ascention desired.

Example 2.

The Sun in 28 degr. 42 min. of ♋, that is, 61 degr. 18 min. from the Equinoctial Point ♎; the Rule is as before. As the Radius is to the Co-Sign of the greatest Decli∣nation, so is the Tangent of the Suns distance from the next Equinoctial-Point 61 degr. 18 min. to the Tangent of 59 degr. 09 as before, which taken from 180, is 130 deg. 51 min. which is the Suns Right-Ascention in 28 degr. 42 min. of Cancer.

Example 3.

The Sun in 1 degr. 18 min. of ♐, that is, 61 degr. 18 min. from the next Equinoctial-Point ♎, the work is the same as before; therefore to the Arch found, I add 180 degr. a Semi-Circle; so 59 deg. 09 min. and 180, makes 239 deg. 09 min. the Right-Ascention of the Sun sought in 1 deg. 18 min. of Sagitarius ♐.

Example 4.

The Sun in 28 degr. 42 min. of Capricorn, 61 deg. 18 min. from the next Equi∣noctial Point ♈, the operation is the same with the former Example; wherefore Sub∣stract the Arch found 59 deg. 09 min. from the whole Circle 360 deg. and there will remain 200 deg. 51 min. which is the Suns Right-Ascention in 28 deg. 42 min. of ♑ Capricorn.

PROBL. IV. The Elevation of the Pole, and Declination of the Sun being given; to find the Ascentional-Difference.

THis is represented in the Figure by SL in the Parallel of Declination, and it is therefore to be reduced into the common Radius, therefore take the Radius of the Parallel ST, and prick it from C to X, as before; then take the extent SL, and setting one Foot upon X, with the other draw the part of an Arch at a, lay a Ruler from C, that it may just touch the outside thereof, and it cuts the Circle in d, and take the Chord or Extent Hd; and you will find it 28 deg. 0 min. which being con∣verted into Time, is an Hour 52 min. and so much doth the Sun Rise before, and Set after Six in Summer; but so much doth he Rise after, and Set before Six in Winter, when he hath the same Declination South.

In the Right-Angled Spherical Triangle SLC are known. 1. SCL the Com∣plement of the Poles Elevation 38 deg. 32 min. 2. The Suns Declination 20 deg. 30 min. hence to find the Ascentional Difference SL.

As the Radius 90 deg. is in Proportion	10
to the Tangent of the Latit. 51 deg. 28 min. SCL	1009887
So is the Tangent of the Suns Declination 20 deg. 30′ SC	957273
to the Sign of SL the Ascentional-difference 28 d. 00 m.	967160

Extend the Compasses on the Artificial-Lines of Signs and Tangents, and you will find it, as before; or if you take the distance NS, and prick it from S to K, and lay the Ruler from Cover L, it will cut the Arch of the Meridian in d; then Measure the distance Nd on the Line of Chords, and it will be 28 deg. 00 min. as before found, that is one Hour 52 min.

PROBL. V. The Suns Right-Ascention, and his Ascentional-difference being given; to find his Oblique-Ascention, and Descention.

TO perform this, you must observe these 2 following Rules. 1. If the Suns Decli∣nation be North, you must Substract the Ascentional-difference from the Right-Ascention, and the Remain will be the Oblique-Ascention; but if you add them toge∣ther, the sum will be the Oblique Descention. 2. If his Declination be South, add the Ascentional-difference, and the Right Ascention together, the sum will be the Oblique-Ascention; but if you make Substraction, the Remainer will be the Oblique-Descention.

Admit the Sun is in the 1 deg. 18 min. of Gemini by the second Problem, his Right-Ascention is 59 deg. 09 min. and his Ascentional-difference by the 4 Problem, is 28 deg. 0 min. therefore according to the first Rule, because his Declination is North, the difference thereof 31 deg. 09 min. is the Suns Oblique-Ascention, and the sum of them 87 deg. 09 min. his Oblique-Descention.

PROBL. VI. To find the time of Sun-Rising, and Setting, with the length of the Day and Night.

YOu must find the Ascentoinal-difference by the 4 Problem, which converted into Time, allowing 4 min. of an hour for every degr. and 4″ seconds for every min. and the sum of Hours and Minutes, is his difference of Rising and Setting before or after the hour of Six, which was found before to be 28 deg. or 1 hour 52 min.

Therefore when the Sun is in Northern Signs, add the sum to Six, and the Total is the Semi-diurnal Arch, as in this Example, is 7 hours 5′2, or time of Sun-setting, and Substract it from Six, and the Remain is 4 h. 8′. m the time of Sun-Rising; double 7 ho. 5′2 m, it is 15 ho. 4′4, the length of the Day; Substract it from 24 ho. 00 m. and the Remain is 8 ho. 16 m. the length of the Night the 12 of May, in Latitude 51 deg. 28 min. at Bristol.

But if the Sun is in Southern Signs, make Substraction, as in this Example; the Sun having 20 deg. 30 min. South Declination, or in 1 deg. 18 min. ♐; Substract 1 ho. 5′2 from 6, the Remain is 4 ho. 08 m. for the time of Sun-Setting, double it, and it is 8 ho. 16 m. the length of the day; add 1 ho. 52 m. to 6, the sum is 7 ho. 52 m. is the time of Sun-Rising; double it, it is 15 ho. 44 m. the length of the Night, in Latitude 51 deg. 28 m. in 1 deg. 18 m. of Sagitarius.

PROBL. VII. The Elevation of the Pole, and Declination of the Sun being given; to find his Amplitude.

MEasure the extent CL with the Compasses in the Line of Signs, and it will reach to 34 degr. 40 min. and so much doth the Sun Rise and Set to the Northward of the East and West in the Latitude of Bristol, when his Declination is 20 deg. 30 min. North; but he Riseth and Setteth so much to the Southward of the East and West, When his Declination is so much South.

Now on the Concave-Sphere, the extent CL on the Horizon, applyed to the Line of half-Tangents, is 34 deg. 40 min. the Amplitude, as before.

If the Suns Parallel of Declination doth not meet with the Horizontal-Line HO, as in Regions far North, the Sun doth not Rise nor Set.

In the Right. Angled Spherical-Traingle LOC of the 4 Problem, having the Angle LCO, the Complement of the Latitude 38 degr. 32 min. and LO the Suns Decli∣nation 20 degr. 30 min. in 1 d. 18 m. ♊ his Amplitude by Calculation may be found thus.

The Artificial Lines by this Rule answers the same.	As the Co-Sign of the Latitude 51 deg. 28 min. LCO	979446
is to the Radius 90 degr.	10
So is the Sign of the Declination 20 degr. 30′ LO	954432
to the Sign of the Amplitude CL 34 deg. 40 min.	974986

This Rule is the common Rule Mariners make use of for the finding of the Variation of the Compass at Sea, by comparing the Coast, or bearing of the Sun, observed by an Amplitude or Azimuth-Compass at the Suns Rising or Setting, and by his bearing, found by these Rules beforegoing, the difference sheweth the Variation.

As for Example.

Admit you observed the Suns Amplitude of Rising or Setting by your Compass in the first Chap. and fifth Book of the Art of Surveying described.

And by the Compass found, the Magnetical Amplitude	45 d. 55′ compl. 44 d. 05′ N.
By the Rules beforegoing find the true Amplitude,	34 d. 40 compl. 55 d. 20′ N.
Substract the less out of the greater, the difference	11 d. 15 m. Variation.

And by reason the Magnetical Amplitude is more than the true Amplitude; there∣fore the Variation is 11 degr. 15 min. which is one Point Westerly; and if you are bound to a place that bear North of you, you must Sail upon the North by West Point; or if you bare West, you must Sail W and by S; and if South, the Course must be South and by East; or if you bear East, then the Course must be East and by North, to make good a North, or West, or South, or East Course; and so of all the rest of the Points you must allow in like manner.

2. But admit the Magnetical Amplitude observed by the Compass, were but 23 deg. 25 min. and the true Amplitude by the former Rules found to be 34 degr. 40 min. the upper Substract from the lower, the difference is 11 degr. 15 min. and by reason the Magnetical Amplitude is less than the true Amplitude, and the difference 11 degr. 15 m. which is one Point Variation Easterly; and so the North Point is the N by E Point, and NE is the NE by E, and E is E by S, and South is S by W, and W is W by N Point of your Sailing Compass, when you have such a Variation, and the Complement of the Amplitude is the Suns Azimuth from the North or South part of the Meridian, according as your Declination is. And this is sufficient for an Example to find the Vari∣ation of the Compass in any place or time.

As likewise by his Oblique-Ascension, and Ascensional difference; or by the Suns being East and West by the following Rules, or by the Suns Azimuth at the hour of Six; as likewise his Azimuth at any other time or place observed, as shall be shewn for the help and benefit of young Mariners.

PROBL. VIII. Having the Latitude of the Place, and the Suns Declination, to find the time when the Sun cometh to be due East and West.

IN the Parallel of Declination, the hour from Six is represented by SM; with that extent upon the Point X draw the Arch b the Ruler laid from C to the outward edge of the said Arch, cuts the Circle at (e,) the distance O e applyed to the Line of Chords sheweth 17 deg. 20 min. it converted into Time is 1 h. 9 m. 20 sec. and so much after Six in the Morning, and before Six in the Afternoon, will the Sun be due East and West; by the Concave-Sphere, if you lay a Ruler from A over M, it cuts the Limb in (e,) measure N e on the Line of Chords, and it is the same 17 deg. 20 min. the Rule by Artificial Signs and Tangents holds as by Calculation. viz.

Suppose the Latitude 51 degrees 28 min. and Declination North 20 degrees 30 min. therefore in the Right-Angled Spherical Triangle ZNM are given (1) Z N the Com∣plement of the Latitude 38 degr. 30 min. (2) N M the Complement of the Sum Decli∣nation 69 degr. 30 min. Then I say.

As the Radius 90 is in proportion	10
To the Co-Tangent of the Declination 69 degr. 30 min. NM	957273
so is the Tangent of ZN compl. of Latitude 38 degr. 32 min.	990112
is to the Co-Sign of ZNM 17 deg. 20 min. which Reduced is 1 h. 9 m. of time, as before.	947385

Which 1 h. 9 m. added to 6 h. is 7 h. 9 m. the moment in the Morning, the Sun will be due East; and if you Substract 1 h. 9 m. 20 sec. from 6 h. 00 m. and the Remain will be 4 h. 50 m. 40 sec. the moment in the Afternoon the Sun will be due West.

PROBL. IX. The Elevation of the Pole, and the Declination of the Sun being given; to find the Suns Altitude when he is due East and West.

MEasure the extent CM on the Vertical-Circle, and apply it to the Line of Signs, and it will reach to 26 degr. 36 min. or the same distance taken of the Concave-Sphere, and applyed to the Line of ½ Tangents, shews the same number, and so much is his Altitude sought in Summer; but when he hath the like Declination South, then so much is his Depression under the Horizon in Winter, when he is East and West; if the Suns Parallel of Declination TM doth not meet with the prime Vertical Circle CZ, the Sun cometh not to the East and West, as it happeneth many times in small Latitudes, or Countreys betwixt the Tropicks.

In the former Diagram, the Suns Altitude when he is due East and West, is shewed by the Arch CM, wherefore in the Triangle CVM we have given, (1) the Suns De∣clination VM 20 degr. 30 min. (2) the Angle of the Poles Elevation MCV 51 deg. 28 min. to find his Altitude CM; I say,

This Rule will hold by the Artificial Lines, of Signs and Tangents.	As the Sign of the Angle of Latit. 51 d. 28 m. UCM	989334
is to the Sign of the Declin. 20 degr. 30 min. UM	954432
So is the Radius 90 degr.	10
to the Sign of the Altitude 26 deg. 37 min. CM	965098

PROBL. X. The Elevation of the Pole, and Declination of the Sun being given, to find the Suns Altitude at the Hour of Six.

TAke the nearest distance from S to the Horizon CL, and apply it to the Line of Signs, sheweth the Altitude to be 15 degr. 54 min. or the same taken of the Con∣cave-Sphere, and measured on the Line of ½ Tangents, sheweth the same; and so much is his Depression under the Horizon at Six, when he hath South-Declination 20 degr. 30 min.

In the Concave-Sphere, you may see all the Triangles plain, and we have known in this Triangle ZSN, (1.) The Complement of the Latitude ZN 38 degr. 32 m. (2.) the Complement of the Suns Declination NS 69 degr. 30 min. to find the Hypotenase ZS; therefore I say,

As the Radius 90 degr. is in Proportion	10
To the Co-Sign of 69 degr. 30 min. NS	954432
so is the Co-Sign of 38 degr. 32 min. ZN	989334
To the Sign of the Altitude 74 degr. 6 min. S 00	943766

Whose Sign is 15 degr. 54 min. is the Suns Altitude at the hour of Six, when he is 1 degr. 18 m. of ♐ in Latitude 51 deg, 28 m. Extend the Compasses from 90 degr. to 20 d. 30′, the same extent will reach from the Latitude 51 deg. 28 min. to 15 deg. 54 m. as before.

PROBL. XI. Having the Latitude of the Place, and the Declination of the Sun given; to find the Suns Azimuth at the Hour of Six.

THis is represented in the Convex-Sphere by VZ in the Parallel of Altitude of the Sun VSB; Prick VB from C to W, and with the distance VS draw the Arch upon W at h, and lay the Ruler just touching the said Arch, cuts the Circle in Y; the distance HY measured on the Chords, sheweth the Azimuth, or the distance G 00, on the Concave-Sphere, applyed only to the Line of ½ Tangents, shews the Azimuth to be 13 deg. 07 min. and so much is the Sun to the Northwards of the East and West of the hour of Six.

In the Right-Angled Spherical-Triangle ZNS of the general Diagram, we have known first, ZN, the Complement of the Latitude 38 degr. 32 min. (2.) NS the Complement of the Suns Declination 69 degr. 30 min. to find the Azimuth of the hour of Six, represented by the Angle NZS.

I say,

As the Radius 90 is in proportion	10
to the Compl. Sign of the Latitude 38 deg. 32′ ZN	979446
So is the Co-Tangent of NS 69 deg. 30 min.	957273
to the Co-Tangent of the Azimuth NZS 76 d. 53′	936719

Or extend the Compasses from the Co-Sign of the Latitude to the Radius; the same extent will reach from the Tangent of the Declination, to the Azimuth 76 deg. 53 min. as before; the Suns Azimuth from the North part of the Meridian in the Latitude of 51 degr. 28 min. and Declination 20 degr. 30 North, (13 degr. 07 min is from the West.)

PROBL. XII. Having the Latitude of the Place of the Suns Declination, and his distance from the Meridian being given, to find the Suns Altitude at any Time assigned.

BY this Case may be found the Suns Altitude on all hours, and the distance of Places, in the Arch of a great Gircle; for the Suns Altitude on all hours thereby is meant, that if the hour of the Day, the Declination and Latitude be given, the Suns Altitude proper to the hour, or his Depression may be found.

Take the Chord of 60 degr. and describe the Arch HTPOD, draw the Hori∣zontal-Line HCO, and from O to P prick of the Chord of the Latitude 51 degr. 28 min. and from P to T and D set of the Complement of the Suns greatest Declination, 66 degr. 29 min. and draw the Parallel of Declination TD, and the Axis CSP, or the Meridian of the hour of Six; then draw the Radius TC, which is the Ecliptick-Line, and take off the Line of Signs, and prick down,

15	1	from 6 before it, and after it.
30	degr. 2
45	for the 3
60	Hours of 4
75	5

Then take the nearest distance from 15 degr. to CS, the Meridian of the hour of Six, or Axis, and prick it from S to 5 and 7; and likewise take the nearest distance fro 30 to CS, and lay it from S to 8 and 4; and in like manner do with the rest, then will the nearest distance from 4 5 6 7 8 9 10 11 to the Horizontal-Line HCO be the Signs of their respected Altitude;

so the Altitude	4	In Summer will be	1 deg.	34 min.
	5	9	30
for the hours	6	18	12

[illustration] geometrical diagram

And so much is the Suns Depression under the Horizon at the hours of 8, 7, and 6 in Winter; as you may soon trie by the same Division on the Parallel of the Suns greatest Declination Southward DST

The Summer Alt. for the h. of	7	are	27:23
8	36:42
9	45:42
10	53:45
11	59:42

And the Winter Altitude for

the Hours of	9	are	5 d. 13
10	10:28
11	13:48

And so much is the Sun deprest under the Horizon in Summer at the hours of 3, 2, 1, from Mid-night, as you may soon find it by the nearest distance from 9, 10, 11, in the Line LD to the Line LO; if you apply that distance to the Line of Signs, you may draw the Parallel of Altitude through each hour, as the Example is through 9 ho. e f, and measure He, or Of on the Line of Chords, and it is 45 degr. 42 min. the Suns Altitude at 9 a clock the 11 of June.

2. But admit the Latitude were 51 deg. 28 min. and his Declination 00 deg. 00 m. and suppose, for Example, seek his distance from the Meridian is 2 ho. 44 min. or 41 deg. 0′0 and the Sun upon the Equinoctial the 11 of March, and 13 of September.

Upon the Equinoctial at K is the Sun represented the 11 of March, or 13 of Septem∣ber, and distance from the Meridian 2 ho. 44, or 41 degr. 00 min. If in the Convex-Sphere you lay S ♁, from C to K the Right-Ascension 59 deg. 9 min. and through K you draw a Parallel to the Horizon, as dM, it will cut the Meridian in d, and mea∣sure the distance Hd on the Line of Chords, and it is 28 degr. 03 min. the Suns Alti∣tude required, or the nearest distance from K to the Horizon-Line HCO applyed to the Line of Signs, would have shewed the same. In the Concave-Sphere, if you take the distance CK, and draw a small Arch at r; and take the nearest distance to the Azi∣muth-Line of East and West ZC from K, and with that distance turn the other Foot over, and cross the Arch at (r,) and through K and r draw a Circle of Altitude, it will cut the Limb in E and F measure HE, or OF on the Line of Chords, sheweth the Altitude 28 degr. 03 min. as before.

In the Concave-Sphere, you may see the Triangle ZAEK; and we have given first AEZ the distance from the Zenith to the Equator, equal to the Latitude of the place, 51 degr. 28 min. Secondly, AEK the Suns distance from the Meridian 41 deg. to find the Suns Altitude from K to the Horizon.

I say, as the Radius 90 deg.	10
is to the Co-Sign of the Suns distance from the Merid. 42 d. 0′ AEK	987777
So is the Co-Sign of the distance from the Equ. to the Zen. 51 d. 28 m. AEZ	979446
to the Co Sign of ZK 61 degr. 53 min.	967223

Whose Complement is the Altitude Kp 28 deg. 03 min. which was required; you may in your practice draw a particular Figure for the Question, and shew every Tri∣angle by it self, by the Line of Chords, and half Tangents.

Secondly, when the Sun is in North Signs, ♈ ♉ ♊ ♋ ♌ ♍.

Let it be required to find the Suns Altitude at 10 a clock and 2 m. past before Noon, when the Sun is in entrance of Gemini, in Latitude 51 degr. 28 min.

First, By the Convex-Sphere, the nearest distance taken from the Suns place, to the Horizon HC, applyed to the Line of Signs, sheweth 51 degr. 12. min. the Suns Alti∣tude required; Or to find his distance from the Meridian, take ST, and prick it from C to X; then with the distance S ♁ on X as a Centre, draw the touch of an Arch at K, a Ruler laid over the Centre, over the outward edge of the Arch, cuts the Arch of the outward Meridian in (n;) then measure On on the Line of Chords, and it is 60 deg. 2 min. S ♁, whose Complement is 29 deg. 58 min. the Suns distance from the Me∣ridian OT; Wherefore in the Triangle NZ ♁, we have known, (1,) ZN the Complement of the Latitude 38 deg. 32 min. (2,) N ♁, the Complement of the Suns Declination 69 degr. 30 min. (3,) the comprehended Angle ZN ♁, the distance of the Sun from the Meridian 29 deg. 58 min. to find Z ♁, and hereby the Suns Altitude 92 ♁, I say,

As the Radius 90	10
is to the Co Tangent of the Latitude 51 deg. 28 min. NZ	990112
So is the Co-Sign of the Angle from the Meridian 29 deg. 58 m. ZN ♁	993767
to the Tangent of the fourth Ark 34 deg. 36 min. N q	983879

From the Complement of the Suns Declination N ✶ 69 deg. 30 min. Substract N q 34 deg. 36 min. there remains 34 deg. 54 min.

As the Co-Sign of the fourth Ark. 34 deg. 36 min. N q	991547
is to the Compl. Sign of the Latitude 38 deg. 32 min. ZN	989334
So is the Co-Sign of the fifth Ark 34 deg. 54 min. q ♁	991389
	1980723
to the Co-Sign of Z ♁ 38 deg. 48 min.	989176

Whose Sign is 51 deg. 12 min. p ♁ the Suns Altitude above the Horizon at 10 a clock, and 02 m past, and 1 a clock 58 min. past in the Afternoon, when he is in 1 deg. 18 m. of ♊ in Latitude of 51 deg. 28 min. North.

Now if you follow the Rules before-going, you may find the Suns Altitude by the Line of Chords, and half-Tangents by this Figure to be the same.

[illustration] geometrical diagram

Suppose the Sun in the Southern Signs ♎ ♏ ♐ ♑ ♒ ♓ in the opposite Point to the former, having South-Declination 20 degr. 30 min. and be also distant from the Meri∣dian 29 d. 58 m. take the Declination 20 deg. 30 min. and prick it from AE to B, and Q to R in both the Spheres, and draw the straight Line in the Convex-Sphere BMR, and take from the Suns place in the opposite Sign in the Parallel of Declination, his distance from the Meridian YR, and prick it on the other side of the Parallel from B to ☉; and the nearest distance to the Horizon-Line HC, applyed to the Line of Signs, shews the Altitude to be 13 d. 23 m. and in the Concave-Sphere take of the Line of half-Tangents, the Declination 20 degr. 30 min. and lay it from the Centre C to M, and the Axis CS continued; take the Complement of the Declination of the Line of Se∣cants, and place it from C on the continued Line or Axis, and that will be the Centre of the Parallel of Declination; or if you take the like Complement 69 degr. 30 min. of the Line of Tangents, and put it from M on the Axis, it will be the Centre of the Parallel of Declination; therefore draw it BMR, and it will cut the Meridian in ♁, the place where the Sun is.

Now, to find the Suns Altitude or Ark (Z ☉) or Z ♁; therefore to find how much it is; you must find the Pole of the Circle N ♁ Z, which is done after this manner.

Lay a Ruler from Z to h, and it will cut the Circle in ♃; then take 90 deg. and prick it from ♃ to ♂, then lay a Ruler over from Z to ♂, and it shall cut the Horizon in ♀, which Point ♀ is the Pole of the Circle Z ♄ n.

To measure the Ark Z ♁, you must lay a Ruler upon ♀ and ♁, which will cut the out∣ward Circle in the Point X, so shall XZ measured on the Line of Chords, give you the quantity of d. contained in the Arch XZ, which will be 76 d. 37 equal to the Comple∣ment of the Suns Altitude. I have been the larger in this precept, that it may be a Rule of Direction, to shew how the Ark of any great Circle of the Sphere; the sides of all Spherical Triangles being such, may be measured whatsoever, by his operation in the Concave-Sphere.

Observe the Figure we have given in the Oblique-Angled Triangle ZN ♁. 1. NZ as before, the complement of the Latitude 38 deg. 32 min. 2. N ♁ 110 deg. 30 min. the same distance from the North-Pole. 3. the Angle ZN ♁ 29 deg. 58′ to find the Altitude ♄ ♁.

As the Radius 90	10
is to the Tangent of NZ 38 deg. 32 min.	990112
So is the Co-Sign of the Angle, ZN ♁ 29 deg. 58′	993767
to the Tangent of Nq 34 deg. 36 min. 4 Ark, as before,	983879

From the Ark N ♁ 110 deg. 30 min. Substract the Arch Nq 34 deg. 36 min. and there remains 75 deg. 54 min.

As the Co-Sign of Nq 34 deg. 36 min.	991547
is to the Co-Sign of ZN 38 deg. 32 min.	989334
So is the Co-Sign of the fifth Ark q ♁ 75 deg. 54 min.	938670
	1928004
to the Co-Sign of Z ♁ 76 deg. 37 min.	936457

Now the Complement of Z ♁, is ♁ ♄ 13 deg. 23 min. which is the Suns Altitude required.

PROBL. XIII. The Suns Altitude, and his Distance from the Meridian, and Declination being given; to find his Azimuth.

DEmonstrate the Question by the Line of Signs and Chords, with 60 degr. draw the Semi-Circle, draw the Horizon-Line HO, draw the Parallel of Altitude ef, and of ☉ B Parallel to the Horizon; then draw the Axis CN, and the Equinoctial CAE at Right-Angles or 90 degrees from N, as NAE; then draw the Parallel of De∣clination TL and (BR) by the Line of Chords, and Signs; then take the extent IE, and set from C to P, and upon P with the extent I ♁, draw the Arch (00) a Ruler laid from C, just touching that Arch, cuts the Limb ({oil}) the Arch H {oil} measured on the Line of Chords, is 41 degr. 42 min. and so much is the Sun to the Southward of the East and West, the Complement is 48 deg. 18 min. and so much is the Suns Azimuth from the South part of the Meridian Westward. Now by the Concave-Sphere, if you find the Centre of the Aximuth-Circle on the Horizontal-Line at ☉, and draw the Circle from Z through ♁ N; measure the distance C ☿ on the Line of half-Tangents, and it will be the Suns Azimuth from the East and West, as before, 41 deg. 42 min. (Complement 48 deg. 18 min. from the South.)

[illustration] geometrical diagram

In the Oblique Spherical-Triangle Z N ♁, we have known. 1. Z ♁ the Com∣plement of the Suns Altitude 38 deg. 48 min. 2. The Angle ZN ♁ 29 deg. 58 m. the Suns distance from the Meridian. 3. The Complement of the Suns Declination N ♁ 69 deg. 30 m.

Now I work thus;

As the Complement-sign of the Altitude 38 degr. 48 min. Z♁	979699
is to the Sign of the Angle from the Merid. 29 deg. 58 m. ZN ♁	969859
So is the Compl. Sign of the Declination 69 deg. 30 N ♁	997158
	1967011
to the Sign of the Suns Aximuth NZ♁ 48 deg. 18 min.	987312

Now admit the Altitude were 13 degr. 23 min. his distance from the Meridian 29 deg. 58′,* 1.1 and his Declination South 20 deg. 30′.

By the foregoing Rules, you will find the Arch H ♑, measured on the Line of Chords 61 deg. 15 min. the Suns Azimuth from the East and West, whose Comple∣ment is 28 deg. 45 min. the Azimuth from the South part of the Meridian.

Or on the Concave-Sphere draw the Azimuth Circle Z ♁ ♄ N, and measure on the Horizontal-Line C ♄, applyed to the Line of half-Tangents, is 61 deg. 15′ as before, and the Azimuth 28 deg. 45 min. H ♄ from the South part of the Meridian; and 151 deg. 15 min. his Azimuth from the North part of the Meridian O ♄.

Observe the general Diagram of the Concave-Sphere, you have in the Oblique-An∣gled Spherical Triangle RN ♁. 1. Z ♁ the Complement of the Suns Altitude Z ♁ 76 deg. 37′. 2. The Angle ZN ♁ 29 deg. 58 min. his distance from the Meridian. 3. N ♁ 110 deg. 30′, the Complement of the Suns Declination. The Rule is to find the Azimuth,

As the Complement-Sign of the Altitude Z ♁ 76 deg. 37	998804
is to the Sign of the Angle from the Merid. ZN ♁ 29 d. 58 m.	969853
So is the Complement-Sign of the Declination N ♁ 69 d. 30 m.	997158
	1967011
to the Sign of the Suns Azimuth NZ ♁ 28 degr. 45 min.	968207

PROBL. XIV. The Poles Elevation, with the Suns Altitude and Declination given; to find the Suns Azimuth.

FIrst, draw a Diagram by the Rules beforegoing in the Convex-Sphere, that is, the Elevation of the Pole ON 51 deg. 28 m. the Suns parallel of Altitude ♑ V 13 d. 23 m. and the parallel of the Suns Declination 20 degr. 30 min. BR; then take ♑ V, the parallel of Altitude, and prick •• from C to 00; then take the distance V ♁, and on ☍ as a Centre, draw the Arch at (••) a Ruler laid over C, and the outside of (♂) shall cut the Arch in (n,) then measure the distance (HW,) and the Line of Chords 41 deg. 15 min. as before.

Or draw the Concave-Sphere, and the Axis or Elevation of the Pole ON; and likewise prick from H and O on both sides the Parallel of the Suns Altitude ♑ V 13 deg. 23 min. and take the same number of the Line of half-Tangents, and prick it on CZ, the Prime Vertical Line of East and West from C upwards to (v,) and draw the parallel of Altitude, ♑ v, whose Centre will be upon the Vertical CZ continued or found, by taking the Complement of the Altitude 76 deg. 37 min. of the Line of Secants, and prick it from C on the Vertical-Line continued, and that is the Centre; draw ½ as (♑ v) thence draw the Parallel of Declination 20 deg. 30 min. by the former direction from AE to B, and from Q to R, and take 20 deg. 30 min. of the Line of half-Tangents, and prick it from C on the Axis to (m,) and draw B ♁ M R the Parallel, whose Centre is found upon the Axis continued, as before, by the Complement Secant of the Decli∣nation 69 deg. 30 min.

Then where the Parallel of Altitude and Declination cross each other, which is at ♁, there is the Sun at that time; therefore draw the Azimuth-Circle, as before (12 Probl. directed,) from Z through ♁ to N, and it will cut the Horizon in ♄; then measure C ♄, and it is the Suns Azimuth from the East and West, which applyed to the Line of half-Tangents, shews 61 deg. 15 min. as before, whose Complement is 28 deg. 45 min. the Azimuth from the South; in like manner measure all Azimuths from the Prime Vertical on the Horizon.

By Calculation; First, consider the Declination of the Sun, whether it be towards the North or South, so have you his distance from the Poles; then add this distance, the Complement of his Altitude, and the Complement of your Latitude all three together, and from half the sum Substract the distance from the Pole or Complement of his De∣clination, and note the difference.

Look well and observe the general Diagram in the Oblique-angled Triangle.

Z ♁ N the Complement of the Suns Altitude is Z ♁ 76 deg 37 min. 2. The Complement of the Suns Declination is N ♁ 69 deg. 30 min. and the Complement of the Latitude ZN 38 deg. 32 min. which known, you may frame your operation, thus.

Decl. South 20 d. 30 m.	The distance from the Pole 110 deg. 30 min.
Suns Altit. 13 d. 23 m.	Complement, or Sign is	76:37, Z ♁ 998804
Latit. North 51 d. 28 m.	Complement, or Sign of	38:32, ZN 979446
37 deg. 18 m.	(4) Sign, or sum is	225:39, 1978250

all

The Sign of 67 deg. 11′, or half sum of (3) 112:49	996461
So is the Sign of 2 deg. 19 min. or the difference 2:19	860662
Add the Radius (2)	2000000
From this Sum Substract 1978250 the fourth Sign	2857123
Take the half with the Radius (it is the 7th. Sign)	1878873
The half doth give the Mean-proportional Sign 14 degr. 22 min.	939439

And the double of 14 degr. 22 min. is 28 degr. 44 min. the Azimuth of the Sun from the South part of the Meridian; and it taken from 180 degr. 00 min. leaves 151 degr. 15 min. the Suns Azimuth from the North part of the Meridian, as before.

An Example for North Declination answerable to the first in (13 Probl.)

By the former Rules, you may find the Azimuth in each Sphere by Instrument. The Suns Declination North 20 degr. 30 min. his Altitude 51 deg. 12 min. the Latitude 51 degr. 28 min.

Observe the Diagram in the Oblique-Triangle Z ♁ N the Complement of the Suns Altitude Z ♁ 38 deg. 48 min. the Complement of the Suns Declination, N ♁ 69 deg. 30 min. the Complement of the Poles Elevation ZN 38 deg. 32 min. which known, the operation may be thus framed.

Suns Decl. North 20 d. 30 m. Compl.	69 d. 30′	As the Radius
Altitude 51:12 Compl.	38:48:979699	is to the Co-Sign of Alt.
Latitude 51:28 Compl.	38:32:979446	so is the Co-Sign of Lat.
The Sum	146:50:1959145	to a fourth Sign 22 d. 59′
The 1/1 Sum.	73:25:998154	as 4 S. is to the S. of ½ sum
The difference	3:55:883445	so is the Sign of the differ.
Add the Base 69 deg. 30 N ♁ (2)	2000000
	3881599
The half of it	1922454	to a 7th. Sign 9 d. 39 m.
So is the Co-Sign 24 deg. 09 min.	961227

The double of it is 48 deg. 18 min. the Suns Azimuth from the South part of the Meridian, as before found, or 131 deg. 42 min. from the North part of the Meridian.

I have set down these two Examples thus particularly, to shew the agreement with the former two; this note, that generally in all Spherical-Triangles where three sides are known, and an Angle required, make that side which is opposite to the Angle required, to be the Base, and gather the Sum, the half Sum, and the difference, as before.

Having these means to find the Suns Azimuth, we may compare it with the Magne∣tical Azimuth, and so find the variation.

As for Example.

Admit the Magnetical Azimuth by the Needle, is	59 deg. 33′
And the Suns Azimuth found, as before, is	48 - 18
The Variation is — the difference West	11 deg. 15

So the Magnetical Azimuth being more than the true Azimuth by 11 deg. 15 min. which is one Point of the Compass; therefore it shews the Variation to be one Point, or 11 deg. 15 min. Westerly.

And suppose the Course by the Compass be East 8 Points from the North or South, or 90 degrees; and let the Variation be 11 deg. 15 min. to the Westward.

I demand the true Rumb;

Mr. Borough observed in 1580 d. 11 deg. 15 min. Variation Easterly in Line, House, Fields.	The Magnetical Rumb	90 d. 00 m.
Substract the Variation Westerly	11 d. 15
there remains the true Rumb NE	78 d. 45

Mr. Gunter 1662 found 6 deg. 15′

So that if the Variation be Westerly, you may concieve by looking upon the North-Point; by the Variation one Point, that it being Westerly, it is always accounted to the left hand, so the North-Point, is one Point to the right hand of his true place; and you must Sail N by W, to make good a North way, and W by S to be a good West, and S by E to be a direct South, and E by N to make good an East Course, which will make an Angle with the Meridian of 78 deg. 45 min.

2 Example.

But suppose the Magnetical Azimuth by the Needle had been	37 d. 03′
And the Suns Azimuth found, as before, to be	48-18
Substract the lesser out of the greater, the difference NE	11 d. 15

And in regard the Magnetical Azimuth is less than the true Azimuth by 11 deg. 15 min. therefore the difference and variation is Easterly one Point, which is 11 deg. 15 min. and consequently all the Points, stand 11 deg. 15 min. or one Point to the left hand out of their true Places; and therefore, to make good a North Course, you must Sail by your Compass N by E; and an East Course, Sail E by South, and South Sail S by W; and to make good a West Course, Sail W by N; and so it is to be understood of all other Courses or Points; for in this Example, the true Course makes an Angle with the Meridian of 48 deg. 18 min.

The Year 1666 at Bristol, in Rownam Meadows, My self and Mr. Phillip Stainard, and some other friends Masters of Ships, took with us a Quadrant described in the 16 Chapter of the Second Book of 20 Inches Semi-diameter, and one Needle, and one A∣zimuth Compass, described in the First of the Fifth Book, the Needle about 9 Inches long, the Chard 8 Inches; and in the Afternoon we made these Observations follow∣ing.

Declin. 23 d. 30′ dist.	66 d. 30
Altitude 44:20 Com.	45:40	As the Radius is in proportion—10
Latit. 51 d. 28 m. Com.	38:32	To the Co-Sign of the Latit. 38 d. 32′ 970446
The Sum	150 42	So is the Co-Sign of the Alt. 45 d. 40 985447
The half Sum	75 21	To the fourth Sign 26 d. 27 m. — 964287
The difference	8:51

Add the Radius to the Sum; Take the half is the Sign of half the Ark.	As the fourth Sign 26 degr. 27 min.	964893
is to the Sign of the half Sum 75 d. 21′	998564
So is the Sign of the difference 8 deg. 51′	918709 add
	1917273 sum
	to the seventh Sign 19 deg. 31 min. (add Rad.)	1952380
the half is the Mean proportional Sign of	35:19976190

☉ Obser. Altitude.	Magnetical Azimuth.	Suns true Azimuth.	Variation VVesterly.
Gr. M.	Gr. M.	Gr. M.	Gr. M.
44 20	72 00	70 38	01 22
39 30	80 00	78 24	01 36
31 50	90 00	88 26	1 34
37 42	95 00	93 36	1 24
23 20	103 00	101 23	1 23

Which doubled, is 70 deg. 38 min. the Suns Azimuth from the South part of the Meridian, or 54 degr. 41 min. the Complement of 35 deg. 19 min. doubled, is 109 deg. 22 min. the Suns Azimuth from the North part of the Meridian; and so of the rest, as they are set down in this Table, viz. from the South part of the Meridian.

PROBL. XV. To find the Altitude of the Sun by the Shadow of a Gnomon set Perpendicular to the Horizon by Scale and Compasses; as also by Calculation.

VVIth your Compasses on a piece of Board, describe the Circle ABCD, place it Horizontal with a Gnomon in the Centre O, cross it with 2 Diameters; then turn the Board, until the shadow be upon one of the Diameters, at the end of the shadow make a Mark, as here at E; lay down also the length of the Gnomon-Pin or Wire from the Centre on the other Diameter from O to F, draw a right-Line from E to F, as EFH; then with the Chord of 60 deg. sweep the Arch GH upon E as a Centre; apply the distonce GH the Arch to your Line of Chords, and that will give you the Altitude of the Sun required, as in this Example will be 52 deg. 53 min.

[illustration] geometrical diagram

As the parts of the shadow 28	144715
are to the parts of the Gnomon 37	156820
So is the Radius 90 deg.	1000000
to the Tangent of 52 deg. 53 min.	1012105

So the Pin or Gnomon OF being 37 parts, and th•• shadow OE 28, such equal parts, the Altitude will be found to be 52 degr. 53; or ••••e Gnomon being 28, and the shadow 37 parts, the Altitude will be IK 37 d. 07 m. or the shadow being 83, the Gnomon or Staff 100, the Tangent of the Angle will be 50 deg. 18 min. 20 the Alti∣tude of the upper edge of the Sun or Angle HEG; from which, taking the Semi-dia∣meter of the Sun 16 m. 27″, there remains 50 d, 1′ 59″ the true Altitude of the Centre of the Sun.

After this manner, if you observe the greatest Meridian-Altitude of the Sun the 11 of June, and 10 of December, you shall by the difference of them find the distance of the Tropicks; the greatest Declination of the Sun, and Elevation of the Equator, and La∣titude of the Place.

As for Example.

At London the greatest Meridian-Altitude of the Sun is 61 deg. 59″ 30″, and the least 14 deg. 56′ 30″.

The Suns greatest Meridian-Altitude taken June 11, is	61 d. 59:30″
The Suns least Meridian-Altitude taken December 10	14:56:30
The distance of the Tropicks, take the half of	47:03:00
And it is the Suns greatest Declination, Substracted from the Alt.	23:31:30
Leaves the Elevation of the Equator,	38:28:0
Whose Complement is the Latitude of the Place,	51:32:0

PROBL. XVI. Having the Latitude of the Place, the Suns Declination, and the Suns Alti∣tude; to find the Hour of the Day.

BY the Line of Chords and Signs, by the Convex-Sphere, set the extent ST, from C to X, and upon X as a Centre, with the extent S ♁, draw the Arch K, a Ruler laid from C just touching the Arch, finds the Point (n,) the Arch (on) measured on the Chords, sheweth 61 deg. 10 min. the Suns distance from the hour of Six, viz. 4 ho. and almost one min.

The Rule is by the Tables; Add the Complement of the Suns Altitude, and the Complement of the Suns Declination, or Distance of the Sun from the Pole, and the Complement of your Latitude, all three together, and from half the sum; Substract the Complement of the Altitude, and note the difference.

Thus in our Latitude of Bristol 51 deg. 28 min. the Declination of the Sun, being 20 deg. 30 min. Northward, and the Altitude 51 deg. 12′. I find the Sun to be 29 deg. 50 min. from the Meridian, as by this Example.

Altitude of the Sun 51 d. 12, the Compl. 38 d. 48, As the Radius 90	10
Declination North 20:30 the distance from the Pole 63 30. to the Sign of the Suns distance from the Pole.	997158
Latitude North 51:28 the Compl. is 38:32 so is the Sign Compl. of Lat.	979414
The sum of all three 146:50 to a fourth Sign	976572
The half Sum — 73:25 as the 4 Sign is to the Sign of half sum	998154
The difference— 34:37 so is the Sign of the difference	975441
	1973595
To a seaventh Sign add the Radius	1997023
Take the half, is the Sign of 14 deg. 55 min.	998511

The Mean Proportional between this seventh Sign, and the Sign of 90; that is, add the Radius to the seventh Sign, and take the half, and it will be the Sign of the Complement of half the Hour from the Meridian, which in this Example is found to be 14 deg. 55 min. the double of that is 29 deg. 50 min. which converted into Hours, doth give almost two Hours, it wants but 40 Seconds.

PROBL. XVII. Having the Azimuth of the Sun, the Altitude of the Sun, and the Declina∣tion; to find the Hour of the Day.

THus the Declination being 20 deg. 30′, the Altitude 51 deg. 12 min. the Azi∣muth from the South, found by the 14 Problem, to be 48 degr. 18 min. I might find the time to be 29 deg. 58′, that is, almost 2 Hours wanting 8 Seconds; so the difference is 32 Seconds; that is, by reason of the several Operations, which is near enough for the Mariners use.

As the Co-Sign of the Declination 69 deg. 30 min.	997158
is to the Sign of the Azimuth 48 deg. 18 min.	987311
So is the Co-Sign of the Altitude 38 deg. 48 min.	979699
to the Sign of the Hour 29 deg. 58 min.	1967010
	969852

PROBL. XVIII. How to find the Right-Ascension of a Star, and the Declination of a Star; having the Longitude and Latitude of that Star given.

PRoject the Sphere Geometrically; that is, draw the great Meridian with a Chord of 60 deg. you may draw the Horizontal-Line HO, and Vertical ZN; then set the Latitude 51 deg. 28′ from O to N, and draw ZAE, and from H to S, and from N to Q▪ and draw the Equinoctial-Line AECQ, then take the distance of the Pole of the Ecliptick, from the Pole of the World 23 deg. 31 min. and prick it from N to P, from AE to ♑, from S to A, and from Q to ♋; then draw the Ecliptick-Line ♑, C, ♋, on which you must put the Longitude or Distance of the Star, from the next Equinoctial-Point, as in this Example; The Star in the Mouth of the great Dog Sirius, his Longitude is found by the following Rules to be 9 deg. 32 min. of Cancer; and his Latitude is 39 deg. 30 min. South, a Star of the first Magnitude; take 9 deg. 32 min. out of 90 deg. the Remain is 80 deg. 28 min. the distance of the Star from the next Equinoctial Point C, prick that from C to K on the Ecliptick, and draw the Circle of Longitude P, KS, then prick the Latitude 39 deg. 30′ from ♑ to D, and from ♋ to F by the Chords; then take the same number of the half-Tangents, and prick it from C to M, and draw the Circle of Latitude of the Star parallel to the Ecliptick, as DMF, and where this Parallel cuts the Circle of Longitude, as at ✶ that is, the place of the Star; then draw the Meridian-Circle from the North-Pole through ✶ the Intersection to the South-Pole, and it cuts the Equinoctial in R; measure CR on the Line of half-Tangents, and it gives the Right-Ascension Complement to 180 deg. which is 82 deg. 21 min. the Complement is 97 deg. 39 min. the Right-Ascension desired.

Now to find the Declination of the Star; lay a Ruler over P and the Ecliptick, at K, and it will cut the Arch in L; take a Quadrant 90 deg. and prick it from L to P; lay a Ruler over p and P, and it will cut the Ecliptick in O; lay a Ruler over O, and ✶, and it cuts the Lamb in e; measure Qe on the Line of Chords, and it is 16 deg. 14 m. the Stars Declination required, By the Concave-Sphere; the Convex-Sphere, will not so conveniently shew the true Scituation and Place of the Stars as this; and there∣fore it is omitted.

[illustration] geometrical diagram

By Calculation.

The Stars have little or no alteration in their Latitude; but in their Longitude they move forward about 1 degr. 25 min. in a hundred years, which is 85′.

By Noble Ticho, his Tables of Longitude and Latitude of the Stars, rectified by himself, to the beginning of the year 1601.

The Latitude of the most bright Star Sirius in the Mouth of the great Dog, is 39 deg. 30 min. and his Longitude is 8 degr. 35 min. 30″ of ♋; I desire the Stars true Longitude, or to be rectified for this present year 1667.

You must work by the Rule of Proportion, thus, if 100 give 85 m. what shall 66 the difference in years betwixt 1601, and 1667 give? Multiply, and Divide, and the Quotient will be 56 m. 6″ added to the Longitude found in the Tables of the Stars in the second Book on the back-side the Nocturnal, which 8 deg. 35 min. 30″ makes 9 d. 31′ 36″ of Cancer, the Longitude of the Star Sirius this year 1667; and so work to find the Longitude of any other Star in any other year, past, or to come.

Take 9 deg. 32′ the Longitude of the Star, out of 90 deg. there remains 80 degr. 28′, his distance from the next Equinoctial Point; which being known, the First Rule is,

As the Radius 90	10
is to the Sign of the Stars Longitude from the next Equ. Point 80 d. 28	CK 999396
So is the Co-Tangent of the Stars Latitude 39 deg. 30′ A ✶	1008389
to the Tangent of the fourth Ark 50 deg. 06 min.	1007785

Compare this fourth Ark with the Arch of Distance betwixt the Poles of the Eclip∣tick, and the Poles of the World 23 deg. 31 min. if the Longitude and Latitude of the Star be alike, as in North Signs ♈ ♉ ♊ ♋ ♌ ♍, and the Latitude is on the North-side the Ecliptick; or if the Longitude be among the Southern Signs, as ♎ ♏ ♐ ♑ ♒ ♓, and the Latitude Southward; then shall the difference between the fourth Ark found, and the distance of the Poles 23 deg. 31′ be your fifth Ark.

But if the Longitude and Latitude shall be unlike, as it is in this Example; as the Lon∣gitude in a Northern Sign, and the Latitude South; or the Longitude in a Southern Sign, and the Latitude North; then Add this fourth Ark found, to the distance of both Poles 23 deg. 31 min. the sum of both shall be the fifth Ark.

Then the Rule is,

As the Sign of the fourth Ark 50 deg. 06′	988488
is to the Sign of the Fifth Ark 73 deg. 37′	998199
So is the Tan. of the Stars Long. from the next Equin. Point 80 deg. 28 m.	4077484
to the Stars Right-Ascension from the next Equin. Point 82 deg. 21 m.	2075683
82 d. 21 m. Substracted from 90 d. or 180 leaves 7 d. 39′ which added to 90 d. the sum is 79 d. 39′ the Right-Ascen. of Sirius required.	1087195

Then the Rule to find the Declination, is,

As the Co-Sign of the fourth Ark 50 deg. 06′	980716
is to the Co-Sign of the fifth Ark 73 deg. 37′	945034
So is the Sign of the Stars Latitude 39 deg. 30′	980351
	1925385
to the Sign of the Stars Decl. required 16 d. 14′	944669

You have the proof of the Work by the foregoing Geometrical Rules; or you may take this by Calculation, if there be no former errour, the Proportion will hold.

As the Co-Sign of the Latitude 39 deg. 30 min.	988740
is to the Co-Sign of the Right-Ascen. from the next Equ. Point 82 d. 21′	912424
So is the Co-Sign of the Declination 16 deg. 14 min.	998233
	1910657
to the Compl. Sign of Longit. from the next Equinoctial 80 d. 28 m. or Sign of the Longitude 9 deg. 32 min. as was at first given.	921917

By these Rules and directions work, to find the Right-Ascension, and Declination of any Star, which you desire to know.

PROBL. XIX. Having the Declination, and Right-Ascension of a Star; to find the Longi∣tude and Latitude thereof.

IN the former Diagram of the 18th Problem, you have the Right-Ascension of the Glistering-Star in the great Dog's Mouth called Sirius; CR 82 d. 21 m. take it off the Line of half-Tangents, and prick it from C to R; and you have also the Decli∣nation drawn Bye; then draw the Meridian-Circle from N, cutting the Point R, the Stars Right-Ascension from the next Equinoctial Point, and Parallel of Declination in ✶ to S the South-Pole; then take the distance of the Poles of the World, and the Poles of the Ecliptick 23 d. 31′, and prick from N to P, and from AE to ♑, and from S to A, and from Q to ♋, and draw ♑ C ♋ the Ecliptick-Line; then draw the Circle of Longitude through P, through the Intersection of the Parallel of Declination, and Meridian, which is the body of the Star to A, and it will cut the Ecliptick in K; mea∣sure CK on the Line of half-Tangents, and you have the Longitude of the Star from the next Equinoctial-Point 80 deg. 28 min.

And to find the Latitude; if you lay a Ruler over P and K, it will cut the Limb in ♄, prick 90 deg. from ♄ to ♃, and lay a Ruler over P and ♃, and it will cut the Eclip∣tick in the Point ♂; lay a Ruler over ♂ and ✶, and it will cut the Limb in F; apply the distance ♋ F, to the Line of Chords, and it will be 39 deg. 30′, the Latitude re∣quired.

In the Triangle ZRC, by Calculation, we have 1. the Angle ZCR, the distance of the Poles 23 degr. 31 min. 2. the side CR 82 degr. 21′, the Right-Ascension from the next Equinoctial-Point; then reason must guide you, as by these Rules, to find the Longiture and Latitude of a Star.

As the Radius 90 deg.	10
is to the Tang. of the Angle ZCR 23 d. 31 m. the Poles distance	963864
So is the Sign of Right-Ascension CR 82 d. 21′ from the next Point	999611
to the Tangent of ZR 23 deg. 20 min.	963475

Which 23 degr. 20′ add to the South-Declination 16 degr. 14 min. makes 39 deg. 34′, ZR; but if the Declination had been Northerly, you must have Substracted the Arch found out of it, and the Remain had been ZR; but if it had been more than the Declination, Substract it out of the Arch found, and the difference had been ZR; so with reason wave the Rule, as occasion requireth.

Then to find the Angle CZR, and the side CZ, the Rule is,

As the Sign of the fourth Ark ZR 23 deg. 20′	959778
is to the Sign of the Angle of distance of the Poles ZCR 23 d. 31′	960099
So is the Sign of Right-Ascen. from the next Equ. Point CR 82 d. 21 m.	999611
	1959710
to the Sign of CZR 86 deg. 49 min. the fifth Ark	999932

Then;

As the Sign of ZCR 23 deg. 31 min.	960099
is to the Co-Sign of RC 23 deg. 20 min.	959771
So is the Sign of CRZ 9 degr. Radius	10
to the Sign of CZ 83 degr. 03 min.	999679

Which Angle CZR 86 deg. 49′ is equal to the Angle ✶ ZK.

Then to find the Latitude of the Star,

As the Sign of ✶ KZ Radius 90 degr.	10
is to the Sign of 39 deg. 34′ the 4 Ark and Declination R ✶	980412
So is the Sign of ✶ ZK 86 deg. 49 min.	999932
to the Sign of the Latitude of the Star desired ✶ K 39 deg. 30 m.	980344

And lastly, to find the Arch ZK, and by it consequently the Longitude CK.

As the Tangent of the Angle ZK ✶ 86 deg. 49 min.	1125479
is to the Radius 90 deg.	10
So is the Tangent of Z ✶ 39 degr. 34′	991713
to the Sign of ZK 2 deg. 35′	866234

Therefore, if you Substract 2 degr. 35 min. out of the Arch CZ 83 deg. 03 min. the Remain is 80 degr. 28 min. the true Longitude of the Glistering Star Sirius in the Mouth of the great Dog, from the next Equinoctial-Point at the time given.

But if the Meridian-Circle had cut the Ecliptick at K, and the Circle of Longitude at Z; then in such cases, add the Arch found ZK to CZ, and the sum had been the true Longitude from the next Equinoctial-Point C; and so work with reason, to find the Longitude and Latitude of all other Stars, as by reason you did Substract ZK from CK; therefore the Sign was above a Quadrant, if you Substract his distance from the Vernal-Equinox 80 deg. 28 min. from 90 d. the Remain is 9 deg. 32′ of Cancer, the Sign and deg. the Star is in, as before.

PROBL. XX. Having the Meridian-Altitude of an unknown Star, and the distance there∣of from a known Star; to find the Longitude and Latitude of the unknown Star.

IN the 16th Chapter of the second Book of Harmonicon Coeleste, Mr. Vincent Wing hath this Example, and Observation, made by the Phoenix of Astronomy Ticho-Braghe in the year 1577, which we will borrow for an Example; it being a useful Rule for all Ingenious Navigators, for by it they may find the Longitude and Latitude, and consequently, by the foregoing Rules, the Right-Ascension, and Declination of those good Stars for their use, that are in the South Hemisphere, viz. as they have been named by the Portugals; the South-Triangle, which Constellation hath 5 Stars, one of the Eastermost corner, which comes last to the Meridian of the second Magnitude. The Crane, in which there is 13 Stars on the left Wing, and another on the right side the back of the second Magnitude. The Phoenix, 15 Stars, the Water-Serpent hath 15 Stars. The Dorado, or Gilt-head-Fish, situate in the very Pole of the Ecliptick; and in that Constellation is 4 Stars. The Chamelion, with the Flie, in which is 13 Stars; The Bird of Paradise, in which is 12 Stars; the Peacock, in which is 15 Stars; one in the head of the second Magnitude; the Naked Indian, in which is 12 Stars; and also the Bird Taican, or Brasilian Pye, in which Constellation is 7 Stars, two of them of the third Magnitude: Also two useful Stars for Navigators; in one Constellation, which are Noah's Dove, which containeth 11 Stars, of which there are 2 in the back of it, of the second Magnitude, which they call the Good Messengers, or Bringers of good News, and those in the right-Wing are consecrated to the appeased Deity; and those in the left to the retiring of the Waters, in the time of the Deluge; and they come to the Meri∣dian about half an hour before the great Dog; and by the Globes are about 21 degr. 30′ distant, from the nearest in the back; but I would have the Sea-men take him exact, as likewise a good Constellation called the Crane Grus, or the Flamengo, as the Spa∣niards call it; this Astensine consisteth of 13 Stars, and hath 3 Stars of the second Magnitude, that in the head is called the Phaenicopter Eye, and the other are on his Back, and the other in his left Wing; These Stars I would desire those Mariners that Sail to the East or West-Indies; to take the Meridian-Altitude thereof, and their distance from any known Stars, and by it you shall have all the rest; for many times I have been Sailing between the Tropicks, and for 12 days together have had no Meri∣dian-Altitude of the Sun, by reason of close and cloudy weather, which is bad for those that are bound to small Islands, and Cape-Lands; therefore to the Southward, as well as to the Northward; these Stars will stand them in great stead, and serve their turn, as well as the Sun, to find the Latitude thereof.

The Rule is thus; About the end of the Year 1577, Ticho observed the distance of rhe little Star in the breast of Pegasus from the bright Star of the Vulture, to be exactly 45 degrees 31′, and by the Meridian-Altitude thereof, he found the Declination thus.

Ticho observed at Ʋraniburge the Star in the breast of Pegasus, and found his Meridian-Altitude	56 d. 32 m.
The Latitude of Uraniburge is 55 d. 54′ the Compl.	34 06	Substract the Alti∣tude of the Equa∣tor.
The Declination is found to be	22:26
To it add the Complement of 56 deg. 32′, which is	33:28
And the sum is the Latitude of Uraniburge	55:54 m.

So the Declination is found to be 22 deg. 26 min. North, which being given, the Lon∣gitude of the said Star is to be inquired.

Therefore in the Oblique-Angled Triangle (of this Diagram) FOL is known.

[illustration] geometrical diagram

First,

The Complement of the Declination of the bright Star of the Vulture 82 degrees 8 min. FL.

Secondly,

FO, the Complement of the Declination of the Star in the Breast of Pegasus 67 degr. 34′.

Thirdly,

OL the distance of them 45 degr. 31′.

By this Rule the Angle at F, which is the difference of their Right-Ascension, will be found to be 44 degr. 54′, as will be here demonstrated.

The Co-Sign of Declination FL,	is 82 deg. 08 min.	999589
The Co-Sign of Declinat. FO, is	67 − 34	996582
The difference is	14 − 34
(1.) The sum is		1996171
(2.) The Quadrat or (2) the Rad.	2000000

The Base or distance of the Stars LO	45 d. 31
The difference of their Declin. FL and FO	14:34
The Sum	60:05
The difference	30:57
The half of the Sum	30 d. 02 30″ Sign	969951
The half of the difference	15:28 30 Sign	942621
The (3) Sum is	1912572

Then say,

As the first Sum	1996171
is to the double or Quadrat of the Radius	2000000
So is the third Sum	1912572
to the double or Quad. of the Sign of half ☞	1916401

The Angle sought, which Bi-sected, gives the Sign of 22 d. 27′ 17″ 958200

Which doubled, is 44 degr. 54′ 34″, is the Angle LFO, which is equal to the Arch DE, the difference of their Right-Ascension, which I add to the Right-Ascen∣sion of the bright Star of the Vulture, 292 degr. 35′, and the Sum is 337 degr. 29′ 34″, is the Right-Ascention of the little Star in the breast of Pegasus.

Then having the Declination of this Star 22 degr. 20′, and the Right-Ascension 337 degr. 29′ 34″, the Longitude of the said Star by the last Problem (19) will be found to be 18 degr. 36 ♓, and the Latitude thereof 29 degr. 24 min. North.

Now to draw the Diagram by Chords, and half-Tangents Geometrically, with the Chord of 60 degr. draw the Circle; then draw ♑, ♋, the Ecliptick 23 degr. 31′ ♑, S, and by C draw the Equator; then by the Parallel of Declination, and Right-Ascen∣sion of the Vulture, will find LO; therefore if you put the difference of Ascension from ♈ to E 22 degr. 30′ 20″, from the nearest Equinoctial-Point, and draw the Meridian-Circle FES, and it will cut the Parallel of Declination at O; draw through O, as PON the Circle of Longitude, and measure ♈ X on the Line of half-Tan∣gents, and it is 11 degr. 24 min. from the nearest Vernal Equinox, Substracted from 30 degr. leaves 18 degr. 36′ of ♓ for the Stars Longitude; and as before directed, you may find the Latitude ♑ ♄ to be 29 degr. 24′.

PROBL. XXI. To find the Parallax of Altitude of the Sun, Moon, or Stars.

THe true Altitude of the Sun, Moon, or Stars, ought to be observed in the Centre of the Earth, (if possible) whereto the Tables are conformed; but because we dwell upon the Superficies of the Earth 4000 Miles nearest, or 3983 English Miles from the Centre of the Earth; therefore, the Planets seem lower to us, than indeed they be; and therefore to find the true place of the ☉ ☽ ✶, you must draw a Right-Line from the Centre of the Earth, through the Centre of the Sun, Moon, or Stars; but the apparent visible place is determined by a Line drawn from the Eye, through the Centre of the Star; therefore the Parallax of a Star is an Arch of a great Circle, pas∣sing by the Zenith, and the true place of the Star, the Arch of the same Circle inter∣cepted between the true and apparent place.

[illustration] geometrical diagram

A Figure or Scheme shewing what the Parallax, or diversity of Aspects is.

In this Figure, C denotes the Centre of the Earth.

D the Place or Superficies of the Earth, from whence the ☉ ☽ or ✶ is seen.

☽ ☉ and ♂, their Place in their Orbs.

C ♂ I, C ☉ M, C ☽ N. the Lines of their true Place.

D ♂ G, D ☉ K, D ☽ M, the Lines of their visible or apparent Places.

Hence the Angle made by the Intersection of the said two Lines through the Body of the Planet, is the Angle of Parallax, that is to say, in ♂ the Angle C ♂ D, which is equal to the Angle I ♂ G, in the ☉ the Angle of Parallax, is the Angle C O D; and lastly in the Moon it is the Angle C ☽ D, or N ☽ M, which is all one.

By this it is manifest, the nearer a Star is to the Horizon and Centre of the Earth, the greater is the Parallax; and hence it is, that the Orbit of the Moon being nearest to the Earth, her Parallax is greatest, and most perceptible, because the Semi-Diameter of the Earth bears a sensible proportion to the Semi-Diameter of the Moons Orbit, though it be very little, or nothing at all in comparison of the Orbs of ♄ ♃, and the fixed Stars, which is caused by the Interval and vast distance which is between them; but this last Problem hath no relation to the common use of Mariners, therefore I shall not insist any further, but refer the Reader to Harmonicon Coeleste, where there is a full discourse, and Rules relating thereunto.

But these Astronomical Propositions as I know to be useful for Sea-men, I have here inserted; for the first and second will find the Suns Place and Declination, together with the 15 Probl. will find the Latitude.

The third, fourth fifth, and sixth will find the Suns Rising and Setting, as the 7, 8, 9, 10, 11, 13, 14, to find the Variation of the Compass, and the 12 to find the Suns Alti∣tude at any time assigned, and the rest being very useful Rules of the Stars, by which you may have the Hour of the Day, and Night; for having the Latitude of the Place, with the Declination and Altitude of the Sun, or any Star, they may find the Hour of the Sun or Star from the Meridian by the 16 and 17 Problem; then comparing the Right-Ascension of the Sun, with the Right-Ascension of the Star, they may have the Hour of the Night in all these Propositions; I have been as plain and as brief, as the several Resolutions thereof would permit me; and I do wish the Practitioner as much delight in the Practice, as I have had in the Composing of it.

The End of the Sixth Book.

Notes

• * 1.1

  Set VY from C to ☍, and with V ♁ draw the Arch at ♂.