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ID	Title	Author	Corrected Date of Publication (TCP Date of Publication)	STC	Words	Pages
A74684	The scales of commerce and trade: ballancing betwixt the buyer and seller, artificer and manufacture, debitor and creditor, the most general questions, artificiall rules, and usefull conclusions incident to traffique: comprehended in two books. The first states the ponderates to equity and custome, all usuall rules, legall bargains and contracts, in wholesale ot [sic] retaile, with factorage, returnes, and exchanges of forraign coyn, of interest-money, both simple and compounded, with solutions from naturall and artificiall arithmetick. The second book treats of geometricall problems and arithmeticall solutions, in dimensions of lines, superficies and bodies, both solid and concave, viz. land, wainscot, hangings, board, timber, stone, gaging of casks, military propositions, merchants accounts by debitor and creditor; architectonice, or the art of building. / By Thomas Willsford Gent.	Willsford, Thomas.	1660 (1660)	Wing W2876; Thomason E1748_1; Thomason E1748_2; ESTC R209647	103,352	283

timber is commonly in length 2 feet a thing necessary but of no necessity whether longer or shorter for this length will contain a foot of board although but 6 in b●oad and what the length of the Ruler cannot comprehend is usually termed under measure with which I will begin The side of a foot square is divided into 12 equal parts called inches the quadrat of 12 is 144 the number of square inches contained in a foot and if it were demanded what length shall be required at 1 inch broad to be equal unto it being an unite is divider the answer will be 144 inches that is in length 12 feet if the breadth were 1 ½ inch with which according to the last Theorem divide 144 the quotient will be 96 inches that is in length 8 foot at 2 inches broad 72 inches or 6 feet in length at 2 ½ broad 4 F. 9 ● ● inches at 3 inches in breadth 4 feet in length and so proceed in the under measure by half inches if you please untill you come at 6 inches with which divide 144 the quotient will be 24 inches or 2 f. in length the board measure being now upon the two foot Rule containing 24 inches and each inch usually divided into 4 equal parts the board measure commonly proceeds also by quarters of an inch so 144 divided by 6 ¼ in the quotient will be 23 1 25 inches in length then for 6 ½ inches take in length upon the Ruler 22 2 13 inches at 6 ¾ take 21 ⅓ inches at 7 you will find 30 4 7 inches and thus proceeding in quarters to 8 inches which will require 1 ½ foot or 18 inches in length and 9 broad 16 inches at 10 broad 14 ⅖ inches at 11 inches broad 13 1 11 inches and 12 inches is the side of a foot square from hence ascending the square exceeding 12 inches the length will lessen but thus proceed by quarters to 24 inches from thence to 3 feet broad by half inches and after that by whole inches onely the difference growing scarcely sensible and howsoever not considerable in things of this nature for if they should be continued to 4 feet the difference betwixt 47 and 48 inches in this square measure will be but 3 47 parts of an inch these proportionals found you may inscribe them upon a Ruler with figures to them and so made ready and apt for common use if exactnesse be required make use of the Problemes delivered you before Yards are divided after the same manner in their proportional squares to any breadth assign'd but usually such measures are taken in feet one F. being the least in breadth that is commonly measured 9 feet making a yard square 3 being the side frequently without any under measure beginning at 3 feet for the breadth for any such superficies to be measured from thence proceeding by inches with their quarters to 10 feet in breadth and more if need require the side of this square contains 3 feet that is 36 inches whose quadrat is 1296 square inches that divided by 48 which is 4 feet the quotient will be 27 inches that is 2 feet 3 inches in length equal to a yard square 5 feet broad requires 21 ⅗ inches 6 feet or 72 inches must have 18 inches in length 7 feet broad 15 ● 7 8 feet broad 13 ● 2 inches in length 9 feet will have 12 inches in length and a long square 10 feet in breadth every 10 ⅘ in length will be equal unto a yard square and according to these dimensions having found the parts in length answering to the feet in breadth from 3 unto 10 F. by the parts found you may inscribe them upon a Rule 36 in in length then find the quarters in the same manner to place between the feet and inches The pole for Land-measures is onely divided into equal parts is quarters c. and so likewise the chains distinguished usually with brass rings and those again by tenns both ready exact and of excellent use especially in Decimal Arithmetick PROBLEME IX To find the Area of the least Quadrangle or square figure that can comprehend the circumference of any Circle propounded The Theorem The Diameter of any Circle squared makes each side a Tangent to the Peripheria or circumference thereof A Demonstration Definitions and terms A Circle is a Geometrical Figure comprehended with one line as E F G H E termed the Peripheria or Circumference the part contained as in other figures is usually called the Area in the middle of which there is a point denominated the center as at 14 from whence all right lines drawn to the Circumference are equal and infinite and if any one be terminated at both ends with the Circumference it is called the Diameter as the line E G 14 whose square is 196 for the content of the Quadrangle A B C D each side being a Tangent so named from touching and not intersecting the circle as in E F G H being the least square that can be made containing the circle which in this Probleme is the thing required PROBLEME X. The diameter or circumference of any Circle being known to find the greatest circumscribed Quadrangle or square made within that Circle propounded The Theoreme The quadrat root extracted from half the square made of the Diameter shall be the side unto the greatest Quadrangle that can be circumscribed by that circle An exact proportion betwixt the Diameter and the Circle was never yet discovered unto man his knowledge therein confined within a small circumference of his own imagination but as for circles within our capacities accommodated to humane use the proportions are usually these viz. as 7 is to 22 or as 71 unto 223 so will the Diameter of any circle be in proportion to the cicumference of it the last is most exact but the first most in use as in this the Diameter E G being 14 the circumference of the circle will contain 44 of those equal parts so the one being known the other mey be found wonderful near the truth now as for this Probleme draw a line or suppose one drawn from E and G to F these sides are equal by construction and by the 11 and 28 Proposit Trigon lib. 1. will inclose a triangle right angled at F. and by this second Probleme the subtendant side E G squared will in quantity contain the squares of the two equal sides and consequently half the square of E G 196 that is 98 must be equal to the square of F G whose quadrat root extracted in tenths lib. 2. parag 1. examp 5. Arith. there will appear 9 399 1000 for the side of the greatest inscribed Quadrangle the content of the whole square is 98 as before PROBLEME XI To find the nearest quadrature of a Circle that is such a square whose superficial content shall without sensible errour represent the Area of the Circles Peripheria Three Theoremes A squared Diameter multiplied by 11 and the product

by which it is apparent that he gained 12 lb 10 ss in the 100 l. or after that rate for 100 l. thus imployed will return 112 ½ l. If any question of this kinde should depend upon Losse the Price at which 't was sold must be less then that by which the Commodity was bought at so the fourth proportional number will be discovered by the same Rule the state of the Question not differing in any thing either by Whole-sale or Retail so it requires no Precedent or Rule but this which will bring your stock short home as unfortunately true as prosperously with increase PROPOSITION VII By the Price which any Wares or Merchandizes were sold at with the rate of Gain or Loss in one Peece how to discover what the whole Commodity cost The RULE As 1 Peece 1 Hund. 1 Yard or 1 Pound weight c. shall be in proportion unto the price thereof so will the number of Peeces or quantity sold be proportionable to the price of them all together An Explanation of Gain or Loss in one Parcel to finde the rest Lib. 2. Parag. 8. Admit 15 Clothes or Pieces were sold for 340 l then was the price of one Piece 22 l 13 ss 4 d as by the first Proposition in this there was present gain 19 ss 4 d upon every Peece which subtracted from the Price 't was sold at viz. 22 l. 13 ss 4 d. the difference is 21 l. 14 ss for the price it cost then will the proportion be as 1 whole Cloth is to 21 7 10 l so shall 15 Clothes be unto 325 l 10 ss as in the Table appears If this Commodity had been sold to loss the differences betwixt the prices makes it evident and then what one Piece or any pa●t had co●● will be discovered as before with all the whole losse sustained and if it should be required after what rate in the 100 l. the last proposition will unfold it according to the Rule of Trade PROPOSITION VIII To finde the Gain or Losse upon Merchandizes bought and sold with time agreed upon betwixt the Debitor and Creditor for payment of the money at any rate per cent per an The RULES Rule 1. As 100 l sterling is to any interest so a summe given If for 12 Moneths Rule 2. What for the time An explanation of Gain or Losse with time at any rate per Cent. Lib. 2. Parag. 10. Admit a Tradesman had bought a Commodity at 5 d the pound and after 6 Moneths time sold it again for 6 d the l. or suppose the Merchandize was bought at 5 ss the yard and sold it presently again for 6 s the yard but with 6 Moneths given for day of payment or to abate so much as the interest should come unto at 8 l per cent per annum by the sixth Proposition the gain of those Wares will be discovered after the rate of 20 l per cent if present pay but here is to be rebate of money or forbearance of the stock and profit for six Moneths suppose 100 l disbursed for these Wares at first which would make 120 l if paid down on the nail but here use is to be considered for that summe and six moneths time with the encrease to be deducted the interest of which summe is thus found in this Proposition 't is six Moneths and 8 l per centum as in the first row or rule in the Table in the second row under 100 l stands the term for a year in the same denomination with the time given viz. 12 moneths and under the third terme the time limited for payment viz. 6 Moneths the products of them according to the double rule of Proportion in the third line is as 1200 to 8 l. so 720. these are again reduc'd in the operation of the fourth Table as 120 to 8. so 72 unto 4 l. 16 s. and might have been reduc'd again to 5.1.24 which will also produce 4 l. 16 s. that subtracted from 120 l. the remainder will be 115 l. 4 s. which shewes 15 l. 4 s. clear gains in relation to the rate by which t was bought and sold at with the interest for the forbearance agreed upon according to custom and contract but not exactly true PROPOSITION IX By the price of any Wares bought and sold with the time limited for payment to finde the gain made or losse sustained and at what rate per cent per Annum THE RULES Rule 1. As the first price shall be unto 100 l. so the gain or losse If for 12 moneths Rule 2. So the time limit An Explanation in Gain or losse with Time Lib. 2. Parag. 10. A Merchant bought Mace at 6 s. 4 d. the l. ready money and he sold the same again unto a Grocer for 7 s. the l. at this rate the Mace was delivered and upon condition to be payd at the end of 4 moneths next ensuing the receipt thereof and it is required what gain the Merchant made of his money and at what rate per cent per Annum In all questions of this kinde make the price at which t was bought and as 't was sold of one denomination the difference shall be the third terme in the first rule 100 l. the second number and the price for which 't was bought the first term in the second rule under the first number I place the magnitude of a year in that denomination in which the time limited is given as in Moneths Weeks or days in this 't is Moneths as the Letter M denotes the space of time given for payment is 4 Moneths subscribe that under the third number then draw a line from thence towards 19 G and that crosse with another as from 12 M t 2 G in this Example these multiplied crosse-wise the second rule being reverst for the lesse time is given for payment the profit will be the greater in the third row stand the products in the Rule of 3 direct and in the fourth Row or Table is plac'd the form of operation wherein the desired product is discovered to be 31 1● 19 l that is 31 l 11 ss 6 d 1● 19 the profit required at the rate per cent per annum PROPOSITION X. A Grocer bought Cloves at 4 ss 3 d the l. and after 6 Moneths time sold them again for 4 ss the l what losse did the Grocer sustain and how much per cent per ann by the last proposition you will find his losse to be 11 l. 15 ss 3 9 17 d. PROPOSITION XI By the difference of prices in any one Commodity bought and sold by whole-sale or retail to finde what time must be allow'd for to gain after any rate per Centum per annum that shall be assigned The RULES Rule 1. As 100 l sterling is unto 12 Moneths so the rate propounded Unto the 1 price Rule 2. so the gain or loss An Explanation in Gain or Loss with time unknown Lib. 2. Parag. 10. A Tradesman bought Nutmegs at 8 ss

find 9 ss 9 d very near so A must be responsable to B or their heirs at 9 years end for 237 L 9 ss 8 d 3 ● 10 q. This exactnesse was not required nor yet so great a number taken for the fraction of 13 sh 4 d. but these if understood the ingenuous will ease themselves by my labours to which end I will proceed QUESTION XI A was to pay unto B 200 L at the full term of 5 years for which debt A was contented to make B a Lease of a Farm to continue in force the same time whose annual rent was 35 L. which of them gained by this contract interest allowed at 6 L per cent per annum In the Table of Rents forborn under years look 5 the decimal number against it is 5.63709 which multiplied by 35 L the Rent respited the term of 5 years the product will be 197.29815 and reduced into money is 197 L 5 sh 11 ½ d. which subtracted from 200 L the remainder is 2 L 14 sh 0 ½ d. and so much A did gain by the bargain or contract made with B. The fourth Table exemplified in discount of Annuities Rents Pensions or Reversions at 6 L per cent per ann compound interest QUESTION XII What is the present worth of 80 L Rent or Annuity to continue 25 years rebating at 6 L per centum per annum Look in the fourth Table for 25 years against which I find 12.78335 This compound Decimal multiplied by 80 L the Annuity propounded the product proves 1022.66800 which reduced into money will be 1022 L 13 sh 4 ¼ d the true value of 80 L per annum yearly payments rebated for 25 years according to demand QUESTION XIII A man hath a Lease of Lands or Tenements worth 15 L per ann more then the rent and hath a Lease yet 4 years in being the Tenant desires to take another in reversion for 21 years at the same rent what must the Lessee pay for a Fine interest allowed at 6 L per centum per annum   for 4 years   for 25 years   346510   1278335 1 1732550 5 6391675 2 51.97650 6 191.75025 3 51 L 19 sh 6 D 7 191 L 15 sh 0 D 4 139.77375 8 139 L 15 sh 6 D First seek the Decimal for the term of four years 346510. which multiplied by 15 L or by 5 as in the first Table in the margent according to lib. 1. sect 1. parag 4. exam 5. the product in the a row will be 51.97650 in money 51 L 19 sh 6 d. and so much the old lease in being is worth when the new for 21 years enters possession now admit the term of the old Lease and the new added together the summe of years is 25 the profit or overplus of Rent is to continue all the time therefore 1278335 the Decimal for 25 years multiplied by 15 L as in the fifth row of this table produceth in the 6.119.75025 equal in value to 191 L 15 ss the difference of the first Lease and the total time in the 8 row is 139 L 15 ss 6 d. and so the difference of decimals in 4 row reduced is very near without a material error being 139 L 15 ss 5 7 10 d. QUESTION XIV A Tenant hath a Lease of 21 years the present thereof is 41 L per ann during the term of 7 years and after that time shall be expired the Lessee is to pay 50 L rent per ann for the residue of the term what is the value of this Lease in ready money interest discounted at 6 L per cent per annum   for 21 years   for 7 years 1 11.76407 5 5.58238   5   9 2 588.20350 6 50.24142 3 588 L 4 ss 0 ¼ d 7 50 L 4 sh 10 d 4 537 L 19 ss 2 ¾ d 8 537.96208 In the fourth table of Rents rebated the Decimal of 21 years is 11.76407 which multiplied by 50 L the rent of 21 yeares the product is 588.20350 as in the second row of this Table which reduced is 588 L 4 ss 0 2 4 d as in the third row which had been the true value of it at L per ann for the whole term of time but the first 7 yeares of this Lease was but 41 L annual rent therefore the first Decimal was too great by the difference of rent which was 9 L per annum then look into the fourth Table for 7 years and against it you will find 5.58238 which multiplied by 9 L as in the first row of this Table the product in the sixth is 50.24142 and reduced is 50 L 4 ss 10 d very near which subtracted from the third row the remainder is 537 L 19 sh 2 ¾ d as in the fourth row or subtract the Decimals found in the sixth from the second row the difference will be 537.96208 which artificiall number reduced would be 537 L 19 sh 2 ¼ d as before the true value of the Lease required QUESTION XV. There is a Lease to be taken for 21 years at 30 L per ann and 100 L Fine the Lessee likes the bargain but not the condition desiring the annual rent to be but 10 L yearly payments and is willing to give such a Fine as shall be proportionable to the rent abated during the aforesaid term of 21 yeares and here the Fine is demanded In all questions of this kind   The Decimal 1 11.76407 2 L 235.28140 3 S 5.628   1256 4 D 7.536 take the rent abated which is here 20 L per ann for 21 years whose decimal in the 4th Table of Rents rebated is 11 76-407 as in the margent which multiplied by 20 produceth 235.2814 that is 235 L. reduce the fraction neglecting the ciphers the value of 20 L per ann the difference of Rent for 21 years is as in the 2 3 and 4 row in all 235 L 5 sh 7 ½ D. this added unto the former Fine 100 L makes in all 335 L 5 sh 7 ½ D the true summe to be paid for a Fine in lieu of 20 L Rent per ann abated during the Lease of 21 years the thing required QUESTION XVI A had a Lease of 130 L per ann to continue 24 years B had another of 210 L per ann and 11 years to come these 2 men mutually exchanged Leases A upon the contract paid unto B 20 L in ready money which of these had the better bargain and how much A   B 12.55036   788687 3765108 1 1577374 1631.54680 2 1656.24270 1631 L 10 sh 11 D. 3 1656 L 4 sh 10 D. 1651 L 10 sh 11 D. 4 4 L 13 sh 11 D. Against the 24 year of the fourth Table look and you will find the Decimal of it 1255036 for A. secondly the lease of B 11 years hath this decimal 7.88687 these 2 numbers multiplied by their respective rents as in the first row of this table according to lib. 1. sect 1. par 4. exam 6 7. or by the vulgar way In the second

tun or 20 hundred gross from hence continued down to an unite in such an order or series of numbers most frequently used in the Commerce and Trade of this our British Island in the next stands their subtile and gross weights noted with numeral letters distinguished in the post-script by the letter g denoting gross the third and fourth Column shews what summe of money they do make in Shillings in the fifth sixth and seventh Column what so many groats amount unto the three next Columns what those gross or subtile summes do make in pence the three last in the least denomination of money what the value riseth to Pounds Sterling Shillings and Pence as by their titles in the head of these tables do evidently appear The benefit of this Table by sundry Examples illustrated to ease the Art of Memory EXAMPLE I. It is required without calculation what 2240 or 20 C gross amounts unto in shillings look under its title and you will find 112 L 0 ss in groats 37 L 6 ss 8 D. so many pence comes to 9 L 6 ss 8 d. and in farthings 2 L 6 ss 8 d. EXAMPLE II. In things sold by Retail Admit a Commodity vended for 3 half pence the pound and it is required at that rate how much it comes unto by the Tun I take it in the least denomination of Coyn that is 6 farthings and at one farthing the pound under that title I find 2 L 6 ss 8 D. for which you must impose so much on every farthing contained in the price which was 6. then consequently the summe must be 6 times 2 L. secondly 6 times 6 ss and also 6 times 8 D. or for brevity 6 times 2 L is 12 L and 6 Nobles is 2 L. in all 14 L at 1 ½ D the lb as was required EXAMPLE III. At 3 D the pound what comes a tunne unto under the title of Pence I find 9 L 6 sh 8 D at a penny the lb. then 3 times that is 28 L. admit a commodity at 8 D the lb. you may work this as before but being the price is 2 groats under that title you will discover 37 L 6 sh 8 D. and since 2 groats is the price of one lb that doubled is 74 L 13 sh 4 D. if the price had been 3 sh the lb the tunne would have come unto 336 L as by the table is evident EXAMPLE IV. How much comes 10 D a day unto by the year I look down in the table for 365 the number of dayes in a uulgar year and under the title of pence I find 1 L 10 sh 5 D. now 10 times that is 15 L 4 sh 2 D or summe it up in your memory thus viz. 10 L then 10 Angels 10 groats and 10 D. or take it in groats and pence as 6 L 1 sh 8 D and 1 L 10 sh 5 D makes 7 L 12 sh 1 D. this doubled the question depending on 2 groats and 2 pence the summe will be as before 15 L 4 sh 2 d. EXAMPLE V. If one pound of Cheese cost 3 ¾ D what comes the Weigh unto This properly belongs to a certain quantity of Wooll and Cheese cons●sting of 32 Cloves whereof one contains 8 lb so the Weigh is 256 lb. which having found in the table I seek it in the colume of pence and find 1 L 1 sh 4 D. and in the row of farthings 5 sh 4 D. the summe 1 L 6 sh 8 D. and being that there was 3 times so much to be imposed in either denominaation the summe is 4 L for the Weigh at the rate propounded EXAMPLE VI. Wine sold at 2 sh 5 D the gallon how much is that a tun containing 252 gallons Having found the number look against it the column of shillings and you will discover 12 L 12 sh at 12 D the gallon which doubled is 25 L 4 sh to which adde 5 L 5 sh for the 5 D the summe is 30 L 9 S. the column of pence answering that number being but 1 L 1 sh so it is easily multiplied by 5. Or take it as mixt in their severall columns of groats and pence 't will be all one in the total EXAMPLE VII Currants sold at 3 ¾ D the pound how much comes 1 C weight gross unto The farthings contained in the price are 15 and against 112 in the last column I find 2 sh 4 D to be imposed on every farthing that is 15 groats twice 15. in all 1 L 15 S. or take it in both columns of pence and farthings as 3 times 2 sh 4 D. and thrice 9 sh 4 D. if this question had been propounded on 100 lb subtile the answer wil be 15 pence and twice 15 shillings that is 1 L 11 sh 3 D. EXAMPLE VIII Fish sold by the warp or couple at 2 sh 10 ½ D the warp what co●es 4 C unto 60 in this commodity is 120 Fi hes to the C. look 60 in the table against which I find in the column of shillings 3 L then for the 2 sh I set down 6 L or keep it in my memory in the next column I observe 1 L and for the 2 groats in the price 2 L then for 2 D in the column of pence I impose 2 Crowns and for the 2 farthings in the price 2 sh 6 D in all 8 L 12 sh 6 D the price of 60 couple now 4 times that is 34 L 10 sh according to the demand EXAMPLE IX If 1 pound of Indico cost 8 sh 7 D 3 q. what comes a quarter of 100 unto subtile The answer will be 10 L 16 sh 1 D 3 q. look 25 and for the 8 sh in that column impose 10 L. for the 7 D 14 sh 7 D. and lastly for ¾ D take 1 sh 6 ¼ D the summe will be as it was before and according to my rules of Practise lib. 2. parag 9. By these 9 Examples all obscurities in this kind are cleared difficulties made easie and burdens to the memory removed made facile even to common capacities without tedious rules of Art the Numbers Weights and Measures of Commerce and Trade-being known to those who are conversant or Masters in their own occupations and if otherwise this will be a guide to conduct them without deviation to the end of each gross summ and may be accommodated unto the Numbers Weights and Measures of any forreign or transmarine place if occasion requires or necessity urges which I refer to the ingenious Any day of the year assign'd for the receipt or payment of money or other business to finde what day of the week 't will fall upon for any time to come The Julian KALENDER Bis New-years day   Moneths Dayes   1659 Saturd I January 1 8 15 22 29 ✚ 1660 Sunday II February 5 12 19 26   1661 Tuesd III March 5 12 19 26   1662 Wednesd IV April 2 9 16 23 30   1663 Thursd V May 7 14. 21 28 ✚ 1664 Friday VI June 4 11 18

that is 2 ½ feet for the common square which is apparently false custome herein exceeding the truth and will prove 187 ½ feet which is too much 7 ½ F in this squared piece of timber PROBLEME XIII The dimension of Pyramids and Cones either in Timber or Stone and to finde the solid contents of either species in inches feet or yards c. The Theorem The magnitude or solid content of these figures is found by multiplying the superficial basis of either in a third part of the length The Pyramid whose dimension is here required admit represented by the figure M R S T the side of the basis 3 whose square is 9 the Pyramid in length 30 of the same measure the third part of it is 10 which multiplied by 9 the superficiall basis the product will be 90 for the solid content required If the solid content of M P Q were desired the square at P or Q the basis is 4 the side being 2 which multiplied by 6 ⅔ being a third part of the length M P or M Q 20 the product of 4 and 20 3 is 26 ⅔ the solid content required if these dimensions were in yards it would contain 26 such cubes and 18 feet or if found in feet then 26 F. and 1152 inches as in lib. 2. pag. 154. Arithmetick may be proved The little Pyramid M N or M O is but 1 yard or 1 foot at base 10 times that is the length one third part of it or 10 3 the product shews the content to be 3 ⅓ that is in cubicall yards 3 and 9 feet if the dimensions were in feet the solid content would have been 3 feet and 576 cubical inches the thing required All Cones are measured as the Pyramids are as for example in the figure V Y or V Z the Diameter at the base Y Z is 3 the superficial content of that circle by the 11 Probleme will be found 7 1 14 or 99 14 which multiplied by 10 one third part of the length the product proves 990 14 or 70 5 7 the solid content of the Cone in cubical inches feet or yards according to the parts by which it was measured If the solid content of the lesser Cone V X had been required whose length is 20 the diam at X the base is 2 and by the 11 Probleme the superficies of it is 3 1 7 or 22 7 which multiply by 20 the height produceth 440 7 or 62 6 7 the content in cubical parts of 3 times the Cone the third part of 62 6 7 is 20 20 21 the true dimension of the figure V X. To find the lesser Cone V W the diameter at the base is 1 the square 11 14 which multiplied by 10 3 the length the product will be 110 42 or 55 21 that is 2 13 21 the content of the Cone in cubical pars required PROBLEME XIV The dimension of all Segments in tapering Timber or Stone c. as they are the parts of solid Pyramids and Cones The Theorem Unto the squares of the two extreames or bases adde the Geometrical mean square which summe multiply by ⅓ part of the height or ⅓ of the totall by the height the product will be the Segments solid content required Admit the solid content of the Segment P Q R S T were required which is part of the Pyramid M T as in the figure the squares at the 2 ends are 4 and 9 their products 36 the square root of it is 6 for the Geometrical mean square the summe of these 3 squares viz. 4 6 9 is 19. which multiplied by ⅓ part of the length that is by 1● 3 the product will be 290 3 or 63 ⅓ the solid content of that Segment To find the Segment N O P Q the superficial squares of the 2 bases are 4 and 1 their products 4 the square root 2 which 3 number viz. 4 1 and 2 added together makes 7 and multiplied by 10 3 viz. N O P Q by construction the product will be ●0 3 or 23 ⅓ the true content of that Segment and the summe of these two viz. 63 ⅓ and 23 ½ is 86 ● ● to which if you adde the little Pyramid found by the last Probleme 3 ⅓ the totall of this with the 2 Segments compleats the solid content of the whole Pyramid M R S T 90 as before which demonstrates all the several dimensions to be true in the same manner the Segments of Cones are measured having first found the squares at either end as in Prob. 11. so to the ingenious Artist no more examples will be required yet being a thing in controversie and not well understood by mechanicall men for ampler satisfaction I will explain it with one demonstration more Segments being the most frequent form of all and so more diligently to be observed An ocular Demonstration in measuring of all tapering timber whether round or square The Pyramidal Segment here proposed for to be measured is A. whose length is 15 feet the square at the greater basis is 9 and the lesser end 1 foot as at A and according to the last Probleme the mean square 3 the summe is 13 which multiply by ● ● part of the length in this by 5 there will be produced 65 feet the true solid content required which to prove take the Segment A into pieces as B C and D there would be 9 of them and all of one length but severall forms viz. B a foot square taken from the middle of the Segment then will there be 4 pieces like wedges a foot square at the base and ending in a line at A having no thickness as the four corner pieces are perfect Pyramids containing one foot square at the base the other ending in a point as D each of whose dimensions according to the last Probleme must be 5 cubicall feet being the length is 15 and consequently 4 of these will contain 20 feet and as for C 2 such pieces turned end for end will be equall to the figure B containing 15 feet then one of them is 7 ½ feet and 4 of those will make 30 the totall of the 9 pieces in content is 65 cubical feet equall as they are all together in the figure A which is evidently proved as was required But this Segment according to common practise is measur'd in the middle by taking the Arithmeticall mean that is by adding the sides of the two squares together and taking the half of that for the common square as in this 1 and 3 makes 4 the half 2 whose square 4 multiplied into the length 15 the product will be 60 feet for the content according to custome which is apparently erroneous and 5 feet too little in this piece as before was demonstrated divers other errours in measuring of solid bodies are crept in for want of Art and having got possession of ignorant people they plead prescription and custome of the place whereas Custome cannot establish a Law upon a bad

neglecting the fraction multiply this superficies by 8 ⅛ or 65 8 the product will be 14755 8 that is 1844 cubicall inches which divided by 230 ⅖ or 1152 5 the quotient will be 8 gallons for the capacity of this vessel in Wine measure as was required and with the former fraction 1 14 more it will prove 8 7 1000 gall but when a broken number proves considerable it must not be omitted but in such a case as this reject it if you please This former Figure represents a Bushell which was well approved of in the Countrey although I cannot for having measured it as before I filled it with a Wine quart and found it to contain but 8 gallons Wine measure whereas according to Winchest Stat. it should be 8 Ale gallons and one of them to contain 288 inches as Mr. Windgate affirms M. Outhred that excellent Artist M. Briggs allows the content of an Ale gallon but 272 cubicall inches which is generally esteemed on being sealed by the approbation of these 2 famous Geometricians and equal to the Standard yet I will not confine you to their authority nor perswade you to let another measure your corn by their bushel since measures are so various every where the scope I here do aim at is to find the capacity in cubical inches in concave vessels which found apply them as you please or as the place admits PROBLEME II. Gauging of Vessels by finding the capacities or quantities of liquid measures or cubicall inches contained in them from the least Rundlet to a Tunn either of Wine or Beer Any Vessel of Wine or Beer may be thus measured if the Diameter at the Head and Bung be equall it makes a perfect Cylinder for the dimension of which figure I refer you to the 12 Prob. and the last but vessels of this kind are commonly biggest in the middle and from the bung to the head or either end they have circular sides which if continued would end in a point as the Cone does and hath 2 bases yet is no segment of those which are bounded with right lines but rather a Sphaerocide or Sphaeroides which Sphaerall Segment may be thus measured The length of this Butt or Pipe is 50 inches the diameter at the bung 30 and at the head or either end 21 inches This known by the 11 Probleme find the superficiall content at the head and bung which in this will prove 346 ½ and 707 1 7 square inches take ⅓ part of 364 ⅓ that is 693 6 and ⅔ of 707 1 7 will be 9900 21 both which reduced are 231 2 and 3300 7 adde these together their summe is 3217 14 which multiplied by 50 inches the length of the vessel that product wil be 410850 14 and reduced to 205435 7 which is 29346 3 7 inches In this case you may reject the fraction and divide 29346 by 231 the quotient will be 127 gall and 9 cubical inches or make the divisor or dividend a unite more as you see cause or if more exactness and less trouble be desired see 22 Axiom lib. 2. parag 7. as for example 205425 to be divided by 7 and that quotient by 230 ⅖ or 1152 5 the dividers multiplied will produce 8064 5 that is 1612 or 1613 with which divide 205425 the quotient will be 127 gallons 1 quart and 1 pint ferè the content of this pipe in Wine measure which multiplied by 4 produceth 509 G. 2 Qu. and divided by 5 the quotient will be 101 G. 3 Q. 1 P. and ⅕ the content of this Butt in beer measure Such Propositions are exquisitely performed by the Decimall Tables PROBLEME III. With the Diameter of any Circle known to find the Circumference of it in proportional parts unto 100 1000 or 10000 c. In all questions of this kind make choice of some proportion betwixt the Diameter Circumference as in the 10 Probleme and first as 7 to 22 unto 22 annex 3 or 4 ciphers which divided by 7 the quotient will be 31428 or unto 223 annex ciphers at pleasure 71 must be then divisor the quotient will be 31408 which proportionals may be made more numerous if occasion requires but first suppose 17 were a Diameter propounded whose circumference is required in a Decimal fraction the proportion will be as 10000 the supposed Diameter is unto 17 the true Diameter so will 31428 a supposed circumference be proportionable to 53 4·276 which differs but little from the 10 or 11 Prob. PROBLEME IV. With the diameters of two circles known and circumference of the one to finde the circumference of the other or a diameter with 2 circles given to finde the other diameter The Theoreme Circumferences of all circles are in proportion one to another as be their respective diameters Admit the diameters of two given circles were 17 and 21 the circumference of the first in a decimal fraction is in proportion as 7 to 22.53.43 the second number multiplied by the third will be 1122·03 which divided by 17 the quotient proves 66 the circumference of the second circle which was required If the circumferences of any two circles with one given to find the other the manner of operation is the same so it requires no example PROBLEME V. With the diameter and superficial content of one circle to finde the superficies of another whose diameter is known The Theorem All circles have proportion one another as the superficial squares made of their diameters have As for example Suppose the Diameters propounded were 7 and 14 whose squares are 49 and 196 their proportions as 1 to 4 the superficial content of the first circle is 38 ½ but as a decimal in the table it is 38.5 which multiplied by 196 the product is 7546.0 and divided by 49 the quotient will be 154 the superficies of that circle whose Diameter was onely known as by the 11 Probleme may be also proved PROBLEME VI. The superficiall content of any two circles propounded with one of their diameters given to find the other The former squares are here again propounded with one diameter known whereby to find the other in this example I take 14 for the diameter the superficial square of whole circle is 154. the diameter squared is 196 for the second number and the superficial square of the other circle is 38 ½ or 77 2 which multiplied by 196 produceth 15092 ● or 7546 and divided by 154 the quotient will be 49 whose square root is 7 for the diameter required as in the last Probleme PROBLEME VII To find the content or convex superficies of any Sphere or Globe whose diameter or cicumference is found or propounded by 4 several wayes according to Art 1. By the Diameter or Circumference find the Area or superficial Content of that Circle according to the former Problemes which multiplied by 4 produceth the convex superficies required As for example admit 7 were the Diameter of a Sphere the nearest Quadrature of that circle is 38 ½ which suppose

as for example admit there were a Turret in height 45 feet there was a Moat before it in breadth 22 feet the square of 45 is 2025 and the square of 22 is 484 the summe of these squares is 2509 the quadrat root of it is 50 feet for the length of the ladder or rope that will reach unto the summity or top of it if the remainder had been considerable you might have extracted the root with a fraction as in lib. 2. parag 1. examp 4 or 5. Arith. but some where ignorance hath got the upper hand of their reason will say peradventure what care they for this give them rope enough and so say I with all my heart PROPOSITION IX To find the height of an accessible Fort Turret or any other place by a common square or with two sticks of equal length artificial joyn'd together at right angles Admit the height required were the Tower C D I move my station from F towards D holding the triangle or square parallel with the ground-ground-line and perpendicular by help of a plummet as at K where by both ends of the little square I behold the Towers summity as at C. Now by the 19 Proposition of my Geometry A B must be equal to B C and A B or L D is found by measure 48 feet the t●ue height of B C to which adde B D or A L the height of the square above ground viz. 3 feet the summe 51 feet for the altitude of C above the Horizontal plain F D the proposition answered PROPOSITION X. To finde the distance unto any Fort or place although not accessible yet discovered by this square or triangle Erect a staff perpendicular whose height is exactly known as in the last Scheme E G which admit 6 feet or 72 inches upon the top of it cut a notch so that the square may fall down in it something straight yet so as to turn at E. suppose the distance required were G D place your eye at E then turn the square upwards or downwards untill by the edge of it you see the basis of the Tower or place at D. the square being fixt look down from E to F. at which place a mark upon the ground and measure the distance F G which is he●e 8 inches here you have 2 equiangled triangles viz. G E F and G E D and by the 19 pr●position of my Geometry the sides are proportional now admit this little triangle were delineated in the greater as G H L then is G L equal to G E and G H to G F. thus are they in the rule of proportion as G H 8 inches is to G L 72 inches so will E G 72 be in proportion unto G D 648 inches or 54 feet the true distance required PROPOSITION IX To finde the height of any place approachable by the shadow which it makes with the help of a Pike erected perpendicular to the horizontall plane or by any Turret whose height is directly known or by the height of any Tower to finde the distance though not approachable The height of the Tower A B is required to be found by the shadow which it makes upon the Horizontal place as in this figure suppose B D by measure found to be 12 per 6 3 10 in upon the same horizontall plane I measure the shadow of some other body or erect a Pike perpendicular as C D whose height above ground is 7 feet and the length of the shadow which it makes extends it self from D to E by measure 12 feet 3 ⅓ inches this known the proportion is as E D 12 feet 3 ⅗ inches is to C D 7 feet so the shadow B D 198 feet 6 ● 10 inches unto 112. 98 for the height of the Tower A B which caused the shadow that is 113 feet 3 inches and more much exactness is required in questions of this nature or else little truth to be expected and shadows commonly falling in broken parts which made me herein use the Decimals yet with more exactness performed by Natural Arithmetick and vulgar fractions and so found 112 F. 11 inches 28 100 if the height A B had been known 113 feet and the distance or extent of the shadow B D required the proportion would have been viz. as C D 7 feet to E D 12.3 so will A B 113 feet be unto 198 feet 5 4 7 that is 198 feet 6 inches and ● 7 for B D the distance required which is very near the truth PROPOSITION XII To discover the altitude of an accessible place by a mirrour or looking-glass or by a Towers height known to find the distance unto it Let the position of your Glass or Mirrour be horizontally plac'd at some convenient distance from thence go backward into a direct line untill you can descry in the glass the top of the Tower or object whose height is required then will the distance from your body to that part of the glass where the summity of the Turret was represented be in proportion to the distance from the glass to the ●er●endicula● basis of the Tower or Sconce as the height of your eye is to the perpendicular height required for by the Optick Science it is an apparent Maxime that the angles of Incedence and Reflection are equal as A D B and F D E and your body being parallel with the Tower the Radius of your sight incloseth a triangle equiangled with that of the Turrets shadow as by the 8 9 or 10 Proposition of my Geometry and consequently by the 19 proposition of the same book those triangles are proportional in all their sides this is so visible that it needs no explanation if they can see themselves from their shadows or shall ever behold my Trigonometry to which I refer them for 2 more ample satisfaction and this to their impartiall and judicious censures yet wishing a legal trial to answer unto my charge if there shall be any fomented in the mean time hopes of a candid construction from a serene verdict free from all obstructions of malice to obtenebrate my intentions bids me with comfort to proceed PROPOSITION XIII A Captain of a Castle expecting to be beleagured makes good his out-works and having fortified those best where he conceived most danger of being stormed he over-looks the inventory of his Magazine and takes a list of his Souldiers with the supernumerary persons in all 800. by which he findes his provision of victuals good but for 3 moneths 3 weeks that is 105 dayes having more men then were necessary and expecting no relief under 6 moneths or 168 dayes the question is ' how many men must be dismiss'd this fort before the enemies approach whereby the same victuals might last the just time required The Solution The rule thus stated according to lib. 2. parag 8. Can. 9. Arith. in a reverss'd proportion viz. as the provision of victuals for 105 dayes allowed for 800 men what number will 168 dayes require allowed in the