Elementary mechanical drawing, for school and shop

Front matter

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Gcfectic ^erieA.

Preface

To the Teacher.

In my endeavors to develop in the minds of pupils in grammar and high school classes a clear comprehension of the elementary principles of mechanical drawing, it was found that something more was needed than the copying of a few isolated problems into drawing-books. The student gained a better grasp of this subject by solving a limited number of written problems, illustrating a principle or method of mechanical drawing that had been previously explained by the teacher from the blackboard.

The problems thus accumulated have been arranged in the following pages with a view to economizing the teacher's time as well as to furnish the student with facilities for independent work.

To the Mechanic.

It has been my aim in the following pages to illustrate only one new principle or method of mechanical drawing in the same problem, and to

6 PREFACE.

make no more problems in any given case than are necessary to sufficiently illustrate the particular point under discussion. By so doing, I have been enabled to present short problems, and thereby to make each point of interest stand out in bold relief, and at the same time greatly to economize the time of the student. But this plan has prevented making the problems, as a whole, to apply to any particular trade or profession. This is no detriment to the work, however, although the individual who is seeking to learn that part of mechanical drawing which most especially applies to a particular trade may at first glance be disappointed, and think there is nothing here for him. To such let me say that the principles of mechanical drawing are every-where the same. Master these, and their application to ship-building, stairbuilding, architecture, sheet-metal work, etc., etc., is an easy matter.

Frank Aborn.

Cleveland, O., July, 1886.

Contents

Draughting Tools

Part 1.— Geometrical Drawing. I. — Straight Lines

.

IL— The Circle .

III. — Angles .

IV. — Triangles

Y. — Perpendicular Lines

.

YI. — Tangent Lines

.

VII. — Parallel Lines

.

fill. — To Divide a Right Line into a Number of Equal Parts

IX.— The Ellipse .

.

Part II. — Constructive Drawing. Introduction I. — Scale Drawing

43 45

CONTENTS.

CHAPTER. PAGE

II.— Plans 48

III.— Elevations 61

IV. — Points in Inclined Surfaces ' . . 77 V. — The Description of Points in a Surface of Regular Curvature . . 85 VI. — The Intersection of Horizontal Lines

with Inclined Surfaces ... 91

VII.— Conic Sections 94

VIII.— The Helix 96

IX. — Section Drawing ..... 103 X. — Foreshortened Dimensions . . . 106 XI. — Development of Surfaces . . . 109 XII. — Development of Oblique Conical Surfaces 119

ELEMENTARY

MECHANICAL DRAWING.

DRAUGHTING TOOLS.

The Pencil.

For Geometrical and Constructive drawing, the pencil must be of such quality and hardness that it will take and hold a fine point.

The H grade Faber pencil is the best for school work.

Sharpening the Pencil. — For ruled lines, the pencil should be sharpened to a flat edge (Fig. 1), as with the common round point it is impossible to make the lines fine and of uniform width, which they must be if accuracy is to be secured.

For free-hand lines, nothing will do but a keen, common, round point.

As there are free-hand lines to be drawn, even in mechanical drawing, the student should have two pencils, one of which is sharpened to a flat, and the other to a round, point.

Fig. 1.

The Scale.

A Scale is a graduated ruler.

For school work, a common box-wood ruler, with beveled edge, and graduated to sixteenths of an inch, will serve the purpose very well.

Dividers.

There is no tool that is so irredeemably bad and entirely worthless as the dividers that are cheap imitations of the better class of instruments. They look well to the uninitiated, but the set screws do not fit, the joints are rough and soon grind themselves loose. Then, nothing will keep its place, and the instrument becomes unreliable and therefore useless. But the Prang School Divider, made of sheet metal, and costing twenty-five cents, is thoroughly reliable in every way for school work. If, however, one wishes to have something better

than this, there is nothing in the market, that it is economy to buy, at a less cost than three dollars; and a set of instruments at this price should consist only of a pair of dividers, with needle-point, pencil-point, pen, and lengthening-bar (Fig. 2) ; and any set of instruments consisting of a greater number of pieces than this must of necessity be correspondingly higher in price or poorer in quality.

DRAUGHTING TOOLS.

T-Square.

A T-Square is an instrument, usually made of wood, consisting of a blade and head at right angles to each other (Fig. 3).

For school purposes, a T-square should have a blade not more than twelve inches in length.

Head

Fig. 3.

Tig. 4.

Fig. 5.

Set-Square.

A Set-Square is an instrument in the form of a right-angled triangle (Figs. 4 and 5).

Protractor.

A Protractor is

an instrument usually in the form of a semicircular disk, having its periphery Fi s- 6 «

graduated into 180 degrees (Fig. 6).

For school work, a protractor made of horn, and three inches in diameter, is the best.

Summary.

Two H Faber pencils.

One pair Prang School Dividers.

One T-square, twelve inch blade.

One set-square, four inches on longest edge.

One box-wood foot rule, beveled edge, graduated to sixteenths of an inch.

One horn Protractor, three inches in diameter.

One slate, eight inches by ten inches; or, drawing board, ten inches by twelve inches, and suitable paper and thumb tacks.

PART I.

GEOMETBICAL DRAWING.

Geometrical Drawing is the description of lines arranged in conformity to some general rule called a geometrical law.

Lilies are of two kinds, — straight and curved.

Straight Lines are horizontal, vertical, and oblique, according to their direction with reference to the plane of the earth's surface.

Curved Lines are either regular or irregular.

A curve is Regular when its curvature follows an established law.

A curve is Irregular when its curvature is not governed by any known rule or law.

Section I. — Horizontal Lines

All level lines, i. <?., lines parallel to the plane of the earth's surface, are horizontal. >

Horizontal Lines are drawn with the help of the T-square.

Prob. 1. — Draw a horizontal line 4 in. long.

MECHANICAL DRAWING.

Fig. 7.

Fig. 8.

Solution. — Place the head of the T-square firmly against the left-hand edge of the slate or drawing board (Fig. 7,) and draw a line that is 4 in. in length along the edge of the blade.

Prob. 2. — Draw a horizontal line 2- 1 in. long.

Prob. 3.— Draw four horizontal lines 2>\ in. long and \ in. apart.

Prob. 4. — Draw three horizontal lines, 1 in., 2 in., and 3 in. in length, and f in. apart.

Prob. 5. — Draw six horizontal lines, \ in. apart and 2\ in. long, with their ends in a plumb line.

Section II. — Vertical Lines

All plumb lines, i. e., lines that are perpendicular to the plane of the earth's surface, are vertical.

Tertical Lines are represented with the aid of the T-square and set-square.

Prob. 1. — Draw a vertical line 3-| in. long.

Solution. — Place the head of the T-square firmly against the left-hand edsre of the slate or drawing

£1 o o

board, and while in this position set one of the shorter edges of the set-square against it (Fig. 8.) Now, along the upright edge of the set-square, draw a line 3^- in. long, which will be the line required in the problem.

Prob. 2. — Draw a vertical line 41 in. long.

Prob. 3. — Draw two vertical lines 2f in. long and J in. apart.

Prob. 4. — Draw" four vertical lines \ in., 1 in., 2 in., and 3J in. long, and f in. apart.

CHAPTER II.— The Circle. Section L — Definitions

A Circle is a plane figure bounded by a curved line, called its circumference, every part of which is equally distant from a point within it called its center.

A Diameter of a circle is a straight line joining two points in the circumference, and passing through the center. Every circle may have an infinite number of diameters.

A Radius is a line extending from the center to the circumference. It is one half of a diameter.

Prob. 1. — Describe a circle -^2" m - m radius.

Solution. — Set the dividers so that the distance between the needle-point and the pencil-point is -^ in. Place the needle-point in the position of the center C, and holding the needle-point leg as nearly upright as possible, revolve the pencil leg about it, so that the pencil shall describe a continuous line ABD, every part of which is equally distant from its center C.

Prob.

2.— Describe a circle 1 in. radius.

Prob.

3. — Describe a circle ^ in. radius.

Prob.

4. — Describe a circle 1^ in. radius.

Prob.

5. — Describe a circle -^ in. radius.

Prob.

6. — Describe a circle 1^ in. diameter.

Prob.

7. — Describe a circle 1 in. diameter.

Prob.

8. — Describe a circle 1-J- in. diameter.

Prob.

9. — Describe a circle If in. diameter.

Prob.

10. — Describe a circle ^g 5 - in. diameter.

Section II — Arcs of Circles.

Any portion of a circumference less than the whole is called an arc.

Every circumference is considered as consisting of 360 equal arcs.

Each of these 360 arcs is called an arc of 1 degree.

The name of an arc depends upon the number of degrees that it contains.

One fourth of a circumference is an arc of 90 degrees, and is written 90°.

One third of a circumference is an arc of 120 degrees, and is written 120°.

Three fourths of a circumference is an arc of 270 degrees, and is written 270°.

GEO ME TRICA L DBA WING.

One half of a circumference is an arc of 180 degrees, and is written 180°, etc., etc.

A Protractor used in school work is usually a semicircular disk, and -therefore its arc contains 180°. Usually these degrees are marked in two lines. One of these lines gives the number of degrees, counting from the left-hand end of the diameter, and one gives the number of degrees counting from the right-hand end of the diameter.

Prob. 1. — Describe an arc of 60°, with a radius of in.

/b

Solution. — Describe a circle ABD, | in. in radius.

Draw a diameter AD. Place the protractor on the circle ABD, so that its center and diameter coincide with the center and diameter of the circle, and mark the 60° point, 6, at its edge. (Fig. 11.) Remove the protractor, and draw the line bBC to the center. (Pig. 12.) The point B, where the line crosses the circumference, will be one end of the required arc; the

MECHANICAL DRA WING.

other end is at D, the end of the diameter from which it was measured. BD is the required arc.

Prob. 2. — Describe an arc of 120°. Radius, -| in. Prob. 3. — Describe an arc of 90°. Radius, 1-J- in. Prob. 4. — Describe an arc of ,30°. Radius, If in. Prob. 5. — Describe an arc of 270°. Radius, 1^- in.

Prob. 6. — Describe an arc of 45°. Radius

> 32

Section III. — To (/escribe an arc with a given radius that will be equal in length to the circumference of a given circle.

The length of a degree varies with the length of the radius of the circle.

Prob. 1. — With a radius of | in., describe an arc that will be equal in length to the circumference of a circle having a radius of J in.

Solution. — To find what part of the circumference of the larger circle is equal to the whole circumference of the smaller, we divide the radius of the smaller circle hy the radius of the larger : !=-§-. Hence, Fig. 13. 8 I 1 '

an arc which is f of the circumference of a circle whose radius is f in. will be equal to the entire circumference of a circle ha vine: a radius of in.

As there are 360° in an entire circumference, \ of it would be \ of 360° = 51.43°, and f of it would be three times 51.43° =

154.29°. Describe an arc, ADJB, having a radius of | in., and on it lay off AB containing 154.29°. This arc will be equal in length to the circle having a radius of f in. (Fig. 13.)

Prob. 2. — With a radius of 1\ in., describe an arc of a circle which is equal to the circumference of a circle of J in. radius.

Prob. 3. — With a radius of 2\ in., describe an arc of a circle which is equal to the circumference of a circle of J in. radius.

Prob. 4. — Describe an arc of a circle having a radius of 1\ in., which is equal to the circumference of a circle, radius If in.

Prob. 5. — Describe an arc, radius If in., equal to the circumference of a circle, radius 1^ in.

Prob. 6. — Describe an arc, radius If in., equal to the circumference of a circle, radius -^ in.

Eule. — Divide the radius of the given circle by the radius of the required arc. Reduce the resulting fraction to degrees, and lay off the number of degrees thus found on the circumference of a circle having the radius of the required arc. The arc so laid off will be equal to the circumference of the given circle.

Section IV. — To find the Circumference of a Circle

The Circumference of a circle is equal to the product of the diameter multiplied by 3.1416.

Note. — The exact ratio between the diameter and the circumference of a circle can not be given in figures, but 3.1416 is near enough for all ordinary purposes. M. D.-2.

Prob. 1. — What is the length of the circumference of a circle that is 11 in. in diameter?

Solution. — As the diameter is 1.5 in., the length of the circumference must be 3.1416 times 1.5 in., which equals 4.7124 in. Hence, 4.7124 in. is the length of a circle 1J in. in diameter.

Prob. 2. — What is the length of the circumference of a circle having a diameter of 2 in. ?

Prob. 3. — What is the length of the circumference of a circle 41 in. radius ?

Prob. 4. — What is the diameter of a circle having a circumference of 6.2832 in.?

Prob. 5. — What is the length of the circumference of a circle having a radius of 3 in.?

Prob. 6. — What is the radius of a circle having a circumference of 10 in. ?

Prob. 7. — Describe a circle having a circumference of 5.4978 in.

Prob. 8. — Describe a semi-circumference 2.7489 in. long.

Prob. 9. — Describe a quarter circumference 7.854 in. long.

Prob. 10. — Describe an arc of 45°, 5.8901 in. long.

Section V. — Bisection of Circular Arcs

To bisect an arc is to divide it into two equal parts.

Prob. 1. — Bisect an arc of 60°; radius, ^-| in.

Solution. — Describe a circle if in. in radius, and on its circumference lay off an arc of 60°, AB. With

A and B as centers, describe ^ _ A

arcs of equal radii intersecting

a straight line, aud this line, \

where it cuts AB in a, bisects

the arc AB. Aa and a^ are ^—''

equal. lg '

Prob. 2. — Bisect an arc of 90°, 1J in. radius. Prob. 3. — Bisect a semi-circumference } in. radius. Prob. 4. — Divide an arc of 120° into four equal arcs. Diameter, 1| in.

Prob. 5. — Divide an arc of /^I^>X 170°, 1J in. radius, into eight

// ^\\ equal arcs.

If \\ Prob. 6. — Represent the face

"" of a semicircular arch consisting

of eight equal blocks. Radius of inner circle, 1J in., and radius of outer circle, 2 in. (Fig. 15.)

Prob. 7. — An arch is 4 in. in s^V^lh^

diameter on the inner circle. /S^ ^\\

The key-stone is 20° wide and by \\

-J in. long measured on the /_] r □

radius, and on each side of the Fi s- 16 -

key are eight equal blocks f in. long measured on

the radii. Represent the face of the arch. (Fig. 16.)

Section VI. — Trisection of Circles

To trisect a circle is to divide it into three equal arcs.

MECHANICAL DRAWING.

A Chord of an arc is a line joining its extremities. (See Fig. 17.)

The chord of an arc of 60° is equal to the radius of the arc.

Prob. 1. — Draw a circle f in. in diameter, and divide its circumference into six equal arcs.

Solution. — Describe a circle AF, f in. in diameter, and, without changing the dividers, set the needle-point at any point in the circumference, A, and describe arcs, cutting the circumference in B and F. Set the needle-point at F and B, and describe arcs cutting the circumference in E and C. So proceed, and the last arcs will meet in D, and the circumference will be divided into six equal arcs. The chord of any of these arcs is equal to the radius of the circle.

To divide the circumference of a circle into three equal divisions: Divide the circumference into six equal arcs, and two of these arcs will constitute one third of the whole circumference.

To divide the circumference of a circle into twelve equcd divisions: Divide the circumference into six equal arcs, and then divide each of these into two equal arcs.

Prob. 2. — Describe two circles, 1J in. and f in. in diameter, on the same center, and divide the annular space between them into twelve equal parts.

Prob. 3. — Describe a semi-circumference, diameter 2f in., and divide it into three equal arc3.

Prob. 4. — Describe a circle ; divide its circumference into three equal arcs, and join the division points. The inscribed figure will be an equilateral triangle.

Prob. 5. — Describe a circle; divide its circumference into six equal arcs, and join the division points. The inscribed figure will be a hexagon (regular six sided polygon).

Section VII. — Concentric or Parallel Circles

Circles are concentric when they have a common center : a, b, d, and e are concentric circles because they have the common center C.

Note. — In describing circles, there is a strong temptation to incline the dividers to one side. In doing this, the center of the circle is so torn and enlarged that it

can not be used again. It becomes necessary, therefore, for the student to practice making concentric circles until he has learned to hold the dividers erect.

Prob. 1. — Describe six concentric circles, J in. apart, the radius of the largest circle being 2 in.

Prob. 2. — Describe six concentric circles, } in., 1 in., 1J in., 2 in., 2| in. in radius.

Prob. 3. — Describe three concentric circles, 3J in., 3 J in., and 3 in. in diameter.

Prob. 4. — Describe four concentric circles, ^g in., -j^- in., \^ in., and f| in. in radius.

Prob. 5. — Represent the top of your ink well.

Prob. 6. — Describe two concentric circles, 2 in. and 2J in. in diameter. Divide the circumference of the smaller one into six equal parts, and with each of the division points as a center, describe six pairs of concentric circles, similar and equal to the first.

Section VI II.— Tangent Circles.

Circles are tangent when their circumferences touch each other in one point only. Circles may be tangent externally or internally.

Externally Tangent Circles. — Circles are externally tangent when the distance between their centers is equal to the sum of their radii.

Prob. 1. — Describe two circles externally tangent to each other; radii, ^ in. and -^ in.

Solution. — As these circles are externally tangent, the distance between their centers will be equal to the sum of their radii : T 3 g in.-f- ^ in. = -J-J in. Draw a line CO J-f in. long (Fig. 20.) With Cas a center, and with a radius of T 3 g in., describe Fig - 20 - the circle A ; with O as a center,

and with a radius of ^ in., describe the circle B.

These circles, A and B, will be externally tangent at b.

Prob. 2. — Describe two externally tangent circles ; radii, J in. and J in.

Prob. 3. — Describe two externally tangent circles ; radii, ^f in. and f in.

Prob. 4. — Describe two externally tangent circles ; the sum of the radii to be 3 in., and one radius to be twice as long as the other.

Prob. 5. — Describe two concentric circles, J in. and f in. in radius, externally tangent to two similar and equal concentric circles, with the larger of one pair of circles tangent to the smaller of the other pair.

Rule.— Draw a line equal in length to the sum of the given radii, and, with the ends of this line as centers, describe the given circles. The circles thus described will be externally tangent, and the point of contact will be on the line joining their centers.

Internally Tangent Circles.— Circles are internally tangent when the distance between their centers is equal to the difference of their radii.

Prob. 1. — Draw two internally tangent circles ; radii, -^ in and -^ in.

Solution. — Since ^ — tq = ^-, the distance between the

centers of the given circles is T 6 g- in. Draw a line, Cc f (Fig. 21,) T 6 g- in. in length, and with c' and C as centers, describe two circles, abd and ABD, -^ in. and a>_ i n . radii. ABD and abd are internally tangent at a.

Prob. 2. — Describe two internally tangent circles, 1 in. and | in. radii.

Prob. 3. — Describe two internally tangent circles, If in. and 1J in. in diameter.

Prob. 4. — Describe two arcs of 90°, T 9 g in. and -J in. radii, internally tangent at one end, and two other similar arcs similarly tangent, having a common point of contact, and all the centers on the same right line.

Rule. — Draw a line equal to the difference of the radii of the tangent circles, and with the ends of this line as centers describe circles with the given radii. The circles so drawn will be internally tangent to each other.

CHAPTER III.— Angles

An angle is formed by the meeting of any two lines. (See Fig. 22.)

A

The point of meeting of two lines is called the apex of the angle.

If the apex of an angle be made the center of a circle, the arc intercepted by the sides of the angle is said to subtend the angle.

GEOMETRICAL DRAWING.

An angle is measured by the subtending arc. (See (Fig. 23).

-?»6, and 1J in. in

'Prob. 1. — Draw an angle of 30°.

Solution. — With any radius describe a circle ABB (Fig. 24).

Lay off on this circle an arc of 30°, AB. From the center, 0, draw \ radial lines through A and B. The arc AB, 30°, subtends and therefore measures the angle ACB.

Prob, 2. — Draw two lines, 1 in. length, forming an angle of 45°.

Prob. 3. — Draw three lines, 2 in. in length, meeting in a common point, and forming three angles of 120° each.

Prob. 4. — Draw three lines, 1J in. in length, meeting in a common point, and forming three angles of 210°, 60°, and 90°.

Prob. 5. — Draw a horizontal line, 2 in. long, bisecting two lines 1J in. in length, one of which makes an angle of 45°, and the other an angle of 135°, with the connecting line.

Rule. — Describe an arc containing the number of degrees in the required angle. Draw the radii at the ends of this arc. The angle formed by the radii thus drawn will be the angle required.

CHAPTER IV.-Triangles

A Polygon is a plane figure having many sides and an equal number of angles.

A Triangle is a polygon that has three angles.

Section I. — To draw a triangle token its three sides are given

Prob. 1. — Draw a triangle, the sides of which are •|| in., !~J in., and -^ in. in length.

Solution.— Draw a line, AB, -|| in. in length, for one of the sides of the required triangle. With one of the ends of this line, A, as a center, describe an arc, ac, T 7 g in. in radius, and with Flg ' 25 ' the other end of the line, B, as a

center, describe an arc fj in. radius, bd, cutting ac in C. Draw AC and BC, and the figure thus formed, ABC, is the required triangle, of which the sides are -|f in., |~J in., and -^ in. (Fig. 25.)

Prob. 2.— Draw a triangle, the sides of which are 1J in., 1J in., and 1 in. long.

Prob. 3.— Draw a triangle, the sides of which are 2 in., | in., and If in. long.

Prob. 4. — Draw an equilateral triangle, the sides of which are 1J in. in length.

Knle.— Draw a line equal in length to one of the given sides of the required triangle, and, with the ends of this line as centers, describe intersecting arcs, the radii of which are equal to the other sides. From the point of intersection of these arcs, draw lines to the ends of the line already drawn. The three lines thus drawn will be the three sides of the required triangle.

Section II. — To draw a triangle when one angle and two sides are given

Prob. 1. — Draw a triangle, of which one angle is 30°, and two of the sides are -^ in. and 1-^- in. in length.

Solution.— Draw any line AB, l T 3 g in. in length (Fig. 26.) At A draw a line AC, -fg in. in length, and forming an angle of 30° with AB. Draw OB. The triangle Fig. 26.

ABC, thus formed, is the required triangle.

Prob. 2. — Draw a triangle, of which one angle is 45°, and two sides are 2 in. and If in. in length.

Prob. 3. — Draw a triangle, of which one angle is 90°, and two sides are 3 in. and 2J in. in length.

Prob. 4. — Draw a triangle, of which one angle is 60°, and two of the sides are If in. and if in. in length.

Rule. — Draw two lines, equal in length to the given sides of the required triangle, and forming an angle equal to the given angle. Join the ends of these lines by a right line, and the three lines thus drawn will form the required triangle.

Section III. — To draw a triangle having given one side and two angles

Prob. 1. — Draw a triangle, one side of which is l T 5 g- in. and two angles of which are 30° and 45°.

Solution. — Draw a line, AB (Fig. 27), l T 6 g- in. in

length, for one of the sides

of the required triangle.

At A draw A C, making an

angle of 45° with AB, and

at B draw BC, making an

Fig ' 27 * angle of 30° with AB.

Produce these lines until they meet. Then ABC,

having one side 1^- in. in length, and two angles

of 30° and 45° is the required triangle.

Prob. 2. — Draw a triangle, of which one side is 1J in. in length, and two angles are 45° and 90°.

Prob. 3. — Draw a triangle, of which one side is 1 in., and two angles are 60° each.

Prob. 4. — Draw a triangle, of which one side is 2 J in. in length, and two angles are 30° and 22 J°.

Prob. 5. — Draw a triangle, of which one side is f in. in length, and two angles are 75° each.

Rule. — Draw a line equal in length to the given side of the required triangle. At the ends of this line, and on the same side, draw lines forming the given angles of the required triangle. Produce these lines until they meet. The figure formed by the three lines thus drawn will be the required triangle.

CHAPTER V. — Perpendicular Lines

Two lines are perpendicular to each other when the angles at their point of meeting are each 90°.

Section I. — To draw a line perpendicular to another line at a point in the middle of the given line.

Prob. 1. — Draw a line 1 in. long, and erect a perpendicular to it at a point midway between the ends.

Solution. — Draw a line, AB (Fig. 28), 1 in. in length. With A and B as centers, and the same radius, describe two arcs of circles, cutting

each other in b and b' on both A

sides of the line AB. Join b and b f , and the line bb' will be perpendicular to the line AB and cut it in the middle. Fig> 2 8.

Prob. 2. — Draw a line 2f in. in length, and cross it in the middle by a line perpendicular to it. The cross line to be If in. in length and equally divided by the first line.

Rule. — Draw a line equal to the given line, and with the ends of this line as centers, describe two arcs, with the same radius, cutting each other on both sides of the given line. Join the points of intersection of these arcs, and the line thus drawn will be perpendicular to the given line at a point in the middle.

Section II. — To draw a line perpendicular to another at a point not in the middle of the given line.