Table of Contents
Chapter One: Algebra
Basic Law of Natural Numbers
Laws of Equality
Inequality
Laws of Exponents
Radicals
Logarithms
Polynomials
Special Products and Factoring
Division of Polynomials
Factor & Remainder Theorem
Binomial Theorem
Pascal’s Triangle
Sum of Coefficients
Proportion
Quadratic Formula
Partial Fraction
Variation
Arithmetic Progression
Geometric Progression
Harmonic Progression
Work, Age, Digit, Number, Clock, Mixture
Problems
Motion Problem (Uniform Motion)
Permutation
Combination
Probability
Repeated Trials
At least One Condition
Matrices and Determinants
Operation of Matrices
Determinant
Solutions to Linear Equations by Determinant
Complex Numbers
Complex Equation
Polar or Trigonometric Form
De Moivre’s Theorem
Exponential Form
Venn Diagram
Chapter Two: Plane and Spherical
Trigonometry
Functions of Right Triangle
Pythagorean Theorem
Trigonometric Identities
Basic Identities
Pythagorean Relations
Sum and Difference of Angles
Double Angle Formulas
Half-Angle Formulas
Powers of Functions
Product of Functions
Sum and Difference of Functions
Oblique Triangle
Sine Law
Cosine Law
Law of Tangents
Mollweide’s Equation
Angles
Properties of Triangle
Area of Triangle
Median of Triangle
Altitudes of Triangle
Angle Bisectors of Triangle
Spherical Trigonometry
Spherical Triangle
Right Spherical Triangle
Napier’s Rules
Oblique Spherical Triangle
Law of Sines
Law of Cosines
Napier’s Analogies
Terrestrial Sphere
Chapter Three: Plane Geometry
Definitions
Theorems and Properties of Triangles
Area of Triangle
Rectangle
Square
General Quadrilateral
Parallelogram
Rhombus
Trapezoid
Cyclic Quadrilateral
Ptolemy’s Theorem
Polygons
Theorems in Polygons
Regular Polygons
Circle
Theorems on Circles
Area of Circle
Sector
Segment
Parabolic Segment
Spandrel of Parabola
Ellipse
Radius of Circles
Circumscribed about a Triangle
Inscribed in a Triangle
Escribed in a Quadrilateral
Circumscribed about a Quadrilateral
Inscribed in a Quadrilateral
Area by Approximation
Trapezoidal Rule
Simpson’s One-Third Rule
Area by Coordinates
Chapter Four: Solid Geometry
Polyhedrons
Regular Polyhedron (Platonic Solids)
Euler’s Polyhedron Theorem
Prism
Rectangular Parallelepiped
Cube
Truncated Prism
Pyramids
Frustum of a Pyramid
Cylinders
Right Circular Cylinder
Cone
Right Circular Cone
Frustum of a Cone
Frustum of Right Circular Cone
Sphere
Spherical Segments
Spherical Cone
Spherical Lune and Wedge
Spherical Polygons
Spherical Pyramid
Solids of Revolution
Pappus’ Theorems
Ellipsoid
Prolate and Oblate Spheroids
Paraboloid of Revolution
Prismatoid
Prismoidal Formula
Volume of Some Prismatoid
Similar Solids
Chapter Five: Analytic Geometry
Distance between Two Points
Straight Line
Slope of the Line
Standard Equations
Angle between Two Lines
Distance from a Point to a Line
Distance between Two Parallel Lines
Division of Line Segment
Midpoint of Line Segment
Conic Sections
Circle
Parabola
Ellipse
Hyperbola
Variations of Problems in Conics
Tangents and Normal to Conics
Polar Coordinate system
Distance between two points
Relationship between Polar and Cartesian
Polar Curves
Space Analytic Geometry
Rectangular Coordinate System
Cylindrical Coordinate System
Spherical Coordinate System
3D Graphs
Chapter Six: Differential Calculus
Limits
L’ Hospital’s Rule
Short Technique on Limits
Differentiation Formulas
Algebraic Functions
Logarithmic and Exponential Functions
Trigonometric Functions
Inverse Trigonometric Functions
Hyperbolic Functions
Inverse Hyperbolic Functions
Slope of the Curve
Rate of Change
Curvature and Radius of Curvature
Circle of Curvatrure
Graph of a Function
Relative Maximum and Relative Minimum
Points of Inflection
Applications of Maxima and Minima
Steps in Solving Maxima & Minima Problems
Common Variable Relationship for Maximum
and Minimum Values
Time Rates
Chapter Seven: Integral Calculus
Integration Formulas
Algebraic, Exponential, & Logarithmic
Functions
Trigonometric Functions
Inverse Trigonometric Functions
Hyperbolic Functions
Integration by Parts
Trigonometric Substitution
Wallis Formula
Examples
Wallis Formulas
Double Integration
Triple Integration
Integration by Parts
Algebraic Substitution
Trigonometric Substitution
Integration by Partial Fraction
Plane Areas
Using Horizontal Strip
Using Vertical Strip
By Polar Coordinates
Area of Some Polar Curves
Length of Arc
Centroid of Plane Areas
Moment of Inertia
Polar Moment of Inertia
Product of Inertia
Mass Moment of Inertia
Properties of Common Shapes
Solids of Revolution
Using Circular Disk
Using Hollow Cylindrical Shell
Surface Area
Volume of Other Solids of Known Cross Section
Centroid of Volume
Work
Work to Stretch a Spring
Work in Winding Up Load
Chapter Eight: Differential Equation
Variable Separable
Homogeneous First Order Differential Equation
Linear First Order Differential Equation
Differential Equation
Bernoulli’s Equation Type
Finding the Differential Equation from a General
Solution
Applications of Differential Equation
Population Growth
Exponential Growth and Decay
Cooling and Heating
Flow Problems
Continuous Compound Interest
Motion Problems
Newton’s Second Law of Motion
Chapter Nine: Engineering Mechanics
Statics
Force Systems
Resultant of Forces
Resultant of Forces in Space
Resultant of Parallel Forces
Resultant of Non-coplanar forces
Equilibrium of Forces
Cables
Parabolic Cables
Catenary
Cables Under Concentrated Loads
Friction
Belt Friction
Properties of Sections
Centroid
Center of Gravity of Flat Plates
Centroids of Composite Figures
Moment of Inertia
Polar Moment of Inertia
Moment of Inertia With Respect to Inclined
Axis
Mohr’s Circle
Dynamics
Kinematics
Translation
Rectilinear Translation
Uniform Motion
Variable Acceleration
Constant Acceleration
Free-falling Body
Curvilinear Motion
Motion Curves
Rotation
Kinetics
Newton’s Laws of Motion
D’ Alembert’s Principle
Centrifugal Force
Conical Pendulum
Banking of Curves
Ideal Angle of Banking
Horizontal Rotating Platform
Work and Energy
Work and Energy Equation
Impulse and Momentum
Impulse and Momentum Equation
Law of Conservation of Momentum
Coefficient of Restitution
Chapter Ten: Strength of Materials
Simple Stress
Normal Stress
Shearing Stress
Bearing Stress
Thin – Walled Pressure Vessels
Cylindrical Vessel
Spherical Vessel
Thick – Walled Cylinders
Simple Strain
Stress –Strain Diagram
Axial Deformation
Shearing Deformation
Poissons Ratio
Biaxial Deformation
Triaxial Deformation
Thermal Stress
Torsion
Helical Spring
Spring in Series
Spring in Parallel
Shear and Moment in Beams
Shear and Moment Diagrams
Moving Loads
Stresses in Beams
Radius of curvature
Flexure Formula
Shearing Stress
Superimposed Beams
Spacing of Rivets or Bolts in Built-up Beams
Economic Sections
Combined Stresses
Combined Axial and Flexure
Kern of a Section
Combined Axial and Shearing Stress
Mohr’s Circle
Combined Torsional and Flexural Stresses
Beam Formulas (Moment, Deflection, and
Rotation)
Simple and Cantilever Beam
Propped Beamasasas
Fully Restrained Beam
Dynamic (Impact) Loading
Chapter Eleven: Fluid Mechanics and
Hydraulics
Properties of Fluids
Unit Weight and Density
Specific Volume
Visocosity
Surface Tension
Capillarity
Bulk Modulus of Elasticity
Compression of Gases
Pressure Disturbances
Unit Pressure
Total Hydrostatic Pressure
On Planes Surfaces
On Curved Surfaces
Buoyancy
Statical Stability of Floating Bodies
Relative Equilibrium of Fluids
Rotating Vessels
Fluid Flow in Pipes
Reynold’s Number
Energy Equation
Bernoulli’s Energy Theorem
Head Lost in Pipe Flow
Pipes in Series
Pipes in Parallel
Equivalent Pipe
Orifice and Tubes
Unsteady Flow
Weir
Cipolletti Weir
Triangular Weir
Suttro Weir
Unsteady Flow
Hydrodynamics
Drag Force
Chapter Twelve: Engineering Economics
Cash Flow Diagrams
Simple Interest
Ordinary and Exact Interest
Compound Interest
Continuous Compounding
Nominal and Effective Rate
Equivalent Nominal Rates
Annuity
Ordinary Annuity
Deferred Annuity
Annuity Due
Perpetuity
Uniform Gradient
Arithmetic Gradient
Geometric Gradient
Capitalized and Annual Costs
Cost Comparison of Different Alternatives
Depreciation
Straight-Line Method
Sinking Fund Method
Sum of the Year’s Digit Method
Declining Balance Method
Double Declining Balance Method
Capital Recovery (Depletion)
Bond
Break-Even Analysis
Chapter Thirteen: Conversion Factors,
Constants
Commonly Used Conversion Factors
Commonly Used Constants
Constants in Physics and Mathematics
Factors for Conversion to S.I. Unit
Prefixes
Chapter One: Algebra
Basic Law of Natural Numbers
Let a, b, and c be any number.
1.
Law of Closure for Addition
a+b
2.
Commutative Law for Addition
a+b = b+a
3.
Associative Law for Addition
a+(b+c) = (a+b)+c
4.
Law of Closure for Multiplication
a×b
5.
Commutative Law for Multiplication
a×b = b×a
6.
Associative Law of Multiplication
a ( bc ) = ( ab ) c
7.
Distributive Law
a ( b + c ) = ab + ac
Basic Laws of Equality
1.
Reflexive Property
a = a
2.
Symmetric Property
If a = b, then b = a
3.
Transitive Property
If a = b and b = c, then a = c. That is, things
equal to the same thing are equal to each
other.
4.
If a = b and c = d, then a + c = b + d.
That is, if equals are added to the equals, the
results are equal.
5.
If a = b and c = d, then ac = bd. That is,
if equals are multiplied to the equals, the
results are equal.
Inequality
A statement that one quantity is greater than or less
than another quantity.
Symbols used in inequality:
a
a
a
a
>
<
≤
≥
a is greater than b
a is less than b
a is less than or equal to b
a is greater than or equal to b
b
b
b
b
Theorems of Inequalities
1.
2.
3.
4.
5.
6.
a > b if and only if -a < -b
If a > 0 , then -a < 0
If -a < 0 , then a < 0
If a > b , c < 0 , then ac < bc
If a > b ,c > d , then ( a + c ) > ( b + d )
If a > b , c > d , and a, b, c, d > 0, then
ac > bd
7.
If a > 0 , b > 0, a > b, then
1
a
<
1
b
Other Important Properties in Algebra
1.
2.
3.
4.
5.
a ×0=0
If a × b = 0, then either a = 0 or b = 0 or both
a and b are zero.
0
a
a
0
a
∞
= 0 if a = 0
= undefined
=0
Laws of Exponents (Index Law)
1.
2.
3.
an = a × a × a ⋯ (n factors)
am × an = am + n
am
an
= am -
n
n
4.
5.
( am ) = a m × n
( abc )n = an ∙ bn ∙ cn
6.
(
7.
an
8.
a-m =
a
m
b
n
) =
an
bn
n
= √am
1
am
and am =
1
a-m
9. a0 = 1
10. If am = an , then m = n provided a ≠ 0
Properties of Radicals
1
1.
2.
3.
4.
a
m
n
= √am = � √a�
n
n
n
n
√a × √b = √ab
n
√a
n
√b
m
n
� √a� = a
n
n
5.
n
an = √a
n
a
= � , provided that b ≠ 0
b
Properties of Logarithms
1.
loga MN = loga M + loga N
2.
loga
3.
4.
5.
6.
7.
8.
9.
loga Mn = n loga M
loga a = 1
loga ax = x loga a = x
loga 1 = 0
If loga M = N, then aN = M
If loga M = loga N , then M = N
loge M = ln M
e = 2.71828 ⋯ ( Naperian ) logarithm
logm 10 = log 10 (Common) logarithm
logn M = log M / log n = ln M / ln n
If logb x = a , then x = anti logb a
ax = anti loga x
log10 4250 = log10 ( 1000 × 4.25 )
= log 1000 + log 4.25
log10 4250 = 3 + 0.6284 = 3.6284
3, the integral part, is called the
characteristic.
0.6284, a non-negative decimal fraction
part, is called the mantissa.
10.
11.
12.
13.
14.
M
N
= loga M - loga N
Polynomials
Expanding Brackets
By multiplying the brackets together, each term in one
bracket is multiplied to each term of the other bracket.
( a + b + c )( d + e ) = ad + ae + bd + be + cd + ce
Factorization
Factorization is the opposite process of expanding
brackets. The usual process incudes a changing a long
expression without any brackets to a shorter
expression that includes brackets.
2x2 - 6x + 4 = 2 � x2 + 3x + 2 �
= 2(x-2)( x -1)
Special Products and Factoring
1.
2.
3.
4.
5.
6.
( x + y )( x - y ) = x2 - y2
( x + y )2 = x2 + 2xy + y2
( x - y )2 = x2 - 2xy + y2
( x + y + z )2 = x2 + y2 + z2 + 2xy + 2xz + 2yz
x3 + y3 = ( x + y ) � x2 - xy + y2 �
x3 - y3 = ( x - y ) � x2 + xy + y2 �
3
3
2
7. x6 - y6 = �x2 � - �y2 � = ( x2 - y2 ) [ �x2 � +
2
� x2 ) ( y2 � + � y2 � ] = ( x + y ) ( x - y ) ( x4 +
x2 y2 + y4 )
Division of Polynomials
Carrying out the division of polynomials in so different,
in principle, to numerical division. Consider the
following example.
Example 1-1
Divide x4 - 10x2 - 9x - 20 by x – 4
Solution A: (By long division)
x3 + 4x2 + 6x + 15 remainder 40
x-4
x4 - 10x2 - 9x - 20
x4 - 4x3
4x3 - 10x2
4x3 - 16x2
1. x4 ÷ x = x3
2. 4x3 ÷ x = 4x2
6x2 - 9x
6x2 - 24x
15x - 20
3. 6x2 ÷ x = 6x
4. 15x ÷ x = 15
-15x - 60
Remainder → 40
Solution B: (By Synthetic Division)
Write the coefficients of the terms, supplying zero as
the coefficient of the missing power of x.
1
0
4
-10
16
-9
24
-20
60
1
4
6
15
40
4
The quotient is x3 + 4x2 + 6x + 15 remainder 40.
Factor Theorem
Consider a function f(x). If f(1) = 0 then ( x - 1) is a
factor of f(x). If f(-3) = 0 then ( x + 3) is a factor of f(x).
Use of factor theorem can produce the factors of a
expression in a trial and error manner.
Example:
Factorize 2x3 + 5x2 - x - 6
Solution:
f(x) = 2x3 + 5x2 - x - 6
f(1) = 2(1)3 + 5(1)2 - (1) - 6 = 0,
hence (x-1) is a factor
f(-1) = 2(-1)3 + 5(-1)2 - (-1) - 6 = -2,
hence (x+1) is not a factor
f(2) = 2(2)3 + 5(2)2 - (2) - 6 = 28,
hence (x-2) is not a factor
f(-2) = 2(-2)3 + 5(-2)2 - (-2) - 6 = 0,
hence (x+2) is a factor
f(-3/2) = 2(-3/2)3 + 5(-3/2)2 - (-3/2) - 6 = 0,
hence (2x+3) is a factor
Thus, 2x3 + 5x2 - x - 6 = (x-1) (x+2) (2x+3)
Remainder Theorem
If a polynomial f (x) is divided by (x - r) until a
remainder which is free of x is obtained, the remainder
is f(r). If f(r) = 0, then (x - r) is a factor of f(x).
Example 1-2
Find the remainder when x4 - 10x2 - 9x - 20 is divided
by (x - 4).
Solution:
f(x) = x4 - 10x2 - 9x - 20
x-r = x-4,r = 4
Remainder = f(4) = (4)4 - 10(4)2 - 9(4) - 20 = 40
Example 1-3
Find k such that x-3 is a factor of kx3 - 6x2 + 2kx - 12.
Solution:
Remainder = f(3) =k(3)3 - 6(3)2 + 2k(3) - 12 = 0
k=2
Binomial Theorem, Expansion of (a+b)n
Properties:
1. The number of terms in the expansion n + 1.
2. The first term is an and the last term is bn ,
3. The exponent of a descends linearly from n
to 0,
4. The exponent from b ascends linearly from
0 to n,
5. The sum of the exponents of a and b in any
of the terms is equal to n,
6. The coefficient of the second term and
second from the last term is n,
Pascal’s Triangle
(Used to determine the coefficients of the terms in a
binomial expansion)
(a + b)0
(a + b)1
(a + b)2
(a + b)3
(a + b)4
(a + b)5
1
1
1
1
1
1
1
2
3
4
5
1
3
1
6 4 1
10 10 5 1
rth term of ( a + b )n
rth term=
n!
an - r+1 br+1
( n - r + 1 ) ! ( r - 1)!
To get the middle term set
r =
n
+ 1
2
Example 1-4
5
Find the 3rd term in the expansion of � x2 + y � .
Solution 1:
(Using the properties and Pascal”s Triangle)
5
5
4
3
�x2 + y� = ( x2 ) + 5 � x2 � y + 10 � x2 � y2
= x10 + 5x8 y + 10x6 y2
Solution 2: (Using the Formula)
n!
an - r + 1 b r - 1
(n-r+1)!(r-1)!
r = 3, a = x2, n = 5, b = y
rth term =
5!
5-3+1 3-1
(x2 )
y
(5-3+1)!(3-1)!
6 2
=10x y
3rd term =
To expand completely a given binomial, one may use
the following procedure:
Example 1-5
Expand (x + y)8
Solution:
By principle, the first term is x8, the second term is
7
8x y. The variable part of the third term is x6y2. To get
the coefficient:
C=
(Coefficient of previous term)(exponent of x)
(exponent of y)+1
(8) (7)
= 28
1+1
rd
3 term = 28 x6 y2
(28) (6) 6 2
4th term =
x y = 56x3 y3
2+1
(56) (5) 4 4
x y = 70x3 y3
5th term =
3+1
C3 =
(x + y)8 = x8 + 8x7 y + 28x6 y2 + 56x5 y3 + 70x4 y4
+ 56x3 y5 + 28x2 y6 + 8xy7 + y8
Sum of Coefficient of Variables
To get the sum of the coefficients in the expansion of
(ax+by+ ⋯)n , substitute 1 to each of the variables x,
y…
Example 1-6
Find the sum of the coefficient of the variables in the
expansion of (2x + 3y - z)8.
Solution:
Sum = [ (2)(1) + (3)(1) - 1 ]8 = 48 = 65536
Proportion
Proportion is a statement of equality between two
ratios.
a : b = c : d or
a
c
=
b
d
where:
b and c are called the means
a and d are called the extremes
d is the fourth proportional to a, b and c
In the ratio a / b, a is called the antecedent and b is
called the consequent
The mean proportional to a and b is
Mean Proportional = √ab
Properties of Proportion:
1.
Proportion by Inversion
If
2.
a
b
=
c
d
then
Proportion by Alteration
b
a
=
d
c
If
3.
=
c
d
then
a
=
c
b
d
a
b
=
c
d
then
a+b
b
=
c+d
d
Proportion by Division
If
5.
b
Proportion by Composition
If
4.
a
a
b
=
c
d
then
a-b
b
=
c-d
d
Proportion by Composition and Division
If
a
b
=
c
d
then
a+b
a-b
=
Quadratic Formula
For the quadratic equation Ax2 + Bx + C = 0
x=
- B ± �B2 - 4AC
2A
Where B2 – 4AC is called the discriminant
If B2 = 4AC, the roots are equal
If B2 > 4AC, the roots are real, unequal
If B2 < 4AC, the roots are imaginary
c+d
c-d
Properties of Roots
If the roots of the quadratic equation Ax2 + Bx + C = 0
are x1 and x2, then,
Sum of the roots, x1 + x2 = Product of the roots, x1 x2 =
B
A
C
A
Partial Fraction
Functions of x that can be expressed in the form
P(x) ⁄ Q(x), where both P(x) and Q(x) are polynomials
of x, is known as rational functions.
A rational function is known as a proper fraction if the
degree of P(x) is less than the degree of Q(x).
Proper Fraction:
2x2 + 4x - 5
5X3 + 6x2 - 2x - 1
A rational function is said to be an improper fraction
if the degree of P(x) is greater than or equal to the
degree of Q(x).
Improper Fraction:
3x2 - 2x + 1 4x2 - 2x + 3
;
2x2 + 6
3x + 2
Improper fractions may be expressed as the sum of a
polynomial and a proper fraction.
For example,
12x2 - 13x - 9
5
= 3x + 2 +
4x - 7
4x - 7
x-4
can be expressed as
2x2 - 4x
the sum of partial fraction, provided that the
denominator will factorize.
Proper fractions such as
Consider the following example:
3
2 (2x - 4) - 3x
x-8
2
=
=
x (2x - 4)
x (2x - 4)
x 2x - 4
If we reverse the process,
2
3
x-8
= x 2x - 4
x (2x - 4)
x-8
Thus, the fraction
can be expressed or
x (2x - 4)
3
2
.
resolved into partial fractions 2x - 4
x
The following are the different cases of fractions that
can be resolved into partial fraction.
Case I. Factors of the denominator all linear, none
repeated.
3x2 + 32x - 51
A
B
C
=
+
+
(x - 1) (x - 2) (x + 3) x - 1 x - 2 x + 3
3x2 + 32x - 51 = A(x - 2) (x + 3) + B (x - 1) (x + 3) +
C (x - 1) (x - 2)
This equation is an identity, hence it is true for any
value of x.
To solve for A, set x = 1, A = 4
To solve for B, set x = 2, A = 5
To solve for C, set x = -3, A = -6
Case II. Factors of the denominator all linear, some
repeated.
4x2 + 7x + 8
A
B
C
D
+
+
+
2
3
x
x
+
2
x (x + 2)
(x + 2) (x + 2)
3
2
2
4x + 7x + 8 = A (x + 2) + Bx (x + 2) +
Cx (x + 2) + Dx
3
=
Expand and equate the coefficients of like powers to
solve for A, B, C and D.
Case III. Some factors of the denominator are
quadratic, none repeated.
x4 - x3 + 14x2 - 2x + 22
A
Bx + C
Dx + E
=
+ 2
+ 2
(x + 1)(x2 + 4)(x2 - 2x + 5)
x+1
x +4
x - 2x + 5
x4 - x3 + 14x2 - 2x + 22 =
A(x2 + 4)(x2 - 2x + 5) + (Bx + C)(x + 1)(x2 - 2x + 5)
+ (Dx + E)(x + 1)(x2 + 4)
Expand and equate the coefficients of like powers to
solve for A, B, C, D and E.
Case IV. Some factors of the denominator quadratic
some repeated.
3x4 - 19x3 + 60x2 - 91x + 64
=
x (x2 - 3x + 4)2
A
Bx + C
Dx + E
+ 2
+
(x - 3x + 4) (x2 - 3x + 4)2
x
3x4 - 19x3 + 60x2 - 91x + 64 =
A (x2 - 3x + 4)2 + (Bx + C)(x)(x2 - 3x + 4) + (Dx + E)(x)
Expand and equate the coefficients of like powers to
solve for A, B, C, D and E.
Partial fractions are often used to help simplify a
separate problem such as one involving integration.
�
1
2
4x + 1
dx = �
dx +
dx
2x - 1
x+3
2x2 + 5x - 3
Variation
Direct Variation
x is directly proportional to y:
x ∝ y or x = ky
k = constant of proportionality
Inverse Variation
x is inversely proportional to y:
x∝
1
y
or x =
k
y
Joint Variation
x is directly proportional to y and inversely
proportional to the square of z:
x∝
y
z2
or x =
ky
z2
Arithmetic Progression (A.P.)
A sequence of numbers in which the difference of any
two adjacent terms is constant.
Ex. 4, 7, 10, 13, 16, … (common difference = 3)
Elements:
a1 = first term
an = nth term
am = any term before an
d = common difference
d = a2 - a1 = a3 - a2
s = sum of all terms
nth term of an A.P.
an = a1 + (n - 1) d or an = am + (n - m) d
Sum of n terms of an A.P.
S=
n
n
(a + an ) or S = [ 2a1 + (n - 1)d ]
2 1
2
Geometric Progression, G.P.
A sequence of numbers in which any two adjacent
terms has a common ratio.
Ex. 2, 6, 18, 54, …. (common ratio, r = 3)
nth term of a G.P.
an = a1 rn-1 or an = am rn-m
a2 a5
Common ratio, r =
=
a1 a4
Sum of n terms of a G.P.
S=
S=
a1 (rn - 1)
, when r >1
r-1
a1 (1 - rn )
, when r < 1
1-r
Sum of an Infinite Geometric Progression, I.G.P.
For a geometric progression of 0 < r < 1 and n = ∞
“infinity”
Harmonic Progression
A sequence of numbers in which their reciprocals
forms an Arithmetic Progression.
Example 1-8
Find the 12th term of the series 6, 3, 2.
Solution
The reciprocals are 1/6, 1/3, ½ which forms an A.P.
with a common difference d of 1/6.
In A.P., the 12th term is:
Sum of an IGP =
a1
1-r
a12 = 1⁄6 + (12 - 1) (1⁄6) = 2
Therefore, in H.P. the 12th term is ½.
Work Problem
Rate=
1
Time to finish the work
Work done=Rate ×Time
Example 1-9
If A can do a job of 4 hours and B can do the same job
in 8 hours, working together from start (a) what part of
the job have they done in 2 hrs? (b) how many hours
can they finish the job?
Solution:
Rate of A = ¼
Rate of B = 1/8
Work done in 2 hours =
( 1⁄4)(2) + ( 1⁄8)(2) = 3⁄4
This means ¾ or 75% of the work was done.
Time to finish the job:
1� t + 1� t = 1 complete job
4
8
t = 2.667 hours
Work Problem with n Persons with the Same Rate
Doing the Job
Principle:
•
If 8 persons can do a job in 6 days, the
number of man-days to finish the job is (8)(6)
= 48 man-days
•
Thus, if 12 persons will do the job, it will take
them 48/12 = 4 days to finish it.
Example 1-10
A job could be done by eleven workers in 15 days. Five
workers started the job. They were reinforced with four
more workers at the beginning of the 6th day. Find the
total number of days to finish the job.
Solution
Let t = number of days the four workers has to work
with five workers to finish the job.
Number of man-days to finish the job:
(11)(15)=165 man-days
Five workers started the job for 5 days and (5 + 4)
workers continued the job at the beginning of the 6th
day for t days until completion.
(5)(5) + (5 + 4)t = 165 ; t = 15.56 days
Total number of days = 15.56 + 5 = 20.56 days
Principles in Age Problem
The difference of the ages of two persons is constant.
If x is the age of Peter now:
His age 5 years ago is x – 5
His age 7 years hence is x + 7
Principles in Digit Problem
For a three-digit number:
Let: h = hundreds digit
t = tens digit
u = units digit
The number is 100h + 10t + u
The reverse number is: 100u + 10t + h
The sum of the digits is h + t + u
Principles in Number Problem
Let x be the first number and y be the second number
First number is twice the other: x = 2y
First number is five (5) more than thrice of the other:
x = 5 + 3y
First number is six (6) times less than one-half of the
y
other: x = �2 - 6
The sum of their squares: x2 + y2
The cube of their difference: (x - y)3
Clock Problem
If the minute hand moves a distance of x, the hour hand
moves x/12.
If the second hand moves a distance of x, the minute
hand moves x/60 and the hour hand moves x/720.
Example 1-11
How many minutes after 2 o’clock will the hands of the
clock be perpendicular for the first time?
Solution:
x = 10 + x⁄12 + 15
x = 27.273 min.
Mixture Problem
Example 1-12
How many grams of gold must be added with 500
grams of an alloy containing 30% gold and 70% silver
in order to produce another alloy analyzing 40% gold
and 60% silver?
Solution
Represent the different mixtures by boxes
500
x
500 + x
30% G
70% S
100% G
40% G
60% S
(500)(30%) + x(100%) = (500 + x)(40%)
1500 + 10x = 2000 + 4x
x = 83.33 grams
Motion Problem
Uniform Motion or Constant Speed
S = vt
Where: S= distance, v = velocity, t = time
If x = speed of airplane (or boat) in still air (or still water)
and y = speed of wind (or water current) in the same
direction then the speed of the airplane (or boat) with
the wind (or current) is x + y and its speed against the
wind (or current) is x – y.
Permutation
Permutation refers to the arrangement of objects in a
definite order.
The permutation of n different things taken r at a time
is:
P(n , r) =
n!
(n - r) !
and P(n , n) = n !
Note: 0! = 1
Example 1-13
How many permutations can be made out of the letters
on the word DIEGO taken 3 at a time?
Solution:
n=5,r=3
P(5 , 3) =
5!
(5 - 3)!
= 60 ways
The permutation of n things of which q are alike, r are
alike and so on is:
P=
n!
q! r!
Example 1-14
How many permutations can be made out of the letters
in the word ENGINEERING?
Solution:
n = 11 ; (3 E's, 3 N's, 2 I's, 2 G's )
P=
11!
3! 3! 2! 2!
= 277, 200 ways
Permutation of n things in a Circle
P = (n - 1) !
Combination
Combination refers to a collection of objects without
regard to sequence or order of agreement.
Combination of n things taken r at a time:
C(n , r) =
P(n , r)
n!
=
(n - r)! r !
r!
and C(n , n) = 1
Example 1-15
How many ways can you draw 3 Queens and 2 Kings
from a deck of 52 cards?
Solution:
A deck of 52 cards has 4 Queens and 4 Kings, thus
C = C (4,3) × C (4,2) = 24 ways
Combination of n things taken 1, 2, 3, …, n at a time
C = 2n - 1
Example 1-16
How many ways can you invite any one or more of your
five friends to your birthday party?
Solution: C= 25 -1=31 ways
Probability
Single Event
Probability =
number of favorable ways
total number of ways
If an event can happen in h ways can fail in f ways,
then the probability that the event will happen is:
p=
h
h+f
And the probability of the event will fail is:
q=
f
h+f
And
p+q=1
Example 1-17
For a single question in the board exams, there are
four choices and only one of which is correct. By
guessing, what is the probability that you will get the
correct answer?
Solution:
The event here is to get the correct answer and there
are four trials. Out of four trials, the event (correct
answer) can happen only once, and can fail 3 times.
Thus, the probability that the event will happen is
1
1
=
p=
1+3 4
Multiple Events
Dependent and Independent Events
Two or more events are said to be dependent if the
happening of one affects the probability of the
happening of the others, and independent if the
happening of one does not affect the probability of the
happening of the others.
The probability of happening of two or more
independent or dependent events is the product of
their individual probabilities.
P = P1 × P2 × P3 × …
Mutually Exclusive Events
Two or more events are said to be Mutually Exclusive
if it is impossible for more than one of them to happen
in a single trial. The probability that some one, two or
more mutually exclusive events to happen is the sum
of their individual probabilities.
P = P1 + P2 + P3 + …
Example 1-18
A box contains 4 blue chips and 5 red chips.
a. If one chip is drawn at random, what is the
probability that it is blue?
b. If two chips are drawn at random, what is the
probability that both are red?
c. If two chips are drawn at random, what is the
probability that one is blue and the other is
red?
Solution:
a. Single Event. There are four blue chips out
of nine chips. P = 4/9
b. Multiple Events. The events (getting red) are
to occur twice.
First draw red: There are five red chips out
of nine chips. P1 = 5/9
Second draw red: There are now only four
red chips out of eight chips. P2 = 4/8 or ½
Thus P = P1 × P2 = (5⁄9)(1⁄2) = 15/18.
c. Mutually Exclusive. The event here is to get
a red and a blue ball in two draws. This can
happen in two ways (first draw red and
second draw blue) and another is (first draw
blue and second draw red), but these two
cannot happen in the same time, hence they
are mutually exclusive events.
First draw Red, Second draw Blue
5
4
5
9
8
18
P1 = � � � � =
First draw Blue, Second draw Red
4
5
5
9
8
18
P2 = � � � � =
5
5
10
18
18
18
Thus, P = P1 + P2 = � � + � � =
Repeated Trials
The probability that an event can occur exactly r times
in n trials is:
P(n, r) = C (n, r) pr qn-r
Where p is the probability that the event can happen
and q is the probability that the event will fail.
Example 1-29
There are ten questions in an examination. The
probability that an examinee will get the correct
answers is 0.25. What is the probability that he will get
(a) exactly 7, and (b) at least 7 correct answers?
Solution
a.) There are 10 questions, n = 10 with p = 0.25 and q
= 0.75.
The probability of getting exactly 7 is
P(10,7) = C (10, 7) 0.257 0.7510 - 7
= 0.00309 or 405/131072
b.)
Pat least 7
“At least seven” means can be exactly
7, 8, 9, or 10.
= P(10,7) + P(10,8) + P(10,9) + P(10,10)
Pat least 7 = C (10, 7) 0.257 0.7510 - 7 + C
(10, 8) 0.258 0.7510 - 8 +C (10, 9) 0.259
0.7510 - 9 +C (10, 10) 0.2510 0.7510 - 10 =0.00351
The ‘At Least One’ Condition
The probability that the event can happen at least
once in a trial is:
P=1-Q
Where Q is the probability that the event will totally fail.
Example 1-20
The probability of getting a credit in each of the three
examinations is 0.65. What is the probability of getting
at least one credit?
Solution:
There are three trials (n=3) with p = 0.65 and q = 0.35.
Thus, the probability of getting no credit at all is Q =
(0.35)(0.35)(0.35) = 0.042875. Thus,
P = 1 - Q = 0.957125
Matrices and Determinants
Matrix
A matrix is a rectangular collection of variables or
scalars contained within a set of square [ ] or round ()
brackets. A matrix consist of m rows and n columns.
Classification of Matrices
Square Matrix – a matrix whose number of rows m is
equal to the number of columns n.
Diagonal Matrix – a diagonal matrix is a square matrix
with all zero values except for aij value for all i = j.
2 0 0
�0 3 0�
0 0 1
Identity Matrix – an identity matrix is a diagonal matrix
with all non-zero entries equal to 1.
1 0 0
�0 1 0�
0 0 1
Scalar Matrix – a scalar matrix is a diagonal matrix
with all non-zero entries equal to some other constant.
8 0 0
�0 8 0�
0 0 8
Triangular Matrix – A triangular matrix has zeros in all
positions above or below the diagonal.
Important Algebraic Operations in Matrices
1. Equality of Matrices
Two matrices are equal if they have the same
number of rows and columns and their
corresponding entries are also equal.
2. Addition and Subtraction of Matrices
Addition (or subtraction) of matrices can be
accomplished by adding (or subtracting) the
corresponding entries of two matrices which
have the same shape.
Example 1-21
Add
1
�7
-3
1+3
= � 7+2
-3+9
4
1
0
4+0
1+5
0+1
1
3
6� + � 2
4
9
0
5
1
2
6�
1
1+2
4
6+6� = �9
4+1
6
4
6
1
3
12�
5
3. Multiplication of Matrices
Multiplication of matrix can be done only if the
number of columns of the left-hand matrix is
equal to the number of rows of the right-hand
matrix. Multiplication is accomplished by
multiplying the elements in each right-hand
matrix column, adding the products, and then
placing the sum at the intersection point of the
involved row and column.
Example 1-22
2
(2)(2)+(1)(4)+(5)(1)
13
2 1 5
�
� × �4� = �
�=� �
(1)(2)+(4)(4)+(7)(1)
25
1 4 7
1
4. Division of Matrices
Division of matrices can be accomplished only
by multiplying the inverse of the denominator
matrix.
Other Operations on a Matrix
1.
The Transpose of a Matrix
The transpose is an (n x m) matrix formed
from the original (m x n) matrix by taking the
ith row and and making it the ith column. The
diagonal is unchanged in this operation. The
transpose of a matrix is indicated as A’.
Example 1-23:
Determine the transpose of A
1
A = �2
7
2.
6
3
1
1
9
'
4� ; Hence A = �6
9
5
2
3
4
7
1�
5
The Determinant of a Matrix
The determinant D, is a scalar calculated
from a square matrix. The determinant of a
matrix is indicated by enclosing the matrix by
vertical lines.
Properties of Determinants
A.
If a matrix has a row or column of zeros, the
determinant is zero.
1
�4
1
B.
0
0� =0
0
If a matrix has two identical rows or columns,
the determinant is zero.
1
�4
1
C.
4
5
7
2
6
2
5
1� = 0
5
If a matrix is triangular, the determinant is
equal to the product of the diagonal entries.
2
�0
0
D.
0
0� = (2)(3)(5) = 30
5
The value of the determinant is not changed
if corresponding rows and columns are
interchanged.
1
�2
1
E.
0
3
0
4
5
7
6
1
2� = �4
9
6
2
5
2
1
7�
9
If each of a column or row of a determinant
is multiplied by m, the value of the
determinant is multiplied by m.
1
�4
2
4
6
8
5
1
1� = �4
4
2
2×2
3×2
4×2
5
1
1� = 2 �4
4
2
2
3
4
5
1�
4
By properties defined in B and E, the
following can be applied:
1
�5
3
4
1
6
2
10� =0
6
(since the elements of column 1 and 3 are
exact multiples)
F.
If two columns or rows of a determinant are
interchanged, the sign is changed.
2 1
�5 4
1 3
G.
6
7
9
1
4�
3
The value of a determinant is not changed if
each element of a column (or row) is
multiplied by a number k and added (or
subtracted) in the corresponding elements
of a column (or row).
1
�4
2
H.
6
2
7� = - �5
9
1
4
6
8
1
5
1� = �4
2
4
4 5 + (1)(3)
1
6 1 + (4)(3)� = �4
8 4 + (2)(3)
2
4
6
8
8
13�
10
If each element of the column (say the kth
column) of a matrix is expressed as the sum
of two terms, the determinant is equal to the
sum of the two determinants, where (a) the
elements of each of the two determinants
are identical to the corresponding elements
of a given determinant except for the
elements of the kth column, and (b) the first
term of the kth column of the given
determinant form the kth column of one of
the two determinants and the second term
form the kth column of the other determinant.
4
�6
2
3.
1
2
5
2
2
3� = �5
1
1
1
2
5
2
2
3� = �1
1
1
1
2
5
2
3�
1
Example 1-24
Solve for x: (2nd order)
4 5
x= �
� = (4)(3) - (2)(5) = 2
2 3
The Cofactor of any Entry in a Matrix
The cofactor of an entry in a matrix is the
determinant of the matrix formed by omitting
the entry’s row and column in the original
matrix. The sign of the cofactor is
determined from the following positional
matrices.
+ - +
�- + -�
+ - +
Or, the sign of the cofactor can be
determined by the relation (-1)i+j , where i is
the column number and j is the row number.
Example 1-25
Find the cofactor of -2 in the following matrix.
2
�-2
3
4.
7
5
4
3
6�
7
Solution:
-2 is at column 1 row 2. The resulting matrix
is
(-1)1 + 2 �7 3�
4 7
The cofactor is:
7 3
-1 �
� = - [ (7)(7) - (4)(3) ] = - 37
4 7
The Classical Adjoint
The classical adjoint is a matrix formed from
the transposed cofactor matrix with the
conventional sign arrangement. The
resulting matrix is represented as Aadj.
Example 1-26
Determine the classical adjoint of
2 3 -4
�0 -4 2 �
1 1 5
Solution:
After Solving the cofactors of each entry, the
matrix of the cofactors is
-18
�-11
-10
2
14
-4
4
5�
-8
The classical adjoint is
-18 -11
Aadj = � 2
14
4
5
5.
-10
-4 �
-8
The Inverse Matrix
The inverse, A-1 of a matrix A is a matrix such
that (A)(A-1) = I where I is a square matrix
with ones along the left-to-right diagonal and
zeros elsewhere.
Example 1-27
Determine the inverse of �
4
2
5
�.
3
Solution:
The determinant is
4 5 (4)(3) (5)(2)
D= �
�=
=2
2 3
The inverse is
1 3 -5
3⁄2 -5⁄2
�
�
�=�
2 -2 4
-1
2
Solution to System of Linear Equations using
Determinants (Cramer’s Rule)
For a system of linear equations,
x=
Ny
Nx
;y=
D
D
Where:
D = determinant of the coefficient of the variables
Nx = determinant taken from D replacing the
coefficients of x by their corresponding constant terms
leaving all other terms unchanged.
Ny = determinant taken from D replacing the
coefficients of y by their corresponding constant terms
leaving all other terms unchanged.
Example 1-28
Solve for x, y, and w in the following equations:
3x - 2y + w = 11
x + 5y - 2w = - 9
2x + y - 3w = - 6
Solution:
3 -2 1 3 -2
D = �1 5 -2 � 1 5 � = - 46
2 1 -3 2 1
11
Nx = � -9
-6
-2
5
1
1 11
-2 � -9
-3 -6
11
-9
-6
1 3
-2 � 1
-3 2
3
Ny = �1
2
3
Nw = �1
2
-2
5
1
11 3
-9 � 1
-6 2
-2
5 � = - 92
1
11
-9 � = 46
-6
-2
5 � = - 138
1
x=
Nx - 92
=
=2
D
- 46
y=
Ny
46
=
=-1
D
- 46
x=
Nx - 138
=
=3
D
- 46
Complex Numbers
Algebraic or Rectangular Form
a + bi
Where a = real part, b= imaginary part, i =
√-1 and i2 = -1
Example 1-29
1.
2.
3
i7 = i6 × i = � i2 � × i = ( - 1 ) 3 × i = - i
i245 = i244 × i = � i2 �
122
Addition or Subtraction
× i = ( -1 )122 × i = i
Addition or subtraction of complex numbers is obtained
by combining similar terms and applying i2 = -1
Example 1-30
Simplify i30 - 2i25 + 3i17
Solution:
15
12
8
=�i2 � - 2 �i2 � i + 3 �i2 � i
=(-1)15 - (2) (-1)12 i + (3) (-1)8 i
=-1 - 2i + 3i = -1 + i
Multiplication of Complex Numbers
Multiplication of complex numbers is similar to
multiplication of polynomials.
Example 1-31
(3 + 2i) (4 - 3i) = 12 - 9i + 8i - 6 = 18 – i
Conjugate of a Complex Number
The conjugate of a complex number is obtained by
changing the sign of the imaginary part.
Number
2 + 3i
3 - 5i
- 5 + 2i
Conjugate
2 - 3i
3 + 5i
- 5 - 2i
The product of a complex number and its conjugate is
always a real number.
Example 1-32
(2 + 3i)(2 - 3i) = 4 - 9i2 = 4 - 9 ( - 1) = 13
Division of Complex Numbers
Division of a complex number is obtained by
multiplying the numerator and denominator by the
conjugate of the denominator.
Example 1-33
3 + 4i 2 + i 6 + 3i + 8i + 4i2
×
=
2-i
2+i
2- i2
=
6 + 11i - 4
2 + 11i
2 11
=
=
+
i
4 - ( - 1)
5
5
5
Complex Equation
Two complex numbers are equal if their real parts are
equal and their imaginary parts are equal.
(a + bi) is equal to (c + di) if a = c and b = d
Example 1-34
Solve for x and y if 3x - 2yi = 6 + 8i
Solution:
3x = 6 , - 2y = 8 ; Hence x = 2 and y = - 4
Polar Form or Trigonometric Form
The polar form of a complex number is used to find the
roots of a complex number.
Argand Diagram (Complex Plane) :
Imaginary axis, b
a
r
a + bi
b
Ɵ
Real Axis, a
In the Argand chart shown:
r = absolute value or modulus
θ = argument or amplitude
a = r cos θ
b = r sin θ
r = � a2 + b2
b
tan θ =
a
a + bi = r cos θ + r sin θ i
a + bi = r (cos θ + sin θ i) = r cis θ
Multiplication
a + bi = r ∠ θ
r1 ∠ θ1 × r2 ∠ θ2 = r1 r2 ∠ (θ1 + θ2 )
Example 1-35:
5 ∠ 30° × 6 ∠ 45° = 30 ∠ 75°
Division
r1 ∠ θ1
r1
=
∠ (θ1 − θ2
r2
r2 ∠ θ2
Example 1-36:
45 ∠ 67°
=3 ∠ 50°
15 ∠ 17°
De Moivre’s Theorem
[r ∠ θ]n = rn ∠ nθ
True for all values of n.
Example 1-37:
[5 ∠ 15°]3 = 53 ∠ (3)(15°) = 125 ∠ 45°
To obtain the mth root of a complex number there
are m solutions. The modulus r is always the same
and the argument θ are symmetrically spaced at
(360°/m) apart, where m is the number of root
required.
Example 1-38
Find √5 + 12i
Solution
There are two roots, each are 360/2 = 180° apart.
First convert 5+12i to polar form. (a=5, b=12i)
r = �52 + 122 = 13
12
, θ = 67.38°
5
[13
Thus, √5 + 12i =
∠ 67.38°]1/2
=131/2 ∠ (1/2)(67.38°)
= 3.61 ∠ 33.69° and 3.61 ∠ 213.69°
tan
Change to rectangular form:
3.61 ∠ 33.69° = 3.61 ( cos 33.69° + i sin 33.69°) = 3 + 2i
3.61 ∠ 213.69° = 3.61 ( cos 213.69° + i sin 213.69°)
= -3 - 2i
Thus, the roots are (3 + 2i) and (-3 - 2i).
Exponential Form
The exponential form of a complex number is used
when finding the logarithms of a complex number.
a + bi = r ( cos θ + i sin θ)
eiθ = r ( cos θ + i sin θ)
e-iθ = cos θ - i sin θ
a + bi = reiθ
Example 1-39
Find ln (3 + 4i)
Solution:
Convert 3 + 4i to polar form (a = 3, b = 4)
r= �32 + 42 = 5, tan θ = 4/3, θ = 53.13° = .9273 rad
3 + 4i = 5 ∠ 53.13° = 5e0.9273i
ln (3 + 4i) = ln 5e0.9273i = ln 5 + ln e0.9273i
ln(3 + 4i) = 1.609 + 0.9273i
Venn Diagram
Example 1-40
An engineering professor conducted a survey
regarding the favorite subjects of the students. The
following data were gathered: 60 students like the
subject algebra, 50 like the subject calculus, and 45
likes the subject physics. Thirty students like both
algebra and calculus subjects, 25 students like both
calculus and physics subjects, and 20 students like
both algebra and physics subjects. Only 15
students like all the three subjects. How many
students were surveyed?
30
Algebra
60
Calculus
50
1
25
10
15
5
20
10
15
25
Physics
45
Solution:
The diagram shows the following information:
25 like the subject algebra only
15 like the subject calculus only
10 like the subject physics only
15 like the subject algebra and calculus
subjects
10 like the subject calculus and physics
subjects
5 like the subject algebra and physics
subjects
15 likes all three subjects
Thus, the total number of students surveyed is:
25 + 15 + 10 + 15 + 10 + 5 + 15 = 95 students
No. of students surveyed = 95
Chapter Two – Plane and Spherical
Trigonometry
Plane Trigonometry
Functions of a Right Triangle
c
a
θ
b
From the right triangle shown (soh-cah-toa)
opposite side o
=
hypotenuse
h
adjacent side a
=
=
h
hypotenuse
opposite side o
=
=
adjacent side a
adjacent side a
=
=
opposite side o
hypotenuse
h
=
=
adjacent side a
hypotenuse
h
=
=
opposite side o
sin θ =
(soh)
cos θ
(cah)
tan θ
cot θ
sec θ
csc θ
(toa)
(tao)
(cha)
(sho)
Pythagorean Theorem
“In any right triangle, the square of the largest side
(hypotenuse) equals to the sum of the squares of
other two sides.”
From the triangle shown below:
c 2 = a2 + b2
Trigonometric Identities
Identity is a type of equation which is satisfied with
any value of the variable or variables. Equations
that are satisfied by some value or values of the
variable are called conditional equation.
Consider the following equations:
x2 - 4 = 0………………………
Conditional
equation
True only for
x = ±2
(x + 2)2 = x2 + 4x + 4………...
Identity
sin θ = 0.5……………………...
Conditional
equation
True only if θ
= 30°, 150°
sin2 θ + cos2 θ = 1…………….
Identity
Basic Identities
c
a
θ
b
From the triangle shown:
a a⁄c sin θ
=
=
b b⁄c cos θ
b b⁄c cos θ
cot θ = =
=
a a⁄c sin θ
c ⁄c
1
c
=
sec θ = =
b⁄c cos θ
b
c c ⁄c
1
=
csc θ = =
a a⁄c sin θ
tan θ =
Pythagorean Relations
From the Pythagorean Theorem:
a2 + b2 = c2
Dividing both sides by c2:
a2
b2
c2
+ 2 = 2
2
c
c
c
Then;
or
a 2
b 2
� � + � � = 1
c
c
sin2 θ + cos2 θ = 1
Dividing a2 + b2 = c2 by b2, we get,
tan2 θ + 1 = sec2 θ
Dividing a2 + b2 = c2 by a2, we get,
1 + cot2 θ = csc2 θ
Sum and Difference of Two Angles
sin ( x + y ) = sin x cos y + cos x sin y
sin ( x - y ) = sin x cos y - cos x sin y
cos ( x + y ) = cos x cos y - sin x sin y
cos ( x - y ) = cos x cos y + sin x sin y
tan ( x + y )=
tan ( x - y )=
tan x + tan y
1 - tan x tan y
tan x - tan y
1 + tan x tan y
Double Angle Formulas:
Double angle formulas can be derived using the
sum of the angle formulas.
Consider the following sample:
sin 2x = sin ( x + x ) = sin x cos x + cos x sin x
Thus,
sin 2x = 2 sin x cos x
We can apply similar procedure to the rest of the
formulas:
cos 2x = cos2 x - sin2 x
= 1 - 2 sin2 x
= 2 cos2 x - 1
2 tan x
tan 2x =
1 - 2 tan2 x
Half-Angle Formulas
x
1 - cos x
sin � � = �
2
2
1 + cos x
x
cos � � = �
2
2
x
1 - cos x
sin x
tan � � =
=
2
sin x
1 + cos x
=�
1 - cos x
1 + cos x
Powers of Functions
sin2 x =
1 - cos 2x
1 + cos 2x
; cos2 x =
2
2
tan2 x=
1 - cos 2x
1 + cos 2x
Product of Functions
sin x cos y =
1
[ sin(x + y) + sin (x - y)]
2
sin x sin y =
1
[ cos(x - y) - cos (x + y)]
2
cos x cos y =
1
[ cos(x + y) - cos (x - y)]
2
Sum and Difference of Functions (Factoring
Formulas)
x+y
x-y
� cos �
�
2
2
x+y
x-y
sin x - sin y = 2 cos �
� sin �
�
2
2
x+y
x-y
cos x + cos y = 2 cos �
� cos �
�
2
2
x+y
x-y
cos x - cos y = -2 sin �
� cos �
�
2
2
sin ( x + y )
tan x + tan y =
cos x cos y
sin ( x - y )
tan x - tan y =
cos x cos y
sin x + sin y = 2 sin �
Oblique Triangles
An oblique triangle is any triangle that is not a right
triangle. It could be an acute triangle (all three
angles of the triangle are less than the right angles)
or it could be an obtuse triangle (one of the three
angles is greater than the right angle). Actually, for
the purposes of trigonometry, the class of oblique
triangles might just as well include right triangles
too. Then the study of oblique triangles is really the
study of all triangles.
C
a
b
A
B
c
Sine Law
In any triangles, the ratio of any one side to the sine
of its opposite angle is constant. (This constant ratio
is the diameter of the circle circumscribing the
triangle.)
a
b
c
=
=
sin A
sin B
sin C
Cosine Law
In any triangle, the square of any one side equals
the sum of the squares of the other two sides,
diminished by twice their product to the cosine of its
included angle.
a2 = b2 + c2 - 2bc cos A
b2 = a2 + c2 - 2ac cos B
c2 = a2 + b2 - 2ab cos C
Law of Tangents
A-B
tan
a-b
2
=
A+B
a+b
tan
2
;
B-C
tan
b-c
2
=
B+C
b+c
tan
2
C-A
tan
c-a
2
=
C+A
c+a
tan
2
Mollweide’s Equations
A-B
sin
a-b
2
=
C
c
cos
2
;
A-B
cos
a+b
2
=
C
c
sin
2
How to get the other trigonometric functions with
one function known.
Example 2-1
If sin θ = 1⁄k, find the other functions.
k
1
θ
�k2 - 1
From the right triangle shown,
�k 2 - 1
�k 2 - 1
cos θ =
; cot θ =
;
k
1
�k 2 - 1
csc θ =
k
tan θ =
Angles
1
�k 2 - 1
; sec θ =
k
�k 2 - 1
The concept of angle is one of the most important
concepts in geometry. The concepts of equality,
sums and differences of angles are important and
used throughout geometry, but the subject of
trigonometry is based on the measurement of
angles.
Angle is the space between two rays that extend
from a common point called the vertex.
Angle, θ
An acute angle is an angle < 90°
A right angle is an angle = 90°
An obtuse angle is an angle > 90°
A straight angle is an angle = 180°
A reflex angle is an angle > 180°.
Complementary angles
whose sum is 90°
are
angles
Supplementary angles are angles whose
sum is 180°
Explementary angles are angles whose
sum is 360°
Units of Angle
90° = π⁄2 radians = 100 grads = 1600 mils
1 radian is the angle subtended by an arc of a circle
whose length is one radius.
r
r
1 radian
r
Other Elements and Properties of a Triangle
Area of a Triangle
B
c
a
h
θ
C
A
b
Given base b and altitude h,
Area =
1
bh
2
Given two sides a and b and included angle θ
Area =
1
ab sin θ
2
Given three sides a, b, and c (Hero’s Formula)
Area = �s ( s - a ) ( s - b ) ( s - c )
s =
a + b + c
2
The area under this condition can also be solved by
finding one angle using cosine law and apply the
formula for two sides and an included angle.
Given three angles A, B, and C and one side a,
Area =
a2 sin B sin C
2 sin A
The area under this condition can also be solved by
finding one side using sine law and apply the
formula for two sides and an included angle.
Median of a Triangle
The median of a triangle is the line drawn one
vertex to the midpoint of its opposite side. The
medians of a triangle intersect at a common point
called the centroid of the triangle.
A
Median
to side a
Median
to side c
b
ma
Centroid
c
Median
to side b
mb
mc
C
B
a
With all sides and angles already known, the
median can be solved using cosine law or by the
following formula:
4ma 2 = 2b2 + 2c2 - a2
4mb 2 = 2a2 + 2c2 - b2
4mc 2 = 2a2 + 2b2 - c2
Altitudes of a Triangle
The altitude of the triangle is a line drawn from one
vertex perpendicular to its opposite side. The
altitudes of a triangle intersect at a point called the
orthocenter of the triangle.
A
Altitude
to side
b
Orthocenter
b
Altitude
to side
c
c
aa
ac
Median
to side
a
ab
C
B
a
With all sides and angles already known, the
altitudes of the triangle can be solved from the right
triangles formed by these altitudes. If the area of the
triangle AT is known, the altitudes can be solved
using the following formulas:
aa =
2AT
2AT
2AT
; ab =
; ac =
a
b
c
Angle Bisectors of a Triangle
The angle bisector of a triangle is the line drawn
from one vertex to its opposite side bisecting the
angle included between the two other sides. The
angle bisectors of a triangle intersect at a point
called the incenter of the triangle.
Incente
A
Angle
bisector to
side c
θ
b
θ
bc
θ
C
Angle
bisector
to side a
ba
θ
Angle
bisector to
side b
c
bb
θ
θ
B
a
With all sides and angles already known, the angle
bisectors of a triangle can be solved using sine law
or using the following formulas:
ba =
bb =
bc =
s =
2
� bcs ( s - a )
b+c
2
� acs ( s - b )
a+c
2
� abs ( s - c )
a+b
a + b + c
(semi-perimeter)
2
Spherical Trigonometry
Spherical Triangle
a
C
a
B
b
b
c
c
A
A spherical triangle is the triangle enclosed by arcs
of great circles of a sphere.
The sum of the interior angles of a spherical is
greater than 180° but less than 540°.
Area of Spherical Triangle
540° ≥ ( A + B + C ) > 180°
The area of a spherical triangle of a sphere of radius
R is
A =
πR2 E
180°
Where E is the spherical excess in degrees and is
given by the following equation
E = A + B + C - 180°
Or
tan
E
s
s-a
s-b
s-c
= � tan tan
tan
tan
2
2
2
4
2
Where
a + b + c
(semi-perimeter)
2
s =
For an arc of a great circle of the earth, the distance
equivalent to 1 minute (0°1’) of the arc is one (1)
nautical mile (6080 feet).
Right Spherical Triangle
B
c
a
C
A
�
A
�
B
c�
b
a
b
NAPIER’S CIRCLE
Napier’s Rules
1.
In the Napier’s Circle, the sine of any
middle part is equal to the product of the
cosines of its opposite parts (SIN-COOP
RULE).
If we take b as the middle part, its
� , then
opposite parts are c� and B
�
sin b = cos c� × cos B
but cos c� = cos (90-c) = sin c
� = sin B
and cos B
then sin b = sin c sin B
2.
In the Napier’s circle, the sine of any
middle part is equal to the product of the
tangents of the adjacent parts (SINTAAD RULE).
� as the middle part, its
If we take A
adjacent parts are c� and b, then
� = tan c� × tan b
sin A
or cos A = cot c tan b
Oblique Spherical Triangles
Law of Sines
sin a
sin b
sin c
=
=
sin A
sin B
sin C
Law of Cosines for Sides
cos a = cos b cos c + sin b sin c cos A
cos b = cos a cos c + sin a sin c cos B
cos c = cos a cos b + sin a sin b cos C
Law of Cosines for Angles
cos A = - cos B cos C + sin B sin C cos a
cos B = - cos A cos C + sin A sin C cos b
cos C = - cos A cos B + sin A sin B cos c
Napier’s Analogies
1
1
(A-B)
(a-b)
tan
2
2
=
1
1
tan c
sin
(A+B)
2
2
sin
1
1
(a-b)
(A-B)
tan
2
2
=
1
1
sin
cot C
(a+b)
2
2
sin
1
1
(A-B)
(a+b)
tan
2
2
=
1
1
cos
tan c
(A+B)
2
2
cos
1
1
(a-b)
(A+B)
tan
2
2
=
1
1
cos
cot C
(a+b)
2
2
cos
The Terrestrial Sphere
A Meridian is a great circle passing throughout the
North and South Poles.
The Equator is a great circle perpendicular to the
meridians.
The Parallels or Latitudes are small circles parallel
to the Equator. Its measure is from 0° to 90°.
The Prime Meridian is the meridian passing through
Greenwich, England.
GMT – Greenwich Mean Time
The earth rotates 360° in 24 hours or 15° every
hour. Therefore, every 15° interval of longitude has
a time difference of one hour.
The mean radius of the earth is 6373 Km (3959
miles), usually taken as 6400 Km (4000 miles).
One Nautical Mile is= 6080 feet. This is the length
of arc on the surface of the earth subtended by one
(1) minute of an arc of the great circle.
The Philippines (Manila) is located 123° 05’ E
Longitude and 64° 36’ N Latitude with time zone of
GMT +8:00.
CHAPTER 3: PLANE GEOMETRY
Definitions:
Altitude of a Triangle – An altitude of a triangle is
perpendicular from any vertex to the side opposite
produced if necessary.
Angle – A plane angle is the opening between two
straight lines drawn from the same point.
Apothem – The apothem of a polygon is the radius
of the inscribed circle.
Area – The area of a plane figure is the number
which expresses the ratio between its surface and
the surface of the unit square.
Center of Polygon – The center of a regular polygon
is the common center of its inscribed and
circumscribed circles.
Circle – A circle is a closed plane curve every point
of which is equally distant from a point in the plane
of the curve.
Complementary Angles – Two angles are called
complementary when their sum is equal to a right
angle and each is called the complement of the
other.
Concurrent Lines – Three or more lines which have
one point in common are said to be concurrent.
Definition of π – The number π used in calculations
in the circle, is the number obtained by dividing the
circumference of a circle by its diameter, that is, π =
C/D. Hence, C = πD or C = 2πr. π = 3.1416 (to 4
decimal places).
Diagonal – A diagonal of a polygon is a line joining
any two nonconsecutive vertices.
Hypotenuse – the hypotenuse of a right triangle is
the side opposite the right angle.
Isosceles Triangle – An isosceles triangle is a
triangle which has two equal sides.
Locus – A locus is a figure containing all the points,
and only those points, which fulfill a given
requirement.
Parallel Lines – Parallel Lines are lines that lie in
the same plane and do not meet however far they
are produced.
Parallelogram – A parallelogram is a quadrilateral
whose opposite sides are parallel.
Perpendicular – If one straight line cuts another so
as to make any two adjacent angles equal, each
line is perpendicular to each other.
Quadrilateral – A quadrilateral is a portion of a plane
bounded by four straight lines.
Rectangle – A rectangle is a parallelogram whose
angles are right angles.
Regular Polygon – A regular polygon is a polygon
all of whose angles are equal and all of whose sides
are equal.
Similar Polygon – Two polygons are similar if the
corresponding angles are equal and their
corresponding sides are proportional.
Supplementary Angles – One angle is the
supplement of another if their sum of angles equals
to two right angles (or 180°).
Tangent – A tangent in a circle is a straight line
which however far it may be produced, has only one
point in common with the circle.
Trapezoid – A trapezoid is a quadrilateral two and
only two of whose sides are parallel.
Triangle – A triangle is a portion of a plane bounded
by three straight lines.
Vertical Angles – When two angles have the same
vertex, and the sides of one are the prolongations
of the sides of the other, they are called vertical
angles.
Triangle
Theorems and Properties of a Triangle:
1.
2.
3.
4.
5.
6.
7.
The sum of the three angles of a triangle
is equal to two right angles or 180°.
The sum of the two sides of a triangle is
greater than the third side, and the
difference is less than the third side.
If two sides of the triangle are unequal,
the angles opposite are unequal, and the
greater angle is opposite the opposite
side, and conversely.
If two sides of the triangle are equal (an
isosceles triangle), the angles opposite
these sides are equal; and conversely.
The perpendicular bisectors of the sides,
and the bisectors of the angles of a
triangle, meet in points which are the
centers of the circumscribed circle and
the inscribed circle, respectively.
The altitudes of a triangle meet in a point
(called orthocenter).
The medians of a triangle are concurrent
at a point which is two-thirds of its
distance from any vertex to the midpoint
of the opposite side. The point of
concurrency is the centroid of the
triangle.
8. Two triangles are congruent if two angles
and the included side of the one are
equal, respectively, to two angles and the
included side of the other.
9. Two triangles are congruent if two angles
and the included side of the one are
equal, respectively, to two sides and the
included side of the other.
10. Two triangles are congruent if the three
sides of the one are equal, respectively,
to the three sides of the other.
Right Triangles
1.
2.
3.
Theorem of Pythagoras. In any right
triangle, the square of the hypotenuse is
equal to the sum of the squares of the
other sides.
Two right triangles are equal if a side and
the hypotenuse of the one are equal,
respectively, to a side and the
hypotenuse of the other.
Two right triangles are equal if a side and
the hypotenuse of the one are equal,
respectively, to the hypotenuse and the
adjacent angle of the other.
4.
If a perpendicular is drawn from the
vertex of the right angle to the
hypotenuse of a right triangle: (1) the two
triangles formed are similar to each other
and to the given triangle; (2) the
perpendicular is a mean proportional
between
the
segments
of
the
hypotenuse; and (3) the square of either
side about the right angle equals the
product of the whole hypotenuse and the
segment adjacent to that side.
B
C
(1)
(2)
(3)
D
A
∆ABC, ∆BDC, and ∆ADC are
proportional.
BD:CD = CD:AD or
(CD)2 = BD/AD
(BC)2 = (AB)(BD)and (AC)2
= (AB)(AD)
Similar Triangles
1. Two triangles are similar if the
angles of one are respectively equal
to the angles of the other, or if two
angles of one are respectively equal
to two angles of the other.
2.
3.
4.
Two triangles are similar if an angle
of one equals an angle of the other
and the sides including these
angles are proportional.
Two triangles are similar if their
sides are in the same ratio.
If two triangles have their sides
respectively parallel, or respectively
perpendicular, each to each, they
are similar.
Area of a Triangle
B
c
h
θ
C
A
b
Given the base b and altitude h
Area =
1
bh
2
Given two sides a and b and included angle θ
Area =
1
ab sin θ
2
Given three sides a, b, and c (Hero’s Formula)
Area = �s ( s - a ) ( s - b ) ( s - c )
Semi-perimeter, s =
a + b + c
2
The area under this condition can be solved by
finding one angle and using cosine law and apply
the formula for two sides and an included angle.
Given three angles A, B, and C and one side a:
Area =
a2 sin B sin C
2 sin A
The area under this condition can also be solved by
finding one angle using cosine law and apply the
formula for two sides and an included angle.
B
B
Y
A
X
Y
C
A
Area of ABC (AB)(AC)
=
Area of AXY (AX)(AY)
X
C
Rectangle
d
Area = ab
Perimeter, P = 2(a + b)
Diagonal, d = �a2 + b2
a
b
Square
d
Area = a2
Perimeter, A = 4a
Diagonal, d = √2
a
a
General Quadrilateral
C
b
B
d
θ
a
d1
d2
A
D
c
Given diagonals d1 and d2 and included angle θ
Area =
1
d d sin θ
2 1 2
Given four sides a, b, c, d and sum of two opposite
angles
Area = �(s - a) (s - b) (s - c) (s - d) - abcd cos2 θ
a+b+c+d
2
∠A + ∠C ∠B + ∠D
θ=
=
2
2
Semi-perimeter, s =
Given four sides a, b, c, d, and two opposite angles
B and D:
Area =
1
1
ab sin B + cd sin D
2
2
Divide the area into two triangles, ∆ABC and ∆CAD:
Parallelogram
1.
2.
3.
The opposite sides of a parallelogram are
equal, and so also are the opposite
angels.
The diagonals of a parallelogram bisect
each other.
If two sides of a quadrilateral are equal
and parallel then the two other sides are
equal and parallel, and the figure is a
parallelogram.
B
C
d1
θ
A
a
d2
b
D
Area of a Parallelogram
Given diagonals d1 and d2 and included angle θ
Area =
1
d d sin θ
2 1 2
Given two sides a and b and one angle A
Area = ab sin A
Rhombus
A rhombus is a parallelogram with
four equal sides. The diagonals of a
rhombus bisect each other at an
angle of 90°.
B
C
D
d1
a
d2
90°
a
A
Area of a Rhombus
Given diagonals d1 and d2:
A=
1
d d
2 1 2
Given side a and one angle A:
A = a2 sin A
Trapezoid
a
A trapezoid is a quadrilateral with
two and only two whose sides
are parallel.
h
b
Area =
Length of the dividing
line parallel to the
two parallel sides.
a+b
h
2
a
c
A1
h
A2
b
c =�
ma2 + nb2 m A1
; =
m+n
n
A2
Cyclic Quadrilateral
A cyclic quadrilateral is a quadrilateral whose
vertices lie on the circumference of a circle.
C
b
c
B
d1
a
D
d2
d
A
∠A + ∠C = 180° ; ∠B + ∠D = 180°
Area = �(s - a) (s - b) (s - c) (s - d) - abcd cos2 θ
Semi-perimeter, s =
a+b+c+d
2
Ptolemy’s Theorem
“For any cyclic quadrilateral, the product of the
diagonals equals the sum of the products of the
opposite sides.”
d1 d2 = (a)(c) + (b)(d)
Polygons
There are two basic types of polygons, a convex
and a concave polygon.
β
θ
A convex polygon is one
β
in which no side, when
θ
θ
extended, will pass
β
θ
θ
inside the polygon,
otherwise, it is called a
aθ
β
β
concave polygon. The
following figure is a convex polygon.
4
4
3
5
3
2
2
6
1
1
6
Polygons are classified according to the number of
sides. The following are some names of polygons.
No. of Sides
3
4
5
6
7
Name
Triangle (trigon)
Quadrilateral (tetragon)
Pentagon
Hexagon
Heptagon
β5
8
9
10
11
12
13
14
15
16
17
18
19
20
30
40
50
60
70
80
90
100
1000
10000
Octagon
Nonagon (enneagon)
Decagon
Undecagon (hendecagon)
Dodecagon
Tridecagon (triskaidecagon)
Tetradecagon (tetrakaidecagon)
Pentadecagon (pentakaidecagon)
Hexadecagon (hexakaidecagon)
Heptadecagon (heptakaidecagon)
Octadecagon (octakaidecagon)
Enneadecagon (enneakaidecagon)
Icosagon
Triacontagon
Tetracontagon
Pentacontagon
Hexacontagon
Heptacontagon
Octacontagon
Enneacontagon
Hectogon
Chilliagon
Myriagon
Theorems in Polygons:
1.
The sum of the angles of a convex
polygon of n sides is 2 (n - 2) right angles.
2.
3.
The exterior angles of a polygon, made
by producing each of its sides in
succession, are together equal to 4 right
angles or 360°
Homologous parts of congruent figures
are equal.
Sum of Interior Angles
The sum of interior angles θ of a polygon of n sides
is
Sum, Σθ = (n - 2) × 180°
Sum of Exterior Angles
The sum of exterior angles β is equal to 360°
Σβ = 360°
Number of Diagonals, D
The diagonal of a polygon is the line segment
joining two non-adjacent sides. The number of
diagonals is given by
D=
n
(n - 3)
2
Regular Polygons
Polygons whose sides are equal are called
equilateral polygons. Polygons with equal interior
angles are called equiangular polygons. Polygons
that are both equilateral and equiangular are called
regular polygons. The area of a regular polygon can
be found by considering one segment, which has
the form of an isosceles triangle.
Circumscribing Circle
x
Inscribed Circle
x
R
Apothem
θ
x
R
θ
θ
x
θ
θ
β
x
α
x
x=side
θ=angle subtended by the side from the
center
R=radius of inscribed circle, also called
the apothem
n=number of sides
360
θ=
n
n
Area = R2 sin θ
2
Perimeter = P = (n)(x)
n-2
× 180°
Interior angle = β =
n
360
Exterior angle = α = θ =
n
Tangent
Circle
Secant
Radius
Chord
Diameter
Arc
Theorems:
1.
Through three points not in a straight
line, one circle and only one can be
drawn.
2.
A tangent to a circle is perpendicular to
the radius at the point of tangency and
conversely.
tangent
r
3.
The tangents to a circle drawn from an
external point are equal and make equal
angles with the line joining the point to
the center.
r
L
θ
θ
r
L
4.
An inscribed angle is measured by onehalf of the intercepted arc.
Inscribed Angle
Central Angle
A
Intercepted Arc
C
θ
O
α
B
θ=
5.
1
α or α = 2θ
2
An angle inscribed in a semicircle is a
right angle.
90
90
90
Thus, if a right triangle is inscribed in a
circle, its hypotenuse is the diameter of
the circle.
6.
An angle formed by two chords
intersecting within the circle is measured
by half the sum of the intercepted arcs.
7.
If two chords intersect in a circle, the
product of the segment of one is equal
to the product of the segments of the
other.
(AE)(BE) = (CE)(DE)
1
θ = (ArcAC + ArcBD )
2
∠ADC = ∠ABC
∠BAD = ∠BCD
D
A
Ɵ
θ
E
B
C
8.
An angle included by a tangent and a
chord drawn from the point of contact is
measured by half the intercepted arc.
Chord
Arc
θ
Tangent
Α=½Ɵ
α
9.
An angle formed by two secants, two
tangents, or a tangent and a secant,
drawn to a circle from an external point,
is measured by half the difference of the
intercepted arcs.
10. If from a point outside a circle a secant
and a tangent are drawn, the tangent is
the mean proportional between the whole
secant and its external segment.
Intersecting Secants
A
θ
O C
D
B
(OA)(OB) = (OC)(OD)
1
θ = (ArcAC - ArcBD )
2
∠ADC = ∠ABC; ∠BAD = ∠BCD
Intersecting Tangent and Secant
B
A
α
θ
O
C
(𝑂𝑂𝑂𝑂)2 = (OA)(OB)
1
θ = (ArcBC - ArcAc )
2
1
α=
ArcAC
2
11. A perpendicular from a point on the
Circumference to a diameter of a circle is
a mean proportional between the
segments of the diameter.
Diameter
h
a
b
h2 =ab
12. The circumference of two circles are in
the same ratio as their radii, and the arcs
of two circles subtended by equal central
angles are in the same ratio as their radii.
Area of Circle
r
D
Circumference = 2πr = πd
π
Area = πr2 = D2
4
Sector of a Circle
C
πrθ
180°
1
πr2 θ
=
= Cr
360° 2
Length of arc, C = rθradians =
r
θ
r
Area =
O
1 2
r θ
2 radians
Note: 1 radian is the angle θ such that C = r
Segment of a Circle
C
r
θ
O
α = 360 - θ
r
r
θ
r
Area = Asector - Atriangle
1
1
A = r2 θ - r2 sin θ
2
2
1
A = r2 ( θ - sin θ)
2
Area = Asector + Atriangle
1
A = r2 ( α + sin θ)
2
Where θ and α are angles in radians.
Parabolic Segment
B
Area =
h
Length of ABC =
A
C
b
m=
2
bh
3
b2
[ me + ln (m + e)]
8h
4h
, e = �1 + m2
b
Spandrel of a Parabola
B
b
h
A=
C
B
h
h
F
A
C
E
D
L
1
bh
3
Areashaded = AADB - AADE + ABFC + ACEF
2
Areashaded = (AC) (h)
3
a
Ellipse
a
V2
a
b
c
F2
c
F1
V1
b
Area = πab
Perimeter = 2π�
𝑎𝑎2 + 𝑏𝑏 2
2
a2 = b 2 + c 2
d3 c
= > 1.0
d4 a
c
Second eccentricity, e' =
b
c
Angular eccentricity, α =
a
a-b
Ellipse flatness, f =
a
a-b
Second flatness, f' =
b
Eccentricity (first eccentricity), e =
Radius of Circles
Circle Circumscribed
(Circumcircle)
About
a
Triangle
A circle is circumscribed about a triangle if it passes
through the vertices of the triangle.
Circumcenter
of the triangle
abc
r=
4AT
a
r
c
b
Where AT = area of the triangle
Circle Inscribe in a Triangle [Incircle]
A circle is inscribed about a triangle if it is tangent
to the three sides of the triangle.
B
c
r
r
a
r
C
A
b
r=
At
a+b+c
; s=
s
2
Circles Escribed about a Triangle [Excircles]
A circle is escribed about a triangle if it is tangent to
one side and to the prolongation of the other two
sides. A triangle has three escribed circles.
ra
rc
c
rb
a
b
ra =
At
At
At
;r =
;r =
s-a b s-b c s-c
Circle Circumscribed about Quadrilateral
A circle is circumscribed about a quadrilateral if it
passes through the vertices of the quadrilateral.
b
a
r
c
d
r=
�(ab + cd)(ac + bd)(ad + bc)
4Aquad
Aquad = �(s - a)(s - b)(s - c)(s - d)
a+b+c+d
Semi-perimeter, s =
2
Circle Inscribed in a Quadrilateral
b
A circle is inscribed in a quadrilateral
if it is tangent to the three sides of the
quadrilateral.
r
a
c
d
ra =
Aquad
a+b+c+d
; Aquad = √abcd ; s =
s
2
Area by Approximation
The area of any irregular plane figure (such as the
one shown) can be found approximately by dividing
it into a number of strips or panels by a series of
equidistant parallel chords (offsets) h1, h2, …, hn the
common distance between the chords being d.
h1
h2
h3
h4
h5
h6
h7
h8
d
d
d
d
d
Area by Trapezoidal Rule
Assume each strips as a trapezoid, then the area
is:
Area =
d
[ h + 2(h2 + h3 + …) + hn ]
2 1
Area by Simpson’s One-Third Rule
This method is more accurate than the previous
because it considers the curved side. Using this
rule, there must be an odd number of offsets, thus
n must be odd.
Area =
d
[ h1 + 2Σhodd +4Σheven + hn ]
3
Area by Coordinates
The area of a planar (convex or concave) with
vertices (x1, y1), (x2, y2), (x3, y3), (xn, yn) is:
(x1,y1)
(x3,y3)
(x8,y8)
(x2,y2)
(x4,y4)
(x6,y6)
(x5,y5)
1 x1
A= �y
2 1
(x7,y7)
x2
y2
x3 xn - 1
y3 ⋯ yn - 1
xn
yn
x1
y1 �
1
A= [�x1 y2 + x2 y3 + ⋯� - �x2 y1 + x3 y2 + ⋯�]
2
The area of a polygon is defined to be positive if the
points are arranged in a counterclockwise order,
and negative in a clockwise order.
CHAPTER 4: SOLID GEOMETRY
Polyhedrons
Polyhedrons are solids whose faces are plane
polygons
Regular polyhedrons are those which have
identical faces. There are only five known regular
polyhedrons, namely tetrahedron, hexahedron,
octahedron, dodecahedron and icosahedrons.
These solids are known as Platonic solids.
Regular Polyhedrons
Regular
Tetrahedron
Regular
Hexahedron
Regular
Dodecahedron
Regular
Octahedron
Regular
Icosahedron
Let m = number of polygons meeting at a vertex
n = number of vertices of each polygon
f = number of faces of the polyhedrons
e = number of edges of the polyhedron, and
v = number of vertices of the polyhedron
Name
f
e
v
m
Tetrahedron
4
4
4
3
Hexahedron
6
12
6
3
Octahedron
8
12
8
4
Dodecahedron
12
30
20
3
Icosahedron
20
30
12
5
Surface Area
2
𝑎𝑎 3
6𝑎𝑎2
2𝑎𝑎2 √3
15𝑎𝑎2 �
3 + √5
5 − √5
5𝑎𝑎2 √3
Where a is the length of the edge
For any polyhedron:
nf
Number of edges, e=
2
Number of vertices, v=
nf
m
Radius of circumscribing sphere, r=
3V
As
Volume
𝑎𝑎3
6√2
𝑎𝑎3
𝑎𝑎3 √2
3
𝑎𝑎3 �15 + 7√5�
4
5𝑎𝑎3 �3 + √5�
12
Euler’s Polyhedron Theorem
For all convex polyhedra:
f = 2f=2v
+e-v
Solids for which Volume = Base area X Height
Prism
Cylinder
Miscellaneous
Volume = Ab h
Prism
Prisms are polyhedron whose bases are equal
polygons in parallel planes and whose sides are
parallelograms.
Prisms are classifieds according to their bases.
Thus, a hexagonal prism is one whose base is a
hexagon, and a regular hexagonal prism has a
base of a regular hexagon.
The axis of a prism is the line joining the centroids
of the bases. A right prism is one whose axis is
perpendicular to the base. The altitude of “h” of a
prism is the distance between the bases.
Volume, V = Ab h = AR L
Lateral area, AL = PR 𝐿𝐿
Where:
AR= area of the section
L= lateral edge
Ab = area of the base
PR= perimeter of right section
Rectangular Parallelepiped
c
d2
d1
a
b
Volume, V = Ab h = abc
Lateral area, AL = AR h=2(ac + bc)
Total surface area, As = At + 2Ab = 2(ab + bc + ac)
Face diagonal, d1 = �a2 + c2
Space diagonal, d2 = �a2 + b2 + c2
Cube (Regular hexahedron)
c
d2
d1
b
a
Volume, V = Ab h = a3
Truncated prism
Lateral area, AL = 4a2
Total surface area, A s = 6a2
Face diagonal, d1 = a√2
Space diagonal, d2 = a�3
Truncated Prism
AR =area of the right section
n=number of sides
Volume AR
Σh
h1 +h2 + ….+hn-1 +hn
= AR
n
n
Pyramids
Pyramids are polyhedron with a polygonal base
and triangular faces that meet at a common point
called the vertex.
Like prisms, pyramids are classified according to
their bases.
h
𝐴𝐴𝑏𝑏
1
Where ;
Volume = Ab h
3
Ab = area of the base
H= altitude, perpendicular distance from the vertex
to the base
Frustum or Pyramid
Frustum of a pyramid is the portion of the pyramid
between the base and the cutting plane parallel to
the base.
Volume =
where:
A1 =lower base area;
A2 =upper base area
h=altitude
h
(A +A +�A1 A2 )
3 1 2
Cylinders
A cylinder is the surface generated by a straight
line intersecting and moving along a closed plane
curve, the directrix, while remaining parallel to the
plane of the directrix.
Like prisms, cylinders are classified according to
their bases
Directrix
Directrix
𝐴𝐴𝑅𝑅
L
h=L
Fixed Straight
Axis
Fixed Straight
Axis
Volume, V = Ab h = AR L
Lateral Area, AL = PR L
where:
AR = area of right section
L = lateral edge
Ab = area of base
PR =perimeter of right section
90°
h
Right Circular Cylinder
r
h
Volume, V = Ab h = πr2 h
Lateral area = AL
AL = Base perimeter*h = 2πrh
Cone
Cone is the surface generated by a straight line,
the generator, passing through a fixed point, the
vertex, and moving along a fixed curve, the
directrix.
Like pyramids, cones are classified according to
their bases
1
Volume, V= Ab h
3
where:
Ab =base area
h=altitude
Right Circular Cone
Slant height, L = �r2 + h2
1
1
Volume, V = Ab h = πr²L
3
3
Lateral area = AL = πrL
where:
r = base radius
h=altitude
Frustum of a Cone
where:
h
Volume, V = (A1 +A2 +�A1 A2 )
3
A1 =lower base area
A2 =upper base area
h=altitude
Frustum of Right Circular Cone Sphere
r
r
L
R
Slant height = L�h2 + (R - r)2
πh 2 2
(R + r + Rr)
3
Lateral area, AL = π(R + r)L
Volume, V =
where:
R=lower base radius
r=upper base radius
h=altitude
Sphere
Surface Area = 𝐴𝐴𝑠𝑠 = 4πr²
4
Volume, V = 𝜋𝜋𝜋𝜋³
3
Spherical Segment of One base
Azone = 2πrh
πh2
Volume, V =
(3r - h)
3
Spherical Segment of Two Bases
Lateral area, As = 2πrh
πh
(3a2 + 3b2 + h2 )
6
Volume, V =
Spherical Cone or Spherical sector
1
A
r
3 zone
2 2
Volume = πr h
3
Volume =
Spherical Lune and Wedge
Alune
4πr2
=
θ
360°
Vwedge 4 πr3
= (
)
θ
3 360°
Spherical Polygons
A spherical polygon is a polygon on the surface of
a sphere whose sides are arcs of great circles.
n = no. of sides
r = radius of sphere
E= spherical excess
π r2 E
180°
E = sum of angles - (n-2)180°
Area =
Spherical Pyramid
B
A
D
C
r
r
Volume =
πr2 E
540°
Solids of Revolution (Pappus theorems)
First proposition of Pappus
The surface area generated by a surface of
revolution equals the product of the length of the
generating arc and the distance travelled by its
centroid.
As = L C
C = R × (Angle of rotation in radians, θ)
If θ = 360°, As = L (2πR)
Second Proposition of Pappus
The volume area generated by a solid of
revolution equals the product of the generating
area and the distance travelled by its centroid.
Volume = AC = A (RƟ𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 )
If Ɵ = 360°, Volume = A(2πR)
Ellipsoid
c
b
Volume =
a
4
π abc
3
Prolate Spheroid
Prolate spheroid is formed by revolving the ellipse
about its major (X) axis. Thus, from the figure
above, c= b, then,
Volume =
As = 2πb2 + 2πab (
4
πab2
3
arcsin 𝑒𝑒
√𝑎𝑎2 − 𝑏𝑏 2
); 𝑒𝑒 =
𝑒𝑒
𝑎𝑎
Spheroid
Prolate spheroid is formed by revolving the ellipse
about its minor (Z) axis. Thus, from the figure
above, c = a, then,
4
πa2 b
3
2
πb
1+e
As = 2πa2 +
( ln
)
Volume =
e
1-e
Paraboloid of Revolution
h
r
Volume =
A=
4πr
3h2
1
π r2 h
2
3�
2
r2
�� + h2 �
4
r 2
- � � �
2
Prismatoid
General Prismatoid
A general prismatoid is a solid such that the area
of any section, say A parallel to and distant y from
a fixed plane can be expressed as a polynomial of
y of the degree not higher than the third.
Ay = ay3 + by2 + cy + d
Where a, b, and c are constant which may be
positive, negative, or zero.
Prismoidal Formula
Volume =
L
[ A1 + 4Am + A2 ]
6
One criteria of knowing if a certain solid is a
prismatoid is that all section parallel to a certain
base are all similar.
Prismatoid Theorem
The volume of a prismatoid is equal to the
algebraic sum of the volumes of a pyramid, a
wedge and a parallelepiped.
Volume of Some Prismatoid
Volume common to two equal cylinders with their
axis meeting at right angles.
Volume =
16 3
r
3
Solid with circular base of radius r and every
cutting plane perpendicular to a certain diameter is
a SQUARE.
Volume =
16 3
r
3
Solid with circular base of radius r and every
cutting plane perpendicular to a certain diameter is
an EQUILATERAL TRIANGLE.
Volume =
4r3
√3
Solid with circular base of radius r and every
cutting plane perpendicular to a certain diameter is
an ISOSCELES RIGHT TRIANGLE with one leg in
the plane of the base
Volume =
8 3
r
3
Solid with circular base of radius r and every
cutting plane perpendicular to a certain diameter is
an ISOSCELES RIGHT TRIANGLE with
hypotenuse on the plane of the base.
Volume =
4 3
r
3
CHAPTER 5: PLANE ANALYTIC GEOMETRY
Cartesian or Rectangular Coordinate System
Second Quadrant (-, +)
First Quadrant
Abscissa
(+, +)
Ordinate
O
Third Quadrant (-, -)
Fourth Quadrant
(+, -)
Distance Between Two Points
The distance between point P1(x1,y1) and P2(x2,y2) is
d = �(x2 - x1 )2 + �y2 - y1 �
2
Straight Line
A straight line is a line that does not change in
direction. Thus, it has a uniform slope.
General Equation of a Line
Ax + By + C = 0
P2 (x2, y2)
P1 (x1, y1)
d
(0, b)
y2 – y1
x2 – x1
O
(a, 0)
The general equation of a straight line is
Ax + By + C = 0
To solve a line, either two points or one point and a
slope must be known.
Slope of the Line
The slope of the line passing through points P1(x1,
y1) and P2(x2, y2) is:
Slope, m =
rise y2 - y1
=
run x2 - x2
Where:
m is positive if the line is inclined upwards to
the right.
m is negative if the
downwards to the right.
line is inclined
m is zero for horizontal lines.
Standard Equations of Lines
1.
Point-slope form
Given a point P1(x1, y1) and its slope m:
y - y1 = m(x - x1 )
2.
Slope-intercept form
Given a slope m and y-intercept:
y = mx + b
3.
Intercept form
Given x-intercept a and y-intercept b:
x y
+ =1
a b
4.
Two-point form
Given two points P1(x1,y1) and P2(x2,y2):
y - y1
y -y
= 2 1 =m
x - x1
x2 - x2
Note: All these four standard equations can be
reduced into the point-slope form.
Angle Between Two Lines
The angle between two lines L1 and L2 is the angle
θ that L1 must be rotated in a counterclockwise
direction to make it coincide with L2.
Line 2, slope 2
α1 = arctan m1
α 2 = arctan m2
α2
θ
α1
θ = α2 - α1
tan θ =
m2 - m1
1 + m1 m2
Lines are parallel if m1 = m2
Lines are perpendicular if m2 = -1/ m1
Line 1, slope 1
Distance from a Point to the Line
Y
Line = Ax + By +C = 0
d
(x1, y1)
O
X
The distance (nearest) from a point P1(x1,y1) to a
line Ax + By + C = 0 is:
d=
Use of Sign:
Ax1 + By1 + C
± � A2 + B2
(+) if B is a positive number
(-) if B is a negative number
(+) if the point is above the line or to the right of the
line for a vertical line
(-) if the point is below the line or to the left of the
line for a vertical line
That is,
If B is positive and the point is above the line, then
use (+) (+) = (+)
If B is positive and the point is below the line, then
use (+) (-) = (-)
If B is negative and the point is above the line, then
use (-) (+) = (-)
If B is negative and the point is below the line, then
use (-) (-) = (+)
If only the distance is required, use the absolute
value:
d= �
Ax1 + By1 + C
�
�A 2 + B 2
Distance between Two Parallel Lines
Y
d
Line = Ax + By + C1 = 0
O
X
Line = Ax + By +C2 = 0
d= �
C2 - C1
�A 2 + B 2
�
Division of Line Segment
L
P2 (x2, y2)
r2
L2
r1
P1 (x1, y1)
P (xp, yp)
r1 = L1 / L
r2 = L2 / L
L1
xp =
x1 r2 + x2 r1
r1 + r2
yp =
y1 r2 + y2 r1
r1 + r2
Midpoint of a Line Segment
The midpoint Pm (xm, ym) of a line segment through
from P1(x1,y1) and P2(x2,y2) is
xm =
x1 + x2
2
ym =
y1 + y2
2
Area of Polygon by Coordinates
Let (x1, y1), (x2, y2), (x3, y3)… (xn-1, yn-1), (xn, yn) be
the consecutive vertices of a polygon arranged in
counterclockwise sense. The area is:
1 x1
A= �y
2 1
x2
y2
x3 xn-1
y3 ⋯ yn-1
xn
yn
x1
y1 �
1
A= [�x1 y2 + x2 y3 + ⋯� - �x2 y1 + x3 y2 + ⋯�]
2
Conic Sections
Conic sections a locus (or path) of a point that
moves such that the ratio of its distance from a fixed
point (called the focus) and a fixed line (called the
directrix) is constant. This constant ratio is called
the eccentricity (e) of the conic.
The term conic section was based on the fact that
these are sections formed if a plane is made to pass
through a cone.
If the cutting plane is parallel to the base of the
cone, the section formed is a circle. If it is parallel
to the element (or generator) of the cone, the
section formed is a parabola. If it is perpendicular
to the base of the cone, the section formed is a
hyperbola. If it is oblique to the base or element of
the cone, the section formed is an ellipse.
General Equation of Conics
Ax2 + Bxy + Cy2 + Dx + Ey + F = 0
If B is not equal to 0, the axis of the conic is oblique
with the coordinate axes (i.e. not parallel to X or Y
axes). Thus if the axis is parallel to either X or Y
axes, the equation becomes
Ax2 + Cy2 + Dx + Ey + F = 0
From the foregoing equations:
If B2 < 4AC, the conic is an ellipse
If B2 = 4AC, the conic is an parabola
If B2 > 4AC, the conic is an hyperbola
Also, a conic is a circle if A = C, an ellipse if A ≠ C
but have the same sign, a parabola if either A = 0
or C = 0, and a hyperbola if A and C have different
signs.
Y
Hyperbola, e > 1
Parabola, e = 1
Vertex of the cone
F
Circle, e = 0
X
Ellipse, e < 1
The parabola antenna
used in communication is
our example.
Other Important Properties of Conic Sections
Imagine that the surface of a conic is a mirror that
reflects light (or any material that reflects sound
beam). The following illustration shows how these
rays are reflected from its source.
Circle
Circle is the locus of a point that moves such that is
always equidistant from a fixed point called the
center. The constant distance is called the radius of
the circle.
Y
r
(h, k)
O
X
r = radius
(h , k) = center
General Equation of Circle (A=C)
Ax2 + Ay2 + Dx + Ey + F = 0 or
x2 + y2 + Dx + Ey + F = 0
To solve a circle, either one of the following two
conditions must be known.
a. Three points along the circle.
Solution: Use the general form
b. Center (h, k) and the radius r.
Solution: Use the standard equation.
Standard Equations of a Circle
Center at (h, k)
Center at (0, 0)
(x - h)2 + (y - k)2 = r2
x2 + y2 = r2
For circle Ax2 + Ay2 + Dx + Ey + F = 0
h=
-D
-E
D2 + E2 - 4AF
;k=
;r=�
2A
2A
4A2
Parabola
The locus of the point that moves such that its
distance from a fixed point called the focus is
always equal to its distance from a fixed line called
directrix.
Vertex
d1
Directrix
2A
a
a
Latus Rectum =4A
Focus
d2
2A
2A
a = distance from the vertex to the focus
LR = length of the latus rectum
General Equation of Parabola (A or C is zero)
C=0
Ax2 + Dx + Ey + F = 0 or
x2 + Dx + Ey + F = 0
A=0
Cy2 + Dx + Ey + F = 0 or
y2 + Dx + Ey + F = 0
To solve for the parabola, either one of the following
conditions must be known:
1. Three points along the parabola and an
axis (either horizontal or vertical)
Solution: use the general form
2. Vertex (h, k), distance from the vertex to
the focus a, and axis.
Solution: use the standard equation
3. Vertex and the location of the focus.
Solution: use the standard equation.
Eccentricity
The eccentricity of a conic is the ratio of its distance
from the focus (d2) and from the directrix (d1).
For a parabola, the eccentricity is equal to 1.
e=1
Latus Rectum, LR
Latus rectum is a chord passing through the focus
and parallel to the directrix or perpendicular to the
axis.
LR = 4A
Standard Equations of a Parabola
Vertex at (0, 0)
y2 = 4ax
x2 = 4ay
y2 = - 4ax
x2 = -4ay
Vertex at (h, k)
(y-k)2 = 4a(x-h)
(x-h)2 = 4a(y-k)
(y-k)2 = - 4a(x-h)
For the
vertical)
parabola
h=
(x-h)2 = -4a(y-k)
Ax2 + Dx + Ey + F = 0
(axis
-D
D2 - 4AF
-E
;k=
;a=
4AE
2A
4A
For the parabola Cy2 + Dx + Ey + F = 0 (axis
horizontal)
h=
E2 - 4CF
-E
-D
;k=
;a=
4CD
2C
4C
Ellipse
The locus of the point that moves such that the sum
of its distances from two fixed points called the foci
is constant. The constant sum is the length of the
major axis, 2a. It can also be defined as the locus
of the point that moves such that the ratio of its
distance from a fixed point, called the focus and a
fixed line, called the directrix, is constant and is less
than one (1).
y
d
Directrix
d
d4
Directrix
Minor Axis, 2b
α
V2 F2
d3
b
F1
a
c
V1
c
d1
d2
P(x, y)
a
a
Major axis, 2a
LR
x
Elements of an Ellipse
a2 = b2 + c 2
d3 c
= < 1.0
d4 a
a
Distance from the center to the directrix, d =
e
2b2
Length of Latus rectum, LR =
a
c
Second eccentricity, e' =
b
c
Angular eccentricity, α =
a
a-b
Ellipse flatness, f =
a
a-b
Second flatness, f' =
b
Eccentricity (first eccentricity), e =
General equation of an Ellipse ( A ≠ C but have
the same sign)
Ax2 + Cy2 + Dx + Ey + F = 0 or
x2 + Cy2 + Dx + Ey + F = 0
To solve the parabola, either one of the following
conditions must be known:
1.
2.
Four points along the ellipse
Solution: use the general form
Center (h, k), semi-major axis a, and
semi-minor axis b
Solution: use the standard equation
Standard Equations of an Ellipse
Center at (0, 0)
x2 y2
+
=1
a2 b2
x2
b
2
+
y2
=1
a2
+
(y - k)2
=1
a2
Center at (h, k)
(x - h)2 (y - k)2
+
=1
a2
b2
(x - h)2
b2
Note: a > b
For the ellipse Ax2 + Cy2 + Dx + Ey + F = 0
h=
-D
-E
;k=
2A
2C
Hyperbola
The locus of the point that moves such that the
difference of its distance between two fixed points
called the foci is constant. The constant difference
is the length of the transverse axis 2a. It may also
be defined as the locus of the point that moves such
that the ratio of its distance from a fixed point called
the focus, and a fixed line, called the directrix is
constant and is greater than one (1).
Directrix
Directrix
y
Conjugate axis, 2b
d4
d2
b
d3
d1
F2
V2
b
b
c
c
x
LR
a
d
d
Transverse Axis, 2a
a
a
d2 - d1 = 2a
V1
F1
c
Asymptote
Elements of a Hyperbola
c 2 = a2 + b2
d3 c
= > 1.0
d4 a
a
Distance from the center to the directrix, d =
e
2b2
Length of Latus rectum, LR =
a
Eccentricity (first eccentricity), e =
Equation of Asymptote
The asymptotes of a hyperbola has the following
equations:
y-k =±m(x-h)
Where (h, k) is the center of the hyperbola and m is
the slope. Use (+) for upward asymptote and (-) for
downward asymptote.
m = b⁄a (if the axis is horizontal)
m = a⁄b (if the axis is vertical)
General Equation of Hyperbola (A and C have
opposite sign)
Ax2 - Cy2 + Dx + Ey + F = 0 or
x2 - Cy2 + Dx + Ey + F = 0
Standard Equations of Hyperbola
Center at (0, 0)
x2 y2
=1
a2 b2
(x - h)2 (y - k)2
=1
a2
b2
x2 y2
=1
b2 a2
(x - h)2
b
2
-
(y - k)2
=1
a2
Note: “a” may be greater, equal, or less than “b”.
For the hyperbola Ax2 - Cy2 + Dx + Ey + F = 0
h=
-D
-E
;k=
2A
2C
Variation of Problems in Conics
1.
2.
3.
Given the equation of the conic, find the
elements (center, eccentricity, focus,
latus rectum, vertex, etc.)
Solution: Reduce the equation to
standard form and apply the necessary
formulas.
Find the equation of the conic through
given points (3 points for circle, 3 points
and axis for parabola, 4 points for ellipse
and hyperbola).
Solution: Substitute the points to the
general equation and solve for A, B, and
C, etc.
Find the equation of a conic, given the
elements (center and radius for circle,
vertex and a for parabola; center, a, b for
ellipse and hyperbola).
Solution: Reduce the equation to
standard from.
Tangents and Normals to Conics
Point of Tangency
Tangent
Y
(x2, y2)
Normal
Conic
O
X
Unlike other curves, the tangent to any conic will
pass only through one point. To find the equation of
the tangent to a conic, we may use differential
calculus or make use of the following substitution
for the variables in the equation:
x2
y2
x
→
→
→
y
→
replace with xx1
replace with yy1
x + x1
replace with
2
y+y
1
replace with
2
Where (x1,y1) is the point of tangency and (x,y) is
any point on the line.
CASE I. Find the equation of the tangent through a
given point (x1,y1) on the conic.
•
•
To find the equation of the tangent we
simply replace the variables in the conic
with the above expressions and
substitute for (x1,y1) the given point.
We can also use differential calculus to
solve for the slope 𝑚𝑚 = 𝑑𝑑𝑑𝑑⁄𝑑𝑑𝑑𝑑 of the
curve at the given point (x1,y1), then use
the point-slope form of the line, y y1 = m (x - x1 ).
CASE II. Find the equation of the tangent that
passes through a given point (x2,y2) outside the
conic.
•
To find the equation of the tangent, we
apply the necessary replacements of
variables leaving x1 and y1. Another
equation relating x1 and y1 can be found
by substituting (x1,y1) to the equation of
the conic. By expressing y1 in terms of x1
in either equation and substituting the
other equation, a quadratic equation is
derived in the form Ax21 + Bx1 + C = 0.
With (x1,y1) known, the tangent can now
be solved.
CASE III. Find the equation of the tangent, given
the slope m of the tangent.
•
•
To find the equation of the tangent, we
use the slope-intercept form y = mx + b
for the tangent line with m and b
unknown. Since the line and the conic
process, we can substitute this value of y
in the conic resulting to a quadratic
equation in the form Ax2 + Bx + C = 0 with
A, B, and C as a function of m and b.
Since the tangent passes through one
point only, we set B2 = 4AC and solve for
b. With b known and m given, the tangent
can now be solved.
Or, for this case, the value of (x1, y1) can
be found by differential calculus knowing
that dy / dx = slope = m. After solving m,
the equation of the line can be found
using the point – slope form y, y1 = m(x,
x1)
POLAR COORDINATE SYSTEM
In this system, the location of a point is expressed
by its distance r from a fixed point called the pole
and its angle θ from a fixed line Ox.
(r, θ)
r
rθ
Pole
Axis
Sign Convention
•
θ is positive (+) if measured
counterclockwise.
•
θ is negative ( - ) if measure clockwise.
•
r is positive ( + ) if laid off at the terminal
side of θ.
•
r is negative ( - ) if laid off prolongation
through O from the terminal side of θ.
Distance between Two Points
The distance between two points (r1, θ 1) and (r2, θ
) can be found using cosine law.
2
(r1, θ1)
d
r1
(r2, θ2)
α = θ2 - θ1
r2
O
X
Pole
d = �r1 2 + r2 2 - 2r1 r2 cos ( θ2 - θ1 )
Relationship between Polar and Cartesian
Coordinate Systems
Equations in rectangular form may be converted to
polar form or vice versa. The following relationships
can be found from the figure shown.
x
(r, θ)
r
y
θ
Pole
Axis
r2 = x2 + y2
x = r cos θ ;
y = r sin θ ; tan θ =
Polar Curves
Cardioid
r=a (1 - sin θ)
y
x
y=1/x
y =| 1 / x|
y = ax3
y3 = ax2
SPACE ANALYTIC GEOMETRY
SPACE COORDINATE SYSTEMS
There are three coordinate systems used in space
analytic geometry. The rectangular, cylindrical, and
spherical coordinate systems.
Rectangular Coordinate System in Space
In rectangular coordinate systems, a point P(x, y, z)
in space fixed by its three distance x, y, and z from
the three coordinate planes.
d = �(x2 - x1 )2 + (y2 - y1 )2 + (z2 - z1 )2
Cylindrical Coordinate System
A point P in space may be imagined as being on the
surface of a cylinder perpendicular to the XY –
plane. P(r, ϕ, z) is fixed by its distance z from the
XY – plane and by the polar coordinate(r,ϕ) of the
projection of P on the XY – plane.
Z
P (r, θ, z)
r
θ
X
Y
Relations between Rectangular and Cylindrical
Coordinates System
The relations between rectangular coordinates (x,
y, z) and cylindrical coordinates (r, θ, z) are:
x = r cos θ
y
x
r2 = x2 + y2 , z = z
y = r sin θ
θ = arctan
Spherical Coordinates System
Z
A point P in space may be
P(r, ϕ, θ)
imagined as being on the
ϕr
surface of a sphere with
O
center at the origin O and
radius r. P(r, ϕ, θ) is fixed
θ
Y
by its distance r from O,
Y
the angle ϕ between OP
and the Z – axis and angle
ϕ which is the angle between
the X – axis and the projection of OP on the XY –
plane.
Relations between Rectangular and Spherical
Coordinates Systems
The relations between rectangular coordinates (x,
y, z) and spherical coordinates (r, ϕ, θ) are:
X
x = r sin ϕ cos θ ; y= r sin ϕ sin θ
r2 = x2 + y2 + 𝑧𝑧 2
�x2 + y2
y
; θ = arctan
ϕ = arctan
z
x
z=r cos ϕ
CHAPTER 6 – DIFFERENTIAL CALCULUS
Limits
Theorems of Limits
1.
If f(x) = c, a constant, then lim f(x) = c
x→a
Let lim f(x) = A and lim g(x) = B, then;
x→a
2.
3.
4.
5.
6.
x→a
lim k f(x) = kA, k being constant.
x→a
lim [f(x) ± g(x)] = lim f(x) ± lim g(x) = A ± B
x→a
x→a
x→a
lim [f(x) g(x)] = lim f(x) lim g(x) = A B
x→a
x→a
lim
f(x)
x→a g(x)
=
lim f(x)
x→a
lim g(x)
x→a
=
A
B
x→a
, provided B ≠ 0
n
lim �f(x)= � lim f(x) = √A , provided
x→a
n
n
x→a
n
√A is a real number
L’ Hospital’s Rule (Intermediate type 0/0)
If a is a number, if f(x) and g(x) are differentiable
and g(x) ≠ 0 for all x on some interval 0 < | x - a | <
δ, and if lim f(x) = 0 and lim g(x) = 0, then, when
lim
f'(x)
x→a g'(x)
x→a
x→a
exists or is infinite,
lim
f(x)
x→a g(x)
lim f'(x)
=
x→a
lim g'(x)
x→a
Short Technique on Limits
To evaluate the lim f(x), substitute for x a value
x→a
that is very close to a and use your calculator
That is;
For x → 2, substitute x = 1.9999 or x = 2.0001
For x → -5, substitute x = -4.9999 or -5.0001
For x → 0, substitute x = 0.00001
For x → ∞, substitute x = 99999
Consider the following examples:
Example 6-1: lim
x4 - 81
;
x→3 x - 3
(2.9999)4 - 81
(2.9999) - 3
Substitute x = 2.9999
= 107.99 ≈ 108
1
1
x
ex - 1
Example 6-2: lim � x→0
�;
Substitute x = 0.0001
�
1
0.0001
-
1
e0.0001 - 1
Example 6-3: lim
cotx
� = 0.49999 ≈ 0.5
x→0 cot2x
cot ( 0.0001 )
cot [2 (0.0001)]
=2
;
Substitute x = 0.0001°
Example 6-4: lim �
ex + e-x - x2 - 2
�;
sin2 x- x2
x→0
Substitute x = 0.0001 radian
Note: set your calculator to radian mode
e0.0001 + e-0.0001 - ( 0.0001 )2 - 2
2
sin2 ( 0.0001 ) – ( 0.0001 )
= 0.25
Differential Formulas
In the following formulas u, v, and w are
differentiable functions of x and a and n are
constants.
Algebraic Functions
1.
2.
3.
4.
5.
6.
7.
d
dx
d
dx
d
dx
d
dx
d
dx
d
dx
(c)=0
( cu ) = c
d(u)
( uv ) = u
d(v)
(u+v)=
(u )=
n
� √u �=
u
d� � =
v
dx
d(u)
dx
dx
+
d(v)
+v
dx
d(u)
dx
d(u)
nun - 1
dx
d(u)
dx
√u
;u≠0
d(u)
d(v)
v
-u
dx
dx
2
v
;v≠0
8.
c
d� � =
v
d(v)
-c dx
v2
Logarithmic & Exponential Functions
9.
10.
11.
12.
13.
14.
d
dx
d
dx
d
dx
d
dx
d
dx
d
dx
�loga u� =
d(u)
dx
u ln a
( log10 u) = log10 e
( ln u) =
d(u)
dx
u
=
u
d(u)
dx
d(u)
eu
dx
(uv ) = vuv - 1
d(u)
+ uv ln u
dx
Trigonometric Functions
15.
16.
17.
18.
d
dx
d
dx
d
dx
d
dx
= 0.43429
;u≠0
(au ) = au ln a
(eu )
d(u)
dx
( sin u) = cos u
d(u)
dx
( cos u) = -sin u
2
( tan u) = sec u
2
( cot u) = -csc u
d(u)
dx
d(u)
dx
d(u)
dx
d(v)
dx
d(u)
dx
u
19.
20.
d
dx
d
dx
( sec u) = sec u tan u
d(u)
dx
( csc u) = -csc u cot u
d(u)
dx
Inverse Trigonometric Functions
21.
22.
23.
24.
25.
26.
d
dx
d
dx
d
dx
d
dx
d
dx
d
dx
( arcsin u) =
d(u)
dx
�1 - u2
( arccos u) = -
( arctan u) =
�1 - u2
d(u)
dx
1 + u2
( arccot u) = ( arcsec u) =
d(u)
dx
d(u)
dx
1 + u2
d(u)
dx
u�u2 - 1
( arccsc u) = -
d(u)
dx
u�u2 - 1
Hyperbolic Functions
27.
28.
d
dx
d
dx
( sinh u) = cosh u
( cosh u) = sinh u
d(u)
dx
d(u)
dx
29.
30.
31.
32.
d
dx
d(u)
( tanh u) = sech2 u
dx
2
d
dx
d
dx
d
dx
( coth u) = -csch u
d(u)
dx
( sech u) = - sech u tanh u
( csch u) = - csch u coth u
d(u)
dx
d(u)
dx
Where:
sinh x =
tanh x=
sech x=
ex - e-x
cosh x=
2
sinh x
coth x=
cosh x
1
csch x=
cosh x
Inverse Hyperbolic Functions
33.
34.
35.
36.
d
dx
d
dx
d
dx
d
dx
( arcsinh u)=
d(u)
dx
�u2 + 1
( arccosh u) = -
( arctanh u) =
( arccoth u) =
d(u)
dx
�u2 - 1
d(u)
dx
1 - u2
-
d(u)
dx
u2 - 1
=
d(u)
dx
2
1- u
ex + e-x
2
1
tanh x
1
sinh x
37.
38.
d
dx
d
dx
( arcsech u) =
( arccsc u) =
-
d(u)
dx
2
u�1 - u
-
d(u)
dx
u�1 + u2
Where:
arcsinh x = ln�x+ √x2 + 1�
arccosh x = ln �x+ �x2 - 1�
1 1+x
ln
2 1-x
1 x+1
arccoth x = ln
2 x-1
arctanh x =
arcsech x = ln
arcsech x = ln
1 + √1 - x2
x
1 + �1 + x2
x
x >0
Slope of the Curve
The slope of the curve y=f(x) at any point is
dy
identical to the derivative of the function or y.
dx
Slope at any point = y’ =
dy
dx
Rate of Change
The derivative of a function is identical to its rate of
change. Thus, the rate of change of the volume
dV
(V) of a sphere with respect to its radius (r) is .
dr
Curvature and Radius of Curvature
Curvature
Curvature refers to the rate of change of the
direction of the curve. Thus, a circle with a smaller
radius has great curvature, or is sharply curved.
The curvature, K, of the curve y = f(x) is:
K=
y"
[1 + (y')2 ]
3�
2
Where y” is the second derivative of the function.
Radius of Curvature
The radius of curvature, ρ, is the reciprocal of the
curvature, k, or:
ρ=
3�
2
1 � 1 + (y')2 �
=
| y" |
k
Circle of Curvature
At any point on a curve y = f(x), where y’ and y”
exist and y” ≠ 0, there is associated with the curve
a circle, which is called the circle of curvature with
the following equation:
( x – h )2 + ( y – k )2 = ρ²
The center (h, k) and radius ρ of the curve is:
h=x-
y' � 1 + (y')2 �
y"
k=y+
ρ=
1 + (y')2
y"
3�
2
�1 + (y')2 �
| y" |
Graph of a Function y = f(x)
The graph of a function y = f(x) may be plotted
using calculus. Consider the graph shown below:
y
B
A
E
C
y = f(x)
D
x
As x increases, the curve rises if the slope is
positive, as of arc AB; it falls if the slope is
negative, as of arc BC.
Relative Maximum and Minimum Points
At a point such as B, where the function is
algebraically greater than that at any neighboring
point, the point is said to have a maximum value,
and the point is called a maximum point (relative
to adjacent points). Similarly, at D the function has
a minimum value (relative to adjacent points). At
maximum or minimum points, the tangent is
horizontal or the slope is zero.
dy
= y' = 0
dx
This does not necessarily mean that at these
points the function is maximum or minimum. It
does only mean that the tangent is parallel to the
x-axis, or the curve is either concave up or
dy
= 0 are
concave down. The points at which
dx
called critical points, and the corresponding values
of x are critical values.
The second derivative of a function is the rate of
change of the first derivative or slope. It follows
that as x increases and y” is positive, y’ is
increasing and the tangent turns in a
counterclockwise direction and the curve Is
concave upward. When y” is negative, y’ is
decreases and the tangent turns in a clockwise
direction and the curve is concave downward.
If y’ = 0 and y” is negative (i.e. y” < 0), the point is
a maximum point (concave downward).
If y’ = 0 and y” is positive (y” > 0), the point is a
minimum point (concave upward).
Points of Inflection
A point of inflection is a point at which the curve
changes from concave upward to concave
downward or vice verse (see point E from the
figure). At these points, the tangent changes its
rotation from clockwise to counterclockwise or vice
versa.
At points of inflection, the second derivative of y is
zero (y” = 0).
Applications of Maxima and Minima
As an example, the area of a rectangular lot,
expresses in terms of its length and width, may
also be expressed in terms of cost of fencing.
Thus, the area can be expressed as A = f(x). The
common task here is to find the value of x that will
give a maximum value of A. to find this value, we
set dA�dx = 0.
Steps in Solving Maxima Minima Problems:
1.
Identify the variable to be maximized or
minimized, say
2. Express this variable in terms of the other
relevant variable(s), say A = f(x, y).
3. If the function shall consist of more than one
variable, expressed it in terms of one variable
(if possible) using the conditions in the
problem, say A = f(x).
4. Differentiate and equation to zero, dA�dx = 0.
Common Variable Relationships for Maximum
or Minimum Values
Minimum length of ladder leaning
against a wall with one end on
the ground outside the wall
L
2�
3
=h
2�
3
2�
3
+x
L
h
x
For maximum area (to admit the
most light) of a Norman window
of given perimeter
r
x
x
;r=
2
2
h=x
y
y=
x
For maximum area of a rectangular
window surrounded by a right
isosceles triangle of known
perimeter
y
y
h
h=y
For maximum viewing angle θ of an
object
x = √ab
b
θ
a
x
Minimum length L of a line
tangent to an ellipse
L
b
L=a+b
For the minimum perimeter
of a rectangular lot of
known area to be fenced on
three sides only
a
b
a
y
y
x
x = 2y
For minimum perimeter of a
rectangle of known area or
for maximum area for known
perimeter
y
x
x = y (a square)
For a given area of a rectangle, the square has the
least perimeter, OR for a given perimeter of
rectangle, the square has the largest area.
Maximum volume of a
CLOSED (both ends)
cylindrical tank of given
surface area, or minimum
surface area for given volume
h
D
Diameter = 2 × height
D = 2h
Maximum volume of an OPEN (one end)
cylindrical tank of given surface area, or minimum
surface area for given volume
Diameter = 2 × height
D = 2h
For maximum volume of a
closed rectangular box with
given sum of all edges or given
total area
z
y
x
x = y = z (a cube)
Maximum volume of open (one end)
rectangular box of square base and
given surface area, or minimum
surface area for given volume.
h
y
x
x = 2y
For a rectangle of maximum
area or perimeter that can be cut
from a circle of radius r
r
y
x = y (a square)
x
For the strongest rectangular
beam that can cut from log of
radius r
2r
2r = b �3
For maximum capacity of a
trapezoidal gutter or canal of
known perimeter, or minimum
perimeter of known capacity
d
b
x
y
y
60°
60°
y
x = 2y (a half-rectangular hexagon)
For maximum volume of a
right circular cone with known
lateral area, or minimum
lateral area for known volume
h = r √2
L
h
r
For the largest rectangle that
can be inscribed in an ellipse
y
y
b
a
x
x
x = a�
√2
y = b�
√2
Strongest beam that can be cut
from an elliptical section
x=
y=
2b
√3
2√2a
√3
Weight of heaviest cylinder that
can be cut from a sphere of
weight W, or largest cylinder that
can be cut from a sphere of
volume V
Wcyl = W�
√3
Vcyl = V�
√3
x
y
a
b
Largest rectangle that
can be cut from a
given triangle
h
x
y
x = b�2
b
y = h�2
Largest beam that can pass
through a corridor
2�
3
L
=x
2�
3
2�
3
+y
Smallest cone that can inscribe
a sphere of radius r
y
L
x
θθ
h
sin θ = 1�3
Largest cylinder that can be
inscribed in a given cone
y=
h
3
r
x
For a minimum cost of closed
cylindrical tank of known volume or
maximum volume for known cost
h
D
Height=
Cost of ends
x Base Diameter
Cost of sides
Largest rectangle that can be
inscribed in a semi-circle
b
h
b = 2h
For a minimum length of wire
running from the top of one post
to the stake on the ground and to
the top of another post
L2
a
L1
x
x
d
=
a a+b
b
d
For maximum product of n numbers whose sum
is A, each number is
n
is �A�n�
b
A
n
and the maximum product
For maximum product xn × ym where x + y = K
x=
K
n
m+n
y=
K
m
m+n
Time Rates
If a quantity x is a function of time t, the time rate
of change of x is given by
dy
dx
.
When two or more quantities, all functions of t, are
relates by an equation, the relation between their
rates of change may be obtained by differentiating
both sides of the equation with respect to t.
Basic Time Rates
Velocity, v =
dS
dt
Where S is the distance
Acceleration, a =
Discharge, Q =
dv
dt
dV
dt
Where V is the volume at any time
Angular Speed, ω =
dθ
dt
Where θ is the angle at any time
CHAPTER 7: INTEGRAL CALCULUS
Integration Formulas
Algebraic, Exponential and Logarithmic
Functions:
1. � adu = a � du = au + C
au
+ C,
ln a
1
un+1 + C
3. � un du =
n+1
2. � au du =
4. � eu du = eu + C
5. � u-1 du = �
a > 1, a ≠ 1
for n ≠ -1
du
= ln|u| + C
u
6. � ln udu = u ln|u|- u + C
Trigonometric Functions
7. � sin u du = - cos u + C
8. � cos u du = sin u + C
9. � tan u du = ln|sec u| + C
10. � cot u du = ln|sin u| + C
11. � sec u du = ln|sec u + tan u| + C
12. � csc u du = ln|csc u- cot u| + C
= - ln|csc u+ cot u| + C
13. � sec u tan u du = sec u + C
14. � csc u cot u du = - csc u + C
15. � sec2 u du = tan u + C
16. � csc2 udu = - cot u + C
Inverse Trigonometric Functions
du
u
= arcsin + C
a
√a2 -u2
du
1
u
18. � 2 2 = arctan + C
a
a
a +u
du
1
u
19. �
= arcsec + C
2
2
a
a
u√u -a
17. �
20. � arcsin u du = u arcsin u + �1-u2 + C
21. � arctan u du=u arctan u- ln �1+u2
Hyperbolic Functions
22.
23.
24.
25.
26.
� sinh u du = cosh u + C
� cosh u du = sinh u + C
� tanh u du = ln cosh u + C
� coth u du = ln |sinh u| + C
� sech2 u du = tanh u + C
28.
� csch2 u du = -coth u + C
29.
� csch u coth u du = -csch u + C
27.
30.
31.
32.
33.
� sech u tanh u du = -sech u + C
u
du
= sinh-1 + C
z
√u2 +a2
du
-1 u
�
= cosh
+ C,
u>a>0
a
√u2 -a2
u
du
1
� 2 2 = tanh-1 + C, u2 < a2
a
a
a +u
1
du
-1 u
+ C, x2 > a2
� 2 2 = - coth
a
a
a -u
�–
Other functions
du
= ln �u+ �u2 ±a2 � + C
√u2 ±a2
1
a+u
du
ln �
� + C,
u2 < a 2
� 2 2 =
2a
a-u
a -u
1
u–a
du
ln �
� + C, u2 < a2
� 2 2 =
2a u + a
u -a
u
a2
u
� �a2 -u2 du = �a2 -u2 +
arcsin + C
2
a
a
2
u+
u
a
ln � 2 2 � + C
� �u2 ±a2 du = �u2 ±a2 +
�u +a
2
2
34. � -
35.
36.
37.
38.
39. � udv = uv- � vdu
Trigonometric Substitution
Some integrations may be simplified with the
following substitutions:
1.
If an integrand contain �a-x2 ,
Substitute x2 = asin2 θ
2.
If an integrand contain �a+x2 ,
Substitute x2 = atan2 θ
3.
If an integrand contain �x2 -a,
Substitute x2 = asec2 θ
More generally, an integrand that contains one of
the forms �a-bx2 , √a+bx2 or �bx2 -a but no
other irrational factor may be transform into
another involving trigonometric functions of a new
variable as follows.
Use
�a-bx2
�a+bx2
�bx2 -a
a 2
sin θ
b
a
x2 = tan2 θ
b
a
2
x = sec2 θ
b
x2 =
To obtain
2
√a�1-sin θ = √a cos θ
√1�1+tan2 θ = √a sec θ
√a�sec2 θ-1 = √a tan θ
Wallis Formula
π/2
� sinm θcosn θdθ=
0
[(m-1)(m-3)…(2 or 1)][(n-1)(n-3)…(2 or 1)]
(m+n)(m+n-2)…(2 or 1)
Where:
α = π/2 when both m and n are even
α = 1 if otherwise
m & n = positive integer, not equal to 1
Examples
Wallis Formula
π/2
�
Ex.7-1:
sin5 xdx =
0
Double Integration
1
Ex.7-2:
3x
(4)(2)
8
×1 =
(5)(3)(1)
15
1 4
� � y3 dydx = �
0
0
0
1 (3x)4
=�
4
0
=
y 3x
� dx
4 0
1
dx = �
0
81 4
x dx
4
81
81x5 1
� =
4(5) 0
20
Triple Integration
3
Ex. 7-3:
2
y
3
2
0
0
0
0
3
0
3 2
2
= � � ydydx = �
0
3
0
0
3
= � 2dx = 2x| = 6
0
Integration By Parts
Ex.7-4:
y
� � � dxdydz = � � [x] dydx
� xsin xdx
Let u = x, du = dx
0
0
y 2
� dx
2 0
dv = sinxdx,v = -cos x
� udv = uv- � vdu
� xsin xdx = x(-cos x) - cos xdx
= -xcos x + sin x + C
Algebraic Substitution
1
xdx
Ex. 7-5:
�
8
0 (x+1)
Let z=x+1
dx=dz
x=z-1
Change the limits
When x = 0; z = 1
When x = 1; z = 2
2 (z-1)dz
�
1
= �
z8
2
= � �z-7 -z-8 � dz
1
1 -6 1 -7 2
z - z � =0.0223
-7
1
-6
Trigonometric Substitution
Ex.7-6 :
Find � -
let x2 = 4tan2 θ
x = 2tanθ
dx=2sec2 θdθ
�
dx
x2 √4+x2
= �
dx
x2 √4+x2
2sec2 θdθ
�4tan2 θ��4+4tan2 θ
1 sec θ
= �
dθ
4 tan2 θ
1
1
+C
= � sin-2 θcosθdθ= 4sinθ
4
From the right triangle:
c = �22 +x2
x
x
sinθ = =
c √4+x2
�
dx
x2 √4+x2
= -
c
x
θ
1
1
+C = x
4sin θ
4
√4+x2
2
�
dx
x2 √4+x2
=-
√4+x2
+C
4x
Plane Areas
Using Vertical Strip
y = f(x)
y = g(x)
Yu
yL
x2
A = � �yu - yL �dx
x1
yU = f(x) ; yL = g(x)
Using Horizontal Strip
y = g(x)
XL
XU
y = f(x)
y2
A = � (xR - xL )dy
y1
xR = g(y); xL = f(y)
By Polar Coordinates
θ2
dθ
r
θ1
θ2
1
A=
� r2 dθ
2
θ1
Length of arc
2
ds
1
dy
dx
(ds)2 = (dx)2 +(dy)2
x2
s = � �1 + �
x1
y2
dx 2
dx 2
� dx or s = � �1 + � � dy
dy
dy
y
1
Centroid of Plane Areas
x
dA
xG
cg
xC
yc
yG
y
Using Horizontal Strip
x2
A XG = � xc dA
x1
y2
A YG = � ydA
y1
Using Vertical Strip
x2
A XG = � xdA
x1
y2
A YG = � yc dA
y1
Centroid of Parabolic Segment and Spandrel
x1
y1
Parabolic
Segment
h
hbx
Spandrel
Spandrel
y
b
3
b
8
1
x1 = b
4
2
Aparabola = bh
3
2
h
5
3
y1 =
h
10
x=
y =
Aspandrel =
Moment of Inertia for Plane Areas
First Moment of Area
x
dA
dA
y
1
bh
3
y2
Ix = � y2 dA
y1
x2
Iy = � x2 dA
x1
Polar Moment of Inertia
J = Ix +Iy
Product of Inertia
Ixy = � xydx
Mass Moment of Inertia
Problems concerned with the rotation of solid
bodies involve the mathematical expression
∫ ρ2
dW
g
, which is known as the moment inertia of
thr body, also called the mass moment of inertia
because the ratio W/g is widely known as the
mass of the body.
Solid Right Circular Cylinder
I=
1
2
Mr2
M = mass; r = radius
Hallow Right Circular Cylinder
I=
1
2
M(R2 - r2 )
R = outer radius
r = inner radius
Uniform Slender Rod
1
Iy = M L2
Iy' =
3
1
12
ML
y
2
Solid Sphere
2
I = M r2
5
Spherical Shell
I=
2
3
M r2
r = mean radius
L/2
y’
L/2
Right Circular Cone
I=
3
10
M r2
r = base radius
Right Elliptical Cylinder
I=
1
4
M�a2 + b2 �
b
a
Properties of Common Shapes
Triangle
y
xc =
a
h
xc
cg
yc
b/2
b/2
a+b
3
yc =
h
3
1
Area= b h
2
bh3
bh3
Ix =
Igx =
12
36
Rectangle
Area = bd
bd3
db3
Iy =
Ix =
3
3
y
d
cg
Igx =
x
bd3
12
Igy =
db3
12
b
Circle
y
1 2
πD
4
4
πr
πD4
=
Igx = Igy =
4
64
Area = πr2 =
r
cg
x
D
Quarter Circle
1 2
πr
4
4r
xc = yc =
3π
πr4
Ix = Iy =
16
Igx = Igy = 0.055r4
Area =
y
r
xc
cg
yc
r
x
Semi - Circle
1 2
πr
2
4r
yc =
3π
πr4
Ix = Igy =
8
Igx = 0.11r4
Area =
y
cg
yc
x
r
r
Ellipse
y
a
b
cg
b
a
x
Area = πab
πab3
Igx =
4
πba3
Igy =
4
Half Ellipse
1
πab
2
4b
yc =
3π
Area =
y
b
cg
a
yc
x
a
Ix =
πab3
8
Igx = 0.11ab3
Igy =
Quarter Ellipse
πba3
8
1
πab
4
4a
4b
xc =
yc =
3π
3π
πab3
πba3
Iy =
Ix =
16
16
Igx = 0.055ab3
Igy = 0.055ba3
Area =
y
b xc cg
yc
a
x
Sector of a Circle
1 2
r (2θ) = r2 θ
2
2 rsin θ
xc =
3 θ
r4
1
�θ - sin2θ�
Ix =
4
2
r4
1
Iy =
�θ + sin2θ�
4
2
Area =
y
r
θ
θ
r
xc
x
Parabolic Segment
y
Area =
x = ky2
3
b
8
2
2
hb3 Iy = bh3
Ix =
15
7
xc =
cg
yc
xc
2
h
5
2
bh
3
yc =
x
h
Spandrel
1
bh
n+1
1
b
xc =
n+2
n+1
yc =
h
4n + 2
Area =
y
y = kxn
h
xc
yc
x
b
Segment of an Arc
length of arc = 2rθ
y
r
θ
θ
xc
xc =
cg
x
rsinθ
θ
When
θ = 90°(semicircle)
2r
xc =
π
Solids of Revolution
Volume
xR
xL
yL
yR
Using Circular Disk
y2
V = π � �xR 2 -xL 2 � dy
y1
x2
or V = π � �yR 2 -yL 2 � dx
x1
Using Hollow Cylindrical Shell
dx
xR – xL
xL
yu – yL
y
dy
y2
V = � 2πy(xR - xL ) dy
y1
x2
or V = � 2πx�yR - yL � dx
x1
Surface Area
r
ds
d2
As = � 2πrds
d1
ds = �1 + (dx/dy)2
or
ds = �1 + (dy/dx)2
Volume of other Solids known Cross- Sections
y2
V = � Ady
y1
Centroid of Volume
y2
VYG = � ydV
y1
Where
V = Total volume of the body
dV = Ady
A = f(y)
Work
Constant Force
The work done by a constant force F acting over a
directed distance s along a straight line is:
Work = Force × Distance = F × s
Work under Variable Force
x2
Work = � (dF)(x)
x1
x2
or
Work= � F(x)dx
x1
Work required to stretch a spring
The work done in stretching a spring of natural
length L from x1 to x2 is:
Work =
1
k�x2 2 - x1 2 �
2
Where, k = spring constant or stiffness in N/m
Work in winding up a load
If a load P is to be wind up by a flexible rope or
cable having a uniform mass and length L, the
work done is:
Work = PL + W(L/2)
W = weight of the rope or cable
P = load at the end of the rope
or cable
CHAPTER 8 – DIFFERENTIAL EQUATIONS
Differential equations are an equations that
contain differential coefficients. Example "dy" ⁄"dx"
" = 12x" and "y\" + 2xy' – y = 0" .
Differential equations are classified according to
the highest derivative that occurs in them. The
differential equation "dy" ⁄"dx" " = 12x" is a first
order differential equation and the equation ("d"
^"2" "y" )⁄("d" ^"2" "x" ) " + " "4dy" ⁄"dx " "- 3y =
0" is a second order differential equation.
A solution to a differential equation that contains
one or more arbitrary constants of integration is
called general solution. When additional
information is given so that these constants may
be calculated the particular solution of differential
equation is obtained. The additional information is
called boundary condition.
Variable Separable
A.
dy
Differential equations of type �dx = f(x)
dy
Differential equations of type �dx = f(x) can
be solved direct integration by writing it in the
form
dy = f(x) dx
Example 8-1
Solve the differential equation dy⁄dx = 2x + sin 3x
Solution
dy = (2x + 3 sin x) dx
y = x2 – (1/3) cos 3x + C (general solution)
dy�
dx = f(y)
B. Differential equations of type
dy�
dx = f(y) can
be solved direct integration by writing it in the
form
Differential equations of type
dx =
dy
f(y)
Example 8-2
Solve the equation (y2 – 1) dy/dx = 3y,
given that y = 1 when x = 13/6.
Solution
Rearrange ∫ dx = ∫
x=
1
6
When x =
13
16
y2 - 1
3y
1
y
dy = ∫ � 3
1
3y
� dy
y2 - ln y + C (general solution)
3
13
, y = 1 then:
16
=
x=
(1)2
6
1
6
1
- ln(1) + C ;
3
1
C=2
y2 - ln y + 2 (particular solution)
3
C. Differential equation of type
dy�
dx = f(x) g(y)
dy
Differential equations of type �dx = f(x) g(y)
can be solved direct integration by writing it in
the form
dy
= f(x)dx
g(y)
D.
Differential equations of type dQ�dt = kQ
The general solution of an equation of the
form dQ�dt = kQ is
Q = C ekt
Where C is constant
Example 8-3
Solve the equation dy⁄dx = 3y
Solution
Q=y
t=x
then y = Ce3x
dQ = dy,
k=3
Homogeneous First Order Differential Equation
Some first order differential equation are not of the
variable-separable type but can be made
separable by changing the variable. Differential
equations in the form of P dy/dx = Q, where P and
Q are functions of both x and y of the same
degree throughout, is said to be homogeneous in
x and y.
To solve these types of equation, the following
procedures may be done:
1. Rearrange the equation into the form
dy
Q
=
dx
P
2. Substitute y = vx from where
dy = vdx + xdv
3. Substitute both y and dy to the original
equation and separable variables.
4. Solve using the previous methods and
substitute v = y / x to solve in terms of
the original variables.
Example 8-4
Solve the equation; �x2 – xy + y2 �dx - xy dy = 0
Solution
The coefficients are of the second degree.
Let y = vx
dy = v dx + x dv
then �x2 - x2 v + v2 x2 �dx - x2 v (v dx + x dv) = 0
Removing x2
(1 – v + v2) dx – v (v dx + x dv) = 0
dx – v dx + v2dx – v2dx – vx dv = 0
(1 –v) dx – vx dv = 0
Separating variables
dx
x
Or
+
dx
x
vdv
v-1
=0
+ �1 +
1
v-1
� dv = 0
Integrating:
ln x + v + ln(v - 1) = ln C
C
=v
x(v - 1)
c
= ev
x(v - 1)
ln
Finally, substituting v = y / x
y
y
x � - 1� ey/x = C or (y - x)ex = C
x
Linear First Order Differential Equation
The equation in the form dy / dx + Py = Q where P
and Q are functions of x only is called linear
differential equation since y and its derivatives are
of the first degree.
The solution for dy / dx + Py = Q is obtained by
multiplying throughout by integrating factor e∫ Pdx
to become
y e∫ Pdx = e∫ Pdx Q dx
Example 8-5
Solve the equation dy + 4xy dx = 2x dx
Solution
Rearranging:
dy / dx + 4xy = 2x
P = 4x and Q = 2x
Integrating factor = e∫ Pdx
= e∫ 4xdx = e2x
Then from y e∫ Pdx = e∫ Pdx Q dx
2
2
y e2x = e2x 2x dx
Integrate the right side to get
2
2
y e2x =
1 2x
e +C
2
Linear Second Order Differential Equation
Equation is in the function a(d2y / dx2) + b(dy / dx)
+ cy = 0 where a, b, and c are constants, is called
a linear second differential equation with constant
coefficients.
Setting D = d / dx and D2 = d2 / dx2, the following
procedures may be followed:
1. Write the equation in D-operation form
(aD2 + bD + c)y = 0, substitute m for D and
solve the auxiliary equation am2 + bm + c =
0 for m.
A. If the roots are real and different
�b2 > 4ac� say m = α and m = β, then
the general solution is
y = A eαx + B eβx
Where A and B are constants.
B.
C.
If the roots are real and equal (b2 =
4ac), say m = α twice, the general
solution is
y = (Ax+B)eαx
If the roots are imaginary �b2 < 4ac�,
say m = α ± βi, the general solution is
y = eαx [ C cos βx + Dsinβx ]
Example 8-6
Solve the equation 2 �
d2 y
dy
dx2
dx
� + 5 � � - 3y = 0
Solution
Writing in D-operator form:
�2D2 + 5D - 3�y = 0
Substituting m for D gives the auxiliary
equation 2m2 + 5m – 3 = 0 which can be
factored as (2m -1)(m +3) = 0, and the
roots are m = 1�2 and m = -3.
Since the roots are real and different the
general solution is y = A eαx + B eβx with α
= 1�2 and β = -3 then the general solution
is y = A ex⁄2 + B e-3x
Exact Differential Equation
Differential equations of form
M(x, y)dx + N(x, y)dy = 0 are said to be exact if
∂M ∂N
=
∂y ∂x
Example 8-7
Solve the equation (2x + 4y + 6)dx +
�4x - 2y - 5�dy = 0
Solution
Check for exactness
M = 2x + 4y +6
∂M
Since
∂M
∂y
=
=4
∂y
∂N
∂x
N = 4x -2y – 5
, the equation is exact
∂N
∂x
=4
(2x + 4y + 6)dx + (4x – 2y – 5) dy = 0
2x dx + 4y dy + 6 dx + 4x dy – 2y dy – 5 dy = 0
2x dx – 2y dy + 6 dx – 5 dy + 4(y dx + x dy) = 0
Note: y dx + x dy = d(xy)
Integrate:
x2 - y2 + 6 – 5 + 4 xy = C
Bernoulli’s Equation Type
Equation of type
dy
dx
+ P(x) y = Q(x) y" has a
general solution of
ve(1 - n) ∫ pdx = (1 - n) � Q e(1 - n) ∫ pdx dx
Where v = y1 – n
If n = 1, the solution is:
ln y = �(Q - P) dx + C
Finding the Differential Equation from a
General Solution
To find the differential equation when the general
solution is given, differentiate the general solution,
differentiate the derived solution, differentiate the
second derived solution, etc.
Example 8-8
Find the differential equation of x2 + y2 = cx.
Solution
Isolate c: c =
x2 + y2
x
= x+
y2
x
Differentiate with respect to x:
x �2yy' � - y2
0=1+
x2
0 = x2 + 2yy’ – y2;
2xyy’ = y2 – x2
Basic Law of Natural Numbers
Let a, b, and c be any number.
1. Law of closure for addition
a+b
2. Commutative law for addition
a+b=b+a
3. Associative law for addition
a + (b + c) = (a + b) + c
4. Law of closure for multiplication
a×b
5. Commutative law for multiplication
a×b=b×a
6. Associative law for multiplication
a(bc) = (ab)c
7. Distributive law
a( b + c ) = ab + ac
Basic Law of Equality
1. Reflexive property
a=a
2. Symmetric property
If a = b, then b = a
3. Transitive property
If a = b and b = c, then a = c. that is, things
equal to the same thing are equal to each
other.
4. If a = b and c = d, then a + c = b + d. That is, if
equals are added to equals, the results are
equal.
Some Applications of Differential Equation
Population Growth
The rate of population growth is proportional to the
present population P.
dP
=kP
dt
P = Po ekt
Where Po is the population at time t = 0, k is
constant.
Exponential Growth and Decay
dQ
=kQ
dt
Q = Qo ekt
Cooling and Heating Problems
Newton’s Law of Cooling: The surface
temperature of a cooling body changes at the rate
proportional to the difference between the surface
and ambient temperatures.
dT
= k (T - ts )
dt
T = ts + (To - ts )e-kt
Where T = temperature of the body at any time
Ts = ambient temperature
Flow Problems
Qin
h1
h2
Qout
dQ
= Rate of inflow - Rate of outflow
dt
t2
� dt = �
t1
h2
h1
dQ
Qin - Qout
Where Q = Concentration or volume of liquid in the
tank at any time
Continuous Compound Interest
dP
=rP
dt
P = Po ert
Motion Problems
Velocity, v =
ds
dt
Acceleration, a =
v dv = a ds
dv
dt
Newton’s Second Law of Motion
F=M
dv
dt
CHAPTER 9 – ENGINEERING MECHANICS
(PHYSICS)
Engineering Mechanics is the science which
considers the effects of forces on rigid bodies. The
subject is divided into two parts: statics and
dynamics. In statics we consider the effects and
distribution of forces on rigid bodies which are and
remain at rest. In dynamics we consider the motion
of rigid bodies caused by the forces acting upon
them.
Outline of Engineering Mechanics
ENGINEERING MECHANICS
STATICS
Force
System
DYNAMICS
Applications
Kinematics
Kinetics
Concurrent
Trusses
Translation
Translation
Parallel
Centroids
Rotation
Rotation
NonConcurrent
Friction
Plane
Motion
Plane
Motion
Statics
Statics is a branch of mechanics which studies the
effects and distribution of forces of rigid bodies
which are and remain at rest.
Force System
A force system is any arrangement where two or
more forces act on a body or on a group of related
bodies.
Resultant of Two Concurrent Coplanar Forces
Parallelogram Law
F2
Resultant, R
θ
F2
α
F1
R2 = F1 2 + F2 2 - 2F1 F2 cos θ
F2
R
=
sin α sin θ
Resultant of Two or More Concurrent Coplanar
Forces
Resultant, R
F2
θ
F1
Analytical Solution
Rx = � Fx = F1 x + F2 x + …
Ry = � Fy = F1 y + F2 y + …
R = �Rx 2 + Ry 2 ; tan θx =
Ry
Rx
Resultant of Concurrent Forces in Space
z
F2
F1
F2z
F1z
θz
F2y
F1x
F2x
F1y
F3y
θy
θx
F3x
R
F3z
x
F3
y
Rx = � Fx = F1 x + F2 x + F3 x +…
Ry = � Fy = F1 y + F2 y + F3 y + …
Rz = � Fz = F1 z + F2 z + F3 z + …
cos θx =
Rx
R
R = �Rx 2 + Ry 2 + Rz 2
cos θy =
Ry
R
cos θz =
Rz
R
Example 9-1
Given the three concurrent forces which pass
through (1, -3, 4) and the indicated points:
F1 = 150 N (5, -6, 2)
F2 = 340 N (4, 0, -3)
F3 = 280 N (-1, 2,6)
Determine the magnitude of the resultant force.
Solution
2
2
2
F1 = 150 N, L1 = �(5 - 1) + (-6 + 3) + (2 - 4) = 5.385
2
2
F2 = 340 N, L2 = �(4 - 1) + (0 + 3)2 + (-3 - 4) = 8.185
2
2
F3 = 280 N, L3 = �(-1 - 1) + (2 + 3)2 + (6 - 4) = 5.745
Rx = � Fx = F1 x + F2 x + F3 x
Rx = 150
5-1
5.835
+ 340
4-1
8.185
+ 280
Ry = � Fy = F1 y + F2 y + F3 y
Ry = 150
-1 - 1
5.745
= 138.56 N
-6 + 3
0+3
2+3
+ 340
+ 280
= 284.76 N
5.835
8.185
5.745
Rz = � Fz = F1 z + F2 z + F3 z
Rz = 150
2-4
-3 - 4
6-4
+ 340
+ 280
= -248.99 N
5.835
8.185
5.745
R = �Rx 2 + Ry 2 + Rz 2 = 402.84 N
Resultant of Parallel Forces
x4
x3
x2
x1
F2
F1
F3
F4
𝑥𝑥̅
R = � F = F1 + F2 + F3 + …
R x� = � F x = F1 x1 + F2 x2 + F3 x3 + …
Resultant of Non-Concurrent Forces
P1
P2
P5
O
d
P4
R
Rx = ∑ Px
P3
Ry = ∑ Py
R = �Rx 2 + Ry 2
Equilibrium of Forces
General Conditions of Equilibrium
∑ Fx = 0
� Mo = 0
∑ Fy = 0
Equilibrium of Two Forces
Two forces are in equilibrium
if they are equal, collinear,
and oppositely directed.
Equilibrium of Two or More Concurrent Forces
If two or more coplanar forces are in equilibrium,
they must form a closed polygon.
P
Q
Q
R
P
R
Force Polygon
R
P
Q
S
Q
R
T
P
S
T
Equilibrium of Three Coplanar Forces
If a flexible wire cable is suspended over two
supports and where most of the loads are
distributed horizontally as in a suspension bridge,
the cable assumes the shape of a parabolic
segment.
L
x1
x2
B T2
T1 A
y2
y1
C
w (N/m)
x1 2 x2 2
=
y1
y2
The solution is to cut a segment from the support to
the lowest point of the cable.
Cutting segment AC:
x
x/2
x/2
T
A
T
W
W
y
θ
H
θ
H
C
w (N/m)
W=wx
�� MA = 0�
H×y=W×
x
2
From the force polygon;
T2 = W2 + H2
W
y
tan θ =
=
H
x⁄2
T sin θ = W
Symmetrical Cable
L
L/2
L/2
B T
T A
d
C
w (N/m)
Tension at C = H =
w L2
8d
Tension at A = T = �H2 + �
wL 2
�
2
Length of Cable S:
Exact S =
L2
[ m k + ln( m + k ) ]
8d
4d
; k = �1 + m 2
L
8d2 32d4
Approximate S = L +
3L
5L3
m=
Catenary
When the cable sags due to its own weight, the
cable assumes the shape of a catenary which is the
graph of the equation y = cosh x.
T2
T1
S2
S1
y2
y1
w (N/m)
y
c
x
x1
x2
O
Tension at lowest point, H = w × c
T1 = w y1
T2 = w y2
T1 = �H2 + (w S1 )2
y1 2 = S1 2 + c2
x1 = c ln
S1 + y1
c
T2 = �H2 + (w S2 )2
y2 2 = S2 2 + c2
x2 = c ln
S2 + y2
c
Friction
Friction
is
the
contact
resistance by one body when
the second body moves or
tends to move past the first
body. Friction force always
acts opposite to the motion or
to the tendency to move.
W
F
f=μN
ø
N
R
Elements
N = total normal reaction
The sum of all forces perpendicular to the
surface
f = friction force (maximum available friction)
μ = coefficient of friction = tan Φ
R = total surface reaction
The resultant of f and R
Φ = angle of friction = arctan μ
tan Φ =
f
=μ
N
The maximum angle that a plane may be inclined
without causing the body to slide down is:
W
θ
θ = Φ = arctan μ
Belt Friction
β
T1
T2
T1
= ef β
T2
ln
T1
=fβ
T2
Where:
f = coefficient of friction
β = angle of contact in radius
e = 2.71828
T1 = tension in the tight side
T2 = tension in the slack side
Properties of Sections
*see Chapter 7 for properties of common geometric
shapes
Centroid of Plane Area
x3
XG
x1
1
G
3
y1
2
y2
YG
O
x2
A XG = � a x = a1 x1 + a2 x2 + …
A YG = � a y = a1 y1 + a2 y2 + …
y3
Center of Gravity of Flat Plates
z
y
w1
W
w2
XG
w3
cg
O
YG
W XG = � w x = w1 x1 + w2 x2 + …
W YG = � w y = w1 y1 + wy2 + …
Centroid of Composite Figures
z
y
w1
W
w2
XG
cg
O
YG
x
W XG = � w x = w1 x1 - w2 x2
W YG = � w y = w1 y1 - wy2
Moment of Inertia (first moment of area)
dA
x
r
y
O
Ix = ∫ y2 dA
Iy = ∫ x2 dA
Polar Moment of Inertia
*referring to the figure above:
J = ∫ r2 dA = Ix + Iy
Radius of Gyration
x
I
rx = � x
A
I
ry = � y
A
Transfer Formula for Moment of Inertia
A
XO
cg
d
X
Ix = ��Ig + Ad2 �
Where Ig = centroidal moment of inertia
Moment of Inertia with Respect to an Inclined
Axis
In some cases, it is necessary to determine the
moment of inertia with respect to axes, which are
inclined to the usual axes. The moment of inertia of
such cases can be found by the use of a formula
and more conveniently by graphical solution using
Mohr’s Circle especially if the principal moments of
inertia are known.
y
U
V
θ
O
θ
x
Mohr’s Circle for Moment of Inertia
The following procedures apply to moments of
inertia using Mohr’s Circle:
1. On a set of rectangular coordinate axes
choose one axis on which to plot values
of Ix and Iy and the other on which to plot
Ixy. These axes area called principal axes.
2. With Ix, Iy, and Ixy known plot (Ix, Ixy) and
(Iy, Ixy) with Ix and Iy along the horizontal
and Ixy along the vertical.
3. Join the points plotted with a straight line.
This line is the diameter of the Mohr’s
Circle having its center on the principal
axis.
4.
The angle between any two radii on the
Mohr’s Circle is double the actual angle
between the usual axes. The rotational
sense of this angle corresponds to the
rotational sense of the actual angle
between the axes.
Ix
Imin
R
C
Iyx
Ixy
2θ
Ix - C
R
Iy
Imax
The following relationships can be derived from the
above procedures:
Iyx = Ixy
Ix + Iy Imax + Imin
C=
=
2
2
R = Imax – C
2
R = ��Ixy � + (Ix - C)2
Imax = C + R
Imin = C - R
Dynamics
The branch of mechanics that deals with bodies in
motion
BRANCHES OF DYNAMICS
Kinematics
The geometry of motion. This term is used to
define the motion of a particle or body without
consideration of the forces causing the motion.
Kinetics
The branch of mechanics that relates the force
acting on the body to its mass and
acceleration
KINEMATICS
Motion of Particles
I. Translation
The motion of a rigid body in which a straight line
passing through any two of its particle always
remain to be parallel to its initial position
II. Rotation
The motion of a rigid body in which the particles
move in circular paths with their centers on a fixed
straight line called the axis of rotation.
III. Plane Motion
The motion of a rigid body in which all particles in
the body remain at a constant distance from a fied
reference plane
Notations
s = distance
x = horizontal displacement
y = vertical displacement
v = velocity at any time (final velocity)
vo = initial velocity (velocity at time = 0)
a = acceleration
g = acceleration due to gravity (9.81 m/s2, 32.2
ft/s2)
t = time
TRANSLATION
Rectilinear Translation
Motion along a straight line.
A. Uniform Motion (constant velocity)
s=vt
B. Variable acceleration
a=
dv
v=
dt
ds
dt
v dv = a ds
Where a may either be a function of v, t, or s, and v
may either be a function of d or s.
C.
Constant acceleration
v = vo + at
s = vo t +
1 2
at
2
v2 = vo 2 + 2as
D.
Free-falling body (vo = 0, s = h)
v = gt
s=
1 2
gt
2
v2 = 2gh
Curvilinear Translation (Projectile Motion)
Resolve the motion into x and y-components and
use the formulas in rectilinear translation.
vy = 0, v = vx
A
B
vx
vy
vo
H
v
y
voy
θ
O
vy = -voy
vox
x
C
R
D
At any point B:
x-component of motion (ax = 0)
vx = vox
x = voxt
y-component of motion (ay = -g)
vy = voy - gt
1
y = voy t - gt2
2
vy 2 = voy 2 - 2gy
y = x tan θ -
gx2
2vo 2 cos2 θ
At the summit A: (vy = 0)
voy 2
2g
voy
t=
g
H=
At point C:
y=0
vy = -voy
v = vo
2
vo sin 2θ
R=
g
2vo sin θ
2voy
t=
=
g
g
Note: θ is positive if the projectile directed upward
and negative if directed downward
•
•
At any point D below the origin O, the sign of y
is negative.
vy is positive if directed upward and negative if
directed downward
Motion Curves
Motion curves are plot of distance s, velocity v, and
acceleration a versus time.
a
zero acceleration
a-t
diagram
t
A1
zero slope
v
v-t
diagram
zero velocity
A1
A2
t
s
zero slope
s-t
diagram
A2
t
The relationships between these curves are:
v=
ds
dt
Which means that the velocity is the slope of the s-t
diagram and,
a=
dv
dt
which means that the acceleration is the slope of
the v-t diagram
Note: The relationship between these curves is the
same as the load, shear, and moment diagrams.
The velocity and distance may be computed from
the a-t diagram as follows:
Pnet
Pnet
Net force diagram
xG
a = Pnet/M
t
Time where the v
and s are required
Area
cg
t
Acceleration diagram
v = vo + area
S = vo t1 + area × xG
ROTATION
Notation
θ = angular displacement, radians
ω = angular speed, rad/sec
α = angular acceleration, rad/sec2
t = time
ω
s
r
r
Uniform Motion
θ=ωt
θ
Uniform Acceleration
ω = ωo + α t
1
θ = ωo t +
α t2
2
ω2 = ωo 2 + 2α θ
Relationship between Translation and Rotation
ω
r
a
θ
r
θ
r
v
r
s
s=rθ
v=rω
a=rα
KINETICS
Newton’s Laws of Motion
1. A body at rest will remain at rest or in motion
will remain in motion along a straight path
unless acted upon by an unbalanced force.
2. A particle acted upon by an unbalanced force
system has an acceleration in line with and
directly proportional to the resultant of the
force system and inversely proportional to its
mass.
a∝
3.
F
F
or a = k
(where k = 1)
M
M
F=m×a
In every action there is always an equal and
opposite reaction.
D’Alembert’s Principle
W
The resultant of the external
forces applied to a body (rigid or
non-rigid) composed of a system
of particles is equivalent to the
vector summation of the effective
forces acting on all particles.
a
REF
P
f
N
REF = P - f
W
REF = M a =
a
g
Where M is the mass and W is the weight
Problems Involving D’Alembert’s Principle
Friction pulley
w1 + w2
a = w2 - w1
g
a
a
W1
W2
w1 < w2
y
y
W 1x
x
x
a
W1
f
W2
a
w1 + w2
��������������⃗ - Forces
⃖�������������� = W2 – f - W1x
a = Forces
g
Centrifugal Force (Reverse Normal Effective
Force)
Whenever a body rotates about its axis at a speed
ω, there exists a force called centrifugal inertia
force directed away from the axis of rotation.
ω
W
r
an
T
CF
Normal acceleration, an = ω2 r = v2 / r
CF = M an = M ω2 r
W 2
W v2
ω r=
CF =
gr
g
The tangential inertia force (centripetal force), also
known as reversed tangential effective force is
given by the formula:
T = M at =
W
rα
g
Conical Pendulum
ω
T
W
θ
L
h
θ
CF
T
r
W
v
CF
CF ω2 r
v2
=
=
W
g
gr
g
g
cos θ = 2 for ω > �
L
ω L
W
T=
cos θ
tan θ =
Time to complete one revolution, t = 2π�
h
g
As the value of θ decreases, the value of h
approaches the limiting value L so that the
maximum time for a revolution is:
t max = 2π�
L
g
Rod uniform mass of length L rotated about one
end:
ω
cos θ =
θ
L
2g
for ω > �2g/L
ω2 L
W
Banking of Curves
The maximum speed v that a car can round a
highway curve without skidding is given by the
relationship:
tan (θ + Φ) =
The ratio
v2
gR
v2
gR
is also known as the impact factor or
centrifugal ration.
Horizontal Rotating Platform
The maximum speed that the platform may be
rotated so that the block will not slide is given by the
formula:
tan Φ = μ =
ω2 R v2
g gR
WORK AND ENERGY
Work-energy method is used particularly for solving
problems involving force, displacement, and
velocity.
Work = Force × distance
1
1 W 2
Kinetic energy, KE = Mv2 =
v
2
2 g
Potential energy, PE = Mgh = W h
Work-energy Equation for Constant Forces
If the forces acting on a body are constant
KEA ± Work ± Wh = KEB
IMPULSE AND MOMENTUM
Problems involving force, velocity, and time are
conveniently solved by means of the impulsemomentum method.
Impulse = Force × time
W
Momentum = Mv =
v
g
Impulse-momentum Equation
When a body of weight W moving with an initial
velocity vo changes its velocity to vf over a period of
t along a straight line,
�(+) Impulse - �(-) Impulse =
W
(vf - vo )
g
(+) Impulse = impulse in the same direction with the
motion
(-) Impulse = impulse in the opposite direction with
the motion
Law of Conservation of Momentum
If a system is composed of particles of mass M1,
M2, etc., having velocities v1, v2, etc., and after
mutual reaction between the particles they possess
new velocities v’1, v’2, etc., the condition that the
momentum of the system be constant may be
expressed as:
V1
M1
V2
V’2
V’1
M2
M1
Before impact
M2
After impact
+→ M1v1 + M2v2 = M1v’1 + M2v’2
Coefficient of Restitution
The coefficient of restitution is defined as the ratio
of the relative velocities of colliding bodies after
impact to their relative velocities before impact. e is
always positive.
Coefficient of restitution, e =
e=
Relative velocity after impact
Relative velocity before impact
v'2 →v'1 vseparation
=
v1 →v2
vapproach
Note: → denotes vectorial subtraction
Where 0 > e < 1 for elastic or inelastic collision
e = 0 for perfectly inelastic collision
e = 1 for perfectly elastic collision
If a ball is dropped from a height h upon a
pavement and rebounds to a height of h’, the
coefficient of restitution between the ball and the
pavement is:
e=�
h'
h
If a ball is thrown at an angle θ with the normal to a
smooth surface and rebounds at an angle θ’,
e=
tan θ
tan θ '
If two masses M1 and M2 moving opposite
directions with velocities v1 and v2 collide, the
energy loss in direct central impact is:
Loss in KE =
1 M1 M2
�1 - e2 � (v1 - v2 )2
2 M1 + M2
CHAPTER 10 – STRENGTH OF MATERIALS
SIMPLE STRESS
Normal Stress
Stress is defined as the strength of a material per
unit area or unit strength. It is the force on a
member divided by the area, which carries the
force, formerly expressed as psi, now in N/mm2 or
MPa.
σ=
P
A
Where P is the applied normal load in Newton and
A is the area in mm2. The maximum stress in
tension or compression occurs over a section
normal to the load.
P
P
Bar in Compression
Bar in Tension
P
P
Shearing Stress
Forces parallel to the area resisting the force
cause shearing stress. It differs to tensile and
compressive stresses, which are caused by forces
perpendicular to the area on which they act.
Shearing stress is known as tangential stress.
τ=
V
A
Where V is the resultant shearing force which
passes through the centroid of the area A being
sheared.
Single Shear
P
P
Double Shear
P
P
P
Bearing Stress
Bearing stress is the contact pressure between
separate bodies. It differs from compressive
stress, as it is internal
compressive forces.
stress
caused
Pb
by
Pb
Ab
σb =
Pb
Ab
THIN-WALLED PRESSURE VESSELS
Thin-Walled Cylindrical Vessel
A tank or pipe carrying a fluid or gas under a
pressure is subjected to tensile forces, which
resist bursting, developed across longitudinal and
transverse sections.
ST
pressure
p
SL
SL
ST
pD
2t
pD
Longitudinal stress. SL =
4t
p = internal pressure - external pressure
Tangential stress, ST =
Spherical Shell
If a spherical tank of diameter D and thickness t
contains a gas under a pressure of p, the stress at
the wall can be expressed as:
Wall stress, S =
pD
4t
THICK-WALLED CYLINDER
In thin-walled cylinders,
the thickness of the
wall is very small
compared to the tank
diameter. If in the case
of thick-walled
cylinders, the
tangential stress ST
and radial stress SR at
any distance r from the
po
b
pi
a
r
center is given by the following equations:
a2 pi - b2 po
SR =
ST =
2
b - a2
a2 pi - b2 po
2
b -
a2
+
a2 b2 ( pi - po )
2
(b - a2 ) r2
a2 b2 ( pi - po )
2
(b - a2 ) r2
Simple Strain
Also known as a unit deformation, strain is the
ratio of the change in length caused by an applied
force, to the original length.
L
𝛿𝛿
A
ε=
P
δ
L
Where δ is the elongation and L is the original
length, thus ε is dimensionless.
Stress-Strain Diagram
Suppose that a metal specimen be placed in a
tension-compression-testing machine. As axial
load is gradually increased in increments, the total
elongation over the gage length is measured at
each increment of the load and this is continued
until failure of the specimen takes place. Knowing
the original cross-sectional area and length of the
specimen, the normal stress σ and the strain ε can
be obtained. The graph of these quantities with the
stress σ along the y-axis and the strain ε along the
x-axis is called the stress-strain diagram. The
stress-strain diagram differs in form for various
materials. The diagram shown below is that for the
medium-carbon structural steel.
Metallic engineering materials are classified as
either ductile or brittle materials. A ductile material
is one having relatively large tensile strains up to
the point of rapture like structural steel and
aluminium, whereas brittle materials has a
relatively small strain up to the point of rapture like
cast iron and concrete. An arbitrary strain of 0.05
mm/mm is frequently taken as the dividing line
between these two classes.
Ultimate strength
Rapture strength
Stress, σ
U
R
Y
E
P
Yield point
Elastic limit
Proportional limit
Strain, ε
O
Proportional Limit (Hooke’s Law)
From the origin, O to a point called proportional
limit, the stress-strain curve is a straight line. This
linear relation between elongation and the axial
force causing it was first noticed by Sir Robert
Hooke in 1678 and is called Hooke’s Law that
within the proportional limit, the stress is directly
proportional to strain or
σ ∝ ε or σ = K ε
The constant of proportionality K is called the
Modulus of Elasticity or Young’s Modulus and is
equal to the slope of the stress-strain diagram
from O to P.
σ=Eε
Elastic Limit
The elastic limit is the limit beyond which the
material will no longer go back to its original shape
when the load is removed, or it is the maximum
stress that may be developed such that there is no
permanent or residual deformation when the load
is entirely removed.
Elastic and Plastic Ranges
The region in the stress-strain diagram from O to
P is called the elastic range. The region from P to
R is called the plastic range.
Yield Point
The point at which the material will have an
appreciable elongation or yielding without any
increase of load.
Ultimate Strength
The maximum ordinate in the stress-strain
diagram is the ultimate strength or tensile strength.
Rapture Strength
The strength of the material at rapture. This is also
known as the breaking strength.
Modulus of Resilience
Modulus of resilience is the work done on a unit
volume of material as the force is gradually
increased from O to P, in N-m/m3. This may be
calculated as the area under stress-strain curve
from the origin O up to the elastic limit E. The
resilience of a material is its ability to absorb
energy without creating a permanent distortion.
Modulus of Toughness
Modulus of toughness is the work done on a unit
volume of material as the force is gradually
increased from O to R, in N-m/m3. This may be
calculated as the area under the entire stressstrain curve (from O to R). The toughness of a
material is its ability to absorb energy without
causing it to break.
Working Stress, Allowable Stress, and Factor
of Safety
Working stress is defined as the actual stress of a
material under a given loading. The maximum safe
stress that a material can carry is termed as the
allowable stress. The allowable stress should be
limited to values not exceeding the proportional
limit. However, since proportional limit is difficult to
determine accurately, the allowable stress is taken
as either the yield point or ultimate strength
divided by a factor of safety. The ratio of this
strength (ultimate or yield strength) to the
allowable strength is called the factor of safety.
Axial Deformation
In the linear portion of the stress-strain diagram,
the stress is proportional to strain and is given by σ
= E ε. Since σ = P / A and ε = δ / L, then P / A = E
δ / L, or
δ=
PL σL
=
AE
E
To use this formula, the load must be axial, the bar
must have a uniform cross-sectional area, and the
stress must not exceed the proportional limit.
If however, the cross-sectional area in not uniform,
the axial deformation can be determined by
considering a differential length and applying
integration.
L
P
y
y
dx
x
t
dx
δ=
P L dx
�
E O A
Where A = t y and y and t, if variable, must be
expressed in terms of x.
For rod of unit mass ρ suspended vertically from
one end, the total elongation of due to its own
weight is:
δ=
ρ g L2 M g L
=
2E
2AE
Where ρ is in kg/m3, l is the length of the rod in
mm, m is the total mass of the rod in kg, A is the
cross-sectional area of the rod in mm2, and g =
9.81 m/s2.
Stiffness, k
Stiffness is the ratio of the steady force acting on
an elastic body to the resulting displacement. It
has the unit of N/mm.
k=
P
δ
Shearing Deformation
Shearing force cause shearing deformation. An
element subject to shear does not change in
length but undergoes a change in shape.
τ
δs
γ
L
τ
The change in angle at the corner of an original
rectangular element is called the shear strain and
is expressed as:
γ=
δs
L
The ratio of the shear stress τ and the shear strain
γ is called the modulus of elasticity in shear or
modulus of rigidity and is denoted as G, in MPa.
G=
τ
γ
The
relationship
between
the
shearing
deformation and the applied shearing force is:
δs =
VL
τL
=
As G G
Where V is the shearing force acting over an area
Aa.
Poisson’s Ratio
When a bar is subjected to a tensile loading there
is an increase in length of the bar in the direction
of the applied load, but there is also a decrease in
the lateral dimension perpendicular to the load.
The ratio of the sidewise deformation (or strain) to
the longitudinal deformation (or strain) is called the
Poisson’s and is denoted by μ. For most steel, it
lays in the range 0.25 to 0.3, and 0.20 for
concrete.
z
y
x
P
P
μ=-
εy
εz
=εx
εx
Where εx is the strain in the X-direction and εy and
εz are the strains in the perpendicular direction.
The negative sign indicates a decrease in the
transverse dimension when εx is positive.
Biaxial Deformation
If an element is subjected simultaneously by
tensile stresses, σx and σy, in the X and Y
directions, the strain in the X direction is σx / E and
the strain in the Y direction is σy / E.
Simultaneously, the stress in the Y direction will
produce a lateral contraction on the X direction of
the amount –μ εy or –μ σy / E. The resulting strain
in the X-direction will be:
εx =
σy
�εx + μ εy � E
σx
-μ
or σx =
E
E
1 - μ2
And
εy =
σy
�εy + μ εx � E
σx
-μ
or σy =
E
E
1 - μ2
Triaxial Deformation
If an element is subjected simultaneously by three
mutually perpendicular normal stresses σx, σy, and
σz, which are accompanied by strains εx, εy, and εz,
respectively.
1
�σ - μ �σy + σz ��
E x
1
�σ - μ (σx + σz )�
εy =
E y
1
�σ - μ �σx + σy ��
εz =
E z
εx =
Tensile stresses and elongation are taken as
positive. Compressive stresses and contractions
are taken as negative.
Relationship between E, G, and μ
The relationship between modulus of elasticity E,
shear modulus G and Poisson’s ratio μ is:
G=
E
2 ( 1 + μ)
Bulk Modulus of Elasticity or Modulus of
Volume Expansion, K
The bulk modulus of elasticity K is a measure of a
resistance of a material to change in volume
without change in shape or form. It is given as:
K=
E
σ
=
3 ( 1 – 2 μ) ∆V / V
Where V is the volume and ΔV is change in
volume. The ratio ΔV/V is called volumetric strain
and can be expressed as:
∆V σ 3 ( 1 – 2 μ ) σ
=
=
V
K
E
Thermal Stress
Temperature changes cause the body to expand
or contract. The amount of linear deformation, δT,
is given by:
δT = α L ( Tf - Ti )= α L ∆T
Where α is the coefficient of thermal expansion in
m/m-°C, L is the length in meter, Ti & Tf is the
initial and final temperatures, respectively in °C.
For steel α = 11.25 x 10-6 / °C.
If temperature deformation is permitted to occur
freely, no load or stress will be induced in the
structure. In some cases where temperature
deformation is not permitted, an internal stress is
created. The internal stress created is termed as
thermal stress.
For a homogeneous rod mounted between
unyielding supports as shown, the thermal stress
is computed as:
L
Thermal stress, σ = E α ∆T = E α (Ti - Tf )
If the wall yields a distance of “a”, the following
calculations will be made:
L
a
α L ∆T = a +
σL
E
Where σ is the thermal stress
Take note that if the temperature rises above the
normal, the stress in the rod is compression, and if
the temperature drops below the normal, the rod is
in tension.
Torsion
Consider a bar to be rigidly attached at one end
and twisted at the other end by a torque or twisting
moment T equivalent to F x d, which is applied
perpendicular to the axis of the bar, as shown in
the figure. Such a bar is said to be in torsion.
F
F
d
r
θ
T
L
T
Torsional Shearing Stress
For a solid or hallow circular shaft subject to a
twisting moment T the torsional shearing stress Ss
at a distance ρ from the center of the shaft is:
Ss =
Tρ
Tr
and Ss max =
J
J
Where J is the polar moment of inertia of the
section and r is the outer radius.
For solid cylindrical shaft
π 4
J=
D
32
16 T
Ss =
πD3
For hallow cylindrical shaft
π
J=
(D4 - d4 )
32
16 TD
Ss =
4
π (D - d4 )
D
D
d
Angle of Twist
The angle θ through which the bar of length L will
twist is:
θ=
TL
in radians
JG
Where T is the torque in N-mm, L is the length of
shaft in mm, G is hear modulus in MPa, J is the
polar moment of inertia in mm4, D and d are
diameter in mm, and r is the radius in mm.
Power Transmitted by the Shaft
A shaft rotating with a constant angular velocity ω
(in radians per second) is being acted on by a
twisting moment T.
Torsion on Thin-Walled Tubes
t
Centerline of wall
The torque applied
expressed as:
in
thin-walled
tubes
is
T=2Aq
Where T is the torque in N-mm, A is the area
enclosed by the centreline of the tube (as shown
in the stripe-filled portion) in mm2, and q is the
shear flow in N/mm.
The average shearing stress across any thickness
t is:
Ss =
q
T
=
t
2At
Helical Springs
When close-coiled helical spring, composed of a
wire of round rod diameter d wound into a helix of
mean radius R with n number of turns, is
subjected to an axial load P gives the following
stresses and elongation:
Ss =
16 PR
πd3
�1 +
d
�
4R
This formula neglects the curvature of the spring.
This is used for light springs where the ratio d / 4R
are small.
For heavy springs and considering the curvature
of the spring, a more precise formula is given by
(A.M. Wahl Formula).
Ss =
16 PR
πd3
�
4m - 1 0.615
+
�
4m - 4
m
Where m is called the spring index and (4m – 1) /
(4m – 4) is the Wahl factor
The elongation of the bar is
δ=
64 P R3 n
G d4
Notice that the deformation δ is directly
proportional to the applied load P. The ratio of P to
δ is called the spring constant k is equal to
k=
P
G d4
=
in N/mm
δ
64 R3 n
Springs in Series
For two or more springs with spring laid in series,
the resulting spring constant k is given by
1
1
1
=
+
+…
k k1 k2
Where k1, k2 … are the spring constants for
different springs.
Springs in Parallel
k = k1 + k2 +…
SHEAR AND MOMENT IN BEAMS
Definition of Beam
A beam is a bar subject to forces or couples that
lie in a plane containing the longitudinal of the bar.
According to determinacy, a beam may be
determinate or indeterminate.
Statically Determinate Beam
Statically determinate beams are those beams in
which the reactions of the supports may be
determined by the use of the equations of static
equilibrium. The beams shown below are
examples of statically determinate beams.
Cantilever Beam
Simple Beam
Overhang Beam
Statically Indeterminate Beams
If the number of reactions exerted upon a beam
exceeds the number of equations in static
equilibrium, the beam is said to be statically
indeterminate. In order to solve the reactions of
the beam, the statics equation must be
supplemented by equations based upon the
elastic deformations of the beam.
The degree of indeterminacy is taken as the
difference between the numbers of reactions to
the number of equations in static equilibrium that
can be applied. In the case of the propped beam
shown, there are four unknown reactions R1, R2,
R3, and M and three static equilibrium equations
(ΣM, ΣFH, and ΣFv) can be applied, thus the beam
is indeterminate to the first degree (4 – 3 = 1).
R3
Propped Beam
M
R1
R2
Restrained Beam
Constinuous Beam
Types of Loading
Loads applied to a beam may consist of a
concentrated load (load applied at a point),
uniform load, uniformly varying load, or an applied
couple or moment. These loads are shown in the
following figures:
Concentrated Loads
Uniformly Distributed
Load
Varying load
Applied couple
Shear and Moment Diagrams
Consider the simple beam shown of length L that
carries a uniform load of w (N/m) throughout its
length and is held in equilibrium by reactions R1
and R2. Assume that the beam is cut at point C a
distance of x from the left support and the portion
x
x
w (N/m)
w (N/m)
B
A
C
R1
A
CV
R2
L
R1
R1 = R2 = wL/2
M
of the beam to the right of C be removed. The
portion removed must then be replaced by a
vertical shearing force V together with a couple m,
to hold the left portion of the bar in equilibrium
under the action of the force R1 and wx.
The couple M is called the resisting moment or
moment and the force V is called the resisting
shear or shear. The sign of V and M are taken to
be positive if they have the senses indicated
above.
Relationship
Moment
between
Load,
Shear,
and
From the figure shown above, the shear and
moment at point C is as follows:
V=
M=
wL
-wx
2
wL
x2
x-w
2
2
If we differentiate M with respect to x,
dM wL
=
- w x = V (shear)
dx
2
Therefore, the rate of change of the moment with
respect to x is the shearing force, or the shear is
the slope of the moment diagram.
If we differentiate V with respect to x,
dV
= - w (load)
dx
Therefore, the rate of change of the shear with
respect to x is the load, or the load is the slope of
the shear diagram.
Properties of Shear and Moment Diagrams
The following are some important properties of
shear and moment diagrams:
1.
2.
3.
4.
The area of the shear diagram to the left or to
the right of the section is equal to the moment
at that section.
The slope of the moment diagram at a given
point is the shear at that point.
The slope of the shear diagram at a given
point equals the load at that point.
The relative maximum moment occurs at the
point of zero shears. This is in reference to
property number 2, that when the shear (also
5.
6.
the slope of the moment diagram) is zero, the
tangent drawn to the moment diagram is
horizontal.
When the shear diagram is increasing, the
moment diagram is concave up.
When the shear diagram is decreasing, the
moment diagram is concave down.
Sign Conventions
The customary sign conventions for shearing force
and bending moment are represented by the
figures below. A force that tends to bend the beam
downward is said to produce a positive bending
moment. A force that tends to shear the left
portion of the beam upward with respect to the
right portion is said to produce a positive shearing
force.
An easier way for determining the sign of the
bending moment at any section is that upward
forces always cause positive bending moments
regardless of whether they act to the left or to the
right of the exploratory section.
Load
Shear Diagram
Moment Diagram
Shear Diagram
Moment Diagram
Load
Shear Diagram
Moment Diagram
Figure: Shear and moment diagrams of beams
Moving Loads
Beams or girders such as in a bridge or an
overhead crane are subject to moving
concentrated loads, which are at fixed distance
with each other. The problem here is to determine
the moment under each load when each load is in
a position to cause a maximum moment. The
largest value of these moments governs the
design of the beam.
Single Moving Load
For a single moving load, the maximum moment
occurs when the load is at the midspan and the
maximum shear occurs when the load is very near
the support (usually assumed to lie over the
support).
Two Moving Loads
For two moving loads, the maximum shear occurs
at the reaction when the larger load is over that
support. The maximum moment is given by:
Mmax =
(PL - Ps d)2
4PL
P = Ps + Pb
Where Ps is the smaller load, Pb is the bigger load,
and P is the total load, (P = Ps + Pb). This formula
is valid only if both loads are within the beam
when this condition occurs.
Two or More Moving Loads
In general, the bending moment under a particular
load is a maximum when the center of the beam is
midway between that load and the resultant of all
the loads then on the span. With this rule, we
compute the maximum moment under each load,
and use the biggest of these moments for the
design. Usually the biggest of these moments
occurs under the biggest load.
The maximum shear occurs at the reaction where
the resultant load is nearest. Usually, it happens if
the biggest load is over that support and as many
as possible of the remaining loads are still on the
span.
In determining the largest moment and shear, it is
important to check the condition when the bigger
loads are on the span and the rest of the smaller
loads are outside.
Stresses in Beams
Forces and couples acting on a beam cause
bending (flexural stresses) and shearing stresses
on any cross section of the beam and deflection
perpendicular to the longitudinal axis of the beam.
If couples are applied to the ends of the beam and
no forces act on it, the bending is said to be pure
bending. If forces produce the bending, the
bending is called ordinary bending.
Assumptions
In using the following formulas for flexural and
shearing stresses, it is assumed that a plane
section of the beam normal to its longitudinal axis
prior to loading remains plane after the forces and
couples have been applied, and that the beam is
initially straight and of uniform cross section and
that the moduli of elasticity in tension and
compression are equal.
Notations
M = bending moment
fb = bending stress
I = moment of inertia
S = section modulus
fv = shearing stress
Q = static moment of area
Radius of Curvature
ρ=
E I Ec
=
M
fb
Flexure Formula
The stresses caused by bending moment are
known as bending or flexural stresses, and the
relation between these stresses and the bending
moment is expressed by the flexure formula.
At any fiber at distance y from the neutral axis, the
bending (flexural) stress is:
fb =
My
I
The maximum flexural stress occurs at the
outermost fiber whose distance from the neutral
axis is denoted as c.
fb max =
Mc M
=
I
S
Section modulus, S =
I
c
Shearing Stress Formula
VQ
It
Q = A y�
fv =
Shear Flow
If the shearing stress fv is multiplied by the width t,
we obtain a quantity q, known as the shear flow,
which represents the longitudinal force per unit
length transmitted across a section at a level y1
from the neutral axis.
q = fv t =
VQ
I
Application of Flexural and Shearing Stresses
to Rectangular Section
For rectangular beam of width b and depth d:
fb max =
6M
bd2
3V
3V
fv max =
=
2bd 2A
Superimposed Beams
If a beam is composed of two or more thin layers
placed on each other without any attachment, the
separate layers would slide past each other and
the total strength of the beam would be the sum of
the strengths of the various layers. The strength of
this beam is lesser than a solid beam having the
same cross-sectional area.
Neglecting friction between any two adjacent
layers, the following relationships may be used:
M = M1 + M2 + M3
I = I1 + I2 + I3
M
Curvature,
= constant
EI
M
M1
M2
M3
=
=
=
E I E I1 E I2 E I3
Spacing of Rivets or Bolt in built-Up Beam
When two or more thin layers of beams are
fastened together with a bolt or a rivet so that they
act as a unit to gain more strength, it is necessary
to design the two sizes or spacing of these bolts or
rivets so that it can carry the shearing force acting
between each adjacent layers.
Consider the beams shown in the figure
s=
RI
VQ
Where R is the total shearing force to be resisted
by the bolts and is equal to the allowable shearing
stress x area x number of bolts in the group. R
may also be taken as the bearing capacity at the
section.
Economic Sections
From the flexure formula fb = M y / I, it can be seen
that the bending stress at the neutral axis, where y
= 0, is zero and increases linearly outwards. This
means that for a rectangular or circular section a
large portion of the cross section near the middle
section is under stressed.
For steel beams or composite beams, instead of
adopting the rectangular shape, the area may be
arranged to give more area on the outer fiber and
maintaining the same overall depth, and saving a
lot of weight.
When using a wide flange or I-beam section for
long beams, the compression flanges tend to
buckle horizontally sidewise. This buckling is a
column effect, which may be prevented by
providing lateral support such as a floor system so
that the full allowable stresses may be used,
otherwise the stress should be reduced. The
reduction of stresses for these beams will be
discussed in steel design.
Combined Stresses
In the previous sections, we studied the three
basic types of stresses; the axial or normal stress,
torsional shearing stress, and flexural stress, in
which we assume that only one of these stresses
acts on a member. However, these stresses may
act simultaneously and there are four possible
combinations of these stresses: (1) axial and
flexure; (2) axial and torsional; (3) torsional and
flexural; and (4) axial, torsional, and flexural.
Combined Axial and Flexure
The simplest of the four combinations is the axial
and flexural because it combines only normal
stresses, which can be added arithmetically.
A member subject to bending moment M and an
axial load P causes a flexural stress of fb = M y / I
at any point from its neutral axis, and a normal
stress of P / A which is uniformly distributed over
its entire area, respectively. The combination of
these stresses results if the member is
eccentrically loaded as in a column or a
prestressed beam. The combined stress
developed is:
𝑓𝑓1 = 𝑓𝑓𝑎𝑎 + 𝑓𝑓𝑏𝑏
f=±
P
My
±
A
I
Where m = Pe, e is the eccentricity of P from the
neutral axis
Note that the axial stress may be pure tensile or
compressive, this is the reason for the (+) and (-)
signs before P / A because axial stress is uniform
over the section while flexural stress vary with
position.
Kern of a Section
When a member is eccentrically loaded so that the
maximum flexural stress Me / I is larger than the
compressive stress P / A, the resulting stress
diagram will be as shown.
The point of zero stress A can be found by
computing a at which the direct compressive
stress equals the tensile stress
a=
I
Ae
If the maximum flexural stress in tension just
equals the direct compressive stress, there would
be no tensile stress in any section of the member.
If p is applied at any point with respect the
principal axes X and Y, there would be a certain
area in which P should act so that there will be no
tensile stress in the section. The stress at any
section whose coordinates are x and y is
My x
P
Mx y
±
±
A
I
I
�P ey � y (P ex ) x
P
±
±
f=
I
A
I
f=
For rectangular section of dimension b and h, the
kern region is within a diamond-shaped area as
shown.
The kern of a circular section of diameter D is a
circle whose diameter is D / 4
Combined Axial and Shearing Stress
In general when a body is subject to combined
loadings, every element will be subject to
combined stresses fx, fy together with the shearing
stress sxy as shown. For normal stresses, tensile
stresses are considered positive, compressive
stresses negative, and shearing stresses positive
if it creates clockwise rotation about the center.
Stresses on an Inclined Plane
If the stresses fx, fy, and sxy are known, the normal
and shearing stresses on a plane inclined at an
angle θ to the x-axis can be determined using the
following formulas or by Mohr’s Circle.
f=
fx + fy fx - fy
cos 2θ + sxy sin 2θ
2
2
sx - sy
sin 2θ + sxy cos 2θ
s=
2
Principal Stresses
There are certain values of the angle θ that will
lead to maximum and minimum values of f for a
given set of stress fx, fy, and sxy. These maximum
and minimum values that f may assume are
termed principal stresses and are given by
2
fmax =
fx + fy
fx - fy
2
+ ��
� + �sxy �
2
2
2
fmin =
fx + fy
fx - fy
2
- ��
� + �sxy �
2
2
Principal Planes
The angle θp between the x-axis and the planes on
which the principal stresses occur are given by the
equation
tan 2θp =
- 2sxy
fx - fy
There are always two values of θp that will satisfy
this equation. The maximum stress occurs on one
of these planes, and the minimum stress occurs
on the other. The planes defined by the angle θp
are known as principal planes.
The shearing stress on the principal planes is
always zero.
Maximum Shearing Stress
The maximum and minimum values of searing
stress are given by
2
fx - fy
2
smax = � �
� + �sxy �
2
min
The angle θ between the x-axis and the planes on
which the maximum and minimum shearing
stresses occur are given by the equation
tan 2θs =
fx - fy
2sxy
Mohr’s Circle
A visual interpretation of the formulas in the
preceding sections, devised by the German
engineer Otto Mohr in 1882, eliminates the
necessity of remembering them. In this
interpretation, a circle is used; accordingly, the
construction is called Mohr’s Circle.
The following are the rules for applying Mohr’s
circle to combined stresses, given fx, fy, and sxy.
1.
On the rectangular f-s axes (f along x-axis
and s along y-axis), plot points having the
coordinates (fx, sxy) and (fy,syx). In plotting
these points, assume tension as plus,
compression as minus, and shearing stress
as plus when its moment about the center of
the element is clockwise.
Note that syx = -sxy
2.
3.
4.
Join the points plotted by a straight line. This
lie is the diameter of the circle whose center
is on the σ-axis.
The radius of the circle to any point on its
circumference represents the axis directed
normal to the plane whose stress
components are given by the coordinates of
that point.
The angle between the radii to selected
points on Mohr’s circle is twice thr angle
between the normals to the actual planes
represented by these points, or twice the
space angularity between the planes so
represented. The rotational sense of the
angle corresponds to the rotational sense of
the actual angle between the normals to the
planes. i.e. if N axis is actually at a
counterclockwise angle θ from the X axis,
then on Mohr’s circle the N radius is laid off at
a counterclockwise angle 2θ from the Xradius.
Combined Torsional and Flexural Stresses
A shaft as shown is subject to combined torsional
and flexural stresses.
The equivalent bending moment M, and torque T,
due to applied loads is given by
Te = �M2 + T2
1
Me = (M + Te )
2
Where M is the actual bending moment and T is
the actual torque. Once Me and Te is computed,
the formulas for flexural and torsional stresses can
be used
Me c 4Me
=
I
πr3
T ρ 16 Te
Max s =
=
J
πd3
Max fb =
SIMPLE AND CANTILEVER BEAM FORMULAS
w (N/m)
A
B
C
Mmax = Mmid =
δmax = δmid =
θA = θB =
wL2
8
5wL4
384 EI
wL3
24 EI
P
L/2
L/2
B
A
L
Mmax = Mmid =
δmax = δmid =
θA = θB =
PL
4
PL3
48 EI
PL2
16 EI
P
a
b
B
A
C
L
x
Mmax = Mp =
Pab
L
x=
δmax = δC =
δmid =
� L 2 - b2
3
Pb �L2 - b2 �
9√3 E I L
3�
2
Pb
�3L2 - 4b2 � when a > b
48 EI
w (N/m)
y
B
A
L
C
x
δmid =
2.5wL4
384 EI
Location of max δ = 0.481 L
8wL3
7wL3
;θ =
360 EI A 360 EI
wx
�7L4 - 10L2 x2 + 3x4 �
EI y =
360L
θB =
y
w (N/m)
x
A
B
L
Mmax =
wL2
12
δmax = δmid =
θA = θB =
EI y =
wL4
120 EI
5wL3
192 EI
wx
�25L4 - 40L2 x2 + 16x4 �
960 L
for 0 < x < L/2
y
0.577L
M
x
A
L
B
Mmax = M
δmax =
ML3
θA =
EI y =
y
at x = 0.577L
9√3 EI
ML
ML
;θ =
6 EI B 3 EI
Mx
(L - x)(2L - x)
6L
P
x
L
A
B
Mmax = MA = - PI
δmax = δB =
θB =
EI y =
PL3
3 EI
PL2
2 EI
Px2
(3L - x)
6
y
P
a
b
x
L
A
B
Mmax = MA = - Pa
δmax = δB =
θB =
EI y =
EI y =
Pa2
(3L - a)
6 EI
Pa2
2 EI
Px2
(3a - x) for 0 < x < a
6
Pa2
(3x - a) for a < x < 1
6
y
w (N/m)
x
L
A
B
Mmax = MA = -
δmax = δB =
θB =
EI y =
wL2
2
wL4
8 EI
wL3
6 EI
wx2
�256 - 4Lx + x2 �
24
y
w (N/m)
x
L
A
B
Mmax = MA = -
δmax = δB =
θB =
EI y =
wL2
6
wL4
30 EI
wL3
24 EI
Mx2
2
PROPPED BEAM FORMULAS
y
M
x
L
A
B
R=
MA = -
P
L
2
Pa2
�3L - a�
2L3
�b2 a +
a2 b
�
2
P
a
b
A
B
L
R
R=
5P
16
MA = -
3PL
16
w (N/m)
A
B
R
L
7wL
R=
16
MA = -
wL2
8
w (N/m)
A
B
R
L
R=
MA = -
7wL
128
9wL2
128
w (N/m)
A
B
b
a
R
L
3
R=
wb
8L3
(4L - b)
MA = RL -
wa2
2
w (N/m)
A
B
L
R
R=
MA = -
wL
10
wL2
15
w (N/m)
A
B
L
R
R=
11wL
40
MA = -
7wL2
120
w (N/m)
B
A
L
R
R=
11wL
64
MA = -
5wL2
64
MA =
3 EI ∆
L2
FULLY RESTRAINED BEAM FORMULAS
P
a
b
A
B
L
MA = -
Pab2
L
MB = δmid =
Pba2
L
Pb2
(3L - 4b)
48 EI
P
L/2
L/2
A
B
L
MA = M B = δmax =
PL
8
PL3
192 EI
w (N/m)
A
B
L
MA = M B = -
wL2
12
4
δmax =
wL
384 EI
w (N/m)
A
B
L/2
L/2
MA = -
5wL2
192
MB = -
11wL2
192
w (N/m)
A
B
L
δmax =
wL4
MA = -
768 EI
2
MB = δmid =
wL
30
wL4
768 EI
wL2
30
w (N/m)
A
B
L
MA = M B = -
5wL2
96
4
δmax =
7wL
3840 EI
M
a
A
b
B
L
MA =
MB = -
Mb 3a
� - 1�
L
L
Ma 3b
�
- 1�
L
L
MA = MB =
6 EI ∆
L2
6 EI ∆
L2
P = y dx
Any loading
y
A
a
b
L
MA = - �
x2
x1
x2
MB = - �
x1
Pab2
L2
Pba2
L2
B
a=x;b=L-x
P = y dx
For varying load, y = f(x)
P
L
δ
For uniform load, y = w (N/m)
Dynamic (Impact) Loading
The deformation produced in elastic bodies by
impact loads caused them to act as spring,
although that is not their designed function.
The spring constant of a beam can be calculated
from the following formula:
k=
P
(N/mm or kN/mm)
δ
Where δ is the deformation due to static load P
Consider the cantilever beam shown
P
h
L
δ
PL3
3 EI
,k= 3
3 EI
L
If a load P is dropped from a height of h, the
resulting deformation δ can be computed from:
Static deformation, δst =
δ
2h
=1+ �1+
δst
δst
Impact Stress
σ
2h
=1+ �1+
δst
σst
Where σst = stress under static load
CHAPTER 11 – FLUID MECHANICS AND
HYDRAULICS
Properties of Fluid
Unit Weight or Specific Weight, 𝛄𝛄
The weight per unit of a volume of a fluid.
γ=
Weight of Fluid
Volume
For water , γ = 9810 N/m3 = 62.4 lb/ft3
Mass Density or Density, 𝛒𝛒
The mass of fluid per unit volume
ρ=
Mass of Fluid
Volume
For water, ρ = 1000 kg/m3
Density of Gases
ρ=
p
RT
where:
p = absolute pressure of gas in Kpa
R = gas constant in joule/ kg-°K
For air, R = 287 joule/ kg-°K
T = absolute temperature in degree
Kelvin
°K = °C +273
Specific Volume, Vs
VS
Specific Gravity, s
s =
γfluid
=
γwater
1
ρ
=
ρfluid
ρwater
Viscosity
The property of a fluid which determines the
amount of its resistance to shearing force. A
perfect would have no viscosity.
Dynamic or Absolute Viscosity, 𝝁𝝁 (mu)
μ =
σ
dV/dy
(Pascal-second or poise)
Note: 1 poise = 0.01 Pa/s
Kinematic Viscosity, ν (nu)
ν=
μ
ρ
(m2 /s or stokes)
Note: 1 stoke = 1 cm2/s = 0.0001 m2/s
Surface tension 𝛔𝛔 (sigma)
The surface tension of a fluid is the work that must
be done to bring enough molecules from inside the
liquid to the surface to form a new unit area of that
surface in ft-lb/ft2 or N-m/m2.
Pressure inside a droplet of a liquid
p=
4σ
d
where:
σ = surface tension in N/ m
d = diameter of the droplet in m
p = gage pressure in Pascal
Capillarity
The rise or fall of a fluid in a capillary tube which is
caused by surface tension and depend on the
relative magnitudes of the cohesion of the liquid
and the adhesion of the liquid to the walls of the
containing vessel. Liquids rise in tubes they
wet( adhesion >cohesion) and fall in tubes they do
not wet (cohesion > adhesion). Capillarity is
important when using tubes smaller than about 3/8
inch ( 9.5 mm) in diameter.
d
d
θ
h
h
θ
h=
4 σ cos θ
𝛾𝛾d
Use θ = 140 ° for mercury on clean glass
For complete wetting, as with water on clean
glass, the angle θ is 0°. Hence the formula
becomes
h=
where:
4σ
𝛾𝛾d
h = capillary rise or depression
d = diameter of the tube
γ = unit weight
σ = surface tension
Bulk Modulus of Elasticity, E
The bulk modulus of elasticity of the fluid
expresses the compressibility of the fluid. It is the
ratio of the change in unit pressure to the
corresponding volume change per unit volume.
E=
dp'
( lb /in2 or Pa)
-dv/v
where:
dp’ = change in pressure
dv = change in volume
v = volume
Compression of Gases
For perfect gas:
pvn = p1 v1 n = constant
where p is the absolute, v is the specific volume
infinity; depending upon the process to which the
gas is subjected. If the process is at constant
temperature (isothermal), n=1
pv = p1 v1
If there is no heat transfer to and from the gas, the
process is known as adiabatic.
p1 v 1 k = p2 v 2 k
A frictionless adiabatic process is called an
isentropic process and n is denoted by k, where
k = Cp/Cv, the ratio of the specific heat at constant
pressure to that at constant volume.
Boyle’s Law (perfect gas)
If the temperature of a given mass of gas remains
constant, the absolute pressure of the gas varies
inversely with the volume.
p=
k
or pV = k
V
p1 V 1 = p 2 V 2
Charle’s or Gay-Lussac’s Law (perfect gas)
If the given mass of gas can expand or contract
with the pressure remaining constant, the volume
V of the gas varies directly as the absolute
temperature T. i.e. V/T is constant.
Pressure Disturbances
Pressure disturbance imposed on a fluid moves in
waves. The velocity or celerity is expressed as:
c=�
EB
(m/s or ft /s)
ρ
where:
c = celerity or velocity of pressure wave
in m/s or ft/s
EB = bulk modulus of elasticity if the fluid
in Pa or lb/ft2
Unit Pressure
Variations in Pressure
The difference in pressure between any two points
in a homogeneous fluid at rest is equal to the
product of the unit weight of the fluid and the
vertical distance between the points.
1
h
2
p2 - p1 = γh
The pressure at any point below the free surface
of a liquid equals the product of the unit weight of
the liquid and the depth of the point.
h
p = γh
Pressure below layers of different liquids
Air Pressure = p
h1
h2
h3
Liquid 1
Liquid 2
Liquid 3
pbottom = Σγh +p
pbottom = γ1 h + γ2 h2 + γ3 h3
1
Total Hydrostatic Pressure
Total Pressure on Plane Surface
Free Liquid Surface
θ
h
Y
F
e
cg
cp
F = pcg × A
Ig
e=
AY
or
F = γhA
y=
h
sin θ
where :
pcg = pressure at the centroid of the
plane
Ig = centroidal moment of inertia of the
plane
A = area of the plane surface
θ = angle that the plane makes with the
horizontal
Total Pressure on Curved Surface
D
C
FV
B
cg
θ
FH
A
FH = pcg A
Fv = γVABCD
F = �FH 2 + Fv 2
tan θ =
FV
FH
where:
FH = total force acting on the vertical
projection of the curved surface.
Fv = weight of imaginary or real fluid
directly above the curved surfaces.
Note: For cylindrical and spherical surfaces, the
total force F always passes to the center of the
circle defined by its surface.
Bouyancy
Archimedes’ Principle - Any body immersed in a
fluid is acted upon by an unbalanced upward force
called the buoyant force, which is equal to the
weight of the fluid displaced.
VD
VD
BF
BF
For a homogenous body floating on
homogeneous liquid, the volume displaced is:
VD =
γbody
γliquid
Vbody =
a
Sbody
V
Sliquid body
Statical Stability of Floating Bodies
νS
VD sin θ
RM or OM = W(x) = W ( Mg sin θ)
MBo =
Where:
ν = volume of the wedge of immersion
s = horizontal distance between the
centroid of the wedges
VD = volume displaced
θ = angle of tilting
If the body has the shape of the rectangular
parallelepiped
MBo =
tan2 θ
B2
�1 +
�
2
12D
where:
B = width
D = draft
Metacentric Height
Metacentric height is the distance from the
metacenter to the center of gravity of the body
measured along the axis of the body.
MG= MBo ±GBo
Value of MBo in the Upright Position (Initial
Value)
MBo =
Where:
along the
I
VD
I = moment of inertia of the body
water section
Horizontal Acceleration
a
θ
tan θ =
a
g
Inclined Acceleration
ah
g + av
aH = a cosα
aV = a sinα
tan θ =
Use (+) if the acceleration is upward and (-) if
downwards
Vertical Acceleration
a
h
a
p = γh �1 ± �
g
Rotation
ω2 x2
2g
dy
tan θ =
dx
ω2 x
tan θ =
g
y=
Volume of Paraboloid
V = 1/2πr2 h
Flow rate
Volume Flow Rate,
Mass Flow Rate,
Weight Flow Rate,
Q = Aν
M = ρQ
W = γQ
Continuity Equation
Q
1
3
2
Q
Incompressible Fluid
Q1 = Q2 = Q3 ……
A1 V1 = A2 V2 = A3 V3…..
Compressible Fluid
ρ1 Q1 = ρ2 Q2
Where: A = cross- sectional area of flow
ν = mean velocity of flow
Reynold’s Number
Reynolds Number R is the ratio of inertia forces to
viscous forces
R=
νDρ νD
=
v
μ
Where:
v = mean velocity of flow, m/s
D = pipe diameter, m
μ = (mu) dynamic viscosity (Pa-sec)
ν = (nu) kinematic viscosity (m2/s) = μ/ρ
ρ = density, kg/m3
For non-circular pipes, use D = 4R, where R is the
hydraulic radius, R= A/P
For R< 2000, the flow is laminar.
Laminar flow in circular pipes can be maintained
up to values of R as high as 50, 000. However, in
such cases this type is inherently unstable, and
the least disturbance will transform it instantly into
turbulent flow. On the other hand, it is practically
impossible for turbulent flow in a straight pipe to
resist at all values of R much below 2000,
because any turbulence that is set up will be
damped out by the viscous friction.
Energy Equation
Total Energy of Flow
E=Kinetic Energy + Potential Energy
v2 p
E=
+ +z
2g γ
v2
= velocity head ( K.E.)
2g
p
= pressure head (P.E.)
γ
z = elevation head (P.E.)
Bernoulli’s Energy Theorem
Between any two points (1 and 2) along the
stream:
B
ZB
A
ZA
Datum
E1 + HA – HE – HL = E2
Where:
E1 = Total Energy (head) at section 1
HA = head added (by the pump)
HE = head extracted (by the turbine or
any other device)
HL = total head lost
Head Lost in Pipe Flow
Major Head Lost (Friction at Losses)
Darcy-Weisbach Formula
hf =
fL v2
in ft or meter
D 2g
For Laminar Flow, f=
64
R
For non-circular pipe, use D=4R
For circular pipes ( S.I.)
v2
8Q2
=
2g π2 gD4
hf =
For S.I. units, hf =
fL 8Q2
D π2 gD4
0.0826fLQ2
D5
Manning’s Formula (S.I.)
v=
where
1 2/3 1/2
R S
n
R = hydraulic radius = A/P
S = slope of EGL = hf/L
a
hf =
6.35n2 Lv2
D4/3
Use D = 4R for non- circular pipes
For circular pipes (S.I.)
hf =
10.29n2 LQ2
D16/3
Hazen- William’s Formula (S.I.)
Q = 0.2785C1 D2.63 S0.54
where
C1 = Hazen – William’s coefficient
S = slope of EGL = hf/L
hf =
10.67LQ1.85
C1
1.85
D4.87
Minor Head Lost
Minor losses are due to changes in direction and
velocity of flow, and is expressed in terms of the
velocity head at the smaller section of the pipe in
case of constrictions.
hm = K
where
v2
2g
K = coefficient of minor loss
Head Lost through Nozzles
1
vn 2
hn = � 2 - 1�
2g
Cv
For a horizontal pipe with uniform diameter, the
head lost between any two points is equal to the
difference in pressure head between the points.
HL =
p2 -p1
γ
For a pipe or system of pipes connecting two
reservoirs, the total head lost is equal to the
difference in water surface elevation of the
reservoirs.
H
HL = H
Pipes in Series
A
1
B
Q1
2
C
3
D
Q3
Q2
Q1 = Q2 = Q3
HL = hL1 + hL2 + hL3
Pipes in parallel
1
Q1
2
Q2
3
Q3
Q = Q1 + Q2 + Q3
HL = hL1 = hL2 = hL3
Equivalent pipe
For a pipe or system of pipes (O), the equivalent
single pipe (E) is must satisfy the following
conditions.
QE = Qo
and HLE = HLo
Orifice and Tubes
An orifice is an opening with a closed perimeter
through which fluid flows.
The velocity and discharge through an orifice is
given by:
v = Cv �2gH
Where:
Q = CAo �2gH
C = Cc Cv
Cv = coefficient of velocity
C = coefficient of discharge
CC = coefficient of contraction
H = total head in meter or feet of the
fluid flowing
Value of H
H=head upstream-head downstream
p
v a 2 pu
+
- hD - D
H = hu +
2g
γ
γ
Where
va = velocity of approach
pu= pressure at the upstream side
pD= pressure at the downstream side
Unsteady Flow(Variable head)
If water flows into a tank at the rate of Q and at
the same time leave: at Q the time for the level to
the change from h1 to h2 is:
h2
t= �
h1
As dh
Qi -Qo
If Qi = 0
h1
t= �
h2
As dh
Qo
If the flow is through an orifice under a variable
head H:
Qo = CAo �2gH
If the cross-sectional area As is constant and the
flow is through an orifice, the formula becomes
t=
where
2As
CAo �2g
��H1 - �H2 �
H1 = initial head (at level 1)
H2 = final head (at level 2)
If the water flows through the pipe connecting the
two tanks shown, the time for the head to change
from H1 to H2 is:
t=
2
As1 As2
��H1 - �H2 �
As1 + As2 CAo �2g
Weir
Weirs are structures built across an open channel
or on top of a reservoir for the purpose of
measuring or controlling the flow of water.
Rectangular Weir (Suppressed)
L
H
General Formula
2
Q = C�2gL�(H + hv )3/2 - hv 3/2 �
3
or Q = Cw L�(H + hv )3/2 - hv 3/2 �
where
hv =
Va 2
2g
velocity head of approach
C = coefficient of discharge
Cw = weir factor
Neglecting va:
Q=
2
C�2gLH3/2
3
or
Q = Cw LH3/2
Francis Formula (Cw =1.84) S.I.
Considering va:
Q=1.84L�(H + hv )3/2 - hv 3/2 �
Neglecting va:
Q=1.84LH3/2
Cipolleti weir
Q = 1.859 L H3/2
θ = 75.9637° = 75°57' 50"
β = 14.0363° =14°2'10"
Triangular V-notch Weir
L
H
θ
Q=
8
θ
C�2gtan H5/2
15
2
Suttro Weir (Proportional Flow weir)
L
X
H
Y
q = CπK�2gH
K = x �y
Unsteady flow weir (Variable Head)
H1
t= �
H2
As dH
Qo
If the flow through a suppressed rectangular
weir:
t=
2As 1
1
�
�
Cw L �H2 �H1
Where Cw = weir factor, H1 = initial head, H2 =
final head
When H2 = 0
t=
2As 1
�
�
Cw L �H1
Hydrodynamics
Force against a fixed flat plate held normal to
the jet
F=
Qγ
v = ρQv
g
Force against a fixed curved vane
V2
R
RY
RX
θ
V1
Qγ
(v - v )
g 1x 2x
Qγ
�v - v �
Fy =
g 1y 2y
Fx =
F = �Fx 2 + Fy 2
where:
v1 = velocity of the jet before hitting the
vane
v2 = velocity of the jet as it leaves the
vane
Force against the Moving Vane
v
u
RY
v
'
RX
v
u
u
θ
v1
θ
v
y
v'
v
2x
v'
Fx =
Q'γ
(v1x -v2x )
g
Fy =
Q' = Au
u = v1 -v'
Q'γ
�v1y -v2y �
g
u = relative velocity of the jet as it moves along the
vane
Q’ = amount of fluid deflected by the vane
Force on Bends and Pressure Conduits
Dynamic Force
Fx =
Q'γ
(v1x - v2x )
g
Fy =
Q'γ
�v1y - v2y �
g
Total Force
Q'γ
(v1x - v2x ) = F1x + Rx - F2x
g
Q'γ
Fy =
�v1y - v2y � = F1y + Ry - F2y
g
Fx =
Drag Force
DF = 𝐶𝐶𝐷𝐷 𝜌𝜌𝐴𝐴
𝑣𝑣 2
𝑣𝑣 2
= 𝐶𝐶𝐷𝐷 𝛾𝛾𝐴𝐴
2
2𝑔𝑔
Where:
𝐶𝐶𝐷𝐷 = Drag Coefficient
𝜌𝜌 = Density of the Fluid
A = Area normal to the direction of motion
Chapter Twelve
Engineering Economics
Interest
Interest is the amount of money earned by a given
capital. From the borrower’s viewpoint, interest is
the amount of money paid for the use of a
borrowed capital. From the lender’s viewpoint, it is
the income generated by the capital that was lent.
Cash Flow Diagrams
Cash flow diagrams may be drawn to help
visualize and simplify problems having diverse
receipts and disbursements.
Conventions used in cash flow diagram:
•
•
The horizontal (time) axis is marked off
in equal increments, one per period, up
to the duration or horizon of the project.
All disbursement and receipts (cash flow)
are assumed to take place at the end of
the year in which they occur. This is
known as the year-end convention. The
exception of the year-end convention is
the initial cost(purchase cost) which
occur at t = 0
To or more transfers in the same period
placed end-to-end may be combined
into one.
Expenses incurred before t = 0 are
called sunk cost and are not relevant to
the problem.
Receipts
and
disbursements
are
represented by arrows on the opposite
sides of the horizontal time axis.
•
•
•
Consider the following example: An electronic
equipment costs P30,000. Maintenance cost is
P3,000 each year. The device will generate
revenues of P15,000 each year for 5 years after
which the salvage value is expected to be P12,000
The following shows the cash flow diagram for
each transaction:
P12T
P15T
0
1
P3T
P30T
P15T
P15T
2
3
P3T
P3T
P15T
4
P3T
P15T
5
P3T
The simplified cash flow diagram is as follows
P12T
0
1
P12T
2
P12T
3
P12T
4
P24T
5
P30T
Simple Interest
In simple interest, the interest earned by the
principal is computed at the end of the investment
period , and thus, it varies directly with time.
Ordinary and Exact Simple Interest
In ordinary simple interest, the interest
computed on the basis of one banker’s year
is
1 banker’s year = 12 months
(30 days each month) = 360 days
In exact simple interest, the interest is based on
the exact number of days in a year , where there
are 365 days for an ordinary year and 366 days for
leap years.
Leap years occurs every four years for years that
is exactly divisible by four, except century marks
(1800, 1900, etc.) but not including those are
exactly divisible by 400(2000, 2400, etc.)
Elements of Simple Interest
P = principal or present worth
I = interest earned
F = future worth
F=P+I
r = simple interest rate(per year)
t = time in years or fraction of the year
Note: P may stand for the amount borrowed or
invested while F may stand for the amount
accumulated.
I = Prt
F = P +I = P+ Prt
F = P(1 + rt)
Value of t
Example:
•
4 years; t=4
•
•
3 months; t=3/12 or ¼
90 days
Ordinary simple interest, t= 90/360
Exact simple interest, t=90/365 or
90/360 for leap years
2 years and 4 months; t=2+4/12
=2.3333
•
Compound Interest
In compound Interest , then interest is computed
every end of each interest period (compounding
period) and the interest earned for that period is
added to the principal( interest plus principal).
To demonstrate this, consider an investment of
1000 pesos to earn 10 percent per year for two
years. The following diagram shows how the
money grows.
0
1
3
2
P1000
P1100
I = 1000 x 0.1
I= P100
P1210
I = 1100 x 0.1
I= P110
P1331
I = 1210 x 0.1
I= P121
Elements of Compound Interest
P = present worth or principal
F = future worth or compound amount
i = effective interest per compounding period
(per interest period)
i = r/m
n = total number of compounding
n=t×m
I = interest earned
I=F-P
r = nominal interest rate
ER = effective interest
t = no. of years of investment
m = no. of compoundings per year
After n periods, the compound amount F is:
F = P(1 + i)n
The term (1+ i)n, also denoted as (F/P,I,n) is called
the single payment compound-amount factor
The present worth of F is:
P=
F
(1 + i)n
The term
1
(1 + i)n
, also denoted as (P/F, i, n) is called
the single payment present-worth factor.
Values of i and n
The following examples show how to get the
values of i and n.
Nominal Interest rate, r = 12%
Number of years of investment, t = 5 years
•
Compounded annually ( m = 1)
i = 0.12/1= 0.12
n = 5(1) = 5
• Compounded semi-annually (m = 2)
i = 0.12/2 = 0.06
n = 5(2) = 10
•
Compound quarterly (m = 4)
I =0.12/4 = 0.03
n = 5(4) = 20
•
Compounded monthly (m = 12)
i = 0.12/12 = 0.01
n = 5(12) = 60
•
Compounded bi-monthly
i = 0.12/6 = 0.02
n = 5(6) =30
Continuous Compounding (m→ ∞)
Interest may be compounded daily, hourly, per
minute, etc. As a limit, interest may be considered
to be compounded an infinite number of times per
year (m=∞).
The future worth of P at an interest rate of r
compounded continuously for t years is:
F = Pert
Nominative and Effective Rates of Interest
Nominal rate is the rate quoted in describing a
given variety of compound interest. Consider a
bank deposit of P1000 to earn 6% compounded
quarterly. After one year, the compound amount of
F is:
F = P(1 + i)n = 1000(1 + 0.06/4)1/4
F = P 1061.36
Notice that the interest earned is P61.36
representing 6.136% of P1000 (not 6% of P1000).
For this case, 6% (compounded quarterly) is
called the nominal rate and 6.136% is the effective
rate.
Thus the effective rate of interest(ER) is the actual
interest earned in one year period. This can be
computed by either of the following:
ER =
Interest earned in one year
Principal at the beginning of the year
r m
ER = �1 + � - 1
m
Thus, the effective rate of 6% compounded
quarterly is, ER=(1+0.06/4)4-1= 0.06136 or 6.136%
as computed previously
The effective rate r(%) compounded continuously
is:
ER = er - 1
Equivalent Nominal Rates
From the previous discussion we see that 6%
compounded quarterly is not the same as 6%
compounded monthly, for the reason that they
have different effective rates.
Two nominal rates are equal if they have the same
effective rates.
Consider a nominal interest rate of 10%
compounded quarterly. The equivalent nominal
rate compounded monthly is:
ERM = ERO
(1 + r/12)12 - 1 = (1 + 0.10/4)4 - 1
r = 0.09918 = 9.918 %
Thus, 10% compounded quarterly will have the
same interest as 9.918% compounded monthly.
Annuity
Annuity is a series of uniform payments made at
equal intervals or time.
Annuities are established for the following
purposes:
1. As payment of a debt by a series of equal
payment at equal time intervals, also known as
amortization.
2. To accumulate a certain amount in the future by
depositing equal amounts at equal intervals.
These amounts are called sinking funds.
3. As a substitute periodic payment for a future
lump sum payment.
Elements of Annuity
A=periodic payment
P= present worth of all periodic payments
F or S= future worth or sum of all periodic
payments after the last payment is made
i= interest rate per payment
n=number of payments
Types of Annuity
A.
Ordinary Annuity
In ordinary annuity, the payment is made at the
end of each period starting from the first period, as
in the diagram shows below
0
1
2
3
4
n
A
A
A
A
A
F
P
The future worth if A is:
F=
The factor
(1 + i)n - 1
i
A[(1 + i)n - 1]
i
is called equal-payment-series-
compound-amount factor and is denoted as (F/A, i,
n)
The value of A if F is known is:
A=
The factor
i
(1 + i)n - 1
Fi
(1 + i)n - 1
is known as equal- payment
sinking- fund factor and is denoted as (A/F, i, n)
The present worth of A is:
P=
The factor
A[(1 + i)n - 1]
F
=
(1 + i) n i
(1 + i)n
(1 + i)n - 1
(1 + i)n i
, is known as equal- payment-
series- present- worth factor and is designated as
(P/A, i, n)
The value of A with known P is:
A=
The factor
(1 + i)n i
(1 + i)n - 1
P(1 + i)n i
(1 + i)n - 1
, is known as the equal-
payment- series- capital- recovery factor and is
designated as (A/P, i, n)
B.
Deferred annuity
In this type , the first payment is deferred a certain
number of periods after the first. Consider the
cash flow diagram below
0
1
2
3
4
5
A
A
A
A
F
P
n=5
For the cash flow diagram shown above, the
following calculations can be made for solving P
and F
So solve for the future worth F:
A�(1 + i)4 - 1�
F=
i
To solve for the present worth P
F
A�(1 + i)4 - 1�
P=
=
5
(1 + i)
(1 + i)5
C.
Annuity due
If the payment is made at the beginning of each
period starting from the first period, the annuity is
called annuity due.
0
1
2
3
4
c
A
S
A
A
A
A
A
F
n=6
P
n=5
From the diagram shown:
A�(1 + i)6 -1�
F=
i
F
A�(1 + i)6 - 1�
P=
=
(1 + i)5
(1 + i)5 i
From the previous examples, the formula of P can
be generalized as :
P=
A[(1 + i)n - 1]
(1 + i)nˈ i
Where n is the number payments and n1is the
number of periods from zero (0) period up to the
last payment.
Perpetuity
Perpetuity is an annuity where the payment
periods extend forever or the periodic payments
continue indefinitely.
If the payment is made at the end of each period
starting from the first period, the present worth of
perpetuity is:
P=
A
i
Uniform gradient
Arithmetic Gradient
0
1
2
A
3
A+G
4
5
n
A+2G
A+3G
A+4G
A+nG
The present worth is:
P=
A[(1 + i)n -1] G (1 + i)n - 1
n
+ �
�
(1 + i)n i
i i(1 + i)n (1 + i)n
The future worth is:
F = P(1 + i)n =
A[(1 + i)n - 1] G (1 + i)n - 1
+ �
- n�
i
i
i
Geometric Gradient
0
1
2
A+G
Let 𝑤𝑤 =
3
A+G(1+r)
A+G(1+r)2
4
n
A+G(1+r)3
A+G(1+r)n
1+𝑟𝑟
1+𝑖𝑖
Present worth
If w ≠ 1
P=
If w = 1(for r = i)
A[(1 + i)n - 1] G 1 - wn
+
�
�
(1 + i)n i
1+i 1-w
P=
A[(1 + i)n - 1] Gn
+
(1 + i)n i
1+r
The future worth is :
F = P(1 + i)n
Capitalized Cost and Annual Cost
Capitalized cost is an application of perpetuity.
The capitalized cost of a project or structure is the
sum of the first cost (FC) and the present worth of
all future payments and replacements which is
assumed to continue forever.
If a project requires a first cost of FC, annual
operation and maintenance of OM for n years a
salvage value of SV after every n years, and a
replacement cost of RC after every end of n
years , then the capitalized cost is:
SV
0
1
2
3
4
5
OM
OM
OM
OM
OM
n
OM
FC
RC
Capitalized cost, K
K = FC +
OM
RC - SV
+
(1 + i)n - 1
i
If RC is not specified, use RC = FC
Capitalized cost may also be defined as the first
cost plus the present worth of annual maintenance
and operation cost plus the present worth of
depreciation assumed to continue forever.
Annual Cost, AC
The annual cost (AC) of a project is:
AC = Annual interest on investment + Annual
operation and maintenance +Annual
In relation to capitalized
cost,
depreciation
costAC is:
Annual Cost, AC = K i
(RC -SV)i
AC = (FC)i + OM +
n
(1 + i) - 1
Example 12-1
A machine costs P300, 000 new, and must be
replaced at the end of each 15 years. If the annual
maintenance required is P5, 000, find the
capitalized cost, if money is worth 5% and the final
salvage value is P50, 000
Solution:
OM RC-SV
+
n
i
(1-i) -1
5,000 300,000 - 5,000
+
K = 300,000 +
15
0.05
(1+0.05) -1
K = FC +
K = P 631,711.44
Cost Comparison of different alternatives
If two or more different articles are available for
the same purpose, they are equally economical if
the corresponding present worth, annual cost or
capitalized costs are the same.
Example 12-2
A certain equipment costs P150, 000.00, lasts for
6 years, and has a salvage value of P30, 000.
How much could an investor afford to pay for
another machine for the same purpose, whose life
is 10 years and salvage value is 40,000.00, if
money is worth 5%?
Solution: (Capitalized cost method)
For the first machine:
RC - SV
OM
+
K = FC +
n
i
(1 + i) - 1
150,000 - 30,000
K = 150,000 +
6
(1 + 0.05) - 1
K = P502,841.92
For the other machine
FC - 40000
K = FC +
= 502,841.92
10
(1 + 0.05) - 1
FC = 218,696.41
Depreciation
Depreciation refers to the decrease in the value of
an asset, due to usage or passage of time. An
asset may depreciate physically or functionally.
Elements of Depreciation
FC = first cost
SV = salvage value or trade-in value
d = depreciation charge
n = economic life of the property in years
m = any time before n
BVm = book value after m years
Dm = total depreciation for m years
The following diagram shows the cost of the
property plotted versus time.
Cost
Dm
D
Cost
Curved
FC
BVm
SV
m
time
n
The book value of the property at any time m is:
BVm = FC - Dm
Methods of Computing Depreciation
A.
Straight Line Depreciation (SLD)
This is the most common method used in
computing depreciation. In this method, the cost of
the property is assumed to vary linearly with time.
The following formulas are used.
FC - SV
d=
n
Dm = d × m
B.
C.
Sinking Fund Method
(FC + SV)i
d=
n
(1 + i) - 1
d[(1 + i)m - 1]
Dm =
i
Sum of the Years Digit Method
n
Sum of the year' s digit, SUM = (1 + n)
2
n-m+1
dm = (FC - SV)
SUM
m(2n - m + 1)
Dm = (FC - SV)
2 × SUM
D.
Declining Balance Method (Constant
percentage method)
n
Constant Percentage, K = 1- �
SV
FC
BVm = FC(1 + K)m
dm = FC(1 - K)m - 1 K
E.
Double Declining Balance Method
2
× BV at the
n
beginning of the year
2 m
BVm = FC �1 - � ≥ SV
n
2
dm = BVm - 1 and BVn = SV
n
Depreciation charge to date =
Capital Recovery Method
If you invest FC now and desires a rate of return r
for n periods, and if you can deposit to an account
earning an interest of I for n periods to recover an
amount of RC, and will also receive salvage value
of SV from your invested property at the end of n
periods, then the periodic dividend or income D
required is:
(RC - SV)i
D = (FC)r +
n
(1 + i) - 1
Note: If RC is not specified, RC = FC
Example 12-3
A mine costs P21M, and will last for 20 years. Its
plant has a salvage value of P1M, at the end of
the time. The mine will yield an equal dividend at
the end of each year. What is the annual dividend,
if it is sufficient to pay interest annually at the rate
of 6% on the original investment and to
accumulate a replacement fund, invested at 4%?
D = (FC)r +
(RC - SV)i
n
(1 + i) - 1
D = (21,000,000)(0.06) +
D = P1,931,635.00
(21,000,000-1,000,000)(0.04)
(1 + 0.04)20 - 1
Bond
A bond is a written contract to pay a certain
redemption value C on a specified redemption
date and to pay equal dividends D periodically.
Elements
F = face value or par value of the bond
C = redemption value on a specified redemption
date
r = bond rate or dividend rate
D = periodic dividend
D=F×r
i = investor’s interest rate of return
P = price of the bond at a given interest i
•
A bond is said to be redeemable at par if
the redemption value C equals the face
value F.
•
A bond is said to be redeemable at a
premium if C > F.
•
A bond is said to be redeemable at a
discount if C < F.
Price of a bond at given i:
P=
C
n
(1 + i)
=
D[(1 + i)n - 1]
n
(1 + i) i
Example 12-4
A P100, 000.00, 6% bond, pays dividends semiannually and will be redeemed at 110 % on July
19, 1999. Find its price if bought on July 1, 1996,
to yield an investor 4%, compounded semiannually.
Solution:
Face value of the bond, F = P100, 000.00
Redemption Value, C = 10 %( 100,000) = P110,
000.00
Bond Rate, r = 0.06/2 = 0.03
Periodic Dividend, D=F x r=100,000(0.03)
= P3, 000.00
Investor’s rate of return (per semi- annual),
I = 0.04/2 = 0.02
Number of dividends, n = 3(2) = 6
D[(1 + i)n - 1]
C
+
P=
n
(1 + i)n i
(1 + i)
3000�(1 + 0.02)6 - 1�
110,000
+
P=
(1 + 0.02)6 (0.02)
(1 + 0.02)6
P=P114,481.14
Break Even Analysis
Break even Analysis is a method of determining
when costs exactly equal revenue. If the
manufactured quantity is less than the break- even
quantity, a loss is incurred. If the manufactured
quantity
is
greater
than
the
breakeven quantity, a profit is incurred.
Elements
f = fixed cost which does not vary with production
a = an incremental cost which is the cost to
produce one additional item. It may also be called
the marginal cost or differential cost.
N = break-even point or quantity produced and
sold for break-even
P = incremental revenue or selling price per unit
R = total revenue
R = pN
C=total cost
C = f + aN
Assuming there is no change in inventory, the
break-even point can be found from:
Cost, C = Revenue, R
f + aN = pN
f
N=
p-a
Amount
Revenue =pN
Profit
Cost = f + aN
Break-even Point
Loss
Break-even
Quantity
Example 12-5
The cost of producing a computer diskette is as
follows: Material cost is P7.00 each, labor cost is
P2.00 each, and other expense is P1.50 each. If
the fixed expenses is P69, 000.00 per month, how
many diskettes must be produced each month for
break- even if each diskettes is worth P45.00?
Solution:
Given f = fixed cost = P69, 000 per month
a = marginal cost = P7.00 +P2.00 + P1.50
a = P10.50
p = marginal revenue = P45.00
f
N=
p-a
69, 000
N=
45-10.5
N=2000 diskettes per month
CHAPTER 13
Commonly Used Conversion Factors
To convert from
To
Square foot
Foot per second2
Cubic foot
Pound per in3
Gallon per in2
Pound force
Acre foot per day
Kip per foot2
Acre foot per day
Acre
Cubic foot per sec
Inches
Foot
Meter
Square meter
Meter per sec2
Cubic meter
Kilogram per m3
Liter per sec
KPa
Newton, N
Pascal, Pa
m3 per sec
Square meter
m3 per sec
Millimetre
Meter
Foot
Multiply by
9.290304 x 10-2
3.048 x 10-1
2.831685 x 10-2
2.76799 x 104
6.309 x 10-2
6.894757
4.448222
4.788026 x 101
1.427541 x 10 -2
4,046.873
2.831685 x 10-2
25.4
0.3048
3.28
Commonly Used Constants
Acceleration due to gravity, g = 9.81 m/s2 (32.166 ft/s2)
Atmospheric pressure (standard) = 101.325 KPa (14.7 psi)
Density of water, p = 1000 kg/m3 (62.4 lb/ft3)
Speed of light in Aircraft STP, v = 340 m/s
Unit weight of water, y = 9.81 kN/m3 (62.4 lb/ft3
Constant in Physics and Mathematics
Name
Symbol
Absolute Entropy
Constant
Absolute Zero
Acceleration of free
fall on earth
(acceleration due to
gravity)
Acre
Air, Critical
Temperature
Air, Density
Air, viscosity of
(200C)
Allandi – Grinstead
Constant
Angstrom
Annee – Lumiere
Aspery’s Constant
Aposthecaries’s
ounce
So/R
Archimedes’
constant
Atsronomical Unit
Atmospheric
Pressure
Value
-1.1517048 (44, 1.1648678,44)
-273.15 0C
g
9.80665 m / s2
32.1740 ft / s2
4046.85642240 m2
-190 0C
183K
0.7734 m3 k-1
𝜂𝜂 o
Α
ly-
𝜍𝜍(3)
Oz(apot
h)
Oz(ap.)
𝜋𝜋
AU
1.8 x 10-5 N s m
0.809394020534
10-10 m
9.46052973 x 1015 m
1.202056903
3.110347680 x 10-10 kg
3.141592653589793
2384626433832795
1.4959787 x 10 11 m
1.010325 x 105 N m2
1.01325 bar
Atomic Mass Unit
Avogrado constant
Avoirdupois Ounce
Backhouse’s
Constants
Base of Natural
Logarithms
Bernsteins’s
Constant
Board foot (timber)
Bohr Magnetron
amu
mu
u
Na
oz avdp
oz (av.)
oz
E
𝛽𝛽
fbm
uB
Bohr Raduis
Boltzmann Constant
Briggsian (common)
Logarithm of 2
Brun’s Constant
Cahen’s Constant
ao
K
lcg102
Carbon – 14 half-life of
T
G
ch
Z0
Catalan’s Constant
Chain
Characteristics
Impedence of
Vacuum
Charge to mass
Quotient, electron
Charge to mass
Quotient, proton
B
14.7 lbf in2
1.66053873(13) x 10-27 kg
931.494013(37) MeV
1.49241778(12) x 10-10 J
6.02214199(47) x 1023 mol
2.834952313 x 10-2 kg
0.0625 lb (av.)
16 drachms (av.)
1.456074948582689671399
59535111654356
2.718281828459045235360
2874713526
0.2801694990
2.359737216 x 10-3 m3
9.27400899(37) x 10-24 JT4
5.788381749(43) x 10-5 eVT
1.39964624(56) x 1010 HzT
5.291772083(19) x 10-11 m
1.3806503(24) x 10-23 J K-1
0.30102999566398119521
1.9021605824
0.6294650204
5570 years
0.915965594
20.1168
376.730313461 ohms
-e/me
-1.758820174(71) x 107 C
kg-4
e/me
9.57883408(38) x 107 C kg-1
Charge Electron
Charge Electron
Specific
Charge, Elementary
Circulation,
Quantum of
Classical Electron
Radius
Common logarithm
of 2
Compton Wave
length of electron
Compton Wave
length, Muon
Compton Wave
length Neutron,
h/mnc
Compton Wave
length Proton, h/mpc
Compton Wave
length Tau
Conductance
Quantum 2e2/h
Conductance,
Quantized Hall
Constant, Absolute
Entropy
Constant, Apery’s
Constant,
Boltzmann
Constant, Catalan’s
Constant, Dirac’s
E
e/h
-e/h
1.602176462(63) x 10-19 C
2.417989491(95) x 1014 A J-1
E
e/h
h/2me
h/me
re
1.602176462(63) x 10-19 C
2.417989491(95) x 1014 A J-1
lcg102
-1.758819(62) x 1011 C kg-1
3.636947516(27) x 10-4 m2-s-1
7.273895032(53) x 10-4 m2-s-1
2.817940285(31) x 10-15 m
0.301029999566398119521
𝜆𝜆e
2.426310215(18) x 10-12 m
𝜆𝜆cen
1.319590898(10) x 10-15 m
𝜆𝜆cep
1.326409847(10) x 10-15 m
G0
7.748091696(28) x 10-5 S
𝜆𝜆ceu
𝜆𝜆cu
11.73444197(35) x 10-15 m
0.69770(11) x 10-15 m
3.87404614 x 10 -5 S
-1.1517048(44)
-1.1648678(44)
1.202056903
1.3806503(24) x 10 -23 J K-1
8.6177342(15) x 10 -5 eV K-1
2.0836644(36) x 10 10 Hz K-1
0.915965594
1.054571596(82) x 10 -34 J s
6.58211899(26) x 10 -16 eV s
Constant, Electric
(1/𝜇𝜇0c2)
Constant, Eulers’s
Constant, Faraday
Constant,
Feigenbaum’s
Constant, Fermi
Constant, Fermi
Coupling
Constant, Fine
Structure
Constant, First
Radiation 2𝜋𝜋hc2
Constant, Gas
Constant,
Gravitational
Constant, Inverse
Fine Structure
Constant,
Khintchine’s
Constant, Loschmidt
Constant, Loschmidt
(T=273.15K p=
100kPa)
Constant, Magnetic
Constant, Planck (h)
Constant,
Pythagoras
Constant, Sackur –
Tetrode
Constant, Solar
Constant, Stefan –
Boltzmann (𝜋𝜋2/60)
8.854187817 x 10 -12 F m-1
F
𝛿𝛿
Gf / (hc)3
𝛼𝛼
0.57721566490153286061
96485.3415(39) C mol-1
4.669210609102990
1.4 x 10-50 J m-3
1.16639(1) x 10-5 GeV-2
7.297352533(27) x 10-3
c1
3.74177107(29) x 10-16 W m2
R
G
8.314 J K-1 mol-1
𝛼𝛼
K
6.673(10) x0-11 N m2 kg 2
137 03599976(50) x 10-3
2.685452001
𝜂𝜂 o
Vm
2.6867775(47) x 1025 m-3
𝜇𝜇 0
4𝜋𝜋 x 10-7 N A-2
12.566370614 X 10-7 N A
6.626068776(52) x 10-34 J s
4.13566727(16) x 10-15 e V s
H
22.710981(40) X 10-4 m3
mol4
1.4142135624
So / R
O
-1.1517048(44)
-1.1648678(44)
1400 W m-2
5.670400(40) x 10-8 W m-2
K-1
k4/h3c2
Constant, Verdet’s
(light at 589 mm at
water)
Constant, von
Klitzing
Constant, Wien
Displacement law
Constant, Planck
(h/2 𝜋𝜋)
Constant, Rydberg
Constant, Linear
Expansivity of
Copper, Specific
Heat Capacity of
Copper, Thermal
Conductivity of
Copper, Young
Modulus for
Cord (timber)
Critical Temp of Air
Cross Section,
Thomson
Cube Foot
Cube Inch
Cube Yard
Curie
Dalton
Rk
0.000477 rad A-1
B
H
(h bar)
R, R,c
R,hc
𝛼𝛼
25812.807572(95) ohms
2.8977686(51) x 10-4 m K
1.054571596(82) x 10-34 J s
6.58211889(26) x 10-16 e V s
10973731.568549(83) m-1
3.289841960368(25) x 1015 Hz
2.179877190(17) x 10-18 J
C
1.7 x 10-5 K-1
K
885 J kg-1 K-1
E
885 W m-1 K-1
cd (UK)
1.8 x 1011 Pa
𝜎𝜎
ft3,
cu. Ft.
in3
cu. in.
yd3
cu. yd
Ci
amu
m
u
3.624556364 m3
-190 oC
183 K
0.665245854(15) x 10-28 m2
2.8316846592 x 10-2 m3
1728 in3
1.6387064 x 10-5 m3
7.64554858 x 10-5 m3
27 ft3
46656 in3
3.7 x 1010 Bq
1.66053873(13) x 10-27 kg
931.494013(37) MeV
1.49241778(12) x 10-10 J
Daniel Cell, emf of
Density of Air (at
stp)
Density, Earth’s Ave
Deuteron Magnetic
Moment
Deuteron Molar
Mass
Deuteron – Electron
Magnetic Moment
Ratio
Deuteron – Electron
Rest Mass Ratio
Deuteron – Proton
Magnetic Mass
Ratio
Deuteron – Proton
Rest Mass Ratio
Dirac’s Constant
Drachm
(Avoirdupois)
Dram (Avoirdupois)
Dry Pint
e
Earth’s Average
Density
Earth’s Average
Radius
𝜇𝜇 d
Md
𝜇𝜇 d/𝜇𝜇 e
md/me
𝜇𝜇 d/𝜇𝜇 p
md/mp
H
(h bar)
dr
(avdp)
dr. (av.)
dr
(avdp)
dr. (av.)
pt (US,
dry)
E
1.08 V
0.7734 m3 kg-1
1.2929 kg m-3
5.517 x 103 kg m-3
0.43307375(15) x 10-26 J T-1
0.4669754479 x 10-3 𝜇𝜇𝜇𝜇
0.857438230(24) 𝜇𝜇 N
3.3435869(20) x 10-27 kg
2.103553214(24) u
1875.61339(57) MeV e-2
0.4664345460(91) x 10-3
3670.483014(75)
0.3070122035(51)
1.999007496(6)
1.054571596(82) x 1034 J s
6.58211889(26) x 10-16 eVs
1.771845195 x 10-3 kg
0.00390625 lb
0.0625 oz
1.771845195 x 10-3 kg
0.00390625 lb
0.0625 oz
5.500610469 x 10-4 m3
2.718281828459045235602
874713526
5.517 x 103 kg m-3
R
6.37 x 106 m
Earth’s Magnetic
Field, Horizontal
Earth’s Mass
Earth – Moon Mean
Distance
Electric Constant
Electrical Power
Specification (UK)
Electrical Power
Specification (US)
Electron Charge
Electron Charge to
Mass Quotient
Electron – Factor
Electron
Gyromagnetic Ratio
Electron Magnetic
Moment
Electron Magnetic
Moment Anomaly
Bo
1.8 x 10-5 T
M
5.972 x 1024 kg
3844 x 108 m
𝜀𝜀 o
E
e/h
-e/me
ge
Ye
Ye/2𝜋𝜋
𝜇𝜇 e
𝜇𝜇 e/ 𝜇𝜇𝛽𝛽
𝜇𝜇 e/ 𝜇𝜇 N
ae
me
Electron Radius,
Classical
Electron Specific
Charge
Electron to Shielded
Helion Magnetic
Moment Ratio
Electron to Shielded
Proton Magnetic
230 V ac, 50Hz
115V ac, 50Hz
Electron Mass
Electron Molar Mass
8,854187817 x 10-12 F m-3
M(e) Me
1.602176462(63) x 10-19 C
2.417989491(95) x 1014 A J1
-1.758820174(71) x 1011 C
kg-19
2.0023193043737(82)
1.760859794(71) x 1011 s-1
T-1 , 28024.954(11) MHz T-1
-928.476362(37) x 10-26 J T-1
-1.0011596521869(41)
-1838.2819660(39)
1.1596521869(41) x 10-3
9.10938188(72) x 10-31 kg
5.485799110(12) x 10-4 u
0.510998902(21) MeV
5.48579911(12) x 10-7 kg
mol-1
re
2.817940285(31) x 10-15 m
-e/mo
-1.758819(62) x 1011 c kg-1
𝜇𝜇 e/𝜇𝜇’h
864.058255(10)
𝜇𝜇 e/𝜇𝜇’p
-658.2275954(71)
Moment Ratio
Electron – 𝛼𝛼 –
Particle Mass Ratio
Electron – Deuteron
Magnetic Moment
Ratio
Electron – Deuteron
Mass Ratio
Electron – Muon
Magnetic Moment
Ratio
Electron – Neutron
Mass Ratio
Electron – Proton
Magnetic Moment
Ratio
Electron – Proton
Mass Ratio
Electron – Proton
Moment Ratio
Electron – Proton
Ration
Electron – Tau Mass
Ratio
Electronvolt
Electronvolt, Million
Elementary Charge
emf of Daniell Cell
emf of Lechlanche
Cell
emf of Nife Cell
emf of Weston Cell
Energy Production,
Sun’s
me/md
𝜇𝜇 e/𝜇𝜇 d
me/md
𝜇𝜇 e/𝜇𝜇 d
me/md
𝜇𝜇 e/𝜇𝜇 p
me / m0
1.3709335611(29) x 10-4
-2143.923498(23)
2.7244371170(58) x 10-4
206.7669720(63)
4.83633210(15) x 10-3
960.92050(23)
5.438673462(12) x 10-4
𝜇𝜇 e/𝜇𝜇 p
-658.2106875(66)
me/mp
5.446170232(12) x 10-4
me / m1
2.87555(47) x 10-4
MeV
eV
E
e/h
1.60217733 x 10-19 J
1.6 x 10-13 J
1.602176462(63) x 10-19 C
2.41798949(95) x 1014 A J-1
1.08V
1.46V
1.40V
1.0186V
3.90 x 1026 W
Energy, Hartree
Euler’s Constant
Faraday Constant
Eh
YC
F
fath
Fathom
Feigenbaum’s
Constant
Fermi Constant
Fermi Coupling
Constant
Fire Structure
Constant
First Radiation
Constant
First Radiation
Constant for
Spectral Radiance
Fluid Ounce, UK
Fluid Ounce, US
Foot
Foot Board (Timber)
Fransen – Robinson
Constant
Free Nuetron, Half
Life
Free Space,
Permeability of
Free Space,
Permeability of
g – factor, Electron
g – factor, Muon
g – Factor, Neutron
𝜎𝜎
Gf/(hc)3
𝛼𝛼
4.35974381(34) x 10-18 J
27.2113834(11) eV
0.5772156690153286061
96485.3415(39) C mol-1
1.3288 m
72 in
6 ft
4.6692106091012990
1.4 x 10-50 J m-3
1.16639(1) x 10-5 GeV-2
7.297352533(27) x 10-3
c1
3.74177107(29) x 10-16 W
m2
cu
1.191042722(93) x 10-16 W
m2 sr-1
floz
floz
ft
ftm
2.841 x 10-5 m3
2.957 x 10-5 m3
0.3048 m
2.359737216 x 10-3 m3
2.8077702420
T
650 s
uo
4 𝜋𝜋 x 10-7 N A-2
12.566370614 x 10-7 N A-2
8.8541187817 x 10-12 F m-4
𝜀𝜀 0
ge
gu
gn
-2.00231930437337(82)
-2.00023318320(13)
-3.82698548(90)
g – Factor, Proton
gp
ng
5.585694675(57)
4.546 x 10-3 m3
1.201 US Gallon
3.785 x 103 m3
0.833 UK gallon
8.314 J K-1 mol-1
8.314472(15) J K-1 mol-1
14.59390294 kg
1.50
kg
1.0 W m-1K-1
𝜑𝜑
𝜑𝜑
𝜆𝜆
1.6180339887498948420
1.6180339887498948420
0.6243299885
Gallon, UK
Gallon, US
Gas Constant
Gas Constant, Molar
Gee pound
Glass, Refractive
Index of
Glass, Thermal
Conductivity of
Golden Number
Golden Mean
Golomb – Dickman
Constant
Grain
Gravitational,
Newtonian Constant
of
Gravitational,
Acceleration,
Moon’s
Gravitational
Constant
Gravity, Acceleration
due to
Gunter’s Chain
Gyromagnetic Ratio,
Electron
Gyromagnetic Ratio,
Neutron
Gyromagnetic Ratio,
Proton
Gyromagnetic Ratio,
R
R
G
G/hc
0/06479 g
6.673(10) x 10-11 m3 kg-1 s-1
6.707(10) x 10-39 (GeV/c2)-2
1.619 m s-2
G
6.673(10) x 10-11 N m2 kg-2
g
9.80665 m s-2
32.1740 ft s-2
20.1168 m
ch
ye
ye/2𝜋𝜋
yn
yn/2𝜋𝜋
yp
yp/2𝜋𝜋
y’p
1.760859794(71) x 1011 s-1 T-1
20024.9540(11) MHz T-1
1.83247188(44) x 108 s-1 T-1
29.1646958(70) MHz T-1
2.675222121(11) x 108 s-1T-1
42.5774825(18) MHz T-1
2.67515341(11) x 108 s-1 T-1
shielded proton
Hafner – Sarnak –
McCurley Constant
Half – Life of Carbon
– 14
Half – Life of Free
Neutron
Hard Square
Entropy constant
Hartee Energy
Hectare
Horizontal
Component of
Earth’s Magnetic
Field
Hundredweight (UK)
Hundredweight short
(Us)
Hydrogen Rydberg
number
Impedence of
Vacuum,
Characteristic
Imperial Pint
inch
International
Nautical Mile
Inverse
Conductance
Quantum
Inverse Fine
Structure Constant
Josephson
Frequency – Voltage
Quotient
y’p/2𝜋𝜋
42.57638888(18) MHz T-1
0.3532363719
T
5570 years
T
650 s
k
1.503048082
Eh
4.35974381(34) x 10-18 J
27.2113834(11) eV
10,000 m2
1.8 x 10-5 T
ha
B0
cwt
short
cwt
RH
5.08 x 104 g
4.535 x 104 g
1.0967758 x 107 m-1
Zo
376.73031461 ohms
in ”
5.682 x 10-4 m3
0.254 m
1852 m
Go-1
12906.403786(47) ohms
𝛼𝛼
137.03599976(50) x 10-3
Ki
2e/h
4.83597898(19) x 1014 Hz V1
Khintchine’s
Constant
Landau –
Ramanujan
Constant
Laplace Limit
Constant
Lechlanche Cell,
emf of
Length, Planck
Lengyel’s Constant
Lieb’s Square Ice
Constant
Light year
Light, Speed of
Linear Expansivity of
Copper
Linear Expansivity of
Steel
Loschmidt constant
Loschmidt constant
Madelung’s
Constant
Magnetic Constant
Magnetic Flux
Quantum
Magnetic Moment
Anomaly, Electron
Magnetic Moment
Anomaly, Muon
Magnetic Moment
Ratio Deuteron –
Electron
K
2.68542001
K
0.764223653
𝜆𝜆
0.6627434193
lp
1.6160(12) x 10-35 m
1.0986858055
1.5396007178
Λ
ly
c
𝛼𝛼
𝛼𝛼
no
Vm
M2
uo
𝜙𝜙
1.46 V
9.46052974 x 1015 m
299792458 m s-1
1.7 x 10-5 K-1
1.2 x 10-5 K-1
2.6867775(47) x 1025 m-3
22.710981(40) x 10-3 m3
mol-1
-1.6155426267
4𝜋𝜋 x 10-7 N A-2
12.566370614 x 10-7 N A-2
2.067833636(81) x 10-15 Wb
ae
1.1596521869(41) x 10-3
au
1.16591602(64) x 10-3
ud/ue
0.45554345460(91) x 10-3
Magnetic Moment
Ratio, Deuteron –
Proton
Magnetic Moment
Ratio, Electron to
Shilded Helion
Magnetic Moment
Ratio, Electron to
Shilded Proton
Magnetic Moment
Ratio, Electron –
Deuteron
Magnetic Moment
Ratio Electron –
Muon
Magnetic Moment
Ratio, Electron –
Neutron
Magnetic Moment
Ratio, Muon –
Proton
Magnetic Moment
Ratio, Neutron to
Shielded Proton
Magnetic Moment
Ratio Neutron –
Electron
Magnetic Moment
Ratio, Proton –
Neutron
Magnetic Moment
Deuteron
Magnetic Moment
Electron
Magnetic Moment,
ud/up
0.3070122035(51)
ue/u’h
864.058255(10)
ue/u’p
-658.2275954(71)
u/ud
-21.43.923498(23)
ue/un
206.7669720(63)
ue/un
960.92050(23)
uu/up
-658.2106875(66)
un/u’p
-/068499694(16)
un/ue
1.04066882(25) x 10-3
up/un
-1.45989805(34)
ud
ue
uu
0.43307375(15) x 10-26 J T-1
0.4669754479 x 10-3 uB
0/857438230(24) un
-928.476362(37) x 10-36 J T-1
-1.0011596521869(41)
-1.838.2819660(39)
-4.49044813(22) x 10-26 J T-1
Muon
Magnetic Moment,
Neutron
Magnetic Moment,
Proton
Magneton, Bohr
Magneton, Nuclear
Mass Ratio,
Deuteron – Electron
Rest
Mass Ratio,
Deuteron – Proton
Mass Ratio Electron
– 𝛼𝛼 – Particle
Mass Ratio Electron
– Deuteron
Mass Ratio Electron
– Proton
Mass Ratio Electron
– tau
Mass Ratio, Muon –
Electron
Mass Ratio, Muon –
Neutron
Mass Ratio, Muon –
Proton
Mass Ratio, Muon –
Tau
Mass Ratio, Neutron
– Muon
un
up
uB
un
-4.84197085(15) x 10-3
-8.89059770(27)
-0.9662364(23) x 10-26 J T-1
-1.04187563(25) x 10-3
-1.91304272(45)
1.41060663(58) x 10-26 J T-1
1.521032203(15) x 10-3
2.792847337(29)
9.27400899(37) x 10-26 J T-1
5.788381749(43) x 10-5 eV T-1
1.399624624(56) x 1010 Hz T-1
5.05078317(20) x 10-27 J T-1
3.1524541238(24) x 10-8 eV T-1
7.62259396(31) MHz T-1
md/me
3670.483014(75)
md/mp
1.999007469(6)
me/ma
1.3709335611
me/md
2.724437117(58) x 10-4
me/mp
4.83633219(15) x 10-4
me/mt
2.87555(47) x 10-4
mu/me
206.7682657(63)
mu/mn
0.1124545079(34)
mu/mp
0.1126095173(34)
mu/mt
5.94572(97) x 10-2
mn/mu
1.8386836550(40)
Mass Ratio, Neutron
– Proton
Mass Ratio,
Neutron- Tau
Mass Ratio, Proton
–electron
Mass Ratio, Proton
– Muon
Mass Ratio, Proton
– Neutron
Mass Ratio Proton –
Tau
Mass Ratio Tau –
Electron
Mass Ratio Tau –
Muon
Mass Ratio Tau –
Neutron
Mass Ratio Tau –
Proton
Mass, Deuteron
Mass, Earth’s
Mass, Electron
Mass Muon
Mass Neutron
Mass Planck
Mass Proton
Mass, Sun’s
Mass Tau
Mile
mile, square
Million Electronvolts
Mill’s Constant
mn/mp
8.89248278(27)
mn/mt
0.528722(86)
mp/me
1836.1526675(39)
mp/mu
8.88024408(27)
mp/mn
0.99862347855(58)
mp/mt
0.527994(86)
mt/me
3.47760(57)
mt/mu
16.8188(27)
mt/mn
1.89135(31)
mt/mp
1.89396(31)
md
M
me
mu
mn
mp
mp
mt
MeV
8.34358360(29) x 10-27 kg
2.013553214(24) u
5.972 x 1024 kg
9.10938188(72) x 10-33 kg
1.8835109(16) x 10-28 kg
1.67492716(13) x 10-27 kg
2.1767(16) x 10-8 kg
1.6726158(13) x 10-27 kg
1.99 x 1030 kg
3.16788(52) x 10-27 kg
1609.344 m
2.589 x 106 m2
1.6 x 10-13 J
1.3063778838
Molar Gas Constant
Molar Mass
Deuteron
Molar Mass Electron
Molar Mass Muon
Molar Mass Neutron
Molar Mass Proton
Molar Mass Tau
Molar Planck
Constant
Molar Volume
Moon’s Gravitational
Accelration
Moon’s Mean
Distance From Earth
Moon’s Mean Mass
Moon’s Mean
Radius
Muon Compton
Wave lenght
Muon g – Factor
Muon Magnetic
Moment
Muon Magnetic
Moment Anomaly
Muon Mass
Muon Molar Mass
Naperian (natural)
logarithm of 10
Naperian (natural)
R
Md
8.314472(15) J mol-1 K-1
2.013553214(24) u
Me
5.48579911(12) x 107 kg
mol-1
0.113428917(34) x 10-3 kg
mol-1
1.0066491578(55) x 10-3 kg
mol-1
1.00727646688(13) x 10-3
kg mol-1
1.90774(31) x 10-3 kg mol-1
3.990312689(30) x 10-10 J s
mol-1
22.413996 x 10-3 m3 mol-1
3.33 x 103 kg m-3
Mu
Mn
Mp
Mt
NAh
Na hc
Vm
3.844 x 108 m
7.33 x 1022 kg
1.738 x 106 m
𝜆𝜆cu
11.73444197(35) x 10-15 m
gu
uu
-2.0023318320(13)
-4.49044813(22) x 1026 J T-1
au
1.116591602(64) x 10-3
mu
Mu
ln 10
1.883353109(16) x 10-28 kg
0.113428916(34) x 10-3 kg
mol-1
2.3025809299404568404
ln 2
0.69314718055994530942
logarithm of 2
natural logarithm of
10
natural logarithm of
2
Number Gold
ounce troy
Paraffin, refrective
Index of
Pi
Porter’s Constant
Pounds per acre
Pounds per Foot
Pounds per Inch
Pounds per Square
Foot
Pounds per Square
Inch
Pounds per yard
Pythagoras’
Constant
Quantized Hall
Conductance
Quantized Hall
Resistance
Quantum of
circulation
Quantum
Conductance
Short ton
Siegbahn Unit
Sierpinski’s
Constant
ln 10
2.3025809299404568404
ln 2
0.69314718055994530942
𝜑𝜑
oz
np
1.618033988874989484820
31.103 g
1.42
𝜋𝜋
3.141592653589793238462
6433832795
1.4670780794
0.112 g m-2
C
lb acre1
lb ft-1
lb in-1
lb ft-2
1.488 kg m-1
17.85 kg m-1
4.882 x 103 g m-2
lb in-2
7.03 x 105 g m-2
lb yd-1
0.496 kg m-1
1.412135624
e2/h
3.87404614 x 10-5 S
RH
25812.8056 ohms
h/2me
3.63947516(27) x 10-4 m2 s-1
Go1
12906.403786(47) ohms
UX
K
2000lb
1.0002077897 x 10-13 m
2.5849817596
Sigma
Siriometre
Siriusweit
Slug
Solar Constant
Specific Heat
Capacity of Copper
Specific Latent Heat
of Fusion of Water
Specific Latent Heat
of Vapourisation of
Water
Ton
Van der Corput’s
Constant
Verdet’s Constant
Viscosity of Air at
20oC
Viscosity of Water at
20oC
Weak Mixing Angle
cc
1 x 1012 m
1.49597870 x 1015 m
1.5428387847 x 1017 m
14.59390294 kg
1400 W m-2
385 J kg-1 K-1
cw
4200 J kg-1 K-1
2.28 x 106 J kg-1
m
106 kg
3.3643175781
0.000477 rad A-1
1.8 x 10-3 N s m-2
1.002 x 10-3 N s m-2
sin2𝜃𝜃 w
0.2224(19)
FACTORS FOR CONVERSION TO S.I. UNITS OF
MEASUREMENTS
To convert from
acre foot
acre
angstrom
atm (standard)
atm (technical)
To
Multiply by
m3
m2
meter, m
Pascal, Pa
Pa
1,233489
4,046.873
1 x 10-10
1.013250 x 105
9.80665 x 104
bar
barrel
board ft
Btu
Btu (British
Thermal Unit) (k –
int’l)
Btu ( int’l table) h
Btu (C, Thermal
Conductance)
Btu
Btu
bushel (U.S.)
Calorie
cm of hg (0oC)
cm of hg (4oC)
chain
circular mil
Cubit
Day
day (sidereal)
degree (angle)
degree Celsius
degree
Fahrenheit
degree
Fahrenheit
degree Rankine
Dyne
Fathom
Foot
Pa
m3
m3
Joule, J
W/(m-K)
1 x 105
1.589873 x 10-1
2.359737 x 10-3
1,055.87
1.442279 x 10-1
Watt, W
W/(m2-k)
2.930711 x 104
5.678263
J/kg
J/m3
m3
J
Pa
Pa
m
m2
m
Second, s
s
rad
K
o
C
2,326
3.725895 x 104
3.523907 X 10-2
4.19002
98.0638
20.11684
20.11684
5.067075 x 10-10
4.57 x 10-4
8.64 x 104
8.616409 x 104
1.745329 x 10-2
Tk = tc + 273.15
tc = (tf-32)/1.8
K
K
N
m
m
Tk = (tf +
459.67)11.8
Tk = TR / 1.8
1 x 10-5
1.828804
3.048006 x 10-1
foot, ft, of water
ft2
ft3
footcandle, fc
footlambert, fL
ft – lbf
ft – lbf / min
ft pundal
gallon UK
gallon US (dry)
gallon US (liquid)
Grad
Grad
grain, gr
gram, g
hectare, ha
Hp
hp (boiler)
hp(electric)
hp (water)
hp (UK)
hour, h
Inch
inch squared
inch cube
in/s
Kgf
kgf-s2/m
kg – m
Pa
m2
m3
lux, lx
cd/m2
J
W
J
m3
m3
m3
2
Rad
Kg
Kg
m2
W
W
W
W
W
second, s
M
m2
m3
m/s
N
Kg
N–m
2,988.98
9.290304 x 10-2
2.831685 x 10-2
10.76391
3.426259
1.355818
2.259697 x 10-2
4.214011 x 10-2
4.546092 x 10-3
4.404884 x 10-3
3.785412 x 10-3
0.09
1.570796 x 10-2
6.479891 x 10-5
0.001
0.0001
745.69999
9,809.5
746
746.043
745.70
3600
0.00254
0.00064516
1.638706 x 10-5
2.54 x 10-2
9.80665
9.80665
9.80665
kgf/cm2
kgf/m2
kWh
kip
Ksi
knot, kn
lambert, L
Liter
maxwell
Mho
micro inch
Micron
mil, mi
mile, mi (int’l)
mi (US)
mi (int’l nautical)
mi (US nautical)
mi2
mi2 (US)
mi/hr
mi/hr (US)
Mbar
mm of Hg
minute (angle)
minute
minute (sidereal)
Oz
oz (troy)
oz (UK fluid)
Pa
Pa
J
N
Pa
m/s
cd/m2
m3
weber, Wb
S
M
M
M
M
M
M
M
m2
m2
m/s
m/h
Pa
Pa
Rad
Sec
Sec
Kg
Kg
m3
9.80665 x 10-4
9.80665
3600000
4,448.22
6.894757 x 106
5.1444444 x 104
3,183.099
0.001
1 x 10-8
1
2.54 x 10-8
0.000001
2.54 x 10-5
1.609344 x 10-3
1,609.347
1,852
1,852.00
2.589988 x 106
2.589988 x 106
4.474 x 106
1.609344
100
133.322
2.908882 x 10-4
60
59.83617
2.834952 x 10-2
3.110348 x 10-2
2.841307 x 10-5
oz (US fluid)
Ozf
ozf – in
poise p
pound, lb
Rod
Second
second (sidereal)
square foot
(100ft2)
Ton
ton(long)
ton(metric)
ton force (200 lbf)
tonne, t
Vara
Wh
yard, yd
yd2
yd3
year (365 days)
year (sidereal)
Prefixes
m3
N
N–m
Pa – s
Kg
M
Rad
S
m2
2.957353 X 10-5
2.780139 x 10-1
7.061552 x 10-3
0.01
4.535924 x 10-1
5.02921
4.848137 x 10-6
9.972696 x 10-1
9.290304
Kg
Kg
Kg
N
Kg
M
J
M
m2
m3
second, s
S
2.916667 x 10-2
1,016.047
1,000
8,896.444
1,000
8.38 x 10-4
3,600
9.144 x 10-4
8.361274 x 10-4
7.645549 x 10-4
3.153600 x 107
3.155815 x 107
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
Yocto (y) ……………….…10-24
Zepto (z) ……………….…10-21
Atto (a) …………………....10-18
Femto (f) …………………10-15
Pico (p)…………………....10-12
Nano (n)…………………..10-9
Mirco (u)…………………..10-6
Centimilli (cm)…………….10-5
Decimille (dm)…………….10-4
Milli (m)…………………….10-3
Centi (c)……………………10-2
Deci (d)…………………….10-1
Deca (D)…………………...101
Hecto (h)…………………...102
Kilo (K)……………………..103
Mega (M)…………………..106
Giga (G)……………………109
Tera (T)…………………….1012
Peta (P)…………………….1015
Exa (E)……………………..1018
Zetta (Z)……………………1021
Yotta (Y)…………………...1024