The Sea Gunner Published By John Seller a Cascable deck b Base Ring c Touchole d The Chamber e Reinfourd Ring f Trunions g Cornish Ring h Trunion Ring THE Sea-Gunner Shewing the PRACTICAL PART OF GUNNERY As it is used at SEA AND As an Introduction thereto there is Exhibited two Compendiums one of Vulgar the other of Decimal ARITHMETICK With necessary Tables relating to that ART To which is added An APPENDIX Shewing the Use of a Proportional Scale for the ready working of any Question in Gunnery And the Use of the Sea-Gunners Rule of an excellent Contrivance containing an Epitome of the Art of Gunnery in it self Composed by 〈◊〉 Sâ—â—â—â—R Senior LONDON Printed by H. Clerk for the Author and are to sold by him at the Hermitage in Wapping 1691. THE PREFACE TO THE READER Courteous Reader HAving observed for several Years that there hath been a great want of a Piece of Sea-Gunnery that has been principally adapted for Sea-Service in a Treatise by it self for those Books that were Extant were chiefly intended for Land Service and at this time most of them being out of Print I judg'd this a fit opportunity for the publishing this small Treatise hoping it will be gratefully accepted by our Sea-Gunners And in regard that those who would be Students in that Art ought in some competent measure to be acquainted with Arithmetick for the sake of such I have exhibited two Compendiums thereof one in Vulgar and the other in Decimal Arithmetick as a necessary preparation for the working those Questions that are incident to that Art And for the ease of such as are not fully acquainted therewith I have furnished them with a Proportional Scale whereby they may perform all the Operations that are useful in Gunnery as also to extract the Square and Cube-Roots and how to perform the same by Logarithms and by Gunter's-Scale To which I have added several necessary Tables useful in Gunnery with proper Questions and their Answers and useful Observations and Instructions And for the better accomplishing the Design of this Book I have consulted with the best approved Authors that have written on this Subject Also at the end of this Treatise I have presented you with a small Tract as an Appendix particularly of the use of the Proportional Scale and of the use of a Rule of a new contrivance fit for the Pocket that hath upon it an Epitome of the Practical part of Gunnery in it self which I call the Sea-Gunner's Rule All which I submit to the favourable construction of the Judicious And rest your Friend to serve you John Seller A TABLE Of the Principal Matters contained in this Book CHAP. I. Of Vulgar Arithmetick Page 1 Of Notation of Numbers ibid. Numeration and Addition 3 Subtraction 5 Multiplication 7 Division 15 The Rule of Three Direct 19 The Rule of Three Reverse 22 Double Rule of Three 23 CHAP. II. Of Decimal Arithmetick 28 Notations of Fractions ibid. Addition of Decimals 33 Subtraction of Decimals 34 Multiplication of Decimals 35 Division of Decimals 37 A Decimal Table of Pence and Farthings 46 A Table of Decimals of one pound Sterling 48 A Table of the Decimals of an English Foot to every Inch and eight parts of an Inch 49 CHAP. III. The Extraction of the Square-Root by Arithmetick 52 To extract the Sqaure-Root by Logarithms and Gunter's-Scale 56 To extract the Cube-Root by Arithmetick 57 To prepare a Cube Number for extraction 59 To extract the Cube-Root by the Logarithms 63 To extract the Cube-Root by Gunter's-Scale 64 A Table of Square-Roots 65 A Table of Cubick-Roots 66 To make the Tables of Square and Cube-Roots 67 A Table of Logarithms 8 A Description and use of the Table of Logarithms 82 Multiplication by Logarithms 86 Division by Logarithms 87 Of a Circle 88 1. The Diameter being given to find the Circumference by the Logarithms ibid. 2. The Circumference being given to find the Diameter 89 3. The Diameter of a Circle being given to find the Area or Superficial Content thereof 90 4 The Circumference being given to find the Area 91 CHAP. IV. Containing Geometrical Rudiments useful in the Art of Gunnery 93 To raise a Perpendicular from the middle of a Line given ibid. To let a Perpendicular fall from a point assigned to the middle of a Line given 94 To raise a Perpendicular upon the end of a Line given 95 To let fall a Perpendicular from a point assigned unto the end of a Line given 96 To draw a Line parrallel to a Line given 97 A Geometrical way to find the Diameter of a Bullet that weigheth twice as much as a known Bullet 98 The weight of a Shot given to find the Diameter Geometrically 100 Chap. V. Geometrical Theorems and Problems 102 Arithmetical Problems appertaining to the Art of Gunnery and wrought by Decimal Arithmetick by Logarithms and Gunter 's Scale Prob. 1. The Diameter of a Circle being given to find the Circumference ibid. Prob. 2. The Circumference of a Circle being given to find the Diameter 105 Prob. 3. The Diameter of a Circle being given to find the side of a Square equal to it 107 Prob. 4. The Circumference of a Circle being given to find the side of a Square equal in Content to that Circle 108 Prob. 5. The Diameter of any Spherical body being known to find the Circumference 109 Prob. 6. The Circumference of any Spherical body being known to find the Diameter 110 Prob. 7. The Diameter and Circumference of any Spherical body being known to find the Superficial Content 111 Prob. 8. The Axis or Diameter of a Globical body being known to find the solid Content 112 Prob. 9. The Diameter of a Bullet being given with the weight to find the weight of another Bullet of the same Metal but of another Diameter either greater or lesser 113 Prob. 10. Having the weight of a Bullet of one kind of Metal to find the weight of a Bullet of another kind of Metal being equal in Magnitude 116 Prob. 11. A Bullet of Iron that weighteth seventy two Pound what will a Bullet of Lead weigh that is equal to it in bigness 118 Prob. 12. The Diameter and Weight of any one Cylender of a Piece of great Ordnance taken at the base Ring being known to find the weight of any other Piece of the same Metal and shape either greater or lesser its Diameter being only known 119 Prob. 13. Having the Diameter and Weight of any Piece of great Ordnance to find the Weight of another Piece of Ordnance of another 〈◊〉 that is of the same shaâ—â— 121 â—â—ob 14. To find the Supâ—â—fiâ—â—l Content of the Convex face of any Piece of Ordnance and also the solid Content of the Concavity thereof 123 Prob. 15. To know how much of every kind of Metal is contained in any Brass Piece of Ordnance 126 Prob. 16. By knowing what quantity of Powder will load some one Piece of Ordnance to find how much of the same Powder will
Sum of them will be the Logarithm of the Number produced by that Multiplication Example Let it be required to Multiply 48 by 5 First set down the Two Numbers to be Multiplied One under another and to them set their respective Logarithms as in the Margin which being added together the Sum of them which is the Logarithm of the Product being sought in the Table the Absolute Number answering thereto is 240 the Product of those Two Numbers Multiplied together 48 1,68124 5 0,69897 240 2,38021 Division by the Logarithms AS Multiplication by the Logarithms was performed by Addition so Division is performed by Subtraction Wherefore to perform Division you must Subtract the Logarithm of the Number by which you are to Divide from the Logarithm of the Number which is to be Divided and the Number which remains shall be the Logarithm of the Quotient Example Let it be required to Divide 228 by 12 228.2,35793 12.1,07918 19.1,27875 First set down the Logarithm of 228 and under it set the Logarithm of 12 and Subtract the Lesser from the Greater the Remainder is the Logarithm of the Quotient which being sought in the Table you will find 19 to be the Answer of the Question being the Quotient sought And so many times is 12 contained in 228. Of a CIRCLE 1. The Diameter being given to find the Circumference by the Logarithms THe Proportion is as 7 to 22 so is the Diameter to the Circumference Wherefore to find the Circumference of any Circle whose Diameter is given Add the Logarithm of the Diameter given to the Logarithm of 22 and from the Sum of them Subtract the Logarithm of 7 the Remainder shall be the Logatithm of the Circumference sought Example If the Diameter of a Circle be 113 what is the Circumference First set down the Logarithm of 22 which is 1,34242 2,05308 3,39550 0,84510 2,55040 Add the Logarithm of 113 which is from which Subtract the Logarithm of 7 which is which being sought in the Tables is the nearest Logarithm of 355 and so much is the Circumference of a Circle whose Diameter is 113. 2. The Circumference of a Circle being given to find the Diameter The Proportion is as 22 is to 7 so is the Circumference to the Diameter Wherefore to the Logarithm of 7 add the Logarithm of the Circumference given and from the Sum Subtract the Logarithm of 22 the Remainder shall be the Logarithm of the Diameter Example If the Circumference of a Circle be 355 what is the Diameter thereof First set down the Logarithm of 7. which is 0,84510 2,55023 3,39533 1,34242 2,05291 and to it add the Logarithm of 355 from which Subtract the Logarithm of 22 and the Remainder is the nearest Logarithm of 113 which is the Diameter required 3. The Diameter of a Circle being given to find the Area or Superficial Content thereof The Proportion is as 28 is to 22 so is the Square of the Diameter to the Area Wherefore to the Logarithm of 22 add the Logarithm of the Diameter doubled and from the Sum subtract the Logarithm of 28 the Remainder shall be the Logarithm of the Area required Example If the Diameter of a Circle be 12 what is the Area or Superficial Content thereof First set down the Logarithm of 22 which is 1,34242 1,07918 1,07918 3,50078 1,44716 2,05362 and to that the Logarithm of 12 the given Diameter set down Twice Add all Three together from which Subtract the Logarithm of 28 The Remainder is the nearest Logarithm to the Number 113 and some small matter more is the Area of that Circle 4. The Circumference of a Circle being given to find the Area The proportion is as 88 is to 7 so is the Square of the Circumference to the Area Wherefore to the Logarithm of 7 add the Logarithm of the Circumference Twice and from the Sum Subtract the Logarithm of 88 the Remainder shall be the Logarithm of the Area required Example If the Circumference of a Circle be 38 what is the Area thereof First set down the Logarithm of 7 which is 0,84510 To which add the Logarithm of 1,57978 the Circumference Twice 1,57978 The Sum 4,00466 Subtract the Logarithm of 88 1,94448 the Remainder is the nearest Logarithm of 115 the Area sought 2,06018 CHAP. IV. CONTAINING Geometrical Rudiments Useful in the Art of GUNNERY How to raise a Perpendicular from the middle of a Line given LEt the Line given be A. B. and let C be a Point therein given from which it is required to raise a Perpendicular First therefore open the Compasses to any convenient distance and setting one Foot in the Point C with the other set off on either side thereof the equal distances C A and C B then opening the Compasses to any convenient wider distance setting one Foot in the Point A with the other strike the Occult Arch at F theâ— with the same distance set one Foot in the Point B and with the other draw the Arch F crossing E in the Point D from whence draw the Line DC which Line is a Perpendicular unto the given Line A B as was required To let a Perpendicular fall from a Point assigned to the middle of a Line given Let the Line given whereupon you would have a Perpendicular let fall be the Line B C D and the Point A to be the Point assigned from whence you would have the Perpendicular let fall from the given line B C D First set one Foot of your Compasses in the Point A and opening them to any convenient distance so that it be more than the line A C Describe one Arch of a Circle with the other Foot so that it may cut the line B C D twice that is at E and at F then find the middle between these which will be the Point C from which Point draw the line at C which is the Perpendicular which was to be let fall To raise a Perpendicular upon the end of a Line given Suppose the line whereupon you would have the Perpendicular raised be the line A B first open your Compasses to a convenient distance and set one Foot in the Point B and let the other Foot fall any where above the line as at the Point D and in that Point let one Foot of your Compasses remain turning the other about until it touch the line A B in the Point E then turn the same Foot of the Compasses towards C and draw an Occult Arch and lay the Edge of a Ruler to those Two Points E and D and where the same edge of the Ruler doth cut the Arch C from that Point draw the line C B which shall be a Perpendicular at the end of the line A B. To let fall a Perpendicular from a Point assigned unto the end of a Line given Let the line A B be given unto which it is required to let a Perpendicular fall from the assigned point D unto the end A. First from the assigned point D draw a line unto any point of
the given line A B which may be the line D C E then find the middle of the line D E which is at C place one foot of your Compasses in that point and extend the other foot unto D or E with which distance draw the Semicircle D A E which shall cut the given line A B in the point A from which point draw the Line D A which is the Perpendicular let fall from the assigned point D on the end of the given line A B as was required To draw a Line Parallel to a Line given Let A B be a Line given whereunto it is required to draw a Parallel First set one Foot of the Compasses in the point C and opening the other Foot at pleasure draw the Arch E then with the same distance set one Foot in the point D and draw the other Arch F. Lastly lay a Rule to the convexities of both those Arches and draw the line G H which shall be a Parallel to A B as was required A Geometrical Problem useful in the Art of Gunnery A Geometrical way to find the Diameter of any Bullet that weighteth twice as much as a known Bullet TAke the Diameter of the lesser Bullet whose weight you know and square that Diameter viz. Make a Geometrical Square each side to be equal to the Diameter of the Bullet given then draw a Diagotal line from either of the Two opposite Angles and that Diagonal shall be the Diameter of a Bullet twice the weight of the other then divide the said Diagonal into Two equal parts setting one Foot of the Compasses in the midst of that Diagonal and with the other Foot describe a Circle and that Circumference will represent a Bullet twice as much weight as the other The sight of the Annexed Figure is a sufficient Explanation A B is the Diameter of the lesser Bullet A C the Diameter of the greater Performed by Arithmetick Suppose the Diameter of the lesser Bullet be Five Inches the Square thereof is Twenty Five the Double of it is Fifty the Root thereof is 7 1 7 and so much is the Diameter of the greater Bullet The weight of any Shot given to find the Diameter Geometrically Suppose a Shot be One Two or Three Pound weight of Metal or Stone assigned if one Pound divide the Diameter into Four parts and Five such parts will make the Diameter of a Shot of the said Metal or Stone that shall weigh just Two Pound Divide the Diameter of a Shot weighing just Two Pound in Seven equal parts and Eight such parts will make a Diameter of a Shot of Three Pound And divide the Diameter of a Shot of Three Pound into Ten equal parts and Eleven such maketh a Shot of Four Pound Divide the Diameter of a Shot of Four Pound into Thirteen parts Fourteen such parts will make a Diameter for a Shot of Five Pound And so dividing each next Diamter into Three equal parts more the next Lesser was divided into and it will with one part added from a Diameter of a Shot that will weigh just one Pound more So you may proceed infinitely increasing or decreasing by taking one part less than it is appointed to be divided into CHAP. V. Geometrical Theorems AND PROBLEMS Theorem 1. ALL Circles are equal to that Right Angled Triangle whose containing sides the one is equal to the Semidiameter and the other to the Circumference thereof Theorem 2. The proportion of the Diameter of a Circle to the circumference is as 1,000000 to 3,141593 fere or as Archim 7 to 22. Theorem 3. The proportion of the Diameter to the side of the Square equal to the Circle is as 1,000000 to 886227 fere Theorem 4. The proportion of the Diameter to the side of the inscribed Square is as 1,000000 to 707107 fere Theorem 5. The proportion of the Circumference to the Diameter is as 1 to .318310 fere or as 22 to 7. Theorem 6. The proportion of the Circumference to the side of the Square equal to the Circle is as 1 to .282095 Theorem 7. The proportion of the Circumference to the side of the inscribed Square is as 1 to .225078 Arithmetical Problems appertaining to the Art of Gunnery and wrought by Decimal Arithmetick by the Logarithms and Gunter 's Scale PROB. 1. The Diameter of a Circle being given to find the Circumference The Analogy AS 1 is to the Diameter so is 3.142 to the Circumference or as 7 to 22 so is the Diameter to the Circumference If the Diameter of a Circle be 15 Inches what is the Circumference by Gunter 's Scale By the Logarithms As the Log. of 15 the Diameter 1,17609 is to the Logarithm of 3,142 0,49720 so is the Logarithm of 0,00000 to the Logar of the Answer 47,1367329 Extend the Compasses upon the Line of Numbers from 1 to the Diameter the same extent will reach from 3.142 to 47.13 the Circumference PROB. 2. The Circumference of a Circle being given to find the Diameter The Analogy AS 3,142 is to 1 so is the Circumference 47 13 to the Diameter 15 Inches If the Circumference of a Circle be 47 Inches and 13 parts of a 100 supposing every Inch to be divided into 100 parts what is the Diameter or as 22 to 7 so is the Circumference to the Diameter By the Logarithms As the Logarithm of 3,142 0,49720 is to the Logarithm of 1 0,00000 so is the Logarithm of 47.13 1,67329 to the Logar of the Answer  1,67329  15 1,17609 By Gunter's Scale Extend the Compasses upon the line of Numbers from 47.13 the Circumference the same extent the same way shall reach from 3,142 to the Diameter 15. PROB. 3. The Diameter of a Circle being given to find the side of a Square equal to it If the Diameter of a Circle be 15 Inches what shall be the side of a Square equal to it The Analogy AS 1 is to 15 so this Number 8862 to 13.29 the side of a Square equal in content to that Circle By the Logarithms As the Logarithm 1 0,00000 is to the Logarithm 15 1,17609 so is the Logarithm of 8862 0,94753 to the Answer 13,29 2,12362 By Gunter 's Scale Extend the Compasses from 1 to 8862 the same extent shall reach from 15 to 13.29 PROB. 4. The Circumference of a Circle being given to find the side of a Square equal in content to that Circle If the Circumference of a Circle be 47 13 the side of a Square equal to it is required The Analogy AS 1 is to 47.13 so is this Number 2821 to 13.29 the side of the Square required By the Logarithms As the Logarithm of 1 0,0000 is to the Logarithm of 4713 0,67329 â—o is the Logarithm of 2821 0,45040 to the Answer 13,29 1,12369 By Gunter's Scale Extend the Compasses upon the Line of Numbers from 1 to 2812 the same extent shall reach the same way from 47.13 to 13.29 the side of the Square required PROB. 5. The Diameter of any
under every one in their due place Example  l. A Man oweth to a Merchant 9758 And he hath paid of that Debt 3514 There Remains due 6244 CASE II. When some of the inferiour Numbers are greater than the superiour Numbers RVLE Set your Numbers in order as before draw a line under them and begin at the Right-hand and according to the Numbers respective value borrow one of the next to the Left-hand above out of which Subtract what remains add to the superiour and set their Sum under the line then what you borrow pay to the next Number on the Left-hand below and so proceed throughout the work according to this or the former Rule  l. s. d. As from 529 13 4 Take 347 16 7 Rests 171 16 9 Proof Add the two inferiours their Sum is equal to the superiour MVLTIPLICATION MVltiplication serveth to perform that at once which Addition doth at many times And to multiply readily it is necessary that the ensuing Table should be perfectly learned Pythagora's Table 1 2 3 4 5 6 7 8 9 2 4 6 8 10 12 14 16 18 3 6 9 12 15 18 21 24 27 4 8 12 16 20 24 28 32 36 5 10 15 20 25 30 35 40 45 6 12 18 24 30 36 42 48 54 7 14 21 28 35 42 49 56 63 8 16 24 32 40 48 56 64 72 9 18 27 36 45 54 63 72 81 The Vse of this Table is to muliply any Number in the outer Column to the left hand by any Figure at the top and in the common Angle of meeting is the Answer to the Question as 7 times 9 you will find to be 63. In Multiplication Note that the uppermost Number is always the Multplicand and the lower the Multiplier and the Figures which remain when the Work is done is called the Product Multiplication may be divided into six Cases CASE I. If the Multiplicand have dovers Figures and the Multiplier but one Rule Draw the Multiplier into the first Figure of the Multiplicand and subscribe the Units of the Product but carry the Tens to the next place then draw the Multiplier into the second Figure of the Multplicand and add the Tens you carried to the Units of that Product subscribe the Units of their Sam and carry the Tens to the third place accordingly proceed to the end of the work As if 5436 is to be mutiplied by 6 according to the following Example 5436 Multiplicand 6 Multiplier 32616 Product CASE II. If the Multiplicand and the Multiplier have each of them more than one Figure Rule For the first Figure do as before and having drawn the second Figure of the Multiplier into the first Figure of the Multiplicand 〈◊〉 the Units of the Product under that second Figure of the Multiplier and carry the Tens setting all the rest of the Multiplication as by the former Rule and this directeth making so many particular Rows of Products as you Figures in your Multiplier at last add them together for a total Product Example 4532 Multiplicand 32 Multiplier 9064 Particular 13596 Products 145024 Total Product CASE III. If Cyphers are in the Multiplicand and Multiplier or either of them Rule Set down to the right hand of the first Product as many Cyphers as are in the Multiplicand and Multiplier so that the first Unit of the Product of the first Multiplier may stand under the first Figure of the Multiplicand and work the rest according to the other Rule Example As CASE IV. If a Cypher or Cyphers be in the middle of the Multiplicand Rule Work according to the former Rules till you come to the Cyphers then under the first o subscribe the Tens you carried but under the rest of the Cyphers set Cyphers except under the last where subscribe the Units remaining of the Product of the next Figure of the Multiplier drawn into the Multiplicand the rest is according to the other Rules Example As CASE V. If a Cypher or Cyphers be in the middle of the Multiplier Rule Multiply as before is taught until you come to the Cyphers in the Multiplier which subscribe in order before the particular product of the next Multipler drawn into the Multiplicand then set the Units of its Product under that Multiplier and observe the other Rules for the rest As CASE VI. If Cyphers be both in the middle of the Multiplicand and also in the Multiplier Rule When you come to the Cyphers in the Multiplicand then under the first Cyphers place set the Tens you carried if any be and after that as many Cyphers as are in the Multiplier no Figure intervening then multiply into the next Figure of the Multiplicand subscribe the Units of the Product and carry the Tens in the same Row and so do in every Row of the particular Products according as this or some of the other Rules require Example You may abbreviate Multiplication by the help of Subtraction especially when to be multiplied by 5 or 9 As CASE I. To multiply any Number by 5. Rule Subtract half the Number and to it add a Cypher Example As 45276 Product 226380 being to be multiplied by 5 halve the Number and add a Cypher at the latter end and the Work is done CASE II. To multiply any Number by 9. Rule Add a Cypher to the Number given to be multiplied by 9 and subtract the first Number out of it and the Remainer is the Product or Answer of the Question Example Let the Number be 6789 to which add o Cypher and the Number is thus 67890 out of which subtract the first Number and the Remainder is 61101 the Product or Answer of the Question DIVISION DIvision serveth to divide any Number into as many parts as you please and consisteth of three Numbers the Divisor the Dividend and the Quotient for see how often the Divisor is contained in the Dividend so many Figures it produceth in the Quotient or see how often 1 is contained in the Divisor so many times the Quotient is contained in the Dividend which is all one If you were to divide 888 pound amongst 4 men the Question is what each man must have Order your Work as in this Example The first demand is how many times 4 can you have in 8 The answer is 2 which 2 place in the Quotient then multiply the 2 in the Quotient by 4 the Divisor and that makes 8 place 8 under the 8 on the lest Figure of the Dividend and draw a line under it and subtract 8 from 8 and there remains o. Then take down the next 8 and demand how many times the Divisor is contained in the Dividend 8 which is 2 times set that 2 in the Quotient and multiply the Divisor 4 by that 2 which makes 8 set that 8 under the second Figure of the Dividend and draw a line as before and subtract it from the 8 in the Dividend and there remains 0. Proceed in the same manner as you have done with the rest and you