THE MEASURER's GUIDE: OR, THE Whole ART OF Measuring MADE Short, Plain and Easie.  
SHOWING,  

1. How to Measure any Plain Superficies.  
2. How to Measure all sorts of Regular Solids.  
3. The Art of Gaugeing.  
4. How to Measure Artificers Work, viz. Carpenters, Joiner's, Plasterers, Painters, Paviers, Glaziers, Bricklayers, Tylors, etc. of Singular use to all Gentlemen, Artificers and others.   
By JOHN BARKER.  
LONDON: Printed for Tho. Salusbury, at the King's Arms, next Clifford's Inn-Lane, by St. Dunstan's Church in Fleetstreet.   

TO EDW. HUBBALD Gent.  
THIS Small Treatise of Measuring, is Humbly presented, and Dedicated; By,  
SIR, Your Kinsman, and Humble Servant, J. BARKER,   

To the READER.  
THE Art of Measureing is of such Singular Use to all Sorts of Persons, that to say any thing of it, would be to as little Purpose, as to tell any Person it is day, when the Sun is in the Meridian. As to this Small Treatise, the Method I have used therein is both Short, Plain, and Easy; So that the Meanest Capacity may Attain to Measure, without the help of a Master; There being more Variety in this than in any other Book that is yet Extant, which Treats upon this Subject. In the first two Chapter, I have showed bow to Measure any sort of Superficies, and Solids of what Form soever. Wherein I have set forth the great Errors committed by the Common Measurers, in Measuring of Round Timber. In the Third Chapter, I have Taught how to Gauge any sort of Vessel. And in the Fourth Chapter I have given Instructions how to Measure all Artificers Work, where I also show how to find the Superficial Content of any Regular Solid; All which is Performed without the help of Tables: For many Times he that uses himself to Measure by the help of Tables, can Perform nothing without 'em; farther, if he leaves his Tables behind him, or that they come by any Mischance, he is Infinitely at a Loss, insomuch as he can do nothing; and such a Person had as good leave his Brains behind him. Whereas any Person may easily carry all the Rules here Delivered in his Head, and thereby Measure any thing with as much ease and expedition as by most Tables. All that I shall Add is this, that as I have been persuaded, after many Importunities, to Publish these few Sheets for Public Good, So if the Reader Reap any Benefit thereby I have my Ends, in proveing Serviceable to the General Advantage of my Country.   

THE CONTENTS.  


HOW to Measure any Plain Superfices, pag. 1.  
How to find the Superficial Content, of a Geometrical Square pag. 1  
How to find the Superficial Content of a Paralleogram, or Long Square pag. 2  
How to find how much in Length of any Long Square Superfices, goes to make a Superficial Foot pag. 9  
How to find the Superficial Content of a Right lined Triangle pag. 7  
How to find the Superficial Content of a Poligon, or any Equal sided Superfices pag. 9  
How to find the Superficial Content of a Trapezium, or Table pag. 11  
How to Reduce Irregular Figures into Trapeziums and Triangles, and to find the Content thereof pag. 14  
By the Diamitre of a Circle, to find the Circumference pag. 15  
How by the Circumference of the Circle, given to find the Diamitre pag. 16  
How to find the Content of a Circle, having the Diamitre and Circumference given pag. 17  
How to find the Superficial Content of a Circle, having only the Diamitre given pag. 18  
How to find the Superficial Conetnt of a Circle, having the Circumferance only given pag. 20  
How by the Content of a Circle given, to find the side of a Square, the Superficial Content of which Square shall be equal to the Superficial Content of the Circle pag. 21  
How by the Circumference of a Circle given to find the side of an Inscribed Square pag. 22  
By the Diamitre of a Circle given, to find the side of an Inscribed Square pag. 23  
How to find the Superficial Content of an Oval pag. 23  
How to Measure all sorts of Regular Solids pag. 25  
How to find the Solid Content of a Cube pag. 25  
How to find the Solid Content of a Cube pag. 25  
How to find the Solid Content of Parallelepidon or Long Square pag. 27  
How to find the Solid Content of Cilinder pag. 28  
How to know how much in Length of any Solid Body, having equal Bases, goes to make a Solid Foot thereof pag. 34  
How to find the Solid Content of a Cone pag. 35  
How to find the Solid Content of a Pyramid pag. 37  
How to find the Solid Content of the Segment of a Cone or Pyramid pag. 39  
How to find the Sold Content of a Globe, or Sphere, the Diamitre being given pag. 42  
How to find the sold Content of a Globe or Sphere, the Circumference being given pag. 43  
How to find the Solid Content of any Solid Body of what form soever, such as Geometry can give no Rule for the Measuring thereof pag. 44  
Of GAUGING pag. 46  
How to Gage a Cubical Vessel pag. 46  
For Ale, Beer, or Wine pag. 49  
How to find the Content of a Square Vessel in Gallons of Ale or Wine pag. 49  
How to Perform the same by the Line of Numbers pag. 52  
How to find the Content of a Cylinder Vessel in Gallons of Ale, Beer or Wine pag. 53  
How to perform the same by the Line of Numbers pag. 56  
How to find the Content of any Pipe, Butt, Barrel, Hogshead or Cask, pag. 37  
How to Perform the same by the Line of Numbers pag. 61  
How to Measure Brewer's Tuns or March Fatts pag. 62  
How to Perform the same by the Line of Numbers pag. 67  
How to Measure an Oval Tun pag. 68  
How to Perform the same by the Line of Numbers pag. 70  
How to turn Barrels into Gallons, Beer Measure pag. 71  
How to Perform the same by the Line of Numbers pag. 72  
How to turn Gallons of Beer or Ale into Wine pag. 72  
How to Perform the same by the Line of Numbers pag. 73  
How to turn Gallons of Ale into Barrels pag. 75  
How to Perform the same by the Line of Numbers pag. 76  
How to Gage a Ship, and thereby to find how many Tuns her Burden is pag. 76  
Of Measuring Artificers Work viz. Carpenters, Joiner's, Plasterers, Painters, Paviors, Glasiers, Bricklaiers etc. pag. 77  
How to Measure Carpenter's Work pag. 82  
How to Measure the Roof of any Building pag. 87  
How to Measure the sides of any Timber Building pag. 88  
How to Measure Paving, Painting, Wainscoting and Plastring, etc. pag. 91  
How to find the Superficial Conten● of a Cylinder pag. 91  
How to find the Superficial Content of a Paralleleppidon or Square Pillar pag. 98  
How to find the Superficial Content of a Cone. pag. 98  
How to find the Superficial Content of a Pyramid pag. 99  
How to find the Superficial Content of a Globe or Sphere pag. 99  
How to Measure Glaziers Work pag. 99  
How to Measure Bricklaiers Work pag. 104  
How to Measure Chimneys pag. 125  
How to Measure Tyling pag. 128     

ERRATA.  
Pag. 1. in the Title, Instead of Superfices, Read Superficies. P. 54. L. 5. for 44. 64 Inches, Read 4464. P. 25. L 5. for Superfices Read Superficies.   

THE MEASURER's GUIDE: OR, THE Whole ART OF Measuring MADE Short, Plain and Easie.  

CHAP. I. How to measure any Plain Superfices.  

PROP. I.  
HOW to find the Superficial Content of a Geometrical Square.  

Example.  
Admit there is a Square represented by the Figure A B C D, whose sides are each Nine Feet, whose Area, or Superficial Content is required.   square ABCD 
To find which, multiply its side 9 by its self, and the product will be 81 Feet, which is the Content of that Square.    

PROP. II. How to find the Superficial Content of a Parallelogram, or long Square.  
Multiply the length by the breadth, and the Product will be the Content thereof.  

Example.  
Admit there is a long Square represented by the Figure B C D E, whose length B C is Twenty Two Inches, and breadth B D Twelve Inches, whose Content is required.   long square (rectangle) BCDE 
To find which, multiply the length 22, by the breadth 12, and the Product will be 264 Inches, which is the Content of the long Square required, which you may reduce into Feet, by dividing the same by 144, the Number of Inches contained in a superficial Foot, and the Quotient will be 1 Foot, and 120 will remain with 120/144 of a Foot, which is 10 Inches.  
But this way of Working being very tedious and troublesome, by reason of the Reduction; for your Multiplier, and Multiplicand, must be reduced both into the Lowest Denomination, and afterwards reduced into Feet. I would advise the young Artist to work by Decimals; and for that end, to provide himself of a Twofoot Rule, on which each Foot shall be divided into Ten equal parts, and each of those parts into Ten other equal parts: So is the Foot divided into 100 equal parts, and thereby is fit to take the Dimensions in Feet, and parts of a Foot; by which Rule, if the Dimensions of the Square in the last Example, was taken, the length B C would be 1. 83 Feet, and the breadth, B D, 1 Foot. Then if you multiply the length by the breadth, as before, the Product will give 1. 83, or 1 Foot 10 Inches very near. So is the Content found without any more trouble; which Method I have made use of in some of the Examples in the following Discourse of Measuring of Superfices and Solids; but in such sort, as any person that understands not Decimal Arithmetic, may easily apply them to Vulgar Arithmetic. But generally I have given the Examples in Integers or Feet.  
If it happens, that any long Square Superfices, be broader at one end than at the other, whose Content is required, first add the two ends together, and take half their Sum for the true breadth, and multiply it into the length, and the Productwill be the Content thereof.   

Example.  
Admit there is a piece of Glass, or any other Superfices, that is One Foot broad at one end, and Two Feet broad at the other, and Six Feet long, whose Content is required.  
To perform which, add the two ends together, viz. One Foot and Two Feet, and their Sum will be Three Feet, the half of which is 1. 5 Foot, or One Foot Six Inches; which multiply into the length Six Feet, and the Product will give Nine Feet for the Content required.    

PROP. III. How to find how much in Length of any Long Square Superfices goes to make a Superficial Foot.  
Divide 1 by the given breadth, and the Quotient will give how much of the length of any Superfices goes to make a Superficial Foot thereof.  

Example.  
Admit there is a long square piece of Board, or any other Superfices, whose breadth is 1. 32 feet, and it be required, to know how far one ought to measure along the same to make a superficial foot thereof.  
To perform which, divide 1 by the breadth, 132, and the Quotient will give 757, which is Nine Inches and something more; and so much of the Length goes to make a superficial foot, as in the following Work.    

PROP. IU. How to find the superficial Content of a Right lined Triangle.  
Although Right lined Triangles are of several Kind's and Forms; as, first, in respect of their Angles, they are either Right Angled or Obliqne Angled, Acute Angled or Obtuse Angled. Second, in respect of their sides; They are either an Equilateral, Isosceles, or Scalenium Triangle. But seeing they are all measured by one and the same Rule, I shall add but One Example for all; and take this for a general Rule.  
Multiply half the Perpendicular into the Base, and the Product will be the Content of the Triangle; or Multiply half the Base into the whole Perpendicular, and the Product will give the same thing: Or if you Multiply the whole Base into the whole Perpendicular, half the Product is the Content of the Triangle.  

Example.  
Admit there is a Triangle A B C, whose Base A C is 72 feet, and the Perpendicular B Q is 36 feet, whose Content is required.   triangle ABC 
To find which, multiply half the Perpendicular, which is 18 feet, into the Base 72 feet, and the Product will be 1296 feet, which is the Content of the said Triangle.  
Or thus,  
Multiply half the Base, which is 36 feet, into the Perpendicular 36 feet, and the Product will give 1296 feet, as before, for the Content of the said Triangle.    

PROP. V How to find the Superficial Content of a Polygon, or many equal sided Superfices.  
Multiply the Length of the Perpendicular let fall from the Centre to the middle of one of the sides, into half the Sum of the sides, or half the Perpendicular into the Sum of the Sides, and the Product shall be the Content of Polygon.  

Example.  
Admit the Polygon given be a Hexagon, represented by the Figure A B C D E F (or a Six equal sided Figure), whose side is 19, and the Perpendicular, or ☉ G, is 16 feet.   hexagon ABCDEF 
Multiply half the Sum of the sides, which is 57, by the length of the Perpendicular 16, and the Product will be 912 feet, which is the Content of the Polygon required.  
Or,  
Multiply the Sum of all the sides, which is 114, by half the Perpendicular 8, and the Product will give 912, as before, for the Content of the said Polygon.  
The reason of this manner of Working is very plain, if from the Centre you draw the Lines ☉ E and ☉ D, thereby making the Triangle E ☉ D, whose Content (by the third Proposition) is found by Multiplying the Perpendicular ☉ G 16 into half the side E D, which is 9.5, and there being in the Hexagon 6 such Triangles, therefore the Perpendicular ☉ G being Multiplied into 6 times half, the side D E produceth 912 for the Content as before.    

PROP. VI How to find the Superficial Content of a Trapezium or Table.  
A Trapezium is a Figure consisting of four unequal sides, and as may unequal Angles, as is the Figure following, A B C D.   irregular quadrilateral or trapezium ABCD 
To Measure which Trapeziums you must first reduce it into two Triangles, by drawing the Diagonal Line A D, and so is your Figure reduced into two Triangles, viz. the Triangle A B D and A C D, then if you let fall Perpendiculars from the two Points B and C, if you find the Content the two Triangles, and add their Content together, the Sum will be the Area or Content of the whole Trapezium.  

Example.  
Having drawn the line A D, and so Reduced the Figure into two Triangles, and let fall the Perpendiculars from B and C (the Base A D is common to both Triangles) admit the Base A D to be 24, and the Perpendicular at C to be 8, and the Perpendicular at B to be 10, then, if according to Proposition the 3d, you Multiply 24, the Base by 4, half the Perpendicular, the Product will give 96 for the Content of the Triangle A C D.  
Likewise, if you Multiply 24, the Base by 5, half the other Perpendicular, the Product will give 120 for the Content of the Triangle A B D.  
Then if you add the Content of these two Triangles together, viz. 96 and 120, their Sum is 216, which is the Content of the whole Trapezium A B C D.  
Or thus,  
Add the two Perpendiculars, viz. 10 and 8 together, and Multiply half the Sum, which is 9, into the common Base A D 24, and the Product will give 216, for the Content of the Trapezium A B C D, as before.  
Or thus,  
Multiply half the common Base A D, which is 12, into the Sum of the two Perpendiculars, which is 18, and the Product will give the same as before, viz. 216, for the Content of the Trapezium.    

PROP. VII. How to Reduce Irregular Figures into Trapeziums and Triangles, and to find the Content thereof.  

Example.  
Let the Figure given to be Measured be A B C D E F G H.   irregular polygon ABCDEFGH 
To perform which, Divide the Figure into Trapeziums and Triangles, according to your Fancy, and as the nature of the Plain will bear, then find the Content of the Trapeziums and Triangle, as is taught in the third and fifth Propositions, and add the several Contents together, so will the Sum be the Content of the Irregular Figure desired. As in this Example, draw the lines B G and D F, so is the whole Figure divided into two Trapeziums and one Triangle, viz. the Trapeziums A B G H, and C D F G, and the Triangle D E F, the Content of which being severally found by the third and fifth Propositions, and added together, their Sum will be the Content of the whole Figure.    

PROP. VIII. By the Diameter of a Circle to find the Circumference.  
To find which, the Analogy, or Proportion, is as followeth.  
As 7 is to 22, 
 So is the Diameter of the Circle, 
 To the Circumference.  

Example.  
Admit there is a Circle represented by the Figure A B C D, whose Diameter A C is 28, whose Circumference is required.   circle ABCD with diameter AC 
To find which, work by the foregoing Proportion, so will you find the Circumference to be 88, as in the following work.    

PROP. IX. How by the Circumference of a Circle given to find the Diameter.  
To find which, the Analogy, or Proportion, is as followeth.  
As 22 is to 7 
 So is the Circumference of any Circle 
 To the Diameter.  

Example.  
Admit the Circumference, the foregoing Circle A B C D be given, which was found to be 88, whose Diametre is required.  
Work by the foregoing Proportion, and so will you find the Diametre of the said Circle to be 28, as in the following Work.    

PROP. X. How to find the Content of a Circle having the Diameter and Circumference given.  
To find which, Multiply half the Circumference by half the Diameter, and the Product is the Superfitial Content of the Circle.  

Example.  
Admit there is a Circle represented by the foregoing Figure, whose Circumference A B C D is 88, and the Diameter A C 28, whose Superficial Content is required.  
To find which, Multiply half the Circumference, which is 44, by half the Diameter, which is 14, and the Product will give 616 which is the Content of the said Circle required.    

PROP. XI. How to find the Superficial Content of a Circle, having only the Diameter given.  
To perform which, first square the Semy-Diamiter, and then work by the following Proportion.  
As 7 is to 22, 
 So is the Square of the Semy-Diamiter 
 To the content of the circle.  

Example.  
Admit there is a circle whose Diameter is 28, and it be required to find the content of the said circle.  
To perform which, first halve the Diameter, the half of which is 14, which 14 Multiply into its self, and the Product will give 196, which is the Square of the Semy-Diamiter, then work by the foregoing Proportion, and so will you find the superfitial content of the said circle to be 616, as in the following work.    

PROP. XII. How to find the Superficial Content of a Circle, having the Circumference only given.  
To find which, first Square the Circumference, then work by the following Proportion.  
As four times 22, which is 88, is to 7, 
 So is the Square of the circumference 
 To the content of the circle.  

Example.  
Admit there is a circle whose circumference is 88, and it be required to find the content of the said circle.  
To find which, first square the circumference 88, whose square you will find to be 7744, then work by the foregoing proportion, so will you find 616 to be the content of the said circle, as in the following work.  
   

PROP. XIII. How by the Content of a Circle given to find the side of a Square, the superficial Content of which Square shall be equal to the superficial Content of the Circle.  
To perform which, extract the square Root of the Content of the Circle, which Root is the side of the Square desired.  

Example.  
Admit the Content of the given Circle be 144, and it be required to find the side of a Square, the superficial content of which Square shall be equal to the superficial content of the Circle.  
To find which, extract the square Root of the content of the Circle, 144, whose Root you will find to be 12, which is the side of the Square desired.    

PROP. XIV. How by the Circumference of Circle given, to find the side of an Inscribed Square.  
To find which, work by the following Proportion:  
As 1.000 is to 0.225, 
 So is the circumference of any circle 
 To the side of the inscribed Square.   

PROP. XV. By the Diameter of a Circle given, to find the side of an inscribed Square.  
To find out which, work by the following proportion. 
 As 1.0000 is to 0.7071, 
 So is the Diameter of any Circle 
 To the side of the Inscribed Square.   

PROP. XVI. How to find the Superficial Content of an Oval.  
Multiply the length by the breadth, and that Product Multiply by Eleven, and that Product Divide by Fourteen and the Quotient shall give the Content of the Oval required.  

Example.  
Admit there is an Oval represented by the following Figure, whose length A C is 50, and breadth B D 40, whose superficial content is required.   oval ABCD 
To find which, Multiply the length 50 by the breadth 40, and the Product will be 2000, which Multiply by 11, and the Product will be 22000, which Divide by 14, and the Quotient will be 1571, which is the content of the Oval required.     

CHAP. II. How to Measure all sorts of Regular Solids.  
HAving in the last Chapter shown you how to Measure any Plain Superfices, which if well understood, the content of all Solid Bodies, whose ends are of the same Dimensions, will be found by the following Rule.  
First find the superficial content at the end by some of the Propositions foregoing, and Multiply the same by the length, and the Product will give the solid content.  

PROP. I. How to find the solid Content of a Cube.  
Multiply its side into its self, and that Product Multiply by the side again, and that Product will be the solid content of the cube desired.  

Example.  
Admit there is a Cube whose side is 6 Feet, represented by the Figure A B C D E F G, whose solid content is required.   cube ABCDEFG 
To find which, Multiply the side 6 by its self, and the Product will be 36, which again Multiply by 6 and the Product will be 216 Feet which is the solid content of the said Cube.    

PROP. II. How to find the Solid Content of a Parallelepipedon, or long Square.  
First find the superficial content at the end by Propositions, the first or second of the last Chapter, which Multiply into the length, and the Product will be the solid content.  

Example.  
Admit there is a Parallelepipedon, represented by the Figure H I K L M N O, whose Square at the end is 3 feet long, and 2 feet broad, and its length, K M, is 39 feet, whose solid content is required.   parallelepiped HIKLMNO 
To find which, Multiply the length at the top H I 3, by the breadth H K 2, and the Product will give 6, which is the superficial content at the top, which Multiply into the length K M 39 feet, and the Product will be 234 feet, which is the solid content of the said Parallelepipedon.    

PROP. III. How to find the solid Content of a Cylinder.  
First find the superficial content of the circle at the top or base (by Proposition the 10, 11 or 12 of the last Chapter) then Multiply it by the length, and the Product will be the solid content thereof.  

Example.  
Admit there is a Cylinder represented by the Figure G, whose circumference of the circle at the end or base is 44 Inches, and the length of the Cylinder 16 Inches, whose solid content is required.   cylinder, Figure G 
To find which, first find the superficial Content of the Circle at the Top or Base, by Proposition the 11th of the last Chapter, which you will find to be 154 Inches; which multiply by the length 16 Inches, and the Product will be 2464 Inches, which is the Content of the Cylinder required; which you may reduce into feet, by dividing the same by 1728, the number of Inches contained in a solid Foot, and the Quotient will give 1 Foot, and there will remain 636, which is 736/1728 of a Foot. But, as I said before, the best way is to work by Decimals; but I have added this Example for their sakes that understand not Decimals: But this, as well as any other Question, may be better resolved by Cross Multiplication than by this Common Method; to perform which, if the Learner understands it not already, I refer him to the last Chapter of this Book; and you will find the Answer to be 1 Foot 4 Inches, and 10 Seconds, as in the following Work:  
 
But here I presume it may not be improper to show the great Error that common Measurers are guilty of in the measuring of Round Timber, whose Method is thus: they girt the Tree, or piece of Timber about, and take the one fourth part of the Circumference for the true Square, which is very erroneous, as may appear by the last Example, where the Circumference, or Girt of the Cylinder was 44, and the Content of the Circle 154. Whereas, if you had taken the one fourth of the Circumference 44, which is 11, for the Square Root of the Circle, which multiplied into its self, the Product is but 121; whereas the Content of the Circle is found to be 154, from which subtract 121, and the Remainder will be 33. So much doth the common-way want of the true Content of the Circle; from whence it is evident, that according to the common way of measuring, there is lost something above the ●/5 part of any Round Timber so measured.  
But all that can be said in Defence of this Custom, is, That all Trees growing round, must be hewed square, before it can be fit for any use; so this Advantage in measuring may be very well allowed for what goes to waste in chips, they being good for nothing but the fire; and although Measurers think not of this Excuse, but take their Rule for an absolute Truth, I presume was the first Occasion of the Rule, though it is a very Unreasonable Allowance, as may thus appear; the inscribed Square of the circle of the Cylinder, in the last Example, will be 10 Inches, as you may prove by the following Proportion.  
As 1.000 is to 0.225, so is the circumference 44 to the inscribed Square 9.900, which wants but very little of 10 inches; and this is the greatest Square such a Round piece of Timber can be hewn into, and this multiplied into its self, gives 98 inches and .010 parts for its content. Now if you add the content of the inscribed Square 98, and the true content of the circle 154, together, their Sum will be 252, and the Middle, or Mean thereof (which is half their Sum) is 126, and the content, after the common way, was but 121. So that you may see this Rule gives an indifferent Allowance both to the Buyer and Seller. But I presume, notwithstanding all that hath been said, that the true content aught to be given, and the Measure exactly known. And as to the Waste thereof, Goodness or Badness of the Timber, there ought to be an allowance made in the Price.    

PROP. IU. How to know how much in length of any Solid Body, having equal Bases, goes to make a Solid Foot thereof.  
First find the superficial content at the End or Base, by which divide 1 (to wit, 1 solid foot), and the Quotient will be the Length that goes to make a foot solid thereof.  

Example.  
Admit there is a piece of Timber, or any other solid body, that is terminated at each end by two equal Geometrical Squares, whose side is 1. 55, and it be required to know how much of the Length thereof goes to make a foot solid.  
To perform which, first find the superficial content at the End or Base, by Proposition the 1st, of the last Chapter, which you will find to be 2. 4025, by which divide 1, and the Quotient will give .416 foot, or 5 inches, as in the following Work.  
   

PROP. V How to find the solid Content of a Cone.  
First find the superficial content of the circle at the Base by (Proposition the 10th, 11th or 12th of the last Chapter, which multiply by ½ of the height, and the Product will be the solid content of the Cone.  

Example.  
Admit there is a Cone represented by the Figure B, whose Diametre at the Base is 5 feet, and its Height, or Altitude, is 18 feet, whose solid Content is required.   cone, Figure B 
To find which, first find the superficial content of the Base (by Proposition the 11th of the last Chapter), which you will find to be 19 64, which multiply by ⅓ of the height, which is 6, and the Product will give 117. 84 feet, which is the solid content of the Cube required.    

PROP. VI How to find the Solid Content of a Pyramid.  
A Pyramid is a Silid comprehended under plain Surfaces, and forms a Triangular, Quadrangular, or any Mutangular Base, diminishing equally less and less, till it diminishes in a Point at the Top as a Cone.  
To find the solid content of any such Figure, first find the superficial content of the Base, by some of the Propositions of the last Chapter; which multiply into ⅓ of its Altitude, or Height, and the Product will be the solid content thereof.  

Example.  
Admit there is a Pyramid represented by the Figure M, whose Base is a Hexagon, the side of which is 25 feet, and its Perpendicular height is 60 feet, whose solid content is required.   pyramid, Figure M 
To perform which, first find the superficial content of the Base, by Proposition 5. of the last Chapter; viz. by multiplying half the Sum of the sides, which is 75 feet, by the Perpendicular (let fall from the Centre to the midst of one of the sides), which will be 13 feet, and the Product will give 975 feet for the superficial content of the Base; which multiply by ⅓ of the height 20, and the Product 19500 feet for the solid content of the Pyramid required.    

PROP. VII. How to find the Solid Content of the Segment of a Cone or Pyramid.  
Here note, that all tapering Timber, or Stone, etc. whose Bases are Regular Figures, are Segments of either a Cone or Pyramid.  
To find the solid content of any such Segment, first find the superficial content of both the Bases, by some of the Propositions of the last Chapter, then multiply the superficial content of the greater Base by the superficial content of the lesser, and extract the square Root of that Product; then add that square Root, and the superficial content of the Two Bases together, and their Sum multiply by ⅓ part of the Length, and that Product will give the solid content of the Segment required.  

Example.  
Admit there is a Segment of a Pyramid, whose Bases are Squares, represented by the Figure P G, and the side of the great Square at G is 1.5 foot, or 1 foot 6 Inches, and the side of the lesser at P is .5 foot, or 6 Inches, and the Length of the said Segment is 30 feet, whose solid content is required.   segment of a pyramid, Figure PG 
To find which, first find the superficial content of both the Ends by Proposition the first of the last Chapter, the greater of which you will find to be 2.25 feet, and the lesser 0.25 foot, then multiply the superficial content at the greater end at G 2.25, by the superficial content of the lesser at P .25, and the Product will be .5625 feet; then extract the square Root of that Product, which you will find to be .75 feet, to which add the superficial content of the Two Ends, and their Sum will be 3.25, which multiply by ⅓ part of the Length, which is 10 feet, and the Product will▪ be 32.50 feet, which is the solid content of the said Segment required, as in the following work,  
If you work the same by Vulgar Arithmetic, you will find the content to be 32 feet 6 Inches.    

PROP. VIII. How to find the Solid Content of a Globe, or Sphere, the Diametre being given.  
To find which work by the following Proportion,  
As 6 times 7, which is 42, 
 Is to 22, 
 So is the Cube of the Diametre of any Globe 
 To the solid content thereof.  

Example.  
Admit there is a Globe or Sphere represented by the Figure following, whose Diametre is 24 Inches, and be required to find the solid content thereof.   globe or sphere with diameter 24 inches 
To perform which, first cube the Diametre, 24 Inches, whose Cube you will find to be 1.3824 Inches; then work by the before-going Proportion, and say, As 42 is to 22, so is 13824 to the solid content of the said Globe, which you will find to be 7241 6/42 Inches.    

PROP. IX. How to find the Solid Content of a Globe, or Sphere, the Circumference being given.  
To find which, work by the following Proportion,  
As 1.0000 
 Is to 0.16887, 
 So is the Cube of the Circumference of any Globe 
 To the solid content thereof.   

PROP. X. How to measure any Body that is hollow.  
Admit it were required to find the solid content of a hollow Tree, or of any other hollow Body whatsoever; To perform which, first you must find the solid content thereof, as though it were not hollow; then find the solid content of the Concavity, as though it were Massy, and subtract it from the whole content, and the Remainder will be the solid content of the hollow body.   

PROP. XI. How to find the Solid Content of any Solid Body, of what Form soever, such as Geometry can give no Rule for the measuring thereof.  
To perform which, take some convenient Vessel, and fill the same to a convenient height with clean water, and make a Mark just how high the water reacheth; then put the Body (whose solid content is required) into the water, which will cause the water to rise above the Mark; then take so much of the water out of the Vessel, as is raised above the Mark, and put the same into an hollow Cube, which have in readiness for that purpose; then find the solid content of the said Cube, so high as the water reacheth, which shall be the solid content of the said body required.    

CHAP. III. Of GAUGING.  
GAuging of Vessels is no other than finding the solid content of the concavities in Inches, as taught in the last Chapter, and Dividing the same by the number of cubick Inches, contained in a Gallon of the Liquor contained in the same, which according to the Establishment of the Excise for Ale, is 282, and for Wine, 231 Inches.  

PROP. I. How to Gauge a Cubical Vessel.  
Admit there is a cube-Vessel represented by the Figure D, whose side is 16 Inches, and it be desired to know how many Gallons of Wine or Ale the same will hold.   cube, Figure D 
To find which, first find the solid content of the said cube in Inches, as taught in Proposition the first of the last Chapter, which you will find to be 4096 Inches, which reduce into Gallons, by dividing the same by 282 for Ale or Beer, and the Quotient will be 14.5 which are the number of Gallons of Ale or Beer the said Cube will hold, and for Wine, Divide the solid content of the said cube by 231, and the Quotient will give 17.73 for the number of Gallons of Wine the said cube will hold, as in the following Work.  
 
To perform the same by the Line of Numbers.  

I. For Ale or Beer.  
Extend your Compasses always from the Gauging Point, which for Ale is 16, into the side of the cube, which in this Example is 16, and the same extent will reach from the said 16.8, being turned twice over unto 14.5 Gallons of Ale or Beer.   

II. For Wine.  
Extend your Compasses from the Gauging Point for Wine, which is always 15 2/10, into the side of the ●ube 16, and the same extent being ●urned twice over, will reach from ●he said 16, unto 17.73 Gallons of Wine.    

PROP. II. How to find the Contents of a Square Vessel in Gallons of Ale or Wine.  
Admit there is a Square Vessel represented by the figure A B C D E, whose Length A B, is 86 Inches, and Breadth A C, 41 Inches, and ' its Depth C D, 9 Inches; and it be Required to know how many Gallons of Ale, Beer or Wine, the said Vessel will hold.   square vessel ABCDE: length AB, breadth AC, depth CD 
To find which first find the Solid Content of the said Vessel, as before Taught in the Measuring of Solids; which you will find to be 31734 Inches, which Reduce into Gallons, by Dividing the same by 282 for Beer or Ale, and 231 for Wine, as in the last Proposition, so will you find the said Vessel will hold 112.53 Gallons of Beer or Ale, or 137.37 Gallons of Wine. As in the following Work.  
 

How to perform the same by the Line of Numbers.  
To perform which you must first find a mean Number, between the Length A B, 86 Inches, and the Breadth A C, 41 Inches, by Multiplying the Length 86, by the Breadth A C 41; and the Product will be 3526, and then Extracting the Square Root of the said Product, which Root is the Number required, which you will find to be 59.38 Inches  
Or Thus, by the Line of Numbers.  
Take the Middle way between 86, and 41, and you will find it the same as before, viz. 59.38 inches.   

Then for Ale.  
Extend your Compasses from the Gauging point 16.8, to the mean Number 59.38 Inches, and the same Extent will reach from the Depth 9 Inches, turned twice over unto 112.53 Gallons, of Ale or Beer, as before.   

For Wine,  
Extend your Compasses from the Gauging point for Wine 51.2, to the mean Number before found 59.38 Inches and the same Extent will reach from the Depth 9 Inches, turned twice over, to 137.37 Gallons of Wine, as before.    

PROP. III. How to find the Contents of a Cylinder Vessel in Gallons of Ale, Beer or Wine.  
Admit there is a Cylinder Vessel represented by the figure A B C G, whose Diamitre A B is 14 Inches, and the Length thereof G C 16 Inches, and it be required to know how many Gallons of Ale, Beer of Wine this said Vessel will hold.   cylindrical vessel ABCG: diameter AB, height GC 
To find which first find the Solid Contents of the said Cylinder as taught in Proposition the Third of the last Chapter, which you will find to be 24.64 Inches, which Reduce into Gallons, by Dividing the same by 282, for Beer and Ale, and 231 for Wine; and you will find the said Vessel to hold 8.73 Gallons, of Beer or Ale, and 10.66 Gallons of Wine, as in the following Work.  
 
Or thus,  
Square the Diametre A B, 14 whose Square is 196, which Multiply by the Length G C 16, and the Product will be 2136 which Divide by 359, for Beer or Ale, and by 294 for Wine, so will you find as before; the said Vessel will hold 8.73 Gallons of Beer or Ale, or 10 66 Gallons of Wine.  

How to perform the same by the Line of Numbers.  

I. For Ale or Beer.  
Extend your Compasses from the Gauging point for Ale or Beer 18.95, unto the Diametre 14, and the same Extent will reach from the Length 16, being twice turned over to 8.73 Gallons of Beer or Ale.   

II. For Wine.  
Extend your Compsses from the Gauging point for Wine 17.15 unto the Diametre 14, and the same Extent will reach, being twice turned over from the Length 16, to 10.66 Gallons of Wine.     

PROP. IU. How to find the Content of any Pipe, Butt, Barrel, Hogshhead or Cask.  

Example.  
Admit there is a Cask Represented by the Figure A B C D, whose Diametre at the Head A B is 18, and the Diametre at the Bung 32, and the Length is 40 Inches; and it be required to know how many Gallons of Ale or Wine the same will hold.   ale or wine cask, Figure ABCD 
To find which there are several ways, but I shall only mention two, the first is Mr. Oghtreds' Method, which is thus, first by Proposition the 11th. of the 3 d. Chapter of Measuring, find the Contents of the Circles at the Bung and Head, which you will find to be 254.57 Inches for the Contents of the Circle at the Head, and 804.57 for the Aria, or Contents of the Circle at the Bung, then take two Thirds of the Contents of the Circle at the Bung, which is 536.38, and one Third of the Contents of the Circle at the Head, which is 84.85, which add together, whose Sum will be 621.23, which Multiply by the Length of the Vessel 40, and the Product will be 24849.20 Inches, which is the Solid Content of the said Vessel, which Reduce into Gallons by Dividing the same by 282 for beer or Ale, and by 231 for Wine, so will you find the said Vessel to hold 88.1 Gallons of Ale, or 107.57 Gallons of Wine, as in the following Work.  
 
 
Or thus,  
Square the Diametre at the Bung, 32, whose Square is 1024, which double, and it makes 2048, than you must Square the Diametre at the Head 18, whose Square is 324 which you must add to the double of the Square of the Diametre at the Bung 2048, and their Sum will be 2372, which Multiply by the Length of the Vessel 40, the Product will be 94880, which you must Divide by 1077 for Ale, 880 for Wine, so will you find as before, the said Vessel to hold 88 104/1077 Gallons of Ale or 107 720/880 Gallons of Wine.   

How to perform the same by the Line of Numbers.  
To perform which you must find a mean Diametre, between the Diametre at the Head and Bung, thus take the difference between the Diametre at the Bung 32, and the Diametre at the Head 18, which is 14, which you must always Multiply by▪ 7, and Divide that Product by 10, and the Quotient you must Add to the Lesser Diametre and so will you have 27.8 for a mean Diametre then.  

I. For Ale.  
Extend your Compasses from the Gauging point 18.95 unto the mean Diametre 27.8 and the same distance will reach from the Length 40, being Twice turned over to 88.11 Gallons of Ale.   

II. For Wine.  
Extend your Compasses from the Gauging point 17.15 to the mean Diametre and the same Extent will reach from the length 40 being twice turned over to 107.57 Gallons of Wine.     

PROP. V How to Measure Brewer's Tuns or Marsh-Fatts  
Admit there is a Tun or Marsh-Fat, Represented by the Figure A B C D, whose Diametre at the bottom C D, is 88 and the Diametre at the top A B 80, Inches, and its Depth L I 96 Inches, and it be required how many Barrels of Ale the same will hold.   brewer's tun or marsh-fat (marsh-vat?), Figure ABCD 
To find which, first find the Solid Content of the same, as taught in Proposition the 6th. of the last Chapter which you will find to be 532625.92 Inches and 6/10, which Divide by 282 the Number of Cubick Inches contained in a Gallon, and the Quotient will be 1888.74, the Number of Gallons of Ale the same will hold, which Divide by 36, the Number of Gallons in a Barrel, and the Quotient will be 52.46, the Number of Barrels of Ale the same will hold, as in the following Work.  
 
 
 
Or thus,  
Square the Diametres of the Top and Bottom, then Multiply one Diametre by the other, and Add that Product and the Square of the Two Diametres together, and Multiply their Sum by the Depth of the said Vessel, and that Product Divided by 1077, and the Quotient will give 1888, which is the Number of the Gallons of Ale the said Vessel will hold; which Reduce into Barrels as before.  

How to perform the same by the Line of Numbers.  
First find a mean Diametre, thus Add the Two Diametres 80 and 88 together, and their Sum will be 168, then take half thereof, which is 84, for the mean Diametre: Then Extend your Compasses from the Gauging Point for Barrels (of Ale or Beer), which is 113 7/10 unto the mean Diametre 84, and the same distance will reach from the Height 96, to turned twice over unto 52.44 Barrels.    

PROP. VI How to Measure an Oval Tun.  
Admit there is an Oval Tun Represented by the Figure A B C D, whose Length at the Bottom C D is 120, and the Breadth K W 90, and the Length at the Top A B 112 and the breadth I L 84 and the Depth of the Vessel L K 40 Inches; and it be required to know how many Barrels of Ale the same will hold.   oval tun, Figure ABCD 
To find which you must first find a mean Number between the Length and Breadth of the Oval at the Bottom, viz. 120 and 90, by Multiplying the Length 120 by the Breadth 90, and the Product will be 10800, and then Extract the Square Root of that Product, and it will give 103 92 for the mean Number required, then likewise by the very same Method you must find a mean Number, between the Length at the Top 112, and the Breath 84, which you will find to be 97, which Two mean Numbers you must Add together, and their Sum will be 200.90, then take half thereof, which is 100.46, which Multiply by the Depth of the Tun 40, and the Product will be 4018 40, which Divide by 359, and the Quotient will be the Number of Gallons of Ale the same will hold, which you will find to be 1119.331, which Reduce into Barrels by Dividing the same by 36, the Number of Gallons, in a Barrel and the Quotient will give 31.09 Barrels.  

How to perform the same by the Line of Numbers.  
To perform which first find the mean Numbers between the Length and the breadth of the Two Ovals. Thus Extend your Compasses from the Breadth of each Oval to its Length; then your Compasses being opened to half that distance, will reach from the Breadth to the mean Number desired, by which Method you will find the mean Numbers as before, to be 103.92 and 97, whose Sum is as before 200.92, the half of which is 100.46 which is the mean Diametre; then Extend your Compasses from the Gauging Point for Barrels, which is 113.7 to the mean Diametre 100.46, and the same will reach from the the Depth 40, turned twice over unto 31.09 the Number of Barrels of Ale the said Vessel will hold.    

PROP. VI How to turn Barrels into Gallons, Beer-measure.  
To perform which, work by the following Proportion.  
As 1 is to 36, the Number of Gallons contained in a Barrel, so is the Number of Barrels given, to the Number of Gall●ns contained therein.  

Example.  
Admit it be required to know, how many Gallons of Beer are contained in 6 Barrels.  
To perform which by Arithmetic, work by the Rule of Three, as followeth:   

To perform the same by the Line of Numbers.  
First, set one Foot of your Compasses in 1, and extend the other to 36; then, with your Compasses at that distance, set one Foot in 6, and the other will reach to 216, the Number of Gallons contained in the said 6 Barrels.    

PROP. VIII. How to turn Gallons of Beer or Ale into Wine.  
To perform which, say, by the Rule of Three,  
As 9 is to 11, 
 So is the Number of Ale-Gallons 
 To the Number of Wine.  

Example.  
Admit there is a Vessel that contains 66 Gallons of Beer, and it be required to know, how many Gallons of Wine the same will hold.  
To perform which, by Arithmetic, work by the Rule of Three, as followeth:   

To Perform which by the Line of Numbers.  
Set one Point of your Compass in 9, and extend the other to 11; then with your Compasses at that distance set one Point in 66, and the other will reach to 80.6.  
☞ Here Note that the Proportion between 9 and 11 was found by Multiplying 282, the Number of Cubick Inches in an Ale Gallon by 9, and Dividing that Product by 231, the Cubick Inches in a Wine Gallon, by which Method you will find 11 something too much, though it is near enough for ordinary use.    

PROP. IX. How to turn Gallons of Wine into Gallons of Ale.  

Example.  
Admit there is a Vessel that will hold 44 Gallons of Wine, and it be required to know how many Gallons of Ale the same will hold.  
To perform which by Arithmetic, work by the Rule of Three, as followeth.   

To Perform the same by the Line of Numbers.  
Set one Foot of your Compasses in 11, and Extend the other to 9, then with your Compasses at that distance, set one Foot in 44, and the other will reach to 36, the Gallons of Ale the said Vessel will hold.    

PROP. X. How to turn Gallons of Ale into Barrels.  

Example.  
Admit there is a Vessel that holds 216 Gallons, and it be required to know how many Barrels are contained therein.  
To perform which by Arithmetiek, say by the Rule of Three, If 36, the number of the Gallons contained in one Barrel will have one Barrel, what will 216 Gallons have, and you will find the said Vessel to contain just 36 Barrels.   

To Perform the same be the Line of Numbers.  
To find which, set one foot of your Compasses in 36, and Extend the other to 1, then with your Compasses at that distance set one Foot in 216, and turning the other towards 1, it will cut the Line in 36, the number of Barrels contained in the said Vessel.    

PROP. XI. How to Gauge a Ship, and thereby to find how many Tuns her Burden is.  
To perform which, Measure the Length of the Keel, the Breadth of the Midship-Beam, and the Depth of the Hold; which three Numbers Multiply one into the other, and Divide the Product 95, and the Quotient will give the true Burden of any Merchant's Ship; but for M●● of War, divide the Product by 100, because in Men of War there is allowance made for Ordnance, Masts, Sails, etc.    

CHAP. IU. Of Measuring Artificers Work, viz. Carpenters, Joiner's, Plasterers, Painters, Paviers, Glasiers, Bricklayers, etc.  
BEcause the method I have used in Multiplying, may be new to many, although it varies but very little from the method used by the common Measurers; for I Work by Cross Multiplication, as they do; all the difference is, I have supposed the ●nch to be Divided into 12 equal parts, which I call Seconds, and each of those Parts to be Divided into 12. other equal Parts, which I call Thirds, etc. So is the Foot Divided Duodecimally, so that 12 Thirds makes 1 Second, 12 Seconds 1 Inch or Prime, 12 Inches 1 Foot.  
Before I proceed I shall show you how to Multiply after the Method I have used in this Chapter; to perform which, observe the following general Rules.  

General Rules.  

1. If you Multiply Feet into Feet, the Product will be Feet.  
2. If you Multiply Inches into Feet, the Product will be Inches.  
3. If you Multiply Inches into Inches, the Product will be Seconds.  
4. If you Multiply Seconds into Feet, the Product will be Seconds.  
5. If you Multiply Seconds into Inches, the Product will be Thirds.  
6. It you Multiply Seconds into Seconds, the Product will be Fourths.    

Example.  
Admit there is given 17 Feet, 11 Inches, 10 Seconds, to be Multiplied by 6 Feet, 9 Inches, 4 Seconds.  
To perform which, first set down the Multiplicand and Multiplier, as in the following Work; then Multiply the Feet in the Multiplicand 17 by the Feet, in the Multiplier 6, and the Product will be 102 Feet; then Multiply the Feet in the the Multiplicand 17, by the Inches in the Multiplier 9, and the Product will be 153 Inches, which is 12 Feet 9 Inches, which set down as in the following work, viz. the Feet under the Feet, and Inches under the Inches; likewise Multiply the Feet in the Multiplier 6, into the Inches in the Multiplicand 11, and the Product will be 66 Inches, which is 5 Feet 6 Inches, which likweise set down as you see in the work; then Multiply the Inches in the Multiplicand 11, into the Inches in the Multiplier 9, and the Product will be 99 Seconds, which is 8 Inches, 3 Seconds; set down the Inches under the Inches, and Seconds under the Seconds, as you see done; Then Multiply the Seconds in the Multiplicand 10, into the Feet in the Multiplier 6, and the Product will be 60 Seconds, which is just five Inches, which set down under the Inches; likewise Multiply the Seconds in the Multiplyer 4, into the Feet in the Multiplycand 17, and the Product will be 68 Seconds, which is 5 Inches, 8 Seconds, which set down as you see in the Work; then Multiply the Second in the Multiplier 4, into the Inches in the Multiplicand 11, and the Product will be 44 Thirds, which is 3 Seconds 8 Thirds; likewise Multiply the Seconds in the Multiplicand 10, into the Inches in the Multiplier 9, and the Product will be 90 Thirds, which is 7 Seconds and 6 Thirds, which likewise set down as in the Work; then Multiply the Seconds in the Multiplicand 10, into the Second in the Multiplier 4, and the Product will be 40 Fourths, which is 3 Thirds and 4 Fourths, which set down as in the Work, then add these several Products all up together, and the Sum will be 121 Feet, 10 Inches, 10 Seconds, 5 Thirds and 4 Fourths, which is the Product desired.  
  

Example 2.  
Admit it be required to Multiply 165 feet, 6 inches, 4 seconds by 79 feet, 2 inches, 3 seconds.  
To perform which, work in every respect as in the last Example, and you will find the Product to be 13107 feet 8 inches 9 seconds and 3 thirds, as in the following work.  
  

How to Measure Carpenters Work.  
Carpenter's Work, which is Flooring, Roofing, Partitioning, Joysting, etc. is Measured by the Square that is 10 feet every way, so that a square containeth 100 superficial feet, in Measuring of which, the thickness is not considered, but an allowance 〈◊〉 〈◊〉 〈◊〉, according to the strength or slightness of the work; so that the Measuring thereof is no other than the Measuring of a Plain Superficies, as before taught; nor indeed is any other work belonging to Building, Except Bricklayers work.   

Example.  
Admit there is a Floor or Partitioning which is 156 Feet 6 Inches in the Length, and 55 Feet 4 Inches in Breadth, whose Content is Required.  
To perform which Multiply the Length by the Breadth, and the Product will be the Content in Feet and Inches, and then Divide the Feet by 100, and the quotient will give the Content in Squares, and the Remainder will be parts of a Square.  
To do which, by Cross Multiplication, first set down the Length and the breadth, as you see in the following Work. Then Multiply the Feet in the Length 165, by the Feet in the Breadth 55, and the Product will be 9075 Feet, then Multiply the Feet in the Length 165 by the Inches in the Breadth 4, and the Product will be 660 Inches, which is 55 Feet, which Set down under the Feet as you see done; then Multiply the Feet in the Breadth 55, by the Inches in the Length 6, and the Product will be 330 Inches, which is 27 Feet 6 Inches, which Set down as in the Work, viz. the Foot under the Feet, and the Inches under the Inches, then Multiply the Inches in the Length 6, by the Inches in the Breadth 4, and the Product will be 24, which are seconds, 12 of which make an Inch, so that it is Just two Inches, which Set down as you see, then Draw a Line and Add these several Products together, and the Sum will be 6157 Feet 8 Inches; then Divide the Feet by 100, the Number of Feet contained in the Square, which is done by only cutting off the two first Figures towards your right hand, viz. 75, which is 57/100 of a Square and the rest Squares, which 57 Multiply by ●0, and divide that Product by 100, and the Quotient will give 5 Feet, and there will remain 70 which is 70/100 of a Foot, which 70 likewise Multiply by 12, Divide by 100, and the Quotient will give 7 Inches and 40/100 of an Inch, so have you found the Content of the said floor to be 91 Squares 5 Feet 7 Inches and 40/100 of an Inch, as in the following Work.   

Example. 2.  
Admit there is a Floor of Joysting that is 65 Feet 8 Inches Long, and 15 Feet 4 Inches Broad, whose Content is Required.  
To find which work in every Respect as Taught in the former Example, and you will find the Content of the same to be 10 Squarse, 7 Inches, as you see in the Following Work.   

How to Measure the Roof of any Building.  

Example.  
Admit there is a Building that is 93 Feet Long, from the outside of the Wall at one end, to the outside of the Wall at the other end, and the Breadth from the outside of one Wall, to the outside of the other Wall 36 Feet; and it be required to know how many Squares of Roofing there is in that Building.  
To perform which, if the Building be true Roofed, you need not Measure the Length of the Rafters or the sidepieces, but take this for a General Rule, viz. that the Length of each Rafter is three Fourths of the Breadth of any Building; so in this Example, the Breadth of the Building being 36 Feet ¾, thereof is 27 Feet, which is the Length of each Rafter, or Breadth of one side of the Roof, which being doubled, for so it must be, for both the sides is 54 Feet, which you must Multiply by the Length of the Building 93, and the Product will be 5022 feet for the Content of the said Roof, which reduce into Squares by dividing the same by 100, so will you find 50 Squares, and 2 feet, and 20/100 of a foot, to be the Content of the Roof required, as in the following work.    

How to Measure the sides of a Timber Building.  

Example.  
Admit there is a Timber Building that is 36 feet broad at each end, and 93 feet long, and the height to the Rafters, or side of the Roof, is 69 feet, whose Content is required.  
To perform which, because the two sides are equal, and likewise the two ends are equal, you need not stand to find the Content of each side severally, but add all the four sides together, whose sum will be 258 feet, which Multiply by the height of the Building 69, and the product will be 17802, which reduce into squares, so will you find the Content of all the four sides of the said would- Building to be 178 squares, and 2/100 of a square, as in the following Work.    

How to Measure the Gable-end of a Timber Building.  

Example.  
Admit it be required to find the Content of the Gable-end of a Building whose breadth is 72 feet from the outside of one Wall, to the outside of the other, the Gable-end of which Building is represented by the following Triangle ABC, whose Base A C, is 72 feet, equal to the Breadth of the Building, and the Perpendicular B Q is 36 feet.   triangle ABC 
To find the Content of which, Multiply 72, the Base by 18, halve the Perpendicular, and the Product will give 1296 feet, which reduce into squares, so will you find the Content of the said Gable-end to be 12 squares, and 9 feet, and 6 Inches.    

How to Measure Paving, Painting, Wainscoting and Plastering, etc.  
These are all Measured by the Yard Square, containing 9 feet, so that in Measuring of this sort of Work, all that it differs from the former, is only instead of Reducing the Content into Squares by dividing the same by 100, you must Reduce it into Square Yards by dividing the feet by 9, the number of feet in a square Yard.  

Example 1.  
Admit there is a Court that is Paved, containing 76 feet 6 Inches in Length, and 17 feet and four Inches in Breadth, whose Content is required.  
To find which, Multiply the length of the said Court 76 feet and 6 inches by the breadth 17 feet and 4 Inches, as before Taught, and the Product will be 136 feet, which feet Divide by 9, the number of feet in a Square Yard, and the Quotient will be 147, and 3 will Remain, which is of a ● Yard so is the Content of the said Court found to be 147 Square Yards and 3/9 of a Yard or one foot, as in the following Work.   

Example 2.  
Admit there is a piece of Wainscot that is 6 foot long, and 3 feet 6 inches Broad, whose Content is required.  
To perform which, work in every respect as in the foregoing Example, and the Answer will be 2 square Yards, and ● of a Yard, or 1 foot, as in the following Work.   

Example 3.  
Admit there is a Room that is 9 feet 6 inches long, and 8 feet Broad, and 7 feet high, which is Painted, whose Content is required.  
To perform which, first Multiply the breadth by the height, and the Product will be the Content of one end of the Room, which because the two ends are alike, double the same, and it will give the Content of both the ends; then also Multiply the length of the Room by the height, and the Product will be the Content of one of the sides, which being doubled, because the sides are both alike, gives the Content of both sides, which being added to the Content of both the ends before found, the sum is the Content of the 4 sides of the Room required, which reduce into square yards, as before taught, by Dividing the same by 9, and the Answer will be 27 square yards, and 2/9 of a yard. * ⁎ *  
* ⁎ * Note that the Dimensions of a Room that is Painted, is best taken by a String, because of the Mouldings, which you bend as you please, and so Measure the Mouldings with the rest, which must be done as being part of the Painting.   

Example.  
Admit there is a Room that is Plastered, that is 16 Feet 8 Inches Long, and 10 Feet 6 Inches Broad, and 7 Feet, 11 Inches High whose Content is Required.  
To perform which, find the Contents of the 4 sides of the Rooms, as directed in the last Example, which you will find to be 430 Feet, 1 Inch and 8 Seconds, then find the Content of the Top of the Room, which is done by Multiplying the Length of the Room by the Breadth, viz. 16 Feet and 8 Inches, by 10 Feet and 6 Inches, and the Product will be 175 Feet, which Add to the Content of the sides, and their Sum is the Content of the said Room which you will find to be 605 Feet 1 Inch and 8 Seconds, which Reduce into Yards, as before taught and so will the Answer be 67 Square Yards and 2/9 of a Yard. As in the following Work.  
Note that in Measuring of Plasterers Work, there is no allowance made for the Windows, Doors or Chimneys.  
But here I judge it may not be improper to add, how to find the Superficial Content of Solid Bodies, because Painters do very frequently Paint such Bodies.    

How to find the Superficial Content of any Solid Body.  

First, How to find the Superficial Content of a Cylinder.  
Multiply the Length by the Circumference, and the Product will be the Superficial Content desired.  

Example.  
Admit there is a Round Pillar that is Painted, whose Circumference is 5 Feet, 10 Inches, and the Length thereof is 12 Feet, 8 Inches, and it be required to know how many Yards Square of Painting there is in the same.  
To perform which, Multiply the Length 12 Feet, 8 Inches, by the Circumference, 5 Feet, ″ 10 Inches, and the Product will give, 73 Feet, 10 Inches, 8 Seconds, for the Superficial Content of the said Pillar, which Reduce into Yards, Dividing the Feet by 9, and the Quotient, will give 8 1/9 Yards, for the Content of the said Painting.    

Secondly, How to find the Superficial Content of a Parallelepipedon or Square Pillar.  
Add the four Breadths together, and Multiply their Sum by the Length, and the Product will be the Superficial Content of the Square Pillar desired.   

Thirdly, How to find the Superficial Content of a Cone.  
Multiply half the Circumference at the Base by the Height of the Cone, or Multiply half the Height of the Cone by the Circumference at the Base, and the Product shall be the Superficial Content desired.   

Fourthly, How to find the Superficial Content of a Pyramid.  
Add all the Dimensions of the Sides at the Base together, and Multiply half their Sum by the Height of the Pyramid, or Multiply half the Height of the Pyramid, by the Sum of all the sides at the Base, and the Product, will give the Superficial Content thereof.   

Fifthly, How to find the Superficial Content of a Globe or Sphere.  
Multiply the Diametre, by the Circumference and the Product, will be the Superficial Content thereof.    

How to Measure Glaziers Work.  
Glaziers' Work is Measured by the Foot, and therefore Multiply the Length, by the Breadth, as before, and the Product is the Content required. But because the Dimensions of Glaziers' Work is taken more exact than any other sort of Work, viz. in Feet, Inches, and part of an Inch; which makes the resolution of questions of that kind, for the most part, something more difficult, than any of the foregoing; therefore I will show you how to perform the same here, although I shown the same thing in the beginning of this Chapter.  

Example.  
Admit there is a Pane of Glass that is 6 Feet 8 Inches (or Primes) 6 Seconds in Length, and 5 Feet 9 Inches (or Primes) and 3 Seconds in Breadth, whose Content is required.  
To perform which, first set down the Dimensions, as you see in the following Work, then Multiply the Feet in the Length 6, into the Feet in the Breadth 5, and the Product will be 30, which set down under the Feet, as in the Work; then Multiply the Inches in the Breadth 9, into the Feet in the Length 6, and the Product will be 54, which are Inches, which is 4 Feet 6 Inch which likewise set down, as in the Work, then Multiply the Inches in the length 8, into the Feet in the Breadth 5, and the Product will be 40, which likewise are Inches, which is 3 Feet 4 Inches, which likewise set down as you see in the Work, then Multiply the Inches in the Breadth 9, into the Inches in the Length 8, and the Product will be 72, which are parts of Inches or Seconds which is Just 6 Inches, then Multiply the Seconds in the Breadth 3, into the Feet in the Length 6, and the Product is 18, which are Seconds, which is 1 Inch 6 Seconds; then Multiply the Seconds in the Length 6, into the Feet in the Breadth 5, and the Product is 30, which are likewise Seconds, which is 2 Inches 6 Seconds, all which set down as in the following Work. Then Multiply the Seconds in the Breadth 3, into the Inches in the Length 8, and the Product is 24, which are parts of Seconds or Thirds, which is 2 Seconds, then Multiply the Seconds in the Length 6, into the Inches, in the Inches, in in the Breadth 9, and the Product is 54, which likewise is Thirds, which likewise is 4 Seconds 6 Thirds, both which set down, as you see in the following Work Lastly, Multiply the Seconds in the Breadth 3, into the Seconds in the Length 6, and the Product is 18, which are Parts of Thirds or Fourths, which is 1 Third, and 6 Fourths; which set down as in the Work, then Add all these Products together, and their Sum is the Product of 6 Feet, 8 Inches, 6 Seconds, Multiplied by 5 Feet, 9 Inches, and 3 Seconds; which is 38 Feet, 8 Inches, 6 Seconds, 7 Thirds, and 6 Fourths, which is the Content of the Pane of Glass required, as in the following Work.  
  

Example.  
Admit there is a Pane of Glass that is 13 Feet, 6 Inches, and 5 Seconds Long, and 6 Feet, 3 Inches, and 9 Seconds Broad, whose Content is required.  
To perform which Work in every respect as in the last Example, and its Content will be found to be 85 Feet, 5 Inches, 3 Seconds, 9 Thirds, as in the following Work.  
   

How to Measure Bricklayers Work.  
The Measuring of which is more difficult, than any sort of Work that belongs to Building.  

In Measuring of which, Observe these General Rules following.  
First, Find the Superficial Content of each Wall, then find the Superficial Content of all the Windows and Doors in each Wall, and Subtract them from the Superficial Content of the Wall in which they are, and the Remainder will be the Content of the Brickwork of each Wall, in its given thickness.  
Secondly, That all Walls of what thickness soever, must be Reduced into a Brick and a half thick, by Multiplying the Content by the number of half Bricks, contained in the thickness of the Wall, and the Product will be the Content of the said Wall in half a Brick thick, which Divide by 3, the number of the half Bricks, in a Brick and a half, and the Quotient will be the Content of the Wall, in a Brick and a half.  
Thirdly, That whereas Carpenter's Work is Measured by the Square, that is 10 Feet, every way, so that the Square contains 100 Feet; so Brickwork is Measured by the Square Rod, that is, 16 ●/2 Feet every way, so that the Rod contains 272 Feet.  
Fourthly, Having found the Content of all the Walls of any Building, in a Brick and half thick, Add them all into one Sum, and Reduce the same into Rods, by Dividing it by 272 the number of the Feet in a Rod, and the Quotient will be the number of Rods, of Brickwork contained in the said Building.   

Example.  
Admit there is a Building, in which the Walls to the first Floor are as followeth, viz. the Wall on the Front, and the Backside, are each 16 Feet, 9 Inches Long, the Wall on the Front being 3 Bricks thick, and the Wall at the Back 3 Bricks and a ½ thick, having in each a Door of the same Dimensions viz. 2 Feet 6 Inches Wide, and 6 Feet 4 inches High, and in the Front Wall are 2 Windows alike, each 3 Feet 6 Inches Wide, and 6 Feet 4 Inches High; and the Wall on the Back, hath 2 Windows, one of which is 2 Feet 4 Inches Wide, and 5 Feet 6 Inches High, and the other 3 Feet 2 Inches Wide, and 5 Feet 11 Inches High; and the side Walls are each 24 Feet 9 Inches Long, and 2 Bricks thick, and the Floor is 9 Feet 6 Inches High; and it be required to know how many Rods of Brickwork there is in all the Walls to the first Floor.  
To perform which, first find the Superficial Content of the Front Wall, which is done by Multiplying its Length 16 Feet 9 Inches, into the Height of the Wall, or Height of the Floor, which is the same, viz. 9 Feet 6 Inches, and the Product will be 159 Feet, 1 Inch, 6 Seconds, which is the Superficial Content of the Front Wall, and because the Front and Back Wall are of the same Length, and the Walls of one Height, it is likewise the Superficial Content of the Back Wall, as in the following Work.  
 
Then find the Superficial Content of the Door and Windows in the Front Wall.  

I. To find the Content of the Door.  
Multiply its Breadth 2 Feet 10 Inches, and the Product will be 26 Feet 11 Inches, for the Content of the Door, as in the following Work.  
 
Now because the Door in the Front Wall, and the Door in the Back Wall, are alike having found the Content of the one, the Content of the other is likewise found.   

II. To find the Content of the Windows.  
Which are 2, both a like, viz: each 3 Feet, 6 Inches Wide, and 6 Feet, 4 Inches High; so that Multiply the Breadth, 3 Feet, 6 Inches, into the Height, 6 Feet, 4 Inches, and the Product will be 22 Feet, 2 Inches, for the Content of one of the Windows, which because the Wall in which they are placed, is all of one thickness, therefore double the same, and you will have 44 Feet, 4 Inches, for the Content of both Windows in the Front Wall, as in the following Work.  
Thus having found the Content of the Front Wall, to be 159 Feet, 1 Inch, 6 Seconds, and the Content of the Door therein to be 26 Feet, 11 Inches, and likewise the Content of the Windows in the said Wall to be 44 Feet, 4 Inches.  
Then set down the Content of the Door and Windows, one under the other, and Add them together as in the following Work, and their Sum will be 71 Feet, and 3 Inches, which Subtract from the Content of the Wall, before found to be 159 Feet, 1 Inch, 6 Seconds, and the Remainder will be 87 Feet, 10 Inches, 6 Seconds, which is the Content of the Brick work in the Front Wall, in the given thickness, viz. 3 Bricks thick, as in the following Work.  
Then by the same method, find the Contents of the Brickwork in the Back-Wall, whose Superficial Content is already found to be 159 Feet, 1 Inch, 6 Seconds, and likewise the Door in the said Wall, is likewise found to be 26 Feet, 11 Inches, being both the same with those in the Front, so that there Remains only the Windows in this Wall to be found.  
To perform which, Multiply their Width into their Height, as before, so will you find the Content of the one to be 12 Feet, 10 Inches, and the Content of the other, to be 18 Feet, 8 Inches, 10 Seconds, as in the following Work.  
Then Add the Content of the Door and Windows together, and Subtract their Sum from the Content of the Wall, as before, and the Remainder will be 100 Feet, 7 Inches, 8 Seconds, which is the Content of the Brickwork, in the Back Wall, in the given thickness of 3 ½ Bricks thick.   

Then find the Content of the side Walls, which are each 24 Feet, 6 Inches, Long.  
To perform which Multiply the Length 24 Feet, 9 Inches, into the Height 6 Feet, 6 Inches, and the Product will be 235 Feet, 1 Inch, 6 Seconds, for the Content of one of the side Walls, which are a like in Length, Height, and Thickness, therefore double the same and you will have 470 Feet, 3 Inches, for the Brickwork of both the side Walls, in the given thickness, viz. 2 Bricks thick, as in the following Work.  
 
Thus having found the Content of all the Walls, to the first Floor, to be as followeth, viz. the Wall on the Front 87 Feet, 10 Inches, 6 Seconds, in 3 Bricks thick, and the Wall on the Back 100 Feet, 7 Inches, 8 Seconds, in 3½ Bricks thick and the 2 side Walls 470 Feet, 3 Inches, in 2 Bricks thick.  
After the same manner find the Content of all the Walls, to the Second Floor, and so on till you have found the Content of all the Brick Walls in the Building you are to Measure.  
Having so done there Remains nothing more to be done, only that you Reduce the Content of each Wall, into a Brick and a half thick, and Add them all into 1 Sum, and Reduce, your Sum into Rods, by the former Rules. But because in all Building you find several Walls of one thickness, therefore the best way to perform the same, is as followeth, Set down the Content of all the Walls of one thickness by themselves, and all the Walls of another thickness by themselves, in Columns Ruled for that purpose, as in this Example.  
2 Bricks. 3½ Bricks 3 Bricks F. I. S. F. I. S. F. I. S. 209″ 3″ 0 100″ 7″ 8 87″ 10″ 0  
The Content of the Front Wall was found as before, to be 87 Feet, 10 Inches, in 3 Bricks thick, which set down a Column marked at the Top with its thickness, as you see is done here, and under the same, in the Column, place the Content of all the rest of the Walls of the same thickness, and the like is to be understood of all the rest, of the Wall of any Building whatsoever, then Add each Column up, and Reduce the Sum of each, Column into a Brick and ½ thick, and when so done, Add them all together, and Divide their Feet in their Sum by 272, the Number of Feet in a Rod, and the Quotient will be the Content of the said Building, in Rods. but in this Example, the trouble of Adding each Column is Spared, because there is but one Row of Figures in each Column, which, that the foregoing Rules may be the better understood, we will call the Sum of each Column; to Reduce which Columns into 2 ½ Brick.  
1. Take the Sum of the first Column towards your Right Hand, which is 207 Feet, 3 Inches, which is the Content of all the Walls in this Example, which are 2 Bricks thick, and Multiply the same by 4, the Number of half Bricks Contained in the thickness, and the Product will be 829 Feet, which is the Content of the said Walls, in ½ a Brick thick, which Divide by 3, the Number of ½ Bricks, contained in 1 ½ Brick, and the Quotient will be 276 Feet, 4 Inches, which is the Content of all the Walls of 2 Bricks thick, in 1 ½ Brick thick, as in the following Work.  
Then take the Sum of the Second Column, 100 Feet, 7 Inches, 8 Seconds, which Multiply by 7, the number of the ½ Bricks contained in the thickness, and the Product will be 704 Feet, 5 Inches, 8 Seconds, which Divide by 3, the Number of ½ Bricks contained in 1 Brick and ½, and the Quotient will be 234 Feet, 9 Inches, 10 Seconds, the Content of all the Walls of 3 ½ Bricks thick, in 1 ½ Brick thick, as in the following Work.  
Then likewise Reduce the Sum of the Third Column 87 Feet, 10 Inches, after the same Method, into 1 Brick and ½ thick, which you will find to be 175 Feet, 8 Inches, as in the following Work.  
Thus having Reduced the Sum, of each Column in this Example, into 1 Brick and ½ thick, Add them all together, and their Sum will be 686 Feet, 9 Inches, 10 Seconds, which is the Content of all the Brickwork Required, in 1 Brick and ½ thick, which Reduce into Rods, by Dividing it by 272, the Number of Feet contained in a Rod, so will you have 2 Rods, and 142/272 of a Rod, which is something above ½ a Rod, for the Content of the Brickwork, to the first floor required, as in the following Work.  
What hath been said, I suppose is sufficient to Explain the beforegoing Rules, so that to give you any more Examples of that kind will be needles.  
But because there is some Walls that have Watet-tables, that is Built some 3, some 5 Feet High, which Water-table is many times ½ a Brick thick, and some times more, I shall show you how to Measure any such Wall.  
To Measure which, first find the Content of the Wall, as before taught, and Reduce the same into Brick and half Thick, then find the Content of the Water-Table, after the same manner, and likewise Reduce it into 1 ½ Brick Thick, then Add the Content of the Water-Table, to the Content of the Wall, and the Sum is the Content of the said Wall with the Water-Table, which Reduce into Rods, as before Taught.  

Example.  
Admit there is a Wall that is 2 Bricks Thick being 91 Feet, 6 Inches Long, and 17 Feet, 3 Inches high, which hath a Water-Table ½ a Brick Thick, and 3 Feet High, whose Content is required.  
To find which, first find the Content of the Wall by Multiplying the Height into the Length, and the Product will be 1578 Feet, 4 Inches, 6 Seconds, which Reduce into 1 ½ Brick Thick, by Multiplying it by 4, the Number of half Bricks, contained in the given Thickneses, and Dividing the Product by 3 the Number of half Bricks contained in 1½ Brick, and so will you find the Content of the Wall in Brick and ½ Thick to be 2104 Feet, 6 Inches, as in the following Work.  
 
 
Then find the Content of the Water-Table, by Multiplying its Height 3 Feet, into the Length of the Wall 91 Feet, 6 Inches, and the Product will be 274 Feet, 6 Inches, which Reduce into 1½ Brick Thick, which because the Thickness of the said Water-Table is a ½ Brick Thick is done by the Dividing by 3, and the Quotient will be 91 Feet, 6 Inches, for the Content of the said Water-Table in Brick and ½, as in the Following Work.  
 
Then Add the Content of the Wall in a Brick and half Thick, before found to be 2104 Feet, 6 Inches, to the Content of the Water-Table in 1½ Brick thick and their Sum will be 2196 Feet, for the Content of the Wall Required, which you must Reduce into Rods, as before taught, so will you find 8 Rods, and 20 Feet, of Brickwork, to be in the said Wall, as in the following Work.  
    

How to Measure Chimneys.  
The Common way allowed by all Measurers, is thus, gird the Chimney Round, below the Mantletree; if it be in a Wood Building where the Wall of the House, doth go to make the Back of the Chimney, and take the girt of the Chimney for the length, and the height of the Room, for the Breadth, and then Multiply the one into the other, and the Product take for the Content, in the same thickness as the Jaumes are on, which must be Reduced into 1½ Brick.  
But if the Chimney stands against a Brick Wall, than the Wall at the Back being the same with the Wall of the House, and therefore being Measured, as Part of the Wall, they only girt it Round to the Wall, and Multiply that into the height of the Room, and take the Product for the Content of the Chimney in the same thickness as the Jaums are on, which you must Reduce as before into 1 ½ Brick thick.   

Example.  
Admit there is a Chimney whose girt round below the Mantle Tree, is 10 Feet, 6 Inches, and the height of the Room where the same stands, 9 Feet, the Jaums of which Chimney, is 2 Bricks thick, whose Content is required.  
To perform which, Multiply the girt 10 Feet, 6 Inches, into the height of the Room 9 Feet, and the Product will give 94 Feet, 6 Inches, for the Content of the said Chimney, in the same thickness as the Jaumes is off, which ●s 2 Bricks thick, which you mu●t Reduce into 1 ½ Brick thick, as before directed, so will you find the Content thereof in Brick and one half to be 129 Feet, as in the following Work.  
Having by some of the before going methods, found the Content of all the Chimneys, you are next to find the Content of Shafts, to perform which, you must gird them about in the narrowest place and take that for the breadth, and their height, for the Length, and multiply one into the other, and the Product will be the Content, the Shaft or Shafts in the same thickness as their breadth is of, for you must understand that they are Measured as if they were a Solid Peice of Brickwork which you must Reduce into 1 ½ Brick thick.    

How to Measure Tyling.  
Tyling is Measured by the Square, Containing 100 Feet, and is in every respect the same with that of Roofing, taught in Measuring Carpenters Work.  

Example.  
Admit there is a Roof of a Building that is Tiled the length of which Building, from the outside of the Wall at one End, to the out side of the Wall at the other End is 15 Feet, and the Breadth from the outside of the one Wall, to the outside of the other, 9 Feet, 6 Inches, and it be required to find the Content of the said Tyling which you will find to be 2 Squares, 1 Foot, 3 Inches, and 81/100 of an Inch.  
 
 
Thus much of Measuring of Artificers Work relating to Building, I presume what hath been said, is sufficient for the meanest Capacity to understand, how to Measure any sort of Building Timber, Stone, or any sort of Body whatsoever.  
FINIS.      

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