the two Figures noted II and III you may plainly see 15. All Angles Plain and Spherical are either Acute Right or Obtuce 17. A Right Angle is alwaâ—â—â— just 90 degrees as you may 〈◊〉 in the Figures II and III by the Angles at A in both of them 18. An Obtuce Angle is alwayes more than 90 degrees as the Angles at D in both Figures shew 19. A Parallel Line is any Line drawn by another Line in such a way that though it were infinitely produced yet they would never meet or cross one another as the Lines AB CD 20. A Perpendiculer Line is when one Line so falleth on another Line that the Angles on each side are equal as CA falls of the Line BA Figure VI. 21. All Triangles are either with three equal sides as Figure IIII or two equal sides as Figure V or all unequal sides as Figure VI the first of which is called Equilateral the second Isosceles the third Scalenum 22. Again they may be sometimes named from their Angles thus Orthigonium with one Right Angle and two Acute Angles Ambligonium with one Obtuce Angle and two Acute Angles Oxigonium with three Acute Angles only 23. The three Angles of every Plain Triangle are equal to two Right Angles 24. All Four-sided Figures are either Squares with four sides and four right Angles all equal or long Squares or Oblongs with the two opposite sides equal or the same crushed together or not Right-Angled as the Rombus and Rombords or else with four unequal Sides called Trapeziaes 25. Lastly many sided-Figures are some Regular having every side alike as 5 6 7 8 9 10 c. Or else unlike as Fields and Woods and Meadows which being infinite cannot be comprehended under any Regular Order or Rule 26. Multiplicator is a term used in Multiplication by which any Number is to be multiplied as is saying 5 times 6 5 is the Multiplicator of 6. 27. Multiplicand is the Number to be multiplied as 6 by 5 as above named 28. The Product is the Issue or Result of two Numbers multiplied one by the other as 30 is the Product of 6 multiplied by 5 for 6 times 5 is 30. 29. Divisor is a term used in Division and is the Number by which another Number is to be divided as to say How many times 5 in 30 5 here is the Divisor 30. Dividend is the Number to be divided as 30 abovesaid 31. Quotient is the Answer to the How many times as in the abovesaid 5 is in 30 6 times 6 then is the Quotient 32. Square is the Product of two Numbers multiplied together as the Square of 6 multiplied by 6 is 36. 33. Square-root of any Number is that number which being multiplied by it self shall have a Product or Square equal to the given Number thus the Square-root of 36 is 6 for 6 multiplied by 6 is 36 equal to the first given Number But if it be a Number that cannot be squared as 72 the content of half a Foot of Board whose near Square-root is 8 4852811 of 10000000 then is the Square-root to be exprest as near as you may or care for as here the Square-root of 72 which is called a Surd Number that will not be squared 34. Cube is a second Product or power of two Numbers increasing or multiplied together as thus the Square of 6 is 36 the first power and the Cube of 6 is 216 that is to say 6 times 36 the second power In Mr. Windgate's Book of Arithmetick is the way of doing it by Numbers or Figures being one of the hardest Lessons in Arithmetick CHAP. III. Certain Geometrical Propositions fit to be known as Preparatory Rudiments for the following Work 1. To draw a Right Line between two Points EXtend a Thred or Hair from one Point to the other and that shall be the Line required But if you use a Rule being the fittest Instrument to try your Rule do thus apply one end to one Point as to A and the other end to the other Point at B and close to the edge draw the Line required then turn the Rule and lay the first end to the last Point yet keeping the same side of the Rule toward the Paper and draw the Line again and if the two Lines appear as one the Rule is streight or else not Note the Figure I. 2. To draw a Line Perpendiculer to another on the middle of a Line On the Point E on the Line AB I would raise the Perpendiculer Line CE set one point of the Compasses in E and open them to any distance as EB and EA and note the Points A and B then open the Compasses wider and setting one Point in A make the part of the Arch by C upwards and if you have room do the like downwards near D Then the Compasses not stirring set one Point in B and with the other cross the former Arks near C D a Rule laid and a Line drawn by those two crossings shall cut the Line AB perpendiculerly just in the Point E which was required 3. To let fall a Perpendiculer from a Point to a Line But if the Point C had been given from whence to let fall the Perpendiculer to the Line AB do thus First set one Point of Compasses in C open the other to any distance as suppose to A and B and then if you have room upon A and B strike both the Arks by D which finds the Point D if not the middle between A and B give the Point E by which to draw CED the Perpendiculer from C desired Note Figure 2. Note That if you can come to find the Point D by the crossing it doth readily and exactly divide the Line AB in two equal parts by the Point E. 4. To raise a Perpendiculer on the end of a Line On the end of the Line AB at B I would raise a Perpendiculer First set one Point of the Compasses in B open them to any distance as suppose to C and set the other Point any where about the middle between D and E as suppose at C then keep that Point fixed there turn the other till it cut the Line as at D and keep both Points fixed there and lay a streight Rule close to both Points and there keep it then keep the middle-Point still fixed at C and turn the other neatly close to the other end and edge of the Rule to find the Point E then a Rule laid to the Points E and B shall draw the Perpendiculer required Or else when you have set the Compasses in the Point C prick the Point D in the Line and make the touch of an Ark near to E then a Rule laid to DC cuts the Ark last made at or by E in the Point E There are other wayes but none better than this Note the Figure 3. 5. From a Point given to let fall a Perpendiculer to the end of a Line being the converse of the former First from
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-given-Line viz. B with this distance set one foot in A at the other end of the given-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
like to the first Superfecies given First find a mean proportion between the unequal sides of the given Superfecies that you are to make one like and find the mean proportion also between the unequal sides of the Figure that you are to make one equal to As thus for Example I have a Romboides whose base is 5 and perpendiculer is 3 and side is 3-55 the mean proportion between is 3-866 Also I have a Triangle whose half-base is 8 and the perpendiculer 4 the mean proportional is 5-6552 and I would make another Romboides as big as the Triangle given whos 's Area is 32 Then by the Line of Numbers say As 3-866 the one mean proportion is to 5-6552 the other mean proportion so is the Sides of the Romboides whose like I am to make to the sides and perpendiculer of the Romboides required to make a Romboides equal to a Triangle given and like to another Romboides first given As thus for Example As 3-866 is to 5-6552 so is 5 the base of the Romboides given to 7-30 the base of the Romboides required And so is 3 the given perpendiculer of the Romboides to 4-38 the perpendiculer of the Romboides required So also is 3-55 the side of the Romboides given to 5-19 the side of the Romboides required for if you multiply 7-30 the base thus found by 4-38 the perpendiculer now found it will make a Romboides whose Area is equal to 32 the Area of the Triangle that I was to make the Romboides equal to and making the side to be 3-55 it will be like the first Romboides propounded If it had been a Trapesia or other formed Figure it might have been resolved into Triangles and then brought into Squares as before Then all them Squares added into one sum whose Square-root is the mean proportional or side of a Square equal to that many-sided Figure whose like or equal is desired to be made and produced 5. One Diameter and Content of a Circle given to find the Content of another Circle by having the Diameter thereof only given The Extent from one Diameter to the other being twice repeated the right-way from the given Area shall reach to the Area required If the Area's of two Circles be given and the Diameter required then the half-distance on the Numbers between the two Area's shall reach from the one Diameter to the other Sect. VIII To find the Square-root of a Number To do this by the Line of Numbers you must first consider whether the Figures whereby the Number whose Root you would have is expressed be even or odd figures that is consist of 2 4 6 8 or 10 or 1 3 5 7 or 9 figures For if it be of even figures then you must count the 10 at the end for the unite and the Root and Square are backwards toward 1. But if it consist of odd Figures then the 1 in the middle of the Line is the unite and the Root and Square is forwards towards 10 for the Square-root of any Number is alwayes the mean proportional or middle space between 1 and the Number propounded counting the unite according to the Rule abovesaid So that the Square-Root of 1728 consisting of four figures it is at 41 and â— â— counting 10 for the unite for the Number 42 â— â—â— is just in the middest between 1728 and 10. And to find the Square-root of 144 consisting of three figures divide the space between the middle 1 and 144 counted forwards into two equal parts and the Point shall rest at 12 the Square-root required To do this by the Line of Lines or Sector First find out a Number that may part the Number given evenly or as even as may be then the Divisor shall be one extream and the Quotient another extream the mean proportional between which two shall be the Square-root required working by the last Rule Example To find the Square-root of 144. If you divide 144 by 9 you shall find 16 in the Quotient Now a mean proportion between 9 the Divisor and 16 the Quotient is 12 the Root found by the last Rule viz. the 7th Sect. IX To find the Cubick-Root of a Number The Cubick-root of a Number is alwayes the first of two mean proportionals between 1 and the Number given counting the unite with the following cautions Set the Number down and put a Point under the 1st the 4th the 7th and the 10th figure and look how many Points you have so many figures shall you have in the Root Then if the last Point fall on the last Figure then the middle 1 must be the unite and the Root the Square and Cube will fall forwards toward 10. But if the last Point fall on the last but one then the unite may be placed at either end viz. at 1 at the beginning or at 10 at the end and then the Cube will be one Radius beyond the unite either forwards or backwards But if it fall on the last but two then 10 at the end of the Line must be the unite and the Root the Square and Cube will alwayes be in the same Radius that is between 10 at the end and the middle 1. So that by these Rules the Cubick-root of 8 is 2 for puttiâ—g a Point under 8 being but one figure it hath but one Point therefore but one figure in the Root Secondly the Point being under the last figure the middle 1 is the unite then dividing the space between 1 and 8 into three equal parts the first part ends at 2 the Root required So likewise in 1331 there is two Points therefore two figures in the Root and the last Point being under the last Figure the middle 1 is the unite and the space between 1 and 1331 being divided into three equal parts the first part doth end at 11 the Cubick-root of 1331. Again for 64 there is one Point and it falls on the last figure but one therefore the Root contains but one figure and 1 at the beginning or 10 at the end which you please may be the unite But yet with this Caution That the Cube must be in the next Radius beyond that which belongs to the unite so that dividing the space between 10 and 64 beyond the middle 1 towards the beginning into three equal parts the first part falls on 4 the Cubick-root required Or if you divide the space between 1 and 64 near the 10 into three equal parts the first part falls on 4 also Again for 729 there is but one Point therefore but one figure Again it falls on the last but 2 therefore 10 at the end is the unite and between 10 and the middle 1 backwards you shall have both Root Square and Cube for the Number required which will be at 9 For if you divide the space between 10 and 729 into three equal parts the first part will stay at 9 the Cubick-root required Note if it be a surd Number that cannot be cubed exactly yet the
the Sphear inscribed and to the Circle circumscribing being measured by Compasses Callipers and Line of Lines very carefully and exactly Then divide the Circumference of the two Ends of the Cillender into 10 equal parts and draw Lines Perpendiculer from end to end and plain all away between the Lines flat and smooth so that the two Plains on both ends will become a regular ten-sided Figure Then making the whole Diameter abovesaid 10000 in the Line of Lines take out 0-309 and with this measure as a Radius on the Center at both ends describe a Circle and if you draw Lines from every opposite Line of the 10 first drawn you shall have Points in the last described circle to draw a Pentagon by which is the Base of one of the 12 Pentagonal Pyramids contained in the body This Work is to be done at both Ends but be sure that the Angle of the Pentagon at one end be opposite to a side of the Pentagon at the other end then these Lines drawn the two ends are fully marked Then to mark the 10 Sides do thus Count the first length 1000 viz. the measure from the top to the bottom or from Center to Center and fit this length in 10 and 10 of the Line of Lines the Sector so set take out 0-3821 and lay it from the two ends and either draw or gage Lines round about from each end and in the midst between the two Lines will remain 0-2358 then Lines drawn Diagonally on the 10 sides will guide to the true cutting of the Dodecahedron If you set a Bevel to 116 deg 33 min. 54 sec. and apply it from the two ends you may try the truth of your Work The Declination and Reclination of all the 10 Pantagonal Plains are as followeth First You have 1 North reclining 26 deg 34 min and 1 South inclining as much Secondly You have 2 North declining 72 and reclining 26 34 and 2 South declining 72 and inclining 26 34. Thirdly You have 2 North declining 36 and inclining 26 34 and 2 South declining 36 and reclining 26 34 And 1 Horizontal Plain and his opposite Base to stand on As for the measuring of this Body the Plain and Natural way is thus First find the Superficial Content of the Base of one of the Pentagons by multiplying the measure from the Center to the middle of one of the Sides which is the contained Circles semi-diameter and half the sum of the measure of all the sides put together and then to multiply this Product by one third part of half the Altitude of the body and the Product shall be the Content of one Pentagonal Pyramid being one twelft part of the Dodecahedron and this last multiplied by 12 gives the solid Content of the Dodecahedron or 12 times the Superficial Content of one side is the Superficial Content thereof Example Suppose the side of a Dodecahedron be 6 then the sum of the sides measured is 30 the contained Circles semi-diameter is 4-12 then 15 the half of 30 and 4-12 multiplied together make 61-80 and 12 times this makes 741-60 for the Superficial Content of the Dodecahedron Then for the Solid Content multiply 61-80 the Superficial Content of one side by 2-233 one 6th part of 13-392 the whole Altitude of the body the Product is 137-99940 Again this multiplied by 12 the number of Pyramids makes 16â—â— 9928 the Solid Content as near as may be in such a Decimal way of Computation 5. For the Icosahedron Which is a regular solid body made up of or contained under 20 Trianguler Pyramids whose Base or one of whose Sides is an equilatteral Triangle and the perpendiculer Altitude of one of these 20 Pyramids is equal to half the perpendiculer Altitude of the Icosahedron from any one side to his opposite side or equal to the semi-diameter of the inscribed Sphear To cut this body take any round Piece and if the Diameter thereof be 10000 let the length thereof be turned flat and even to 8075 or if the true Round and Cillendrical Form in Diameter be 4910 let the true length when the ends are plain and flat be 3964 then divide the Cillendrical part into 6 equal parts and plain away all to the Lines so that the two ends may be two 6-sided-figures then making 5000 the former semi-diameter 1000 in the Line of Lines take out 616 and on the Center at each end describe a Circle and by drawing Lines to each opposite Point make a Triangle whose circumscribing Circle may be the Circle drawn at each end but be sure to mark the side of one Triangle opposite to the Point of the other Triangle at the other end as before in the Dodecahedron thus both the ends shall be fully and truly marked Then making the length a Parallel in 1000 of the Line of Lines take out -379 and -095 and prick those two measures from each end and by those Points draw or gage Lines round about on the 6 sides Then Diagonal Lines drawn from Point to Line and from Line to Point round about shews how to cut the Body at 12 cuts Note That if you set a Bevel to 138-11-23 and apply it from each end it will guide you in the true plaining of the sides of the Icosahedron And a Bevel set to 100 degrees will fit being applied from the midst of one side to the meeting of two sides The Reclination of the three Triangles whose upper sides are adjacent or next to the three sides of the upper Horizontal-Triangle is 48 11.23 from the Perpendiculer or 41 48 37 from the Horizontal and when one corner stands South the Declination of one of these 3 viz. that opposite to the South-corner a direct North th' other two decline 60 degrees one South-east the other South-west the other 6 about the corners of the Horizontal-plain do all recline 19 deg 28 min. 16 sec. the two that behold the South decline 22 deg 14 min. 29 sec. and those two that behold the North decline 37 deg 45 min. 51 sec. toward the East and West the other two remaining recline as before and decline one North-east and the other North-west 82 deg 14 min. 19 sec. The other Nine under-Plains opposite to every one of these decline and incline as much as the opposite did recline and decline as by due consideration will plainly appear For the measuring of this body do as you did by the Dodecahedron find the Area of one Triangle and multiply it by 20 gives the Superficial Content and the Area of one Triangle multiplied by one sixt part of the Altitude of the body gives the solid Content of the Trianguler Pyramid and that Product multiplied by 20 the number of Pyramids gives the whole Solid Content of the Icosahedron Example Suppose the side of an Icosahedron be 12 first square one side viz. 12 which makes 144 then multiply that Square by 13 and then divide the Product by 30 the Quotient and his remainder is the Superficial
184 22  0 2 10 0 2 57 0 321  1 0 3 10 0 3 80 0 475   0 4 33 0 5 30 0 663   0 6 00 0 7 35 0 920 21  0 7 60 1 1 29 1 161  2 1 1 80 1 4 00 1 500   1 3 90 1 6 56 1 821   1 6 10 2 1 22 2 153 20  2 0 66 2 4 34 2 543  3 2 3 50 2 7 98 2 998   2 6 16 3 3 10 3 388   3 0 70 3 6 20 3 772 19  3 3 80 4 2 00 4 250  4 3 6 50 4 5 30 4 663   4 1 80 5 1 35 5 169   4 5 25 5 5 60 5 700 18  5 0 42 6 1 45 6 182  5 5 3 90 6 5 70 6 713  5 7 20 7 1 70 7 213   6 2 80 7 6 20 7 777 17  6 6 50 8 2 65 8 333  6 7 2 20 8 7 20 8 900   7 5 50 9 3 20 9 400   8 1 10 9 7 70 9 960 16  8 4 80 10 4 20 10 525  7 9 0 70 11 1 00 11 125   9 4 50 11 5 40 11 806   10 0 40 12 2 20 12 275 15  10 4 30 12 7 00 12 876  8 11 0 50 13 4 10 13 513   11 4 30 14 0 80 14 110   12 0 30 14 5 80 14 725 14  12 4 29 15 2 80 15 350  9 13 0 30 15 7 70 15 926   13 4 30 16 4 60 16 577   14 0 40 17 1 60 17 200 13  14 4 60 17 6 60 17 827  10 15 0 50 18 3 40 18 425   15 4 48 19 0 30 19 037   16 0 80 19 6 50 19 815 12  16 5 50 20 3 25 20 40◠ 11 17 2 20 21 1 00 21 225  17 7 90 22 0 00 21 000   18 5 49 22 6 98 22 644 11  19 2 00 23 4 31 23 391  12 19 6 16 24 1 48 24 184   20 3 00 24 7 30 24 961   20 7 40 25 4 60 25 575 10  21 3 10 26 1 36 26 170  13 21 7 40 26 6 38 26 799   22 3 00 27 3 36 27 130   22 7 00 28 2 18 28 174 9  23 3 00 28 5 18 28 648  14 23 7 30 29 2 19 29 275   24 3 70 29 7 40 29 926   24 7 40 30 3 90 30 488 8  25 3 60 31 1 00 31 125  15 25 7 50 31 5 80 31 726   26 3 30 32 2 60 32 325   26 7 00 32 7 00 32 875 7  27 3 00 33 3 80 33 475  16 27 6 40 34 0 30 34 037   28 2 20 34 4 80 34 600   28 5 80 35 0 80 35 100 6  29 1 40 35 5 34 35 668  17 29 4 80 36 1 80 36 225  30 0 40 36 6 29 36 788   30 4 10 37 2 29 37 287 5  30 7 50 37 6 54 37 820  18 31 3 00 38 2 39 38 299   31 6 10 38 6 64 38 833   32 1 80 39 2 70 39 338 4  32 5 00 39 6 00 39 752  19 32 7 80 40 1 80 40 225   33 2 10 40 4 90 40 614   33 4 80 41 0 10 41 012 3  33 7 40 41 3 65 41 457  20 34 2 00 41 6 77 41 848   34 4 30 42 1 43 42 180   34 6 20 42 4 00 42 500 2  35 0 10 42 6 70 42 840  21 35 2 00 43 0 64 43 055   35 3 60 43 2 70 43 338   35 4 80 43 4 19 43 524 1  35 6 00 43 5 42 43 678  22 35 6 80 43 6 51 43 816   35 7 40 43 7 50 43 938   36 0 00 44 0 00 44 000  CHAP. XIII The use of the Line of Numbers in Questions of Interest and Annuities Problem I. At any rate of Interest per annum for a hundred pounds to find what the Interest of any greater or lesser sum comes to in one year EXtend the Compasses from 100 to the increase of 100 l. in one year the same Extent shall reach from the sum propounded to its increase for one year at that rate propounded Example What is the increase or profit of 124 l. 10 s. for one year at 6 per cent per annum The Extent of the Compasses from 100 to 6 being laid the same way from 124 l. 10 s. which is at 124-5 shall reach to 7-47 which is 7 l. 9 s. 4 d. the profit of 124· 10 s in one year Problem II. Any sum of Money and the rate of Interest propounded to find what it will increase to at any number of years counting Interest upon Interest The Extent of the Compasses from 100 to the increase of 100 being turned as many times from the sum propounded the same way as there be years propounded shall at last stay at the Principal and Interest required Example To what sum shall 143 pounds 10 shillings amount to in 10 years counting Interest upon Interest at 6 per cent The Extent of the Compasses from 100 to 106 being turned 10 times from 143 ½ shall reach to 257 l. 0 s. the sum of Principal and Interest at 10 years end Note That in doing this you ought to be very precise in taking the first Extent from 100 to 106 but to cure the uncertainty thereof you have this very good remedy If you have a Diagonal Scale equal to the Radius of the Line of Numbers then use that if not use the Line of Lines on the Sector-side which should be made fit to or the double or the half of the Radius of the Line of Numbers As thus Take the Extent from the Line of Numbers between 100 and 106 this Extent measured on the Line of Lines will be 0253058 could you see so many Figures but 02531 will serve your turn very well which Number you must note is the Logarithm of 106 neglecting the Caracteristick then this Number multiplied by 10 the Number of years is 25310 this Extent taken from the Center on the Line of Lines and laid increasing from 143 ½ shall reach to 257 l. 0 s. 0 d. the true Number of the Use and Principal of 143 l. 10 s. put out or forborn for ten years Problem III. A sum of Money being due at any time to come to find what it is worth in ready Money to be paid presently at any rate propounded This Problem is the contrary to the last
stor'd As first with Gunters Sector and his Quadrant eke also By Foster altred after and with Gunters Rule and Bow The Traviss Quadrant and Cross-staves the Davis Quadrant too Their uses all to more than halfs this Instrument will do With this advantage more beside of lying in less room A fault that Saylors must abide when they on Ship-board come In the next place the Rudiments of Geometry exact The right Sines â—heir complements and how they lie compact Within a Circle and the rest the Chords and versed Sines About a Circle are exprest the Tangents Secants Lines And how their use and place is seen in Round and Plain Triangles Which serve to deck Urania Queen as Iewels Beads and Spangles In the next place Arithmetick by Numbers and by Lines In wayes that won't be far to seek by them that use their times Because the Precepts are explain'd by things of frequent use That for the most part are contain'd in City Town or House As Land and Timber Boards Stones Roofs Chimneys Walls and Floor Computed and reducâ—d at once in Thickness Less or More The cutting Platoe's Bodies five which are not yet made six And them the best way to contrive and Dials on them fix Their Measure and their Magnitude in Circle circumscribed Whose Properties by old Euclide and Diggs have been described Then also in Astronomy are many Propositions Which fitly to th' Rule I apply avoiding repetitions And after in the pleasant Art of Shadows I do wander To draw Hour-lines in every part both upright over and under And all the usual Ornaments that on Sun-Dials be Which are describ'd to the intent Sol's travels for to see As first his Place and Altitude his Azimuth likewise His Right Ascention Amplitude and how soon he doth Rise The same also to Moon and Stars is moderately appli'd Whereby the time of Night appears the Moons Age and the Tide Then Heights and Distances to take at one or at two Stations Performed by those wayes that make the fewest Operations And also ready Rules to use the Logarithmal Table Which may prove ready Hints to these that are in those most able And many other useful Thing is scattered here and there Which formerly by Me hath been accounted very rare And lastly for the Saylors sake I have spent many an Hour Th' Trianguler-Quadrant for to make more useful than all other Sea Instruments that they do use at Sea for Observation And sure I am it won't abuse them in their Operation As in the following Discourse to them that willing be It will appear with easie force if they have eyes to see The Method and the Manner us'd as neer as I was able To follow the old Wayes still us'd and counted warrantable And in this having done my best 〈…〉 up my male Ascribing to my self the least would have the Truth prevail And give the honour and the praise to him that hath us made Of willing minds his Fame to raise by his assisting aid To whom be honour now and eke henceforth for evermore Ascribed by all them that seek the Truth for to adore J. B. ERRATA PAge 28. line 8. for Rombords read Romboides P. 73. l. last f. 337 r. 247. p. 75. l. 1. f. 7. r. 8. p 87. l. 14. r. multiplied by p. 89. l. 14. f. 5 371616. r. 538.1616 l. 21. f. 537 r. 538. p. 90. l. 4. f. 537 r. 538. l. 5. add being better done with a parallel answer p. 100. l. 2 add the Thred p. 128. l 2. dele 10 min. p. 133. l. 6. f. 60 r. 16. p. 143. l. 10 11. f. from 12 to 7 r. from 7 to 12. p. 146. l. 22. f 12 Section r. 13 Section p. 158 l. last dele and. p. 160. l. 11. f. 72 r. 720 also in line 15 23. p. 164. l. 19. f. Diameter r. Area p. 165. l. last add to 707. p. 184. l. 10 f. foot r. brick l. 20. f. ½ r. 1 ½ p. 187. l. 17. f. Ceiling r. Tileing p. 201. l. 11. f. 52 Links r. 55 Links l. 12. f. 48 Acres r. 4 Acres 3 Roods 8000 Links p 102. l. 5. f. 21 Acres 42 Links r. 2 Acres 0 Roods but 14760 Links read so likewise in l. 11. of the same page p. 204. l. 1. f. 16 ½ r. 18 ½ p. 205. l. 8. f. 55 r. 50. r. 50 f. 55 in l. 21 22. p. 206. l. 19. f 4-50 r. 4-50000 l. 21. f. 1 Chain 25 r. 11 Chains 23. p. 229. l. 16. f. 8-10 th r. 8-100 p. 231. l. 15. f of r. at p. 234. l. 22. f. 1 of a foot r. 1.10 th of a foot p. 236 the 3 lines over 134-5 are to come in after 134-5 Also the two lines over 3-545 should come in after 3-545 p. 257. l. 13. f. â—496 r. 249-6 p. 370. l. 3. f. sine r. Co-sine p. 383. l. 22. add by the general Scale p. 384. l. 14. f. = S. ☉ r. = Co-sine p. 414. l. 11. f. or r. on p. 420. l. 22. f. 71 r. 31. p. 429. l. 15. f. Declination r. Suns Right Ascention The Description and some Uses of the Triangular Quadrant or the Sector made a Quadrant being an excellent Instrument for Observations and Operations at Land or Sea performing all the Uses of the Fore-staff Davis-Quadrant Gunter's-Bow Gunter's-Cross-staff Gunter's-Quadrant and Sector with far more conveniency and as much exactness as any or all of them will do The Description thereof 1. FIrst it is a joynted Rule or Sector made to what Length or Radius you please as to 6 9 12 18 24 30 or 36 inches Length when it is folded or shut together the shorter of which Lengths is big enough for Land uses or Paper draughts the four last for Sea uses or Observations To which is added a third Piece of the same length of the Sector with a Tennon at each end to fit into two Mortice-holes at the two ends of the inside of the Sector to make it an Aequilateral Triangle from which shape and its use it is properly called a Triangular Quadrant 2. Secondly as to the Lines graduated thereon they may be more or less as your use of them and as the cost you will bestow shall please to command But to make it compleat for the promised Premises these that follow are necessary to be inscribed thereon as in the Figure thereof And first you are in order hereunto to consider The outer-edges of the Sector or Instrument the inner-edges the Quadrantal-side the Sector-side and the third or loose-piece also the fixed or Head-leg the moving-leg the head and the end of each leg also the head and leg center of which more in its proper place 1. And first on the outer-edge is placed the Lines of Artificial Numbers Tangents Sines and versed Sines to as large a Radius as the Instrument will bear 2. Secondly on the in-side or edge on short Rules is placed inches foot measure the line of 112
or such-like But on larger Instruments a Meridian line to one inch or half an inch more or less for one degree of the Aequinoctial for the drawing of Charts according to Mercator or any other more useful Line you shall appoint for your particular purpose 3. Thirdly in decribing the Lines on the two sides first I shall speak to the Sector-side where the middle Lines all meet at the Center at the head where the Joynt is the order of which went the head or joynted end lyeth toward your left hand the Sector being shut and the Sector-side upermost is thus 1. The first pair of Lines and lying next to you is the Line of Sines and Line of Lines noted at the end with S and L for Sines and Lines the middle Line between them that runs up to the Center and wherein the Brass center pricks be is common both to the Sines and Lines in all Parallel uses or entrances 2. The Line next these and counting from you is the Line of Secants beginning at the middle of the Rule and proceeding to 60 at the end and noted also with Se for Secants one of which marginal Lines continued would run to the center as the other did 3. The next Lines forward and next the inner-edge on the moving-leg are the Lines of Tangents the first of which and next to you is the Tangent of 45 being the largest Radius as to the length of the Rule the other is another Tangent to one fourth part of the length of the other and proceeds to 76 degrees a little beyond the other 45 the middle Line of these also is common to both in which the Center pricks must be At the end of these Lines is usually set T. T. for Tangents 4. On the other Leg of the Sector are the same Lines again in the same order counting from you wherein you may note That as the Lines of Sines and Lines on one Leg are next the outward-edge on the other Leg they are next the inward-edge so that at every or any Angle whatsoever the Sector stands at you have Lines Sines and Tangents to the same Radius and the Secants to just half the Radius and consequently to the same Radius by turning the Compasses twice Also any Tangent to the greater Radius above 45 and under 76 by turning the Compasses four times as afterwards will more appear Which contrivance is of excellent convenience to avoid trouble and save time and happily made use of in this contrary manner to the former wayes of ordering them 5. Fifthly without or beyond yet next to the greater Line of Tangents on the head-leg is placed the first 45 degrees of the lesser Tangents which begin from the Center at 45 degrees because of the straitness of the room next the Center where they meet in a Point yet this is almost of as good use as if it had gone quite to the Center by taking any parallel Tangent from the middle or common Line on the great Tangents right against the requisite Number counted on the small Tangent under 45. 6. Sixthly next to this will not be amiss to adde a Line of Sines to the same Radius of the small Tangent last mentioned and figured both wayes for Sine and co-Sine or sometimes versed Sines 7. Seventhly next to this a Line of Equal Parts and Chords and the Secants in a pricked line beyond the little Tangent of 45 all to one Radius To which if you please may be added Mr. Fosters Line Soll and his Line of Latitudes but these at pleasure 8. Eighthly on the outermost-part of both Legs next the out-side in Rules of half an inch thick and under is set the Line of Artificial versed Sines laid next to the Line of Artificial Sines on the outer-edge but if the Rule be thick enough to bear four Lines then in this place may be set the Meridian Line according to Mr. Gunter counting the Line of Lines as a Scale of Equal Parts Thus much as for the Sector-side of the Instrument 4. Fourthly The last side to be described is the Quadrantal-side of the Instrument wherein it chiefly is new Therefore I shall be as plain as I can herein To that purpose I shall in the description thereof imagine the loose piece or third piece to be put into the two Mortise-holes which position makes it in form of an Aequilateral Triangle according to the Figure annexed noted with ABCD where in AB is for brevity and plainness sake called the Moveable-leg DB the Head or Fixed-leg DA the loose-piece B the Head A and D the ends C the Leg-center at the beginning of the general Scale the center at B the head-center used only in large Instruments and when you please on any oother For the Lines graduated on this side First On the outer-edge of the moveable-Leg and loose-piece is graduated the 180 degrees of a semi-circle C being the center thereof And these degrees are numbred from 060 on the loose-piece toward both ends with 10 20 30 40 c. and about on the moveable-leg with 20 30 40 50 60 70 80 and 90 at the head Also it is numbred from 600 on the moveable-leg with 10 20 toward the head and the other way with 10 20 30 40 50 60 on the loose-piece and sometimes also from the Head along the Moveable-leg with 10 20 30 c. to â—0 on the loose-piece and the like also from the end of the Head-leg and sometimes from 60 on the loose-piece both wayes as your use and occasion shall require Secondly On the Quadrantal-side of the loose-piece but next the inward-edge is graduated 60 degrees or the Tangent of twice 30 degrees whose center is the center-hole or Pin at B on the Head or Joynt of the Sector Which degrees are numbred three wayes viz. First from D to A for forward Observations and from the middle at 30 to A the end of the Moving-leg with 10 20 30 and again from D the end of the Head-leg to A with 40 50 60 70 80 90 for Observations with Thred and Plummet Thirdly Next to these degrees on the Moving-leg is the Line of the Suns right Ascention numbred from 600 on the degrees with 1 2 3 4 5 6 toward the Head and then back again with 7 8 9 10 11 12 c. 1 2 3 4 5 on the other side of the Line as the Figure annexed sheweth The divisions on this Line is for the most part whole degrees or every four minutes of time Fourthly Next above this is the Line of the Suns place in the Zodiack noted with ♈ ♉ ♊ ♋ toward the Head then back again with ♌ ♠♎ over 600 in the degrees and 12 and 24 in the Line of the Suns right Ascentions then toward the end with ♠♠♑ then back again with â™’ and ♓ being the Characters of the 12 Signes of the Zodiack wherein you have exprest every whole degree as the number of them do shew
there being 30 degrees in one Sign Fiftly Next above this is a Kalender of Months and Dayes every single Day being exprest and three or more Letters of the name of every Month being set in the Month and also at the beginning of each Month and every 10th day noted with a Prick on the top of the Line representing it as is usual in such work Sixtly Next over the Months is the Line to find the Hour and Azimuth in a particular Latitude Put alwayes on smaller Instruments and very rarely on large Triangular Quadrants for Sea Observations the lowest Margent whereof and next the Months is numbred from the end toward the Head with 10 20 30 40 50 60 70 80 90 100 110 120 130 near the Head Center For the Semi-diurnal Ark of the Suns Azimuth and in the Margent next above this with 4 5 6 7 8 9 10 11 12 near the end for the Morning hours then the other way viz. toward the Head on the other-side the Hour Line with 1 2 3 4 5 6 7 8 for the Afternoon hours Seventhly On the same Quadrantal-side and Moveable-leg on the spare places beyond the Months toward the end is set an Almanack and the Names of 12 or more Stars to find the hour of the Night which 12 Stars are noted with 1 2 3 4 5 6 7 8 9 10 11 12. among the degrees in small Figures as in the Figure Eightly Next of all to the in-side is the Line of Natural versed Sines drawn to the Center with his correspondent Line on the other or Head-leg Exprest sometimes in a pricked Line for want of room Ninthly On the Head-leg and next to the versed Sines last mentioned is first the Line of Equal Parts or Line of Lines and on the same common Line wherein is the Center is the Line of Natural Sines whose length is equal to the measure from the center at C to 600 on the moveable-leg so that the Line of degrees is a Tangent and the measure from C to any Tangent a Secant to the same Radius of the Natural Lines of Sines and Lines Also beyond the Center C on the same common middle Line is another smaller Line of Natural Sines whose length is equal to the measure from C to 60 on the loose-piece then if you count from the Center pin at 60 on the loose-piece toward the end of the movable-leg they shall be Tangents to the same Radius and the measure from the Center C to those Tangents shall be Secants to the same Radius which may be well to be ordered to a third or fourth part of the former from the Center downwards These two Lines of Sines are best figur'd with their Sines and Cosines the other way with a smaller figure and the Line of Lines from the Center downward from 1 to 10 where 90 is which Lines of Sines may be called a general Scale for all Latitudes Tenthly Next to this toward the outer-edge is another Line of Natural Sines fitted to the particular Line of Hour and Azimuths for one particular Latitude noted Pert. Scale of Altitudes or Sines Eleventhly Next to this is the Line of 29 ½ for so many dayes of the Moon 's age in short Rules of the whole length but in longer not being easily known by the single strokes and Figures annexed to those strokes Twelfthly Next the outer-edge is a Line of 24 hours 360 degrees or 12 Signs or in most Rules inches also used together with the former Line of 29 ½ and as a Theory of the Sun and Moon and ready way of finding the Hour by the Moon or fixed Stars Thirteenthly To this Instrument also belongs a Thred and Plummet and Sights as to other Quadrants and a pair of Compasses as to other Sectors a Staff and Ball socket also if you will be curious and accurate And for large Instruments for Sea a Square and an Index which makes it a perfect sinical Quadrant and two sliding sights also which makes it a fore and back-staff and bow as will appear more at large afterward Some Uses of the Trianguler Quadrant for Land and Sea Observations and Operations CHAP. I. Numeration on the Lines graduated on the Instrument IN the first place it will not be amiss to hint a few words as to the reading the Lines or more properly Numeration on the Lines wherein take notice That all Lines of Equal Parts or Lines applicable to Arithmetick as the Line of Lines the Line of Numbers the Line of Foot-measure and the like wherein Fractions of Numbers are requisite they are most commonly accounted in a Decimal way and as much as may be the small divisions are numbred and counted accordingly But in the Lines of Sines Tangents Secants and Chords being Lines belonging properly to a Circle in regard that the Sexagenary Fraction is still in use the intermediate Divisions are as much as may be fitted to that way of account viz. by whole degrees where they come close together or the Line of no great use And if more room is to half degrees or 30 minuts and sometimes to quarters of degrees or 15 minuts but toward the beginning of the Line of Natural Sines or the end of the Natural Tangents and Secants where the degrees are largest they are divided to every 10th minute in all large Rules as by considering and accounting you may plainly perceive Take two or three Examples of each kind 1. First On the Line of Lines to find the Point that represents 15. In the doing of this or any the like you must consider your whole Scale Radius or length of the Line may be accounted as 1 as 10 as 100 as 1000 or as 10000 and no further can be applicable to any ordinary Instrument Wherein observe That if the whole Line be one then the long stroke by every Figure doth represent one tenth of that Integer and the next shorter without Figures are hundredth parts of that one Integer and a 1000th part is estimated in smaller Instruments and sometimes exprest in larger But the hundredth thousand part is alwayes to be estimated by the eye in all Instruments whatsoever 2. But if the whole Line of Lines shall represent 10 as it usually doth and as it is figured then the long stroke at every Figure is 1 and the next longer are tenths and the shortest are hundred parts and the thousand parts as near as can be estimated 3. But if the whole Line represents a hundred as here in our present Example then the long stroke by every Figure represents 10 and every shorter stroke is one and the shortest strokes are tenths and the hundredth parts as much as can be estimated 4. But if the whole Line shall represent a 1000 then the long stroke by the Figure shall represent a hundred and every shorter 10 and every of the shortest strokes is one Integer and a 10th part as near as can be estimated 5. But lastly if the whole Line represent 10000 then
every long stroke is 100 and every shortest cut is ten and every single Integer is as near as can be estimated by any ordinary Instrument Now our present Example will properly come under the third Rule by conceiving the whole Line to represent 100 then the first long stroke by 1 is 10 then the next shorter is for 11 the next 12 c. to 15 which is cut up a little above the Line for the more ready reckoning without telling the parts which 15 is the Point required to be found Example the second to find out 1550 on the Line This will come under the Notion of the 5th Rule wherein the whole Line is conceived to represent 10000 then the first 1 is for the 1 thousand then the fifth longer stroke next is for the 500 and lastly the middle between the 500 stroke and the 6000 stroke is for the 50 being a little beyond the Point for 15 in the first Example A third Example of 5025. This third Example may suâ—fice for this work being so plain after a little due consideration For first the whole Line is conceived to represent 10000 then the long stroke by 5 is for 5000 then there is no hundreds therefore the Point required must be short of the next longer stroke which signifies hundreds and being it is just 25 which is ¼ of an hundred the true Point readily sheweth it self If you require a more plainer and larger wording oâ— this matter I refer you to the third Chapteâ— of Mr. Windgates Rule of Proportion 〈◊〉 the first Chapter of the Carpenters Rule bâ— I. Brown Lastly In nameing of any Point found out on the Line great care and respect musâ— be had as to the true value of the Number according to the rate of the question propounded for the same Point that represents 15 doth represent 150 and also 1500 or 15000 increasing above the bounds before mentioned also it signifies one and a half or 15 of one hundred which is usually exprest thus in a Decimal Fraction 15 100 or more readily 0.15 Also if it should be a Number with a digit two ciphers and another digit as 2.005 this Number would be found close to the long stroke by the figure 2 and may represent either two thousand and 5 of 1000 more or 20 and 5 of a hundred or 2 hundred and 5 of another 10 more or plainly as it is set down two thousand no hundred but five Thus you see the manner of expressing whole Numbers or whole Numbers and Decimal Fractions which on the Lines is one and the same thing and thus all Decimal Scales are to be accounted and in the same manner is the Line of Numbers to be read as you may see more at large in the two Books before mentioned SECT II. But for Numerateon on all Circular Lines it is much easier For first very few Instruments unless at one part of the Line can express nearer than minutes of a degree Secondly The whole Radius or Line of Lines is but 90 degrees or but 45 of the Tangents or 60 of the Chords or Secants So that in Instruments of 12 or 18 inches Radius you may express very well every tenth minute to 60 on the Line of Sines and every half degree to 75 and whole degrees to 90. And on the Tangents or Chords every 10th minute quite through and the Secants as the Sines So that any degree or minute being named tâ— find the same on the respective Line count thus First every 10th degree is noted with a long stroke and figures set thereunto Secondly every whole degree is cut between two or three Lines and sometimes with a Point or Mark on the end of the stroke and every 5th degree cut up higher than the rest and sometimes with three Points on the end of the Line or some other convenient distinction for readiness sake and every 10th 15th or 30th minute is cut only between two Lines and no more as will appear very plain with a little practice Example to find the Sine of the Latitude being at London 51 degrees 32 minutes 1. First look on any Line of Sines on the Quadrantal or Sector side according aâ— you have occasion till you see 50 which iâ— 50 degrees then one degree forward toward 60 is 51 degrees then count threâ— 10ths of minutes more for 30 minutes and then for the odde two minutes estimate onâ— fifth part of the next 10 minutes forwarder and that is the precise Point for the Sine 〈◊〉 51 degrees 32 minutes the latituâ—e of London where sometimes is set a Braâ—s Center-Pin Example the second 2. To find the Cosine of the Latitude there are two wayes to count the Compleâ—ment of any Ark or Angle First by substracting the Ark or Angle out of 90 by the Pen and count the residuâ— from the beginning of the Line of Sines and that shall be the Sine Complement of the Latitude required Example 51 32 taken from 90 the remainder is 38 28 now if you count so much from the beginning of the Line of Sines according to the last Rule that shall be the Point for the Sine of 38 28 the Complement of 51 32 or the Sine Complement of the Latitude Or Secondly If you count 51 32 from 90 calling 80 10 and 70 20 and 60 30 50 40 40 50 c. wheresoever the Number whose Complement you would have shall end that is the Sine Complement required which will be at 38 28 from the Center or beginning for the Co-sine of 51 32 The like work serves for any other Number or on any other Line as on the Degrees Tangents or Secants Natural or Artificial as by practice will more plainly appear to the willing Practitioner SECT III. To find the versed Sine of an Ark or Angle or the Sine of an Ark or Angle above 90 degrees or the Chord above 180 degrees observe these Rules 1. First a right Sine is the measure on the Line of Sines from the center or beginning of that Line to the Point that doth represent the Ark or Angle required 2. The right Sine of an Ark or Angle above 90 degrees is equal to the right Sine of the Complement thereof to 180 degrees being readily accounted thus Count the excess above 90 backwards from 90 toward the Center then the measure or distance from the end of the account to the Center is the Sine of the Ark above 90 required Example Let the Sine of 130 be required first if you take 130 from 180 the remainder is 50 then I say that the right Sine of 50 is also right Sine of 130 for if you count backwards from 90 calling 80 100 and 70 110 and 60 120 and 50 130 the measure from thence to 00 or the Center is the right Sine of 130 degrees 3. The versed Sine of an Ark or Angle is the measure on the Line of Sines from 90 toward the Center counted backwards as the small
that point at the end of the Natural-Sines on the Sector-side and at 90 and 45 on the edge of the Rule for the Artificial Sines and Tangents or 10 on the Line of Numbers and 10 and 90 on the Line of Lines and Sines on the Quadrantal-side of the Instrument 2. A Right Sine of any Ark or Angle is the measure from the beginning of the Line of Natural-Sines to that Point on that Line of Sines which represents the degrees and minutes contained in that Ark or Angle required But on the Artificial-Sines we respect not any measure but the Point only 3. The same account is used both for the Right-Tangent and Secant also the Natural-Tangent taken from the beginning to the degree and minute required the Artificial respecting the Point only 4. In the same manner count for the Secants and Chords Lines or versed Sines 5. A Cosine or Sine Complement of any Ark or Angle is the measure from the Point representing the Ark or Angle counted from 90 to the beginning of the Line of Sines being in effect the Right-Sine of the Cosine of the Ark or Angle required As for Example I would take out the Cosine of the Latitude of London which is 51 32 Count 51 32 from 90 toward the beginning and you shall find your account to end at the Right-Sine of 38 28 which is the Complement of 51 32 for both put together makes 90 the whole Sine or Radius But on the Artificial-Lines count backward to the Point required without minding any distance or measure till you come to Proportion 6. A Lateral Sine Tangent or Secant or Scale of Equal Parts is any Sine Tangent or Secant taken along the length of any Line from the beginning onwards being a term used only in operation with a Sector or one Line and a Thred and opposed to a parallel-Sine Tangent or Secant the thing next to be explained 7. A Parallel Sine Tangent or Secant is any Sine Tangent or Secant taken across from one Leg to the other of a Sector or from any degree and minute on one Line to a Thred drawn streight with the other hand or any other fixed Line whatsoever at the nearest distance 8. The Nearest Distance to any Line is thus taken When one Point of the Compasses stands in any one Point and the Line being laid I open or close my Compasses till the other moveable-Foot being turned about will but just touch or cleave the Thred But if you are to lay the Thred to the nearest distance then one Point of the Compasses being set fast the other is to be turned about and the Thred also slipped to and fro till the Compass-point shall just cleave the Thred in the middest 9. To adde one Sine or Tangent to a Sine or Tangent is to take the Right-Sine or Tangent of any Ark or Angle between your Compasses and setting one Point of the Compasses in the Point of the other Number and then to see how far the other Point will extend Laterally Example To adde the Sine of 20 to the Sine of 30 take the Sine of 20 between your Compasses and then putting one Point in 30 the other shall reach to the Sine of 51 21 therefore the distance from the beginning to 51 21 is the sum of the Sine of 30 and 20 added together The like way is to add Tangents 10. To Substract a Sine from a Sine or a Tangent from a Tangent is but to take the Lateral least Right-Sine or Tangent between your Compasses and setting one Point in the term of the greatest turn the other toward the beginning and note the degree and minute that the other Point stayes in for that is the difference or remainder Example Suppose I would take the Sine of 10 degrees from 25 Take the distance 10 between your Compasses and setting one foot in 25 and the other turned toward the beginning shall reach to 14 23 the residue or difference required Or you may sometimes take the distance between the greater and the less and lay this from the beginning shall give the remainder in distance on the Sines as before 11. The Rectifying-Point is a Point or Hole on the Head of the Trianguler Quadrant in the inter-secting of the hour and Azimuth-line and the common Line to the Lines and Sines on the Head-leg in which Point you are when the Rule is open to stick a small Pin to look to the object whose Altitude above the Horizon you would have in degrees and minutes Of Terms used in DIALLING PLain is that Board Glass or flat Superficies you intend to draw the Dial upon either single of it self or joyned to some other Pole of the Plain is an imaginary Point in the Horizon for all upright Dials directly opposite to the Plain or in all Plains a Point every way 90 degrees from the Plain Declination of a Plain is only the number of degrees and minutes that the Pole-point of the Plain is distant from the North and South-points of the Horizon The Perdendiculer-Line on the Plain is a Line Square to a Horizontal-line being part of a Circle passing through the Zenith and Nadir and Pole-point of the Plain The Horizontal-line is a Line drawn on any Plain exactly parallel to the true Horizon of the place you dwell in Reclination is when a Plain beholdeth the Zenith-point over our heads But Inclination is when a Plain beholdeth the Nadier as in a Roof of a House the Tiled-part reclines and the Celid-part inclines The Meridian-line on all Plains is the Hour-line of 12 but the Meridian of the Plain is the great Circle of Azimuth perpendiculer to the Plain bing the same with the Perpendiculer-line on the Plain passing through the Points of Declination The Substile-line on all Dials is that Line wherein the Stile Gnomon or Cock of the Dial doth stand usually counted from 12 the Meridian-line or from the Perpendiculer-line which in all erect Dials is 12. The Stile of a Dial is the Angle between the common Axis of the World and the Plain upon the Substile-line on the Plain on all Dials The Angle between 12 and 6 is onely the number of degrees and minutes contained between the Hour-line of 12 and the Hour-line of 6 a clock on any kind of Plain especially those having Centers The Inclination of Meridians is the number of degrees and minutes counted on the Aequinoctial between the Meridian or Hour-line of 12 and the Substile being the distance between the Meridian of the place viz. 12 a clock and the Meridian of the Plain but counted on the Aequinoctial and doth serve to make the Table of Hour-Arks at the Pole and to prove your work The Lines Parallel to 12 are two Lines peculiar to this way of Dialling by the Sector and are only
Almicanters were parallel to the Horizon so that the Latitude of a Star is counted from the Ecliptick toward the Poles of the Zodiack but the Sun being alwayes in the Ecliptick is said to have no Latitude 6. Degrees of Longitude in respect of the Heavens are measured by the degrees on the Ecliptick from the first point of Aries forward according to the succession of the 12 Signs of the Zodiack But Longitude on the Earth is counted on the Equinoctial Eastwards from some principal Meridian on the Earth as the Isles of Azores or the Peak of Tenneriff or the like 7. Right Ascention is an Arch of the Equinoctial counted from the first Point of Aries that cometh to the Meridian with the Sun Moon or Stars at any day or time of the year being much used in the following discourse noted in the Analemma by EH or the like but counted as afterward is shewed 8. Oblique Ascention is an Arch of the Equinoctial between the beginning of Aries and that part of the Equinoctial that riseth with the Center of a Star or any portion of the Ecliptick in an Oblique-Sphear 9. Ascentional Difference is the difference between the Right and Oblique Ascention to find the Sun or Stars rising before or after 6. 10. Amplitude is an Arch of the Horizon between the Center of the Sun and the true East-point at the very moment of Rising represented by ♋ F in the particular Scheam and GE and FE in the Analemma useful at Sea 11. A Circle of Position is one of the 12 Houses in Astronomy or Astrology 12. An Angle of Position is the Angle made in the Center of the Sun between his Meridian or Hour and some Azimuth as the Prime Vertical or the Meridian or any other Azimuth being useful in Astronomy and sometimes in Calculation represented bp P ☉ Z in the Analemma Thus much for Astronomical terms CHAP. V. Some Uses of the Trianguler Quadrant Use I. And first to rectifie the Rule or make it a Trianguler Quadrant FIrst open the Rule and put in the loose piece into the two Mortice-holes which putting together makes it a Trianguler Quadrant but if you do not use the loose-piece then open it to an Angle of 60 degrees which is thus exactly done Measure from the Rectifying-point to any Number on the Sines or Lines then keeping the Point of the Compasses still fixed in the Rectifying-point turn the other to the Common-Line of the Hour and Azimuth-Line that cuts the Rectifying-point and there keep it then removing the Point of the Compasses from the Rectifying-point open or close the Rule till the other Point shall touch the distance first measured in the Line of Sines or Lines then shall you see the Lines on the Head and Moveable-leg to meet and also see quite through the Rectifying-point to thrust a Pin quite through and thus is it set to an Angle of 60 degrees without the help of the loose-piece or to an Angle of 45 or whatsoever else the Rule is made for Use II. To observe the Sun or a Stars Altitude above the Horizon Put a Pin in the Center-hole on the Head-Leg and another in the Rectifying-point and a third if you please in the end of the hour-Hour-line on the Moving-leg Then on the Pin in the Leg-center hang a Thred and Plummet then if the object be low viz. under 25 degrees high Look along by the two Pins in the Rectifying-point and the Moving-leg and see that the Plummet playeth evenly and steady then the degrees cut by the Thred shall be the Altitude required counting from 60 0 toward the Head as the smaller Figures shew But if the Object be above 25 degrees high then look by the Pin in the Rectifying-point and that on which the Plummet hangeth and observe as before and the Thred shall shew the Altitude required as the Figures before the Line sheweth If you have Sights use them instead of Pins and by Practice learn to be accurate in this Work the ground and foundation in every Observation and according to your exactness herein is the following Work also Note also that this looking up toward the Sun is only then when the Sun is in a cloud and may be seen in the Abiss but will not give a clear shadow Or else you must use a piece of Red or a Blue or Green Glass to darken the luster that it offend not the eyes But if the Sun be clear and bright then you need not look up toward it but hold the Trianguler Quadrant so that the shadow of the Pin in the Center may fall just on the shadow of the Pin in the Rectifying-point and both those shadows on your finger beyond them and the Plummet being somewhat heavy and the Thred small and playing evenly by the Rule then is the Observation so made likely to be near the very truth Note also further That the shaking of the hand you shall find will hinder exactness therefore when you may find some place to lean your Body or Arm or the Instrument against that you may be the more steady But the surest and best way is with a Ball-socket and a Three-leg-staff such as Land Surveighers use to support their Instruments withal then you will be at liberty to move and remove it to and fro till the Sights or Pins and Plummet and Thred play to exactness without which care and exactness you cannot certainly and knowingly attain the Sun's or a Star's Altitude to a minute either by this nor any other Instrument whatsoever though they be never so truly made Yet I dare affirm to do it or it may be done as well by this as by any other graduated Instrument whatsoever The Line of degrees on this being only two thirty degrees of a Tangent laid together of which that on the in-side of the loose-piece is the largest and consequently the best to distinguish the minuts of a degree withal Use III. To try if any thing be Level or Upright Set the Moving-leg of the Trianguler Quadrant on the thing you would have to be Level then if the Thred play just on 60 degrees or the stroke by 600 then is it Level or else not But to try if a thing be upright or not apply the Head-leg to the Wall or Post and if it be upright the Thred will play just on the common Line between the Lines and Sines on the Head-leg and cut the stroke by 90 on the Head of the Instrument or else not Use IV. To find readily what Angle the Sector stands at at any opening First on the Sector side about the Head is 180 degrees or twice 90 graduated to every two degrees so that opening the Rule to any Angle the in-side of the Moving-leg passing about the semi-circle of the Head sheweth the Angle of opening to one degree But to do it more exactly do thus The two Lines of Sines that issue from the Center in Rules of
Right Ascention and the Ascentional-difference is the Oblique-Ascention But in Southern declinations the sum of the Right Ascention and difference of Asscentions is the Oblique Ascention Example On or between the 25 and 26 of Iuly the Oblique-Ascention is by Substraction 112 15 On the 30th of October the Oblique-Ascention is 337 45 by Addition Use VIII The Day of the Month or Sun's Declination and Altitude being given to find the Hour of the Day Take the Suns Altitude from the particular Scale of Altitudes setting one Point of the Compasses in the Center at the beginning of that Line and opening the other to the degree and minute of the Sun's Altitude counted on that Line then lay the Thred on the Day of the Month or Declination and there keep it Then carry the Compasses set at the former distance along the Line of Hours perpendiculer to the Thred till the other Point being turned about will but just touch the Thred the Compasses standing between the Thred and the Hour 12 then the fixed Point in the Hour-Line shall shew the hour and minute required but whether it be the Fore or Afternoon your judgment or a second observation must determine Example On the first of August in the morning at 20 degrees of Altitude you shall find it to be just 52 minutes past 6 but at the same Altitude in the afternoon it is 7 minutes past 5 at night in the Latitude of 51 32 for London Use IX The Suns Declination and Altitude given to find the Suns Azimuth from the South-part of the Horizon First by the 4th Use find the Suns Declination count the same on the particular Scale and take the distance between your Compasses then lay the Thred to the Suns Altitude counted the same way as the Southern-Declination is from 600 toward the loose-piece and when need requires on the loose-piece then carry the Compasses along the Azimuth-line on the right-side of the Thred that is between the Thred and the Head when the Declination is Northward and on the left-side of the Thred that is between the Thred and the End when the Declination is Southward So as the Compasses set to the Declination as before and one Point staying on the azimuth-Azimuth-line and the other turned about shall but just touch the Thred at the nearest distance then I say the fixed-Point shall in the Azimuth-line shew the Suns-Azimuth required Example 1. The Sun being in the Equinoctial and having no Declination you have nothing to take with your Compasses but only lay the Thred to the Altitude counted from 600 toward the loose-piece and in the Azimuth-line it cuts the Azimuth required Example At 25 degrees high you shall find the Suns Azimuth to be 54 10 at 32 degrees high you shall find 38 20 the Azimuth Again At 20 degrees of Declination take 20 from the particular Scale and at 10 degrees of Altitude lay the Thred to 10 counted as before then if you carry the Compasses on the right-side for North-Declination you shall find 109 30 from South but if you carry them on the left-side for South-Declination you shall find 38 30 from South The rest of the Vses you shall have more amply afterwards CHAP. VI. The Use of the Line of Numbers on the Edge and the Line of Lines on the Quadrantal-side or on the Sector-side being all as one HAving shewed the way of Numeration on the Lines as in Chapter the first Also to add or substract one Line or Number to or from one another as in Chapter 4th Explanation the 9th I come now to work the Rules of Multiplication and Division and the Rule of Three direct and reverse both by the Artificial and Natural-Lines and first by the Artificial being the most easie and then by the Natural-lines both on the Sector and Trianguler Quadrant being alike and I work them together First because I would avoid tautology Secondly because thereby is better seen the harmony between them and which is best and speediest Thirdly because it is a way not yet as I know of gone by any other And last of all because one may explain the other the Geometrical Figure being the same with the Instrumental-work by the Natural way Sect. I. To multiply one Number by another 1. By the Line of Numbers on the Edge Artificially thus Extend the Compasses from 1 to the Multiplicator the same extent applied the same way from the Multiplicand will cause the other Point to fall on the Product required Example Let 8 be given to be multiplied by 6 If you set one Point of the Compasses in 1 either at the beginning or at the middle or at the end it matters not which yet the middle 1 on the Head-leg is for the most part the most convenient and open the other to 6 or 8 it matters not which for 6 times 8 and 8 times 6 are alike but yet you may mind the Precept if you will the same Extent laid the same way from 8 shall reach to 48 the Product required which without these Parenthesis is thus The Extent from 1 to 6 shall reach the same way from 8 to 48. Or The Extent from 1 to 8 shall reach the same way from 6 to 48. the Product required By the Natural-Lines on the Sector-side or Trianguler Quadrant with a Thred and Compasses the work is thus 1. For the most part it is wrought by changing the terms from the Artificial way as thus The former way was as 1 to 6 so is 8 to 48 or as 1 to 8 so is 6 to 48 but by the Sector it is thus As the Latteral 6 taken from the Center toward the end is to the Parallel 10 10 set over from 10 to 10 at the end counted as 1 so is the Parallel-distance between 8 8 on the Line of Lines taken a-cross from one Leg to the other to the Latteral-distance from the Center to 48 the Product required Or shorter thus As the Latteral 8 to the Parallel 10 So is the Parallel 6 to the Latteral 48. See Figure I. 2. Another way may you work without altering the terms from the Artificial way as thus by a double Radius Take the Latteral-Extent from the Center to 1 or from 10 to 9 if the beginning be defective make this a Parallel in 6 6 then the Latteral-Extent from the Center to 8 of the 10 parts between Figure and Figure shall reach across from 48 to 48 as before See Fig. II. The same work as was done by the Sector is done by the Line of Lines and Thred on the Quadrant-side that if your Sector be put together as a Trianguler Quadrant you may work any thing by it as well as by the Sector in this manner or by the Scale and Compass as in the Figure I. and first as above Sector-wise Take the Extent from the Center to 6 latterally between your Compasses set one Point in 10 and with the other lay the Thred in the nearest distance
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.
Or else thus As 5 ½ counted for 11 is to = 6 counted for 12 So is = 9 to 8 ¼ near the end and as large as may be Thus you may many times vary the manner of work to get the Answer latterally and as large as may be on the Scale of Lines by doubling or halfing the Numbers or taking the whole Number of quarters or using a less or a bigger Scale as hath been hinted and shall be more in places convenient in the following Discourse to attain exactness and ease â—s much as may be as time and practice will demonstrate to the willing Practitioner in these Operations Problem IV. Having the length and breadth given in Inches to find the Content in Superficial-Square Inches As 1 inch to the breadth in inches so is the length in inches to the Content in Superficial inches Example 20 inches broad and 36 inches long The Extent of the Compasses from 1 on the Line of Numbers to 20 shall reach the same way from 36 to 72 the true Number of Superficial Square inches in that Oblong By the Line of Lines As 36 to = 5 counted as 1 So is = 10 counted as 20 to 72 at the largest Extent For Note The reason that the Latteral 72 and 36 are from the same Scale in account and the Parallel 1 and 20 counted Decimally are from the same Scale also or else according to the Proportion by the Line of Numbers As 1 to = 20 So is 36 to = 72. Here also is the same advance Decimally from 1 to 20 as before Problem V. Having the length and breadth given in Inches to find the Content in Feet Superficial As 144 to the breadth in inches so is the length in inches to the Content in Feet Superficial Example at 40 broad and 60 long For the Extent on the Line of Numbers from 144 the number of inches in one foot to 40 the breadth shall reach from 60 inches the length to 16 foot ½ and 26 inches To count so many inches on the Line observe with me this way of Reduction the 16 foot and ½ is very plainly seen And Note That there is 10 cuts in this place between 16 and 17 and 10 times 14 is 140 which is near 144 the inches in a foot so the Point of the Compasses staying at near 2 10ths beyond the half-foot I count almost twice 14 which is 26 inches for the Fraction above 16 foot and ½ By the Line of Lines As 144 found between 1 and 2 near the Center is to = 40 at the figure 4 So is half = 60 or the measure from the Center to 3 to = 8-35 which is the half of 16 foot ½ and 26 inches For if you had taken all 60 it would have exceeded the whole Parallel Radius where the Answer would have been right 16 ½ and better but taking the half it gives the half also Or else work thus with a Latteral Answer As ½ 60 to = ½ 144 So is all = 40 to 16 7 10. Or As all 40 to ½ = 144 So is all = 60 to 33 and 4 10 the double of 16 and 7 10. Note That in both these two last workings the 144 is at 72 which is the half of 144 to make the work the larger By these the excellency of the Line of Numbers over the Line of Lines is evident in these kind of Proportions And for discovering the Reason of these Proportions read the beginning of the 6th Chapter Section 3d. Problem VI. The length and breadth of an Oblong Superficies being given to find the side of a Square equal to it by the Line of Numbers Divide the space between the length and breadth into two equal parts and the middle Point shall be the side of the Square equal to the Oblong given in quantity Example If a long Square or Oblong be 18 foot one way and 12 foot the other way the middle Point between 18 and 12 is 14 and 7 10 ferè for 18 multiplied by 12 makes 216 and 14-7 multiplied by 14-7 is near 216 also To do this by the Line of Lines is shewed at large in the 7th Section of the 6th Chapter Problem VII Having the Diameter or Circumference of a Circle to find the Circumference or Diameter or Squares equal or Inscribed and Content For this purpose there are certain Proportional Numbers found out As thus If the Diameter of a Circle be 10 then the Perifera or Circumference is 31 42 the side of the Square equal to the Circle is 8-862 the side of the Square inscribed is 7-071 and the Superficial Content is 78-54 so that any one of these being given you may find out any of the rest by the Line of Numbers Thus having the Diameter to find the Circumference As 10 to 31 42 so is the given Diameter to the Circumference required Or As 10 to 8-862 so is the given Diameter to the Square equal As 10 to 7-071 so is the given Diameter to the Square inscribed As 10 to 30 so is 78-54 to the Square of the Area of that Circle whose Diameter is 30. Or To the Diameter turning the Compasses twice As for Example Let the given Diameter of a Circle be 30 The Circumference is 94 26 Diameter 10 00 The qu are equal is 26 58 ½ Circumference â—1 42 The Square within is 21 21 Square equal 8 862 The Content or Area is 707 00 Square within 7 071 The Diameter is 030 00 Area or Content 78 54 Also Note If the Circumference be first given then say As 31 42 is to the Number 10 for the Diameter or 8-862 for the Square equal or to 7-071 for the inscribed Square so is the given Circumference to the rest But to find the Area say As the fixed Diameter 10 is to the given Diameter 30 so is the fixed Area for 10 viz. 78-54 to 707 by turning the Compasses two times Or if the Circumference be given and I would find the Area As 31-42 the fixed Circumference is to 94-26 the given Circumference so is 78-54 the fixed Area answering the fixed Circumference 707 the Area required turning the Compasses two times the same way Thus by having five Centers at the five fixed Numbers or four Centers answering to the four fixed Numbers for a Circle whose Diameter is 10 having any one of those 5 given you may find any of the other required Thus you have eight Problems couched in one therefore be the more diligent to understand it To work these by the Line of Lines observe the former directions which for brevity sake I now omit Problem VIII The Content of a Circle being given to find the Diameter Divide the distance on the Line of Numbers between the fixed Content or the Point 78-54 and the given Content into two equal parts that distance laid the same way from the fixed Diameter shall reach to the required Diameter Example The Content being 707-00 The half distance between 78-54 and 707 shall reach from 10
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 Product on the second For Division alwayes thus As the Divisor found on any one side is to 1 on the second or other-side So is the Dividend on the same first-side to the Quotient on the second For the Rule of Three As the first term on the first-side to the second on the other For Superficial Measure by Inches and Feet As 12 to the breadth in inches on the second So is the length in feet to the Content on the second For any thing else the same Rules and Precepts you find in Chap. VII will give you ample and plain directions The Lines being fitted as much as may be to speak out the Answer to the Question as by well considering the Figure you may see CHAP. XI To make and measure the Five Regular Platonical Bodies with their Declinations and Reclinations 1. For the Cube being the Foundation of all other IT is a Square Solid Body every way alike and spoken of largely before as to the Mensuration thereof and obvious enough to every indifferent Workman as to the making thereof and needs no repetition in this place 2. For the Tetrahedron It is a Figure comprehended of 4 equilatteral plain Triangles or a Trianguler Pyramid last mentioned the best and nearest way as I conceive of making is thus According to directions of Mr. Iohn Leak On any rough Piece make one side plain and flat so large as to contain the Triangle which you intend shall be one side of the Tetrahedron then set a Bevel to 70 degrees 31 minutes and 42 seconds and plain another side to fit the former side and the Bevel secundum Artem then mark this last plained side according to the former and cut away the residue plaining them away just to the strokes and fit to the Bevel formerly set and you shall constitute the Tetrahedron required The Superficial Content is the Area of the 4 equilatteral Triangles mentioned before and the solid Content is found by multiplying the Area of one Triangle by one third part of the Altitude of the Pyramid or Tetrahedron from the midst of one Plain to the Apex or top of the opposite Solid Angle If the measuring the sides Perpendiculer and Altitude of the Tetrahedron with Compasses Callipers and Scale serve not to exactness then proceed thus First for the Perpendiculer the Triangle being equilatteral Multiply one side given by 13 and divide the Product by 15 the Quotient is the Perpendiculer Example If the side of the Tetrahedron be 12 that multiplied by 13 gives 156 which divided by 15 leaves 10-4 for the length of the Perpendiculer in the equilatteral Triangle whose side is 12. Then for the Perpendiculer Altitude work thus by the Artificial Numbers and Sines As the Sine of 90 to the sine of 70 deg 31 min. 42 sec So is 10-4 the Perpendiculer to 9-80 the perpendiculer Altitude required Or by the Sector work thus Take 12 the side of the Tetrahedron from any Scale or the Line of Lines and set the Sector to 60 degrees by making the Latteral 12 a Parallel 12 then the nearest distance from 12 to the Line of Lines is the true Perpendiculer which measured on the same Line of Lines will be found to be 10-4 as before then make this 10-4 a Parallel Sine of 90 and 90 the Sector so set take out the Parallel-sine of 70-31-42 and measure it on the same Scale and it shall be 9-8 as before Then Lastly This perpendiculer Altitude being multiplied by the Area of the Base gives a Number whose third part is the Solid-Content of the Tetrahedron required For 12 the side and 5-2 the half Perpendiculer makes 62-4 the Superficial-Content of one Triangle or Base then 62-4 the Base multiplied by 9-8 the perpendiculer Altitude gives 611-52 and a third part of 611-52 is 203-86 the solid Content required The three Triangles recline from the Perpendiculer upright 19 degr 28 min. and 18 sec. and decline when the edge is South 60 South East and South West and the opposite Plain a just North but if you make one South then the other two are are North-east and North-west 60 deg 3. For the Octahedron Whis is a solid body comprehended under 8 equilatteral Triangles The way of making which is thus Make a plain Parallelepipedon or long-Cube if the breadth both wayes be 1000 let the length be 1-414 or if the length be 500000 the breadth both wayes must be 3-53553 then to these Measures square it exactly then divide the length and breadth just in the midst and draw Lines both wayes on all the 6 sides then draw the Diagonal-Lines from the midst of the length to the midst of the breadth and cut by these Diagonal-Lines and the Octahedran will appear to be truly made For the Mensuration thereof it is the same as in the Tetrahedron For supposing the side of one of the Triangles 12 the Base is 144 in Content and the Triangles Perpendiculer is 10-4 as before But the Perpendiculer Altitude is just half the length viz. 8-49 for if the breadth be 12 then the length must be 16-98 whose halfs are 6 and 8-49 Then if you multiply 144 the Base by 8-49 the perpendiculer Altitude the Product will be 1222-56 whose third part is 407-52 the half of the Tetrahedron and 815-04 is the whole solid Content of the Tetrahedron required as near as we can see by Instrumental Operation but if you work to a Figure more you shall find the total Area to be but 814-656 more exact To find this Perpendiculer Altitude by the Sector work thus First The Triangles Perpendiculer being 10-4 as before Take the Latteral 10-4 from the Line of Lines make it a Parallel in 90 lay the Thred exactly to the nearest distance and there keep it then the Parallel-distance from the Sine of 54 deg 44 min. 45 sec. the Reclination shall be 8-486 the true Perpendiculer Altitude required Then if the Octahedron stand on one Triangle you have one Horizontal Plain and one South and North Reclining and Inclining 19 deg 28 min. 18 sec. as the Tetrahedron was and two South and two North declining 60 and reclining and inclining 19 deg 28 min. 18 sec. as afore But if it stand on a Point then you have 4 direct or declining 45 and reclining 54 44 45 and 4 incliners inclining as much and direct or declining as you shall please to set them 4. For the Dodecahedron Which is a regular solid body contained under or made up of 12 Pentagonal Pyramids or Pyramids whose Base hath 5 equal sides and the perpendiculer Altitudes of those 12 Pyramids equal to half the Dodecahedrons Altitude standing on one side or equal to the semi-diameter of the inscribed Sphear To cut this Body take any round Piece and if the Diameter be 100000 the length must be 0-81005 or as 4-906 to 3-973 then the Piece being turned round and the two Ends flat to the former measures of Length and Diameters which are near according to
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
large and plain in this first Use I shall be I hope as plain though far more brief in all the rest for if you look back to Chapt. VI. Probl. I. Sect. 3. you shall there see the full explaining of Latteral and Parallel and Nearest-distance and how to take them the mark for Latteral being thus The mark for Parallel thus = Nearest-distance thus ND c. Use II. The Sine of any Ark or Angle given to find the Radius to it Take the Sine between your Compasses and setting one foot of the Compasses in the given Sine and with the other Point lay the Thred to the nearest-distance and there keep it then the nearest-distance from the Sine of 90 to the Thred shall be the Radius required Make the given Sine a Parallel Sine and then take out the Parallel Radius and you have your desire The Artificial Sines and Tangents are not proper for this work further then to give the Natural Number thereof as before therefore I shall only add the use of them when it is convenient in the fit place Use III. The Radius or any known Sine being given to find the quantity of any other unknown Sine to the same Radius Take the Radius or known Sine given and make it a Parallel in the Sine of 90 for Radius or in the Sine of the known Angle given and lay the Thred to ND Then take the unknown Sine between your Compasses and carry one Point along the Line of Sines till the other foot being turned about will but just touch the Thred then the place where the Compasses stayes shall be the Sine of the unknown Angle required to that Radius or known Sine Make the given Radius a Parallel Radius or the given Sine a = Sine in the answerable Sine thereof Then taking the unknown Sine carry it parallelly along the Line of Sines till it stay in like parts which parts shall be the Numerator to the Sine required Use IV. The Radius being given by the Sines alone to find any Tangent or Secant to that Radius Take the Radius between your Compasses and set one Point in the Sine complement of the Tangent required and lay the Thred to the ND then the ND from the Sine of the Tangent required to the Thred shall be the Tangent required And the ND from 90 to the Thred shall be the Secant required Make the given Radius a = in the co-Sine of the Tangent required then the = Sine of the inquired Ark or Angle shall be the Tangent required and = 90 shall be the Secant required to that Radius Use V. Any Tangent or Secant being given to find the answerable Radius and then any other proportionable Tangent or Secant by Sines only First if it be a Tangent that is given take it between your Compasses and setting one foot in the Sine thereof lay the Thred to ND then the = Co-sine thereof shall be Radius But if it be a Secant take it between your Compasses and set one foot alwayes in 90 lay the Thred to the ND then the nearest distance from the Co-sine to the Thred or the = Co-sine shall be the Radius required Take the given Tangent make it a = in the Sine thereof then the = Co-sine thereof shall be Radius Or if it be a Secant given then Take the given Secant make it a = in 90 then the = Co-sine thereof shall be the Radius required Then having gotten Radius the 4th Use shewes how to come by any Tangent or Secant by the Sines only Use VI. To lay down any Chord to any Radius less then the Sine of 30 degrees Take the given Radius set one Point in the Sine of 30 lay the Thred to the ND and for your more ready setting it again note what degree and minuit the Thred doth stay at on the degrees and there keep it Then the ND from the Sine of half the Angle you would have shall be the Chord of the Angle required Take the given Radius and make it alwayes a = in 30 and 30 of Sines the = Sine of half the Chord shall be the Chord required Use VII To lay down any Chord to the Radius of the whole Line of Sines Take the Radius between your Compasses and setting one Point in 90 of the Sines lay the Thred to the ND observing the place there keep it Then taking the = Sine of the Angle required with it set one Point in the Line to which you would draw the Angle as far from the Center as the Radius is then draw the Convexity of an Ark and by that Convexity and the Center draw the Line for the Angle required Example Let AB be a Radius of any length under or equal to the whole Line of Sines Take AB between your Compasses and setting one Point in 90 lay the Thred to ND then take out the = Sine of 38 or any other Number you please and setting one Point in B the end of the Radius from A the Center and trace the Ark DC by the Convexity of which Ark draw the Line AC for the Angle required Take the given Radius AB make it a = in 90 and 90 of Sines then take out = 38 and setting one foot in B draw the Ark DC and draw AC for the Angle required Or else work thus Take AB the given Radius having drawn the Ark BE and make it a = in the Co-sine of half the Angle required and lay the Thred to ND or set the Sector Then Take the = ND from the right-sine of the Angle required and it shall be BE the Chord required to be found Note That the contrary work finds Radius Use VIII To lay off any Angle by the Line of Tangents or Secants to prove it Having drawn the Ground-Line AB at the Point B raise a Perpendiculer as the Line BC extended at length then make AB the Radius a = Tangent in 45 and 45 then take out the = Tangent of the Angle required and lay it from B to C in the Perpendiculer and draw the Line AC for the Angle required Also If you take out the Secant of the Angle as the Sector stands and lay it twice in the Line AE it will reach just to C the Point required Also Note That if you want an Angle above 45 degrees as the Sector stands take the same from the small Tangent that proceeds to 75 and turn that Extent 4 times from B and it shall give the Point required in the Line BC. Use IX To lay down or protract any Angle by the Tangent of 45 only First make a Geometrical Square a ABCD and let A be the Anguler Point then making AB Radius make AB a = Co-sine of the Angle you would have and lay the Thred to
the nearest distance then the ND from the right Sine of the Angle to the Thred shall be the Tangent required Example I make AB Radius a = in 50 the co-Sine of 40 then the = Sine of 40 shall be BE. Again If I make AD equal to AB the = co-Sine of 30 viz. 60 and then take out the = Sine of 30 and lay it from D to F it shall be an Angle of 60 from AB or 30 from D to F. But by the Sector this is more easie Use X. To take out readily any Tangent above 45 by the Tangent to 45 on the Sector-side Take the given Radius make it a = in the co-Tangent of the Tangent required then the = Tangent of 45 shall be the Tangent required Example I would have a Tangent to 80 degrees take the given Radius make it a = in 10 the complement of 80 then the = Tangent of 45 shall be the Tangent of 80 required But if your Radius be so big that you cannot enter it then take the half or a quarter of your Radius and then = 45 will be the half or the quarter of the Tangent required Use XI How to work Proportions in Sines alone by the Natural Sines There are 4 Varieties in this Work that include all Proportions viz. 1. When the Sine of 90 is the first term then the work is thus Lay the Thred to the second term counted on the degrees from the Head toward the loose-piece and count the third term on the Line of Sines from the Center downwards and taking the nearest distance from thence to the Thred and that distance measured from the Center downwards on the Line of Sines gives the 4th term required Example As Sine 90 to Sine 23-30 So is 30 to 11-31 Take the Latteral second term make it a = Sine of 90 then take out the = third term and measuring it from the Center it gives the 4th term required 2. When the Sine of 90 is the third term then work thus Take the Sine of the second term from the Center downwards and make it a = Sine in the first term laying the Thred to ND then on the degrees the Thred shall give the 4th term required Example As the Sine of 30 to 23-31 So is the Sine of 90 to Sine of 52-56 But by the Sector Take the Sine of the second make it a = Sine of the first term then take out = 90 and measure it from the Center and it shall give the 4th term required Example as before 3. When the Radius or Sine of 90 is in the second place work thus Take 90 from a lesser Scale as the uppermost Sine above the Center or the Line of Right-Ascentions or the Azimuth-Scale or the like and make it a = in the Sine of the first laying the Thred to ND then the = third term taken and measured on the same Scale that 90 was taken from shall give the 4th term required Example As 90 on the Line of Right-Ascention is to = 30 So is = 20 to 43-12 measured on the same Line that 90 was taken from Or else secondly work thus As 30 to = 90 So is 20 to = 43-12 By carrying the Compasses till it so stayes as that the foot turned about will but just touch the Thred at the nearest distance Or else thus thirdly By transposing the terms when the third is not greater than the first thus As the first to the third So is the second term to the 4th Where the Radius being brought to the third place it is wrought by the second Rule as before By the Sector Take a smaller Sine of 90 make it a = in 30 then the = Sine of 20 taken and measured on the small Sine gives 43-12 as before Again As 90 to = 30 So is = 20 to 43-12 Again As 30 to = 90 So is = 20 to = 43-12 Lastly by transposing As 20 to = 30 So is = Radius to 43-12 as before 4. When Radius is none of the given terms Then when the first term is greater than the second and third work thus Take the second term make it a = in the first laying the Thred to the ND then the nearest distance from the third term to the Thred measured from the Center downward give the 4th Sine required Example As 20 to 12 so is 18 to 10-50 By the Quadrant As 12 to = 20 So is = 18 to 10-50 When only the second term is greater than the first then transpose the terms and work as before Or else use a double Radius which is on this Instrument very easily done having several Radiusses Or Lastly use a Parallel entrance or answer rather as before which being carefully wrought will do very well By the Sector The same manner of work is as before by the Quadrant and the setting the Sector is all one to the laying the Thred as will be largely seen in all the following Propositions wrought both by the Artificial and Natural Lines of Numbers Sines and Tangents as followeth Use XII Having the day of the Month or Suns place given to find his Declination Lay the Thred on the day of the Month in the Kalender and in the Line of degrees on the Moving-leg you have his Declination either Northward or Southward according to the time of the year counting from 600 toward the Head for North-declination or toward the End for South-declination By the Artificial Sines and Tangents on the Edge of the Instrument Extend the Compasses from the Sine of 90 to the Sine of 23 degrees 31 minuts the Suns greatest Declination The same Extent applied the same way from the Sine of the Suns place or the Suns distance from the next Equinoctial-point shall cause the Moving-point to fall on the sine of the Suns declination This being the general way of working Example The Extent from the sine of 90 to the sine of 23-31 shall reach from the sine of 30 to 11 deg 31 min. the Suns declination in ♉ Taurus 30 degrees from ♈ Aries the next Equinoctial-point and from 60 degrees the Suns distance in ♊ Gemini 60 degrees from ♈ 20 deg 12 min. the Suns declination then This being the manner of working by these Lines by extending the Compasses from the first to the second term I shall for the rest wave this large repetition of extending the Compasses and render it only thus by the words of the Cannon-general in all Books As Sine 90 to Sine 23-31 So is the Sine of 30 to Sine 11-31 Lay the Thred to 23-31 on the degrees on the Moveable-piece counted from the Head toward the End then count the Suns place from the next Equinoctial-point on the Line of Sines from the Center downwards and take the ND from thence to the Thred then this distance being measured from the Center downwards
Azimuth-line and lay the Thred to the ND and on the degrees it sheweth the Altitude required counting from 600 toward the End As S. latitude S. of 90 So S. of Suns declination to S. Suns height at East or West Take the sine of the Suns declination make it a = in the sine of the latitude and lay the Thred to ND and on the degrees it shall shew the Suns Altitude at East and West required Example Declination 10. Latitude 51-32 the Altitude is 12 degrees and 50 minuts As S. of the Suns Declination to = S. of Latitude So is the = S. of 90 to S. of Vertical Altitude Use XIX Having the Latitude and Suns Declination to find the time when the Sun will be due East or West Having gotten the Altitude by the last Rule take it from the particular Sine then lay the Thred to the Suns declination counted on the degrees then setting one Point in the Hour-line so as the other turned about shall but just touch the Thred and the Compass-point shall stay at the hour and minuit of time required As Tangent latitude to Sine 90 So is the Tangent of the Suns declination to Co-sine of the hour Or As sine 90 to Tangent Suns declination So is Co-tangent-latitude to Co-sine of the hour from noon Example Latitude 51-32 declination 10 the Sun will be due East at 6-32 and West at 5-28 Take the Tangent of the Latitude on the loose-piece counting from 60 towards the moveable-leg or else from 600 on the moving-leg or degrees according as the Latitude is above or under 45 degrees and lay it from the Center downwards and note the Point where it ends Then take from the same Tangent the Tangent of the Suns declination and setting one foot in the Point last noted lay the Thred to ND then the = sine of 90 shall be the sine of the hour from 6. Or by the Sines only work thus Take the sine of the Suns declination make it a = in sine of the latitude lay the Thred to ND then take ND from the Co-sine latitude to the Thred then set one foot in the Co-sine of the Suns declination lay the Thred to ND and on the degrees it gives the hour from noon as it is figured or the hour from 6 counting from the head counting 4 minuts for every degree Make the small Tangent of the Latitude if above 45 taken from the Center a = sine of 90 then the Tangent of the Suns declination taken from the same small Tangent and carried Parallely till it stay in like Sines shall be the Sine of the hour from 6. Or as before by Sines only Make sine Declination a = sine Latitude then take = Co-sine Latitude and make it a = Co-sine of the Suns Declination then take = 90 and lay it from the Center it gives the Sine of the hour from 6. Use XX. Having the Latitude and Suns Declination to find the Ascentional Difference or the Suns Rising and Setting and Oblique Ascention Lay the Thred to the Day of the Month or to the Suns Declination or true Place or to his Right Ascention for the Thred being laid to any one of them is then also laid to all the rest then in the Azimuth-line it cuts the Ascentional difference if it you count from 90 or the Suns Rising as you count the morning hours or his Setting counting the afternoon hours The Oblique Ascention is found out for the six Northern signs or Summer half-year by substracting the Suns difference of Ascentions out of the Suns Right Ascention But for the other Winter-half year or six Southern signs it is found by adding the Suns difference of Ascentions to his Right Ascention this sum in Winter and the remainder as above-said in Summer shall be the Suns Oblique Ascention required As Co-tangent Lat. to sine 90 To is the Tangent of the Suns declination to the sine of the Suns Ascentional difference Take the co-tangent latitude from the loose or moveable-piece as it is above or under 45 degrees make it a = in sine 90 lay the Thred to ND then take the Tangent of the Suns declination from the same Tangents and carry it = till it stay in the parts that the other foot turned about will but just touch the Thred which parts shall be the Sine of the Suns Ascentional difference required Or â—hus by Sines only Make the sine of Declination a = Co-sine of the Latitude lay the Thred to ND then take the = sine of Latitude make it a = in Co-sine of the declination and lay the Thred to ND and on the degrees it shall cut the Suns Ascentional-difference required which being turned into time by counting 4 minuts for every degree and added to or taken from 6 gives the Suns Rising in Summer or Winter Make the Co-tangent Latitude a = sine of 90 then take Tangent of the Suns declination and carry it = till it stay in like parts viz. the Sine of the Suns Ascentional difference required Example otherwise As sine 90 to = Tangent 38-28 So is = Tangent of 23-31 the Suns greatest declination to the sine of the Suns greatest Ascentional difference 33 deg and 12 min. Use XXI The Latitude and Suns Declination given to find the Suns Meridian Altitude When the Latitude and Declination is both alike viz. both North or both South then substract the Declination out of the Latitude or the less from the greater and the remainder shall be the complement of the Suns Meridian Altitude But when they be unlike then add them together and the sum shall be the complement of the Meridian Altitude The contrary work serves when the complement of the Latitude and Declination is given to find the Meridian Altitude Lay the Thred to the Declination counted on the degrees from 600 the right way toward the Head for North and toward the End for South declination Then Take the nearest distance from the Center-prick at 12 in the hour-Hour-line to the Thred this distance measured on the Particular-line of Sines shall shew the Suns Meridian-Altitude required Use XXI The Latitude and Hour from the midnight Meridian given to find the Angle of the Suns Position viz. the Angle between the Hour and Azimuth-lines in the Center of the Sun As Sine 90 to Co-sine of the Latitude So is the Sine of the Hour from Midnight to the sine of the Angle of Position Example As Sine 90 to Co-sine Latitude 38-28 So is the Co-sine of the Hour from midnight 120 for which you must use 60 to 32-34 the Angle of Position Take the distance from the Hour to the 90 Azimuth on the Hour-line and measure it in the particular sines and it shall shew the Angle of Position required This holds in the Equinoctial Take Co-sine Latitude make it a = in sine 90 then take
or from 12 as it is figured Example On April 20 at 30 deg 20 min. of Altitude Latitude 51-32 the hour will be found to be just 2 hours from 6 or just 8. Again On the 10th of November at 8 deg 25 min. high it is just 3 hours from 6 or 9 a clock in the forenoon or 3 afternoon Or somewhat differing thus Take the sine of the sum or difference of the Suns present Altitude and Altitude at 6 and make it a = in the co-sine of the Latitude and lay the Thred to the nearest distance then take out the = Secant of the declination beyond 90d and make it a = sine of 90 and laying the Thred to the nearest distance on the degrees it shall shew the hour from 6 required First by Use 25 find the Suns height at 6 or depression in Winter then by the former 2d find the sum or difference between the Altitude at 6 and the Suns present Altitude but if you have Tables of Natural Sines and Tangents then in Winter add the Natural Sines of the two Altitudes together and in Summer substract the lesser out of the greater and find the Ark of difference more exactly Then As the Co-sine of the Latitude to the Secant of the Declination counted beyond 90 as much forward as from 90 to the Co-sine of the Suns Declination So is the Sine of the sum or difference to the hour from 6 required Or else â—hus As the Co-sine of the Latitude to the Sine of the sum or difference So is sine of 90 to a 4th Then As the Co-sine of the Suns declination to that 4th So is sine 90 to the hour from 6. By the Sector Take the secant of the Suns declination make it a = in the co-sine of the Latitude then take out the = sine of the sum or difference and turn it twice from the Center latteraâ—ly and it shall be the sine of the hour from 6 required Example April 20 the Suns Declination is 15 degrees and the Suns Height at 6 then is 11 deg 42 min. now the Natural sine of 11-42 20278 taken from the Natural sine of 30 deg 20 min. 50502 the Suns present Altitude the residue is 30224 the sine of 17 deg 35 min. and a half Then The Secant of 15 made a = sine of 38-28 and the Sector so set the = sine of 17-35 ½ turned latterally twice from the Center shall reach to 30 the sine of 2 hours from 6 the hours required Use XXVII Having the Latitude the Suns Declination and Altitude to find the Suns Azimuth Take the Declination from the particular Scale of Sines for the particular Latitude the Instrument is made for Then count the given Altitude on the degrees from 600 toward the loose-piece and sometimes on the loose-piece also and thereunto lay the Thred then carry the Compasses so set along the Azimuth-line on the right-side of the Thred in Northern-declinations and on the left-side in Southern-declinations till the other foot turned about will but just touch the Thred then the fixed-point shall stay at the Suns true Azimuth required Take two or three Examples 1. First When the Sun is in the Equinoctial and hath no Declination then there is nothing to take between your Compasses but just to lay the Thred to the Suns Altitude counted from 600 on the loose-piece toward the End then 〈◊〉 the azimuth-Azimuth-line it cuts the Azimuth from the South required Example At 00 degrees high the Azimuth is 90 from South and at 10 degrees high it is 77-5 at 20 high the Azimuth is 62-45 at 30 degrees high it is 43-30 at 34 degrees high it is 32 degrees of Azimuth from South and at 38-28 degrees high it is just South 2. Secondly at 10 degrees of Declination Northward and 20 degrees of Altitude take 10 degrees from the particular Scale and lay the Thred to the Suns present Altitude as before and carry the Compasses on the right-side of the Thred on the Azimuth-line till the other foot being turned about will but just touch it then shall the Point rest at 80 degrees 42 min. of Azimuth from the South 3. But if the Declimation be the same to the Southwards and the Altitude also the same then carry the Compasses on the left-side of the Thred on the Azimuth-line till the other foot turned about will but just touch it and you shall find the Point to stay at 41 deg 10 min. the true Azimuth from the South required Note That any thing as thick as the Rule laid by the Rule and the Thred drawn over it will keep the Thred steady till you get the nearest distance more truly First by the 18th Use find the Suns Altitude in the Vertical Circle or Circle of East and West thus Take the sine of the Suns Declination and set one foot in the sine of Latitude lay the Thred to ND and in the degrees you shall have the Altitude at East and West required Which Vertical Altitude in Summer or Northern Declinations you must substract out of the Suns present Altitude or take the lesser from the greater to find a difference but in Winter you must add this depression in the Vertical Circle to the Suns present Altitude to get a sum which must be done on a Line of Natural Sines or by the TABLE of Natural Sines as before in the Hour by laying it over or under the Center and taking from that noted Point to the Suns present Altitude all that day Then take the distance from the Center to the Tangent of the Suns present Altitude on the loose-piece which is the Secant of the Suns present Altitude and lay it from the Center on the Line of Sines and note the place then take the distance from 60 on the loose-piece to the co-tangent of the Latitude by counting 10 20 30 c. from 60 toward the moveable-leg between your Compasses then setting one Point on the Secant of the Suns Altitude last found and noted on the Line of Sines and with the other lay the Thred to the nearest distance and there keep it by noting what degree day of the month or hour minut or Azimuth it cuts Then take the distance on the Sines from the sine of the Suns Vertical Altitude to his present Altitude for a difference in Summer Or The distance from a Point made beyond the Center equal to the sine of the Suns Vertical depression to the Suns present Altitude for a sum in Winter Then having this distance of sum or difference for Winter or Summer between your Compasses carry one Point parallelly on the Line of Sines till the other being turned about shall just touch the Thred at the ND the place where the Point stayeth shall be the Azimuth from East or West as it is figured from the Center or from North or South counting from 90. Which work in brief may be sufficiently worded thus As co-tangent
which shall be the versed Sine of the Hour from the North Meridian or mid-night Or If you take the distance from the difference to the Co-altitude and carry that = till it stay in like sines it shall be the hour from noon counting the Center 12 at noon the middle at 90 the two sixes and 180 at the end for 12 at night Use XXXV Having the Latitude the Suns Altitude and Distance from the Elevated Pole to find his true Azimuth from South or North by Natural versed Sines First Of the Co-altitude and Co-latitude find the sum and difference by Addition and Substraction Secondly Count the sum and difference from the Center and take the distance between them with Compasses on the versed Sines Thirdly Make it a = versed sine of 180 and so keep the Sector Fourthly Take the distance between the sum and the Suns distance from the Pole counting the Center the Elevated Pole and 90 the Equinoctial and carry it = till it stay in like parts which shall be the Azimuth from South Or If you take the distance from the difference to the Suns distance from the Pole and carry it as before it shall stay at the versed sine of the Azimuth from the North part of the Horizon These five general wayes of finding the Hour and Azimuth are not all needful to be learned by every one but to delight the ingenious and to hold forth the usefulness of the Instrument and to supply defects that at some times may happen by Excursions and as a four-fold Testimony to shew the harmony in several wayes of Operation the first particular way and this last by versed Sines being most easie and comprehensive of any other Use XXXVI To work the last without the Line of versed Sines Note That if for want of room the versed Sines be set but on one Leg then it is to be laid at the nearest distance instead of like parts after the manner of using the Thred on the General Quadrant Also If you have it not at all then the Azimuth-line for the particular Latitude and if that be too large the little Line of Sines beyond the Center will supply this defect very well thus First Turn the Radius or whole length of that Line of Sines two times from the Center downwards which in Sea-Instruments will most conveniently stay at 30 on the large Line of Sines or general Scale as was hinted in the 28th Use being just 4 times as much one as the other For a Point representing 180 of versed Sines to set the Compasses in when you lay the Thred to ND and to take any versed sine above 90 degrees this being premised the Operation is thus Example Lat. 51-32 ☉ Dist. from Pole 80 ☉ Alt. 25 to find the Hour The sum of Co-lat 38-28 and 80 is 118-28 And the difference is 41-32 Now in regard the sum is above 90 count the Center 90 10 on the smaller Sines 100 and 20 on the same Sines 110 and 28 deg 28 min 118 deg 28 min. turn this distance the other way from the Center downwards and note that place for the Point representing the sum on the versed sines Then The Extent between this sum and the difference 41-32 as the smaller figures reckon it being taken between your Compasses set one Point in 180 the Point first found and lay the Thred to ND and there keep it or observe where it cuts then taking the distance between the versed sine of the difference counted as the small figures are reckoned and the sine of the Suns Altitude 25 as the greater figures are reckoned from the Center toward the End and carrying this Extent parallelly along the greater Line of sines till the other Point will but just touch the Thred at ND Then I say the measure from that Point to the Center measured on the small sines as versed sines shall be the versed sine of the Hour required viz. 62 from South or 7 hours 52 minuts from mid-night This Rule or Use is longer far in wording than the Operation need be in working for if you shall approve of this way the adding of two brass Center-pins will shew you the two Points most used very readily and the Thred is sooner laid than the Legs can be opened or shut and the Instrument keeps its Trianguler form as it is in during the time of Observation Use XXXVII Having the Latitude Suns Declination and Hour to find his Altitude This Problem being not of such use as the contrary viz. having the Altitude to 〈◊〉 the hour it shall suffice to hint only two â—ayes the most convenient And First by the Particular Quadrant Lay the Thred to the Day or Declination then the ND from the Hour to the Thred measured in the particular Scale of Altitudes shall shew the Suns Altitude required Secondly by the versed Sines 1. First of the Co-latitude and Suns distance from the Pole find the sum and difference 2. Take the distance between them and make it a = versed sine of 180 by setting the Sector or laying the Thred to ND 3. Then take the = versed sine of the Hour and lay it latterally from the sum and it shall give the complement of the Altitude required This work is the same both by the Sector or General Quadrant as is shewed in Use â—he 36th and is nothing else but a backward working but the Altitude at any Azimuth is not so to be done To do the same by the Natural-Sines First having the Latitude and the Suns Declination find the Suns Altitude or Depression at 6 and note the Point either below or above or in the Center as is largely shewed in Use the 26th where the Altitude is given to find the hour in any Latitude Then Lay the Thred to the Hour counted in the degrees either from 12 or 6 Then Take the ND from the Co-sine of the Suns Declination and make it a = in the sine of 90 laying the Thred to the ND then the ND from the sine complement of the Latitude to the Thred shall reach from the noted Point for the Suns Altitude or Depression at 6 to the Suns Altitude required Example Latitude 51-32 Declination 23-71 a 8 or 4 viz. 2 hours from 6 Southwards the Altitude will be found to be 36-42′ 1. For the Altitude at 6 at any time of the year say As the sine of 90 to sine of the Latitude So is the sine of the Suns Declination to the sine of the Suns Altitude at 6. 2. For the Suns Altitude at any hour or quarter in Aries or Libra the Equinoctial As sine 90 to Co-sine Latitude So is the sine of the Suns distance from 6 in degrees to sine of the Suns Altitude 3. For the Suns Altitude at all other hours or times of the year As sine 90 to Co-tangent Latitude So is sine of the Suns distance from 6 to the Tangent of a 4th Ark in the Tangents Which
place the Change-day and open the other to the beginning or the end of the Line of 24 hours Then The same Extent applied the contrary way from 6 hours or 7 dayes and a half the Moons Age shall give 28 deg 58 min. ♈ to which you must add 7 degrees and 30 minuts the Suns place between and the sum shall be the Moons true place required viz. 6-28 degrees in ♉ Example If the Moon Change on the 8th day the First Quarter being 7 dayes and a half after will be on the 15th day later at night then the difference between the Sun and Moons Right Ascention will be found to be near 6 hours for the Suns Right Ascention Ianuary 15 is 8-32 and the Moons Right Ascention the same day being about 8 degrees and a half in ♉ is 2 hours and 28 minuts if you take the distance between them on the 24 hours it is near 6 hours which is the difference of time between the Moon and the Suns hour Again For the Full Moon on the 22 day near 4 hours after noon the Moons Age being 14 dayes ¾ if you add 12 hours or 6 signs to the Moons place aâ— the Change you shall find ♋ 29-0 to which if you add 14-45 the dayes between the New and Full you shall find ♌ 13 deg 45 min. for the Moons place the Suns Right Ascention the 22 day is 9 hours and the Moons the same day at 1 afternoon is 9 hours also or rather 12 difference so that the Suns hour and the Moons is equal only one is North and the other South Again For the Last Quarter 22 ¼ dayes or 18 hours added and 22 degrees also together makes â™ 22 deg 11 min. for the Moons place by help of which to find the Moons hour by her Altitude above the Horizon found by observation Or Without regarding the Sun or Moons Right Ascention having her true Age and Hour Say thus As 12 on the Line of 24 hours is to the Moons Age in the Line of her Age So is the Moons hour to the true hour For The Extent from 12 in the middle to the Moons Age under or over the middle shall reach the same way on the same Line from the Moons hour to the true hour The like work serves to find the hour of the night by any Planets as Saturn Mars or Iupiter which are seen to shine very brave and bright in Winter evenings and having learned their Place by their distance from the fixed Stars or by the Ephemerides then their Altitude and Place will find their hour from the Meridian and the comparing their Right Ascentions with the Suns gives the true hour as before in the Fixed-Stars Use XLV To find the Moons Place and Declination without the Ephemerides somewhat near First observe when the Moon is in the Meridian and then find her Altitude and take the same from the particular Scale between your Compasses then set one Point in the hour 12 and lay the Thred to ND and on the degrees it shall shew the Moons declination and in the Line of the Suns Place the Moons present Place counting her Progress orderly from the last Change-day or New Moon when she was with the Sun Otherwise thus Observe what Hour the Moon sheweth on any Sun-dial at the same instance by the Fixed Stars or other wayes find the true Hour Then The Extent from the Moons Hour to the the true Hour shall reach the same way from 12 to the Moons Age right against which is her coming to South at which time you may find her true Altitude and so come by her Declination Yet again for her Age and Place according to Mr. Street and Mr. Blundevil Add the Epact the Month and Day of Month in one sum counting the Months from March by calling March the first Month April the second c. then that sum if under 30 is the Moons Age but if the sum be above 30 then substract 30 and the remainder is the Moons Age when the Month hath 31 dayes but if the Month hath but 30 or less than 30 dayes then substract but 29 and the remainder is the Moons Age. Or thus Add to the Epact for the present year and in Ianuary 0 in February 2 in March 1 in April 2 in May 3 in Iune 4 in Iuly 5 in August 6 in September 8 in October 8 in November 10 in December 10 and the sum if under 30 or the excess above 30 added to the day of the Month abating 30 if need be gives the Moons Age that day but substracted from 30 leaves the day of her Change in that Month or from the 〈…〉 â—onth Example July 10. 1668. The Epact that year is 26 and the Number for Iuly is 5 the Excess above 30 is 1 which added to any day of the Month as to 10 gives 11 for the Moons Age Iuly 10. 1668. Then for the Moons Place Multiply the Moons Age by 4 and the Product divided by 10 the Quotient giveth the signs and the remainder multiplied by 3 gives the degrees which you must add to the Suns place that day to find out the Moons place for that day of her Age. Example On Iuly 10. 1668 the Moons Age is 11 which multiplied by 4 makes 44 and 44 divided by 10 gives 4 signs in the Quotient and 4 the remainder multiplied by 3 makes 12 degrees more which added to Cancer 29 degrees the Suns place on the 10th day of Iuly makes 11 degrees in Sagittarius the Moons place the same day propè verum Or rather by the Rule thus on the Line of 24 hours by particular Scale having the Moons place to find her Age by the Line of 24 hours The Extent from the Suns true place to the Moons true place shall reach the same way from 0 day to the day of her Age. Or contrarily having the Moons true Age to find her true Place The Extent from 0 day old to the Moons true Age shall reach the same way from the Suns true Place to the Moons Or having the Moons true Place at the New Moon to find her Place any day of her Age after The Extent from ♈ to the Moons true Place at the Change shall reach the same way from the day of her true Age to her true Place adding as many degrees to the Number found as the Moon is dayes old Then Having her Place and Age it is easie to find the Moons Hour and then her true Hour but I fear I spend herein too much time on an uncertain subject Use XLVI The Right Ascention and Declination of any Star with the Suns Right Ascention and the Hour of the Night given to find the Altitude and Azimuth of that Star and thereby to know the Star if you knew it not before Set one Point of the Compasses in the Stars Right Ascention found in the Line of twice 12 hours and open the other to the Suns Right Ascention found in the same Line then this
Extent shall reach in the same Line from the true hour of the Night to the Stars hour from the Meridian then laying the Thred to the Stars Declination the ND from the Stars hour in the Line of hours to the Thred measured on the particular Scale of Altitudes gives the Stars Altitude then by his Declination and Altitude you may soon find his Azimuth by Use 27. And if the Instrument be neatly fixed to a Foot to set North and South and turn to any Azimuth and Altitude you may find any Star at any time convenient and visible Use XLVII The Altitude and Azimuth of any Star being given to find his Declination Lay the Thred to the Altitude on the degrees counted from 600 toward the end then setting one Point on the Stars Azimuth counted in the Azimuth Line and take the ND from thence to the Thred which distance measured from the beginning of the particular Scale of Altitudes shall give the Declination If the Compasses stand on the right-side of the Thred then the Declination is North if on the left it is South according as you work for the Suns Azimuth in a particular Latitude Use XLVIII The Altitude and Declination of any Star with the Right Ascention of the Sun and the true Hour of the Night given to find the Right Ascention of that Star First by the 43d Use find the Stars Hour viz. How many hours and minuts it wants of coming to or is past the Meridian then the Extent of the Compasses on the Line of 24 hours on the Head-leg from the Stars hour to the true hour shall reach the same way from the Suns Right Ascention to the Stars Right Ascention on the Line of twice 12 or 24 hours Use XLIX To find when any Fixed-Star cometh to South by the Line of twice 12 or 24 hours In Use 42 Section 4 you have the way by Substraction with its Cautions But by the Line of twice 12 or 24 hours work thus Count the Suns Right Ascention on that Line and take the distance from thence to the next 12 backward viz. that at ♈ at the beginning of the Line when the Suns Right Ascention is under 12 hours or to the next 12 in the middle of the Rule at ♎ when the Suns Right Ascention is above 12 hours which is nothing but a rejecting 12 for more conveniency Then The same Extent laid the same way from the Stars Right Ascention shall reach to the Stars coming to South Or The Extent from the Sun to the Stars Right Ascention shall reach the same way from 12 to the Stars coming to South Example for the Lyons-Heart August 20. The Suns Right Ascention the 20th of August is 10 hours 36 minuts the Right Ascention of the Lions-Heart is 9 hours and 50 min. Therefore The Extent from 10 hours 35 min. to the beginning shall reach the same way from 9 hours 50 min. by borrowing 12 hours because the Suns Right Ascention is more than the Stars to 11 hours 13 min. of the next day viz. at a quarter past 11 or at 11 hours and 13 min. the same day where you may observe that the remainder being above 12 if you add 24 hours the time of Southing is between mid-night and mid-day next following Use L. To find what two dayes in the year are of equal length and the Suns Rising and Setting Lay the Thred on any one day in the upper Line of Months and Dayes and at the same time the Thred cuts in the lower-Line of Months the day that is answerable to it in length rising setting and declination and other requisites Example The 1st of April and the 21 of August are dayes of equal length and the Suns Rising and Setting is the same on both those dayes only in the upper-Line the dayes are increasing in length and in the lower-Line they are decreasing Use LI. To find how many degrees the Sun is under the Horizon at any Hour the Declination and Hour being given Count the Suns Declination on the degrees the contrary way viz. for North Declination count from 600 toward the end and count for Southern Declination toward the Head and thereunto lay the Thred then take the nearest distance from the hour given to the Thred this distance measured in the particular Scale of Altitudes shall shew the Suns Depression under the Horizon at that hour Example January the 10th at 8 at Night how many degrees is the Sun under the Horizon On that Day and Hour the Suns Declination is about 20 degrees South then if I lay the Thred to 20 degrees of Declination North and take the nearest distance from 8 to the Thred that distance I say measured in the particular Scale gives 34 degrees and 9 min. for the Suns Depression under the Horizon of 8 afternoon To do this in other Latitudes you are to find the Suns Altitude at 8 in Northern Declination by Use 37. CHAP. XVII The use of the Trianguler Quadrant in finding of Heights and Distances accessable or inaccessable Use I. To find an Altitude at one Station FIrst The Trianguler Quadrant being rectified and fixed to a Ball and Socket and three-legged-staff being necessary in these Operations to perform them exactly especially for Distances look up to the object as you would to a Star and observe what degree and minut the Thred cuts and set it down Also observe the place where you stand at the time of Observation and the distance from your Eye to the ground and the place on the object that is level with your eye also as the playing of the Thred and Plummet will plainly shew Also you must have the measure from the place where you stood observing to the Point exactly right under the object whose height you would have in Feet Yards Perch or what you please to Integers and Fractions in Decimals if it may be Also Note That in all Right-Angle-Triangles one Acute Angle is alwayes the complement of the other so that observing or finding one by Observation by consequence you have the other by taking that from 90. These things being premised the Operation followes by the Artificial Numbers Sines and Tangents and also by the Natural Note also by the way That in regard the complement of the Angle observed is frequently used if you count the degrees the contrary way that is to say from the Head you shall have the complement required as hath been oftentimes hinted before Then As the sine of the Angle opposite to the measured side is to the measured side counted on the Numbers So is the sine of the Angle found to the Altitude or Height required on Numbers Example at one station Standing at C I look up to B the object whose Height is required and I find the Thred to fall on 41 degrees and 45 minuts but if you count from the Head it is 48-15 the complement thereof as in the Figure you see Also the measure from C to A is found to be
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
Distance Let CA be the Altitude given and AB the distance required Then I standing at C observe the Angle CAB by setting the end of the Head-leg to my eye and the Head-end downwards and set down as the Thread cuts numbring both wayes for the Angle at C and at B his complement Then say As the Angle at B 30 deg 40 minutes counted on the Sines to 105 the height of the Tower So is 59 deg 20 min. the Angle at C on the Sines to 176 the distance required on the Numbers Also note by the way That if you take an Altitude at two stations as suppose at E and at B if the Angle observed at B be found to be the half of the Angle at E as here in Figure VIII the Angle at E being 61-20 and the Angle at B 30-40 the just half thereof then I say that the distance between the two stations is equal to the Hypothenusa EC at the first station viz. EB is equal to EC which being observed say As the sine of 90 to 120 on the Numbers So is 61-20 on the Sines to 105 the height required on the Numbers A further proof hereof take in this following Figure IX Let AB be a breadth of a Wall or Fort not to be approached unto then by the degrees on the in-side of the loose-piece to find that breadth one way is thus Put two Pins into the two holes in the Head and Moving-leg or set the sights there in large Instruments then move nearer or further from the objects till your eye fixed at the rectifying Point can but just see the marks A and B by the two Pins in each Leg which will only be at the mark C at an Angle of 60 degrees for so the Rule is made to that Angle then the Instrument being still fixed at C look backward in a right Line from the middle of the loose-piece and rectifying Point toward D putting up a mark either in or over or beyond the Point D and also be sure to leave a mark at C the first place of observation Then remove the sights to 15 degrees the half of 30 counting from the middle and go back in a right Line from C toward D till you can just see the marks by the two sights set at 15 degrees each way for then I say that the measure between the two stations C and D shall be exactly equal both to AB the breadth required and also to CB or CA the Hypothenusaes then having the sides CB and CD and the Angles BCE and CBE and BDC it is easie to find all the other Sides and Angles by the Rules before rehearsed by the Lines of Artificial Nmmbers and Sines For As the Sine of 15 degrees the Angle at D viz. BDC to 108 on the Numbers So also is the Angle at B viz. DBC 15 to 108 on the Sines and Numbers So also is the Sine of 150 the Angle at C viz. DCB to BD 208 ½ on the Numbers Note also That if the Angles of 60 and 30 be inconvenient then you may make use of 52 and 26 or 48 and 24 or 40 and 20 or any other and the half thereof and then the measured distance and the Hypothenusa BC at the nearest station will alwayes be equal but not equal to the breadth at any other Angle except 30 and 60 as in the Figure But having the Angles and those Sides you may soon find all the others by the Artificial Numbers Sines and Tangents by the former directions The End of the First Part. The Table or Index of the things contained in this Book TRianguler Quadrant why so called Page 2 The Lines on the ouâ—ter-Edg N. T. S. VS Page 2 The Line on the inner-edge I. F. 112 Page 3 The Lines on the Sector-side L.S.T. Sec. Page 3 Lesser Sines Tangents and Secants Page 5 The Lines on the Quadrant-side Page 6 The 180 degrees of a semi-Circle variously accounted as use and occasion requires Page 7 60 Degrees on the Loose-piece as a fore-Staff for Sea-Observation Page 7 The Line of right Ascentions Page 8 The Line of the Suns true Place ibid. The Months and Dayes ibid. The Hour and Azimuth-line for a Particuler Latitude Page 9 Natural versed Sines ibid. Lines and Sines or the general Scale of Altitudes for all Latitudes Page 10 The particular Scale of Altitudes or Sines for one Latitude only Page 11 The Degreees of a whole Circle 12 Signs 12 Inches or 24 Hours and Moons Age ibid. The Appurtenances to this Instrument ibid. Numeration on Decimal-lines Page 12 Three Examples thereof Page 13 Numeration on Sexagenary Circular-lines with Examples thereof Page 17 How Right Sines Versed Sines and Chords are counted on the Rule Page 20 Of a Circle Diameter Chord Right Sine Sine Complement or Co-sine Versed Sine Tangent Secant what it is Page 23 24 Two good Notes or Observations Page 25 Of the division of a Circle ibid. What a Radius is Page 26 What an Angle a Triangle Acute Right or Obtuse Plain or spherical Angle is Page 26 27 Parallel-lines and Perpendiculer-lines what they are ibid. The usual Names of Triangles ibid. Of four sided Figures and many sided Page 28 Terms in Arithmetick as Multiplicator Product Quotient c. what they mean Page 29 Geometrical Propositions Page 31 To draw a right Line ibid. To raise a Perpendiculer on any line ibid. To let fall a Perpendiculer any where Page 33 To draw Parallel-lines Page 34 To make one Angle equal to another Page 35 To divide a Line into any number of parts ib. To bring any 3 Points into a Circle Page 36 To cut any two Points in a Circle and the Circle into two equal parts Page 37 A Segment of a Circle given to find the Center and Diameter Page 38 A Segment of a Circle given to find the length of the Arch Page 39 To draw a Helical-line and to find the Centers of the Splayes of Eliptical arches and Key-stones Page 41 To draw an Oval ibid. Explanation of Terms particularly belonging to this Instrument Radius how taken Page 41 Right Sines how taken and counted Page 42 Tangents Secants and Chords how taken ib. Sine complement or co-sine Tangent complement or co-tangent how taken and counted on this Instrument ibid. Latteral Sine and Tangent Page 43 Parallel Sine and Tangent ibid. Nearest Distance what it means ibid. Addition on Lines ibid. Substraction on Lines Page 44 Of Terms used in Dialling Plain and Pole of the Plain Page 45 Declination Reclination and Inclination of a Plain what it is Page 46 What the Perpendiculer-line and Horizontal-line of a Plain are ibid. Meridian-line Substile-line and Stile-line Angle of 12 and 6 and the Inclination of Meridians what they are Page 47 Parallels and Contingent-lines what Page 48 Vertical-line and Point what ibid. Nodus Apex and foot of the Stile what ibid. Axis of the Horizon what Page 49 Erect Direct what ibid. Declining Reclining or Inclining