The seaman's companion being a plain guide to the understanding of arithmetick, geometry, trigonometry, navigation, and astronomy. Applied chiefly to navigation: and furnished with a table of meridional parts, to every third minute: with excellent and easie ways of keeping a reckoning at sea, never in print before. Also, a catalogue of the longitude and latitude of the principal places in the world with other useful things. The third edition corrected and amended. By Matthew Norwood, mariner.

OF ASTRONOMY.

I Had thoughts to have handled the Doctrine of Spherical Triangles fully, but if I should, I might do that which my Father hath done before, better than I can; the thing which induced me to this thought was that I might make the Application of it to Great Circle sailing; but because that way of sailing (though the best) is seldom used, I have not set it in my Book; if you desire to know it, you may see the way of it in my Fathers works, as plainly as (I think) it can possibly be demonstrated by any man.

I have thought good to make use of his fundamental Axiom, to shew how to resolve those Questions in the Sphere, which I have wrought in this Book by the Plain Scale, though I confess (for Amplitudes of Rising and Setting, and Azimuths at certain given hours, with divers other things that are chiefly to be minded at Sea, for the Variation of the Compass, and other necessary uses) I hold the Plain Scale to be true enough, but that I conceive not to be so satisfactory to all men, and there∣fore I have done this, the Axiom is this:

The Sine of a middle part with Radius, is equal to the Tangents of the Extremes adjacent, or to the Sine Complement of the opposite Extremes.

Now for Brevities sake it is common to use these Characters, for this Character + signifieth with Radius or more, or one Side or Angle with an another, for equal =, for less −; so that the general Rule or fundamental Observation which we purpose to use, may be expressed thus:

The Sine of the middle part + Radius is = to the Tangents of the Extremes adjacent, or to the Sine complement of the op∣posite Extremes.

This Rule being rightly understood gives such full directi∣ons, that a man may resolve any Question in the Sphere by it, which makes me judge those culpable of a fault, that accuse him of Prolixity, that made it in so brief a manner. I will define a spherical right angled Triangle, and so come to the use of this Axiom.

In the spherical right angled Tri∣angle A B C, right angled at B, there are 5 things considerable, beside the right Angle, namely, the three Sides and the two Angles; the two Sides next to the right Angle are not no∣ted to be Complements, but the Side A C, the Angle A, and the Angle C, is noted by their Complements to a Quadrant (which are the three things farthest from the right An∣gle B.)

[illustration] depiction of geometrical figure

If you have two things given beside the right Angle, to find a third (as you always have) the first thing that you are to consider, is, whether it be an adjacent or an opposite Ex∣treme.

That is an adjacent Extreme, when the given parts and the required part joyn.

For Example in Adjacent Extremes.

Suppose the given things were A C and A C B, and the re∣quired thing were C B: Here you see that A C is joyning to C,

and so is C B, therefore they all joyn, and C is the middle part, for it is between B C and C A the Extremes.

That is an opposite Extreme that hath two things lying to∣gether, and the third thing lieth alone, and that part which lieth alone is the middle part.

Example.

Suppose the three things (namely the two given things, and the required thing) were B C, the Angle at A, and C A; then I conclude that C B is the middle part, and that it is an oppo∣site Extreme, because it is alone, and the other thing, namely, C A and A are together: If you ask how this can be, thus:

A C B is between C B and C A, and A B is between C B and the Angle at A, which proves that B C is alone, and A C and the Angle at A joyns: Then is A C and the Angle at A also the two Extremes.

Pray observe this for a Rule, that the right Angle never parts any thing, for if C B, the Side B A, and the Angle at A be the three things, it is nevertheless an adjacent Extreme; for though the right Angle be between the Sides B A, and B C, yet do not count that it parts them, (as another Angle would) so as to make it an opposite Extreme.

Or if the given and required things were the three Sides, you might judge B A the middle part, because it is alone, for the Angle at A parts it from A C, and the right Angle at B is be∣tween B C and it: Now never count that the right Angle at B parts B C from B A, but that they lie together, and A C lies alone, being parted by the Angle C from C B, and by the Angle A from A B, and so is the middle part: These be the first conside∣rations (after you have a Question set you) that you ought to mind, without regarding which of the three things is given or required.

We now come to apply this general Axiom. Suppose I have the Side B C given, the Angle at C, and that the Side C A be required.

From what hath been said I know C to be the middle part and that it is an adjacent Extreme; I will apply the Rule thus, it saith

the Sine of the middle part, which intimates that the middle part must be always so called, as you may take the Sine of it; now because its noted to be a Complement, you must call it Sine Complement, to find the Sine of it. This done, I will go to one of the Extremes, and consider what the Axiom saith for that, (admit it be the Extreme C B) It saith the Sine of the middle part with Radius, is equal to the Tangents of the Ex∣tremes adjacent; from whence I conclude, that I must call C B so, as that I may have the Tangent of it; Now C B is not noted to be a Complement, and therefore I call it Tangent C B. Lastly, for the other Extreme C A, for the same reason I must call it so, as that I may take the Tangent of it: and because it is noted to be a Complement, therefore I call it Tangent Com∣plement, to have the Tangent of it: Now as I go along thus in my thoughts, I will set it down as I consider it, and it frames this Rule:

Sine com. C + Radius is = to Tang. C B + Tang. comp. C A.

My Question being brought into this order, I will consider which is given and which is required: those which are given, I will set down first, the thing required undermost, and look for them in the Tables by the directions which I have framed from the Axiom.

Look for	Sine comp. C
For	Tangent B A
Add them together, casting away Radius, and the Sum look for in the Tangents	take Tan. com. C A
it produceth your desire.

And thus this Axiom sets your business in order; but now that you may know when to take a Complement arithmetical, or let it a lone; and also, that you may know which thing it is that you must take the Complement arithmetical of, observe this Rule, either in adjacent or opposite Extremes.

That if the middle part and an extreme be given, take the Complement arithmetical of the given Extreme; but if the two Extremes be given, to find the middle part, take no Complement arithmetical at all: This may be sufficient for all Questions in

right angled Triangles, will fall out one of these two ways; this Complement arithmetical is what every figure in the Tables wants of 9: it was first invented by my Father, it saves a labour of Subtracting, and so abbreviates the work much.

To many which I have taught, it hath seemed strange that a thing is called Sine comp. or Tan. comp. (when one is to find the Sine or Tangent of it) but the reason of it is this; if a thing be a Complement already (as three things in every right angled spherical Triangle is noted to be) the comp. of that Comple∣ment must be the Sine or Tangent of that Complement.

But pray mind me; I have noted that there are five things in every right angled Triangle considerable, besides the right Angle.

Now observe this for a Rule, (when you are framing your Rule from the general Axiom to work by) that if you call any of these five things by their names, you shall not have the Com∣plements of them, but call any of them out of their names, and you shall; I mean thus. Suppose you have a mind to take the Sine of A B, it is noted not to be a Complement, therefore call it Sine A B, to have the Sine of it, or Tangent A B to have the Tangent of it, the like for C B.

Again, if you have a mind to take the Sine of C A, I consider that C A is noted by the name of a Complement, call it Sine comp. so shall you have the Sine of it, or Tangent Complement, and you shall have the Tangent of it; the like for the two Angles contrariwise: If you have occasion to take the Complements of them, call them out of their names.

Example in opposite Extremes.

Suppose A B be given, and B C, and the Side A C be required.

From what hath be said, I consider this is an opposite Ex∣treme, and that A C is the middle part; (for that is alone) call A C Sine Complement to find the Sine of it: and because the Axiom saith, that the Sine of the middle part with Radius, is = to the Sines Complement of the opposite Extremes (for the former Reasons) call A B Sine comp. and B C Sine comp. which is out of their names, and you have this Rule:

Sine com. A C + Radius is = to Sine com. A B + Sine com. B C.

This done, set the things down by the order

Sine com. C B

Sine com. A B

Sine com. A C

of this Rule, with the required thing under most thus:

Take the figures answering the number of degrees and minutes, which the given things may contain, (out of the Tables) add them together, casting away Radius, (if it comes to be above it) and look for the Remainder amongst the Sines, and take the Complement of that Sine that answers it, to answer your demand.

Suppose it were A B and A C that were given, to find the Angle at C; this is opposite Extreme, and A B is the middle part, call A B (the middle part) Sine, (which is by its name) to find the Sine of it, and A C, and the Angle at A out of their names (which is Sine A C, and Sine A) to find the Sine Com∣plement of them, and thus for any else.

Be sure to remember the Rule that I propounded about the Complement arithmetical, that so you may know when to take it, and when to let it alone. This is in brief.

My Father hath shewed the proportions that holds, and the reasons of them, which I purposely omit, knowing that this Rule works those Proportions: We will now come to the work it self; and I shall do only those Questions which I wrought by the Plain Scale, as being most necessary for our use that go to Sea.

Latitude 50 deg. 00 min. Northerly, Declination 13 deg. 15 m. Northerly, I demand the Meridian Altitude of the Sun.

LEt M G n R represent the Meridian, M n the Horizon, K y the Equinoctial, I ♈ the Parallel of the Suns Decli∣nation, G the Zenith, R the Nadir, subtract the Latitude G K 50 deg. 0 min. (= to n f) from G K M 90 deg. and the Re∣mainder is K M, the height of the Aequinoctial above the Hori∣zon, 40 deg. 0 min. to which add the Suns Declination K I 13 deg. 15 min. and the Sum is I M the thing required, 53 deg. 15 min. as you may see in this Example.

[illustration] depiction of geometrical figure

This, if the Suns Declination be

50 00

90 00

40 00

13 15

53 15

Northerly

But if the Suns Declination were Southerly, namely, so as that her Parallel of Declination cut the Meridian in s; subtract the Suns Declination K s from the height of the Aequinoctial above the Horizon K M, and the Remainder is s M, the Suns Meridian Altitude.

In any Latitude, subtract the Latitude from 90 deg. and so you have the Aequinoctials height above the Horizon; then see whether the Meridian Altitude be greater or lesser than that; if greater, add the Suns Declination to the Aequinoctials height above the Horizon, if lesser subtract it, and the Remainder is the Suns Meridian Altitude: and this is evident, because the Suns Declination is his Distance from the Aequator, and so he can be but as much higher or lower than the Aequator, as his Declination is.

Latitude 50 deg. 0 min. Declination 13 deg. 15 min. Northerly, I demand the Suns Amplitude of Rising and Setting.

NOte, that the sides of a Spherical Triangle are three Ar∣ches of Great Circles, every Arch being less than a Se∣micircle; and therefore the Parallels or other lesser Circles of the Sphere must not be taken as the Sides of a Triangle.

F O P is an Arch of the Meridian, cutting the Center of the Sun at his Rising, which is at O (you have been told that all Me∣ridians cut the Aequinoctial at right Angles) then must of ♈ be a right Angle, and in it I have given s 0 the Suns Declination (= to t c) and s ♈ 0 = 40 deg. 0 min. the measure of f ♈ 0, is E K = to t A the Complement of t s to 90 deg. (t s being the Latitude 50 deg. 0 min.) and I am to find ♈ 0, the Suns Am∣plitude of Rising; I consider it is an opposite Extreme, and that s 0 is the middle part.

Sine f 0 + Radius is = to Sine ♈ + Sine ♈ 0.

[illustration] depiction of geometrical figure

Sine f 0 the Suns Declination 13 deg. 15 min.	9,36021
Sine f ♈ 0 the com. of the Poles Elev. 40 d. co. ar.	9,19193
Sine ♈ 0 the Suns Amplitude of Rising or Setting 20 deg. 53 min. 24 sec.	9,55214

Use this way to find the Suns true Amplitude of Rising, upon occasion of finding the Variation of the Compass.

The Suns place of Rising from the East towards the North, namely, East 20 deg. 53 min. 24 sec. Northerly; O ♈ is the Complement of O E, the Suns Azimuth of Rising from the North: so that if you subtract it from 90 deg. ♈ E the remain∣der is 69 deg. 6 min. 36 sec. O E; but if you have a desire to find the Suns Azimuth by the things you have given, here it is in the Triangle F E O, and is done as in this following Question.

Latitude 50 deg. Declination 13 deg. 15 min. I demand the Suns Azimuth of Rising or Setting.

LEt F O f P be an Arch of the Meridian, cutting the Sun at his Rising (which is at O) f O is the Suns Declination 23 deg. 15 min. which subtracted from f F 90 deg. the remain∣der is O F 76 deg. 45 min. the Suns distance from the North Pole, and F E is the Poles Elevation 50 deg. 0 min. and thus you have two sides given in the right angled Triangle O E F, to find the third side O F; its an opposite Extreme, and F O is the middle part: The general Axiom produceth this:

Sine com. F O + Radius is = to Sine com. F E + Sine com. O E.
Sine com. O F 76 deg. 45 min.	9,360215
Sine com. F E 50 deg. 0 min.	comp. arith.	0,191932
Sine com. O E 69 deg. 6 min. 36 sec.	9,552147

Thus I find the Sun riseth to the Eastwards of the North 69 deg. 6 min. 36 sec. which is E b N 9 deg. 38 min. 24 sec. Northerly; the other Question shews the same, when you find the Suns Amplitude.

Latitude 50 deg. Declination 13 deg. 15 min. Northerly, I demand the Suns height at six of the Clock.

LEt G P m N be an Azimuth passing through the hour of 6: Now in the Triangle ♈ m P, right angled at m, you have given P ♈ the Suns Declination 13 deg. 15 min. and the Angle at ♈, namely P ♈ m = to the Arch f O (for f O is the measure

of it) it is the Poles Elevation 50 deg. 0 min. to find P m the Suns height at six of the Clock: I consider it is an opposite Extreme, and the required thing P m is the middle part.

Sine P m + Radius is = to Sine P ♈ + Sine P ♈ m.

[illustration] depiction of geometrical figure

Sine P ♈ the Suns Declination 13 deg. 15 min.	9,360215
Sine P ♈ m the Poles Elevation 50 deg. 0 min.	9,884254
Sine P m the Suns height at six of the Clock 10 deg. 6 min. 51 see.	9,244469

The same things given, to find the Suns Azimuth at six of the Clock.

THe Suns Azimuth at 6 of the Clock is m ♈, and the reason is, for that P u in the Heavens (being parallel to the Ho∣rizon) is as many degrees in that Circle, as ♈ m is in the Hori∣zon; for ♈ is under m, and m under P.

So that this Question will fall in the same Triangle as the other did, and we will use the same things to find it.

The Angle P ♈ m is 50 deg. 0 min. the Poles Elevation.

P ♈ is the Suns Declination 13 d. 15 m. and ♈ m is required.

I consider that this is an adjacent Extreme, and that P ♈ m is the middle part; The general Axiom produceth.

Sine com. P ♈ m + Radius, is = to Tang. ♈ m + Tang. com. P

Tan. com. P ♈ the Suns Decl. 13 d. 15 m. com. arith.	9,371933
Sine com. P ♈ m the Poles Elevation 50 d. 0 m.	9,808067
Tang. ♈ m the Suns Azimuth 8 deg. 36 min 24 sec.	9,180000

Latitude 50 deg. 0 min. Declination 13 deg. 15 min. to find the Suns height being due East or West.

LEt O L I u be an Arch of the Meridian, cutting the Sun in the East and West Azimuth, and it helps to make the right angled Triangle L I ♈, in which lieth our business: For L ♈ is the Suns height being due East; and to find it we have I L the Suns Declination 13 deg. 15 min. and L ♈ I the Angle of the Latitude, besides the right Angle; I consider it it is an opposite Extream and I L is the middle part, there will be this inference produced from the general Axiom.

Sine I L + Radius, is = to Sine I ♈ L + Sine L ♈.

[illustration] depiction of geometrical figure

Sine I ♈ L the Latitude 50 deg. comp. arith.	0,1157459
Sine I L the Suns Declination 13 deg. 15 min.	9,3602154
Sine L ♈ Suns height being due East 17 d 24′ 34″	9,4759613

The same height that the Sun is being due East in the morning, the same height he is when he is due West in the afternoon.

The Latitude being 50 deg. 0 min. Declination 13 deg. 15 min. I desire to know the Difference of Ascension.

I Have shewed what the Difference of Ascension means in this Book before; and if you look in the Question to find the Amplitude of the Suns rising before, there is the same Scheme that this is, but for your better uuderstanding I have here set it.

In the Triangle B D ♈ right angled at D, you have given D B the Suns Declination 13 deg. 15 min. the Angle at ♈ which is the Complement of the Arch I Z to 90 deg. or equal to O A the Complement of the Latitude 40 deg. 0 min. to find D ♈, which is as many degrees and minutes as B 6 is: It is an adjacent Ex∣treme and ♈ D is the middle part: The conclusion from the general Axiom is

Sine Y D + Radius, is = to Tan. D B + Tan. com. D Y B.

[illustration] depiction of geometrical figure

Tangent D B Suns Declination 13 deg. 15 min.	9,371933
Tan. com. D ♈ B com. Poles Elevat. 40 d. 0 m.	10,076186
Sine ♈ D the Difference of Ascension 16 d. 17′ 40″	9,448119

Which converted into time is 1 hour 5′ 2/15 and ⅔ of 1/15 of a minute.

The same things given, to find the time of Sun Rising.

FOr the doing of this, first find the Difference of Ascension (as hath been shewed in the last Question) and convert it into time; which done, subtract it from 6 hours, and you have your desire (in this case where the Declination and Latitude is both one way) but in any case take this in general, that if the Difference of Ascension be before 6 of the Clock, subtract it; if after 6, add it to 6 hours, and you have the time of Sun Rising.

This stands to good reason, forasmuch as the Difference of Ascension is the portion of time that the Sun riseth before or after 6 of the Clock.

In this Example you see the Difference of Ascension is before 6 of the Clock 1 hour 5′ 3/15 (we will omit the part of a part of a minute) I would know the time of the Suns Rising.

Example.

Subtract the Difference of Ascension	1 h. 5 m. 3/15
From 6 hours	5 60
The remainder is the time of Sun Rising	4 54 12/15 or ⅘

Thus I conclude the Sun riseth at 4 a Clock 54 min. ⅘

If you have a desire to find the time of Sun setting, subtract the time of Sun rising from 12 hours, and you have it; for as many hours and minutes as the Sun riseth before 12, so many hours and minutes he sets after 12.

Example.

Here the Sun riseth at 4 of the Clock, 54 m. ⅘ which we express thus	4 h. 54 m. ⅘
This subtracted from	11 60
Leaves the time of Sun setting	7 05 ⅕

Which is 5 min. ⅕ past 7 of the Clock in the afternoon.

This doubled is the length of the whole day, which is 14 hours 10 min ⅖.

This subtracted from 24 hours, is the length of the night, 9 hours 49 min. 8/5.

To find the length of the longest Day in that Latitude before proposed.

VVHen the days are at the longest in any North Latitude, the Sun is in the Tropick of Cancer; in a Southern Latitude, in the Tropick of Capricorn: This is a Northern Lati∣tude, therefore make the Tropick of Cancer the Parallel of the Suns Declination as here, O R is the Parallel of the Suns Decli∣nation, then must I 6 be the Difference of Ascension, which is = to F ♈. In the right Angled Triangle I F ♈ right angled at F, you have I ♈ F the Complement of the Latitude 40 deg. given, and I F the Suns Declination 23 deg. 30 min. to find F ♈; find it as you was shewed to find the Difference of Ascension before, convert it into time, and find the length of the day (as was shew∣ed before) This note, that Sine s I F L is an Arch of the Meridian, cutting the Horizon in that place of it where the Sun riseth.

Sine ♈ F + Radius is = to Tang. I F + Tang. comp. I ♈ F.

[illustration] depiction of geometrical figure

Tang. I F Suns Declination 23 deg. 30 min.	9,638301
Tang. com. I ♈ F Poles Elevation 40 deg. 0 min.	10,076186
Sine F ♈ Difference of Ascension 31 d. 12 m. 39 sec.	9,714487

But you may ask all this while how this Fraction is found, it is thus found: Admit I would find the sine of the Arch answer∣ing to 9714487, I look in the Sines, and find the nearest less than it to be

	9,714352
I take the next greater than it and find it to be	9,714560
I subtract the lesser from the greater, the remainder is	208
Then from the figures that came forth	9,714487
I subtract the nearest in the Tables less	9,714352
And the Remainder is	135

Then say,

As the Difference between the nearest less and the nearest greater 280	comp. arith.	7,681936
Is to the Difference between the same and that less 135	2,130333
So is 60 min.	1,778151
To 39 sec.	1,590420

The like is to be understood of any Fraction else; if you have occasion for the artificial Sine or Tangent of an Arch that hath a Fraction to it: Say,

As 60 seconds

Is to the seconds in the Fraction, which is here	39
So is the Difference between the nearest lesser and the nearest greater	208
To the Fraction	135
And this added to the lesser, makes the artificial Sine of the Arch required	9,714352
9,714487
The like is to be understood of a Tangent.

To find the hour of the Suns being due East or West.

Latitude 50 deg. 0 min. Declination 13 deg. 15 min. Northerly, I demand the time of the Suns being due East or West.

IN the following Scheme I O is the East and West Azimuth, R X is the Parallel of the Suns Declination, and cuts that Azimuth in C; from whence I conclude, that when the Sun is at C, she is over the East point of the Horizon; D C B A is an Arch of the Meridian, cutting the Sun in C: So that in the

[illustration] depiction of geometrical figure

Triangle C B r, you have given B C the Suns Declination, C r B the Angle of the Latitude 50 deg. 0 m. to find B r, which is as many degrees and minutes in a great Circle, as C 6 is in a lesser; find it and convert it into time, and in this case add it to 6 hours, and it gives the hour of the Suns being due East.

And for the time of the Suns being due West, you are to sub∣tract it from 6 hours, (in this case) and the remainder is your desire: The reason is, because as many hours as the Sun is due East after 6 of the Clock in the morning, so long time is he due West before 6 of the Clock in the Afternoon. I have left the working of this to your own practice, only I have set down the Resolution of it:

The Sun is due East 45 min. 9/15 past 6 of the Clock.

The Sun is due West 45 min. 9/15 before 6 of the Clock.

Which is at 5 of the Clock 14 min. 6/15.

Note, That if the Suns Declination be Southerly, then will he be East before 6; so that whereas here you add, there you subtract from 6 hours, to find the hour of the Suns being due East, or add to 6 hours for the time of its being due West; but your own Reason (if you look well on the Scheme) will guide you to know this; and also to know that, it is useless in such cases, for then he is not above the Horizon.

Latitude 50 deg. 0 min. Declination 13 deg. 15 min. I demand the Continuance of Twilight.

AS I have noted before, the Sun is accounted 17 deg. under the Horizon when the day breaks; therefore in the fol∣lowing Scheme let 17 I, be an Arch parallel to the Horizon, 17 deg. under it: Let D B C A be an Azimuth cutting the Aequi∣noctial, and the line of 17 deg. in the place of their intersection, which is at C: then in the Triangle r B C (right angled at B) you have given B C 17 deg. the Angle B r C the Complement of the Poles Elevation 40 deg. to find C r the continuance of Twilight (for C r and B u are equal.) I leave it to your own Pra∣ctice: I find the continuance of Twilight to be 27 deg. 4 min. which converted into time is 1 h. 48 min. 4/13 nearest.

[illustration] depiction of geometrical figure

So I conclude that between the Day breaking and Sun rising, it is 1 h. 48 min. 4/15, which is u B: Now if you add this to the Difference of Ascension (before found) B 6, which was 1 h. 5 min. 3/15 of a minute (I omit the smaller Fraction) you have the time between break of day, and 6 of the Clock, which may be termed u B 6 2 h. 53 min. 7/15.

And because the day breaks so much before 6 of the Clock, if you subtract it from 6 hours, you have the hour and minute of day breaking, which is at 3 of the Clock 6 min. ••8/15.

Add it to 6 of the Clock, and you have the time of Twilight ending, which is at 8 h. 53 min. 7/15.

To find the Suns Place and Right Ascension, provided, the Latitude and Declination be given.

Latitude 50 deg. 0 min. Declination 13 deg. 15 min, I demand the Suns Place in the Ecliptick and right Ascension.

RIght Ascension is an Arch of the Aequinoctial between a Meridian, cutting the Aequinoctial and the point of Aries or Libra; the Meridian must also cut the Sun or Star in the Ecliptick as here in this Scheme: A B C is an Arch of the Meridi∣an, cutting the Sun at B in the Ecliptick, and the Aequinoctial in C; then C ♈ is the Suns right Ascension.

The Suns Place (as I shewed formerly) is the Suns distance in the Ecliptick, from the nearest Aequinoctial point here B Y.

To find the Suns place ♈ B, you have given the Angle B ♈ C 23 deg. 30 min. and C B the Suns Declination 13 deg. 15 min. in the right angled Triangle B C ♈; it is an opposite Extreme, and C B is the middle part, I find it to be 35 deg. 5 min. if you propose the month and the day of the month that the Sun hath this Declination, you may tell what Sign the Sun is in, and what degree of that Sign.

[illustration] depiction of geometrical figure

As suppose the Sun hath this Declination here given upon the 14th. of April, I consider what Sign belongeth to that Month, and I conclude it is ♉, then say I, as much as the Sun is distant from the next Aequinoctial point above 30 deg. (which is the Sine of ♈) so much he is entered into ♉, which is 5 deg. 5 min. I omit the Fraction.

For the Suns Right Ascension.

You have the same things given in the same Triangle, to find the Suns right Ascension C ♈; but this is an adjacent Extreme, and the required thing is the middle part.

I find it to be 32 deg. 47 min. some part of a minute more, (but I omit the Fraction.)

This must be converted into time, and then is your demand fully answered, it is 2 h. 11 min 2/15.

These be things necessary and very useful in Navigation for several occasions, and that hath made me insist upon them: If you are unacquainted with this way of working, my Father hath insisted upon it largely, and in his works, commonly called The Doctrine of Triangles; you may be satisfied both in the Doctrine of right lined Triangles, and Spherical. This I judge sufficient for the two kinds of Sailing.

I would have handled something in oblique Spherical Trian∣gles, and so have proceeded to that incomparable way of Sailing by the Arch of a Great Circle; but the goodness of it is not known because it is not practised: If you love that way of Sailing, (Reader) I commend thee to my Fathers Doctrine of Triangles in large, where it is largely treated of.

FINIS.