« ZurückWeiter »
you have only to take 35 laterally from the Live of Chords, and apply it parallel-wise from 60° to 60° in the same lines, and they will be opened to the given angle of 35o.
If to the angle 35o you add the angle 6", which they contain, the sum is 41°; then take 41° laterally from the Line of Chords, and apply it parallel from 00 to 60, the sides or edges of the sector will contain the same angle of 35 degrees. And, in this case, the sector becomes a general Recipe-Angle, which is an instrument for taking the quantity of any angle contained between two Inclining Planes, as those in fortifications, &c.
9. Of Proportional Compasses. Though this sort of Compasses does not pertain to a common case of instruments, yet a short account of their nature and use may not be unacceptable to those who are not acquainted with them. They consist of two parts or sides of brass, which lie upon each other, so nicely as to appear but one when they are shut. These sides easily open, and move about a centre, which is itself moveable in a hollow canal cut through the greatest part of their length. To this centre on each side is affixed a sliding piece of a small length with a fine line drawn on it serving as an Index, to be set against other lines or divi. sions placed upon the Compasses on both sides. These lines are, 1. A Line of Lines. 2. A Line OP SUPERFICES, Arkas, or Plans. 3. A Line of Solids. 4. A LINE of CIRCLES, or rather of Polygons, to be inscribed in circles.
These lines are all unequally divided, the three first from 1 to 10; the last from 6 to 20; their uses are as follow :
By the Line of Lines, you divide a given line into ang number of equal parts; for, by placing the Index against 1, and screwing it fast, if you open the Compasses, then the distance between the points at each end will be equal.-If you place the Index against 2, and open the Compasses, the distance between the points of the longer legs will be twice the distance between the shorter ones ; and thus a line is bisected, or divided into two equal parts. If the Index be placed against 3, and the Compasses opened, the distances between the points will be as 3 to 1, and so a line is divided into three equal parts ; and thus you proceed for any number of parts under 10.
The number of the Line of Planes answer to the Squares of those in the Line of Lines ; for, because Superfices or Planes are to each other as the squares of their like sides, therefore if the Index be placed against 2 in the Line of Planes, then the distance between the small points will be the side of a plane whose area is 1 ; but the distance of the larger points will be the like side of a plane whose area is 2, or twice as large.--If the Index be placed at 3, and the Compasses opened, the distances between the points at each end will be the like sides of planes, whose areas are as i to 3, and so of others.
The numbers of the Line of Solids answer to the Cubes of those in the Line of Lines ; because all solids are to each other as the cubes of their like sides or diameters; therefore, if the Index be
placed at No. 2, 3, 4, &c. in the Line of Solids, the distances beiween the lesser and larger points will be the like sides of solids, which are to each other as 1 to 2, 1 to 3, 1 to 4, &c. For Example, If the Index be placed at 10, and the Compasses be opened, so that the small points may take the diameter of a bullet weighing 1 ounce, then the distance between the larger points will be the diameter of a bullet or globe of 10 ounces, or which is 10 times as large.
Lastly, the numbers in the LINE of Circles are the sides of Polygons, to be inscribed in a given circle, or by which a circle may be divided into those equal parts from 6 to 20. Thus, if the Index be placed at 6, the points of the Compasses at either end, when opened to the radius of a given circle, will contain the side of a Hexagon, or divide the circle into 6 equal parts. If the Index be placed against 7, and the Compasses opened, so that the larger points may take in the radius of the circle ; then the shorter points will divide the circle into 7 equal parts for inscribing a Heptagon.--Again, placing the Index to 8, and opening the Compasses, the larger points will contain the radius, and the lesser points divide the circle into 8 equal parts, for inscribing an Octagon or Square. And thus you proceed for otbers.
Prob. 1.-To bisect a given line AB.
3. Through the points fC, draw the line fC, and it will be the perpendicular required.
When the point is at, or near, the end of the line.
a 1. From the points AB, as cer.tres, with any distance greater than half AB, describe arcs cutting each other in c and d.
A 2. Draw the line cd and the point E, where it cuts AB, will 1. Take any point d above the be the middle of the line required. line, and with the radius or dis
tance dC, describe the arc eCf,
cutting AB in e and C. Prob. 2.- From d given point 2. Through the centre d and C, in a given right line AB, to the point e, draw the line edf, erect a perpendicular.
cutting the arc eCf, in f. When the point is near the middle 3. Through the points f and C, of the line.
draw the line fC, and it will be the perpendicular required.
Prob. 3.-From a given point C, out of a given right line AB, to let fall a perpendicular.
Ad 1. On each side of the point C take any two equal distances Cd and Ce. 2. From d and
e, with dius greater than Cd, or Ce, describe
arcs cutting each other in f.
1. From the point C, with any
radius, describe the arc de, cutting 1. Froin the point B, with any AB in e and d.
radius, describe the arc AC. 2. From the points end with the 2. From A and C, with the same, or any other radius, describe same or any other radius, describe two arcs cutting each other, in f. arcs cutting each other in d.
3. Through the points C,f, draw 3. Draw the line Bd, and it will the line CDs, and CD wül be the bisect the angle ABC, as was reperpendicular required.
Prob. 4.- At a given point D, upon the right line DE, to make an angle equal to a given angle aBl.
Prob. 6.- To trisect, or divide, a right angle ABC into three equal angles.
1. From the point B, with any 1. From the point B, with any radius BA, describe the arc AC, radius, describe the arc al, ciitting cutting the legs BA, and BC, in the legs Ba, Bb, in the points a and b. A and C. 2. Draw the line De, and from
2. From the point A, and C, the point D, with the same radius with the radius AB, or BC, cross as before, describe the arc ef, cut- the arc AC in d, and e. ting DE in e.
3. Through the points e,d, draw 3. Take the distance la, and the lines Be, Bd, and they will apply it to the arc ef, from e to f. trisect the angles as was required.
4. Through the points D,f, draw the line Df, and the angle eDf, will be equal to the angle b Ba, as Prob. 7. -Through a given was required.
point C, to draw a line parallel to
a given line AB. Prob. 5.—To divide a given an. gle ABC into two equal angles.
When the parallel is to be at a given de, upon the circle, from the given distance EF from AB. point B, and draw the chord eB.
2. Upon B, as a centre, with E F
the distance Bd, describe the arc C
D fdg, cutting the chord eB in f.
3. Make dg equal to df, through 8 draw gb, and it will be the tan
gent required. A
B 1. From any two points c and d, in the line AB, with a radius equal
Prob. 10.-A circle ABC being to EF, describe the arcs e and f.
given, and a tangent CH to that 2. Draw the line CD, to touch circle, to find the point of contact. those arcs without cutting them, and it will be parallel to AB as was
Prob. 8.-To draw a tangent to a given circle, that shall pass through a given point A.
1. Take any point e, in the tangent CH, and join eG.
2. Bisect eG in f, and with the radius fe, or fG, describe the semicircle eCG, cutting the tangent and
the circle in C, it will be the point 1. From the centre O, draw the required. radius OA.
2. Through the point A, draw DE perpendicular io OA, and it
Prob. 11.-Given three points will be the tangent required.
A, B, C, not in a straight line, to
draw a circle through them. Prob. 9.-To draw a tangent to a ciicle, or any segment of a circle
B ABC, through a given point B, without making use of the centre of the circle.
1. Bisect the lines AB, and BC, by perpendiculars, meeting at du
2. Upon do with the distance
dA, dB, or dC, describe AB i Take any two equal arcs Bd, will be the circle required.