of one side and the half sum of the measure of all the sides multiplied together shall be the true Area or Content thereof All other Figures whatsoever of how many sides soever they be may be reduced to Triangles or to Trapeziaes and measured as before which kind of Figure Surveyors and Builders oftentimes meet withal in their Operations Problem XI For the measuring of on Oval the best way Ovals is to reduce it to a Circle thus Divide the distance on the Line of Numbers between the length and the breadth of the Oval into two equal parts and the middle Point where the Compass stayeth on shall be the Diameter of a Circle equal in Area to the Oval given Example Suppose an Oval be 10 foot long transverse and 8 foot broad conjugate the mean proportion between 10 and 8 is 8-95 I say that a Circle whose Diameter is 8-95 is equal to an Oval of 8 broad and 10 long And how to measure the Circle is shewed before Of these Figures If the Content be 100 then the sides of these Regular Figures are 〈…〉 and also so in proportion is any other quantity of content required Perpendiculer-Triangle 13. 123. Trianguler-side 15. 2. Square its Side 10. 0. Pantagon of five Sides 7. 62. Hexagon of six 6. 02. Heptagon of seven 5. 26. Octagon of eight 4. 55. Nonagon of nine 4. 03. Decagon of ten 3. 06. Half Diameter or Radius 5. 64. Example as thus I would have a Triangle to contain 200 What must the Sides be The half distance on the Numbers between 100 and 200 shall reach from 15-2 to 21-5 the side required And from 13-123 the fixed perpendiculer for a Triangle whose Area is 100 to 18-6 the perpendiculer of an equilatteral Triangle whose Area is 200. But if the Sides be given and you would find the Area work thus The Extent from the fixed-side to the given-side shall reach at two turnings from the fixed Area to the Area required The Extent from 15-2 to 21-5 shall reach at twice repeating from 100 to 200. Problem XII To make an Oval equal to a Circle having the Diameter of the Circle and the length or breadth of the Oval given Sâ—t one Point of the Compasses in the Diameter of the Circle found out on the Line of Numbers and the other Point to the Ovals length then turn that distance the contrary way from the same Diameter-point and it shall reach to the breadth of the Oval required Example Let the Diameter of a Circle be 10 foot I would have an Oval to contain as much as the Circle and be 12 foot long the Query is how broad must it be Set one Point in 10 and the other in 12 that Extent turned the other way from 10 shall reach to 8-34 the breadth of the Oval required If you please to alter the breadth or length you shall soon find the length or breadth accordingly To work this by the Line of Lines you must work by the Directions in the 7th Section of the 6th Chapter as thus First To find the Content of the Oval joyn the length and breadth in one sum to get the sum the half sum and difference and half difference then open the Sector or lay the Thred on 600 to a Right Angle Then count half the difference from the Center downwards and note the place then take half the sum between your Compasses and setting one Point in the half-difference and extending the other to the other Leg or perpendiculer Line and it shall shew a Point whose distance from the Center is the mean proportional required which is the Diameter of a Circle equal in Area to the Oval or Elipsis given to be measured as before is shewed To make an Oval equal to a Circle Take the guessed half-sum of the length and breadth of the Oval and setting one Point in the Diameter of the Circle and on the other Leg set at a Right Angle the other Point shall shew half the difference between the length and breadth of the Oval then if the mean proportional between them be equal to the Diameter you have wrought right if not then resolving upon the length or breadth of the Oval take more or less for the breadth or length accordingly Herein also is seen the excellency of the Line of Numbers in many operations Problem XIII The length and breadth of any Oblong Superficies given in Feet to find the Content in Yards As 9 foot the number of feet in one yard to the length in feet and parts So is the breadth in feet and parts to the Content in yards Example at 13 Foot 6 Inches long and 7 Foot 6 Inches broad The Extent of the Compasses from 9 to 13 ½ the length shall reach the same way from 7 ½ the breadth to 11 yards and a quarter the Content Note That if yoâ— measure by feet and hundred parts you shall find this way exceeding ready the Answer being given in yards and hundred parts of a yard But if you have a yard divided into a 100 parts to measure withal Then the Rule is thus As 1 to the length oâ— breadth so is the breadth or length to the Content in yards Example at 3 yards 72 parts broad and 5 yards 82 parts long The Extent of the Compasses on the Line of Numbers from 1 to 3-72 shall reach the same way from 5-82 to 21 yards 65 parts the Content in square yards and 100 parts By the Line of Lines As 5-82 to = 1 at 10 the end So is = 3-72 to 21-65 yards Or in the Example before As 13-6 counting 6 ½ for 13 is to = 9 So is = 7 ½ to 11 ¼ as you counted at first Problem XIV The length and breadth of any Wall being given in feet and 100 parts to find how many Rods of Walling there shall be at a Brick and an half thick First you must Note That 272 foot and a quarter makes one Rod or so many feet is in a Rod Secondly That let the Walls be half a Brick one Brick two Bricks two and a half or three Bricks thick it is to be reduced to Brick and a half thick as a standard thickness Thirdly Note That this reducing to a Brick and half thick may be at the measuring or after the casting-up as you please as in the Examples following will plainly appear As thus for Instance A Front or side-Wall of a House is to be measured wherein the Celler-story Wall is 2 Bricks and a half thick The Shop and first Chamber-story is two bricks thicks the other Stories 1 Brick and a half thick and the gable-Gable-ends 1 Brick thick The nearest way to measure this Wall I conceive is thus 1. The Cellar-story is 10 foot high but being 2 bricks and a half thick I make it 16 foot 8 inches high by adding two thirds of 10 foot to the 10 foot high which is 6 foot 8 inches in all 16 foot 8 inches 2. The other two Stories
are supposed 22 foot but in regard they are two bricks thick I add one third part of 22 foot which is 7 foot 4 inches to 22 and it makes 29 foot and 4 inches the height of the Shop and next Story above 3. The other two Stories being a brick and half thick need no alteration which suppose may be 19 foot 4. The gable-Gable-end or Garret-story if â—ny be being but one brick thick you must take away one third part to bring it to a brick and a half Also if it be a gable-Gable-end Note it is a Triangle and you must measure but half the height and the whole breadth to find the Content which here may be 5 foot The Cellar Story 16 8 Two next Stories 29 4 Two next Stories 19 0 The Garret 5 0  70 0 5. Add all these sums of feet high together and they make 70 then measure the breadth which is common to every Room the out-side going upright which in a double House may be 36 or 40 foot 6. Then having gotten the Dimentions right by the Line of Numbers Say As 272 ¼ the feet in one Rod is to 40 foot the breadth of the House so is 70 foot the whole height of every several Story reduced to 10 Rod and 29 parts which 29 parts you may call a quarter of a Rod and 10 foot and a half The reason whereof is apparent thus As 100 is to 272 ¼ so is 29 to near 79 of which 79-68 is a quarter of a Rod or 25 of 100 is a quarter likewise which by the Line of Numbers is apparently seen then every 10th part is 2 foot and 72 of a hundred which is near two and three quarters so that here 25 being a quarter of a Rod there is 4 hundred parts more in 29 Then thus the double of 4 is 8 or twice 4 is 8 and four times three quarters is three foot more of which you must abate somewhat because 72 ¼ is not 75 which is just three quarters and all put together make ten rod one quarter ten foot and a half for if you shall divide the Product of 40 multiplied by 70 which is 2800 by 272 ¼ you shall find the Quotient to be 10 rod 78 ½ which is as before 10 rod 1 quarter and 10 foot and a half But note also by the way That when you come to take out the deductions for the doors and windows if any happen in a Wall of two Bricks and a half or in two Bricks you must add two thirds or one third more to the length or bredth one way and then casting them up severally when they be of several lengths or breadths you shall do no wrong to the Work-master nor Work-man For true Arithmetick and Geometry will lie for no man or use any kind of partiality This I conceive is as near a way as any such business can be performed But if you will measure every Story severally taking account of each Story severally in their thicknesses then after it is cast up the best way by the Rule to reduce it is thus As 3 half bricks for a brick and a half is to any other number of half bricks thick over or under 3 So is the Content at that rate accordingly to his Content at a brick and a half required Example 1269 foot at 5 half bricks thick is 2115 for two thirds of 1269 which is 846 added to 1269 makes 2115 For the Extent on the Line of Numbers from 3 to 5 shall reach the same way from 1269 to 2115 the Number required to be found out Otherwise thus To bring any kind of thickness to one brick and a half thick at one operation by the Line of Numbers For this purpose you must use several Points as so many gage Points as in the short Table following doth appear For half a brick use 3-00000 For 1 brick 1-50000 For 1 brick a half use 1-0000 For 2 bricks 0-7500 For 2 bricks a half use 0-6000 For 3 bricks 0-5000 For 3 bricks a half 0-4285 For 4 bricks 0-3750 For 4 bricks and a half 0-3333 For 5 bricks 0-3000 For 5 bricks and a half 0-2727 For 6 bricks c. ad infinitum 0-2500 Example at the 6 ordinary thicknesses Let a Wall be 30 foot long and 10 foot high and let it be supposed of any of these thicknesses following from half a bricks length to three bricks length in thickness then thus in order increasing c. First at half a Foot For ½ brick As 3 to 30 so is 10 to 100 foot at 1 ½ For 1 brick As 15 to 30 so is 10 to 200 foot at 1 brick For 1 ½ thick As 10 to 30 so is 10 to 300 foot at 1 ½ For 2 bricks As 0-75 to 30 so is 10 to 400 foot at 1 ½ For 2 ½ thick As 0-60 to 30 so is 10 to 500 foot at ½ For 3 bricks As 0-50 to 30 so is 10 to 600 foot at 1 ½ For 3 ½ thick As 0-4285 to 30 so is 10 to 700 foot at 1 ½ For 4 bricks As 0-3750 to 30 so is 10 to 800 foot at 1 ½ And so for any other thickness as far as you please which Points are found thus The Extenâ— from the number of bricks any Wall is thick to 15 or 1 and ½ shall reach the same way from 10 or 1 to the Gage-Point required for that Wall or Walls of that thickness Example As 2 to 1 ½ so is 10 to 0-750 for 2 bricks thick c. Lastly having the Number of Feet in the whole work to find how many Rods there is Say If 272 ¼ be one Rod what shall any other Number of Feet make in Rods The Extent of the Compasses from 172 ¼ to 1 shall reach the same way from the Number of Feet to the Number of Rods and hundred Parts or Rods and Quarters and Feet as by the 6th last mentioned Example In 5269 Feet how many Rods The Extent from 272 ¼ to 1 shall reach the same way from 5269 to 19 Rod and 36 parts of a 100 or 19 Rod 1 quarter and 29 foot and a quarter of a foot The 19 Rod and a quarter is seen plainly on the Rule and 25 being a quarter 36 is 11 parts more for which 11 parts more I say 2 times 11 is 22 foot and 11 3 quarters of a foot is near 8 foot which put together makes 29 foot as before Or as the Compasses stand turn them the contrary way from the Decimal parts above the even quarter and it shall reach to the odd feet above the quarter required Example The Extent from 272 ¼ to 100 or 1 shall reach the contrary way from 10 ½ to 29 foot the feet above ¼ of a Rod. 8. Observe That the Tyling the Roof the Floors and Partitions are measured by the Square which is 10 foot Square every way or 100 foot in Area The Chimneys are usually done by a certain rate for a Chimney or if
to be measured thus are the height and breadths taken c. If a Chimney stand singly and alone not leaning against or in a Wall the usual way is to girt it about and if the Jaumes are but a brick thick and wrought upright over the Mantle-tree to the Floor then I say girt it about for a length and the height of the Story is the breadth at a brick thick because of the gathering together to make room for the next Hearth above in the next Story But if the Chimney-back be a Party-Wall the Wall being first measured then the brest and the depth of the two Jaumes is one side and the height of the Story another side to be multiplied together at a brick and a half thick or a brick thick according as the Jaumes be and nothing to be abated for the want between the Hearth and the Mantle-tree because of the Wit hs and thickning for the next Hearth For measuring the Shafts of the Chimneys Girt with a Line round about the least place of them for one side and the height for the other side at a brick thick in consideration of the Wit hs Pargitting and Scaffolds In measuring of Ceiling a foot broad and the length of the Vallies is alwayes to be allowed more than the whole Roof Also the length of the Rafter feet above or beyond the Roof When Rafters have their usual pitch which is when the breadth of the House is 12 foot the Rafter is 9 foot long which is 3 quarters of the Floors breadth be it more or less then I say that the Content of one Floor and half so much is the Area of the whole Roof in Squares to which is to be added the Vallies and Rafter-Feet or Eves in Tileing And also a Deduction for Chimney-room and Gutters if any be Which work by the Line of Numbers is done at one Operation thus As 6666 is to the length of the House So is the breadth to the Content in the Roof Example A House 30 foot long and 20 foot broad is 900 foot or 9 square For the Extent from 6666 to 20 shall reach the same way from 30 to 900. Also in measuring of the Roof as to Carpenters work by the Square there is to be allowance for those Rafters in the Dormers and gable-Gable-ends on which no Tiles are laid as over-work for a particular use and convenience more than need be in a bare Covering or Roof Also in measuring of Plasterers work in Partitions and Walls the Timbers and Quarters are not to be deducted out of the rendring for Work only except when the Workman finds the Work and Stuff also then substract a 6th part for the Quarters in the rendring Work But in Ceilings the Summers which are seen are alwayes abated and Doors and Windows also unless by a due considerate or an unconsiderate bargain of running measure Thus you have a brief account of the usual order used among Workmen in taking the Dimentions of a House viz. Brick-work by the Rod Tileing and Carpenters-work by the Square Chimneys usually by the Fire And Plasterers and Painters-work by the Yard Glasiers by the Foot There are many other things to be taken notice of in the Carpenters Bill as Lintels Mantle-trees and Tassels Luthern Lights and other Lights both Architrave and Plain Lights Sky-lights or Cubiloes Modillean Cornish and guttering Penthouse Cornish Timber-Front-Story Cellar-doors and Door-cases the Plank and Curb at the Cellar-stairs Dogleg-stairs and Open-Newel-stairs Canted-stairs counted either by the step or pair together with the half Spaces on the Corners of the open Newel-stairs the Rayles and Ballasters small and great Cornish Outside-work and Partitions Ceiling Joysts and the Ashlering Boarded Partitions and Chequer-work back-Doors and Door-cases Window-boards and Wall-timber Planks in the Foundation Palcing Penthouse-floors and Penthouse-roof furring the Platform Centerings for the Chimney Trimmers Girders-ends Ends of Brest-summers and Plate and more the like which will come in Accompt to be remembred and set down according as the Building is Also with due allowance into the Wall that way the ends of the Joysts are entred or laid in the Wall as thus If it be Framing Work is only measured then 9 Inches ought to be allowed into each Wall that way the Joysts ends are laid because every Joyst if well laid should have 9 inches at least hold on the Wall But if it be Timber and Boarding both to be measured then 6 inches only is a competent allowance because the Timber is usually vallued at one third part more than the Boarding is Also As the Workman doth think on this the Work-master may not forget to deduct for Stairs and Chimneys also where Work and Stuff are both measured though for Work only it may be very well allowed unless the better Price make an allowance for it Note also That by the Line of Numbers you may readily find the length of the Hips and Rafters in a Roof of any largeness at true pitch by this following Proportion and Table The Breadth of the House being 40 Feet and the Ends Square the Length and Angles are as in the Table at the usual truâ— pitch  fâ—et 100 par Whole breadth 40 00 Half breadth 20 00 Rafter 30 00 Hip-Rafter 36 00 Diagonal Line 56 57 Half-Diagonal 28 28 Perpendiculer 22 36   deg min. Hip Angles at Foot 38 22 at Top 51 38 on the Outside 116 12 Rafter Angles at Foot 48 10 at Top 41 50 For any other House by the Numbers thus as suppose 18 Foot broad The Extent of the Compasses from 40 the breadth in the Table to 18 the breadth given shall reach the same way from 30 the Rafter in the Table to 13-50 the Rafter required And from 36 the Hip in the Table to 16-22 the Hip required And from 22-36 the Perpendiculer in the Table to 10-06 the Perpendiculer required And from â—6-57 the Diagonal in the Table to â—5-48 the Diagonal required The Angles are alwayes the same in all Rooss small or great as in the Table being Square and true pitch If you would have Directions for Bevel or Taper Frames to find the Lengths and Angles of Raâ—ters and Hips you may have it at large in an Appendix to the Mirrour of Architecture or Vincent Stamoâ—â—i Printed for William Fisher at the Postern-Gate 1669. By which Directions and the Sector you may find any thing that is there set down As also by the Trianguler Quadrant Thred and Compasses Note also That having Inches and Foot-measure together you may presently by inspection find the price of one Foot having the price of the Square and the contrary Also having the 12 Inches on the other Foot divided into 85 parts near and figured at every 8 with 1 2 3 4 5 6 7 8 9 10. shall represent pence and half farthings then at any price the Rod you have the price of one Foot the contrary As thus Let every Inch represent one pound 〈◊〉 â—very 8th
the Point E draw the Line ED of which Line find the middle between E and D viz. the Point C then the extent CE or CD keeping one Point in C shall cross the Ground-line in the Point B by which and E you may draw the perpendiculer Line EB which is but the converse of the former 6. To draw a Line Parallel to another at any distance To the Line AB I would have another Parallel thereto to the distance of AI take AI between your Compasses and setting one Point in one end of the Line as at A sweep the Ark EIF then set the Compasses in the other end as at B and sweep the Ark GDH then just by the Round-side of those Arks draw a Line which shall be the parallel-Line required Or thus Take BC the measure from the Point that is to cut the Parallel-line and one end of the given-Line viz. B with this distance set one foot in A at the other end of the given-Line and draw the Arch at K then take all AB the given-Line and setting one Point in C cross the Ark at K then C and K shall be Points to draw the Parallel-line by Note the Figure 4. 7. To make one Angle equal to another The Angle BAC being given and I would have another Angle equal unto it set one point of the Compasses in A and draw the Arch CB then on the Line DE from the Point D draw the like Ark EF then in that Ark make EF equal to CB then draw the Line DF it shall make the Angle EDF equal to the Angle BAC which was required 8. To divide a Line into any Number of parts Let AB represent a Line to be divided into Eight parts On one end viz. A draw a Line as AD to any Angle and from the other end B draw another Line Parallel to AD as BE then open the Compasses to any convenient distance and from A and B divide the Lines AD and BE into eight parts then Lines drawn by a Ruler laid to every division in the Lines AD and BE shall divide the Line AB in the parts required Note the Figure marked VI. This Proposition is much easier wrought by the Line of Lines on the Sector thus Take AB between your Compasses and fit it over parrally in 8 and 8 of the Line of Lines then the Parallel distance between 1 and 1 shall divide AB into 8 parts required 9. Any three Points given to bring them into a Circle Let ABC be three Points to be brought into a Circle first set one Point on A and open the other above half-way to C and sweep the part of a Circle above and below the Point A as the two Arches at D and E not moving the Compasses do the like on C as the Arks F and G then set the Compass-point in B and cross those Arks in DEF and G then a Rule laid from D to E and from F to G and Lines drawn do inter-sect at H the true Center to bring ABC into a Circle 10. Any two Points given in a Circle to draw part of a Circle which shall cut them and the Circumference first given into two equal parts Let A and B be two Points in a Circle by which two Points I would draw an Arch which shall cut the whole Circumference into two equal parts First draw a Line from A the Point remotest from the Center through the Center and beyond the Circumference as AD then draw another Line from A to a Point in the Circumference perpendiculer to AD and cutting the Center C as the Line AE Then on the Point E draw another Line perpendiculer to the Line AE till it inter-sect AD at D then these three Points ABD brought into a Circle or Arch by the last Rule shall divide the Circumference into two equal parts Note the Figure 8 where the first Circle is cut into two equal parts at F and G by part of a Circle passing through the Points A and B. 11. Any Segment of a Circle given to find the Diameter and Center of the Circle belonging to it Let ABC be the Segment of a Circle to which I would find a Center any where about the middest of the Segment set one point of the Compasses at pleasure as at B on the point B at any meet distance describe a Circle and note where the Circle doth cross the Segment as at D and E then not stirring the Compasses set one point in D and cross the Circle twice as at F and I and again set one point in E and cross the Circle twice in G and H Lastly by the Points GH and FI draw two Lines which will meet in the point O the center required 12. Or else to find the Diameter thus Multiply the Chord or flat-side of the half-Segment viz AK 12 by it self which is called Squaring which makes 144 then divide that Product 144 by 8 the Line KB called a Sine the Quotient which comes out will be found to be 18 then if you adde 8 the Sine and 18 the Quotient together it shall make 26 for the Diameter required to be found 13. Any Segment of a Circle given to find the Length of the Arch of the Segment Lay the Chord of the whole Segment and twice the Chord of half the Segment from one Point severally and to the greatest extent adde one third part of the difference between the Extents and that sum of Extents shall be equal to the Arch. Example 14. To draw a Helical Line from any Three Points to several Radiusses without much gibbiosity useful for Architect Shipwrights and others Let ABCDE be five Points to be brought into a Helical-Line smoothly and even without gibbiosity or bunches as the under-side of an Arch or the bending of a Ship or the like First between the two remote Points of 3 as A and C draw the Line AC then let fall a Perpendiculer from B to cut the Line AC at Right Angles and produce it to F draw the like perpendiculer-Line from the point D to cut the Line CE at Right-Angles produced to F. I say the Center both for the Arches AB the lesser and BC the greater will be found to be in the Line BF the like on the other-side for DE and CD the Helical-Circle or Arch required But if you divide the Arch ABCDE into 24 or more parts the several Centers of the splay-Lines are thus found Take the measure AG and lay it from B or D or C on the Line GF and those Points on GF shall be the several Points to draw the splay-Lines of the Arch and Key-stone by CHAP. IV. Of the Explanation of certain Terms used in this following Book 1. RAdius or Sine of 90 or Tangent of 45 or Secant of 00 are all one and the same thing yet taken respectively in their proper places and is the whole Line of Sines or Tangents to 45 or more particulary
particular Products of every double Number for the third term one after another This done repeating the Rule of Three as often as there be double Numbers the 4th term produced from those Operations shall be Answers to the Questions required viz. the quantity of each mans gain or loss Example 25 30 A's Oxen and time of feeding multiplied is 750 15 40 B's Oxen and time of feeding multiplied is 600 20 40 C's Oxen and dayes of feeding multiplied is 800 The Sum 2150 AS 215 to 50 so is 750 A's Stock to 17-9 A's Rent  600 B's 13-19 B's  800 C's 18-12 C's 9. To work by the Line of Numbers the Extent of the Compasses from 1 to 25 shall reach the same way from 30 to 750 the first Product of a A's double Number or Stock And as 1 to 15 so is 40 to 600 the Product of B's double Number and Stock And as 1 to 20 so is 40 to 800 the Product of C's double Number and Stock Which three Products added make 2150 the first term and 50 is the second term and 750 600 and 800 the three Products severally the third term Then The Extent from 2150 to 50 shall reach the same way from 750 to 17-45 or 17 l. 9 s. And from 600 to 13-95 or 13 l. 19 s. And from 800 to 18-60 or 18 l. = 12 s. the several Answers required which being added together make up 50 l. the whole Rent to be paid among them There be other Rules of Arithmetick as the Rule called Allegation Medial and Alternate and the Rule of Position or Falsehood in the working of which are so many Cautions in ordering the Numbers before you come to the proportional work that it would make the Book more bulky than useful therefore I shall wave it and refer you to the particular Books of Arithmetick as that of Mr. Record Dee and Mellis or that of Mr. Wingate Natural and Artificial having in it plenty of Examples and others also as Iohnfâ—us Iaggers or Moores Arithmetick any of which exceed the bounds I intend for this whole discourse I shall therefore pass on to the Rules of Practice in several kinds as measuring Superfecies and Solids and Rules of double and treble Proportion and Questions of Interest which are tedious by the Pen without the help of particular Tables and very easie by the Line of Numbers as will fully appear in the next Chapters CHAP. VII The use of the Line of Numbers in measuring of any kind of Superficial Measure THe Measure that is commonly used in this Work is a Foot-Rule divided into 100 parts or else into 12 inches and those inches into halves and quarters or 8 parts or inches and 10 parts but in regard that the Numbers do most fitly agree to the 100 parts of a Foot it will be convenient here to shew how to reduce them or any other Fraction from 12 s. to 10 s. or any other whatsoever from one Fraction to the other which by the Line of Numbers is quickly done as thus from 12â— to 10â— Reduction Extend the Compasses from one Denominator to the other the same Extent shall reach the same way from one Numerator to the other Example As 12 to 10 so is 6 half of 12 to 5 half of 10. Again As 120 to 100 so is 30 a 4th of 120 to 25 a 4th of 100. Which two Lines of Inches and Foot-Measure are usually set together on Rules for the ready way of Reduction by Occular inspection only in this manner as in the Figure And the like may be for any thing whatsoever as Mr. Edmond Windgate hath largely shewed in his Arithmetick Which Line being next to the Line of Numbers on your Rule will be very plain and ready in the use of the Line of Numbers for feet and inches or shillings and pence and the same Rule of Reduction serves for all manner of Fractions For as the Denominator of one Fraction is to the Denominator of the other which in the Decimal work is alwayes a unite with one two or more Cyphers so is the Numerator of one to the Numerator of the other And Note That the operation of Decimal Numbers and their Fractions is no other than whole Numbers except only the cutting off so many Figures as there is Fractions in the Multiplicator and Multiplicand after any Multiplication as in the following Examples will appear This being premised I come next to the Work Problem I. The breadth of an Oblong Superficies given in Foot Measure to find how much in length makes one Foot The Extent of the Compasses from the breadth to 1 shall reach the same way from 1 to the length required Example at 7 10th broad As 7 to 1 so is 1 to 1 Foot and 43 parts The breadth given in inches to find how much make a Foot As the breadth in inches to 12 so is 12 to the length of a Foot in inches and 10 parts Example At 8 inches broad you must have 18 inches to make a Foot for the Extent from 8 to 12 shall reach the same way from 12 to 18. To work these two by the Line of Lines By Inches As 1 to = 7 so is = 1 to 1 43 the length in Foot-measure By Inches As 12 to = 8 so is = 12 to 18 Or else By Inches As 8 to = 12 so is 12 to = 18 the length in inches Problem II. Having the breadth of an Oblong Superficies given in Foot-measure to find how much is in a Foot long This is soon wrought for in every Foot long there is just as much as the breadth is either in Foot-measure or inches for a piece of Board half a Foot broad and a Foot long is just half a Foot Problem III. Having the length and breadth in Foot-Measure to find the Content in Feet The Extent from 1 to the length shall reach the same way from the length to the Content in Feet Example As 1 to 1 foot 50 the breadth so is 11 foot 10 parts the length to 16 foot and 65 parts the Coment required The breadth given in inches and the length in feet to find the Content in feet As 12 to the breadth in inches so is the length in feet to the Content in feet required Example at 9 inches broad and 11 foot long The Extent from 12 to 9 shall reach the same way from 11 to 8 foot 3 inches or ¼ By the Line of Lines As 11 to = 12 So is 9 to 8 ¼ But Note That in working this and many such-like it will be convenient to double your Scale in account calling 10 at the end 20 and every single figure as much more as to call 12 and 24 c. So that in this Operation the work runs thus As 11 taken from the Line of Lines counting 1 for 10 as usually To = 6 the half of 12 reckoned double for 12 So is = 4 ½ counted for 9 to 8 ¼ between the Center and 1.
but 43 inches and 1 ◠about whose 4th part is under 11 so that here I may very well abate 1 inch from the 4th part of the Line So consequently if the Rind be thinner and the Tree less a less allowance will serve and if the Rind be thicker and the Tree large there ought to be more as by cutting the Rind away and then taking the true diameter you may plainly see This measuring by the 4th part of the Circumference for the side of the Square and allowance for the Bark being allowed for as before I say will prove to be just one 5th part over-measure Especially considering this That when it is hewed and large wanes left then the Tree is marked for more measure sometimes by 10 foot in 60 than there was before it was hewed the reason is because when the Tree is round and unhewn the girting it and counting the 4th part for the side of the Square is but very little more than the inscribed Square and then being hewen and that scarce to an eight Square and measuring with a pair of Callipers to the extremity of that doth not then allow the Square equal to the Circle for the side of the Square as by the working by those several Squares will very plainly appear which being foretold and warned of let those whom it concerns look to it But this being premised and the Parties agreeing the difference being as 4 to 5 the best way to measure round Timber I conceive is by the Diameter taken with a pair of Callipers and the length which for the just and true measure is largely handled already But if this allowance be agreed on then the Proportion for it is thus As 1-526 to the Diameter So is the length to a 4th and so is that 4th to the Content in feet Example The Extent from 1-526 to 15-26 shall reach being twice repeated from 10 foot the length to 10 foot the Content required being all at one Point Or another Example The Extent from 1-526 to 20 inches the Diameter being twice repeated the same way from 10 foot the length shall reach to 17 foot ¼ the Content Or if you have the Circumference and length Then the Extent from 48 to the inches about being turned twice the same way from the length in feet shall reach to the Content required The Extent from 48 to 62 the inches about being turned twice from 10 the same way shall reach to 17 foot ¼ the Content in that measure Thus you have full and compleat Directions for the measuring of any round Timber by the Line of Numbers by having the Diameter and length given after any usual manner there remains only one general and natural way by finding the base of the middle or one end by the 7 th Problem of Superficial measure and then to multiply that base by the length will give the true Content in feet or inches Thus Having found the Base of the Cillender by the 7th or 10th Problem of Superficial-measure then if you multiply that Base being found in square inches by the length in inches you shall have the whole Content in Cube Inches Example Suppose a Cillender have 10 inches for its Diameter then by the 7th or 10th abovesaid you shall find the Base to be 78-54 then if you multiply 78-54 by 80 the supposed length in inches you shall find 2356-20 Cube Inches which divided by 1728 the inches in a Cube Foot sheweth how many feet there is c. And as to the number of figures and the fractions cutting off you have ample Directions in the first Problem and the third Section of the six◠Chapter Problem VII How to measure a Pyramis or taper Timber or the Section of a Cone 1. First get the Perpendiculer length of the Pyramis or Cone thus Multiply half the Diameter of the Base AB by it self then measure the side AD and multiply that by it self then take the lesser Square out of the greater and the Square root of the residue is the Perpendiculer Altitude required viz. DB. Example Suppose the half Diameter of the Base AC were 10-25 and the side DA 100 AB 10-25 and 10-25 multiplied together called Squaring makes 105 0625 DC 100 multiplied by 100 called Squaring makes 10000 then the lesser Square 105 0625 taken out of 10000 the greater Square the remainder is 9894 9375 whose square Root found by the 8 th Problem of the sixt Chapter is 99-475 the true length of the Line DB the length or height of the Cone Then if you multiply the Area or Content of the Base AC 20-5 which by the 7th or 10th of Superficial measure is found to be 160-08 by 33-158 a third part of 99-475 the whole height makes 5308 cutting off the Fractions for the true Content of the Cone whose length is 99 inches and near a half and whose Base is 20 inches and a half Diameter 2. Then Secondly for the Segment or Section of a Cone the shape or form of all round taper Timber the truest way is thus By the length and difference of Diameters find the whole length of the Cone which for all manner of Timber as it grows this way is near enough As thus As the difference of the Diameters at the the two ends is to the length between the two ends So is the Diameter at the Base to the whole length of the Cone Example The difference between the Diameters AC and EF is 13-70 the length AE is 66-32 then the Extent on the Line of Numbers from 13-70 the difference of the Diameters to 66-32 the length between shall reach the same way from 20-50 the greater Diameter to 99 and better the length that makes up the Cone at that Angle o◠Tapering in the Timber then if by the last Rule you measure it as a Cone of that length and also measure the little end or point at his length and diameter and then lastly this little Cone taken out of the great Cone there remains the true Content of the Taper-piece that was to be measured viz. 5246-71 when 61-30 the Content of the small Cone at the end is taken out of 5308 the Content of the whole Pyramid 3. If this way seem too troublesome for the common use then use this being more brief To the Content that is found out by the Diameter in the midst of the Timber and the length add the Content of a Piece found out by half the difference of Diameters and one third part of the length of the whole Piece and the sum of them two shall be the whole Content required 4. Or else Divide the length of the Tree into 4 or 5 parts and measure the middle of each part severally and that cast up by his proper length shall give the Content of each Piece then the sum of the Contents of all the Pieces put together is the true Content of the whole Taper Piece very near Note That this curiosity shall never need to be
the true Circumference of a round Cillender Fiftly At 4 foot just is set R for the Circumference according to the former allowance Sixtly At 1 foot 5 inches 1 7 fâ—rè is a spot and close to it the letter W as the Gage-point for a Wine-gallon Seventhly At near 19 inches or 1 foot 7 inches is another spot and close to it the letter a as the Gage-point of an Ale-gallon Eightly At 2 foot 8 inches 8 10 is a spot and close to it is set B for the Gage-point of a Beer Barrel and at 2 foot 7 inches is set A for an Ale Barrel The Uses whereof in order follow The Figures on the left-side or fixed-edges are read and counted as those on the right For the small 1 2 3 4 5 6 7 8 9 10 11 are to represent inches and the cuts between quarters of inches Then the 1 2 3 4 5 6 7 8 9 10 Figures next somewhat bigger as to represent so many feet and the cuts between are whole inches Then 20 30 40 50 60 70 80 90 100 150 for tens of feet and the parts between for single-feet for the most part or else whole and half feet as is usual The Uses follow Use I. A piece of Timber being not Square or having its breadth and depth unequal to make it Square or find the Square equal Set the breadth of the Piece counted on the right-side to the same breadth counted on the left-side then right against the depth found on the left-side on the right or single-side is the inches and quarters square required Example at 15 inches broad and 9 inches thick or deep Set 9 inches on the right-side to 9 on the left then right against 15 inches or 1 foot 3 inches found out on the double or left-side on the right or single-side is 11 inches and â… the Square equal required Also if you set 15 to 15 then right against 9 found out on the left-side on the right-side is 11 inches â… the Square equal required Use II. The Side of the Square given to find how much in length will make 1 Foot Set the inches or feet inches found out on the right-side to 1 foot on the left then right against 1 foot on the right is the inches or the feet and inches required to make a Foot of Timber Example at 9 inches square Set 9 inches found out on the right-side to 1 foot on the left-side then right against 1 on the right-side is 1 foot 9 inches 2 â— on the left If the Square be so big that the 1 on the right falls beyond the End at the beginning then right against 10 foot on the right-side is on the left the hundredth part of a foot that makes a Foot of Timber Example at 4 Foot Square Set 4 foot found out on the right to 1 foot on the left then right against 10 foot on the right-side is 0-063 on the left-side or against 12 foot you have 9 12 parts of 1 inch the length that goes to make 1 foot of Timber required Use III. At any bigness or Inches or feet and inches square to find how much is in 1 Foot long Just as the Rule stands even that is 1 foot on the right against 1 foot on the left seek the inches or feet and inches the Piece is square on the right or single-side and just against it on the left or double-side is the Answer required in inches or feet and inches Example at 19 inches square Just against 1 foot 7 inches or 19 inches which is all one found out on the right-side on the left-side is 2 foot 6 inches the quantity of Timber in 1 foot long at 19 inches square which Number of 2 foot 6 inches multiplied by the length in feet gives the true Content of the whole Piece of Timber required Note That this is a most excellent way for great Wood and very exact Also Note That here by inspection you may square a small Number or find the square-root of a small Number As thus The square of 8 ½ is near 72 Or The square-root of 72 is near 8 ½ Use IV. The side of the Square and the length of any Piece being given to find the Content in feet and parts Set the word square or 1 foot alwayes to the length found out on the left-side then right against the inches or feet and inches square counted on the right on the left is the Content required Example at 20 foot long and 15 inches square Set 1 foot on the right to 20 on the left then right against 1 foot 3 inches on the right is 31 foot 2 inches and ½ the Content Note That if the Piece be very small call the feet on the left-side inches and the parts between 12â— of inches then the Answer will be found on the left-side in 144â— of a Foot Example at 2 inches square and 30 foot long how much is there Set 1 foot on the right to 30 foot on the left then right against 2 foot on the right counted as 2 inches is 120 parts of a foot divided into 144 parts being just 10 inches for 10 times 12 is 120. But if it be â— great Piece of Timber then work thus Set 1 foot or the word square to the length on the left counting the single feet 10â— of feet then right against the feet and inches square are the 100â— of feet required Example at 40 foot long and 4 foot square Set 1 foot to 4 foot counted as 40 on the left then right against 4 foot on the right is 640 the true Content increasing the 10â— to 100â— Thus much for square Timber Though there be many other wayes and manner of workings some whereof you may find in a Book set forth under the name of The Carpenters Rule 1666 by I. Brown and well known abroad already Use V. For Round Timber The middle Diameter of any Piece given to find how much is in a Foot long at true measure Set the spot by t.d. to 1 foot on the left then just against the inches or feet and inches Diameter found on the right is the quantity of Timber in 1 foot long on the left-side required Example at 2 foot 9 inches Diameter Suppose a piece of Stone-Pillar or Garden-Roul be two foot 9 inches Diameter set the spot by t.d. just against 1 foot then right against 2 foot 9 inches found on the right on the left is 6 foot the quantity of solid measure in one foot long which being multiplied by the length in feet gives the true Content of the whole Piece Note That if you would have the usual allowance set D to 1 instead of t.d. Use VI. The Diameter of any Piece of Timber given to find how much in length will make one Foot Set 1 foot on the left to the inches Diameter counted on the right then right against t.d. for true measure or D for the usual allowance is the Answer required found on the
the Fragments of a Globe But The Other is by a way found out by Mr. Bennit a Cooper that hath long exercised the way of Gaging which is by comparing a Cask known and its quantity of emptiness to a Cask unknown and its inches of emptiness as followeth First by the Line of Numbers and Artificial Line of Segments to find the quantity of Gallons that any Vessel wants of being full at any number of Inches from the inside of the bung-hole to the superficies of the Liquor which is usually called Inches dry Extend the Compasses on the Line of Numbers from the inches and tenths diameter at the bung to 100 on the Line of Segments the same extent applied the same way from the inches and parts dry shall reach to a 4th Number on the Line of Artificial Segments which 4th Number you must keep Or if you will you may use the inches wet laying the same extent from the inches wet and that also will on the Segments give a 4th Number which you must likewise keep Then secondly As the Extent from 1 to the whole Content of the Vessel in Wine or Ale-gallons So is the 4th Number kept to the Gallons of emptiness or fullness that it wants of being full or the quantity of Gallons in the Vessel Example of a Canary-Pipe whose Diameter at bung is 28 inches and 7 and full Content is Gallons 116 ½ at 12 inches dry or 16 inches and 7 tenths wet The Extent of the Compasses from 28-7 to 100 at the end of the Line of Segments shall reach the same way from 12 the inches dry to 39 ½ on the Line of Segments for a 4th or from 16-7 wet to 60 2 10 on the Segments for his 4th also which two 4ths keep Then secondly The Extent from 1 to 116 ½ the whole Content in Gallons shall reaâ—h from 39 the dry 4th on the Line of Numbers to 46 3 10 for the gallons dry or wanting or the same extent shall reach the same way on the Line of Numbers from 60 2 10 the 4th Number for wet to 70 gallons and 2 tenths in the Vessel at 16 inches and 7 tenths wet which two Numbers put together makes up 116 gallons and a half the full Content The like manner of working serves for any Cask whatsoever and the nearer the Vessel wants of being half empty the more near to the truth will your work be and the most errour in very round and swelling Cask when the emptiness is not above one or two inches but in Vessels near to Cillenders it will give the Answer very true and as readily as any way whatsoever Observe also That if you use the Segments in taking the wants you must abate of the gallons found till you come to the 2 thirds of the half diameter that is to say the Rule sayes there is more wanting than indeed there is and that somewhat considerable about the first 6 inches in a vessel of 30 inches diameter So that I find a Table made as a mean between the Superficial and solid Segments would do the work the truest and best of any other Or else use the mean diameter and mean parts of emptiness found thus Take the equaded diameter from the diameter at the bung and note the difference then half this difference taken from the inches and parts empty gives the mean emptiness then use the mean diameter and mean emptiness instead of the other and the work is more exact The other way of Mr. Bennits invention is thus First you are to fill an ordinary Cask of a competent magnitude as 60 or 100 gallons of a mean form between a Spheriord or roundish form and a Cillenderical form or else fill two Casks of each form and learn the true Content and Diameter of that mean Vessel or rather of both those Vessels and the Vessel being full draw off with a true gallon-measure and on the drawing off every gallon take the exact quantity of inches and 10th parts that the drawing off of every gallon makes in the emptiness or driness of that mean Vessel or rather both those Vessels at least until you have drawn off the half quantity of the Vessel which number of gallons drawn off and the inches and tenth parts of emptiness or fulness or driness or wetness you must draw into a Table or insert them on a Rule making the inches as equal parts and the gallons and his proportional part of a gallon the unequal parts then with the Line of Numbers and this mean Table or rather two Tables or Scales which you may put on a Rule as Mr. Bennit hath done you may find out the wants of any Cask whatsoever either Spherioid or Cillender-like as followeth This measured Cask on the Scale or Table for methods sake and avoiding tautologie I shall call the first Cask and the Vessel or Cask whose wants you would know I shall call the second Cask then the proportion is thus As the Diameter at the bung of the second Cask is to the bung diameter of the first Cask which is always fixed So is the inches dry of the second Cask to a 4th on the Line of Numbers which 4th Number sought on the inches of your Table or Scale on the opposite-part of your Scale or Table gives a 5th Number which you must keep Then As the whole Content of the first Cask is to the whole Content of the second Cask So is the first Number kept to the Number of Gallons the Vessel wants of being full at so many inches dry Example There is such a Scale made on purpose for Victuallers use to measure what they want of a Barrel of Ale being put into a Beer-barrel which Scale I shall here use to try this former Example by Suppose as before a Canary Pipe want 12 inches of being full and the Content 116 ½ gallons and 28 inches and 7 tenths diameter at bung The Extent on the Line of Numbers from 28-7 to 22-5 shall reach from 12 to 9-4 then just against 9 inches and 4 tenths on that Barrel Scale I find 14 gallons of Beer which is 17 gallons and a half of Wine being the 5th Number to be kept Then the Extent from 44 the Content of a Barrel in Wine-gallons to 116 ½ the Content of a Canary-Pipe in the same gallons shall reach the same way from 17 ½ the Number kept to 46 and near a half the gallons wanting at 12 inches dry in the Canary Pipe and 46 gallons and 3 quarts is the Number Mr. Bennit finds in a Canary-Pipe by measuring at 12 inches dry Thus you have an account of the two easie Mechanick wayes to discover the wants of Cask very applicable and ready and experimented to be Propè verum The Gallons wanting in a Barrel at every inch and quarter  Beer Gall. Wine Gall    gal pi 100 gal pi 100 gal 1000   0 0 40 0 0 49 0 0612   0 1 20 0 1 47 0
from A but if you remove to E the Thred falling on 3 on the Shadows the Altitude will be but one third part of the distance EA From hence you may observe that observing at C and at D where the Thred falls on 1 and on 2 the distance between C and D is equal to the Altitude so likewise at D and at E and so by consequence at 1 ½ and 2 ½ and 3 ½ or any other equal parts This is an excellent easie way The like will be if you observe at D and C looking up to F where the Altitude AF is twice the distance AC Use V. Another way by the Line of Shadows at one station Measure any distance as feet yards or the like from any object as suppose from A to D were 200 foot and looking up to B the Thred cuts the stroke by 2 on the Line of Shadows Then by the Line of Numbers say As 2 the parts cut is to 1 So is 200 the distance measured to 100 the height Or Suppose I measured any other uneven Number and the Thred fall between 00 on the Loose-piece and 1 on the Shadows commonly called contrary Shadow The Rule is alwayes thus As the parts cut by the Thred are to 1 So is the measured distance to the height required being less than the measured distance But when the Thred falls between 1 and 90 at the Head called right Shadow then the Rule goes thus As 1 to the parts cut by the Thred So is the measured distance to the height being alwayes more than the measured distance from the foot of the object to the station Use VI. Another way by the Line of Shadows and the Sun shining When the Sun shineth find his Altitude and also as the Thred lies see what division on the Line of Shadows is cut by the Thred and then straightway measure the shadows length on the ground and if the Sun be under 45 degrees high the shadow is longer than the length of that object which causeth the shadow but if the Sun be above 45 degrees high then the object is longer than the shadow and the Operation is thus by the Line of Numbers only with a pair of Compasses The Height of the Sun being under 45 say As the parts cut by the Thred on the Shadows is to 1 So is the Shadow measured to the height required The Height of the Sun being above 45 say As 1 to the parts cut by the Thred on the Line of Shadows So is the measure of the shadow to the height in the same parts Use VII To find an inaccessable Altitude by the Quadrat and Shadows otherwise Observe the Altitude at both stations and count the observed Altitudes at both stations on the Quadrat or Shadows according as it happens to be either above or under 45 degrees and take the lesser out of the greater noting the remainder for the first term and the Divisor to divide the distance between the stations increased with Cyphers if need be and the Quotient is the Answer required But by the Line of Numbers work thus The Extent from the difference to 1 shall reach the same way from the measured distance to the height required Example Figure II. Let ABCDE represent the Object and three Stations let the Line AC represent the Altitude the Point B one station 50 foot from A D another station 100 foot from A or 50 from B and E another station 73 foot from D or 173 foot from A all which measures you need not know before but only BD and DE Also the Angle at B 63-27 and his complement counting the other way being the Angle at C 26 degrees 33 minuts the Angle at D 45 and his complement so also the Angle at E 30 and his complement 60. Now mind the Operation by either of these First lay the Thred on 26-33 and in the Quadrat it cuts 50 lay the Thred on 45 and in the Shadows or Quadrat it cuts 100 or 1 or if you lay the Thred to 60 then in the shadows it cuts 173. The difference between 173 and 100 is 73. Then As 73 the difference in Tangents between the two observations is to the distance in feet 73 So is Radius 100 or the side of the Quadrat to 100 the hight required Again for the two nearest Observations whose difference of Tangents is 50. As 50 the difference in Tangents to 50 foot the measured distance So is 100 the side of the Quadrat to 100 the height Again lastly by the observations at B E the difference of Tangents being 123. As 123 the difference in Tangents to 123 the measured distance So is 100 the Radius or side of the Quadrat to 100 the height required Or In the first Figure the Angles at the top being 33-30 and 48-15 and the measured distance 88 foot and a half the difference in Tangents will be 45-8 Then As 45-8 to 100 the side of the Quadrat So is 88 ½ the measured distance to 194 the Altitude required This way is general for any Station though both of right shadow or both of contrary or mixt of right and contrary and done by the Line of Numbers or by Multiplication and Division Also Note That you may find this difference in Tangents or Secants by the Natural Tangents or Natural Secants on the Sector and the Scale of equal parts belonging to them Thus Take the distance between the compleplement of the two observations on the greater or lesser Line of Tangents as is most convenient and measure this distance in the Line of Lines or equal parts equal to that Radius and that shall be the difference in Tangents required The like for the Secants Also By the Artificial Numbers Sines and Tangents you may come by this differences in Tangents or Secants very well thus Just right against the Tangent of the Co-altitude counted on the Line of Tangents in the Line of Numbers is one Number and against the Tangent of the complement of the other Angle is the other Number only with this Caution That if the Tangent be above 45 then take the distance from 45 to the Tangent as it is counted backward with Compasses and set the same the increasing way from 1 on the Numbers to the other Number required then the lesser taken from the greater leaves the difference in Tangents that was required In the same manner the Sines counted from 90 and laid the contrary way from 1 increasing will give the difference in Secants to measure the Baâ—â— and Hypothenusa by Numbers only Use VIII Another pretty way by Scale and Compass without Arithmetick from T. S. Then draw the Line that the Thred maketh on the Board Then measure from your standing to the foot of the Object and take the number of feet or yards from any Scale and lay it from the right Angle on the other Line and raise a Perpendiculer from thence to the Plumb-line made by the Thred and
that shall be the Altitude required being measured on the same Scale Example Let ABGD represent the Boards end or Trencher and on that let AB be one streight Line and AG another Perpendiculer to it in the Point A knock in one Pin and in B or any where toward the end another On the Pin at A hang a Thred and Plummet and standing at I any convenient station look up by the two Pins at B and A till they bourn in a right Line with the Point H the object whose height is to be measured then the Plummet playing well and even make a Point just therein and draw the Line AD as the Thred shewed Then having measured the distance from G the foot of the Object to I the station take it from any first Scale and lay it from A to G then on the Point G raise a Perpendiculer to AG till it intersect the Plumb-line AD then I say the distance CD measured on the same Scale you took AC from shall be equal to the Altitude GH which was required Use IX The same work at two stations But if you cannot come to measure from I the first station to G then measure from I to K and having observed at I and drawn the Plumb-line AD take the measure between I and K the two stations from any fit Scale of equal parts and lay it on the Line AC from A to C viz. 79 parts and in the Point C knock another Pin and hang the Thred and Plummet thereon and observe carefully where this last Plumb-line doth cross the other as suppose at E then from E let fall a Perpendiculer to the Line AC which Line AC shall be the height GH required or thus the nearest distance from E to AC is the height required viz. 120 of the same parts that IK is 79 Note the Figure and behold that ACFE the small Figure on the Board is like and proportional to AA GH the greater Figure Other wayes there be as by a Bowl of Water or a Glass or a Plash of Water or a Square but these set down are as convenient and ready as any whatsoever As in the next Figure you may see the way by the Glass and Square As thus Let C represent a Glass a Bowl or Plash of Water wherein the Eye at A sees the picture or reflection of the Object E. Then by the Line of Numbers As CB the measure from your foot to the Glass is to AB the height from your eye to the ground at your foot So is the measure from C to D to the height DE. See Figure VI. Again to find a distance by the Square that is not over-long Let C represent the upper-corner of a Square hung on a staff at F then the one part of the Square directed to E and the other to A. The Proportion will hold by the Line of Numbers As FA 11-37 to FC 50 So is FC 50 to FE 220. That is So many times as you find AF in FC So many times is FC in FE and the like Note That you must conceive AFE to be the Ground or Base-line in this Operation by the Square C being the top of an upright Staff 5 foot long called 50 for Fraction sake Use X. To find a Distance not approachable by the Trianguler Quadrant First I plant my Trianguler Quadrant set upon a three legged Staff and Ball socket right over the place A and then bring the Index with two sights in it laid or fastened to the Center of the Trianguler Quadrant right over the Lines of Sines and Lines cutting 90 at the Head the Index and sights so placed hold it there and bring it and the Instrument together till you see the mark at C through the two sights by help of the Ball-socket and then there keep it then remove the Index only to 0-60 on the loose-piece which makes a right Angle and set up a mark in that Line at any convenient distance as suppose at B 102 foot from A then remove the Instrument to B and laying the Index on the Center and 0-60 on the loose-piece direct the sights to A the first station by help of a mark left there on purpose Then remove the sights till you see the mark at C and note exactly on what degree the Index falleth as here on 60 counting from 060 on the loose-piece or on 30 counting from the Head which is the Angles at B and at C. Then by the Artificial Numbers Sines and Tangents on the edge say As the sine of 30 the Angle at C to 102 the measured distance counted on the Numbers So is the sine of 60 the Angle at B to 117 on the Numbers the distance required So also is 90 the Angle at A to 206 the distance from B to C. Or by the Lines and Sines on the Quadrant-side as it lies thus As the measure of 102 taken from any Scale as the Line of Lines doubling to the = sine of 30 laying the Index or a Thred to the nearest distance So is the = sine of 60 to 117 measured latterally on the same Line of Lines And So is the = sine of 90 to 206 the distance from B to C. So also If you observe at B and at D only you must be sure to set your Instrument at one station at the same scituation as at the other as a looking back from station to station will do it and the same way of work will serve For As the Sine of 20 to 110 So is the Sine of 40 to 206. And So is the Sine of 120 to the Line DC 278 c. Use XI To find a Breadth and a Distance at any two Stations Let AB be two marks as two corners of a House or Wall and let the breadth between them be demanded and their distance from C and D the two stations First set up two marks at the two stations then setting up the Instrument at C set the fiducial Line on the Rule to D the other mark then direct the sights exactly to B and to A observe the Angles DCB 45 and DCA 113-0 as in the Figure Secondly Remove the Instrument to D the other station and set the fiducial-Line of the Quadrant viz. the Line of Lines and Sines directly to C then fix it there and remove the Index and sights to A and to B to get the Angles CDA 42-30 and CDB 109-0 Then observe that the 3 Angles of every Triangle being equal to 180 degrees having got the Angles at C 113 and the Angle at D 42-30 by consequence as you take 155 the sum of the Angles at C and D out of 180 then there remains 24-30 the Angle at A. So also Taking 109 and 45 from 180 rests 26 the Angle at B then also taking 45 the Angle BCD out of 113 the Angle DCA rests 68 degrees the Angle BCA in like manner taking 42-30 from 109 the Angles at D rests