Book II. Ch. II.

Of the Use of the Sector.

The Uses we shall here lay down, are only those that most appertain to the Sector, and which by it can be better performed, than by any other Instrument.

Section I. Of the Use of the Line of equal Parts.

Use I. To divide a given Line into any Number of equal Parts; for Example, into seven.

Take between your Compasses the proposed Line, as AB, and carry it, upon the Line of equal Parts, to a Number on both Sides, that may easily be divided by 7, as 70, whose 7th Part is 10; or else the Number 140, whose 7th Part is 20. Then keeping the Sector thus opened, shut the Feet of your Compasses, so that they may fall on the Numbers 10 on each Leg of the Sector, if the Number 70 be used; or upon the Numbers 20, if 140 be taken for the Length of the proposed Line; and this opening of your Compasses will be the 7th Part of the proposed Line.

Note, If the Line to be divided be too long to be applied to the Legs of the Sector, only divide one half, or one fourth of it by 7, and the double, or quadruple, of this 7th Part, will be the 7th Part of the whole Line.

Use II. Several right Lines, constituting the Perimeter of a Polygon, being given, one of which is supposed to contain any Number of equal Parts: to find how many of these Parts are contained in each of the other Lines.

Take that Line’s Length, whose Measure is known, between your Compasses, and set it over, upon the Line of equal Parts, to the Number on each Side, expressing it’s Length. The Sector remaining thus opened, carry upon it the Lengths of each of the other Lines, parallel to the beforementioned Line, and the Numbers that each of them falls on will shew their different Lengths: But if any one of the said Lines doth not exactly fall upon the same Number of the Lines of equal Parts, upon both Legs of the Sector; bat, for Instance, one of the Points of the Compasses falls upon 29, and the other upon 30; the Length of the said Line will be 29 and a half.

Use III. A right Line being given, and the Number of equal Parts if contains; to take from it a lesser Line, containing any Number of it’s Parts.

Let, for Example, the proposed Line be 120 equal Parts, from which it is required to take a Line of 25. First take the proposed Line between your Compasses, and then open the Sector, so that the Feet of your Compasses may fall upon 120, on the Line of equal Parts,. upon each Leg of the Sector: The Sector remaining thus opened, take the Distance from 25 to 25, and that will give the Line desired. It is manifest, from the three aforementioned Uses, that the Line of equal Parts, upon the Legs of the Sector, may very fitly serve as a Scale for all kinds of plane Figures, provided that one of their Sides be known; and that., by means of this Line, they may be augmented or diminished.

Use IV. Two right Lines being given, to find a third Proportional: and three being given, to find a fourth.

If there be but two Lines proposed, then take the Length of the first between your Compasses, and lay it off upon the Line of equal Parts from the Center, in order to know the Number whereon it terminates; then open the Sector, so that the Length of the second Line may be terminated by the Length of the first. The Sector remaining thus opened, lay off the Length of the second Line upon one of the Legs from the Center; and, Note, the Number whereon it terminates, and the Distance between that Number, on both Legs of the Sector, will give the third Proportional required.

Let, for Example, the first Line proposed be AB, 40 equal Parts; and the second CD, 20. First take the Length of 20 between your Compasses, and opening the Sector, set over this Distance upon 40, and 40 on each Leg of the Sector. The Sector remaining thus opened, take the Distance from 20 to 20, which will be the Length of the third Proportional sought; which being measured, on the Line of equal Parts, from the Center, you will find it 10; for As 40 is to 20, So is 20 to 10.

But if three Lines be given, and a fourth Proportional to them be required; take the second Line between your Compasses, and, opening the Sector, apply this Extent to the Ends of the first, laid off from the Center, on both Legs of the Sector. The Sector being thus opened, lay off the third Line from the Center, and the Extent between the Number, whereon it terminates on both Legs of the Sector, will be the fourth Proportional required.

Let the first of the three Lines be 60, the second 30, and the third 50; carry the Length of 30 to the Extent from 60 to 60; and the Sector remaining thus opened, take the Distance from 50 to 50, which is 25, and this will be the fourth Proportional sought: for 60 Is to 30 As 50 to 25.

Use V. To divide a Line into any given Proportion.

As for Example, to divide a Line into two Parts, which may be to each other as 40 is to 70: First add the two Numbers together, and their Sum will be 110; then take between your Compasses the Length of the Line proposed, which suppose 165, and carry this Length to the Distance, from 110 to 110, on both Legs of the Sector. The Sector remaining thus opened, take the Extent from 40 to 40, and also from 70 to 70; the first of the two will give 60, and the latter 105, which will be the Parts of the Line proposed; for 40 Is to 70, As 60 is to 105.

Use VI. To open the Sector, so that the two Lines of equal Parts may make a right Angle.

Find three Numbers, that may express the Sides of a right-angled Triangle, as 3, 4, or 5, or their Equimultiples; but since it is better to have greater Numbers, let us take 60, 80, and 100. Now having taken, between your Compasses, the Distance from the Center of the Sector to 100, open the Sector, so that one Point of your Compasses, set upon 80 on one Leg, may fall upon 60, of the Line of equal Parts, upon the other Leg; and then the Sector will be so opened, that the two Lines of equal Parts make a right Angle.

Use VII. To find a right Line equal to the Circumference of a given Circle.

The Diameter of a Circle is to the Circumference almost as 50 to 157; therefore take, between your Compasses, the Diameter of the Circle, and set it over, upon the Legs of the Sector, from 50 to 50, on both Lines of equal Parts. The Sector remaining thus opened, take the Distance from 157 to 157, between your Compasses, and that will be almost equal to the Circumference of the proposed Circle; I say almost, for the exact Proportion of the Diameter of a Circle to it’s Circumference hath not yet been Geometrically found.¹

Section II. Of the Use of the Line of Planes.

Use I. To augment or diminish any Plane Figures in a given Ratio.

Let, for Example, the Triangle ABC be given, and it is required to make another Triangle similar, and triple to it.

Take the Length of the Side AB between your Compasses, and open the Sector, so that the Points of your Compasses fall upon 1 and 1, on each Line of Planes; the Sector remaining thus opened, take the Distance from the third Plane to the third, on each Leg of the Sector, which will be the Length of the homologous Side to the Side AB. After the same manner may the homologous Sides to the other two Sides of the given Triangle be found, and of these three Sides may be formed a Triangle triple to the proposed one. Note, If the proposed Plane Figure hath more than three Sides, it must be reduced into Triangles, by drawing of Diagonals.

If a Circle is to be augmented or diminished, you must proceed in the same manner with it’s Diameter.

Use II. To similar Plane Figures being given; to find the Ratio between them.

Take either of the Sides of one of the Figures, and open the Sector, so that it may Fall upon the same Number or Division, on the Line of Planes, on both Legs of the Sector. Then take the homologous Side of the other Figure, and apply that to some Number or Division on both Legs of the Sector; and then the two Numbers, on which the homologous Sides fall, will express the Ratio of the two Figures. As suppose the Side ab, of the lesser figure, falls upon the fourth Plane; and the homologous Side AB, of the greater, falls upon the sixth Plane, the two Planes are to each other as 4 to 6. But if the Side of a Figure is applied to the Extent of some Plane, on both Legs of the Sector, and the homologous Side cannot be adjusted parallel to it, so as it may fall on a whole Number on both Legs of the Sector; then you must place the Side of the first Figure upon some other Number, on each Leg; ’till a whole Number is found on both Legs of the Sector, whose Extent is equal to the Length of the homologous Side of the other Figure, to avoid Fractions.

If the proposed Figures are so great, that their Sides cannot be applied to the Opening of the Legs of the Sector, take the half, third, or fourth Parts of any of the two homologous Sides of the find Figures, and compare them together, as before, and you will have the Proportion of the said Figures.

Use III. To open the Sector, so that the two Lines of Planes may make a right Angle.

Take between your Compasses the Extent of any Plane from the Center of the Sector; as, for Example, the 40th: then apply this opening of your Compasses, upon the Line of Planes, on both Sides, to a Number equal to half the precedent one, which, in this Example, is 20; then the two Lines of Planes will be at right Angles: because, by the Construction of the Line of Planes, the Number 40, which may represent the longest Side of a Triangle, signifies a Plane equal to two other similar Planes, denoted by the Number 20 upon the Legs of the Sector: Whence, from Prop. 48. lib. i. Eucl. the aforenamed Angle is a right one.

Use IV. To make a plane Figure similar and equal to two other given similar plane Figures.

Open the Sector (by the precedent Use) so that the Lines of Planes be at right Angles, and carry any two homologous Sides, of the two proposed Figures, upon the Line of Plants, from the Center, the one upon one Leg, and the other upon the other Leg; and then the Distance of the two Numbers found will give the homologous Side of a plane Figure similar and equal to the two given ones.

As, for Example, the Side of the lesser Figure being laid off from the Center, will reach to the fourth Plane; and the homologous Side of the greater Figure, likewise laid off upon the other Leg, will extend to the ninth Plane: then the Distance from 4 to 9 is the homologous Side of a Figure equal to the two proposed ones, by means of which it will be easy to make a Figure similar to them.

By means of this Use may be added together any Number of similar plane Figures, viz. in adding together the two first, and then adding their Sum to the third, and so on.

Use V. Two similar unequal plane Figures being given; to find a third equal to their Difference.

Open the Sector, so that the*two Lines of Planes may make a right Angle; then lay off one Side of the lesser Figure from the Center of the Sector. This being done, take the homologous Side of the greater Figure, and set one Foot of your Companies upon the Number whereon the first Side terminates, and the other Point will fall on the other Leg, upon the Number required.

As, for Example; having laid off the Side of the lesser Figure from the Center, which falls upon the Number 9, take the Length of the homologous Side of the greater Figure, and setting one Foot of your Compasses upon the Number 9, the other will fall on the Number 4 of the other Leg; therefore taking the Distance of the Number 4 from the Center of the: Sector, that will be the homologous Side of a Figure similar and equal to the Difference of the two given Figures, whose Ratio is as 9 to 13.

Use VI. To find a mean Proportional between two given Lines.

Lay off both the given Lines upon the Line of equal Parts, in order to have their Lengths expressed in Numbers; the lesser of which suppose 20, and the greater 45: Then open the Sector, so that the Distance from 45 to 45, of the Lines of Planes, be equal in Length to the greater Line. The Sector remaining thus opened, take the Distance from 20 to 20 of the Line of Planes, which will be the mean Proportional sought; and having measured it upon the Line of equal Parts from the Center, you will find it to be 30: for As 20 is to 30, So is 30 to 45.

But because the greatest Number on the Line of Planes is 64, if any one of the Lines proposed be greater than 64, the Operation must be made with their half, third, or fourth Parts, in the following manner: Suppose the lesser Number be 32, and the greater 72; open the Sector, so that half of the greater Number, viz. 36, may be equal to the Distance from 36 to 36, of the Line of Planes, upon both Legs of the Sector; and then the Distance from 16 to 16 doubled, will be the mean Proportional sought.

Section III. Of the Uses of the Line of Polygons.

Use I. To inscribe a regular Polygon in a given Circle.

Take the Semidiameter AC, of the given Circle, between your Compasses, and adjust it to the Number 6, upon the Line of Polygons, on each Leg of the Sector; and the Sector remaining thus opened, take the Distance of the two equal Numbers, expressing the Number of Sides the Polygon is to have: for Example; take the Distance From 5 to 5, to inscribe a Pentagon; from 7 to 7 for a Heptagon, and so of others: either of these Distances, carried about the Circumference of the Circle, will divide it into so many equal Parts. And thus you may easily describe any regular Polygon, from the equilateral Triangle to the Dodecagon.

Use II. To describe a regular Polygon upon a given right Line.

If for Example, the Pentagon of Fig. 6. is to be described upon the Line AB: Take the Length of the said Line between your Compasses, and apply it to the Extent of the Numbers 5, 5, on the Line of Polygons: The Sector remaining thus opened, take, upon the same Lines the Extent from 6 to 6, which will be the Semidiameter of the Circle the Polygon is to be inscribed in; therefore if, with this Distance, you describe, from the Ends of the given Line AB, two Arcs of a Circle, their Intersection will be the Center of the Circle.

If an Heptagon was proposed, apply the Length of the given Line to the Extent of the Numbers 7 and 7, on both Legs of the Sector, and always take the Extent from 6 to 6, to find the Center of the Circle; in which it will be easy to inscribe an Heptagon, each Side of which will be equal to the given Line.

Use III. To cut a given Line, as DE, into extreme and mean Proportions.

Apply the Length of the given Line to the Extent of the Numbers 6 and 6, on both Fig. 7. Sides, upon the Line of Polygons; and the Sector remaining thus opened, take the Extent of the Numbers 10 and 10, on both Legs of the Sector, which are those for a Decagon. This Extent will give DF, the greatest Segment of the proposed Line, because the greatest: Segment of the Radius of a Circle, cut into mean and extreme Proportion, is the Chord of 26 Degrees, which is the 10th Part of the Circumference.

If the greater Segment is added to the Radius of the Circle, so as to make but one Line, the Radius will be the greater Segment, and the Chord of 36 Degrees will be the lesser Segment.

Use IV. Upon a given Line DF, to describe an Isosceles Triangle, having the Angles at the Base double to that of the Vertex.

Open the Sector, so that the Ends of the given Line may fall upon 10 and 10, of the Line of Polygons, upon each Leg of the Sector. The Sector remaining thus opened, take the Distance from 6 to 6, and this will be the Length of the two equal Sides of the Triangle to be made.

It is manifest that the Angle, at the Vertex of this Triangle, is 36 Degrees, and that each of the Angles at the Base is 72 Degrees; but the Angle of 36 Degrees, is the Angle of the Center of a Decagon.

Use V. To open the Sector so, that the two Lines of Polygons may make a right Angle.

Take between your Compasses the Distance of the Number 5, from the Center, on the Line of Polygons; then open the Sector, so that this Distance may be applied to the Number 6 on one Side, and to the Number 10 on the other, and then the two Lines of Polygons will make a right Angle; because the Square of the Side of a Pentagon is equal to the Square of the Side of a Hexagon, together with the Square of the Side of a Decagon.

Section IV. Of the Uses of the Line of Chords.

Use I. To open the Sector, so that the two Lines of Chords may make an Angle of any Number of Degrees.

First take the Distance, upon the Line of Chords, from the Center of the Joint, to the Number of Degrees proposed; then open the Sector, so that the Distance, from 60 to 60 on each Leg, be equal to the aforesaid Distance, and then the Lines of Chords will make the Angle required.

As to make an Angle of 40 Degrees; take the Distance of the Number 40 from the Center, then open the Sector, ’till the Distance from 60 to 60, be equal to the said Distance of 40 Degrees. If a right Angle be required, take the Distance of 90 Degrees from the Center, and then let the Distance from 60 to 60 be equal to that, and so of others.

Use II. The Sector being opened, to find the Degrees of it’s Opening.

Take the Extent from 60 Degrees to 60 Degrees, and lay it off upon the Line of Chords from the Center; then the Number, whereon it terminates, sheweth the Degrees of it’s Opening.

Sights are sometimes placed upon the Line of Chords, by means of which Angles are taken, in adding to the Sector a Ball and Socket, and placing it upon a Foot, to elevate it to the Height of the Eye: but these Operations are better performed with other Instruments.

Use II. To make a right-lined Angle, upon a given Line, of any Number of Degrees.

Describe, upon the given Line, a circular Arc, whose Center Jet be the Point whereof the Angle is to be made; then lit off the Radius, from 60 to 60, on the Lines of Chords, The Sector remaining thus opened, take the Distance of the two Numbers upon each Leg, expressing the proposed Degrees, and lay it from the Line upon the Arc described. Lastly, draw a right Line from the Center, through the End of the Arc, and it will make the Angle proposed.

Suppose, for Example, an Angle of 40 Degrees be to be made at the End B, of the Line AB; having described any Arc about the Point B, always lay off the said Radius from 60 to 60 on the Line of Chords (because the Radius of a Circle is always equal to the Chord of 60 Degrees), and lay off the Distance of 40 Deg. and 40 Deg. from C to D. Lastly, drawing a Line through the Points B and D, the Angle of 40 Degrees will be had. Vid. Fig. 10.

By this Use a Figure, whole Sides and Angles are known, may be drawn.

Use IV. A right-lined Angle being given; to find the Number of Degrees it contains.

About the Vertex of the given Angle describe the Arc of a Circle, and open the Sector, so that the Distance from 60 to 60, on each Leg, be equal to the Radius of the Circle. Then take the Chord of the Arc between your Compasses, and carrying it upon the Legs of the Sector, see what equal Number, on each Leg, the Points of your Compasses fall on, and that will be the Quantity of Degrees the given Angle contains.

Use V. To take the Quantity of an Arc, of any Number of Degrees, upon the Circumference of a given Circle.

Open the Sector, so that the Distance from 60 to 60, on each Line of Chords, be equal to the Radius of the given Circle. The Sector remaining thus opened, take the Extent of the Chord of the Number of Degrees upon each Leg of the Sector, and lay it off upon the Circumference of the given Circle.

By this Use may any regular Polygon be inscribed in a given Circle, as well as by the line of Polygons, viz. in knowing the Angle of the Center, by the Method and Table before expressed, in the Construction of the Line of Polygons.

For Example; to make a Pentagon by means of the Line of Chords: Having found the Angle of the Center, which is 72 Degrees, open the Sector, so that the Distance from 60 to 60, on each Leg of the Sector, be equal to the Radius of the given Circle; and then take the Extent from 72 to 72, on each Leg, between your Compasses, which carried round the Circumference, will divide it into five equal Parts, and the live Chords being drawn, the Polygon will be made.

Use VI. To describe a regular Polygon upon the given right Line FG.

As, for Example, to make a Pentagon, whose Angle of the Center is 72 Degrees; open the Sector, so that the Distance from 72 Degrees to 72 Degrees, on each Line of Polygons, be equal to the Length of the given Line. The Sector remaining thus opened, take the Distance from 60 to 60, on each Leg, between your Compasses; with this Distance, about the Ends of the given Line, as Centers, describe two Arcs intersecting each other in D; and this D will be the Center of a Circle, whose Circumference will be divided, by the given Line, into five equal Parts.

Section V. Of the Uses of the Line of Solids.

Use I. To augment or diminish any similar Solids in a given Ratio.

Let, for Example, a Cube be given, and it is required to make another double to it. Carry the Side of the given Cube to the Distance of some equal Number, on both Lines of Solids, at pleasure; as, for Example, to 20 and 20. The Sector being thus opened, take the Extent, on both Legs of the Sector, of a Number double to it, that is, of 40 and 40; and this is the Side of a Cube double the proposed one.

If a Ball or Globe be proposed, and it be required to make another thrice as big; carry the Diameter of the Ball to the Distance of some equal Number, on both Lines of Solids, at pleasure, as to 20 and 20; then take the Distance from 60 to 60 (because 60 is thrice 20), and that will be the Diameter of a Ball three times greater than the proposed one, because Balls are to each other as the Cubes of their Diameters.

If, again, a Chest, in figure of a right-angled Parallelepipedon, contains three Measures of Grain, and it be required to make another similar Chest to contain five Measures; open the Sector, so that the Distance from 30 to 30, on each Line of Solids, be equal to the Length of the Base of the Chest; then the Distance from 50 to 50, on each Leg, will be the homologous Side of that Solid to be made. Again, apply the Breadth of the Base to the Distance of the said Numbers 30 and 30, and then the Distance from 50 to 50 will be the homologous Side to the said Breadth. Now having made a Parallelogram with these two Lengths, your next thing will be to find the Depth: To do which, open the Sector, so that the Distance from 30 to 30 be equal to the Depth of the given Chest; then the Distance from 50 to 50 will be the Depth of the Chest to be made. This being done, it will be easy to make the Parallelepipedon, containing the five proposed Measures.

If the Lines are so long, that they cannot be applied to the Legs of the Sector, take any of their Parts, and with them proceed as before; then the respective Parts of the required Dimensions will be had.

Use II. Two similar Bodies being given; to find their Ratio.

Take either of the Sides of one of the proposed Bodies between your Compasses, and having carried it to the Distance of some equal Number, on each Line of Solids, take the homologous Side of the other Solid, and note the Number on each Leg it falls upon; and then the said Numbers will shew the Ratio. of the two similar Solids.

But if the Side of the first Solid be so applied to some Number on each Leg of the Sector, that the homologous Side of the other cannot be applied to the Extent of some Number on each Leg; then you must apply the Side of the first Solid to such a Number on each Line, that the Length of the Side of the second Solid may fall upon some whole Number on each Line of Solids, to avoid Fractions.

Use III. To construct and divide a Line, whose Use is to find the Diameters of Cannon-Balls.

It is found, by Experience, that an Iron Ball, three Inches in Diameter, weighs 40 Pounds; whence it will be easy to find the Diameters of other Balls of different Weights, and the same Metal, in the following manner: Open the Sector, so that the Distance from the 4th Solid to the 4th Solid, on each Line of Solids, be equal to three Inches The Sector remaining thus opened, take upon the Lines of Solids the Distances of all the Numbers, from 1 to 64, on one Leg, to the same Numbers on the other Leg; then lay off all these Lengths upon a right Line drawn on a Ruler, or upon one of the Legs of the Sector, and where the Diameters terminate, denote the Weights of the Balls.

But now to mark the Fractions of a Pound, as $\frac{1}{4}$, $\frac{2}{3}$, $\frac{3}{4}$, open the Sector, so that the Distance of the 4th Solids on each Leg of the Sector, be equal to the Diameter of a Ball of one Pound. The Sector remaining thus opened, the Distance from the 1st Solid to the 1st on each Leg of the Sector, will give the Diameter for $\frac{1}{4}$ of a Pound; from the 2d to the 2d, for $\frac{1}{2}$ of a Pound; and from the 3d to the 3d, for $\frac{3}{4}$ of a Pound, and so of others. When the Diameters of Balls are known, the Diameters or Bores of Cannon, to which they are proper, will likewise be known: but there are commonly two or three Lines given for the Vent of great Balls, and for lesser ones in proportion. The Diameters of Balls are measured with spherick Compasses, as will be more fully explained among the Instruments for Artillery.

Use IV. To make a solid similar and equal to the Sum of any Number of similar given Solids.

Open the Sector, and apply either of the Sides of either of the Bodies to the same Number on each Line of Solids; then note on what equal Numbers, on both Legs of the Sector, the homologous Sides of the other Solids fall. This being done, add together the said Numbers, and take the Extent, on both Lines of Solids, of the Number arising from that Addition; and this Extent will be the homologous Side of a Body, equal and similar to the Sum of the given Bodies.

Example; Suppose the Side chosen of the first Solid be applied to the fifth Solid, on each Leg of the Sector, and the homologous Sides of the others fall, the one on the 7th, and the other on the 8th Solid, on each Line of Solids; add the three Numbers 5, 7, and 8 together, and their Sum is 20; therefore the Distance from 20 to 20, on each Line of Solids, will be the homologous Side of a Body, equal and similar to the three others.

Use V. Two similar and unequal Bodies being given; to find the third similar and equal to their Difference.

Open the Sector, and apply either of the Sides of either of the Bodies to some equal Number on each Leg of the Sector, and see what equal Numbers, on both Legs, the homologous Sides of the other Solids fall upon; then substract the lesser Number from the greater, and take the Distance from the remaining Number, on one Line of Solids, to the same on the other; and this will be the homologous Side of a Body, equal to the Difference of the two given ones.

As, for Example; the Side of the greatest being set over, upon the Line of Solids, from 15 to 15, the homologous Side of the lesser will be equal to the Distance from 9 to 9; then taking 9 from 15, there remains 6: therefore the Distance from 6 to 6 will be the homologous Side of the Solid sought.

Use VI. To find two mean Proportionals between two given Lines.

For Example, suppose there are two Lines, one of which is 54, and the other 16: open the Sector, so that the Distance from 54 to 54, on each Leg of the Sector, be equal to the Length of the longest Line. The Sector remaining thus opened, the Distance from 16 to 16, on each Leg, will be equal to the greater or the mean Proportionals, and will be found to be 36. Again, shutting the Legs of the Sector closer, ’till the Distance between 54 and 54, on each Leg, be equal to 36; then the Distance from 16 to 16 will be the lesser of the mean Proportionals, and will be found to be 24: Whence these four Lines will be in continual Proportion, 54, 36, 24, 16.

If the Lines be too long, or the Numbers of their equal Parts too great you must, takes their halfs, thirds, or fourths, &c. and proceed as before. For Example; to find two mean Proportionals between two Lines, one of which is 32, and the other 256, take the fourth Parts of both the Lines, which are 8 and 64. This being done, open the Sector, so that the Distance from 8 to 8, on each Line of Solids, be equal to 8; then take the Distance from 64 to 64, and that gives 16, for $\frac{1}{4}$ of the first of the two mean Proportionals. Again, open the Sector, so that the Distance from 8 to 8 be equal to 16; the Sector being thus opened, the Distance from 64 to 64 will give 16, for $\frac{1}{4}$ of the second of the mean Proportionals sought: whence the mean Proportionals are 64 and 1283 for 32, 64, 128, 256, are proportional.

Use VII. To find the Side of a Cube equal to the Side of a given Parallelepipedon.

First, find a mean Proportional between the two Sides of the Base of the Parallelepipedon 5 then between the Number found, and the Height of the Parallelepipedon, find the first of two mean Proportionals, which will be the Side or the Cube sought.

For Example, let the two Sides of the Parallelepipedon be 24 and 54, and it’s Height 63; the Side of a Cube equal to it is sought.

Open the Sector, so that the Distance between 54 and 54, on the Line of Planes, be equal to the Side of 54; then take the Distance from 24 to 24 on the same Line, which, measured upon the Line of equal Parts, will give 36 for a mean Proportional. This being done, take 36 between your Compasses, and open the Sector, so that the Points of the Compasses may fall upon 36 and 06, on each Line of Solids; then take the Distance from 63 to 63 on the Lines of Solids, which will be found almost 44$\frac{1}{2}$, for the Side of a Cube equal to the given Parallelepipedon.

Use VIII. To construct and divide a Gauging-Rod to measure Casks, and other the like Vessels, proper to hold Liquors.

The Gauging-Rod, of which we are now going to speak, is a Ruler made of Metal, divided into certain Parts, whereby the Number of Pints contained in a Vessel may be found, in putting it in at the Bung-hole, ’till it’s End touches the Angle, made by the Bottom, with that part of the Side opposite to the Bung-hole, as the Line AC diagonally situated.

The Gauging-Rod being thus posited, the Division, answering to the middle of the Bung-hole, shews the Quantity of Liquor, or Number of Pints the Vessel, when full, holds.

But it is necessary to change the Position of the aforesaid Rod, so that it’s End C may touch the Angle of the other Bottom B, in order to see whether the middle of the Bung-hole be in the middle of the Vessel; for if there is any Difference, half of it must be taken.

The Use of this Gauging-Rod is very easy: for, without any Calculation by it, the Dimensions of Casks may immediately be taken; all the Difficulty consist only in well dividing it.

Now, in order to divide it, a little Cask, holding a Septier, or Gallon, must be made similar to the Vessels that are commonly used; for this Rod will not exactly give the Dimensions of dissimilar Vessels, that is, such that have the Diameters of the Heads, those of the Bungs, and the Lengths not proportional to the Diameters of the Head, Bung, and Length of that which the Divisions of the Rod are made by.

Now suppose the Diameter, at the Head of a Cask, be 20 Inches, the Diameter of the Bung 22, and the interior Length 30 Inches this Vessel will hold 27 Septiers of Paris Measure, and it‘s Diagonal Length, answering to the middle of the Bung-hole, will be 25 Inches, 9 Lines and a half, as is easy to find by Calculation: because in the right-angled Triangle ADC, the Side CD being 15 Inches, and DA 21, by adding their Squares together, you will have (per Prop. 47. lib. 1. Eucl.) the Square of the Hypothenuse AC; and by extracting the Square Root, AC will be had.

According to the same Proportions a Cask, whose Dimensions are one Third of the former ones, will contain one Septier, or eight Pints; that is, if the Diameter of the Head be 6 Inches, and 8 Lines; that of the Bung 7 Inches, 8 Lines; the Length 8 Inches, 8 Lines; and it’s Diagonal 8 Inches, 7 Lines.

Another Cask, whose Dimensions are half of that before-mentioned, will contain one Pint; that is, if the Diameter of the Head be 3 Inches, 4 Lines; that of the Bung 3 Inches, 8 Lines; the interior Length of the Cask 5 Inches; and the Diagonal, answering to the middle of the Bung-hole, 4 Inches, 3 Lines and a half.

Now take a Rod about 3 or 4 Feet long, and chuse either of the three Measures, which you judge must proper: As, for Example; if you will make Divisions for Septiers upon the Rod, make a Point, in the middle of it’s Breadth, distant from one of it’s Ends, 8 Inches, 7 Lines, and there make the Division for one Septier upon it; double that Extent, and there make a Mark for 8 Septiers; triple the same Extent, and there make a Mark for 27 Septiers; quadruple it, and there make a Mark for 64 Septiers; because similar Solids are to each other, as the Cubes of their homologous Sides.

Again, to make Divisions upon it for the other Septiers, take between your Compasses the Length or 8 Inches, 7 Lines; let over this Distance, upon each Line of Solids of your Sector horn the first Solid to the first. The Sector remaining thus opened, take the Distance from the second Solid to the second, which mark upon the Rod for the Division of two Septiers.

Again; take the Distance from the third Solid to the third, which mark upon the Rod for the Length of the Diagonal, agreeing to three Septiers, and so on; by which means the Rod will be divided, for taking the Dimensions of Vessels in Septiers. With the same facility may the Divisions for Pints be made upon the Rod; for half of the Distance of the Division of two Septiers, will give the Division for two Pints; half of the Distance of the Division for three Septiers, will give the Division for three Pints; half of the Distance of the Division for four Septiers, will give the Division for four Pints, and so on.

If the Sector be not long enough to take the Diagonal Length answerable to one Septier, from the first Solid to the first, take the Diagonal Length answerable to one Pint; and having divided the Rod for any Number of Pints, the Diagonal Lengths of the same Number of Septiers, may be had, by doubling the Diagonal Lengths of the Pints. As, for Example; if the Diagonal Length for 6 Pints be doubled, that Distance will be the Diagonal Length of a Vessel holding 6 Septiers: Also if the Diagonal Length of 7 Pints be doubled, the Length of the Diagonal of a Vessel, holding 7 Septiers, will be had; and so of other Diagonal Lengths.

If the Diagonal Length is yet too long to be applied to the Distance of the Division for the first Solid, on each Leg of the Sector, it’s half must be applied to the same; and the Sector remaining thus opened, take the Distance of the Divisions for the second Solid on both Lines of Solids, and double it; then you will have the Diagonal Length of a Vessel holding two pints. Having again taken the Distance of the Division for the third Solid upon each Leg of the Sector, which Distance being double, the Diagonal Length of a Vessel holding three Pints will be had, and may be marked upon your Rod; and so of others.

The Divisions for Septiers go across the whole Breadth of the Rod, upon which are their respective Numbers graved; and the Divisions for Pints are shorter than the others, for their better Distinction.

In order for this Gauging-Rod to serve to take the Quantity of Liquor contained in different dissimilar Vessels, other Divisions may be made upon it’s Faces, according to the different Proportions of their Lengths and Diameters, and at the bottom of the faces must be writ the Diameters and Lengths by which the Divisions were made: For Example; at the bottom of the Face, upon which the precedent Divisions were made, there is wrote, the Diameter of the Head 20, the Diameter of the Bung 22, and the Length 30.

If, for dividing another Face, you use a Vessel, whose Diameter of the Head is 21 Inches, that of the Bung 23, and the interior Length 27 $\frac{1}{2}$ Inches; this Vessel is shorter than that before-named, but contains almost the same Quantity of Liquor, when full, viz. 27 Septiers, and the Length of it’s Diagonal will be 26 Inches.

If another Vessel hath all it’s Dimensions $\frac{1}{3}$ of the precedent ones, this Vessel will hold one Septier, and it’s Diagonal AC will be 8 Inches and 8 Lines in Length. Now by means of this Vessel, and it’s Diagonal Length, you may divide the aforesaid Face in the Manner directed for dividing the first Face, and at the Bottom of this Face you must write, Diameter reduced 22, Length 17$\frac{1}{2}$.

If the four Faces of the Rod are divided, as before-named, you will have four different Gauges for gauging four different kinds of Vessels; and by examining the Proportions of the Diameters of the Heads and Lengths, you must make use of such a Face accordingly.

Instead of using the Sector in dividing the before-mentioned Gauging-Rod, it is better using the Table of Solids.

For having found, by Calculation, that the Length of the Diagonal of a Vessel, holding 27 Septiers, is 6 Inches, it will be easy to find the Diagonals of Vessels of any proposed Bignesses, having the same Proportions to the Diameters reduced, as 22 to 27$\frac{1}{2}$, or as 4 to 5.

As, for Example; it is required to find the Diameter of a Quarteau, or Firkin, which holds 9 Septiers; seek, in the Table of Solids, the Number answering to the 9th Solid, which will be found 520; at the same time find the correspondent Number to the 27th Solid, which will be found 750: then state a Rule of Three, in the following manner; 750 : 520 :: 26 : 18; whence 18 Inches will be the Length of the Diagonal of a Vessel holding 9 Septiers. The Coopers about Paris make their Vessels almost in the Proportion of 4 to 5; as is, for Example, a half Muid, having 19 Inches 2 Lines in Diameter reduced, and 24 Inches in Length; in which Case the Diagonal will be 22 Inches, 8$\frac{1}{2}$ Lines, as you will easily find by Calculation.

But, in general, as soon as the Proportions used in making Vessels are known, the Diagonal of some one of those Vessels, holding a known quantity of Septiers being first found (per Prop. 47. lib. [1]. Eucl.) you may afterwards find the Lengths of the Diagonals of all Vessels made in the same proportion, by means of the aforesaid Table of Solids.

Section VI. Of the Construction and Use of other kinds of Gauging-Rods.

The Gauging-Rod, of which we have already spoken, serves only to find the Quantity of Liquor contained in similar Vessels; but that which we are now going to mention, may be used in taking the Dimensions of dissimilar Vessels.

In order to construct the first Gauge of this kind, the Measure which you use must be determined, by comparing it with some regular Vessel, as a Concave Cylinder, in which a Quart or a Gallon of Water being poured, you must exactly note the Depth occupied by the Water.

As, for Example, if a Gauge is to be made for Paris, where a Pint is 48 Cubic Inches, or 61 Cylindrick Inches, you will find, by Calculation, that a Concave Cylinder, 3 Inches, 11$\frac{1}{3}$ Lines in Diameter, and the like Number in Depth, contains one Pint of Paris; and a Cylinder, whole Dimensions are double the aforesaid ones, that is, 7 Inches, 10$\frac{2}{3}$ Lines, will hold one Septier: for similar Solids ate to each other, as the Cubes of their like Sides.

This being supposed, lay off that Length of 3 Inches 11$\frac{1}{3}$ Lines, upon one Face of the Rod, as often as the Length of the Rod will admit, and mark Points, whereon set 1, 2, 3, 4, 5, &c. each of these Parts may be subdivided into 4 or more. This Face, thus divided, is called the Face of equal Parts, and is used in measuring the Lengths of Vessels.

You must likewise mark, upon another Face of the Rod, the Diameter of the Cylinder of 3 Inches, 11$\frac{1}{3}$ Lines, and then the Diameters of Circles double, triple, quadruple, &c. by any of the Methods before explained for dividing the Line of Planes on the Sector, the easiest and shortest of which is to make a right-angled Isosceles Triangle ABC; each of the Legs about the right Angle of which being 3 Inches, 11$\frac{1}{3}$ Lines, the Hypothenuse BC will be the Diameter of a Circle double to that, whose Diameter is 3 Inches, 11$\frac{1}{3}$ Lines: therefore having produced one of the Legs AB towards D, lay off the said Hypothenuse from A towards D, and at the Point whereon it terminates mark the Number 2; then take the Distance C2, and having laid it off upon the Line AD, mark the Number 3 at the Point whereon it terminates. Again, take the Distance C3, and having laid it off upon the Line AD, there mark the Number 4, &c.

Note, A4, which is the Diameter of a Circle quadruple the first, is double AC, or AB; because Circles are to each other as the Squares of their Diameters: whence since AB is 1, it’s Square is also 1; and the Line A4 being 2, it’s Square must consequently be 4.

To use this Gauge, you must first apply the Face of equal Parts to the exterior Length of the Vessel, from which you must take the Depth of the two Croes, that thereby the true interior Length may be had.

This being done, apply the Face of Diameters to the Diameters of the Heads of the Vessel, and note the Number answering to them, and whether they are equal; for if there be any Difference between the Diameters of the Heads, you must add them together, and take half their Sum for the mean Head-Diameter.

Again; put the Rod downright in at the Bung-hole, in order to have the Diameter of the Bung, which add to the Head-Diameter, and take half the Sum for an arithmetical Mean; this being multiplied by the Length of the Vessel, will give the Number of Pints the Vessel holds.

As suppose the interior Length of a Vessel is 4$\frac{3}{4}$ of the equal Parts of the Rod, the Diameter at the Head 15, and the Bung-Diameter 17; add 15 to 17, and their Sum is 32, half of which is 16; which multiplied by the Length 4$\frac{3}{4}$, and the Product 76 will give the Number of Pints the Vessel holds.

Now to construct the second kind of Rods, it is found, by Experience, that a Cylinder, whose Height and Diameter is 3 Foot, 3 Inches, and 6 Lines, holds 1000 Paris Pints.

Then take upon a Ruler a Length of 3 Feet, 3 Inches, and 6 Lines, which divide into 10 Parts, each of which will be the Height and Diameter of a Cylinder holding one Pint, (because similar Cylinders are to each other as the Cubes of their Diameters). Again, divide each of these Parts into 10 more, which may easily be done by help of the Line of Lines on the Sector; then each of these last Parts will be the Height and Diameter of a Cylinder holding the 1000th part of a Pint: Every five of these small Parts being numbered, your Rod will be made. One of these Rods, of 4 or 5 Feet in Length, will serve to gauge great Vessels, as Pipes, &c.

To use this Rod, you must note how many of the small Divisions of the Rod the Diameters of the Head and Bung, as also the Length, contains.

But, Note, by the Length of a Vessel is understood the interior Length, which is the Distance between the Head and the Bottom; and by the Diameters is understood the interior Diameters included between the Staves.

Note, also, If the Diameters at Top and Bottom are unequal, compare one of them with the Bung-Diameter, and the middle between these two is called the mean Diameter of the Vessel.

But if the Diameters at Top and Bottom are unequal, add them together, and take half of their Sum, which is called the mean Diameter of the Head and Bottom; then compare this mean Diameter with the Diameter at the Bung, add them together, and take half their Sum for the mean Diameter of the Vessel.

Then square the mean Diameter of the Vessel, and multiply the said Square by the Length of the Vessel; then the Product will give you the Quantity of Liquor in 1000th Parts the Vessel holds; and by casting away the last three Figures, you will have the Number of Pints contained in the Vessel, when full.

Let, for Example, the Diameter at the Head be 58 Parts of the Guaging-Rod, and the Bung-Diameter 62; add these two Numbers together, and their Sum will be 120, whose half 60 is the mean Diameter of the Vessel: then the Square of this mean Diameter will be 2600; and if this Square be multiplied by the Length of the Vessel, which suppose 80, the Product will be 288000; and by taking away the three last Figures, the Number of Paris Pints the Vessel holds will be 288.

This way of Gauging is exact enough for Practice, when there is but a small Difference between the Bung and Head-Diameters, as are the Diameters of Paris-Muids; but when the Difference between the Bung and Head-Diameters is considerable, as in the Pipes of Anjou, whole Bung-Diameters are much greater than the Head-Diameters, Dimensions taken in the before-directed manner will not give the Quantity of Liquor exact enough: But to render the method more exact, divide the Difference of the Bung and Head-Diameters into 7 Parts and add 4 of them to the Head-Diameter, and that will give you the mean Diameter: for Example; if the Diameter of the Head is 50, and the Bung-Diameter 57, the mean Diameter of the Vessel will be 54; with which mean Diameter proceed as before.

Having found by the Rod how many Paris Pints a Vessel holds, you may find how many other Measures the same Vessel holds, in the following manner:

A Paris Pint of fresh Water weighs 1 Pound, 15 Ounces; therefore you need but weigh the sought Measure full of Water, and by the Rule of Three you may have your Desire.

As, for Example; a certain Measure of Water weighs 50 Ounces, and it is required to find how many of the same Measures is contained in a Paris-Muid, which holds 288 Pints: Say, by the Rule of Three, As 50 is to 31, So is 288 Pints to a fourth Number, which will be 178$\frac{14}{25}$ of the said Measures.

There may be marked Feet and Inches upon the vacant Faces of the aforesaid Gauging-Rod, each of which Inches may be subdivided into four equal parts, which will be a second means to gauge Vessels; the Feet are marked with Roman Characters, and the Inches with others.

We have already said, that a Paris Pint contains 61 Cylindrick Inches; therefore having the Solidity of a Vessel in Cylindrick Inches, it must be divided by 61, to have the Number of Pints the Vessel holds. An Example or two will make this manifest.

Let the Length of a Vessel be 36 Inches, the Head-Diameter 23, and the Bung-Diameter 25; add the two Diameters together, and their Sum will be 48, half of which is 24 for the mean Diameter. This Number 24 being squared, will be 576; and this Square being multiplied by the Length 36, gives 20736 Cylindrick Inches: which being divided by 61, the Quotient will give 339 Pints, and about $\frac{3}{4}$.

If the Diameters and Lengths of Vessels are taken in fourth Parts of Inches, the last Product: must be divided by 3904, to have the Number of Pints contained in a Vessel, when full.

Let, for Example, the Length of a Vessel be $\frac{1}{4}$ Inches, the Head-Diameter 23 Inches, and the Diameter at the Bung 25$\frac{1}{2}$ Inches; add the two Diameters together, and their Sum will be 48 $\frac{1}{2}$, half of which will be 24$\frac{1}{4}$; which, for ease of Calculation, reduce to 4ths: 97 is the Number to be squared, which will be 9409; which multiply by 141, and that Product again by 35$\frac{1}{4}$, reduced to 4ths of Inches, will give this Product 1326669; which being divided by 3904, the Quotient will (as before) be 339 Pints, and about $\frac{3}{4}$.

The Construction and Use of a new Gauging-Rod.

Mr Sauveur, of the Academy of Sciences, has communicated to us a new Gauging-Rod of his Invention, by means of which may be found, by Addition only, the Quantity of Liquor that any Vessel holds, when full; whereas hitherto Multiplication and Division has been used in Gauging.

To make this Gauging-Rod, you must first chuse a Piece of very dry Wood, as Sorbaple or Pear-tree, without Knots, about 5 Foot long, in Figure of a Parallelepipedon, and 6 or 7 Lines in Breadth; Fig. 17. shews it’s four Faces.

Now upon the first of the four Faces are made Divisions for taking the Diameters of Vessels.

The Divisions of the second Face serves to measure the Lengths of the Diameters.

The Divisions upon the third Face are for finding the Contents of Vessels.

And, Lastly, upon the fourth Face, the Numbers of Septiers and Pints, which the Vessel holds, are marked.

The aforesaid Divisions are made in the following manner:

First, divide the fourth Face into Inches, and each Inch into 10 equal Parts; those Divisions denote Pints, and are numbered 1, 2, 3, 4, 5, 6, &c. every 8 being Septiers, because 1 Septier is 8 Pints: On the End of this fourth Face is written Pints and Septiers.

The Divisions of the Other three Faces are made by help of Logarithms, in manner following.

Note, The Divisions of the fourth Face serve as a Scale to the third, and ought no be contiguous to it.

To divide the third Face of the Rod.

If you have a mind to place any Number upon the third Face of your Rod; for Example, 240: seek in the Table of Logarithms for 240, or the nighest Number to it, which will be found against 251 in your Table; then place 240 upon the third Face, over against 252 Pints on the fourth Face, and, proceeding in this manner, you may divide the third Face.

But because, in the Table of Logarithms, 240 doth not stand against 251, but instead thereof there stands 2.39996, which nighly approaches it, therefore to make the Divisions as exact as possible, you must add 1 to the first Number of the Logarithm 240, and then seek for 3.40, over against which stands 2512; which shews, that the Logar. 240 must be placed not over against 251 of the Divisions of Pints, but against 251 and two Parts of the Divisions of a Pint, supposed to be divided into 10 Parts more. You must write Contents at one End of this third Face.

The Manner of dividing the second Face.

A Cylindrical Vessel, whose Length and Diameter is 3 Inches, 11$\frac{1}{3}$ Lines, holds one. Paris Pint; therefore the first part of the second Face, which is without Divisions, must be of that Length. This said Length must be laid off ten times, and more, if possible, upon the said Face, upon which make occult Marks; then one of these Parts must be divided into 100 more, upon a separate Ruler, serving as a Scale.

This being done, suppose any Number is to be placed upon the second Face; as, for Example, 60: Seek in the Table of Logarithms for 60, which will be found against 39 and 40, or rather against 3981, without having regard to the Numbers 1, 2, 3, that precede it, and which are called Characteristicks: therefore I take 98, or 981, by esteeming one Part divided into 10, upon the small Scale divided into 100, and I place this Distance next to the third occult Point, which denotes three Centesms, or three Thousandths. You must thus mark Divisions from 5 to 5, and every of these 5ths must again be subdivided into 5 equal Parts. Finally, upon the End of this Face, you must write Lengths.

The Manner of dividing the first Face.

The first Part of this Face, which is not divided, represents the Diameter of a Cylindrical Vessel holding one Paris Pint; therefore it’s Length must be 3 Inches, 11$\frac{1}{3}$ Lines.

And for dividing this Face, lay off upon it the Divisions of the second Face; but instead of writing 5, 10, 15, 20, 25, &c. write their Doubles 10, 20, 30, 40, 50, &c. and subdivide the Intervals into 10 Parts, and at the End of this Face write Diameters.

The Use of the New Gauging-Rod

Measure the Length of the mean Diameter of the Vessel with the Face of Diameters of your Rod, which suppose to be 153.00. Likewise take the Length of the Vessel with the second Face of your Rod, which suppose to be; add these two Numbers together, then seek their Sum 245.85 upon the third Face, and over against it, on the fourth Face, you will have 36 Septiers, or 288 Pints.

$$ \begin{array}{r} &153.00\\ +&92.85\\ \hline &245.85 \end{array} $$

But to make the Use of this Rod general; suppose the Weight of a Pint of fresh Water of some Country be 50 Ounces Avoirdupoise; then seek 31, the Number of Ounces Avoirdupoise a Paris Pint of fresh Water weighs, upon the fourth Face of Septiers of the Rod, which will be found against 239.4 on the third Face.

Likewise, against 50 on the fourth Face, answers 260.2 on the third Face.

$$ \begin{array}{r} &\text{Then from }160.2\\ +&\text{Take }239\ \ \ \\ \hline &\text{And there remains }20.8 \end{array} $$

$$ \begin{array}{r} &\text{Again from from }245.85 \text{ before found}\\ +&\text{Take }20.80\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \hline &\text{And there remains }225.05\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \end{array} $$

Now against this Number 225.05, on the third Face, you will find, on the fourth Face, 22 Septiers 2 Pints, or 178 Pints, which is the Number of Pints of that Country a Vessel of the aforesaid Dimensions holds.

Section VII. Of the Use of the Line of Metals.

Use I. The Diameter of a Ball, of any one of the six Metals, being given; to find the Diameter of another Ball of any one of them, which shall have the same Weight.

Open the Sector, and taking the given Diameter of the Ball between your Compasses, apply it’s Extremes to the Characters upon each Line of Metals, expressing the Metal the Ball is made of. The Sector remaining thus opened, take the Distance of the Characters of the Metal the sought Diameter is to be of, upon each Line of Metals, and this will be the Diameter sought. As, for Example, let AB be the Diameter of a Ball of Lead, and it be required to find the Diameter of a Ball of Iron, having the same Weight. Open the Sector, so that the Distance between the Points ♄ and ♄ be equal to the Line AB: The Sector remaining thus opened, take the Distance of the Points of ♂ on each Line of Metals, and that will give CD, the Length of the Diameter sought. If, instead of Balls, similar Solids of several Sides had been proposed, make the same Operation, as before, for finding each of their homologous Sides, in order to have the Lengths, Breadths, and Thicknesses of the Bodies to be made.

Use II. To find the Proportion that each of the six Metals have to one another, as to their Weight.

For Example; It is required to find what Proportion two similar and equal Bodies, but of different Weights, have to one another.

Having taken the Distance from the Center of the Joint of your Sector, to the Point of the Character of that Metal of the two proposed Bodies which is least (and which is always more distant from the Center), apply the said Distance across to any two equal Divisions on both the Lines of Solids. The Sector remaining thus opened, take the Distance on the Line of Metals, from the Center of the Joint to the Point, denoting the other Metal: and applying it to both Lines of Solids, see if it will fall upon some equal Number on each Line; if it will, that Number, and the other before, will, by permuting them, shew the Proportions of the Metals proposed.

As, for Example: To find the Proportion of the Weight of a Wedge of Gold, to the Weight of a similar and equal Wedge of Silver.

Now because Silver weighs less than Gold, open the Sector, and having taken the Distance from the Center of the Joint to the Point ☽︎, apply it to the Numbers 50 and 50 on each Line of Solids. The Sector remaining thus opened, take the Distance from the Center to the Point ☉︎, and applying it on each Line of Solids, and you will find it to fall nearly upon the 27th Solid on each Line. Whence I conclude, the Weight of the Gold to the Weight of the Silver, Is as 50 to 27$\frac{1}{6}$, or as 100 to 54$\frac{1}{3}$; that is, if the Wedge of Gold weighs 100 Pounds, the Wedge of Silver will weigh 54$\frac{1}{3}$ Pounds, and so of other Metals, whose Proportions are more exactly laid down by the Numbers of Pounds and Ounces that a cubick Foot of each of the Metals weighs, as is expressed in the Table adjoining to the Proof of the Line of Metals. If nevertheless their Proportions are required in lesser Numbers, you will find, that if a Wedge of Gold weighs 100 Marks, a Wedge of Lead, of the same Bigness, will weigh about 60$\frac{1}{2}$, one of Silver 54$\frac{1}{2}$, one of Brass 47$\frac{1}{4}$, one of Iron 42$\frac{1}{10}$, and one of Tin 39 Marks.

Use III. Any Body of one of the six Metals being given; to find the Weight of any one of the five others, which is to be made similar and equal to the proposed one.

For Example; Let a Cistern of Tin be proposed, and it is required to make another of Silver equal and similar to it. First weigh the Tin-Cistern, which suppose 36 Pounds. This being done, open the Sector, and having taken the Distance from the Center of the Sector to the Point ☽︎ (which is the Metal the new Cistern is to be made of), apply that Distance to 36 and 36 on each Line of Solids. Then take the Distance, upon the Line of Metals, of the Point ♃, from the Center; and applying that Distance cross-wise on each of the Lines of Solids, you will find it nearly fall upon 50 and 50 on each Line: Whence the Weight of a Silver Cistern must be 50$\frac{1}{4}$ Pounds, to be equal in Bigness to the Tin-Cistern. The Proof of this Operation may be had by Calculation, viz. in multiplying the different Weights reciprocally by those of a Cubic Foot of each of the Metals. As, in this Example; multiplying 720 lib. 12 Ounces, which is the Weight of a Cubick Foot of Silver, by 36 lib. which is the Weight of the Tin-Cistern; and again, multiplying 516 lib. 2 Ounces, which is the Weight of a Cubick Foot of Tin, by 50$\frac{1}{4}$ Pounds, which is the Weight of the Silver Cistern, the two Products ought to be equal.

Use IV. The Diameters, or Sides, of two similar Bodies of different Metals, being given; to find the Ratio of their Weights.

Let, for Example, the Diameter of a Ball of Tin be the right Line EF, and the Line GH the Diameter of a Ball of Silver; it is required to find the Ratio of the Weights of these two Balls. Open the Sector, and taking the Diameter EF between your Compasses, apply it to the Points ♃ on each Line of Metals. The Sector remaining thus opened, take the Distance of the Points ☽︎ on each Leg of the Sector; which compare with the Diameter GH, in order to see whether it is equal to it: for if it be, the two Balls must be of the same Weight. But if the Diameter of the Ball of Silver be lesser than the Distance of the Points D, on each Leg of the Sector, as here KL is, it is manifest that the Ball of Silver weighs less than the Ball of Tin; and to know how much, the Diameters GH and GL must be compared together. Wherefore apply the Distance of the Points D, which is GH, on each Leg of the Sector, to some equal Number on both the Lines of Solids; as, for Example, to the Numbers 60 and 60; then note upon what equal Number, on both Lines of Solids, the Diameter KL falls, which suppose 20: whence the Ball of Silver, whose Diameter is KL, weighs but $\frac{1}{3}$ of the Weight of the Ball of Tin, whose Diameter is EF.

Use V. The Weight and Diameter of a Ball, or the Side of any other Body, of one of the Six Metals, being given: to find the Diameter or homologous Side of another similar Body of one of the other five Metals, which shall have a given Weight.

Let, for Example, the right Line MN be the Diameter of a Ball of Brass, weighing 10 Pounds; and the Diameter of a Ball of Gold is required weighing 15 Pounds. You must first find the Diameter of a Ball of Gold, weighing as much as that of Silver, and then augment it by means of the Line of Solids. To do which, open the Sector, so that the Distance between the Points ♀, on each Line of Metals, be equal to MN. The Sector remaining thus opened, take the Distance of the Points ☉︎ and ☉︎ on each Line of Metals, which suppose to be the Diameter of the Ball of Gold OP; then open the Sector again, and apply this Distance to 10 and 10 of the Line of Solids, on each Leg of the Sector: The Sector remaining thus opened, take the Distance from 15 to 15, on each Line of Solids, and this last Extent QR will be the Diameter of a Ball of Gold weighing 15 Pounds.

¹No, nor never will, it being impossible. The most exact Proportion in small Numbers, is that of Adrian Metius, viz. 113 to 355.