in yards of that pane which is almost 5 yards a quarter and an half or rather 5 yards and one third part of a yard PROBLEM V. The breadth of any Superficies being given in inches or feet to find how much in length will make a superficial foot The Rule is thus AS the breadth in inches to 144 So is 1 to the length in inches to make a foot Example Let the breadth given be 30 inches and the length to make a foot at that breadth be required Set 30 on the first to 144 on the second and then against 1 on the first is 4,8 on the second So much in length makes a foot at 30 inches broad But if the breadth be given in feet Then As the breadth in feet is to 1 So is 1 to the length to make a foot Therefore Set 2,5 on the first to 1 on the second and then against 1 on the first side is 4 on the second which 4 signifieth four tenth parts of a foot So much in length makes a foot at that breadth PROBLEM VI. The length and breadth of a plot of land being given in chains to find the Content in Acres HAving a chain of 4 perches long divided into 100 links that is 25 in the perch measure the length and breadth of the land to be measured in chains and links And then cast up the Content into Acres thus As 10 to the breadth in chains So is the length in chains to the Content in Acres Example Let the breadth given be 7 chains 50 links and the length 45 chains 75 links and the Content in Acres sought Set 10 on the first to 7,5 on the second and then against 45,75 on the first is 34,31 on the second which is the Content in Acres of that plat of land Or Had the breadth been 15 chains 25 links and the breadth 22 chains 50 links then set 10 or 1 on the first to 15,5 on the second and then against 22,50 on the first is 34,31 on the se-second the Content in Acres sought for which Fraction 31 above the 34 Acres contains 1 rood 10 perches So that the true Content is 34 Acres 1 rood and 10 pole If the plot of land be of a Triangle form or any other figure whatsoever the Analogie is this As 10 is to one of the two numbers of chains that are to be multiplyed together So is the other of them to the Content in Acres As if the half perpendicular of a Triangle be 3,75 chains and the Base 45,75 the Content will be found to be 17,15 Acres For set 10 on the first to 3,75 on the second and then against 45 75 on that first is 17,15 the content Or else having the whole Base and whole perpendicular say thus As 20 to the whole perpendicular 7,50 So is the whole Base 45,75 To 17,15 the content in Acres as before For Set 20 on the first to 7,50 on the second and then against 45,75 on the first is 17,15 on the second Or if the Base of a Triangle be 36,83 chains and the perpendicular 17,59 chains the content will be found by either way of working to be 32,39 which is 32 acres 1 rood 22 perches PROBLEM VII The Content of a piece of land being measured by one kind of perch to find the Content thereof after another kind of perch THese kind of proportions are wrought by the Rule of Three reverse after a duplicated proportion and the Analogie is thus As the length of the second perch is to the length of the first perch So the content in acres given to a fourth number and so is that fourth number to a fifth number which is the content sought Suppose a piece of land measured by the 16½ foot-pole do contain 34,3 acres and it be demanded How much it would contain if it were measured by an 18 foot-pole Note that I call the 16½ foot-perch the first perch because by it the land was measured and the 18 foot-perch I call the second perch because according to it the content is sought for Wherefore Set the second perch 18 on the first to 16,5 on the second the perch first used and then against 34,3 the content in acres given on the first is 31,45 on the second and then against that 31,45 on that first is 28,8 on the second So much is the content in acres by the 18 foot-pole which was demanded In like manner were the content given 5 acres 2 roods 20 pole or 5,62 Set 18 on the first to 16,5 on the second and then against 5,62 on the first is 5,15 on the second And lastly against that 5,15 on that first is neer about 4,7 on the second So much is the content of that close by the 18 foot-pole PROBLEM VIII The one side of any piece of land being given to find how much in breadth the other way will make an acre of land LEt the side of a close be 20 pole and it be required How much in breadth will make an acre of land at that length Set the breadth given 20 on the first to 160 on the second and then against 1 or 10 on that first is 8 on the second So much in breadth makes an acre at 20 pole long Or if the side measured be 25 then set 25 to 160 and against 1 on that first is 6,4 on the second the breadth sought for And if the length be 32 pole the breadth to make an acre will be 5 pole PROBLEM IX A plot of land being laid down and cast up by any Scale to find how much it will contain by any other Scale either greater or lesser SUppose a plot of land being laid down and cast up by a Scale of 10 in the inch does contain 28,5 acres and it is required to know how many acres it will contain should it be cast up by a Scale of 12 in the inch Here because 12 the Scale to be used is lesser than 10 the Scale formerly used And so by consequence the content of the given plot by the Scale of 12 will be more acres than it is by the Scale of 10 in the inch Therefore Set the Scale used 10 on the first to the Scale to be used 12 on the second and then against the content known 28,5 on the first is 34,2 acres or neer thereabouts on the second So much is the content by the Scale of 12 in the inch But if the plot had been laid down and cast up by a Scale of 12 and the content required by a Scale of 10 in the inch which is the greater Scale and therefore the content is the lesser Then Set 12 on the first to 10 on the second which done right against the content in acres by the Scale of 12 on the first is the content by the Scale of 10 on the second As if the content by the Scale of 12 be 34,2 acres then the content by the Scale of 10 will be found to be 28,5 acres neer For
VI. The Rule of Proportion Inverse Three numbers being given to find a fourth in an Inversed Proportion IT is to be noted in this Inverse Rule of Proportion that if the third number be greater than the first then will the fourth number be lesse than the second And contrariwise if the third number be lesse than the first then the fourth number is to be greater than the second But in the Rule of Proportion direct If the second number or term be more than the first then the fourth term is also more than the third And if the second term be lesse than the first then is the fourth term lesse than the third This Inverse Rule may be wrought two wayes on our double lines One way is thus Set the first term on the first to the other term of the same denomination on the second And then against the other term of contrary denomination sought out on that second is the fourth number sought for on the first Or else set the third term on the first to the first term on the second and then against the second term on the first is the answer on the second Example 1. If 60 Pioneers can make a trench in 45 hours In how long time can 40 Pioners make it Set 40 on the first to 60 on the second the two numbers of the same denomination and then against 45 on that first the number of contrary denomination is 67,5 on the second which 67,5 is the fourth term in reciprocal proportion to the other three and of the same denomination with 45 viz. hours and is the answer to the question demanded shewing that 40 men can do as much in 67,5 hours as 60 men can do in 45 hours Or set 60 on the first to 40 on the second and then against 45 on that second is 67,5 on the first Another way is thus Set 40 the third term on the first to 45 on the second which is the term of contrary denomination to the other two and then right against 60 on the first which is the number of the same denomination with the 40 is 67,5 on the second the number sought Thus one way of working proves the other Example 2. If 45 men do a work in 30 dayes In how many dayes will 270 men do it Set 270 on the first to 45 on the second and then against 30 on the first is 5 on the second Or Set 270 on the first to 30 on the second and then against 45 on the first is 5 on the second And therefore the fourth number sought for is 5 shewing that 270 men will do as much work in 5 dayes as 45 men can do in 30 dayes This Rule is proved by multiplying together the first and second terms and also the third and fourth And if the two Products be equal the work is truly wrought or else not By the lines it is thus proved Set 1 on the first to 30 on the second and then against 45 on the first is 1350 on the second This done Set 1 on the first to 5 on the second and then against 270 on the first is 1650 on the second here both Products being equal declareth the work to be truly wrought PROBLEM VII Of Duplicated Proportion Three numbers being given to find the fourth in a Duplicate Proportion THis Rule chiefly concerns the proportion of Lines to Superficies or of Superficies to Lines 1 Of the Proportion of Lines to Superficies Example 1. If the diameter of a circle be 14 inches and its Content 154 inches What will the Content be of another circle that is 28 inches in diameter Set 14 on the first to 28 on the second they being the terms of one denomination viz. Lines and then against 154 on the first the content of the circle given is 308 on the second This 308 seek on the first and against it on the second is 616 which 616 is the content of that other circle of 28 inches diameter Example 2. Let the diameter of one circle be 7 foot and the Area of it 38,5 foot and let it be demanded What the superficial Area of another circle is whose diameter is 18 foot Because 7 and 18 be terms of one denomination viz. Lines Set 7 the diameter of the the circle whose Content is known on the first to 18 on the second being the diameter of the other circle whose Content is sought and then against 38,5 the Content known on the first is 99 on the second and then seek this 99 on the first and against it is 254,5 tenths on the second which is the superficial Area or Content in feet of that other circle which was demanded Example 3. If a peece of land that is 20 pole square be worth 30 pounds What is a peece of land of the same goodnesse worth that is 35 pole square Set 20 on the first to 35 on the second and then against 30 on the first is 52,5 on the second and lastly against 52,5 on the first is 91,8 on the second that is 91 pounds and eight tenths of a pound or 16 shillings So much is the worth of that piece of land of 35 pole square Example 4. How many acres of land of our English measure of 16,5 foot to the pole are contained in 30 Irish acres of 21 foot to the pole Place 16,5 on the first to 21 on the second and then against 30 on the first is 38,2 on the second and against 38,2 on the first is 48,6 on the second So many English acres are in 30 Irish acres 2 Of the Proportion of Superficies to Lines If the two terms of like denomination be of superficial Contents and a diameter or a line sought for Example Let two circles be given the Content of the one being 154 and its diameter 14 the Area of the other circle is 616 and its diameter is required Set the Area of the circle known viz. 154 on the first to its diameter 14 on the second and then against 616 on the first is 56 on the second The half whereof viz. 28 is the diameter of that other circle whose Content is 616 which 28 is feet or inches or any other measure such as the diameter of the other circle was measured by PROBLEM VIII Of Triplicate Proportion Three numbers being given to find a fourth in a Triplicated Proportion THis Rule concerneth the proportion betwixt Lines and Solids Example 1. There is a Bullet whose diameter is 4 inches weigheth 9 pounds What will another Bullet weigh whose diameter is 8 inches and of the same metal Set 4 on the first to 8 on the second rhat is the one diameter to the other And then against 9 on the first which is the weight of the Bullet of 4 inches diameter is 18 on the second and against 18 on the first is 36 on the second and thirdly against 36 on the first is 72 on the second This third summe found is the fourth proportional number which was
paid 224 pound is 16 which is 16 pounds So much is the yearly rent sought A man borrowed 666 pounds 13 shillings and 4 pence for 2 years and covenanted that he would repay at the 12 years end 1333 pounds 6 shillings and 8 pence It is desired to know after what rate of Interest by the 100 he paid for his money borrowed Set 1333â— on the first the sum of money to be paid to 666â— on the second the summe lent and then against 12 on the second the term of years it was lent is 5,95 which is 5 pounds 19 shillings So much by the 100 doth he pay for the money lent A Sum of money being due at a certain time to come to find what it is worth in present money to take in There is 402 pounds 2 shillings due at the end of 5 years to come I would know what it is worth in ready money abating Interest for the money received in before due after the rate of 8 in the 100. First set 108 on the first to 100 on the second and then against 402,1 on the first is 37,1 on the second Secondly against 37,1 on the first is 34,46 on the second Thirdly against 34,46 on the first is 31,95 on the second Fourthly against 31,95 on the first is 29,71 on the second Fifthly against 29,71 on the first is 373 on the second So much namely 373 pounds may be received in present money for the 402 pounds and 2 shillings due five years hence as being the present worth thereof Here take notice that as the Principal and Interest of money forborn for many years increaseth in a proportion direct So in this case where money is paid many years before due it decreaseth in the like proportion CHAP. III. The Vse of the double Scale of Numbers in Superficial measure as Board Glasse Land and the like PROBLEM I. The length and breadth of any square or long square Superficies being given to find the Content thereof IF the length and breadth be given in inches Then As 1 to the breadth in inches So is the length in inches To the Content in inches Example Let a plain Superficies as a Board or Plank be given to be measured the breadth is found 30 inches and its length 183 and the Content required Set 1 on the first to 30 on the second and then against 183 on the first is 5490 the Content sought in inches If the Superficies given be a piece of land 30 perches broad and 183 long the Content is 5490 perches Again let a piece of Wainscot be 2,5 foot in breadth and 15,25 foot in length the Content will be found 38,12 foot For Set 1 on the first to 2,5 on the second and then against 15,25 on the first is 38,12 foot the Content sought PROBLEM II. The breadth and length of any Superficies being given in one kind of measure to find the Content in another kind of measure LEt the length and breadth be given in inches and the Content required in feet The Rule is thus As 144 to the breadth in inches So is the length in inches To the Content in feet Example Let the breadth be 30 inches and the length 183 inches and the Content in feet required In this case because 144 inches make a foot of superficial measure Set 144 on the first to 30 the breadth on the second and then against 183 the length on that first is 38 foot and a Fraction of a foot being a little more than one tenth part of a foot So many foot are in that Board or what other platform it be that is given to be so measured If the platform were a piece of land 30 perches broad and 183 perches long then the Analogie is thus As 160 the perches making an acre to the breadth in perches So is the length in perches To the Content in acres And therefore in this case of land measure Set 160 on the first to 30 the breadth on the second and then against 183 the length on that first is 34,31 on the second So many acres of land are contained in that ground If the place whose Content is to be cast up be a Triangle a Trapezia or of any other form whatsoever the Analogie in general is this As 144 or 160 c. Is to one of the numbers given to be multiplyed together So is the other of them to the Content in Feet or Acres c. There is a piece of Wainscot that is 3,5 foot broad and 21 foot long How many yards is in it Seeing that in a yard are contained 9 foot Therefore Set 9 on the first to 3,5 on the second and then against 21 on the first is 8,16 on the second So many yards is in that piece PROBLEM III. The breadth of a Superficies being given in one kind of measure and the length in another to find the Content in the greater measure LEt the breadth of a Superficies given be in inches and the length in feet and the Content in feet required Here the Analogie is As 12 to the breadth in inches So the length in feet to the Content in feet So that if the breadth be 30 inches and the length 15,25 foot the Content will be 38,12 foot For Set 12 on the first to 30 on the second and then against 15,25 on the first is 38,12 on the second being the Content sought for Or else Set 12 to 15,25 and then against 30 on that first is 38,12 on the second By this rule also If the breadth of a plot of land be given in perches and the length in chains being measured by a chain of 4 perches long the Content in Acres is readily had Example Let a piece of land be in breadth 30 perches and in length 15,25 chains measured by a chain of 4 perches in length In this case the Analogie is thus As 4 is to the breadth in poles So is the length in chains to the Content in Acres Therefore Set 4 to 30 and then against 15,25 on that first side whereon the 4 is you shall have 11,4 on the second side So much is the Content in Acres of that piece of land Again let a piece of land be 36 poles broad and the length 23 chains and an half to find the Content Set 4 on the first to 36 on the second and then against 23,5 on that first is 21,1 and better on the second the Content sought PROBLEM IV. The length and breadth of a Superficies being given in feet to find the Content in yards TAke this for a general Rule As 9 is to the breadth in feet So is the length in feet To the Content in yards Example Let the breadth of a pane of Wainscot be 4 foot and the length 12 foot and the Content in yards be sought for Here is no more to be done but to set 9 on the first to 4 on the second and then against 12 on the first is 5,35 on the second The Content
so expressed are all Decimal Fractions whose Denominator is a Vnite with so many Cyphers as there be Fraction figures as the Denominator of the Fraction figures of the 75 belonging to the whole number 46 is 100 being a unite with two cyphers because two figures and the Denominator to 5 is 10 being a unite with one cypher because but one figure And when a Fraction is to be expressed alone without a whole number then the Numerator is first expressed and after it the Denominator right on in the line with a Comma betwixt as 75 100 and 5 10 are thus expressed 75,100 5,10 and so of other the like PROBLEM IV. Of Continual Proportionals Two numbers being given to find a third a fourth a fifth or many numbers in continual proportion Geometrical to them two Example LEt the two numbers given be 2 and 4 and it be required to find several numbers in continual Geometrical proportion to them two Set 2 on the first to 4 on the second and then against 4 on that first is 8 on the second which is the third number in continual proportion Geometrical to them two and then against 8 on the first is 16 on the second which 16 is the fourth number in continual proportion to them two and against 16 on the first is 32 on the second the fifth continual proportional and against 32 on the first is 64 on the second the sixth continual proportional and against 64 on the first is 128 on the second the seventh continual proportional and against 128 on the first is 256 on the second which is the eighth proportional to the two proposed numbers Wherefore 8 16 32 64 128 and 256 are a rank of numbers in continual Geometrical proportion to 2 and 4 the thing that was required Example 2. Let it be required to find a rank of numbers in continual proportion as 2 to 3. Here set 2 on the first to 3 on the second and then without moving the Instrument against 3 on the first you have 4,5 on the second and against the same 4,5 found on the first is 6,75 on the second and against 6,75 on the first is 10,125 on the second Therefore 4,5 6,75 and 10,125 are a rank of numbers in continual proportion to 2 and 3 as is required If it be required to find such a rank of proportionals to the numbers 2 and 4 which may bear the same proportion to one another as 2 bears to 4 Set 4 to 2 and then against 2 on that first side whereon the 4 is you have 1 on the second side which is the third proportional to 4 and 2 bearing the same proportion to 2 as 2 doth to 4. And against 1 on the first is 5,10 on the second the fourth proportional and against 5,10 on the first is 25,100 on the second which is the fifth number in continual proportion inverse or backward If the two numbers given be 10 and 9 and a rank of numbers to them in an inverse proportion Geometrical be required set 10 on the first to 9 on the second and then against 9 on the first is 8,1 on the second which is the third proportional and then against 8,1 is 7,29 the fourth proportional wherefore 10,9 8,1 and 7,29 are numbers in a continual proportion But if the numbers given be 1 and 9 and a third and fourth numbers in proportion to them as 9 is to 1 be required then must the numbers found be accounted 81 and 729 they being the third and fourth numbers in a Geometrical proportion to 1 and 9. In like manner if the two numbers given be 10 and 12 then if you set 10 to 12 you shall see on the first against 12 14,4 which is the third proportional and against 14,4 on the first is 17,28 on the second which is the fourth proportional But had the two numbers given been 1 and 12 then bring 1 on the first to 12 on the second and you shall have against 13 on the first 144 on the second for the third proportional and the fourth will be 1728 and so of all other PROBLEM V. Of the Rule of Proportion direct Three numbers being given to find a fourth the Analogie standeth thus AS the first number is to the second So is the third number to a fourth Therefore work thus Set the first number in the proportion on the first side to the second number in the proportion on the second side And then against the third number on the first is the fourth number sought for on the second Example 1. Let the Diameter of a known circle be 7 and its circumference 22 and it be required to know what the circumference of another circle is whose Diameter is 14. To resolve this quaere Set 7 on the first to 22 on the second and then against 14 the Diameter of the other circle found on the first is 44 on the second This 44 is the circumference of that other circle whose Diameter is 14. Example 2. If 45 yards of Stuff cost 30 pound what will 84 yards of it cost Set 45 the first number in the Rule on the first to 30 the second number on the second and then against 84 the third number in the Rule on the first is 56 on the second which 56 is the fourth number and sheweth that 84 yards will cost 56 pounds Or If 45 acres of land be worth 30 pounds a year what will 84 acres be worth by the year The answer is as before 56 pounds And if 26 of any thing give 64 what will 36 of the same give Set 26 to 64 and then against 36 on the first is 88,615 the answer to the question demanded on the second Note that generally in the Rule of direct proportion If the third number be greater than the first then will the fourth number be greater than the second But if the third number be lesse than the first then the fourth number will be lesse than the second Example 3. If the circumference of a circle be 22 and its Diameter 7 what will the Diameter of another circle whose circumference is 44 Here set 22 on the first to 7 on the second and then against 44 on the first is 14 on the second which 14 is the Diameter of that circle whose circumference is 44 be their measures taken in inches feet or any other measure whatsoever To make proof of the work whether truly wrought or not Multiply the first term in the Rule and the fourth term newly found the one by the other and likewise the second and third terms and if the two Products be equal the work is truly wrought or else not To prove the last question by the lines Set 1 on the first to 22 on the second and then against 14 on the first is 308 on the second Next set 1 to 7 and then against 44 on that first is 308 on the second here both Products being equal proves the work to be truly wrought PROBLEM
lengths known it is required to make another Pentagonal Fort whose quantity or Content in land shall be the just fourth part of the proposed Pentagonal Fort propounded where note that the half of the sides and lines of the Fort made will make another Fort that will contain the fourth part of the ground in that Fort made And that you may readily have half of all the lines of the Fort made do thus Set 2 on the first to 1 on the second and then against the number of any line on the first is its half on the second As against 662 the length of the side of the Pentagon given is 331 the length of the Pentagons side to be made The Lines thus remaining without any moving against 119 on the first the length of the Gorge-line in the made Fort is 59,5 the length of the Gorge-line in that Fort to be made And against 424 the length of the Curtain in the made Fort is 212 on the second the length of the Curtain in that Fort to be made And against 456 the length of the proposed Poligons perpendicular which is made is 228 the length of the perpendicular in that Poligon which is to be made And against the Semidiameter 564 is 282 And against 100 the length of the present Bulwarks flank is 50 the length of the Flank in the Bulwark to be made And against 224 the length of the Bulwarks head-line is 112 the length of the head-line in the Fort to be made And a-against 310 the distance of the shoulders GE or MH in the Fort made is 155 for the distance of the shoulders in that Fort which is to made And in like manner against the length of any Line in the proposed Fort is the length of that same Line in the Fort to be made Thus very speedily is found out the true length of all the Lines in and belonging to such a Fort as is required to be made whose lines and sides shall be just half the length of those in the Fort already made And being drawn on the ground and made up will in all parts be like proportional to the same proposed Fort. The like manner of working is to be used for any other proportion whatsoever either greater or lesser Suppose it were required to make a Pentagonal Fort whose sides and so all the lines thereof should be the one third part in length of the sides and lines in that proposed Fort which in the annexed Diagram is represented by the letters CABSTXKOGFEDNMLPH To which demand to give a ready answer do thus Because one third part is sought for Set 3 on the first to 1 on the second and then against 662 on the first the side of the Pentagon given it 220â…” on the second which is one third part of 662 sheweth that the length of the side of the Pentagon to be made is 220â…” yards and against 119 the length of the Gorge-lines in the proposed Fort is 39â…” that is 39 yards 2 foot which is the third part of 119 and is the just length of the Gorge-lines in the Fort to be made In like manner against any Line or distance in th proposed Fort is the length of the same Line in that Fort to be made For to be short Take any number on the first and right against it on the second is the one third part thereof How to make a Fort greater than the Fort proposed SVppose there be a Fort made such an one as is that represented in the afore-mentioned Diagram and another Fort is to be made like to it so that in all the lines and sides thereof it be double to those in the Fort that is already made In this case and the like we have no more to do but to set 1 on the first to 2 on the second and then against the number of the length of any side or line on the first is the double of it on the second As if you set 1 on the first to 2 on the second then against 456 on the first the length of the perpendicular in the Pentagonal Fort given you shall have 912 on the second the double of 465 which is the length of the perpendicular in that Fort to be made and against 662 the length of the Curtain in the Fort made is 1322 the length of the Curtain in the Fort to be made And so of all the rest of the Lines If it shall be required to make such a Fort as in its sides and lines shall be one third part longer than those in the proposed Fort Then take any one line as for example CI 456 yards and thereof take the one third part which is 152 This add to the same 456 and it makes 608. This done Set 456 on the first to 608 on the second and then against any line of the Fort made on the first is its match-line in the Fort to be made As against DN the Curtain in the made Fort 424 yards is 565â— So many yards will the Curtain be in that Fort to be made and so of all other lines And if a Fort be to be made whose sides and lines must be half so much more in length as are the sides and lines of the proposed Fort Then take the half of any number and add it to the same number and so proceed as before Note that the lines in the Type or Diagram annexed may be accounted in their lengths there set down either so many foot or so many yards or so many perches or what other measure you are to use And so the side of the Pentagon 662 may be accounted 662 yards or 662 foot or 662 perches Thus by supposing a Pentagonal Fort to have lines and sides in length as those in the Diagram you may by these double lines very easily discover all the lines and sides of any other Pentagonal Fort either greater or lesser CHAP. VIII The Vse of the double Scale of Sines in Astronomy PROBLEM I. The Suns greatest declination with his place or distance from the next Equinoctial point being given to find his present declination for the time given THe Suns greatest declination in this our age is found to be 23 degrees 30 minutes This is general his place given on the day proposed is 0 degr of Taurus which point is distant from the next Equinoctial point af Aries 30 degrees The Analogie for Solution of this Proposition is thus As the Radius or Sine of 90 degrees Is to the Suns greatest declination 23 deg 30 min. So is the distance of the Sun 30 deg from the next Equinoctial point which is Aries To the Sine of the declination 11 degr 30 min. sought for Wherefore Set 90 degres on the first to the Suns greatest declination 23 deg 30 min. on the second and then against 30 deg the Suns distance from the next Equinoctial point on the first is 11 deg 30 min. on the second So much is his declination sought