« AnteriorContinuar »
PROBLEM III. To erect a Perpendicular from the end, or any part of a given Line. Fig. 24.
Open the Dividers to any convenient distance, as from D to A, and with one foot on the Point D, from which the Perpendicular is to be erected, describe an Arch, as AEG ; set off the same distance from A to E and from E to G; upon E and G describe two Arches to intersect each other at H; draw a Line from H to D, and one Line will be perpendicular to the other. Note. There are other methods of erecting a Per
pendicular, but this is the most simple. PROBLEM IV. From a given Point, as at C, to flrop a Perpendicular on a given Line AB. Fig. 25.
With one foot of the Dividers in C describe an Arch to cut the given Line in two places, as at F and G; upon F and G describe two Arches to intersect each other below the Line as at D ; lay a Rule from C to D and draw a Line from C to the given Line.
Perpendiculars may be more readily raised and let fall, by a small Square made of Brass, Ivory or Wood.
PROBLEM V. To make an Angle at E, equal to a given Angle ABC. Fig. 26.
Open the Dividers to any convenient distance, and with one foot in B describe the Arch FG; with the same distance and one foot in E, describe an Arch from H; measure the Arch FG, and lay off the same distance on the Arch from H to I; draw a Line through I to E, and the Angles will be equal.
PROBLEM VI. To make an Acute Angle equal to a given number of Degrees, suppose 36. Fig. 27.
Draw the Line AB to any convenient length; from a Scale of Chords take 60 Degrees with the Dividers, and with one foot in B describe an Arch from the Line AB; from the same Scale take the given number of Degrees, 36, and lay it on the Arch from C to D ; draw a Line from B through D, and the Angle at B
PROBLEM VII. To make an Obtuse Angle, supa pose of 110 Degrees. Fig. 28.
Take a Chord of 60 Degrees as before, and describe an Arch greater than a Quadrant ; set off 90 Degrees from B to C, and from C to E set off the excess above 90, which is 20 ; draw a Line from G through E and the Angle will contain 110 Degrees. Note: In a similar manner Angles may be measured ;
that is, with a Chord of 60 Degrees describe an Arch on the angular Point, and on a Scale of Chords measure the Arch intercepted by the
Lines forming the Angle. A more convenient method of making and measuring Angles is to use a Protractor instead of a Scale and Dividers.
PROBLEM VIII. To make a Triangle of three given Lines, as BO, BL, LO. Fig. 29.
Draw the Line BL from B to L; from B, with the length of the Line BO, describe an Arch as at O; from L, with the length of the Line LO, describe another Arch to intercept the former; from O draw the Lines OB and OL, and BOL will be the Triangle required.
PROBLEM IX. To make a Right Angled Triangle, the Hypothenuse and Angles being given. Fig. 30.
Suppose the Hypothenuse GA 25 Rods or Chains, the Angle at C 35° 30' and consequently the Angle at A 54° 30'. See Note after the 39th Geométrical Defi. nition. Note. When Degrees and Minutes are expressed,
they are distinguished from each other by a small Cypher at the right hand of the Degrees, and a Dash at the right hand of the Minutes ; thus 35°
30'is 35 Degrees and 30 Minutes. Draw the Line CB an indefinite length ; at C make an Angle of 35° 30%"; through where that number of Degress cuts the Arch draw the Line CA 25 Rods,
drop a Perpendicular from A to B, and the Triangle will be completed. Note. The length of the two Legs may be found by
measuring them upon the same Scale of equal
parts from which the Hypothenuse was taken. PROBLEM X. To make a Right Angled Triangle, the Angles and one Leg being given. Fig. 31.
Suppose the Angle at C 33° 15', and the Leg AC 285.
Draw the Leg AC making it in length 285 ; at A erect a Perpendicular an indefinite length ; at C make an Angle of 33° 15' ; through where that number of Degrees cuts the Arch draw a Line till it meets the Perpendicular at B. Note. If the given Line CA should not be so long
as the Chord of 60°, it may be continued beyond
A, for the purpose of making the Angle. PROBLEM XI. To make a. Right Angled Triangle, the Hypothenuse and one Leg being given. Fig. 32.
Suppose the Hypothenuse AC 40, and the Leg AB 28.
Draw the Leg AB in length 28 ; from B erect a Perpendicular an indefinite length; take 40 in the Dividers, and setting one foot in A, wherever the other foot strikes the Perpendicular will be the Point C. Note. When the Triangle is constructed the Angles
may be measured by a Protractor, or by a Scale of Chords.
PROBLEM XII. To make a Right Angled Triangle, the two Legs being given. Fig. 33.
Suppose the Leg AB 38, and the Leg BC 46.
Draw the Leg AB in length 38; from B erect a Perpendicular to C in length 46; and draw a Line from A to C..
PROBLEM XIII. To make an Oblique Angled Triangle, the Angles and one Side being given. Fig. 34.
Suppose the Side BC 98 ; the Angle at B 45° 15', the Angle at D 108° 30', consequently the other Angle 26° 15'.
Draw the Side BC in length 98 ; on the Point B make an Angle of 45° 15'; on the point C make an Angle of 26° 15', and draw the Lines BD and CD.
PROBLEM XIV. To make an Oblique Angled Triangle, two sides and an Angle opposite to one of them being given. Fig. 35.
Suppose the Side BC 160, the Side BD 79, and the Angle at C 29° 9'.
Draw the Side BC in length 160 ; at C make an Angle of 29° 9', and draw an indefinite Line through where the Degrees cut the Arch; take 79 in the Di. viders, and with one foot in B lay the other on the Line CD; the point D will be the other Angle of the Triangle.
PROBLEM XV. To make an Oblique Angled Triangle, two Sides and their contained Angle being given. Fig. 36.
Suppose the Side BC 109, the Side BD 76, and the Angle at B 101° 30'.
Draw the Side BC in length 109; at B make an Angle of 101° 30', and draw the Side BD in length 76 ; draw a Line from D to C and it is done.
PROBLEM XVI. To make a Square. PLATE II. Fig. 37.
Draw the Line AB the length of the proposed Square ; from B erect a Perpendicular to C and make it of the same length as AB; from A and C, with the same distance in the Dividers, describe Arches intersecting each other at D, and draw the Lines AD and DC.
PROBLEM XVII. To make a Parallelogram. Fig. 38.
Draw the Line AB equal to the longest Side of the Parallelogram ; on B erect a Perpendicular the length
and from A, with the shortest Side, describe Arches intersecting each other at D, and draw the Lines AD and DC.
PROBLEM XVIII. To describe a Circle which shall pass through any three given Points, not lying in a Right Line, as A, B, D. Fig. 43.
Draw Lines from A to B and from B to D; bisect those Lines by PROBLEM II. and the point where the bisecting Lines intersect each other, as at C, will be the Centre of the Circle.
PROBLEM XIX. To find the Centre of a Circle.
By the last PROBLEM it is plain, that if three Points be any where taken in the given Circle's Periphery, the Centre of the Circle may be found as there taught.
Directions for constructing irregular Figures of four or more Sides may be found in the following Treatise on SURVEYING