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
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-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-parallel-Line required Or thus Take BC the measure from the Point that is to cut the parallel-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-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
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
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
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-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
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
a Foot shut are drawn usually just 5 degrees assunder or rather the two innermost Lines on each Leg are always just one degree from the inside so that if you put a Center-pin in the Line of Tangents just against the Sine of 30 it makes the two innermost Lines that come from the Center just 2 degrees assunder which is easie to remember either in adding or substracting as followeth two wayes 1. Take the Latteral Sine of 30 viz. the measure from the Center to 30 the Compasses so set set one Point in the Center-pin in the Tangents just against 30 and turn the other till it cut the common Line in the Line of Sines on the other Leg and there it shall shew what Angle the two innermost-Lines make counting from the end toward the Head and two degrees less is the Angle the Sector stands at both on the in-side and out-side the Legs being parallel which Number must nearly agree with what the in-side of the Leg cuts on the Head-semicircle or there is a mistake As thus for Example Suppose I open the Rule at all adventures and taking the Latteral Sine of 30 from the Sines on the Sector-side and putting one Point of the Compass in the Center on the Tangents right against the Sine of 30 on the other Leg or the beginning of the Secants on the same Leg and turning the other Point to the Line of Sines on the other Leg it cuts the Sine of 60 on the innermost Line that comes from the Center then I say that the Lines of Sines and Tangents are just 30 degrees assunder and the in-side or out-side of the Legs but 28 viz. two degrees less as a glance of your eye to the Head will plainly shew 2. This way will serve very well for all Angles above 20 and under 80 But for all under 20 and above 80 to 120 this is a better way Open the Rule to any Angle at pleasure and take the distance parallelly that is across from one Leg to the other between the Center-pin at 30 in the Sines and that in the Tangents right against it and measure it latterally from the Center and it shall shew the Sine of half the Angle the Sines and Tangents stand at and one degree less is the Sine of half the Angle the Sector stands at Example Suppose that opening the Sector at adventures or to the Level of any thing I would know the Angle it stands at I take the parallel Distance between the two Centers and measuring it latterally from the Center I find it gives me the Sine of 51 degrees viz. the half Angle the Lines stand at or 50 the Angle the Rule stands at which doubled is 102 for the Lines or 100 for the Legs of the Sector as a glance of the eye presently resolves by the inner-edge of the Moving-leg and the divided semi-circle 3. On the contrary Would you set the Legs or Lines to any Angle take the half thereof latterally or one degree less in the half for the Legs and make it a Parallel in the two Centers and the Sector is so set accordingly Example I would set the Legs to 90 degrees or a just Square take out the Latteral Sine of 44 one degree less than 45 the half of 90 and make it a Parallel in the two Centers abovesaid and you shall find the Legs set just to a Square or Right-Angle as by looking to the Head you may nearly see At the same time if you take Latteral 30 and lay it from the Center according to the first Rule you shall see a great deficiency therein as above is hinted Use V. The Day of the Month being given to find the Suns Declination true place in the Zodiack Right Ascention Ascentional Difference or Rising and Setting 1. Lay the Thred to the Day of the Month in the upper Line of Months where the length of the Dayes are increasing or in the lower-Line when the Dayes are decreasing according to the time of the year then in the Line of degrees you have his Declination wherein note that if the Thred lie on the right hand of 600 then the Suns Declination is Northwards the contrary-way is Southwards Also on the Line of the Sun 's Right Ascention you have his Right Ascention in degrees and hours counting one Hour for 15 degrees as the Months proceed from March the 10th or Equinoctial the Right Ascention being then 00 and so forward to 24 hours or 360 degrees as the Months and Dayes proceed Again on the Line o◠the Sun 's true place you have the sign and degree of his place in the Ecliptick Aries or the Equinoctial-point being the place to begin and then proceeding forward as the Months and Dayes go Lastly on the Hour-line you have the Ascentional-difference in degrees and minutes counting from 6 or the Suns Rising counting as the morning hours proceed or his Setting counting as the afternoon hours proceed Of all which take two or three Examples 1. For March the 12th lay the Thred to the Day and extend it streight then on the Line of degrees it sheweth near 1 degree or 54 minutes Northward 2. The Suns Right Ascention is in time 8 minutes and better or in degrees 2 deg 5 minutes 3. The Suns Place is 2 degrees and 16 minutes in Aries ♈ 4. The Ascentional Difference is 1 degree and 10 minutes or the Sun riseth 4 minutes before and sets 4 minutes after 6. Again for May the 10th the Thred laid thereon cuts in the degrees 20 deg 9 min. for Northern Declination and 57 deg 24 min. or 3 hours 52 min. Right Ascention and 29 37 in ♉ Taurus for his true place and 27 12 for difference of Ascentions or riseth 11 minutes after 4 and sets 49 minutes after 7. Again on the last of October or the 21 of Ianuary near the Declination is 17 22 Southwards the Right Ascention for October 31 is 225 53 for Ianuary 21 314 21 The true place for October 31 is ♠Scorpio 18 deg 22 min but for Ianuary 21 ♒ Aquarius 11 52 according as the Months go to the end at ♑ and then back again but the Ascentional difference and Rising and Setting is very near the same at both times viz. 23 10 and Riseth 32 minutes and more after 7 and Sets 28 minutes less after 4. Use VI. The Declination of the Sun or a Star given to find his Amplitude Take the Declination being counted on the particular Scale of Altitudes between your Compasses and with this distance set one foot in 90 on the Azimuth-Line the other Point applied to the same Line shall give the Amplitude counting from 90. Example The Declination being 12 North the Amplitude is 19 deg 15 min. Northwards Or the Declination being 20 South the Amplitude is 34 deg 10 min. Southwards Use VII The Right Ascention and Ascentional-difference being given to find his Oblique-Ascention When the Declination is North then the difference between the
from A but if you remove to E the Thred falling on 3 on the Shadows the Altitude will be but one third part of the distance EA From hence you may observe that observing at C and at D where the Thred falls on 1 and on 2 the distance between C and D is equal to the Altitude so likewise at D and at E and so by consequence at 1 ½ and 2 ½ and 3 ½ or any other equal parts This is an excellent easie way The like will be if you observe at D and C looking up to F where the Altitude AF is twice the distance AC Use V. Another way by the Line of Shadows at one station Measure any distance as feet yards or the like from any object as suppose from A to D were 200 foot and looking up to B the Thred cuts the stroke by 2 on the Line of Shadows Then by the Line of Numbers say As 2 the parts cut is to 1 So is 200 the distance measured to 100 the height Or Suppose I measured any other uneven Number and the Thred fall between 00 on the Loose-piece and 1 on the Shadows commonly called contrary Shadow The Rule is alwayes thus As the parts cut by the Thred are to 1 So is the measured distance to the height required being less than the measured distance But when the Thred falls between 1 and 90 at the Head called right Shadow then the Rule goes thus As 1 to the parts cut by the Thred So is the measured distance to the height being alwayes more than the measured distance from the foot of the object to the station Use VI. Another way by the Line of Shadows and the Sun shining When the Sun shineth find his Altitude and also as the Thred lies see what division on the Line of Shadows is cut by the Thred and then straightway measure the shadows length on the ground and if the Sun be under 45 degrees high the shadow is longer than the length of that object which causeth the shadow but if the Sun be above 45 degrees high then the object is longer than the shadow and the Operation is thus by the Line of Numbers only with a pair of Compasses The Height of the Sun being under 45 say As the parts cut by the Thred on the Shadows is to 1 So is the Shadow measured to the height required The Height of the Sun being above 45 say As 1 to the parts cut by the Thred on the Line of Shadows So is the measure of the shadow to the height in the same parts Use VII To find an inaccessable Altitude by the Quadrat and Shadows otherwise Observe the Altitude at both stations and count the observed Altitudes at both stations on the Quadrat or Shadows according as it happens to be either above or under 45 degrees and take the lesser out of the greater noting the remainder for the first term and the Divisor to divide the distance between the stations increased with Cyphers if need be and the Quotient is the Answer required But by the Line of Numbers work thus The Extent from the difference to 1 shall reach the same way from the measured distance to the height required Example Figure II. Let ABCDE represent the Object and three Stations let the Line AC represent the Altitude the Point B one station 50 foot from A D another station 100 foot from A or 50 from B and E another station 73 foot from D or 173 foot from A all which measures you need not know before but only BD and DE Also the Angle at B 63-27 and his complement counting the other way being the Angle at C 26 degrees 33 minuts the Angle at D 45 and his complement so also the Angle at E 30 and his complement 60. Now mind the Operation by either of these First lay the Thred on 26-33 and in the Quadrat it cuts 50 lay the Thred on 45 and in the Shadows or Quadrat it cuts 100 or 1 or if you lay the Thred to 60 then in the shadows it cuts 173. The difference between 173 and 100 is 73. Then As 73 the difference in Tangents between the two observations is to the distance in feet 73 So is Radius 100 or the side of the Quadrat to 100 the hight required Again for the two nearest Observations whose difference of Tangents is 50. As 50 the difference in Tangents to 50 foot the measured distance So is 100 the side of the Quadrat to 100 the height Again lastly by the observations at B E the difference of Tangents being 123. As 123 the difference in Tangents to 123 the measured distance So is 100 the Radius or side of the Quadrat to 100 the height required Or In the first Figure the Angles at the top being 33-30 and 48-15 and the measured distance 88 foot and a half the difference in Tangents will be 45-8 Then As 45-8 to 100 the side of the Quadrat So is 88 ½ the measured distance to 194 the Altitude required This way is general for any Station though both of right shadow or both of contrary or mixt of right and contrary and done by the Line of Numbers or by Multiplication and Division Also Note That you may find this difference in Tangents or Secants by the Natural Tangents or Natural Secants on the Sector and the Scale of equal parts belonging to them Thus Take the distance between the compleplement of the two observations on the greater or lesser Line of Tangents as is most convenient and measure this distance in the Line of Lines or equal parts equal to that Radius and that shall be the difference in Tangents required The like for the Secants Also By the Artificial Numbers Sines and Tangents you may come by this differences in Tangents or Secants very well thus Just right against the Tangent of the Co-altitude counted on the Line of Tangents in the Line of Numbers is one Number and against the Tangent of the complement of the other Angle is the other Number only with this Caution That if the Tangent be above 45 then take the distance from 45 to the Tangent as it is counted backward with Compasses and set the same the increasing way from 1 on the Numbers to the other Number required then the lesser taken from the greater leaves the difference in Tangents that was required In the same manner the Sines counted from 90 and laid the contrary way from 1 increasing will give the difference in Secants to measure the Baâ—â— and Hypothenusa by Numbers only Use VIII Another pretty way by Scale and Compass without Arithmetick from T. S. Then draw the Line that the Thred maketh on the Board Then measure from your standing to the foot of the Object and take the number of feet or yards from any Scale and lay it from the right Angle on the other Line and raise a Perpendiculer from thence to the plumb-Plumb-line made by the Thred and
that shall be the Altitude required being measured on the same Scale Example Let ABGD represent the Boards end or Trencher and on that let AB be one streight Line and AG another Perpendiculer to it in the Point A knock in one Pin and in B or any where toward the end another On the Pin at A hang a Thred and Plummet and standing at I any convenient station look up by the two Pins at B and A till they bourn in a right Line with the Point H the object whose height is to be measured then the Plummet playing well and even make a Point just therein and draw the Line AD as the Thred shewed Then having measured the distance from G the foot of the Object to I the station take it from any first Scale and lay it from A to G then on the Point G raise a Perpendiculer to AG till it intersect the Plumb-line AD then I say the distance CD measured on the same Scale you took AC from shall be equal to the Altitude GH which was required Use IX The same work at two stations But if you cannot come to measure from I the first station to G then measure from I to K and having observed at I and drawn the Plumb-line AD take the measure between I and K the two stations from any fit Scale of equal parts and lay it on the Line AC from A to C viz. 79 parts and in the Point C knock another Pin and hang the Thred and Plummet thereon and observe carefully where this last Plumb-line doth cross the other as suppose at E then from E let fall a Perpendiculer to the Line AC which Line AC shall be the height GH required or thus the nearest distance from E to AC is the height required viz. 120 of the same parts that IK is 79 Note the Figure and behold that ACFE the small Figure on the Board is like and proportional to AA GH the greater Figure Other wayes there be as by a Bowl of Water or a Glass or a Plash of Water or a Square but these set down are as convenient and ready as any whatsoever As in the next Figure you may see the way by the Glass and Square As thus Let C represent a Glass a Bowl or Plash of Water wherein the Eye at A sees the picture or reflection of the Object E. Then by the Line of Numbers As CB the measure from your foot to the Glass is to AB the height from your eye to the ground at your foot So is the measure from C to D to the height DE. See Figure VI. Again to find a distance by the Square that is not over-long Let C represent the upper-corner of a Square hung on a staff at F then the one part of the Square directed to E and the other to A. The Proportion will hold by the Line of Numbers As FA 11-37 to FC 50 So is FC 50 to FE 220. That is So many times as you find AF in FC So many times is FC in FE and the like Note That you must conceive AFE to be the Ground or Base-line in this Operation by the Square C being the top of an upright Staff 5 foot long called 50 for Fraction sake Use X. To find a Distance not approachable by the Trianguler Quadrant First I plant my Trianguler Quadrant set upon a three legged Staff and Ball socket right over the place A and then bring the Index with two sights in it laid or fastened to the Center of the Trianguler Quadrant right over the Lines of Sines and Lines cutting 90 at the Head the Index and sights so placed hold it there and bring it and the Instrument together till you see the mark at C through the two sights by help of the Ball-socket and then there keep it then remove the Index only to 0-60 on the loose-piece which makes a right Angle and set up a mark in that Line at any convenient distance as suppose at B 102 foot from A then remove the Instrument to B and laying the Index on the Center and 0-60 on the loose-piece direct the sights to A the first station by help of a mark left there on purpose Then remove the sights till you see the mark at C and note exactly on what degree the Index falleth as here on 60 counting from 060 on the loose-piece or on 30 counting from the Head which is the Angles at B and at C. Then by the Artificial Numbers Sines and Tangents on the edge say As the sine of 30 the Angle at C to 102 the measured distance counted on the Numbers So is the sine of 60 the Angle at B to 117 on the Numbers the distance required So also is 90 the Angle at A to 206 the distance from B to C. Or by the Lines and Sines on the Quadrant-side as it lies thus As the measure of 102 taken from any Scale as the Line of Lines doubling to the = sine of 30 laying the Index or a Thred to the nearest distance So is the = sine of 60 to 117 measured latterally on the same Line of Lines And So is the = sine of 90 to 206 the distance from B to C. So also If you observe at B and at D only you must be sure to set your Instrument at one station at the same scituation as at the other as a looking back from station to station will do it and the same way of work will serve For As the Sine of 20 to 110 So is the Sine of 40 to 206. And So is the Sine of 120 to the Line DC 278 c. Use XI To find a Breadth and a Distance at any two Stations Let AB be two marks as two corners of a House or Wall and let the breadth between them be demanded and their distance from C and D the two stations First set up two marks at the two stations then setting up the Instrument at C set the fiducial Line on the Rule to D the other mark then direct the sights exactly to B and to A observe the Angles DCB 45 and DCA 113-0 as in the Figure Secondly Remove the Instrument to D the other station and set the fiducial-Line of the Quadrant viz. the Line of Lines and Sines directly to C then fix it there and remove the Index and sights to A and to B to get the Angles CDA 42-30 and CDB 109-0 Then observe that the 3 Angles of every Triangle being equal to 180 degrees having got the Angles at C 113 and the Angle at D 42-30 by consequence as you take 155 the sum of the Angles at C and D out of 180 then there remains 24-30 the Angle at A. So also Taking 109 and 45 from 180 rests 26 the Angle at B then also taking 45 the Angle BCD out of 113 the Angle DCA rests 68 degrees the Angle BCA in like manner taking 42-30 from 109 the Angles at D rests