given be AB having placed the one foot of your Compasses at A or B extend the other at pleasure above the half of the Line by estimation and draw two Arches the one above the Line and the other beneath then keeping the Compasses at the same wideness place the one foot at the other end of the Line cutting the former Arches in C and D with the other foot lay a Ruler at the points C and D and where the same doth cut the Line AB as in E so shall the Line AB be cut into two equal halves at E which was required and the Lines CE and ED shall be both perpendicular to the Line AB by the 19 th Definition of this Prob. 2. An Angle being given to divide the same into two halves Let the Angle be ABC to be divided into two equal halves Place the one foot of your Compasses at B and extend the other at your pleasure to D and E describe two Arches cutting each other in the point F So shall you divide the Angle ABC into two equal halves if you draw a Line from B to F with two Angles ABF and FBC as equal to the whole Angle ABC being divided into two halves Prob. 3. From any Point or from the end of a Line to erect or raise a Perpendicular Let there be given the Line AB and the Point in the same given C from which a Perpendicular is to be raised Place the one foot of your Compasses at C and with the other take at pleasure above the Line as D then from the point D describe the Arch of a Circle as ECF cutting the Line AB in F then lay your Ruler at F and D and where it cutteth the Arch as in E draw a Line from C to E which shall be perpendicular to AB from the point C. Again from the point A a Perpendicular is raised the same manner of way placing one foot of your Compasses at A extend the other above the Line to the point G and draw from G the Arch IAH cutting the Line at H then laying your Ruler by G and H and where it cutteth the Arch as in I from A to I draw a Line so shall AI be a perpendicular from the end of the Line AB or from the point A. The same may be performed by the Line of Chords as by the practice is evident Prob. 4. From a Point without a Line to let fall a Perpendicular to a Line given Let the Line given be AB and the Point without the same at C Draw a Line to AB at Oblique Angles as CD which you shall cut by the first Problem in two halves at E then place the one foot of your Compasses at E extend the other unto C drawing the Arch CF and where the same cutteth the Line AB as in F from C to F draw a Line as CF so shall CF be a Perpendicular let fall from the Point C upon the Line AB as was required Prob. 5. A Line being given to draw a Parallel to the same at any Distance required Let the Line given be AB and the Distance from C Extend your Compasses to the length of C and placing one foot of your Compasses in any Point of the Line draw two Arches and by the Compasse of these Arches draw a Line as DE which shall be parallel to AB required as CE. The same another way Let the Line given be AB and the Point at C From the Point C draw a Line at pleasure unto AB which shall touch in F which you shall cut by the first Problem into two halves at E take then a Point betwixt F and A at pleasure which is here G draw then a right Line from the Point G through E and set the Distance DE equal to the Line EG draw then a Line through the Points CD so shall the same Line be parallel to the Line AB Prob. 6. Two Lines being given to find a third proportional Line Let A and B be two Lines given and let it be required to find a third Line in proportion to them First Make any Angle as CDH then set the Line A from D to E and the Line B from D to F and also from D to G then draw EF that done by the Point G draw a parallel to EF as GH So shall DH be the third proportional Line required A. 4. B. 6. DE. 4. DF. 6. DG 6. DH 9. Prob. 7. Three Lines being given to find a fourth in proportion that is to perform the Rule of Three in Lines Let AB and C be three Lines given and it is required to find a fourth proportional Line First as in the last Problem make any Angle at Pleasure as the Angle EFG Then take with your Compasses the Line A and set it from F to H take also the Line B and set it from F to I and draw the Line HI That done take the third Line C and set it upon the Line FG viz. always upon the same Line where the first Line A was placed from F to K then by K draw a Parallel to HI as KL to cut FE in L So shall FL be the fourth Proportional Line required FH 12. FI. 14. FK 18. FL. 21. Prob. 8. To divide a Line given into two parts in proportion one to the other according to two Lines given Let AB be a Line given to be divided into two such parts that the lesser may be in proportion to the greater As the Line C to the Line D. From the end A draw the Line AE making the Angle BAE then set the Line C from A to F and the Line D from F to E and draw the Line EB Lastly By the Point F draw a Parallel to EB as FG to cut AB in G So shall AB be divided in G as C to D which was required AF. 20. FE 30. AG. 16. GB 24. Prob. 9. To cut off from a Line given any part or parts required Let in the last Diagram AB be a Line given and let it be required to cut off from it â…– parts First From the end A draw the Line AE making any Angle as BAE then set on any five equal parts from A to E and also two of the same parts from A to F That done draw the Line EB then by F draw a Parallel thereunto to cut AB in G So shall AG be the â…– parts of AB which was required Prob. 10. To divide a Line in such sort as another Line is before divided This Problem differeth not much from the two last Problems Draw a Right Line at pleasure as the Line OB which Line you shall divide into 100 equal parts as exactly as you can then extend your Compasses to the Distance OB and describe an Arch in which arch you are to include the desired Line A then draw the Line OC and you have the Angle COB which we shall hold for a ground-Rule because the Legs OB
and OC contain each of them 100 or 1000 parts which is in the Line A. For Example I would have 35 parts of the Line A whose whole length making 100 parts as the Line OB doth I number on the Line OB 35 parts from O to F with this distance I draw the Arch FG then is the nearest Distance from F to G a Right Line which is the 35 parts of the Line A. Likewise if it were desired to have 15 parts of the Line A Then draw from O the Arch ED which Distance is the desired 15 parts of the Line A as the Learned Adriani Meti doth teach And also how to find the Parts of a longer Line than the Line of Measure it self is it is done after the same manner and is of great use to a Gunner Vse This may be of great use if the Gunner be to find the first Round Ball in a strange place and the Weight unequal to Ours Prob. 11. Between two Lines given to find a mean Proportion Let A and B be two Lines given between the which it is required to find a mean Proportion Join the Lines A and B so together that they make one Right Line as CD being joined together in the Point E and upon the Line CD describe the Semi-circle viz. CFD Then upon the Point E where the Lines A and B being joined together meet erect a Perpendicular to cut the Limb in F as EF which shall be a mean Proportion between the Lines A and B required The same another way Again in the same Diagram let the Lines A and G be given between the which it is required to find a mean Proportion Take the Line G and lay it down from C to D and draw CD whereupon describe the Semi-circle CFD Then take the Line A and set it from D to E Then upon the Point E erect a Perpendicular to cut the Limb in F Lastly Draw DF which shall be a mean Proportion between DE and DC or between the Lines A and G required And if you draw CF it shall be a mean between B and G that is between CE and CD Prob. 12. To divide a Line given by Extream and Mean Proportion Let AB be a Line given to be divided by Extream and Mean Proportion Increase AB at length to C then upon the Point A erect a Perpendicular as AD of the length of AB That done take half AD or AB and set it from A to E then with the distance ED make the Arch DG So shall AB be divided by Extream and Mean Proportion in G and AG is the greater Segment and GB the lesser The same another way Prob. 13. The greater Segment of a Line divided by Extream and Mean Proportion to find the whole Line Let AB be the greater Segment given and the whole Line is required Increase BA to C then upon the end B erect a Perpendicular of half the length of AB as BD and draw the subtendant side AD from which subtract DB rests AF That done set AF from A to C so shall CB be the whole Line required Prob. 14. To describe a Circle upon any three Points given not being in a Right Line Let ABC be three Prob. 14 Points given Set one foot of the Compasses in the middle Point at B and open your Compasses to any extention that is above one half of the distance between B and the farthest of the other two Points and with that distance draw the blind Arches DE and FH with the same extent set one foot in C draw the Arch FH Again with the same extent setting one foot in the Point A draw the Arch DE then laying a Ruler to the Intersection of these Arches draw the Lines DG and HG which will cross each other in the Point G and there is the Centre of the Circle inquired Where setting one foot of your Compasses and extending the other to any of the three Points you describe the Arch of a Circle which shall pass through the three Points given and give the whole Circumference required which having you may find the Diameter by the 12 th Definition of Chap. 10. of this Book Observation By a Segment or a piece of a Ball being found to know the weight of the whole Ball. Here is to be observed That if any Town Fort or Place be Besieged or Blocked up by an Enemy and the Enemy shooting continually there are divers Segments or Pieces of their Ball found in the Place and it is desired to know what Ordnance they do Batter with The Gunner shall take the Segment or piece of the Ball and lay it on a piece of Paper and set there down by the Circumference three Pricks or Points as ABC in the Figure following and thereby and by the preceding Problem shall find the Diameter Centre and Circumference of the Ball whereby you may know the weight of that Ball as is described in pag. 54. Prob. 15. How to Reduce a Right Line into the Circumference of a Circle The Line given being AB to be reduced into a Circumference The which Line AB you shall first divide into three equal parts and of these three Parts you shall make an Equilateral Triangle as CDE the half of CD is F and the half of the side DE is G and where they cut one another as in H that is the Centre of the Triangle CDE further divide DF into halves with the point I draw then a right Line from the Centre H to I which is HK and divide the Line HI into four equal parts of which you shall add one part unto HI which shall be HK which is the same fourth part five times set set then one foot of your Compasses in the point H and the other in the point K and draw with the same distance a round Circle whose Circumference shall be equal to the Line given viz. the Line AB which was required Prob. 16. How to reduce the Circumference of a Circle given into a Right Line The Mean Proportion betwixt DL and AG the fourth part of the Circumference giveth the side of a Square equal to the Circle From DE subtract the half of AC there remaineth DK which is equal to BF and is the side of a Decagon inscribed in the Circle Prob. 17. The Diameter of a Circle being given to find its Proportion to the Circumference or the Reverse The true knowledg of this Problem is of great use for Gunnders both by Sea and Land Whensoever the Circumference of a Circle is divided into 10000 equal parts then doth the Diameter thereof contain 3183 of the said parts saith Adriani Meti But Archimedes sheweth us the easiest and fittest for our purpose to wit That if the Circumference be divided into 22 equal parts then the Diameter doth contain seven of the like parts as is demonstrated in this Figure where you may see that in the Circle ABC the Circumference is divided into
Multiply the Figure last placed in the Quotient first by it self and then the Product by the triple Number before subscribed this done subscribe the last Product under the said triple Number to wit Unites under Unites Tens under Tens c. So 4 being squared or multiplied by it self the Product is 16 which being multiplyed by the triple Number 15 the Product is 240 this therefore I subscribe under the aforesaid triple Number 15. Subscribe the Cube of the Figure last placed in the Quotient under the Resolvend in such manner that the first place of this Cube to wit the place of Unites may stand under the place of Unites in the Resolvend So 64 being the Cube of 4 I write it under the Resolvend 32464 in such manner that the Figure 4 which is in the place of Unites in the Cube 64 may stand under the Figure 4 which is seated in the place of Unites of the Resolvend Drawing yet another Line under the Work add the three last Numbers together in the same order as they are seated and subtract the Sum of them from the Resolvend placing the Remainder orderly underneath So the Sum of the three last Numbers as they are ranked in the Work is 32464 which if you subtract out of the Resolvend 32464 the Remainder is 0. Thus the whole Work being finished the Cube Root of 157464 the Number propounded is found to be 54. Note 1. When the Sum of the three last Numbers before mentioned is greater than the Resolvend the Work is erroneous and then you are to reform it by placing a lesser Figure in the Quotient Note 2. For every one of the particular Cubes distingushed by the Points except the first Cube on the left hand a Resolvend is to be set apart by bringing down to the Remainder the next Cube And as often as a Resolvend is set apart so often is a new Divisor to be found by adding the triple of all the Root in the Quotient consisting of what number of places soever to the triple of the Square of such Root after they are orderly placed according as is above-mentioned Note 3. The Work of the Table of Simple Cubes in Folio 9 for finding the first Figure of the Root as before declared is but once used in the Extraction of the Root of any Number whatsoever but the Work of all the following Rules is to be used for the finding of every place in the Root except the first The practice of these three Notes will be seen when we describe how to extract the Cube Root of Numbers not Cubical CHAP. IX Another Example wrought by the Genitures SUppose a Number given to be 16387064 of which the Cube Root is required First You must cut the Cube given into Ternaries from the right hand to the left as was declared in Chap. 8. Then find the Root of the first Cube from the left hand 16. Wherof the greatest Root is 2 for 2 being multiplied cubically giveth 8 the which 8 being deducted from 16 the first Cube of the Number propounded there remaineth 8 then set the Root found 2 with the Square thereof above it and by the same the Geniture and then find a second Figure for the Root of the second Cube and you shall have 5 which ye shall set down with its Square and Cube under it right against the Geniture towards the right hand then multiply each one by another and add the Products together there cometh 7625 which being subtracted from 8387 there doth remain 762 In the same manner find a Root for the Numbers remaining to be extracted and it shall be the Root of your third Cube And the Example will stand thus And seeing there remaineth nothing it is manifest that the Number propounded 16387064 is a Cube Number and the Root thereof is 254. By the 4 th Proposition of the second and 20 th Definition of the seventh Book of Euclide When we come to calculate the Table of Cubes by which you may make an Inch Rule for height or Line of Diameters the way shall be described how to extract the Cube Root of Numbers not Cubical or as they are termed Irrational Numbers from which no true Root can be obtained yet many times the Error will not be â— 100000 part of an Unite CHAP. X. Principles of Geometry THere are divers Reasons which make me to give these few Principles of Geometry because the whole Work of this Book is either to be done by Arithmetick or Geometry and besides that a Gunner cannot obtain to know the truth of a Diameter of a Ball except he can Geometrically extract the Wind of the Bore of the Peece and thereby find the Ball fitting such a Peece And in general will be most useful for any Gunner Definitions 1. A Point or Prick is that which is the least of all Materials and it is the beginning of Things as being void of Length Breadth and Depth having neither Part nor Quantity expressible in Numbers and therefore it admits of no Division but that which is mental only This Point or Prick is represented unto you by the Letter A. Thus A. 2. A Line is a Magnitude extending it self in length without breath or thickness whether it be a straight line or crooked and in respect of its length may be divided into Parts but will admit of no other division but in length only As is set forth to you by the Line BC the extremities whereof being Points as B and C. 3. A Right or Straight Line is the nearest Distance that can be betwixt two Points As is the former Line BC. 4. Circular or Crooked Lines are longer though they be extended no further than Right Lines as are the Lines DE and FG. 5. Right-lined Parallels are two straight Lines so drawn as they are equi-distant in all places one from the other so that although they were infinitely extended yet could they never meet as may be seen by the Lines HI and KL 6. A Circular Parallel is a Circle drawn either within or without another Circle upon one and the same Centre as is seen by the two Circles viz. NOPQ and RSTV being both drawn upon the same Centre W and therefore are parallel one to another 7. A Superficies is the second kind of Quantity and to it are attributed two Dimensions Length and Breadth but not Thickness for a Superficies is the term or end of a Body as a Line is the end and term of a Superficies As WXYZ is a Superficies 8. The Extreams of a Superficies are Lines as the Ends Limits or Bounds of a Line are Points confining the Line so are Lines the Limits Bounds and Ends inclosing a Superficies As in the foregoing Figure you may see the Superficies inclosed with four Lines viz. YW WX XZ and YZ which are the Extreams or Limits thereof 9. A plain Superficies is that which lyeth equally between his Lines like as a Right Line is the shortest extention or draught from one