How may this Declination be found for all times, and upon all points of the Compass?

• First, Consider whether the Sun be coming towards the Equinoctial, or going from him:
• That being known, consider the time wherein you seek the Declination, then look for the Suns Declination in your Regiment for that day, and also seek his Declination for the next day, subtract the lessor out of the greater, and the remainder is the whole declination which the Sun declineth in 24 hours, or in his moving through all points of the Compass,
• from which number you may by the Rule of Proportion find his Declination upon every point of the Compass for every hour of the day, as by these Examples may appear:

Examples.

• In the year 1625, the 20 of March, I desire to know the Suns Declination when he is upon the North part of the Meridian of London, I seek the Suns Declination for that day, and find it to be 3 deg. 59 min. the Sun then going from the Equator. (Readers after the change to the Gregorian Calendar in 1754 will appreciate that this is after the equinox)
• I also search his Declination for the next day, being the 21 day of March, and find it to be 4 deg. 22 min. I then subtract 3 degrees 59 minutes from 4 degrees 22 minutes, and there resteth 23 minutes so much the Sun doth decline in 24 hours, or in going through all the points of the Compass.
• Then I say by the rule of Proportion, if 24 hours give 23 minutes of Declination, what will 12 hours give, &c. I Multiply and Divide, and find it to be 11 min. 30 sec. the Suns Declination in 12 hours motion to be added to the Declination of the 20th day being the Suns going from the Equator:
• Or for the points of the Compass, I may say, if 32 points give 22 min. of Declination, what will 16 points give, which is the distance between South and North: I multiply and divide as the rule of proportion requireth, and find that 16 points give 11 min. the Suns Declination in moving through 16 points of the Compass, which is to be added to the Declination of the 20th day, because the Sun goeth from the Equator, so I conclude the Declination to be 3 deg. 52 min. the Sun being North on the 20 of March.

• Being West from the Meridian of London 90 degrees of Longitude, I desire to know the Suns Declination when the Sun is upon the Meridian on the 20 of March 1625,
• I must here consider that 90 deg. of Longitude makes 6 hours of Time, for every hour containeth 15 deg. whereby I know that when the Sun is South at London, he is but East from me, for when it is 12 of the Clock at London, it is but 6 of the Clock in the morning with me: & when it is 12 of the Clock with me, it is then 6 of the Clock in the afternoon at London:
• therefore I must seek for the Declination of the Sun at 6 of the clock in the afternoon, and that is the Meridional Declination which I must use being 90 degrees West from London, which to do the last Example both sufficiently teach you, whereby you may easily gather the perfect notice of whatsoever is requitue in any of these kind of observations, if you read with the eye of Reason, and labour to understand with judgement that you read.

Example.

• The day and year proposed being the 20 of March 1625: Declination 3 deg. 59 min. The next day the 21 of March 4 deg. 22 min. The deduction made there resteth 23 min. the proportional part to be found for 90 deg. West, or 6 hours of time,
• if 24 hours give 23, What does 6 hours give? Facit 5 min. 18 seconds which being that the Declination encreaseth, add 5 min. 18 seconds, to the Declination for the day preferred: that total is the Meridional Declination for 90 deg. of Westerly longitude from the meridian of London. (note that in 1625, 20th March is after the spring equinox)

There is another way most excellent for the finding of the Suns Declinationn at all times, that is to search by the Ephemerides the Suns true place in the Ecliptick for any time proposed whatsoever, and then by the Tables of Sinus the Declination is thus known: Multiply the Sinus of the Suns Longitude from the Equinoctial points of Aries or Libra, to which soever be nearest, by the Sinus of the Suns greatest Declination, and divide the Product by the whole Sinus, and the Arke of the Quotient is the Declination desired: but because Seamen are not acquainted with such Calculations, I therefore omit to speak further thereof, as this plain way before taught is sufficient for their purpose

The Use of this Instrument. (Not currently shown)

By this Instrument you may sufficiently understand, the reasons of whatsoever is before spoken for the finding of the Poles Elevation, or the latitude of your being: into the consideration whereof because the young practiser may the better enter, I think it not amiss by a few examples to express the necessary use thereof.

1. Q. The Sun being 7 degrees of North Declination and the Pole Articke being 45 degrees above the Horizon, I demand what will be the Suns Meridional distance from my Zenith.

1. A. First I turne the Horizon until I bring the North Pole to be 45 deg. above the same, there holding the Horizon not to be moved. I then bring the third* that is falted* to the Center of the Instrument, 7 degrees from the Equinoctial towards the North, because the Sun hath so much North Declination, and the third* both shew me upon the vertical Circle, that the Sun is 38 deg. from the Zenith.

2. Q. The Pole Artick being 50 deg. above the Horizon, and the Suns distance 30 deg. from the Zenith, I demand what is the Suns Declination ?

2. A. As in the first question I place the North Pole 5 deg. above the Horizon, there holding the Horizon not to be moved, then I bring the third* to the 30. deg. upon the vertical Circle, because the Sun is 30 deg. from my Zenith, and the third* sheweth upon the Meridian between the Tropick of Cancer and the Equinoctical, that the Sun hath 20 degrees of North Declination.

3. Q. The Sun having 10 deg. of South Declination, being upon the Meridian, is 53 deg. from my Zenith, I demand what is the Poles height ?

3. A. In the first question the Precise height and the Suns declination are given for the finding the Suns meridional distence from the Zenith, In the second the Poles height is given, and the Suns meridional distance from the Zenith, thereby to find the Suns declination, And in this question the Suns declination and meridional distance, is given for the finding of the Poles height. I therefore bring the third* fa__ned in the Center of the Instrument 10 degrees South from the Equator, between the Equinoctiall and the Tropicke of Capricorne, there holding the third not to be moved, I then turne to the Horizon, untill I bring the 53 degree of the verticall circle under the third, and then the Horizon sheweth me, that the North Pole is 43 degrees above the same.

4. Q. The Sun having 12 degrees of South declination, and being upon the Meridian South from me, is 30 degrees above the Horizon. I demand how far the Sun is from my Zenith, how much the Equinoctall is above the Horizon, and what is the Poles Height ?

4. A. First, I bring the third to the place of the Suns declination, as before, there holding in not to be moved, then I turn the Horizon untill I bring it to be 30 deg. under the third, and then the third sheweth me that the Sun is 60 deg. from my Zenith, and the Horizon sheweth that the Equinoctial is 42 deg. above the Horizon; although these questions are so very easie and plain, as that they may readily be ansered by memorie, yet because the reasons how they are answered may the better appeare, is the cause wherefore they are demanded, and in this sort answered, only for the benefit of such as are not altogether expert in these practices, that thereby they might likewise frame unto themselves questions of other variety, and so gather thereby the more sufficient judgement in this part of Navigation.

What is the Zenith ?

The Zenith is that prick, or point, in the heavens, which is directly over your head, from whence a line falling perpendicularly, will touch the place of your being, and so pass by the Center of the sphere and this line may be called the Aris of the Horizon, and the Zenith the Pole of the same, being 90 d. from all parts thereof, as by the former figure may most plainly appear.

The use of the Tables of the Sunnes declination.

Because the height of the pole (which is equal to the latitude of any place) cannot be exactly known, except you have the Suns true declination; here are Tables of the Suns declination, exactly calculated for the years, 1649,1650,1651,1652. which will serve until the year 1672. without correction, because there will be no sensible error in their use. And these are the plainest of any Tables of the Suns declination, that are extant; the declination is nominated, whether it be North, or South, Between the Columns and the Equinoctial, is signified by three black lines in March, when the declination comes to be North; and it is signified by one black line in September, when the declination comes to the South; and there are the years in every month for which the declination serveth at the top of the Tables, But besides the Tables of declination, the whole book hath not any equal to it, of its quantity, for the direction of young beginners; there being in it the principles of the Sphere, and the ground and foundation of Navigation, with divers other things, that are very useful, commodious, and pleasant. But for a further description of the Tables, you have in every month four great columns, each of them consisting of three lesser columns; in the first column is the dayes of the month, in the second is the degrees of the Suns declination, under the letter G, and in the third is the minutes, under the letter M. The first of the great columns serving for the first year after leap year; the second for the second year; the third for the third year; and the fourth for the leap years, viz. 1652, 1656, 1660, 1664, 1668, 1672.

To find the Suns declination at any time, first seek the month, then the yeare of our Lord, and the day of the month, and right against the day of the month, under the yeare of our Lord is the Suns declination.

Example. the 10th day of March, 1651. I desire to know the Suns declination, in March I look for it, 1651. and find it over the third great column, and right against the 10th day I finde 00 degrees and 02 minutes South declination, and the likes is to be done for any other.

Click Here for the Tables of the Sunnes Declination
The rule of the Regiment (1643)

For as much as the Poles height cannot be observed by the Sunne, unlesse the Sunnes true declination bee knowne, I have therefore carefully calculated these Tables or Regiment, out of Origanus, for the yeares 1625, 6.7.and 8. which will serve untill the yeere 1644. without further correction: and because there may grow no error by mistaking the yeares, I have over every Month written the yeare of the Lord, in which the declination of the same Month is to be used, therefore when in any year and Month you seeke the Sunnes declination, first looke for the month, and there you shall finde 4. of those Months, which are the Months between the leape years, then look over each of those months, until you find the year of the Lord, wherein you seeke the declination, and directly under that year is the Month wherein you must seeke the Suns declination:

Example 1626. the tenth day of Feb. I would know the Suns declination, first I seeke out February, & under the second yeare I see the yeare 1626, therefore this is my Month, against the tenth day of which Month I find that the Sun hath 10 deg. 49 min of South declination, and after the like manner you must do in all the rest as occasion requireth.