The Use of the Table to number.

THis Table signifies thus much, That if there be one Figure alone, it is but so many units as there is in its name, as 7 is seven ones or units. If there be two Figures, the first Figure towards the left hand is as many tens as there is units in its name, and the other Figure is units: As if it were 10, that is ten units: if it were 23, the first Figure is two tens, that is twenty units: and the next Figure is 3, that is three units. The next place is hundreds, and con­sists of three Figures, as 700 is seven hundred: 723 is seven [Page 2] hundred twenty three. And thus you may count any Number by observing the place of any Figure, and in their places gi­ving them their names as you count: As suppose I have a Num­ber, namely 7864319, if you tell from 9 backwards to 7, you will find 7 in the place of millions, then begin and say seven millions: then if you count again, you will find 8 in the place of hundred thousands, 6 in the place of ten thousands, and 4 in the place of thousands, then to give them their proper Number, you will read thus: Seven millions for 7, eight hundred thou­sand for 8, sixty thousand for 6, four thousand for 4, that is, Seven millions eight hundred sixty four thousand; then for 3 three hundred, for 1 ten, and for 9 nine units, that is then altogether 7 millions 8 hundred 64 thousand 3 hundred and ninteen, which is the quantity of units that is in the given Number 7864319. And thus much for Numeration.

Addition of Degrees and Minutes.

NOte that sixty Minutes is one Degree, three Miles is one League, sixty Seconds is one Mile or Minute, sixty Mi­nutes of time is one Hour in time: fifteen Minutes of a Degree is one Minute in time, twelve Hours is an artificial Day, twenty four Hours is a natural Day, fifteen Degrees is an Hour in time, four Minutes of time is one Degree, three hundred and sixty Degrees is the Circumference of a Circle, one hundred and eighty Degrees is half a Circle, or a Semicircle; ninety De­grees is a Quadrant, or a quarter of a Circle, and eleven De­grees fifteen Minutes is one point of the Compass, two and thir­ty times eleven Degrees fifteen Minutes is the Circumference of a Circle; so that there is two and thirty points of the Com­pass in every Circle: one point of the Compass is three quarters of an Hour in time, or 45 Minutes.

Admit I had kept an account of a Ships Difference of Longi­tude in Degrees and Minutes, that had been out eight days, and had made Difference of Longitude each day, as followeth:

• The first day 2 deg. 20′
• The second day 2 deg. 45
• The third day 1 deg. 30
• The fourth day 1 deg. 11
• The fifth day 2 deg. 39′
• The sixth day 1 deg. 29
• The seventh day 2 deg. 19
• The eighth day 2 deg. 10

I desire to know (if it be all one way) what number of degrees and minutes it is in one sum.

Begin as you did in Addition, and say 9 and 9 is 18, and 9 is 27, and 1 is 28, and 5 is 33, set down 3 and carry 3: then say 3 that I carried and 1 is 4, and 1 is 5, and 2 is 7, and 3 is [...]0, and 1 is a 11, and 3 is 14, and 4 is 18, and 2 is 20: cast out all the sixes from 20, and there remains 2, set it down, and as many sixes as you cast away, so many units carry to the place of Degrees (which was three sixes) and say 3 that I carried and 2 is 5 and so forwards in that as in any other Sum in Addition: it being thus added, I find that it comes to 16 d. 23 m. the whole.

Now you may ask a reason why I cast away the sixes, the reason is because 6 tens make a degree: now with the three tens that I carried, and the tens that where in the row added up last, there were 20 tens, that must be 3 degrees and 2 tens over: the 3 degrees I carry to the place of degrees, and the two tens I set down in the place of tens: and thus in any other Sum.

Addition of Hours and Minutes.

ADmit I have these several numbers of hours and minutes to cast up in one sum, namely, then because 60 minutes is one hour, I do just as I did before: be sure in the place of minutes set never above 60, for if you do, it is more than a degree.

The same is to be done in Addition of Money, only remem­bring in the place of pence to go no farther than 12, because twelve pence make a shilling, as you do here to 60, because 60′ makes a degree, and in shillings no farther than twenty, because twenty shillings make a pound. And thus in any thing else (pro­vided that you know how many of the one makes one of the other) but for such things they are not so necessary to the thing we purpose to handle, and therefore I refer you to Mr. Hodder's Book, it being of small price, and much matter, and likewise very plain.

Subtraction.

SUbtraction is the taking a lesser number from a greater, and to see what remains: As suppose a man was bound from Cales to Dover, which is 21 miles: now if he have sailed 11 miles of the way, I demand what he hath yet to sail? Take 11 from 21, and the remainder is 10, so I conclude he hath 10 miles still to sail before he come to Dover: but in greater numbers they are ordinarily set so that you are forced to borrow, so that they are something harder.

Set the greatest number uppermost, and the least under it, and draw a line between them, and a line under them both, and say, 7 from 9 and the remainder is 2, set down 2: then 2 from 9, there remains 7 set it down: then 2 from 7, and the remainder is 5: whatsoever number is underneath so much is still to sail, as here 572 miles.

But now I'll give you an Example wherein the figures in the lesser Sum, are some of them bigger than in the greater, from which they must be taken: now in such a case borrow ten, and add to the upper figure.

Example.

A oweth to B 7822 pound of Tobacco, now he hath paid 4998 l. of it: I demand what is still due.

Set down your sums as in the former di­rections, and say 8 from 2 I cannot, but 8 from 12, and the remainder is 4: set it down, 1 that I borrowed and 9 is 10, from 2 I cannot, but 10, from 12, and the remainder is 2, set down 2 and carry 1, then 9 and 1 that I carried is 10, from 8 I cannot, but 10 from 18, and the remainder is 8 set down 8 and carry 1: then 1 that I carried and 4 is 5: 5 from 7 and the remainder is 2, set it down, and thus you may do any sum of one denomination.

Subtraction of Degrees and Minutes.

THis being of several Denominations, observe as before in Addition (of degrees and minutes) how many of the one makes one of the other: and if the under figures be bigger than the uppermost (in the place of minutes) borrow from 60 the de­nominator of a degree, and subtract from it, and that remainder add to the upper figure, and set it down: but in the place of de­grees, if the under number be bigger, do as you did before.

I have made in my Reckoning —

I demand what way I have differ'd my Lat. most, and how much? I have differed my Latitude and it is Northerly, because my Northing is more than my Southing.

Say 9 from 8 I cannot, but 9 from 18 there remains 9, set it down; 5 and 1 that I borrowed is 6, 6 from 4 I cannot, but 6 from 6, and the remainder is 0, that 0 and 4 is 4: set it down, then 1 that I borrowed and 9 is 10, from 2 I cannot, but 10 from 12, and there remains 2, set it down: 1 that I carried and 2 is 3, 3 from 3 and there remains 0,

The same may be done in Subtraction of pounds, shillings, and pence, observing but how many of the one goes to make one of the other: be sure in setting down a sum, you do not set more minutes in the place of minutes than 60, or pence in the place of pence than 12, or so for any thing else set no more than go­eth to make one in the next denomination.

Subtraction of Time.

You may try this over: it is done as the former, and needs no demonstration (as I said hefore) if you observe in any case how many of one makes one of the other, it is but borrowing from that number, as we have done here.

Multiplication.

IT doth Addition many times, for when you have many Addi­tions it makes one work of it, and also of two given Numbers it increaseth the greater as often as there is units in the lesser, or the lesser as often as there is units in the greater. Now in multiplying there are three things considerable.

First, The Multiplicand, or Number that you would multiply.

Secondly, The Multiplier, or Number you multiply by.

Thirdly, The Product or Sum produced.

Quest. 1. I demand how many Feet there are in 320 Yards, every Yard being three Feet?

If you set down 320 yards one under the other three times, and add them, you will have the sum of 320 yards in feet: but that is a tedious way, for then you must set down your 320 [Page 8] yards an hundred times one under the other, if there were one hundred times so much (of any thing) in it: but now Multipli­cation doth it at once; because there is 3 foot in a yard, there­fere there must be three times 320 foot in 320 yards.

Example. I set down 320 and 3 under the first figure on the right hand, and say 3 times 0 is 0, set down 0, then 3 times 2 is 6, set down 6: then 3 times 3 is 9, set down 9, so the Product is 960 feet, which answers your demand. Note, that when your Multiplication cometh to more than 9, as many tens as are in it, so many you carry to the next place, (as you did in Addition) and set down the rest, as you shall in this Example following.

Quest. 2 I demand how many Minutes is in 360 Degrees?

[...] Note that 60 minutes is one degree: it is less work always to make your greatest Number your Multiplicand, and then it stands as in the Margin: then say 0 no­thing is 0, set down 0; then 0 six is 0, set it down; then 0 three is 0, set it down: then go to 6, and say 6 times 0 is 0, set it down, one figure to the left hand under the former; then say 6 times 6 is 36, set down 6 and carry 3; then say 6 times 3 is 18, and 3 that I carried is 21, set down 1 and car­ry 2; now because there is no more figures, set down the two (as I have shewed in Addition) and thus 360 degrees is 21600 minutes, if both the Products be added together. Then to prove your work, add up your Multiplicand, and cast away the nines from it, and set the remainder down over the Cross; then cast away the nines of the Multi­plier if it comes to 9, but if not, set it down under the Cross; then multiplie 6 by 0, and it makes 0, set down that on the right side of the Cross; then cast up your Product, and cast away the nines from it, and if it comes to the same that the Multi­plicand and Multiplier did, (the nines being cast away) then the sum is done right, and stands as you see in the Margin. I shall here following [Page 9] shew two Examples, and let down two more for your practice: if you do not understand what is said, examine these Examples which are wrought, and they will satisfie you.

Quest. 3. How many units is 612345 times 62453125 units?

Quest. 4. How much is 123456 times 45321672?

Let these be the two Questions which here following are wrought one after the other.

[...]

The Product of the first Question 38242858828125

[...]

The Product of the second Question 55952640636024

Quest. 5. How many units is in 3438239 times 5289231423?

Answ. 18185641758584097.

Quest. 6. How many units is in 289764532172 times 343217812?

Answ. 199452348727277447664.

Division.

DIvision is that by which one may know (at one operation) any part of a whole Number, provided that the part be not under an unit: by part I mean thus, the third of a number, or the fourth of a number, or the twentieth part of a number, or the like.

In Division there are four things considerable: First, the Dividend; Secondly, the Divisor; Thirdly, the Quotient; And fourthly, the Remainder after the Division.

The Dividend is the number which you desire to have divided, or to have a part known.

The Divisor is that which you divide it by, or the name of the part which you would have known.

The Quotient is the number of units that is contained in the part required.

And the Remainder of the Division is that part under an unit, which is wanting to the true portion which is left in the Quotient.

Quest. 1. A man promised to do a Piece of work for me in a Quarter of a Year: Now if there be 365 Days in a Year, I desire to know how many Days is in this Quarter?

The fourth part of 365 days must needs be the number of days which this man will be doing the Piece of Work, there­fore divide 365 days by 4, and the Quotient shall be the whole days, and the remainder of the Division shall be the parts of a day above the whole days. 91 days, and [...] one fourth of a day is a quarter of a year, or the quantity of time he will be doing this Work. We express a Fraction thus: one fourth ¼, two fifths ⅖, or so for any Fraction.

But you will ask how this is done?

Which is thus: First, set down your Divi­dend 365; then to the left hand set down [Page 11] the Divisor under your Dividend. but you will ask why I did not set it under the 3? And the reason is, because 4 is more than 3, so that I could have had nothing from 3 till I had remo­ved it where it is: if 3 had been a number that I could have took my Divisor from, I would have set the 4 under the 3.

Your Sum being set, say how many times 4 can you have in 36? Nine times. Set down 9 in the Quotient, and multiply 4 by 9, saying 9 times 4 is 36, from 36 and the remainder is 0: cancel 4 and 36 as you go (by a dash with your pen) and your work stands thus: [...]

Then set your 4 under the 5, and say how many times 4 can you have from 5? Once. Set down 1 in the Quotient, and multiply it by 4, saying, once 4 is 4, from 5, and the remainder is 1: set 1 over the 5, cancelling your 4 and 5 by a dash with your pen (as you speak:) now because there is no more figures, your sum is done, and stands thus: [...] The 1 which remains is ¼ of a unit (because the Di­vidend is units) for whatsoever the Dividend is, the Quotient and the Remainder shall be a part of it: it hath its Denomination or Name from the Divisor 4.

Quest. 2. A man had an Estate of 78998 l. now he oweth to 23 several men 27821 l. apiece. I demand what each man shall have of the Estate, so that none of them may be wronged?

You will conclude because they were owed alike, they should have alike of that which was left, therefore divide the Estate by 23, and the Quotient with the Remainder will be each mans due.

Each will have 3434 l. 16/23 of a pound for his Debt. [...]

This is Division of two figures when the Divisor is two figures, and it is something harder than the other Example: The work­ing [Page 12] of it is thus, first set your Sum in order thus:

Then I say, how many times 2 can you have from 7? Three times: 3 times 2 is 6, from 7 and there remains 1 (set 3 in the Quotient) and say again 3 times 3 is 9, from 18, and there remains 9, then your Sum stands thus: [...]

You might have said 3 times 3 is 9, from 8 I cannot, but 9 from 18, and the Remain­der is 9, and 1 (for your borrowed one ten) from 1, and there remains 0, which is most easie for memory.

Then set your Divisor on place further towards the right hand in this form: [...]

How many times 2 can you have from 9? Four times: then 4 times 2 is 8, from 9, and there remains 1, and 4 times 3 is 12, from 19 and the Remainder is 7: and 1 (for the ten) from 1, and the Remainder is 0; and then your Sum stands thus: [...]

Then remove your Divisor to the next place, and see how many times 2 you can have out of 7, which is 3 times: 3 times 2 is 6, from 7 and the remainder is 1: and 3 times 3 is 9, from 9 and the remainder is 0; and your Sum stands thus: [...]

Then remove your Divisor to the next place, and see how many times 2 you can have from 10? which is 5 times: now 5 times 2 is 10, from 10 and there remains 0, and 5 times 3 is 15, from 8 I cannot (you see 8 is all that is left, so that you have no­thing to horrow from) therefore you could not take 5 times 23 from 108, try for 4 times, 4 times 2 is 8, from 10, and the remainder is 2, and 4 times 3 is 12, from 28 I can, you see you may take 4: set 4 in the Quotient, and say 4 times 2 is 8, from 0 I cannot, but 8 from 10, and the remainder is 2, and 1 that I borrowed from 1, and the remainder is 0, (cancel as you speak) then 4 times 3 is 12: now 12 from 28, and the remain­der [Page 13] is 16; but it is better to use your self to this method, namely, to take your units from the place of units, and your tens from the place of tens, as here 2 is in the place of units, and 8 above, therefore say 2 from 8, and there remains 6, and 1 (for the ten) from 2 and the remainder is 1: Now if this had fallen out so that the upper number had been 21, and the other 12, you know that 12 will be taken out of 21, but 2 would not be taken out of 1; now in such a case borrow one 10, and say as in Subtraction, 2 from 1 I cannot, but 2 from 11 and the remainder is 9, one 10 that I borrowed and 1 is 2, from 2 and there remains 0, and thus it is done at once: The reason why it is done this way is, because the Sum you are to take from may be big, so that a man cannot tell readily what the remainder will be without Subtraction; for this differs nothing from Subtraction, as it is shewed before, but only that you carry one number in your head, whereas there both the numbers are before you.

The hardest Sum in Division hath but these Difficulties in them:

First, Be sure take no more to set in the Quotient, than your Divisor multiplied by it) will come under the Number you are to take it from, or equal to it.

Secondly, Be sure you take the one as many times as you can from the other.

Thirdly, Be careful in takeing one number from another, to use this way of Subtraction.

This Sum is done, and stands thus: [...] And it signifies that the 23. part of 78998 is 3434 units and 16/23 of a unit.

Here following is a Sum of four figures done, and the way to prove any Sum.

[...]

This Sum is set down in its several Operations. From what hath been said already, I suppose you may be able to examine it; the like is to be understood of any Sum else in Division, there­fore I shall say no more, only shew how to prove them when they are thus wrought.

The best way to prove Division is by Multiplication, for if you multiply your Quotient by the Divifor, and add in the Remainder, that product will be equal to the Dividend, if the Sum be right, and it stands to reason it should be so, for if the Quotient with the remainder be in this Sum the 2345th. part of the Dividend, then 'tis evident that 2345 times that Quo­tient must be the same that the Dividend is (with the remainder which is added in.)

Example.

Here you see the Example of it, and the Product the same that the Dividend of the Sum divided was, and this I say is the best way for young Practitioners to prove Division, be­cause this makes them perfect in Mul­tiplication.

There is a shorter way to prove any Sum in Division, and that is this: Cast up your Quotient, and cast away all the nines from it, (as you do in Multiplication) and what remains set down upon one side of a Cross thus: [...]

Then cast away your nines from your Divisor in like manner, and there remains 5, set it down on the other side against 4 thus: [...]

Multiply them one by the other, and they pro­duce 20, cast away all the nines from 20, and the remainder is 2, keep this 2 in mind, and cast up the remainder of the Division, and cast all the nines from it: which if it be done in this Sum, afterwards there will remain 3, which add to the 2 that was in your mind, and it makes 5, which set down thus: [...]

Then cast away all the nines from your Divi­dend, and the remainder will answer to the last number you set down in the Cross, if the Sum be right, which here it is [...]

Reduction.

REduction sheweth how to bring gross or great Denomina­tions into small, or small into great.

Quest. 1. Suppose a Man had been at Sea twenty five Years, and for every Minute of that time he was to have on Farthing, I demand how many farthings is due to him?

Here I see how many minutes is in that quantity of time, and so many minutes so many farthings there is due, which is, 13140000 farthings. Now to find this, I first multiply the years by the days con­tained in one year, which is 365, and that produceth 9125 days, the number of days in that number of years: this mul­tiplied by 24 gives for its Product 219000 the number of hours in the whole time, because 24 hours is a day. That Product multiplied by 60, is the number of minutes in the whole time (because every hour is 60 minutes) which is the last Product, or the thing required, which I set down as the Answer to my Question thus, Facit 13140000 farthings. [...]

Quest. 2. Pray tell me what these farthings come to in Pounds, Shillings, and Pence, 1740120 Farthings.

[...]

Four farthings make a penny, therefore I divide the whole Sum of Farthings by 4, and the Quotient is ½ of the number of farthings, or the farthings reduced into pence, 435030 d. of which Sum I take the 1/12 by dividing it by 12, which is the next Quotient (the number of shillings that is the farthings) which is 36252 s. 6 d. Twenty shillings make a pound, therefore I take the twentieth part of this last Product, by dividing it by 20, and that Quotient shall be pounds, and the remainder of the Division shillings: 1812 l. 12 s. The Answer of your desire I will set down thus: 1812 l. — 12 s. —6 d.

Quest. 3. Some say that in a Mile there is 1760 Paces, three Feet being one Pace, twelve Inches one Foot, three Barley-corns being one Inch: I pray tell me how many Barley-corns will reach three Mi.

[...]

Because there is 3 miles, 3 times 1760 is the number of paces contained in 3 miles: then because 3 foot is 1 pace, multiply 5280 by 3, and it produceth the number of feet that is in 3 miles; then because 12 inches is a foot, multiply your feet by 12, and it pro­duceth 190080, the number of inches contained in 3 miles: and because 3 Barley-Corns make an inch, multiply your inches by 3, and it gives you the number of barly-corns that are in 3 miles, and the same is to be understood for any other number of miles.

Suppose I would know how many Miles is in the same number of Barley-corns, namely, 570240 Barley-corns.

As before you multiplied to bring Miles into Barley-corns, so now divide to bring things of small Denomination into great, only begin at the lowest, where before you ended, for the de­mand of this is the beginning of the other.

Example.

[...]

Here the Quotient of the first Division is Inches, because three Barly-corns make an Inch, namely, 190080 Inches.

The Quotient of the second Division is the number of Feet contained in those Inches, (and therefore the Inches were di­vided by 12) namely, 15840 Feet.

The Quotient of the third Division is Paces, and therefore it was divided by 3, the number is 5280 Paces.

The Quotient of the last Division is the number of Miles con­tained in either 570240 Barly-corns, or 190080 Inches, or 15840 Feet, or 5280 Paces, which is 3 Miles, the thing required: And after the same manner you may reduce minutes into degrees, or seconds into minutes and degrees, or degrees into hours, pounds into shillings, shillings into pence, and pence into far­things, or the contrary, by knowing how many of the one makes one of the other.

The Proof of Reduction.

The Proof of Reduction is the same thing which we did last, I framed it so for brevities sake. When you would prove any Sum that is brought into a small Denomination (by Multiplica­tion,) divide it backwards, beginning with the Product of your last Multiplication, and compare them in each as you do it, and if the Sum be right they will agree.

When you would prove any Sum that is brought into a great­er Denomination (by Division) then as before you divided, so now multiply; for I have shewed that Multiplication and Divi­sion is the Proof one of the other.

Quest. 4. I will reduce 28 degrees 20′ 40″ into seconds.

First Multiply your degrees by 60, and it brings them into minutes, add your odd 20 minutes to that Product, and it makes [Page 19] the number of minutes in that given number of degrees and minutes: multiply that by 60, and it produceth the number of seconds contained in that given number of degrees and minutes: to that Product add your odd 40 seconds, and it gives the number of seconds contained in the degrees, mi­nutes, and seconds, as here in this Example: and the reason is, because 60 minutes is a degree, and 60 seconds is a minute: The same is to be understood of the reducing of any number of several Denominations into one.

Facit 296440 seconds in 28 degrees, 20 mi­nutes, 40 seconds.

Thus much for what Mariners have need to know in Reduction.

Note, That the Remainder of any Division either in Redu­ction, the rule of Three, or in any Division, is some part of a Unit, namely thus: If the Quotient be shillings, the remainder is some part of a shilling: If the Quotient be pounds, the remain­der is some part of a pound. The remainder is the Numerator, and the Divisor is the Denominator of the Fraction; so that if the Quotient were pounds, and the remainder 12, the Di­visor is 20, and the Fraction is the 12/20 of a pound, which is 12 shillings; the like for any thing else.

The Golden Rule, or Rule of Three.

THis Rule well understood, is the Sum of the rest that fol­low it; for they are all (but as it were this Rule) it is cal­led The Golden Rule, because it is the Foundation of the rest: It is called The Rule of Three, because three numbers are given to find a fourth.

Now there is the Rule of Three direct, and the Back Rule of Three: the Rule of Three direct, is for such questions as are so stated, that the third number and second multiplied together, and that Product divided by the first gives the fourth.

The Back Rule of Three is so stated, that the first number [Page 20] and the second are multiplied together, and that Product divi­ded by the third, finds the fourth.

And this is all the distinction that I can make between the forward and backward Rule of Three, which is all the Rule of Three: I will not make therefore two distinct Rules of them, (but one) therefore pray observe this difference:

In all Questions in the Rule of Three there is three things given to find a fourth, which number will have alike proportion to the third, as the second hath to the first.

In stating your Question, be sure let it be so that you may set your sum in order with the first and third number, speaking both of one thing: I mean if the first be Ells let the third be Ells; if the first be Degrees, let the third be degrees, or some­thing that Degrees can be reduced to.

If your Sum be set with the numbers in several Denominations, before you bring your Sum in order, you must reduce your num­bers into the least Denomination: As suppose your Sum was this, If 5 pound 10 ounces of Tobacco, cost 10 shillings 9 pence, what shall 510 pounds cost? Reduce your 5 pounds 10 ounces into ounces, and your 510 pounds into ounces: likewise your shillings into pence, taking in the odd pence and odd ounces, as I have shewed how to do it in page 16 and 17.

Your Sum being reduced (as I have shewed) into the least Denominations, set it in order thus:

If 90 Ounces cost 129 pence, what costs 81600 ounces?

Your Sum being in order, consider whether your third num­ber require more or less: if it require more, multiply the second number by the greatest of the two extreams, and divide that Product by the lesser, it produceth the fourth number in the same Denomination that the middle number was.

If your third number require less, multiply the middle num­ber by the least of the two extremes, and divide that product by the greatest, and it gives your fourth number in the same denomi­nation that your second number was; then if you desire to reduce it into greater Denomination, do it as I have shewed already.

For your better understanding of this Rule of Three, I have here following done an Example of it in right Lines, and have found it upon this Question.

The Rule of Interest, and Interest upon Interest.

THis is the Rule of Three, for it runs thus:

Quest. 1. I let 478 l. for two years, what shall the Ʋse of it come to, after the rate of 6 l. a year, for the lent of 100 l. which is the common Ʋse of it in a year.

I'll set my Question in order thus:

If 100 l. gives 6 l. what shall the principal 478 give?

Facit 28 l. [...]

Here we have found what the Principle hath gained for a twelve-month, next consider what proportion 12 Months is to the time your money was lent, which was 24 Months: now if [Page 25] it had been three times as long a time as 12 Months, you must have made it 3 times as much as you find it for 12 Months, but being but twice as much, you must make the Use to to be twice 28 68/100 which is 57 l. 36/100 of a pound; the same is to be understood of any number else; but in stead of seeing what proportion the time limited for the lent bears with 12 Months, you may work it as you do the Rule of five Numbers, by saying (after you have abbreviated your Fraction.)

If 12 Months give 28 l. 68/100, what shall 24 Months gain? But I say the other way is best, after your other manner; If it had been one fourth of the time, it would have been ¼ of the use; if one fifth of the time, ⅕ of the use. For it stands to reason, that if 100 l. or any other quantity gains 6 l. in twelve Months. it gains four times as much in four times the time, or five times as much in five times twelve Months, and so forth.

I might say something of Interest upon Interest, but that differs but little from this: At the years end, add in your simple Interest to your Principal, and make it one entire Sum till the next year, then add that Interest to the Principal the year before, and receive the Interest of it: this is all it differs from the former.

The Rule of Fellowship.

THis Rule is the Rule of Three, done several times in one Sum or Question: for here are several Stocks, and several men that own them; now if the principal, or all their Stocks gain so much, how much shall each man gain according to the Stock he put in? I'll say, if the whole Stock gained so much, what shall the first mans Stock gain? and so for the rest.

Quest. 1. Six men make a Stock, the first puts in 30 l. the second 40. l. the third 52 l. the fourth 58 l. the fifth [Page 26] 60 l. and the sixth 78 l. If the whole Stock together gains 200 l. what shall each man have, so that there may be no wrong?

First, find the whole Stock by adding every mans Stock together, which is 318 l.

Then say, If the whole Stock 318 l. gains 200 l. what shall the first mans Stock gain?

For the Second.

If the whole Stock 318 l. gains 200 l. what shall 40 l. gain, which is the second Stock?

The like is to be understood of the rest: I forbear to work them, as being out of my intentions: I leave their operation to your genius who never learned them.

Quest. 2. There where five men made a Stock, the first put in 200 l. the second 59 l. the third 180 l. the fourth 78 l. the fifth 240 l. they lost at the return of the Ship 120 l. I demand each mans Loss proportionable to his Venture?

First find the whole Stock, by adding every mans Ven­ture together: this done, say for the first mans Loss, if the whole Stock lost 120 l. what shall the first mans Stock 200 l. lose?

For the second, If the whole Stock lose 120 l. what shall the second man lose which put in 59 l?

And so for the rest: The way to prove one of these Questions is less trouble than the Rule of Three, for after you have done, add all your Facits together, and it will make the same that was lost in the whole Stock, or gain­ed in the whole Stock, (if the Question be for Gain) and this stands, to good reason; for every mans Loss or Gain must together be equal to the whole, or else they do not [Page 27] contribute to the Loss, or enjoy the Gain that they have lost or gained.

There is more intricacy in this Rule than I have here cited, but the Foundation-work lies on the Rule of Three, and the intricacy that I speak of, will be understood with some con­sideration: I had thoughts to have shewed it, but it is shewed as well as can be (I think) by Mr. Record and Mr. Hodder: be­sides it would take up much paper, and I should digress from my intentions to Navigation.

What I have shewed since I treated of the Rule of Three, is only a shew how all Rules depend upon this Golden Rule. Blame me not for being so large in Arithmetick; but both that, and all faults else, season with the Salt of a charitable Construction, remembring that Navigation is imperfect with­out it, I end.

Geometrical Definitions.

A Point or Prick is this. (.) and is void of length, breadth or thickness.

A Line is length without bredth or thickness, and is properly called the nearest distance (if a streight line) between two pla­ces: but if not streight but circular, it is termed an Arch.

An Angle is when two lines are so drawn that they will meet, not in one streight line, but one to cross the

other: now the place where they cross or intersect, is called the Angular Point, as in this figure A is the Angular Point. All Angles that are bounded by streight lines, are called right lined Angles, as the Angle A.

A Triangle is when three lines meet, ma­king three angular points; now there be two sorts, right lined, and spherical: spherical being all the sides Arches of great Circles. There be three sorts of Angles, namely, obtuse, acute, and right angles.

A right angled Triangle, is when a line is drawn from another line, so exactly upright, that it leans neither one way nor other, [Page 29] as you see the line A B and

B C doth: also the Arch which is bounded between them two sides, will be a quarter of a Circle, as n A is: the line C A being drawn makes it a right an­gled right lined Triangle, whereas before it was but a right Angle.

An acute angled Triangle, is when the sides lean so one to the other, as that none of the three

Angles will contain an Arch of 90 degrees, which is a quarter of a Circle, but will be all less, as you see in the acute angled Triangle V S A.

An obtuse angled Triangle is when the lines are so drawn, that one of the Angles is bounded by above a Quadrant, or the ¼ of a Circle, as is

the Angle at K, see 8 u is above ¼ of a Circle: whatsoever part of a Circle any Angle bounds, that Arch is the measure or con­tents in degrees or points of that Angle, as the Arch u 8 is the measure of the Angle at K, the Center of the Arch is K. When in writing we would express an Angle in letters, we set the middle letter for the Angle that is meant, as here O R K is the obtuse Angle at K. The Complement of an Angle or Arch, is what it wants of 90 degrees or a Quadrant: or sometimes (if need require) what it wants of a Semicircle, which is twice 90 or 180 degrees.

A Circle is a plain figure, swept out by a pair of Compasses [Page 30] from the Center, which is the place where one foot of the Compasses stands fixed, whilest the other describes it; some call it a limb or perimeter: that

point of the Compasses which stands fixed, is at O the Center, the other foot describes it at the extent you set them, as here the Circle ABCD: a part of a Circle is termed a Segment.

If divers Circles be descri­bed having all one Center, they be called Concentricks, but if they have divers Cen­ters, they be called Excen­tricks.

The Circumference of any Circle consists of 360 degrees, every degree being 60 minutes.

The Diameter of a Circle is a line drawn from one side of the Circle to the other through the Center, As A C.

The Semidiameter of a Circle is half the Diameter, or the di­stance from the Circumference to the Center.

[Page 41]

Describe your Circles one within another, as you see here (makeing one center for them all when you have so done, divide it into 32 equall parts, as I have before shewed, and draw every [Page 42] other point to the inner Circle, the rest to the fourth Circle) thus the Compass is made. If you have a desire to divide it into half points, divide the whole ones into two equal parts round about, and draw them to the third Circle; if into quarters, divide your halfs into two equal parts, and draw them to the second Circle: draw none but the North and South, East and West, Northeast and Southwest, Northwest and Southeast points through the center. For drawing the points, it is but observing to make the North point a whole point, and the other seven great points will fall by course (if you observe to make a great point of every other from the North point.) It is common for a Circle without all this to be divided into 360 deg. and then you may see how 11 deg. 15′ is a point of the Compass. But I have proved that in Arithmetick, which saves me a labour here to do it.

Describe a Semicircle upon the line A C (if you are minded to have your Scale large, let your Circle be large, for the Scale will be according to the Circle) and divide it into two equal parts by the Perpendicular B 0: the one of these Quadrants divide into 8 points of the Compass, the other Quadrant, namely B A, divide into degrees, which is done by dividing it into 9 equal parts, and every one of them parts into 10 equal parts, which is 90 degrees for the Points of the Compass, they are shewed sufficiently before. Suppose your Scale were gaged fit for your turn, namely, a place to set the line of Chords, a place to set the line of Sines, and a place to set the Points of the Compasses: (there are Lines of Longitude, and Tangent lines, but we shall meddle with neither of them at this time, be­cause without them these will be sufficient with a Scale of equal parts.)

BEfore I shew how to project the Sphere, it will be necessary to understand the Circles of the Sphere.

A Sphere is a Scheme or Figure which represents the Heavens, and therefore is round exactly, though upon a Plain it doth not seem so; from whence it is evident, that the lines here drawn in it cannot be streight lines, but Circles: Now there are two sorts of Circles, namely, the Greater Circles, and the Lesser; the Greater Circles are such as go round the very Body of the Globe, and so cut through two opposite points, dividing it into two Hemispheres, and are six in number, which are these:

• 1. The Axis of the World. (From South to North Pole.)
• 2. The Horizon.
• 3. The Aequinoctial.
• 4. The Ecliptick.
• 5. The East and West Azimuth.
• 6. The Meridian.

The smaller Circles of the Sphere, are all such Circles as do not divide the Sphere of Heaven into two Hemispheres, and so are less than those that do; and they be four, namely, The two Tropicks, and the two Polar Circles, which to distinguish [Page 47] the Southermost from the Northermost, we call the Souther­most Tropick, Capricorn; the Northermost, The Tropick of Cancer; the Southermost Polar Circle, The Antartick; and the Northermost, The Artick Circle.

All Circles that cut through the two Poles of the World, are called Meridians; and are also Great Circles, because they di­vide the Sphere into two Hemispheres, cutting in two opposite points: it is certain, if a Circle goeth through two opposite points, it is as great a Circle as that globous Body can bear, and must divide that Body into two Hemispheres, as doth the Meridian A B C D.

The Poles are two opposite places in the Heavens, and are the ends (as it where) of that Line called the Axletree, which the Heavens may be imagined to turn upon: here that Line is called the Axis of the World.

The Aequinoctial is a Great Circle of the Sphere, which lieth between the Pole so equally, that its distance from either Pole is 90 degrees (the half of the distance between the Poles) and because it is so equal between them, there begins Latitude; so that whatsoever Latitude you are in, so many degrees and mi­nutes the Aequinoctial is below that part of the Heavens which is right over your head, and is called the Zenith; or from being right under you, which is called the Nadir. All Meridians cut this Aequinoctial at right Angles.

The East and West Azimuth is a Great Circle of the Sphere that cuts through two opposite points of the Heavens, namely, the Zenith and Nadir: it also cuts the Horizon at right Angles in the points of East and West; and therefore is called the East and West Azimuth: all Azimuths are great Circles and cut the Horizon at right Angles.

The Horizon is a Great Circle of the Sphere, and divides that part of the Heavens which we do not see, from that part which we do see: or it is the furthest part of the Heavens which we can see for the Body of the Sea; it is only to be seen at Sea, or upon the shore where there is no land between you and it, and so you have divers Horizons according to your motion, for as you raise one part of the World, you lay the other.

The Zodiack is a great Circle, it is the bounds of the twelve Signs: now in the middle of it is the Ecliptick Line, in which Line the Center of the Sun goeth, and passeth every day its mo­tion of Declination, till it comes to its utmost bounds, which is to the Tropicks: it cuts the Aequinoctial in two opposite points, and makes from these points an Angle of 23 deg. 30 min. which is the Angle S R t, so the Sun is never out of it. The twelve Signs divide this Ecliptick into twelve equal parts, and to every Sign theee is a Name, and a Character for that Name, and a Month to that Sign; which for your better knowledge I have here following inserted.

Six of these be Northern Signs, and are in the North part of the Zodiack; and six of them are Southern Signs, because they be in the South part of the Zodiack. The twelve Signs are twelve Constellations, the Months answering to them are agree­able. Thus much for the great Circles of the Sphere.

The smaller Circles are the Polar Circles and the Tropicks.

The Polar Circles are distant 23 deg. 30 min. from the Poles, and between them and the Poles is counted the Frozen Zones.

The Tropicks are the bounds of the Suns Declination, and they go parallel to the Aequinoctial (and 23 deg. 30 min. from it) the Tropick of Cancer being 23 deg. 30 min. to the North­wards, the Tropick of Capricorn being as far to the South­wards, which is the Suns furthest distance from the Aequator at any time.

The Meaning of the Terms used in the following Work.

THe Latitude of any place is the Distance between the Zenith of that place, and the Aequinoctial in degrees and minutes.

The Elevation of the Pole is always equal to it, and is the Poles height above the Horizon in degrees and minutes.

Altitude signifies the Height of a thing: If it be meant of a Star, or of the Sun (not being upon the Meridian) it is meant the Distance between the Sun or Star and the Ho­rizon.

Meridian Altitude is the Suns height above the Horizon, when he is upon the Meridian of that place, which is at 12 of the Clock; or at any other time when a Star is upon the Meridian either North or South.

The Suns true Amplitude of rising or setting signifies the number of degrees and minutes that the Sun riseth from the East, or sets from the West points of the Compass in the Horizon, as it is found in the Sphere.

The Suns true Azimuth of rising or setting signifies the di­stance (in degrees and minutes) of the Suns rising or setting (in the Horizon) from the North or South points in the Compass, as it is found in the Sphere.

There is the true Amplitude of the Suns rising and set­ting, and the Magnetical Amplitude; the Magnetical Am­plitude is the degree of the Compass that he is observed to rise or set upon, and is Called Magnetical Amplitude from the Load-stone: The Magnetical Azimuth is Called so for the same reason, and is the Suns rising from the South or North.

Longitude is the Easting or Westing between two places in degrees and minutes, if in sailing.

Departure from the Meridian is the Easting or Westing between two places in miles or leagues.

Latitude, as I have said, is the Distance of the Aequi­noctial from the Zenith of that place in degrees and minutes, or the Elevation of the Pole; but Difference of Latitude is the Northing or Southing that a Ship makes in any Di­stance run, or between two places, either in degrees, leagues, or miles.

Declination (if it be meant by the Sun, or any Star that is near the Aequinoctial) is the nearest distance between the Sun or Star and the Aequinoctial.

Stars that are nearer the Poles, their Declination are their nearest Distance to the Poles.

The Suns place in the Ecliptick is his nearest Distance from the place where the Suns Parallel of Declination cuts the Ecliptick to the next Aequinoctial point, which is ♈ or ♎.

There be seven Planets which I have here inserted, with their Characters under their Names.

I have noted already in pag. 3 of this Book, the quantity of degrees that make an hour in time, of minutes that make an hour, the quantity of degrees and minutes that are in a point of the Compass and the like: if you want to know, look there.

I'll now proceed to the Work of the Sphere in Plano, and I will handle no more than is necessary for our use in Navi­gation.

I had thoughts to have made a Table of the hard Terms of Art used (which our vulgar Capacities do not ordinarily com­prehend) with their meaning, but then I considered what my own Method would be; and so I would not trouble my self with that at the Beginning of my Book: There are no intricate Terms used, but the meaning of them is commonly shewed in the Beginning of the Book.

Latitude and Declination given, to find the things following in the several Questions.

Latitude 50 deg. 00 min. Northerly.

Declination 13 deg. 15 min. Northerly.

Rule. All distances of the Sphere that you desire to know, must be extended to the Arch of a great Circle, for by great Cir­cles is the Sphere measured.

The Meridian Altitude of the Sun being given, and the Latitude, to find the Suns Decli­nation, as also all the other things before found.

FOR the doing of this, project your Sphere (as before was shewed) that done, I consider the Meridian Alti­tude of the Sun is given to be 52 deg. 30 min. I will take 52 deg. 30 min. from the Scale of Chords, and fix one foot of my Compasses in the South point of the Horizon at A, and extend the other foot upwards upon the Meridian, which falls at B; then I conclude the Sun is upon the Meridian, in the point B: Take the Distance between B and the Aequinoctial, which is B C, and apply it to the Scale of Chords, and so much as it is, so much is the Declination of the Sun; and you see it is Northerly, for the Sun is to the Northwards of the Aequinoctial. If you have a desire to find any thing else in this Sphere, draw the Parallel of the Suns Declination (thus found) and work with it as before in other Examples; also draw the Ecliptick and Tropick.

Many other things might be done, but for Brevities sake I omit them, judging this to be as much as we have occasion for in Navigation, which is the thing I aim at. If I have committed a fault in being too tedious in these things, pardon me, and remember it was my Beginnings; riper years (which may bring more ability, it is probable) will bring forth some­thing more new and admirable. Be charitable.

Difference of Latitude and Course given, to find the Distance run, and the Departure from the Meridian.

IN this Question we have the Course 5 points from the Me­ridian towards the West, which is 5 times 11 deg. 15 min. or 56 deg. 15 min. for 5 times 15 min. is 75 min. or 1 deg. 15 min. and 5 times 11 deg. is 55 deg. to which add the 1 deg. and 15 min. and the whole is 56 deg. 15 min.

To protract this Question or any other of this nature, I wish you to observe the end of your Book, that lies from you to be the North part of your Book, then that to the right hand will be the East part, that to the left hand the West, and that right to you the South part: The Reason why I wish you to observe this Method, is, that so in any Travis you may set every Angle the right way, which must be observed then, and is good to be always used.

First draw the South line B D, let B be the Lizard, then set off from B 60 miles upon the South line (taking it from a Scale of equal parts) which is your Southing B D. Forasmuch then as B D is the Difference of Latitude that I have made since I set from B, D must stand in the parallel that I am in; and be­cause I have sailed to the Westwards of the Meridian D B, I

will raise a Perpendicular which is a West line from D, and draw it at length, which is D A. This done, take an Arch of 60 deg. from your Scale of Chords, and (making the place you set from, the Center) describe the Arch D O, and upon it set off 56 deg. 15 min. which is D O 5 points; then by B and O draw the sine B O, till it intersect the line A D, which it doth in A: Now Forasmuch as D A is parallel in the Latitude the Ship was in after her sailing, and D B A an Angle of the Course from the Meridian, B A must intersect the West line in the place where the Ship is; then the Distance run is B A, and the Departure from the Meridian is D A, which take off severally, and apply to the Scale of equal parts, and set down what they come to: Here I find them as followeth:

But besides this Geometrical way of resolving Triangles, there is an exacter way by the Tables of Sines, Tangents, and Logarithms, which is done thus,

Course and Difference of Longitude in miles given, to find the Difference of Latitude and Distance.

LEt A represent the place your Ship is in, after she hath sailed; from whence draw the East line A D, then take 90 miles from your Scale of equal parts, and set off upon that line from A, which falls at D.

From D raise a Perpendicular or North line which is D B: Now, forasmuch as it is said, that the Ship sailed S W b W, [Page 77] which is 5 points from the South towards the West backwards, that is N E b E, 5 points from the North towards the East, or 3 points from the East towards the North. I will take an Arch of 60 deg. from my Scale of Chords, and from A as a Center I'll describe the Arch O N, and from N I'll set 3 points upon that Arch which reacheth to O, and by A and O, I will draw

the line A B till it cut the North and South line in B; then is B D the Difference of Latitude, and A B the Distance run, which measure and set down.

The Difference of Latitude is 60 miles.

The Distance run is 108 miles.

Here to find the Distance run, I made use of the things that were given me, but for brevities sake that I might shew the taking of a Complement arithmetical, I used Radius with the Distance before found, to find the Difference of Latitude.

But if you would not do so, then the things given will do as well, and the Proportion holds thus.

As Sine comp. A (which is Sine B) is to A D so is Sine A to B D.

Course and Distance run given, to find the Difference of Latitude and Departure from the Meridian.

LEt A D be a Meridian line, A the Latitude of the Lizard, D A B the Angle of the Course, set off by the Arch of 60 deg. N O s (N O being 3 points) A B a line drawn from the Latitude of the Lizard A by O, and the Distance that the Ship hath sailed set upon it, and no more, namely, 108 miles; then I conclude that the Ship is at B, from B let fall the Perpendicular B D, which cuts the Meridian in D, and leaves D A for the Difference of Latitude, and is it self the Departure from the Meridian; measure them as you have been shewed before, and apply them.

Difference of Latitude D A 90 miles.

Departure D B 60 miles.

Distance run and Difference of Latitude given, to find the Course and Departure from the Meridian.

FIrst draw the North and South line A D at length: Let A be the place in the Latitude of 50 deg. take the Difference of Latitude 90 miles from the Scale of equal parts, and set it off North from A, which falls at D: Now because D is as far to [Page 81] the Northwards of A, as the Difference of Latitude which you made from A is, therefore if you draw the West line D B at length, it must cut through the place where the Ship is: I look upon the Question again, and I find that the Distance run since the Ship set out is 108 miles; take 108 miles from the Scale of equal parts; and because the Ship set from A,

fix one foot in A, and where the other foot intersects, or cuts the parallel that the Ship is in D B, there the Ship must be; and thus if her Difference of Latitude be A D 90, and her Di­stance run A B 108 miles, B D must be the Westing, and n O [Page 82] the measure of the Course from the Meridian of Northwesterly, or the Angle D A B.

Take the Difference of Longitude D B, and apply it to your Scale as before.

Take O N the measure of the Angle of the Course, and apply it to the Scale of Chords (if you are minded to see her Course from the Meridian in degrees and minutes) to the points of the Compass, if you desire to know it in points.

Note that A is the Center of the Arch O n 60 deg. It is plain, that if an Arch of 60 deg. described, and any number of points or degrees that is set upon it, be an Angle of those points set down (as it hath been in every Example before) then whatsoever you take from an Arch of 60 deg. is fit to apply to the Scale.

I find the Course to be North 33 deg. 45 min. Westerly, or N W b N. I find the Departure from the Meridian to be 60 mil.

Departure from the Meridian and Distance run given, to find the Course and Difference of Latitude.

HEre the Course from the Meridian will be the Complement of the Course there, and the Difference of Latitude will be the Difference of Longitude, or Departure from the Meridian there.

But instead of taking the Logarithm answering to the number here, take the Logarithm answering to ten times the number, and so you shall always have your sides come out to the tenth part of a unite; I say it will come out in tenths, for you must divide by 10, to bring it into miles after it is done.

Now instead of multiplying your given sides by 10, do but set a Cipher more to the number and it is done, as now for 60 set down 600, which is ten times 60; for 108 set down 1080, which is ten times 108; and thus for any number that will be comprehended in the Radius of the Tables.

Divide these tens by 10, and it gives the miles con­tained [...] in the side D A the Difference of Latitude, which is 89 miles 8 tenths of a mile, whereas before it was 90 miles, whis is 2 tenths of a mile more: Now when you have the sides thus found in tenths or Centisms; work for the Angles, and you will find them to come out roundly alike: But if you desire more exactness, work then in Centisms, pro­vided, that the numbers in Centisms do not out-run the Tables.

A Centism is the hundredth part of an unite, or of any thing; and if I would put the number 60 into Centisms, I will set two Ciphers behind it, and it is the same as before; I multi­plied [Page 85] 60 by 100, for it is 6000, the like is to be understood of any other number.

I would work to the nearest Fraction for the Angles here, but I conceive it to be no way beneficial, and therefore I'll refer it to the work of the Sphere, where it is of [...]uch more use; this finds the Angles to a minute, which is near enough.

It is like that some will be so curious, that this way of finding things to the tenth or hundredth part of unit will not suffice them, but they would have the real number answering to the Logarithm to the 1000^th. part of a unite; because I find it of no real consideration or use in Navigation, I omit that here, and desire such to look in Book 1. Page 11. of my Fathers Tri­gonometry, and there it is plainly shewed.

Difference of Latitude and Departure from the Meridian given, to find the Course and Distance run

FIrst, draw a North and South line, and upon it set off the Difference of Latitude, which is D B 60 miles; then from the Northermost part of your Difference of Latitude, which is D, raise a Perpendicular, East and West line, and set off your Departure from the Meridian upon it, from D, which falls at A: Then from A draw A B, for the Distance run; and because B is the place you set from, let it be the Cen­ter

[Page 88] of the Arch of 60 deg. s O D, on which the measure of the Course is O D, which must be measured, as you have been shewed, and also the Distance run A B.

The Course I find to be N W b W.

The Distance run 108 miles.

Difference of Latitude given alone, Difference of Longitude and Distance run in one intire Sum, I demand the Course and Sides several.

IN this case (because it would take up more room than my Book would hold) I'll make every 10 upon my Scale of [Page 90] equal parts 20, then every one of the small divisions will be 2. For the protracting of it Geometrically, first draw the North and South line B R at length, and upon it set off your Diffe­rence of Latitude, which is D B, make B the place of your Ships setting out, because it is the Northermost Latitude, and your Difference of Latitude is Southerly, raise a Perpendicular (or

East and West line) from D, and it must cut through the place where the Ship is; because she is in that Parallel or Latitude: upon this East and West line, from D set the Distance run, and Departure together, which is 197 miles 8/10, and it reacheth to E: by E and B draw the line E B, and divide it into two equal parts by a Perpendicular (you see it is a Perpendicular by the Arches h I and f g) draw it from P, till it cuts the Parallel E D, which it doth in A; from A then draw a line to the place you proposed to your self the Ship set from, which is B; now the length of that line, namely, A B, shall be the Distance run, and A D the Departure from the Meridian.

A TRAVIS.

FOr the doing this Geometrically, I have the Distance run, and the Course given; to lay down the Triangle A C B, make A the place you set from, and lay it down as was shewed in an Example of that nature before, Quest. 3.

Then it is said, from thence I sailed S S E, till I altered my Latitude 1 deg. or 20 leagues; which signifies that from C is the place I altered my Course; therefore from C draw a South line, which must be parallel to A B s; Consider what you have given, and work as hath been shewed in an Example of that nature, Quest. 1. be sure to set your Angles the right way: After that Angle is worked,

I find my Ship is at e, then from e, draw the North line e G parallel to the first North and South line (for the Course is Northerly) and work your Tri­angle as hath bin shewed in an Ex­ample of that kind Quest. 4. where the Diffe­rence of Latitude and Distance run is given, and you will find your Ship at F; let fall a Perpendicular from F upon the first Merid. line (which falls at R) and the length of it is the Departure from your first Meridian that you have made; likewise it cuts that Meridian, so that it leaves the Difference of Latitude R A, that you have made since you set out; then if you draw the line A F, it is the Distance that your Ship is now from the place she set from; and if you describe your first Arch of 60 deg. from s to O, and extend your side A F to it, it gives the measure of the Course s e O: Note that your Difference of [Page 94] Latitude is Southerly, and therefore you must cast it into de­grees and minutes, and subtract it from 32 deg. 20 min. (your first Latitude) and the Remainder is the Latitude you are in; whereas if you had been in a Southern Latitude, and had differed your Latitude farther Southerly, you must have made your Latitude so much more, by adding your Difference of La­titude, as here, if you had gone to the Northwards, you must have done so, to have known your increase of Latitude Norther­ly: The like is to be understood in any case else of this nature. Here following I have answered the demands.

The departure from the first Meridian is F R 56 leagues.

The Course is the measure of the Arch s e O, which is E S E 9 deg. 8 min. Easterly.

The Distance that I am from the Port I set is A F 57 leagues.

And thus you see you sailed from A to C, and then altered your Course, and sailed from C to e, and then from E to F; the like is to be understood of any Travis else. And thus if you had a Travis of 24 hours, you might find what Distance your Ship is from the place she was the day before, and he other things, as Difference of Latitude, the Latitude you are in, the Depar­ture, the Course made good upon a streight distance; so that you need not set down every thing to every alteration in your Reckoning which is endless and fruitless.

For their bearing one from the other.

Imagine (as I said before) that A B were the edge of a Scale cutting upon the two places. Look for that Corner [Page 109] that is nearest without the edge (which is S) fix one foot of your Compasses in S, and with the other sweep the edge of your Scale, which will be S G, keep fast your Scale, and from the Corner S run that foot of your Compasses (keeping it at the same Distance from the Scale by the other foot running by the edge of the Scale) till you come to the next side with it (which in this ex­ample fails at F) then take the Distance E L, and apply it to the graduated Square, either from D or E, and as many points as it is there, so many Points doth the Lizard bear to the East­wards of the North, from the Maderas, which is here two points and almost a quarter. The same will be done from a Corner farther from the Scale, and this is evident; for by the Flower de luys you see that S L is a North line from S, and a line drawn from S to F is parallel to B A, or runs upon the same point to the Eastwards of the North from S. Therefore whatsoever line S F is, the same is B A, the like is to be understood of any other example, as, suppose we had not found the bearing of the Lizard from the Maderas, and would find the bearing of the Maderas from the Lizard. Because my Scale runs no farther than B, I cannot use the Corner at S (which is nearest) and the rest are upon the graduated Meridian, therefore I will use the Corner at L, as I did the Corner at S; but because it is the contrary thing, I'll run the contrary way, namely from L to H, then is L H and F S parallel, and because S L is the Meridian to them both, therefore are H S and L F equal, so that if S H be applied to the graduated Square, it is as many points to the Westwards of the South, as L F is to the Eastwards of the North, therefore the Maderas bear from the Lizard South South-west almost a quarter of a point Westerly: if it had been from any other Corner, it would have been the same, the like for any other.

For the Departure from the Meridian between two Places.

Suppose Xwere one Place, and (‖) the other, and I would know the Departure or Difference of Longitude between them.

Fix one foot of your Compasses in (‖), and extend the other to sweep the Meridian nearest short of the other place, which is the Meridian V, that Distance apply to the graduated Meridian, and see how much it is, then take the Distance from the other place (which is X) to sweep the nearest part of the same Meridian, and apply that to the graduated Meridian, see how much it is, and add it to the former, and it makes the whole Departure between the two places, X and (‖), convert it into miles or leagues, according as you desire; in most Plats you have a Scale or Leagues, which is necessary for small Distances.

For the Difference of Latitude between two Places.

The Difference of Latitude between two places is taken af­ter the same manner from a Parallel, as the Departure is from a Meridian Line; as suppose I would take the Difference of Latitude between X and (‖), first, take the nearest Distance between X and the Parallel that T is in, and see how much it is. Then take the nearest Distance between (‖) and the same Paral­lel, and see how much it is, and add them together, and that convert into miles or leagues according as you desire.

To find what Latitude any place lies in, take the nearest Distance between that Place and the next Parallel with your Compasses, and look where that Parallel cuts the graduated Me­ridian, and carry your Compasses with that distance there (one foot) and observe to set the other that way as the place is, either to the Northwards or Southwards of that Meridian, and see upon what degree and minute of a degree it falls, and that is the Lati­tude of the place.

As, suppose I would know the Latitude of the North side of the Maderas, I will fix my Compasses, one foot in B, and extend the other to sweep the Parallel that S is in, that Distance I will carry [Page 111] to the place were the Parallel cuts the graduated Meridian which is at 34, and fixing one foot there, I will see where the other falls to the Southwards towards L (because it is the Southwards of that Parallel) and I find it falls in the Latitude of 32 deg. 30 min. from whence I conclude, that the Northermost part of the Maderas lies in the Latitude of 32 deg. 30 min.

The way that the Square E C D is divided into Points, is on this manner: First, I draw the line O C, then the other lines for half-points, and quarter-points, then I describe the Circles, making the outmost come to the innermost of the lines (but whether it be or no is no great matter) the Circles being divided into Points, half-points, and quarters, I lay my Scale by the Center of these Circles, which is O, and the divisions on the Circles, & where my Scale cuts upon the outer line, from thence I draw my Points, half-points, and quarter-points, as you see here in the following Blank; and the Points upon the sides of the Square thus set off are the Tangents of the Points upon the Arch: for when from the outer­most edge of an Arch a Perpendicular is raised from a side that bounds that Arch (as the side N 4 is,) and the other side that bounds the same Arch be extended without the Arch to intersect it (as the line O C intersects N 4 at 4,) that Perpendicular is the measure of that Arch, either in degrees or points; as here N W is 45 degrees or 4 points, and so is N 4. And thus you may divide a Tangent line for degrees or points, from an Arch divided into degrees or points.

To set off any Course from a Place assigned upon a Blank.

Let the Place assigned be X, from which I would draw a N W b N line.

Take three points from the graduated Square (making D or E the place you take it from) and fix one foot of your Compass in the next convenient Corner to X (which is at M) and set off the three points which fall at K; then from the next Corner upon the same Meridian that M is, (and to the Southwards of it) which is at P, lay your Scale to K, then your Scale lieth N W b N from [Page 112] P, and a line Parallel to the Scale from X must be a N W b N line, the like for any Course else, observing the quarter of the Compass, it is to be set off in, as also whether it be nearer the East or West than the North or South; for if it be nearer the East or West than the North or South, you must do the same from a Parallel, as here you did from a Meridian; for the side of a Square is but the measure of four points, which is but half the points between a Meridian and a Parallel. This way of work­ing may seem hard and tedious at first, but you will soon find that it is free from mistakes, and both exact and easie, if you practise it.

Two Sides with an Angle opposite to one of them given, to find the other Angles and Side.

FIrst draw a North and South line white, and then from that set off the Northermost Ships Course, make the Rock or Place setting out the Place in the North and South line that you draw that Course from, which is C, upon this Course set off 38 leagues (because the Question saith you sailed 38 leagues upon it) and extend the side C A to the Arch of 60 deg. at t, and set off five points from t to S, and draw S C a white line, which is W S W Course: (for it is five points between the W S W and the N W b W) this done, take 58 leagues from the Scale of equal parts, and fix one foot of your Compasses in A, and where the other intersects the Course B C (which it doth in B) there is the other Ship, then to measure the Angle of the Ships bearing one from the [Page 114] other it is B, and B C is an E N E line; extend B A to the Arch of 60 deg. whose Center is at B, and see how many degrees or points it is more Northerly than an E N E line, and so the other Ship (namely, the Ship at A) bears from the Ship at B;

then take the length of B C, and apply it to your Scale, and see how many leagues or miles it is. I have wrought this in leagues, but I will work the rest in miles, because it is more exact. I find that the Ship at A bears from B, N E b N 45 min. Easterly, the Distance run of the Southermost Ship is 70 leagues.

The three Angles of a Triangle given, with one of the Sides, to find the other two sides.

FIrst draw your North and South line B s. Let B be the Head-land. From B then draw a W S W line (by describing the Arch of 60 deg. s D from the Center B, and setting it off as hath been shewed) upon this W S W line set off the Ships run 58 miles B O: From O as a Center describe the Arch of 60 deg. R n, and extend the side O B to it: Now because the Question saith that B O is a W S W line from B, then O B must be an E N E line from O. Also because the Question saith, that the Ship that met her sailed N W, the side A O is a N W line from A, which is a S E line from O to A. Now there are 6 points between an E N E and a S E Course: there­fore from the line D B extended upon the Arch of 60 deg. from B, set off 6 points or 67 deg. 30 min. which is n R, and [Page 117] draw O R a white line: Then because the Question saith that the place the other Ship set from, bears from the Head-land B S S W. Draw a S S W line from B, which is B A, and where

it cuts the line O R, which is in A, there will be the place the other Ship set from, and A B will be the distance of the Head-land and that asunder, A O will be the distance that the Ship hath sailed that came from it, which I find as followeth.

The distance between the two places is A B 58 miles.

The distance that the Ship sailed is A O 44 4/10 miles.

If you have occasion to find the Latitude the place at A lies in, let fall the Perpendicular A N upon the South line that comes from B, and it cuts it in N, and leaves the Difference of Latitude B N, see how many miles it is, and subtract it from 50 deg. and you have your desire, measure A N, and you have the Westing that A lies from the Head-land B.

Two Sides and their contained Angle being given to find the third Side and the other Angles.

FIrst draw the North and South line A S. Let A B repre­sent the Port you set from: Then from A as a Center, describe the Arch of 60 deg. F S D: From S to the East­wards set off one point: for N b E the Eastermost Ships Course draw the line A D at length, and set the first Ships distance 70 miles upon it, which is A B, then from A draw a N W line 4 deg. 45 min. Westerly, and set off the Wester­most Ships distance run upon that, which is 89 miles, and it reacheth from A to C, draw the line C B from the ends of the other two sides, which is the Distance between the Ports (for the ends of each Ships run must be the Ports) then is A C B the Angle of the bearing of the Eastermost Port from the Westermost, or A B C the bearing of the Westermost Port from the Eastermost, C A is a S E line, 1 deg. 45 min. Easterly from C, see how many points are contained between the C B and C A, by the way that is shewed in the Question before, or because B A is N b E [Page 120] line (from B) you may find the Point of the Compass that B C runs upon from B, you need find it but one of these ways, and the opposite point must be the other, as I have shewed before, then take the length of the side C B, which is the distance of the Ports asunder, I find their distance and bearing to be as followeth.

The Distance between the two Ports is C B 79 miles.

The Bearing of them is East 5 deg. 35 min. Northerly, or West 5 deg. 35 min. Southerly.

The three Sides of a Triangle given, to find the Angles.

FIrst draw the North and South line A e, make A the place of the Ships setting out, set off a N b E Course upon the Arch of 60 deg. I e, which is e D (because the Eastermost Ship sails N b E) likewise set off her Distance which she sailed upon that Course, which is A B 70 miles, then take the other Ships

[Page 123] Distance, which is 89 miles, and fixing one foot of your Com­passes in A, with the other describe the Arch at C: Lastly, take the Distance of the two Ships asunder 78 miles 9/10, and fixing one foot of your Compasses in B, cross that Arch, and from the place of intersection draw the sides of the Triangle C A and C B: then measure the Angles from an Arch of 60 deg. as hath been shewed before; and to know them, consider that A B is a N b E line, so that A F will be upon a Course so many points, or degrees, and minutes from a N b E, as there are points, or degrees and minutes in the Arch D F. Secondly, for the bearing of the Ships, which is the Angle at B: consider that B A is a S b W line from B, and then the degrees and minutes, or points contained in the Arch O R, is so many degrees and minutes, or points from the S b W.

QUESTION V.

SUch Questions as these are used in situating of places that are in sight one of the other; or, as a man sails by a land, he may take the Headlands bearing one from the other, and also the Imbays as well as the Head-lands, if he be not too far out.

This Question is wrought as the second Question, for here you have the bearing of the places you note in both Sta­tions, and also the Course that you steer from one Station to the other, with the Distance; which is two Angles and a Side, to find the other two Sides, which is your Distance at each Station from the places.

For the protracting of this Question; from your first Sta­tion draw a line upon those several Courses, as the Question saith the places bear; As, suppose the first Station be A, from A draw A I a N N E line, and (mind) that line must be cut upon the Eastermost Head-land, because the Question saith it bears so.

Next draw a N W line from A, which is A C, and cuts the the Westermost Head-land.

Lastly, draw out a South line from A, which is A D, and cuts the Island.

Then from A draw an East line (for the Course you steered) and set off 8 miles upon it, which is A R, from R draw a line to every place, as the Question saith they bear; Then (namely) from R to the Eastermost Head-land, draw [Page 128] a N E b E line 5 deg. Easterly: from R to the Westermost Head-land draw a N N E line, and from R to the Island an E S E Course. Now because the lines drawn from each Station cut the Head-lands and Island; it is evident that the

[Page 129] Point of their intersecting one another must be the places, (thus) A I and R I are the lines drawn upon the Course that lead from the Stations to the Eastermost Head-land, and cut in I, then is I the Headland. A O and R O are the Courses that lead to the Westermost Head-land from both Stations, and intersect in O, therefore O is the Westermost Head-land.

Lastly, A s and R s are the Courses that lead from the two Stations to the Island; and therefore s is the Island, for there they cut.

Draw a line from every one of them as from s to O, and from O to I, from I to s, and take the distance of them asunder; then for their bearing one from another, draw a North and South line from each place, and by an Arch of 60 deg. measure it, as hath been shewed.

Two Sides and a contained Angle being given, to find the third Side and the Course (that each Side runs upon) pro­vided that no Course be named, only the half of the Compass that you sail in (or the quarter which is most commonly) and the Difference of Latitude that is made between the extent of the two Sides.

THe place the Ship set from, I make to be B, take off 60 deg. from your Scale, and describe an Arch, making B your Center, as the Arch T R: then I take 50 deg. from my Scale, and set off somewhere upon that Arch, namely, from T to R, and from B, upon the Course that leads to T, I set off 40 leagues, my Westermost Ships Distance, and so conclude my Westermost Ship is at C: then from B upon the Course that leads to R, I set off my my Eastermost Ships Di­stance run, which is 50 leagues from B to A. Lastly, draw A C, which is the Distance of the Ships at the end of their sailing.

Then because the Question saith the Eastermost Ship was 15 leagues more Southerly in Latitude, than the Westermost; I will take 15 leagues from the Scale of equal parts, and fix one foot of my Compasses in A, and with the other describe the Arch I S N, and draw the line from C, by the upper edge of the Circle at length, which must be an East and West line, because A is just 15 leagues to the Southwards of the nearest part of it: then let fall a Perpendicular from B upon C N, and extend it as far as the Arch of 60 deg. which you first described [Page 132] T R, and it cuts in O, this must be a South line from B, namely B S; then measure upon the Arch of 60 deg. from O to R, for the Eastermost Ships Course, and from O to T for the Westermost Ships Course, as you do in any other Question. I find them here as followeth,

The Eastermost Ship sailed S b E 4. d. 40 m. Easterly.

The Westermost Ship sailed S W 11 deg. Southerly.

The Distance asunder is C A 39 leagues.

Two Sides of an oblique Triangle being given in one Sum, and the other Side alone, with an Angle opposite to one of the Sides, to find the Sides several, and the other Angles.

LEt B A be the Parallel line the Ports be in. Let A be the Eastermost Port, B the Westermost, then must A B be 92 miles: from B set off a N E b E line, which is B C O, set off both the Ships runs upon it, which is 159 miles B C: then draw C A, which is the other side of the Triangle B C A, and divide it into two equal parts by a Perpendicular (as you see the Perpendicular D S doth, in the point R) draw this Perpendicu­lar from one side of the Triangle to the other, as R I; then from I draw a line to A.

Then forasmuch as A R is the half of A C or equal to R C, and I R is a Perpendicular to A C from R, I A and I C must be equal, and if B I C be the length of the two sides 159, B I and I A are equal to them, so that B I is the Distance that the Westermost Ship sailed, and I A is the Eastermost Ships run, measure them, and set them down; also I A B is the Angle from the West toward the North, that the Easter­most Ship hath steered, I have found them to be as fol­loweth.

The Eastermost Ship sailed North 7 d. 20 m. Westerly.

The Westermost Ship's Distance run is 102 miles.

The Eastermost Ship sailed 57 miles.

The three Sides of a Triangle being given in one intire Sum, and the Angles apart, to find the Sides apart.

FOr the resolving of this Question, I will suppose the side that the Westermost Ship sails upon to be 30 leagues, and so imagine you have the three Angles and a Side given, as you have in Quest. 2. We will protract this after the directions there given, it being so laid down: I find by the Plain Scale, that these sides are as followeth:

The side C N is given to be 30 leagues.

The side N B is found to be 42 leagues.

The side C B is found to be 24 leagues.

Three Sides of a Triangle given, to find the Center.

I Have applied this Question to sailing here, but before in this Book I have used it in a thing which is very proper for it; and indeed it may be of use many times, which makes me give it place here. I shall say nothing of this Question by the Plain Scale, for it is done, as before you see: but we will do it by the Tables of Tangents and Logarithms.

Here I have given the side M er 40 leagues, the side M W 50 leagues, and the Angle at M, namely, its Complement to 180 deg. R M er 78 deg. 45 min. to find the Angles M er W and M W er:

Subtracted from 39 deg. 22 min. and the Remainder is M W er 34 deg. 50 min. added to it makes M er W 44 deg. 34 min.

But now you will demand of me, how we shall find any thing in the Triangle W F er, being there is but one thing known in it, namely, the Distance of the Eastermost Ship from the West­ermost er W, I'll tell you.

Mr. Euclid proves this, which makes me forbear it.Mr. Euclid (Book 3. Prop. 20.) saith, That in a Circle an Angle at the Center is double to an Angle at the Circumference, (provided) that both the Angles have to their Base one and the same part of the Circumference. The Center is F here, the Triangle W F er and W M er have W er to both their Bases: then must B F A be double to I M O, or C B A double to D M I (that is to the Arch I O D) so that if you double D O I, 101 deg. [Page 143] 15 min. it makes C B A 202 deg. 30 min. that subtracted from a Circle 360 deg. leaves the Triangle W F er: It is evident then that W F er is double to R M I because R M I is what I O D wants of a Semicircle, as W F er is what C B A wants of a whole Circle, which is double to it.

QUESTION X.

FOr the doing of this Question, First draw a N W b N line, and set off 6 leagues upon it which is A I; then draw a N E line, and set 8 leagues off upon it, which is A D; draw the side I D, then consider that as A I is to A D, so is the West­ermost Ships run to the Eastermost; so that the sides A D and A I extended so far that a line of 12 leagues long, drawn parallel to D I, will just touch the extended sides which must cut them in the places were the Ports are, namely, A B in B, and A I in C: measure the sides C A and A B, and also the Angles of the Ports bearing A B C, or A C B and count them as hath been shewed before.

I find the side A B the Eastermost Ships run to be 10 leag. 7/10. I find A C the Westermost Ships run to be 8 leagues. The Westermost port bears from the Eastermost West, 4 deg. 14 min. Southerly.