Machinery's Handbook Pocket Companion: Quick Access to Basic Data & More from the 32nd Edition

By Richard Pohanish and Christopher McCauley

Since publication of the first edition more than 100 years ago, Machinery’s Handbook has been acclaimed as the most popular, bestselling engineering resource of all time. Universally considered the principal reference in the manufacturing, mechanical, and metalworking industries, the Handbook is the ultimate collection of essential information.

A concise yet authoritative, highly useful reference that draws its content from the Machinery’s Handbook, the latest Machinery’s Handbook Pocket Companion is an ideal quick resource for students and professionals in manufacturing, metalworking, and related fields for whom convenient access to the most basic data is essential. This expertly curated collection includes tables, charts, and text selected from the Handbook. Much of the information has been reorganized, distilled, or simplified to increase the usefulness of this data-rich resource, while keeping it compact.

Designed as a time saver, the Pocket Companion is not intended to replace the new Machinery’s Handbook, 32nd Edition. Instead, it serves as a handy and more portable complement to the Handbook’s vast collection of text, data, and Standards information.
 
Features
 

• A handy, portable, and time-saving, quick-access complement to the larger compilation of explanatory text, extensive data, and vital Standards information in the Machinery’s Handbook.
• Revised to reflect numerous changes made in the 32nd Edition, the new Pocket Companion includes updated Standards, key revisions, and even more tables.
• Students and professionals will find the Pocket Companion a compact ready-reference to keep nearby while mastering fundamentals and solving problems, working on engineering designs, or seeking essential specifications daily on the shop or factory floor.

In addition to the printed book, the Pocket Companion also is sold as an eBook. For information on this handy format, as well as the Machinery’s Handbook 32 Digital Edition, the eBook version of the Machinery’s Handbook Guide, and other digital products, visit the Industrial Press eBookStore site at ebooks.industrialpress.com.

• Language: English
• Publisher: Industrial Press, Inc.
• Release date: Apr 15, 2024
• ISBN: 9780831199623

Book preview

MATHEMATICAL FORMULAS AND TABLES

Dimensions of Plane Figures

Square:

A = s² = ¼d²

Example: Side s of a square is 15 in. Find the area of the square and the length of its diagonal.

A = s² = 15² = 225 in²

d = 1.414s = 1.414 × 15 = 21.21 in

Example: The area of a square is 625 cm². Find the length of side s and diagonal d.

Rectangle:

Example: Side a of a rectangle is 12 cm, and the area is 70.5 cm². Find the length of side b and diagonal d.

Example: The sides of a rectangle are 30.5 and 11 cm. Find the area.

A = ab = 30.5 × 11 = 335.5 cm²

Parallelogram:

A = ab

a = A/b

b = A/a

Note: The dimension a is the length of the vertical drawn at a right angle to side b. Dimension a is also considered the height of the parallelogram.

Example: Base b of a parallelogram is 16 ft. Height a is 5.5 ft. Find the area.

A = ab = 5.5 × 16 = 88 ft²

Example: The area of a parallelogram is 12 in². The height is 1.5 in. Find the length of the base b.

b = A/a = 12/1.5 = 8 in.

Right

Triangle (one angle is a 90-degree angle):

From the Pythagorean theorem, a² + b² = c2, thus

Example: Side a is 6 in. and side b is 8 in. Find side c and area A :

Example: Side c = 10 and side a = 6. Find side b:

Acute Triangle (all three angles measure less than 90 degrees):

Example: Side b = 7 inches, h = 4 inches, so A = bh//2 = (7 in. × 4 in.)/2 = 28 in²/2 = 14 in²

Example: Side a = 10 cm, b = 9 cm, and c = 8 cm. Find the area.

Obtuse Triangle (one angle measures greater than 90 degrees):

Example: If b = 5 cm and h = 3 cm, then A = bh/2 = (5 cm × 3 cm)/2 = 15 cm²/2 = 7.5 cm²

Example: Side a = 5 in., side b = 4 in., and side c = 8 in. Find the area.

Trapezoid:

Note: In Britain, this figure is called a trapezium and the figure below it is known as a trapezoid, which is the reverse of the US terms.

Example: Side a = 23 meters, side b = 32 meters, and height h = 12 meters. Find the area.

Trapezium:

The area of a trapezium also can be found by dividing it into two triangles, as indicated by the dashed line. Each area is added to give the total area of the trapezium.

Example: Let a = 10 in., b = 2, c = 3 in., h = 8 in., and H = 12 in. Find the area.

Regular Hexagon:

A = 2.598s² = 2.598R² = 3.464r²

R = s = radius of circumscribed circle = 1.155 r

r = radius of inscribed circle = 0.866s = 0.866R

s = R = 1.155 r

Example: The side s of a regular hexagon is 40 millimeters. Find the area and the radius r of the inscribed (drawn inside) circle.

A = 2.598s² = 2.598 × 402 = 2.598 × 1600 = 4156.8 mm²

r = 0.866s = 0.866 × 40 = 34.64mm

Example: What is the length of the side of a hexagon circumscribed on (drawn around) a circle of 50 millimeters radius? In this case, because the hexagon is circumscribed on the circle, the circle is inscribed (drawn within) the hexagon. Hence, r=50 mm and 5 = 1.155 r = 1.155 × 50 = 57.75 mm

Regular Octagon:

A = area = 4.828s² = 2.828R² = 3.314 r²

R = radius of circumscribed circle = 1.307s = 1.082r

r = radius of inscribed circle = 1.207s = 0.924R

s = 0.765R = 0.828r

Example: Find the area and the length of the side of an octagon inscribed (drawn inside) in a circle of 12 inches diameter.

Diameter of circumscribed (drawn around) circle = 12 inches; hence, R = 6 in.

A = 2.828R² = 2.828 6² = 2.828 36 = 101.81 in²

s = 0.765R = 0.765 6 = 4.590 in.

Circle:

Area = A = πr² = 3.1416 r² = 0.7854 d²

Circumference = C = 2πr = 6.2832r = 3.1416 d

Length of arc for center angle of 1° = 0.008727d

Length of arc for center angle of n° = 0.008727nd

Example: Find area A and circumference C of a circle with a diameter of 2¾ inches.

A = 0.7854d² = 0.7854 × 2.752 = 0.7854 × 2.75 × 2.75 = 5.9396 in²

C = 3.1416d = 3.1416 × 2.75 = 8.6394 in

Example: The area of a circle is 16.8 in². Find its diameter.

Sector of a Circle:

Example: The radius of a circle is 35 millimeters, and angle a of a sector of the circle is 60 degrees. Find the area of the sector and the length of arc l.

A = 0.008727 αr² = 0.008727 × 60 × 35² = 641.41mm² = 6.41cm²

l = 0.01745rα = 0.01745 × 35 × 60 = 36.645 mm

Segments of a Circles

See also Segments of Circles starting on page 7.

Example: The radius r is 60 inches and the height h is 8 inches. Find the length of the chord c.

Example: If c = 16, and h = 6 inches, what is the radius of the circle of which the segment is a part?

Cycloid:

Area = A = 3πr² = 9.4248 r² = 2.3562d²

                = 3 × area of generating circle

Length of cycloid = l = 8 r = 4 d

Example: The diameter of the generating circle of a cycloid is 6 inches. Find the length l of the cycloidal curve and the area enclosed between the curve and the base line.

l = 4d = 4 × 6 = 24 in.        A = 2.3562d² = 2.3562 × 6² = 84.82 in²

Circular Ring (Annulus)::

Area = A = π(R²–r²) = 3.1416( R²–r²)

                = 3.1416(R + r)(R – r)

                = 0.7854( D² + d²) = 0.7854( D + d )( D – d)

Example: Let the outside diameter D = 12 centimeters and the inside diameter d = 8 centimeters. Find the area of the ring.

A = 0.7854(D² – d²) = 0.7854(122 – 82) = 0.7854(144 – 64) = 0.7854 × 80

                       = 62.83 cm²

By the alternative formula:

A = 0.7854(D + d)(D – d) = 0.7854(12 + 8)(12 – 8) = 0.7854 × 20 × 4

                           = 62.83 cm²

Sector of

Circular Ring:

Example: Find the area, if the outside radius R = 5 inches, the inside radius r =2 inches, and α = 72 degrees.

A = 0.00873α(R² – r²) = 0.00873 × 72(5² – 2²)

= 0.6286(25 – 4) = 0.6286 × 21 = 13.2 in²

Ellipse:

Area = A = πab = 3.1416 ab

An approximate formula for the perimeter is

Example: The larger, or major, axis is 200 millimeters. The smaller, or minor, axis is 150 millimeters. Find the area and the approximate circumference. Here, then, a = 100, and b = 75.

Spandrel or Fillet:

The shaded region is the spandrel (fillet).

Example: Find the area of a spandrel, the radius of which is 0.7 inch.

A = 0.00873α(R² – r²) = 0.00873 × 72(52 – 22)

Example: If chord c were given as 2.2 inches, what would be the area?

= 0.6286(25 – 4) = 0.6286 × 21 = 13.2 in²

Parabola:

Area = A = ⅔ xy

The area of the shaded portion is equal to two-thirds of a rectangle which has x for its base and y for its height.

Example: Let x in the illustration be 15 centimeters, and y be 9 centimeters. Find the area of the shaded portion of the parabola.

Parabola:

When x is small in proportion to y, the following is a close approximation:

Example: If x = 2 feet and y = 24 feet, what is the approximate length lof the parabolic curve?

Hyperbola:

Example: The half-axes a and b are 3 and 2 inches, respectively. Find the area shown shaded in the illustration for x = 8 inches and y=5 inches.

Inserting the known values in the formula:

Formulas and Table for

Regular Polygons.—The following formulas and table can be used to calculate the area, length of side, and radii of the inscribed and circumscribed circles of regular polygons (equal sided).

Area, Length of Side, and Inscribed and Circumscribed Radii of Regular Polygons

Segments of Circles for Radius = 1 (US Customary or Metric Units)

Formulas for segments of circles are given on page 4. When the central angle a and radius r are known, the tables on these pages can be used to find the length of arc l, height of segment h, chord length c, and segment area A. When angle a and radius r are not known, but segment height h and chord length c are known or can be measured, the ratio h/c can be used to enter the table and find α, l, and A by linear interpolation. Radius r is found by the formula on page 4. The value of l is then multiplied by the radius r and the area A by r², the square of the radius.

Angle α can be found thus with an accuracy of about 0.001 degree; arc length l with an error of about 0.02 percent; and area A with an error ranging from about 0.02 percent for the highest entry value of h/c to about 1 percent for values of h/c of about 0.050. For lower values of h/c, and where greater accuracy is required, area A should be found by the formula on page 4.

Diameters of Circles and Sides of Squares of Equal Area (US Customary or Metric Units)

Propositions of Geometry

Useful Trigonometric Relationships

Signs of Trigonometric Functions

This diagram shows the proper sign (+ or –) for the trigonometric functions of angles in each of the four quadrants of a complete circle.

Examples: The sine of 226° is –0.71934; of 326°, –0.55919.

Useful Relationships Among Angles

Examples: cos (270° – 0) = –sin 0; tan (90° + 0) = –cot 0.

The

Law of Sines.—In any triangle, any side is to the sine of the angle opposite that side as any other side is to the sine of the angle opposite that side. If a, b, and c are the sides, and A, B, and C their opposite angles, respectively, then:

The

Law of Cosines.—In any triangle, the square of any side is equal to the sum of th squares of the other two sides minus twice their product times the cosine of the include angle; or if a, b, and c are the sides and A, B, and C are the opposite angles, respectivel then:

These two laws, together with the proposition that the sum of the three angles equals 180 degrees, are the basis of all formulas relating to the solution of triangles.

Formulas for the solution of right-angled and oblique-angled triangles, arranged in tabular form, are given on the following pages.

Trigonometric Functions and Identities.—On page 15, a diagram, Signs of Trigonometric Functions, is given. This diagram shows the proper sign (+ or –) for the trigonometric functions of angles in each of the four quadrants, 0 to 90, 90 to 180, 180 to 270, and 270 to 360 degrees. Thus, the cosine of an angle between 90 and 180 degrees is negative; the sine of the same angle is positive.

Trigonometric identities are formulas that show the relationship between different trigonometric functions. They may be used to change the form of some trigonometric expressions to simplify calculations. For example, if a formula has a term, 2 sin A cos A, the equivalent but simpler term sin 2A may be substituted. The identities that follow may themselves be combined or rearranged in various ways to form new identities.

Basic

NegativeAngle

sin(–A) = –sinA cos (–A) = cosA tan(–A) = –tanA

Pythagorean

sin²A + cos²A = 1 1 + tan²A = sec²A 1 + cot²A = csc²A

Sum and Difference of Angles

Double-Angle

Half-Angle

Product-to-Sum

Sum and Difference of Functions

Solution of Right Triangles

Solution and Examples of Oblique Triangles (US Customary or Metric Units)

One Side and Two Excluded (Not Between) Angles Known (Law of Sines):

Two Sides and One Included Angle Known

Two Sides and the Angle Opposite One of the Sides Known:

All Three Sides Known:

Rapid Solution of Right and Oblique Triangles

Trigonometric Values of Angles from 0° to 15° and 75° to 90°

For angles 0° to 15° 0′ (angles found in a column to the left of the data), use the column labels at the top of the table; for angles 75° to 90° 0′ (angles found in a column to the right of the data), use the column labels at the bottom of the table.

Trigonometric Values of Angles from 15° to 30° and 60° to 75°

For angles 15° to 30° 0 (angles found in a column to the left of the data), use the column labels at the top of the table; for angles 60° to 75° 0’ (angles found in a column to the right of the data), use the column labels at the bottom of the table.

Trigonometric Values of Angles from 30° to 60°

For angles 30° to 45°0′ (angles found in a column to the left of the data), use the column labels at the top of the table; for angles 45° to 60° 0′ (angles found in a column to the right of the data), use the column labels at the bottom of the table.

Formulas for Compound Angles