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

Geometrical Problems.

PROBLEM I. Ʋpon a Right Line given, how to erect another Right Line which shall be perpendicular to the Right Line given.

THE Right Line given is AB, upon which from the Point E it is required to erect the Perpendicular EH. Opening your Com∣pass at any convenient distance, place one Foot in the assigned Point E, and with the other make the two Marks C and D, equal on each side the Point E; then opening your Compasses again to any other convenient distance wider than the former, place one Foot in C, and with the other describe the Arch GG; also (the Compasses remaining at the same distance) place one Foot in the Point D, and with the other describe the Arch FF. Then from the Point where those two Arches in∣tersect or cut each other (which is at H) draw the Right Line HE, which shall be Perpendicular to the given Right Line AB, which was the thing required to be done.

[illustration] geometrical diagram

PROBL. II. How to erect a Perpendicular on the end of a Right Line given.

LEt AB, be a Line given, and let it be required to erect the Perpendicular AD. First upon the Line AB, with your Compasses opened to any small distance, make five small Divisions, beginning at A, noted with 1, 2, 3, 4, 5. Then take with your Compasses the distance from A to 4, and place one Foot in A, and with the other describe the Arch e e: Then take the distance from A to 5, and placing one Foot of the Compasses in 3, with the other Foot describe the Arch h h, cutting the former Arch in the Point D: Lastly, from D draw the Line DA, which shall be perpendicular to the given Line AB.

[illustration] geometrical diagram

This operation is grounded upon this Conclusion, viz. These three Numbers 3, 4, and 5, make a Right-angled Tri∣angle, which is very necessary in many Mechanical Operations, and easie to be re∣membred.

PROBL. III. How to let fall a Perpendicular upon any Point assigned, upon a Right Line given.

THe Point given is C, from which Point it is required how to draw a Right Line which shall be perpendicular to the gi∣ven Right Line AB.

[illustration] geometrical diagram

First from the given Point C, to the Line AB, draw a Line by chance, as CE, which divide into two equal parts in the Point D. Then placing one Foot of the Compasses in the Point D, with that distance DC, describe the Semicircle CFE, cutting the given Line AB in the Point F. Lastly, If from the Point C you draw the Right Line CF, it shall be a Perpendicular to the given Line AB, which was required.

PROBL. IV. How to make an Angle equal to an Angle given.

LEt the Angle given be ACB, and let it be required to make another Angle equal thereunto.

First draw the Line EF at plea∣sure; then upon the given Angle at C (the Compasses opened to any di∣stance) describe the Arch AB; and also upon the Point F, the Compasses unaltered, describe the Arch DE; Then take the distance AB, and set the same from E to D; Lastly draw the Line DF: So shall the Angle DFE be equal to the given Angle ACB.

[illustration] geometrical diagram

PROBL. V. A Right Line being given, how to draw another Right Line which shall be parallel to the former, at any distance required.

THe Line given is AB, unto which it is required to draw another Right Line pa∣rallel thereunto, at the distance AC or DB. First open your Compasses to the distance AC or BD; then placing one Foot in A, with the other de∣scribe the Arch C; also (at that distance place one Foot in B, and with the other describe the Arch D. Lastly, draw the Line CD, that it may only touch the Arches C and D: So shall the Line CD be parallel to the Line AB, and at the distance required.

[illustration] geometrical diagram

PROBL. VI. To divide a Right Line into any number of equal Parts.

LEt AB be a Right Line given, and let it be required to divide the same into five equal Parts.

[illustration] geometrical diagram

First, From the given Line A, draw the Line AC, making any Angle from the end of the given Line which is at the Point B. Then draw the Line BD equal to the Angle CAB. Then from the Points A and B, set off upon these two Lines any Number of equal parts, being less by one than the Parts into which the Line AB is to be divided, which in this Example must be 4. Then draw small Lines from 1 to 4, from 2 to 3 twice, and from 1 to 4, &c. which Lines crossing the given Line AB, shall divide it into five equal Parts, as was required.

PROBL. VII. A Right Line being given, how to draw another Right Line parallel thereunto, which shall also pass through a Point assigned.

LEt AB be a Line given, and let it be required to draw another Line paralle thereunto, which shall pass through the given Point C. First, Take with your Com∣passes the distance from A to C, and place∣ing one Foot thereof at B, with the other describe the Arch DE; then take in your Compasses the whole Line AB, and place one Foot in C, and with the other describe the Arch FG, crossing the former Arch in the Point H: Then if you draw the Line CH, it shall be parallel to AB, the thing required.

[illustration] geometrical diagram

PROBL. VIII. Having any three Points given which are not scituated in a Right Line, How to find the Center of a Circle which shall pass directly through the three Points given.

THe three Points given are A, B, and C; now it is required to finde the Center of a Circle whose Circumference shall pass through the three Points given.

[illustration] geometrical diagram

First open your Compasses to any distance greater than half the distance between B and C; then place one Foot in the Point B, and with the other describe the Arch FG; then the Compasses remaining at the same distance, place one Foot in the Point C, and with the other turn'd about make the marks F and G in the former Arch, and draw the Line FOG at length, if need be.

In like manner open your Compass at a di∣stance greater than half AB; Place one Foot in the Point A, with the other describe the Arch HK: Then the Compasses remaining at the same distance, place one Foot in the Point

C, and turning the other about, make the marks HK in the former Arch. Lastly, draw the Right Line HK, cutting the Line FG in O, so shall O be the Center, upon which you may describe a Circle at the distance of OA, and it shall pass directly through the three given Points ABC, which was required.

PROBL. IX. How to describe a Circle in a Triangle, that shall only touch the three Sides; and to find the Centre.

LAy down the Triangle ABC, the three Sides equal; then divide the Sides of the Triangle AB in two equal parts, as at E, and draw the Line CE; and likewise di∣vide BC, and draw the Line AD; and where they cross one the other, as at O, that is the Center: Therefore put one Point of the Compasses in the Center O, and extend the other to either side, and describe the Cir∣cle GF, which will only touch the Sides A BC of the Triangle.

[illustration] geometrical diagram

PROBL. X. How to lay down a Triangle in a Circle, and to find the Center of the Circle in the Triangle.

DRaw the three Sides of a Triangle AB C, it is no matter if they be equal or not; then put one Foot of your Compasses in the Point B, open the other to more than half the length of the greatest side, as to C; and with that distance describe the Arch FHDG; and so removing the Compasses to C, cross the former Arch at F and D, and draw the Line DF. Again, the Compasses at the same distance, put one Foot in A, and describing a small Arch, cross the for∣mer Arch at H and G; and laying a Ruler over the Intersections of these two Arches at H and G, draw the Line GH; and where these two Lines cross one the other, as at K, that is the Center of the Triangular Points. From it extend the Compasses to either of the Points, and describe the Circle ABC, and the Triangle will be within the Circle.

[illustration] geometrical diagram

PROBL. XI. Any three Right Lines being given, so that the two shortest together be longer than the third, To make thereof a Triangle.

LEt it be required to make a Triangle of these three Lines A, B, and C, the two shortest whereof, viz. A and B together, are longer than the third C.

[illustration] geometrical diagram

First draw the Line DE equal to the given Line B; then take with your Com∣passes the Line C, and setting one Foot in E, with the other describe the Arch FF: also take with your Compasses the given Line A, and placing one Foot in D, with the other describe the Arch GG, cutting the former Arch in the Point K. Lastly, from the Point K, if you draw the Lines KE and KD, you shall constitute the Triangle KDE, whose Sides shall be equal to the three given Sides ABC.

PROBL. XII. Having a Right Line given, How to make a Geometrical Square, whose Sides shall be equal to the Right Line given.

THe Line given is RI, and it is required to make a Geometrical Square whose Sides shall be equal to the Line RI, First draw the given Line RI, then (by the first and se∣cond Problem) upon the Point B raise the Per∣pendicular BC, making the Line BC equal to the given Line RI also: Then taking the said RI in your Compasses, place one Foot in C, with the other describe an Arch at D; The Compass at the same distance, set one Foot in A, and cross the former Arch at D; then draw the Lines D, C and DA, which shall conclude the Geometrical Square ABCD, which was required.

[illustration] geometrical diagram

PROBL. XIII. Two Right Lines being given, How to find a third which shall be in pro∣portion unto them.

LEt the given Lines be A and B; and it is required to find a third Line which shall be in proportion unto them.

[illustration] geometrical diagram

First draw two Right Lines, making any Angle at pleasure, as the Lines LM and MN, making the Angle LMN: Then take the Line A in your Compasses, and set the length thereof from M to E; also take the Line B, and set the Length

thereof from M to F, and also from M to H: Then draw the Right Line EH, and then from the Point F draw the Line FG parallel to EH. So shall MG be the third Proportional required: For Arithmetically say,

As ME to MH: So is MF to MG 18.

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PROBL. XIV. Three Right Lines being given, To find a fourth in proportion to them.

THe three Lines given are ABC, unto which it is required to find a fourth Pro∣portional Line. This is to perform the Rule of Three. As in the last Problem, you must draw two Right Lines, making any Angle at pleasure, as the Angle EFG; then take the Line A in your Compass, and set it from F to I; then take the Line B in your Compas∣ses, and set that from F to K; then take the third given Line in your Com∣passes, and set that from F to H, and from that Point H draw the Line H L, parallel to IK; So shall the Line FL be the third Proportional required.

[illustration] geometrical diagram

Note, That these Lines are taken off a Scale, that is divided into 20 parts to an Inch: To do it Arithmetically say,

As FI is to FK: So is FH to FI.

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Here note, That in performance of the last Problem, That the first and third Terms, namely the Lines A and C, must be set upon one and the same Line, as here upon the Line FE, and the second Line B must be set upon the other Line FG, upon which Line also the fourth Proportion will be found.

PROBL. XV. How to work the Rule of Proportion by a Scale of equal Parts, and such other Conclusions as are usually wrought in Lines and Numbers, as in Mr. Gunter's 10 Prob. 2 Chap.

The Scale of Inches is a Scale of equal Parts, and will perform (by protraction upon a Flat or Paper) such Conclusions as are usually wrought in Lines and Numbers, as in Mr. Gunter's 10 Prob. 2 Chap. Sector, may be seen, and in others that have writ in the same kind. This way Mr. Samuel Foster hath directed in the I Chap. of his Posthumus Fosteri.

An Example in Numbers like his Tenth Probl.

As 16 to 7: So is 8 to what?

Here because the second Term is less than the first, upon the Line AB, I set AC the first Term 16, and the second Term AD 7, both taken out of the Scale of equal parts: thence also the third Number 8 being taken, with it upon the Center C, I de∣scribe the Arke EF, and from A draw the Line AE, which may only touch the same Arke; then from D, I take DG, the least distance from the Line AE, and the same measured in the same Scale of equal parts, gives 3½, the fourth Term required.

[illustration] geometrical diagram

But if the second Term shall be greater than the first, then the form of working must be changed, as in the following Example.

EXAMPLE.

As 7 to 16: So 21 to what? — 48.

Upon the Line AB, I set the second Term 16, which is here supposed to be AD; then with the first Term 7 upon the Center D, I describe the Arke GH, and draw AG that may just touch it: Again, having taken 21 out of the same Scale, I set one Foot of that Extent upon the Line AB, removing it until it fall into such a place, as that the other Foot being turned about, will justly touch the Line AG before-drawn; and where (upon such Conditions) it resteth, I make the Point C; then measuring AC upon your Scale, you shall find it to reach 48 Parts, which is the fourth Num∣ber required.

The form of Works (although not so Geometrical) is here given, because it is here more expedite than the other by drawing Parallel Lines; but in some Practice the other may be used. I have been the more large upon this, because in the following

Treatise I shall quote some more remarkable Places in Posthuma Fosteri: and the So∣lution of Proportions must be referred thither, the form of their Operations being the same with this. In them therefore shall only be intimated what must be done in ge∣neral, the particular way of working being here directed.

PROBL. XVI. To divide a Right Line given, into two parts, which shall have such pro∣portion one to the other as two given Right Lines.

THe Line given is AE, and it is required to divide the same into two parts, which shall have such proportion one to the other, as the Line C hath to the Line D.

First, From the Point A draw a Right Line at pleasure, making the Angle BAE; then take in your Compasses the Line C, and set it from A to F; and also take the Line D, and set it from F to B, and draw the Line BE: Then from the Point F draw the Line FG, parallel to BE, cutting the given Line AE in the Point G: So is the Line A B divided into two parts in the Point G, in proportion to the other, as the Line C is to the Line D.

[illustration] geometrical diagram

Arithmetically, let the Line AE contain 40 Perches or Foot, and let the Line C be 20, and the Line D 30 Perches; and let it be required to divide the Line AE into two parts, being in proportion one to the other, as the Line C is to the Line D.

First, Add the Lines C and D together, their Sum is 50: Then say by the Rule of Proportion, If 50 (which is the Sum of the two given Terms) give 40, the whole Line AE; What shall 30 the greater given Term give? Multiply and divide, and you shall have in the Quotient 24 for the greater part of the Line AE; which being taken from 40, there remains 16 for the other part AG: For

As AB is to AE: So is BF to EG.

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PROBL. XVII. How to divide a Triangle into two parts, according to any proportion as∣signed, by a Line drawn from any Angle thereof; and to lay the les∣ser part unto any Side assigned.

LEt ABC be a Triangle given, and let it be required to divide the same by a Line drawn from the Angle A, into two parts, the one bearing proportion to the other, As the Line F to the Line G; And that the lesser part may be towards the Side AB.

[illustration] geometrical diagram

By the last Problem divide the Base of the Triangle BC in the Point D, in propor∣tion as the Line F is to the Line G (the lesser part being set from B to D.) Lastly, draw the Line AD, which shall divide the Triangle ABC in proportion as F to G.

As the Line F, is to the Line G:

So is the Triangle ADC, to the Triangle ABD.

PROBL. XVIII. The Base of the Triangle being known, To perform the foregoing Problem Arithmetically.

SUppose the Base of the Triangle BC be 45, and let the Proportion into which the Triangle ABC is to be divided, be as 2 to 4. First add the two proportional Terms together, 2 and 4, which makes 6; then say by the Rule of Proportion, If 6, the Sum of the Proportional Term, give 45 (the whole Base BC) What shall 4 the greater Term given? Multiply and divide, and the Quotient will give you 30, for the greater Segment of the Base DC, which being deducted from the whole Base 45, there will remain 15 for the lesser Segment BD.

As 2/4 is to 45: So is 4 DC 30.

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PROBL. XIX. How to divide a Triangle (whose Area or Content is known) into two Parts, by a Line drawn from an Angle assigned, according to any Proportion required.

LEt the Triangle ABC contain 9 Acres, and let it be required to divide the same into two Parts, by a Line drawn from the Angle A, the one to contain 5 Acres, and the other 4 Acres. First, measure the whole length of the Base, which suppose 45; Then say, If 9 Acres the quantity of the whole Triangle, give 45 the whole Base, What parts of the Base shall 4 Acres give? Multiply and divide, the Quotient

will be 20 for the lesser Segment of the Base BD; which being deducted from 45 the whole Base DC, then draw the Line AD, which shall divide the Triangle ABC according to the proportion required.

If 9 Acres give 45, What shall 4 Acres give?

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PROBL. XX. How to divide a Triangle given into two parts, according to any Propor∣tion assigned, by a Line drawn from a Point limited in any of the Sides thereof; and to lay the greater or lesser part towards any Angle assigned.

THe Triangle given is ABC, and it is required from the Point M to draw a Line that shall divide the Triangle into two parts, being in proportion one to the other, as the Line N is to the Line O; and to lay the lesser part to∣wards B.

[illustration] geometrical diagram

First, from the limited Point M draw a Line to the opposite Angle at A; then divide the Base BC in proportion as O to N, which Point of Division will be at E; then draw ED parallel to AM: Lastly, from D draw the Line DM, which will divide the Triangle into two parts, being in Proportion one to the other, as the Line O is to the Line N.

PROBL. XXI. To perform the foregoing Problem Arithmetically.

IT is required to divide the Triangle ABC, from the Point M, into two parts in proportion as 5 to 2.

First divide the Base BC according to the given Proportion; then because the lesser Part is to be laid towards B, measure the distance from M to B, which suppose 32: Then say by the Rule of Proportion, If MB 32, give EB 16, what shall AR 28 (Perpendicular of the Triangle) give? Multiply and divide, the Quotient will be 14, at which distance draw a Parallel Line to BC, namely D; then from D draw the Line DM, which shall divide the Triangle according to the required Proportion.

PROBL. XXII. How to divide a Triangle (whose Area or Content is known) into two Parts, by a Line drawn from a Point limited, into any Side thereof, according to any number of Acres, Roods, and Perches.

IN the foregoing Triangle ABC, whose Area or Content is 5 Acres 1 Rood, let the limited Point be M in the Base thereof; and let it be required from the Point M to draw a Line which shall divide the Triangle into two parts between Johnson and Powell, so as Johnson may have 3 Acres 3 Roods thereof, and Powell may have 1 Acre and 2 Roods thereof.

First, Reduce the quantity of Powell's, being the lesser, into Perches (Observe, 160 square Poles contains 1 Acre, half an Acre contains 80 Perch, a quarter or one Rood 40 Perch.) which makes 240. Then considering on which side of the limited Point M this part is to be laid, as towards B, measuring the part of the Base from M to B 32 Perch, whereof take the half, which is 16, and thereby divide 240, the Parts be∣longing to Powell, the Quotient will be 15, the length of the Perpendicular DH, at which Parallel-distance from the Base BC, cut the Side AB in D, from whence draw the Line DM, which shall cut off the Triangle DBM, containing 1 Acre 2 Roods, the part belonging to Powell: Then the Trapezia ADMC (which is the part be∣longing to Johnson) contains the residue, namely, 3 Acres 3 Roods.

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PROBL. XXIII. How to divide a Triangle according to any Proportion given, by a Line drawn parallel to one of the Sides given.

The following Triangle ABC is given, and it is required to divide the same into two Parts, by a Line drawn parallel to the Side AC, which shall be in pro∣portion one to the other, as the Line I is to the Line K.

First (by the 16th Problem) divide the Line BC in E, in proportion as I to K; then (by the 27th Problem following) find a mean Proportional between BE and BC, which let be BF, from which Point F draw the Line FH, parallel to AC, which Line shall divide the Triangle into two parts, viz. the Trapezia AHFC, and the Triangle HFB, which are in proportion one to the other, as the Line I is to the Line K.

PROBL. XXIV. To perform the foregoing Problem Arithmetically.

LEt the Triangle be ABC, and let it be required to divide the same into two parts, which shall be in proportion one to the other, as 4 to 5, by a Line drawn Parallel to one of the Sides.

First let the Base BC, containing 54, be divided according to the propor∣tion given; so shall the lesser Segment BE con∣tain 24, and the greater EC 30; Then find out a mean Proportional be∣tween BE 24, and the whole Base BC 54, by multiplying 54 by 24, whose Product will be 1296; the Square Root thereof is 36, the mean Proportional sought, w^ch is BF. Now if BF 36 give BE 24, what shall AD 36? The Answer is HG 24, at which distance draw a Parallel Line to the Base, to cut the Side AB in H, from whence draw the Line HF, Parallel to AC, which shall divide the Triangle as was required.

[illustration] geometrical diagram

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PROBL. XXV. To divide a Triangle of any known Quantity into two Parts, by a Line Parallel to one of the Sides, according to any Number of Acres, Roods, and Perches.

THe Triangle given is ABC, whose Quantity is 8 Acres, 0 Roods, and 16 Per∣ches; and it is desired to divide the same (by a Line drawn up parallel to the Side AC) into two Parts, viz. 4 Acres, 2 Roods, 0 Perches; and 3 Acres, 2 Roods, and 16 Perches.

First, Reduce both Quantities into Perches (as it is hereafter taught) and they will be 720, and 576; then reduce both these Numbers by abbreviation into the least proportional Term, viz. 5 and 4; and according to that proportion, divide the Base BC 54 of the given Triangle in E: then seek the mean Proportion between BE and BC, which Proportion is BF 36, of which 36 take the half, and thereby divide 576, the lesser Quantity of Perches, the Quotient will be HG 32, at which Pa∣rallel-distance from the Base, cut off the Line AB in H, from whence draw the Line HF parallel to the Side AC, which shall divide the Triangle given, according as it was required.

〈 math 〉〈 math 〉

PROBL. XXVI. From a Line given, To cut off any Parts required.

THe Line given is AB, from which it is re∣quired to cut off 3/7 Parts.

[illustration] geometrical diagram

First, draw the Line A C, making any Angle, as CAB; then from A set off any 7 equal Parts, as 1, 2, 3, 4, 5, 6, 7; and from 7 draw the Line 7 B. Now because is to be cut off from the Line B, there∣fore from the Point 3, draw the Line 3 D, parallel to 7 B, cutting the Line AB in D; So shall AD be the 3/7 of the Line AB, and DB shall be 4/7 of the same Line.

As 7 is to AB: So is A 3 to AD.

PROBL. XXVII. To find, a Mean Proportional between two Lines given.

IN the following Figure, let the two Lines given be A and B, between which it is re∣quired to find a Mean Proportional. Let the two Lines A and B be joyned toge∣ther in the Point E, making one Right Line as CD, which divided into two equal Parts in the Point G; upon which Point G, with the distance GC or GD, describe the Semicircle CFD: Then from the Point E, where the two Lines are joyned toge∣ther, raise the Perpendicular EF: So shall the Line EF be a Mean Proportional between the two given Lines A and B. For,

As ED	is to EF:	So EF	to CF.
9	12	12	16

PROBL. XXVIII. How to finde two Lines, which together shall be equal in Power to any Line given; And in Power the one to the other, according to any Propor∣tion assigned.

IN this Figure let CD be a Line given, to be divided in Power, as the Line A is to the Line B.

[illustration] geometrical diagram

First, divide the Line CD in the Point E, in proportion as A to B (by the 16th Probl.) Then divide the Line CD into two equal Parts in the Point G, and on G, at the distance GD or GC, describe the

Semicircle CFD, and upon the Point E raise the Perpendicular EF, cutting the Se∣micircle in F. Lastly, draw the Line CF and DF, which together in Power will be equal to the Power of the given Line CD; and yet in Power one to the other, as A to B.

PROBL. XXIX. How to divide a Line in Power according to any Proportion given.

FIrst, Divide the Line C D in the Point E, in pro∣portion as A to B: Then di∣vide the Line CD in two e∣qual Parts in the Point G, and upon G as a Center, at the di∣stance CD, describe the Se∣micircle CFD, and on E raise the Perpendicular of EF, cutting the Semicircle in F: Then draw the Line CF and DF, and produce the Line CF to H, till FH be equal to FD, and draw the Line D. H. Lastly, draw the Line FK, parallel to DH: Then shall the Line CD be divi∣ded in K; so that the Square of CK shall be to the Square of KD, as CE to ED, or as B to A.

[illustration] geometrical diagram

PROBL. XXX. How to enlarge a Line in Power, according to any Proportion assigned.

IN the Diagram of the 28th Problem, let CE be a Line given, to be enlarged in Power as the Line B to the Line C.

First (by the 16th Problem) find a Line in proportion to the given Line CE, as B is to C, which will be CD; upon which Line describe the Semicircle CFD, and on the Point E erect the Perpendicular EF: Then draw the Line CF, which shall be in power to CE, as C to B.

PROBL. XXXI. To enlarge or diminish a Plot given, according to any Proportion required.

LEt ABCDE be a Plot given, to be diminished in Power as L to K.

Divide one of the Sides, as AB in Power as L to K, in such sort that the Power of AF may be to the Power of AB, as L to K; then from the Angle A draw Lines to the Point C and D. That done, by F draw a Parallel to BC, to cut AC in G, as FG: again, from G draw a Parallel to DC, to cut AD in H. Lastly, from H draw a Parallel to DE, to cut AE in I: So shall the Plot AFGHI be like ABCDE, and in proportion to it, as the Line L to the Line K, which was required.

Also if the lesser Plot was given, and it was required to make it in proportion to it as K to L; then from the Point A draw the Lines AC and AD at length; also increase AF and AI: That done, enlarge AF in Power as K to L, which set from A to B; then by B draw a Parallel to FG, to cut AC in C, as B C: likewise from C draw a Parallel to GH, to cut AD in D: Lastly, a Parallel from D to HI, as DE, to cut AI being increased in E; so shall you include the Plot ABC DE, like AFGHI, and in proportion thereunto, as the Line K is to the Line L, which was required.

[illustration] geometrical diagram

PROBL. XXXII. How to make a Triangle which shall contain any Number of Acres, Roods, and Perches, and whose Base shall be equal to any (possible) Num∣ber given.

LEt it be required to make a Triangle which shall contain 6 Acres, 2 Roods, 25 Perches, whose Base shall contain 50 Perches. You must first reduce your 6 Acres 2 Roods, and 25 Perches, all into Perches, after this manner.

First, Because 4 Roods makes 1 Acre, multiply your 6 Acres by 4. makes 24; to which add the 2 odd Roods, so have you 26 Roods in 6 Acres 2 Roods; then because 40 Perches makes 1 Rood, multiply your 26 by 40, which makes 1040, to which add the 25 Perches, and you shall have 1065, and so many Perches are contained in 6 Acres, 2 Roods, and 25 Perches.—Now to make a Triangle that shall contain 1065 Perches, and whose Base shall be 50 Perches, do thus; double the number of Per∣ches given, namely 1065, and they make 2130; then be∣cause the Base of the Trian∣gle must contain 50 Perches, divide 2130 by 50, the Quo∣tient will be 42 ⅗ which will be the length of the Perpen∣dicular of the Triangle. This done, from any Scale of equal Parts, lay down the Line BC equal to 50 Perches; then upon C raise the Perpendicular CE, equal to 42 ⅗ Perches, and draw the Line AE, pa∣rallel to BC; then from any Point in the Line AE, as from G, draw the Line BG, and GC, including the Triangle BGC, which shall contain 6 Acres, 2 Roods, 25 Perches, which was required.

[illustration] geometrical diagram

PROBL. XXXIII. How to reduce a Trapezia into a Triangle, by a Line drawn from any Angle thereof.

THe Trapezia given is ABCD, and it is required to reduce the same into a Triangle.

[illustration] geometrical diagram

First, extend the Line D C, and draw the Diagonal BC; then from the Point A draw the Line AF, pa∣rallel to CB, extending it till it cut the Side DC in the Point F. Lastly, from the Point B draw the Line BF, constituting the Triangle FBD, which shall be equal to the Trapezia ABDC.

And so I have concluded what I did intend of Geometrical Problems: Neither had I gone so far as I have, in regard Mr. William Leybourn hath ingeniously and very fully demonstrated them in his First Book of his Compleat Surveyor. But no Book (as I remember) now extant of Navigation, hath the foregoing Problems so large. Be∣sides, I shall direct (in the following Treatise) the Mariner to Survey any Plantation or Parcel of Land very exactly and easily, by his Sea-Compass.