The new method of fortification, as practised by Monsieur de Vauban, Engineer General of France with an explication of all terms appertaining to that art / made English.

BOOK II. (Book 2)

AS all Rivers have their particular Bran∣ches, and every general Science its Divisions, and its Parts; so Geometry, ge∣nerally taken, is the Mother of several other Arts and Sciences, whereof we shall treat in this Second Part. These are, Measuring of Heights, of Distances, of Plains, of Solids, of Concaves, and the turning of one Body into another.

CHAP. I. Of Measuring of Heights.

MEasuring of Heights is a Science which shews how they are to be measured•• But because there are several Instruments for that purpose, as a Quadrant, a Panto∣metre, an Astrolabe, a Iacob's Staff, &c. We shall at present only make use of the Astrolabe, because it is the surest and the most convenient: and we shall know how it is to be used by the following Problems.

PROBLEMS.

I. To take the Vertical Height of any thing, when you can come to the bottom.

Let the height of a Tower AB be to be measured.

1. Draw a right Line as you please from the point A, a•• AC.

2. Measure the Line AC with your Scale, or your Chain•• of 26 Feet for instance.

3. Set the Staff, upon which the Astro∣labe is Horizontally fastned, upon C, so that by the fixed Sights, one may see the Line CA, from one end to the other, as fas as the Tower, and that by the two other mo∣vable sights the point B, which marks the height of the Tower, may also be seen.

4. When the Instrument is thus fixed, com∣pute whether the Angle CDE be 40 degrees.

5. Then draw upon Paper a right Line of the length of AC, as FG: But this must be done by a small Scale made upon the Pa∣per, or by a Scale of Wood or La••ten: Af∣terwards erect a Perpendicular upon G.

6. Take the Reporter [the French word is Rapporteur] and set it directly upon FG, so that the middle of it may answer to the point F: then compute an Angle of 40 De∣grees equal to CDE, as FHI.

7. From F draw a right Line thro•• I, ••ill it touches the Perpendicular G in K.

8. With a small Scale measure the height of KG, and add the height of the Staff to which the Astrolabe is fastned, and the Sum will give you the height of the Tower AB.

II. To measure a Vertical Height where you cannot come at the bottom.

Let AB be the height to be measured; the foot inaccessible, because of a River which hinders all approaches.

1. Draw a right Line upon the Ground, on this side of the River, as CD; let it be 22 Foot long: (Take notice, when you can raise a right-angled Triangle upon this Line, opposite to the height•• which you would measure, it is much more convenient, other∣wise the Operation is the same.)

2. Set the Instrument upon C, so that by the immovable ••ights you may see the point D directly, and by the two movable ones you may see the point A beyond the River: When this is done, compute the Angle CEF, which may be of 90 Degrees.

3. Set the Instrument upon D, so that the fixed sights may make one Line with DC; by the two others you will see the point A: this Angle ought to have 28 De∣grees.

4. Draw upon Paper a Line equal to that drawn in the Field (CD) as GH of 2•• Feet.

5. With the Reporter take an Angle at H equal to the Angle CEF, viz. of 90 De∣grees, as IHK, and lengthen the side HK. Afterwards upon the Point G take another Angle of 28 Degrees, equal to the Field-Angle DOP, as GLM, and then lengthen the side GM till it cuts the other HK, in N, and this Point N will mark the foot of the height A.

6. Having the distance from the Instru∣ment to the foot of the Tower, it will be an easie thing to work the rest by the fore∣going Problem.

III. To measure a Perpendicular Height, where you can come at the bottom, with the shadow of a Staff.

1. Fasten a Staff a foot long perpendicu∣larly into the ground, as AB, and measure the shade by the Staff, 3 Foot for Instance.

2. With the shade of the Staff, viz. 3 feet, measure the shade of the House DE, whose height you would know; and as many feet as the shade of the House exceeds the shade of the Staff, so many foot high will the House be: For Instance, if the shade of the

House be 24 times longer than that of the S••aff, then we ought to say that the House is 24 Foot high.

Or thus,

• 1. Erect a Stick, divided into 10 equal Parts, Feet, or Inches, perpendicularly in the ground, and at the same time measure the shade caused by this Staff, 20 Foot for In∣stance.
• 2. Measure also the shadow of the House, whose height you would know, for Instance 38 Feet.
• 3. By the Rule of Three say, If the sha∣dow of 20 Foot gives a height of 10 Foot, what will the shadow of 38 Foot give? Answ. 19 Foot.

IV. To measure the Inclination of a Mountain.

Let the Inclination of the Mountain AB be to be measured.

1. Draw from the Point A, which stands for the foot of the Mountain, a right Line, 36 Foot long, as AC.

2. Set the Instrument upon the Point A, and look thro' the two fixed sights towards the Point C, and thro' the two movable ones towards the Point B, the top of the Moun∣tain; this Angle, for Instance, may be 110 Degrees.

3. Then set the instrument upon C, and look thro' the two fixed sights upon the foot A, and thro' the two other fixed ones upon the top B: then compute this Angle, which may be for Instance 40 Degrees.

4. Afterwards, draw a Line upon Paper, 36 Foot long, because it ought to be equal to the Line AC, as DE.

5. Then by the Reporter set upon D an Angle equal to that of the Figure at A, viz. 110 Degrees, as DFG.

6. Upon the Point E, set another Angle of 40 Degrees, equal to that at C, as EH••.

7. Lengthen the sides DF and EI, and where these lengthened Lines intersect one another, as at K, draw from that Point to D a right Line, which will represent the Incli∣nation which was required, provided the height of the Staff upon which the Astrolabe was fastned, be added.

V. To measure the vertical height of a Moun∣tain.

This Problem must be wrought as the fore∣going one, provided that you let fall a Per∣pendicular from the Point K, which marks the top of the Mountain, which must touch the Line CD lengthned to L; the Line KL will represent the vertical height of the Moun∣tain.

VI. To measure the height of a Tower built upon a Rock.

Let the height BD be to be measured.

1. Seek the inclining height of the Moun∣tain by Prob. 4. and let the Triangle ABC demonstrate its Operation.

2. Make the Triangle HIK upon Paper equal to the Triangle ABC.

3. Measure the Field-Angle CBD, for Instance 38 Degrees.

4. Upon K, which represents the Point B, and so by consequence the top of the Moun∣tain, raise a Perpendicular of what length you please.

5. With the Reporter raise upon the Point I an Angle equal to the Angle of the Figure ABCD, viz. ILM.

6. Lengthen the side IM till it touches the Perpendicular raised upon K in N.

7. Measure the Line NK, it will give you the height of the Tower BD, if you add to it the height of the Staff.

VII. To measure the depth of a Well.

1. Measure its Diameter AB, 9 Foot for Instance; and find the Angle BCD of 90 Degrees.

2. Set the Instrument so upon A, as that by the two fixed ••ights you may see the Point B, and by the two movable ones you may see the bottom of the Pit E, and let this Angle be 60 Degrees, as AHI.

3. Draw a Line 9 Foot long upon Paper, which shall represent the Diameter AB, as FG.

4. From F let fall a Perpendicular which may represent the depth BE.

5. Make an Angle at G equal to the An∣gle AHI, viz. of 60 Degrees, as GKL.

6. Lengthen the side GL, and where this lengthened Line touches the Perpendicular F, as in M, measure from M to F, and you will have the depth of the Well.

CHAP. II. Of Measuring of Distances.

MEasuring of Distances is a Science which teaches to measure the Distance of one place from another, in an open Field.

PROBLEMS.

I. To measure the Distance of one place from another, where they are both accessible.

Let the Distance from A to B, be to be measured.

1. From A draw a right Line, what you please, and measure with a Chain or Scale; suppose 22 feet in length, as AC.

2. Set the Staff upon which the Astrolabe is fixed, upon the Point A, so as that by the two fixed sights, you can only see directly forward towards the Point C, and by the two others the Point B: then count the Angle ADE, suppose 90 Degrees.

3. Remove the Instrument upon C, and look thro' the fixed sights towards A, and thro' the movable ones towards B; then compute this Angle CFG, suppose 22 De∣grees.

4. Draw a Line upon Paper 22 feet long, equal to the Line AC, as HI.

5. With the Reporter upon the Point H draw an Angle equal to the Angle ADE, that is to say, of 90 Degrees, as HKL: do the same thing at I, where make another Angle equal to the Angle CFG, i. e. of 22 Degrees, as IMN.

6. Lengthen the sides HK, TM, and from the Point where the••e two lengthned Lines intersect each other, as O, to the Point H, measure the Distance, which shall be equal to the length AB.

II. To measure the Distance of two places, where∣of one is inaccessible.

Let the Distance from A to B be that which is required, and B inaccessible, because of a River which hinders all approaches.

1. From A draw a right Line 22 Feet long, as AC.

2. Set your Instrument upon A, and look thro' the two fixed sights upon the Point C, and thro' the two movable ones upon B: then compute the Angle ADE of 93 De∣grees.

3. Carry the Instrument to C, and look thro' the two fixed sights upon the Point A, and thro' the two others upon B, and then compute the Angle CFG of 33 Degrees.

4. Draw upon Paper a Line equal to AC, i. e. of 22 feet, as HI.

5. Upon H make an Angle equal to the Angle of the Figure ADE, i. e. of 93 De∣grees, as HKL: So also upon I make ano∣ther Angle equal to the Angle CFG, as IMN.

6. Lengthen the sides HK and TM, and in the place where they intersect, as at O, let fall a Perpendicular upon H, which will give you the Distance AB.

III. To find the Distance of two places which are both inaccessible.

Let the Distance AB be to be measured.

1. Take a fixt Point C, from which draw a right Line CD, which shall be 33 Foot long.

2. Set the Instrument upon C, and look thro' the two fixed sights upon the Point D, and thro' the two others upon the Point A: then compute the Angle CHK for Instance of 90 Degrees.

3. Leave the Instrument as it is, and re∣move the movable sights, till thro' them you may see the Point B, and then compute that Angle, for Instance 36 Degrees, as CHI.

4. Carry the Instrument to D, and ••it it so, as that thro' the fixed sights you may see the Point C, and thro' the two other movable ones the Point B: then compute this Angle, which may be of 100 Degrees, as DEG.

5. Leave the Instrument as it is, and turn the movable sights till you may see the Point A thro' them; then compute that Angle, of 30 Degrees, as DEF.

6. Afterwards draw a Line of 33 foot upon Paper, to represent the Line CD, as LM.

7. With the Reporter set two Angles at L, one of 90 degrees, and the other of 36 de∣grees, so that the first LNO may represent the Angle CHK, and the other LNP the Angle CHI.

8. Lengthen the sides LO and LP.

9. Add two Angles to the other Point N, one of 100 degrees, as MQR, to represent the Angle of the Figure DEG, and the o∣ther of 30 degrees, as MQS, to represent the Angle DEF.

10. Lengthen the sides MR, MS, and observe where these two lengthened Lines touch the two first, as at TV. For this di∣stance TV will give you the distance AB, which was required.

IV. To measure the breadth of a River with a Staff.

1. Fix a Staff perpendicularly upon the bank of a River, as A.

2. Cleave it at the end, and fasten a cross•• staff into the cleft, as BC.

3. Stand behind the small cross-staff B•••• and use it as you would the sights of an A∣strolabe, i. e. move it downwards, till from

BC you can see the other ••ide of the River at PA.

4. Afterwards letting the small Staff stand as it is, turn the large one perpendicularly, till from BC you can see the whole length of the Bank: Then stand behind B, and ob∣serve the last place where from C you can see the Bank on the other side, as at E.

5. Measure the distance EA, it will give you the breadth of the River AD.

V. To describe a Figure taken from the Field upon Paper.

Let the Figure to be described be ABCDE.

1. Set the Staff to which the Astrolabe is fixed, upon the Point E, and look upon the Point D thro' the fixed sights, and thro' the two movable ones upon the Point A. Then compute this single Angle EAD, 60 degrees for Instance.

2. Measure the Line AE of 23 feet.

3. Measure another Line ED of 24 feet.

4. Set the Instrument upon D, and look towards C with the fixed sights, and with the movable ones towards E: Then compute this Angle, of 70 degrees for Instance, as DEC.

5. Let the side DC be of 28 feet.

6. Seek the Angle Cdb after the afore∣said manner, and let it have 50 degrees.

7. Measure the Line CB, for Instance, of 13 feet.

8. The side BA will be found of it self.

When this is done in the Field, then

• 1. Draw a Line upon Paper 23 foot long, to represent the Figure AE, as KF.
• 2. With a Reporter make at the Point F and Angle equal to the Angle of the Figure Eda, i. e. of 60 degrees, as Fkg.
• 3. Lengthen the side of this Angle Fg, and add its length to it, i. e. 24 feet, as FG.
• 4. Upon G make an Angle equal to that of the Figure Dec, i.e. of 70 degrees, as Gfh.
• 5. Lengthen the side Gh, and add the side DC, i. e. of 28 feet, as GH.
• 6. Upon the Point H set another Angle equal to that of the Figure Cdb, of 50 de∣grees, as Hgi.
• 7. Lengthen the side Hi, and add 13 feet to it, as HI, which may represent the side CB.
• 8. Joyn IK together by a right Line, which will give the last side of the Figure BA, and which at the same time will entire∣ly represent the given Figure upon Paper.

VI. To describe upon Paper a Figure taken from the Field, which is inaccessible.

Let the Figure ABC be to be described upon Paper from the Point D.

1. From D draw a right Line DE 32 foot long.

2. Set the Instrument upon D, and look thro' the fixed sights towards E, and thro' the two moveable ones towards A: Then compute the Angle Dfg, of 105 degrees.

3. Leave the Instrument as it is, and re∣move the moveable sights till you can see the Point B, and compute the Angle Dgh of 90 degrees.

4. Leave the Instrument still as it is; look thro' the moveable sights towards the Point C, and compute the Angle Dgi of 60 de∣grees.

5. Set the Instrument upon E, and look thro' the fixed sights towards the Point D, and thro' the moveable ones towards A, and compute the Angle Ekl, for Instance of 50 degrees.

6. Leave the Instrument as it is, and look thro' the moveable sights towards B, and compute this Angle Ekm of 70 degrees.

7. Look thro' the same sights towards the Point C, and then compute the Angle Ekn, ••or Instance of 90 degrees.

This being done in the Field, then

• 1. Draw a Line of 32 feet upon Paper, as OP, equal to the Line DE.
• 2. Take an Angle of 105 degrees at O, as Oqr, which may represent the Angle of the Figure Dfg, and lengthen the side Oq.
• 3. Leave the Reporter as it is, and com∣pute another Angle of 90 degrees, as Ors, to represent the Angle Dgh, and lengthen the side Os.
• 4. Seek the last Angle of 60 degrees Ort, which may represent the Angle of the Figure Dgi, and lengthen the side Ot.
• 5. Set the Reporter upon the Point P, and compute an Angle of 50 degrees, to represent the Angle Ekl, as Pab; lengthen the side Pb till it meets with the side Oq in C: for this point of Intersection will mark the Point A.
• 6. Leave the Reporter as it is, and com∣pute another Angle of 70 degrees, which may represent the Angle Ekm, as Pad; lengthen the side Pd till it cuts the length∣ned Line Os in E, which Point will mark the Point B in the Figure.
• ...

• 7. Compute another Angle of 90 degrees equal to the Angle Ekn as Paf; lengthen the side Pf till it meets with the lengthned Line Ot in G, which answers to the point C in the Field-figure.
• 8. Joyn CE, EG, and GC together, and you shall have your Figure ABC represent∣ed upon Paper.

CHAP. III. Of measuring of Plains.

MEasuring of Plains is a Science which teacheth us to measure all sorts of Sur∣faces of Things, which Geometricians call Area's.

PROBLEMS.

I. To measure the Area of a right-angled Triangle ABC.

1. Measure the Line AB, 12 foot long for the Base.

2. Measure the Perpendicular BC, of 8 foot.

3. Multiply the half of the Base upon the whole perpendicular, or the half of the per∣pendicular upon the whole Base, and the Sum,

(48 foot) will be the Area of the Triangle which was desired.

II. To measure the Area of an Oblique-angled Triangle DEF.

1. Let fall a perpendicular from F to its opposite side DE, which may divide the Tri∣angle into 2 right-angled Triangles, DGF, and FGE.

2. Measure these two Triangles by the foregoing Problem; i. e. multiply the whole Base DG and GE, by the half of the Perpen∣dicular FG, or the whole Perpendicular by the half of the Base DG and GE.

Thus for instance: Let the Base DG be 12 foot, and the Perpendicular FG 10: Take the half of the Perpendicular, i.e. 5 feet, and multiply it by the Base of 12 feet, and the sum will be 60 feet, which gives the A∣rea of the Triangle DFG.

3. Let the Base GE of the Triangle FGE be six feet, and the Perpendicular FG stands as it did, i. e. to feet. Multiply 5 by 6, or or 3 by 10; the sum will be 30 feet, which will give the Area of the Triangle. Then add 30 to 60, the sum 90 will be the number of feet in the whole Area of the given Triangle DEF.

REMARKS.

Sect. 1. If one of the Lines to be multiplied has only fathoms, and the other fathoms and feet; when you multiply, instead of a fa∣thom take 6 foot. For instance: if you are to multiply 7 fathoms by 8 fathoms and 4 feet, instead of the first sum (seven fathoms) you must put 6 fathoms and 6 feet, because these 6 feet make but one fathom: For since there are no feet expressed, you cannot any other∣wise multiply the feet of the second sum: So the number will be 48 fathoms, and 24 feet: for 6 into 8 gives 48: and 6 into 4 gives 24.

Sect. 2. If one of these lines has only fathoms, and the other has fathoms, inches, and feet; for instance, if you are to multiply 6 fathoms by 8 fathoms, 4 feet, and 6 inches, this is the Method you are to take:

• 1. Bring these 8 fathoms, 4 feet, and 6 inches, into one Line.
• 2. Instead of 6 fathoms, take 5 fathoms and 6 feet; and put the first cypher 5 under 8, because they both mark fathoms: take one out of the second cypher 6, so as that there may be but 5 feet to put under the feet of the first sum: divide the foot which you substracted into 12 inches, for 12 inches

• are equal to a foot: then set the inches un∣der the 6 inches of the first sum.
• Sect. 3. Reduce the feet into fathoms, di∣viding by 6; divide also the inches by 12, and it will give you the feet: for instance, 72 inches gives 6 feet, and 20 feet give 3 fa∣thoms and 2 feet.

III. To measure the Area of an equilateral right-angled Square ABCD.

Multiply one side by the other: AB for instance of 12 feet, by BC of 12 feet: be∣cause the sides are equal: the product 144 feet will give you the Area which is sought for.

IV. To measure the Area of a Parallelogram ABCD.

Multiply the shortest side BC of six foot, by the longest side AB of 12 foot: the product 72 foot will give you the Area of the Parallelogram which was required.

V. To find the Area of a Rhomb, ABCD.

1. Let fall a Perpendicular from A to the Line DC, and AE.

2. Let fall another Perpendicular from B, equal to AE, as BF; and joyn CF together. So that instead of a Rhomb you have a Pa∣rallelogram ABEF: then work by the fore∣going Problem. See Prob. 28. Book. 1. Chap. 2.

VI. To find the Area of a Rhomboid ABCD.

1. Let fall a Perpendicular from A to the opposite Line DC, as AF.

2. Let fall another from B, equal to AF, as BE.

3. Joyn CE together, and instead of the Rhomboid ABCD you have a Parallelo∣gram ABEF: You will find its Area by the ••ourth Problem of this Chapter.

VII. To find the Area of any unequilateral Quadrangle.

Let the given Quadrangle be ABCD.

1. Divide this Quadrangle into two Tri∣angles ABD, and CDB.

2. Divide each of these Triangles into two right-angled ones, as ABF, BCE, ADF, DEC.

3. When you have these 4 right-angled Triangles, search their Area's by the two first Problems of this Chapter: then joyn

the Sum of all these Triangles together, and it will give the Area which is required.

VIII. To find the Area of any regular Polygon.

Let the regular Pentagon demanded be ABCDE.

1. Measure one side of the Polygon as AB, for instance, 6 fathoms: multiply this number by the number of the sides of the Polygon, as here by 5, because this Polygon has but 5 sides: the product of this multi∣plication will give you the length of all the sides.

2. Let fall a Perpendicular from any side to F the Center of the Polygon, as FG: then measure this Line, 5 fathoms for in∣stance.

3. Multiply the sum of all the sides, by the half of the Perpendicular: the product will give the Area which is required.

IX. To find the Area of any irregular Poly∣gon ABCDE.

1. Divide the whole Figure into right an∣gled Triangles; as this irregular Pentagon ABCDE into ABG, BCG, ADH, CDH, AEF, and DEF.

2. When you have only right angled Tri∣angles, work by the first Problem of this Chapter; the sum of all the Triangles re∣duced to one, gives the Area which is fought for.

X. To find the Area of a Circle.

Reduce this Circle to a right angled Tri∣angle, by Prob. 21. Book 1. Chap. 2. and then seek the Area by the Rules of the first Problem of this Chapter.

XI. To find the Area of an Oval Figure.

1. Divide the proposed Figure into four equal parts, as ABCD, so as that the line AB may make one Diameter, and CD ano∣ther.

2. Seek for a mean proportional between these two Diameters, which you may thus find:

Set the Diameter AB upon a long line, and upon the same line from the point B, set the other Diameter CD, as BC; then take the half of the distance between A and C, and carry it on to B, as EB; this shall be a mean proportional; then divide EB into two equal parts, as EF, FB; then draw

a Circle from the point F ••hro the point EB, which shall be equal to the given Oval Figure.

3. Having this Circle, find its Area, by the foregoing Problem; i. e•• raise a Perpendicu∣lar upon B, one extream point of the Dia∣meter, which shall have thrice the length of the Diameter EB, and a seventh part over, as BG.

4. Joyn F the Center of the Circle to the point G by a right line, which will give a right angled Triangle equal to the given Circle; its Area may be found by the Rules of the first Problem of this Chapter.

[This Triangle which he says is equal to the Area of the Circle is not expressed in the Scheme.]

XII. To measure the Surface of any Equilate∣ral Pyramid.

1. Let ABCD be an equilateral Square, which shall be the Base of the Pyramid gi∣ven.

1. Find the Area of this Base by Prob. 3. of this Chapter, of 48 foot.

2. Multiply one of its sides, as AB, by the height of the Pyramid EF, that is 12 by 28, whose product will be 336 feet.

3. Multiply 336 by 4, the number of the sides of this Pyramid, and the product will be 1344 feet.

4. To this sum add the Area of the Base, 48 feet, and you will have the total sum of the Area of the Pyramid proposed, viz. 1392 feet.

XIII. To find the Convex Surface of a right lined Cylinder.

1. Set the height of the Cylinder BD upon a long line.

2. Set upon the same line, beginning at the point D, the Diameter of the Base of the Cylinder marked CD, as DE.

3. Divide the line EB into two equal parts, BF, EF.

4. With the distance FD draw a Circle, which shall contain the Convex Surface of the given Cylinder: this you may find by Prob. 21. Chap. 2. of the first Book, which teaches you to make a right angled Tri∣angle equal to a given Circle; when you have this Triangle, find its Area by Prob. 1. of this Chapter.

5. When you have the Area of this right angled Triangle, which represents the Con∣vex Surface of the Cylinder, then find the Area of the two Bases of the Cylinder by

Prob. 1 of this Chapter•• Add the sum of the two Bases to the Area of that Triangle which gives you the Convex Surface of the Cylinder, without counting the Bases, and you will have the general sum of the whole Surface of the Cylinder, with both its Bases. See Archimedes de Sphaerâ & Cylindro, lib. 1. Prop. XIII.

XIV. To find the Convex Surface of a regu∣lar Cone.

Let the given Cone be ABD, and its dia∣meter AD.

1. Set the height of the Cone DB upon a ••ight line.

2. Set its semi-diameter CD upon the same line, as DE.

3. Divide EB into two equal parts, EF, FB.

4. Take the distance FD, and make a circle, as large as the Cone without the Base.

5. Reduce this circle to a right angled Triangle by Prob. 21. Chap. 2. of the first Book, and find its Area by Prob. 1. of this Chapter.

6. By the two forementioned Problems seek the area of the base of your Cone, and add this sum to the former; the sum total

will give the whole surface of the given Cone. See Archimed. Lib. cit. Prop. XIV.

XV. To measure the Convex Surface of a Globe.

1. Me••sure the diameter AB, of nine feet.

2. Find the circumference of the circle, whereof that is the diameter, which is always thrice as long, and a third part over.

3. Multiply the diameter by the circum∣ference; the product will give you the re∣quired surface of the given Globe.

XVI. To find the Area of any Trape••ium, as ABCD.

You must work by the Rules of Problem 7. of this Chapter.

XVII. To find the Area of the Base of a round Tower, where you can only come at one part.

1. Measure the line AB, of fourteen fa∣thoms, and from its middle point draw a per∣pendicular to the arch, as DC, of four fa∣thoms.

2. Multiply the distance AD by the re∣mainder DB; i. e. 7 by 7; then divide by

4 the number of the perpendicular DC.

3. Add to this the length of the perpendi∣cular, i.e. four fathoms, and it will give you the length of the diameter, to compleat the base.

4. When you have found the base, seek for a right angled Triangle, equal to it, by Prob. 21. Chap. 2. Book 1; afterwards you may find the area of that Triangle by Prob. 1. of this Chapter.

CHAP. IV. Of Measuring of Solids.

MEasuring of Solids is a Science which shews how one is to find the Solidi∣ty of Bodies: Solidity is Magnitude with Breadth, Length, and Thickness.

PROBLEMS.

I. To find the Solidity of a Parallelepiped.

1. Measure the height, length and breadth of the Parallelepiped.

2. Multiply the breadth AD, of two feet, by the length AB, of four feet, and

the sum will give you the area of the Base.

3. Multiply the number of this Area by the height of the Parallelepiped, which here shall be eight feet; and the product from thence will shew you the Solidity required, of sixty four feet.

II. To find the Solidity of a Prism.

1. Find the Area of the Base ABC, by Prob. 2. Chap. 3. Book 2. of six feet.

2. Measure the height CF, here of nine feet.

3. Multiply the Base by its height, i. e. 6 by 9, and the product 54 will give the Solidity required.

III. To find the Solidity of a Cylinder.

1. Find the Area of the Base by Prob. 21. Book 1. Chap. 2. and by Prob. 1. Chap. 3. Book 2. for instance five feet.

2. Find the height of the Cylinder AB, here of nine feet.

3. Multiply the Area upon the height, i. e. 5 by 9, and the sum 45 will give the So∣lidity required.

IV. To find the Solidity of Pyramids and Cones.

1. Measure their height FE and AB, for instance, of twelve feet.

2. Find the Area of their Bases; that of the given Pyramid by Prob. 6. Chap. 3. Book 2. of sixteen feet; and that of the Cone by Prob. 10. Chap. 3. Book 2. of six∣teen feet.

3. Multiply the entire height by the third part of the Base; i. e. twelve ••eet by five feet four inches; or the whole Base by the third part of the height, i. e. sixteen feet by four feet; the sum sixty ••our will give the Solidity of the Pyramid and the Cone; for this is equal to the given Pyramid: and be∣cause we are to find its Solidity, that also will be equal to the Solidity of the same Py∣ramid.

V. To find the Solidity of a Globe.

1. Multiply the Semi-diameter AB by the Surface of the Globe, which you shall find by Problem 15. of the foregoing Chap∣ter.

2. Divide this sum by 3, and the Quo∣tient

will give you the required Solidity: for instance, let the Semi-diameter be 28 feet, and the Convex Surface 9856 feet; multiply this by 28, and the product will be 275968; divide that by 3, and your Quotient 91989 1/•• will give you the num∣ber of feet which make up the Solidity of this Globe.

VI. To find the Solidity of a Tetraëdrum.

Since a Tetraëdrum is nothing else but a triangular equilateral Pyramid, its Solidity must be sought by Prob. 4. of this Chapter; i. e. by multiplying the Base by a third part of the height, or the height by a third part of the Base; the product will shew the Solidity which is required.

VII. To find the Solidity of an Octaëdrum.

Since an Octaëdrum is made up of two Pyramids which have one common Base, i e. an equilateral right angled Square.

1. Find the Solidity of one of these Pyra∣mids by Prob. 4. of this Chapter, for in∣stance of sixteen feet.

2. Multiply these 16 feet by 2, the pro∣duct will give the required Solidity of 32 feet.

VIII. To find the Solidity of a Dodecaëdrum.

If one draws from the Center of a Dode∣caëdrum a right line to every angular point, one may from twelve equal Pyramids, eve∣ry one whereof will have a regular Penta∣gon for its Base; upon which account

• 1. Find the Solidity of one of these twelve Pyramids, by the same 4th Problem, and let it be for instance of twelve feet.
• 2. Multiply these twelve feet by the the number of the Pyramids, i. e. 12; the product 144 gives the requir'd Solidi∣ty.

IX. To find out the Solidity of an Icosaë∣drum.

This being formed of twenty equal Te∣traëdra, if one draws right Lines from its Center to every angular Point, one must,

• 1. Find the Solidity of one of these Te∣traëdra by the same fourth Problem, for in∣stance of 8 Feet.
• 2. Multiply 8 by 20, the number of the Tetraëdra; and the product 160 will shew the number of the solid feet in the given Icosaëdrum.

X. To find the Solidity of a Cube.

By the third and fourth Problems of the foregoing Chapter, multiply the Area of the Base by the length, and the product gives the Solidity.

XI. To find the Solidity of a solid Rhomb, and Rhomboid.

Find the Area of the Base of a Rhomb by Prob. 5. of the foregoing Chapter, and the Area of the Base of the Rhomboid by Prob. 6. of the same: when they are found, mul∣tiply them by their respective lengths, as was directed in the foregoing Problem, and the product gives the Solidity required.

XII. To find the Solidity of a Wall, Rampart, Curtain, &c.

This is done by Problem 1. of this Chap∣ter.

CHAP. V. Of Measuring Concave Bodies.

MEasuring Concave Bodies is a Science which teaches us to find their capaci∣ty that so we may know how much they contain: Cylindro-metrical Bodies have need of a Cylindro-metrical Scale: and Cubo-me∣trical Bodies have need of a Cubo-metrical Scale: So that these two Scales are made use of in the measuring of Concave Bodies.

PROBLEMS.

I. To find the Capacity of a Concave Parallele∣piped.

By which I understand all square Bodies; as Chambers, Granaries, Cellars, Towers, Pits, &c. So that if one would know how much they contain, he must,

• 1. With his Cubo-metrical Scale measure, the height, length, and depth.
• 2. Multiply the length by the breadth, and their product by the height.
• 3. Examine how many measures the whole

• sum takes up upon the Scale, and so the requi∣red Capacity may be found.

II. To find the Capacity of a Ditch.

1. Measure its breadth at the top, AF, 15 Fathoms and 1 foot.

2. Measure its breadth at the bottom CD 11 fathoms 5 feet.

3. Measure its depth CG, 9 fathoms.

4. Measure its length, 496 fathoms.

5. Add the 2 breadths together, and mul∣tiply the half of the sum by the length of the Ditch; there will be 6438 fathoms.

6. Multiply these 6438 fathoms by the depth 9 fathoms, and the product will be 57942 fathoms, or 347652 feet, which gives the Capacity requir'd.

III. To find the Capacity of Columns, Towers, and other Prisms.

1. With a Cubo-metrical Scale measure the height of the Prism.

2. Find the Area of the Base.

3. Multiply that number by the height, and you will find the Capacity.

IV. To find the Capacity of any regular Py∣ramid.

1. Measure the Area of the Base with a Cubo-metrical Scale.

2. Measure the height likewise, the same way.

3. Multiply the height upon the Area of the Base.

4. Divide the Product by 3, and the Quotient will give you the Capacity.

V. To find the Capacity of a Cone.

1. Measure the Area of the Base with a Cubo-metrick, or Cylindro-metrick Scale.

2. Multiply this number by the third of the height.

3. Divide the Product by 3, and the Quotient will give you its Capacity.

VI. To find the Capacity of a Cylinder.

1. With a Cylindro-metrical Scale find the Area of its Base.

2. Multiply that by the height of the Cylinder; the Quotient gives its Capaci∣ty.

VII. To find the Capacity of a Cylinder whose Bases are unequal.

1. Find the Area of each Base by the Cy∣lindro-metrick Scale.

2. Add the sums together, and multiply its height by the half of that sum; the Quo∣tient will give its Capacity.

VIII. To find the Capacity of a Barrel whose Heads are equal.

We must take notice that the Rod by which a Vessel is to be gauged, is marked with different measures; and so one side of the Gauging••Rod is for the length of a Bar∣rel, and the other for the depth: In answer therefore to the Question propos'd,

• 1. Measure the Area of the bottom AB.
• 2. Measure the thickness with the Gaug∣ing-Rod, and by that take the Surface of the Belly of the Barrel.
• 3. Add these two sums together, and mul∣tiply the half by the length of the Ves∣sel, so the Product gives you its Capaci∣ty.

IX. To find the Capacity of a Barrel whose Heads are unequal.

1. Measure the Diameter of the Head A, and by this known Diameter, measure its Surface.

2. Measure the other Diameter EF, and thereby find the Area of that Head.

3. Add these two sums together.

4. Find the depth of the Barrel CD, by which you find the Surface of its Circumfe∣rence; by Prob. 22. Chap. 2. Book 1. and by Prob. 10. Chap. 3. Book 2.

5. Add the sum of this Surface to the sum of the two Heads; then multiply the half of all these added together, by the length of the Barrel: this will give you the Capa∣city.

CHAP. VI. Of Transmutation.

TRansmutation is a Science which tea∣ches us to encrea••e or diminish any given Figures, or to give them another form. We ought to take notice, that what is here said of Figures, may also be under∣stood of Bodies, when one would give them another shape which they had not before: But since we produ'd somewhat of this matter in Chap. 2. Book. 1. from Prob. 21. to the end of the Chapter, we shall at present only treat of the Transmutation of Bo∣dies.

PROBLEMS.

I. To turn a Cylinder into a Parallelepiped of the same height.

1. Instead of the Base of the Cylinder CBA, make a Square ABCD, by Prob. 23. Chap. 1. Book 1.

2. Raise upon ABCD four Perpendicu∣lars of the height of AB, which may repre∣sent

the height of the Cylinder, as EF GH.

3. Joyn EF, EH, FG, and GH toge∣ther by right lines; and you have the Pa∣rallelepiped.

II. To turn a Cone into a Pyramid of the same height.

1. Make a Triangle, Square, Pentagon, or what other Polygon you please, equal to the Base of the Cone CBD, and let it be the Base of a Pyramid, as ABCD.

2. Raise from the middle of the Base F a Perpendicular, of the height of the Cone BA, as FE; then draw from ABCD lines to the point, and the Pyramid is made.

III. To turn a Parallelepiped into a Cylinder.

Make a Circle equal to the Base of the Pa∣rallelepiped, and raise this Circle to be as high as the Parallelepiped, and the work is done.

IV. To turn a Pyramid into a Cone.

1. Make a Circle equal to the Base of the Pyramid.

2. Raise upon its Center a Perpendicular as high as the Pyramid, as BA.

3. Draw the extremities of the Diameter CD together till they meet at the top of the Perpendicular A, and the Cone will be finished.

V. To turn a Prism or a Cylinder into a Py∣ramid or a Cone of the same height, or the contrary.

1. Enlarge the Base of the Cylinder three times as much as it is, by Prob. 1. Chap. 2. Book 1.

2. Upon its Center raise a Perpendicular as high as the Cylinder, and draw the ex∣tremities of the Diameter together to the top of the Perpendicular, and the Pyramid will be made: In like manner you may turn a Cylinder or a Prism into a Cone: On the contrary, when you would make a Prism or a Cylinder equal to a Pyramid or a Cone, you must make the Base of the Py∣ramid or of the Cone three times less, and

then upon this Base erect a Cone or Pyra∣mid of the height of the Prism or Cylinder given.

VI. To make a Cube equal ta a Parallepiped.

1. If the Base of the Parallelepiped be square, find a mean proportional between its height and one side of the Base; this mean proportional will be the true mea∣sure of the Cube required.

2. If the Base be only a Parallelogram, turn it into a Square by Prob. 25. Chap. 2. Book 1. then go on as at first, and your work will be done.

VII. To make a Cube equal to a Cylinder given.

1. Make a Parallelepiped equal to a Cy∣linder given, by Problem 1. of this Chap∣ter.

2. Afterwards make a Cube equal to that Parallelepiped, by the precedent Problem, and you have the Cube desired.

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VIII. To make a Cube equal to a given Co••e.

1. Make a Parallelepiped equal to a gi∣ven Cone, by Problem 5•• of this Chap∣ter.

2. Afterwards make a Cube equal to that Parallelepiped by Problem 6. of this Chap∣ter.

IX. To make a Cube equal to a Pyramid.

This Problem must be wrought like the last foregoing.

X. To make a Cone equal to a Globe.

1. Take one of the great Circles of your Globe, and make it four times bigger, by Prob. 11. Book 1. Chap. 2.

2. Raise a Cone upon this Base as high as the Semi-diameter of the Globe AB.

3. The Cone CDE will be equal to the Globe ABD.

XI. To make a Cube equal to a given Globe.

1. Make a Cone equal to a Globe, by the foregoing Problem.

2. Make a Parallelepiped equal to this Cone, by Prob. 5. of this Chapter.

3. Lastly, make a Cube equal to this Pa∣rallelepiped, by Prob. 6. of the same Cha∣pter.

Here we conclude our Treatise of Geome∣try, being perswaded, that the things which we have said, will be enough for those who would make any Progress in Fortification.

FINIS.