Draw Impedance Triangle And Show Various Quantities On It

RC Circuit Analysis

John Clayton Rawlins M.S. , in Basic AC Circuits (Second Edition), 2000

Pythagorean Theorem Applied to Impedance Solutions

The Pythagorean theorem can be applied to circuit problems involving resistance and reactance. Figure 7.15 shows the resistive-reactance phasor diagram for a series RC circuit. The total impedance of the circuit, Z, is the vector sum of the resistance, R, and reactance, X_C. If the values of the resistance and reactance are known, the impedance can be calculated using the Pythagorean theorem.

Figure 7.15. Pythagorean Theorem Can Be Used to Calculate Impedance on Vector Diagram

Notice that the length of the reactance vector is the same as the length between the tip of the resistance vector and the tip of the impedance vector. In fact, the vector diagram can be drawn with the reactance vector placed as shown in Figure 7.16 . Now, the right triangle of the Pythagorean theorem becomes evident. Applying the Pythagorean theorem, the total impedance (the hypotenuse) can be calculated,

Figure 7.16. Vector Diagram

Redrawn with the Reactance Vector Obvious

(7–15). $\text{Z\hspace{0.25em}=}\sqrt{{\text{R}}^{\text{2}}{\text{+X}}_{\text{C}}{}^2}$

Total impedance is equal to the square root of the resistance squared plus the reactance squared.

Figure 7.17 shows a typical series RC circuit. The impedance can be calculated using equation 7–15 . The resistance of this RC circuit is 12 ohms and the capacitive reactance is 16 ohms.

Figure 7.17. Example Circuit for Calculating Impedance Using Equation 7–15

(7–15). $\begin{array}{l}\text{Z\hspace{0.25em}=}\sqrt{{\text{R}}^2+{\text{\hspace{0.25em}X}}_{\text{C}}{}^2}\\ \text{\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=}\sqrt{{\text{12}}^2+{\text{\hspace{0.25em}16}}^2}\\ \text{\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=}\sqrt{144+256}\\ \text{\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=}\sqrt{400}\\ \text{\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=\hspace{0.25em}20}\Omega \end{array}$

Thus, the total impedance of the circuit is 20 ohms.

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Number Theory, Elementary

Robert L. Page , in Encyclopedia of Physical Science and Technology (Third Edition), 2003

I.E Pythagorean Triples

The Pythagorean theorem states that the sum of the squares of the legs of a right triangle equals the square of its hypotenuse, that is, a ²  + b ²  = c ², as shown in Fig. 1. This result was certainly known before the time of Pythagoras, but whether he was the first to actually prove the theorem is unknown because of the Pythagoreans' custom of ascribing all new knowledge to the Master.

FIGURE 1. The Pythagorean theorem: a ²  + b ²  = c ².

Numbers whose values satisfy the Pythagorean theorem, such as 3, 4, and 5(3²  +   4²  =   9   +   16   =   25   =   5²), are permissible values for the sides of a right triangle. In Egypt, men known as rope stretchers made use of the 3, 4, 5 relationship to establish right angles in surveying property or constructing buildings by stretching a rope with knots marking lengths of 3, 4, and 5 units around three stakes, as in Fig. 2.

FIGURE 2. Rope stretching.

Trial and error leads to the discovery of many such Pythagorean triples, for instance, 5, 12, 13 and 8, 15, 17. It is natural to inquire whether formulas exist that will generate sets of Pythagorean triples. One method, which is attributed to Pythagoras, is as follows: Let n be any positive integer. Then a  =   2n  +   1, b  =   2n ²  +   2n, and c  =   2n ²  +   2n  +   1 constitute a Pythagorean triple. If n  =   3, we obtain 7, 24, 25, for which 7²  +   24²  =   49   +   576   =   625   =   25². For n  =   4   we have 9, 40, 41, whereby 9²  +   40²  =   81   +   1600   =   1681   =   41². All triples generated in this way lead to triangles having a hypotenuse that is one unit longer than the larger leg.

A more general method consists of choosing integers u and v, one of which is odd and the other even, such that u and v have no common divisors other than the number 1. We then say that u and v are relatively prime. Then, letting a  =   2uv, b  = u ²  − v ², and c  = u ²  + v ² gives the desired result. The following examples illustrate this method.

Multiplying each member of a Pythagorean triple by the same integer yields another Pythagorean triple.For example,

$2\left(3,4,5\right)=6,8,10,$

and

$6^2+8^2=36+64=100={10}^2.$

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Dynamics of Point Masses

Howard D. Curtis , in Orbital Mechanics for Engineering Students (Second Edition), 2010

1.2 Vectors

A vector is an object that is specified by both a magnitude and a direction. We represent a vector graphically by a directed line segment, that is, an arrow pointing in the direction of the vector. The end opposite the arrow is called the tail. The length of the arrow is proportional to the magnitude of the vector. Velocity is a good example of a vector. We say that a car is traveling east at eighty kilometers per hour. The direction is east and the magnitude, or speed, is 80   km/h. We will use boldface type to represent vector quantities and plain type to denote scalars. Thus, whereas B is a scalar, B is a vector.

Observe that a vector is specified solely by its magnitude and direction. If A is a vector, then all vectors having the same physical dimensions, the same length and pointing in the same direction as A are denoted A, regardless of their line of action, as illustrated in Figure 1.1. Shifting a vector parallel to itself does not mathematically change the vector. However, parallel shift of a vector might produce a different physical effect. For example, an upward 5   kN load (force vector) applied to the tip of an airplane wing gives rise to quite a different stress and deflection pattern in the wing than the same load acting at the wing's mid-span.

Figure 1.1. All of these vectors may be denoted A, since their magnitudes and directions are the same.

The magnitude of a vector A is denoted $\Vert A\Vert$ , or, simply A.

Multiplying a vector B by the reciprocal of its magnitude produces a vector which points in the direction of B, but it is dimensionless and has a magnitude of one. Vectors having unit dimensionless magnitude are called unit vectors. We put a hat $(^)$ over the letter representing a unit vector. Then we can tell simply by inspection that, for example, $\widehat{u}$ is a unit vector, as are $\widehat{B}$ and $\widehat{e}$ .

It is convenient to denote the unit vector in the direction of the vector A as ${\widehat{u}}_A$ . As pointed out above, we obtain this vector from A as follows

(1.1) ${\widehat{u}}_A\text{=}\frac{A}{A}$

Likewise, ${\widehat{u}}_C\text{=}C\text{/}C$ , ${\widehat{u}}_F\text{=}F\text{/}F$ , etc.

The sum or resultant of two vectors is defined by the parallelogram rule (Figure 1.2). Let $C$ be the sum of the two vectors A and B. To form that sum using the parallelogram rule, the vectors A and B are shifted parallel to themselves (leaving them unaltered) until the tail of A touches the tail of B. Drawing dotted lines through the head of each vector parallel to the other completes a parallelogram. The diagonal from the tails of A and B to the opposite corner is the resultant C. By construction, vector addition is commutative, that is,

Figure 1.2. Parallelogram rule of vector addition.

(1.2) $A\text{+}B\text{=}B\text{+}A$

A Cartesian coordinate system in three dimensions consists of three axes, labeled x, y and z, which intersect at the origin O. We will always use a right-handed Cartesian coordinate system, which means if you wrap the fingers of your right hand around the z axis, with the thumb pointing in the positive z direction, your fingers will be directed from the x axis towards the y axis. Figure 1.3 illustrates such a system. Note that the unit vectors along the x, y and z-axes are, respectively, $\widehat{i}$ , $\widehat{j}$ and $\widehat{k}$ .

Figure 1.3. Three-dimensional, right-handed Cartesian coordinate system.

In terms of its Cartesian components, and in accordance with the above summation rule, a vector A is written in terms of its components $A_x$ , $A_y$ and $A_z$ as

(1.3) $A\text{=}{A}_x\widehat{i}\text{+}{A}_y\widehat{j}\text{+}{A}_z\widehat{k}$

The projection of A on the xy plane is denoted $A_{xy}$ . It follows that

$A_{xy}\text{=}{A}_x\widehat{i}\text{+}{A}_y\widehat{j}$

According to the Pythagorean theorem, the magnitude of A in terms of its Cartesian components is

(1.4) $A\text{=}\sqrt{A_x^2\text{+}{A}_y^2\text{+}{A}_z^2}$

From Equations 1.1 and 1.3, the unit vector in the direction of A is

(1.5) ${\widehat{u}}_{\text{A}}\text{=}\cos {\theta }_x\widehat{i}\text{+}\cos {\theta }_y\widehat{j}\text{+}\cos {\theta }_z\widehat{k}$

where

(1.6) $\cos {\theta }_x\text{=}\frac{A_x}{A}\text{undefined}\cos {\theta }_y\text{=}\frac{A_y}{A}\text{undefined}\cos {\theta }_z\text{=}\frac{A_z}{A}$

The direction angles θ_x , θ_y and θ_z are illustrated in Figure 1.4, and are measured between the vector and the positive coordinate axes. Note carefully that the sum of θ_x , θ_y and θ_z is not in general known a priori and cannot be assumed to be, say, 180 degrees.

Figure 1.4. Direction angles in three dimensions.

Example 1.1

Calculate the direction angles of the vector $A\text{=}\widehat{i}\text{-}4\widehat{j}\text{+}8\widehat{k}$ .

Solution

First, compute the magnitude of A by means of Equation 1.4:

$A\text{=}\sqrt{1^2\text{+}{(\text{-}4)}^2\text{+}{8}^2}\text{=}9$

Then Equations 1.6 yield

Observe that θ_x + θ_y + θ_z = 227.3°.

Multiplication and division of two vectors are undefined operations. There are no rules for computing the product $AB$ and the ratio $A/B$ . However, there are two well-known binary operations on vectors: the dot product and the cross product. The dot product of two vectors is a scalar defined as follows,

(1.7) $A·B\text{=}AB\cos \theta$

where θ is the angle between the heads of the two vectors, as shown in Figure 1.5. Clearly,

Figure 1.5. The angle between two vectors brought tail to tail by parallel shift.

(1.8) $A\text{·}B\text{=}B\text{·}A$

If two vectors are perpendicular to each other, then the angle between them is 90°. It follows from Equation 1.7 that their dot product is zero. Since the unit vectors $\widehat{i}$ , $\widehat{j}$ and $\widehat{k}$ of a Cartesian coordinate system are mutually orthogonal and of magnitude one, Equation 1.7 implies that

(1.9) $\begin{array}{l}\widehat{i}\text{·}\widehat{i}\text{=}\widehat{j}\text{·}\widehat{j}\text{=}\widehat{k}\text{·}\widehat{k}\text{=}1\\ \widehat{i}\text{·}\widehat{j}\text{=}\widehat{i}\text{·}\widehat{k}\text{=}\widehat{j}\text{·}\widehat{k}\text{=}0\end{array}$

Using these properties it is easy to show that the dot product of the vectors A and B may be found in terms of their Cartesian components as

(1.10) $A\text{·}B\text{=}{A}_x{B}_x\text{+}{A}_y{B}_y\text{+}{A}_z{B}_z$

If we set B = A, then it follows from Equations 1.4 and 1.10 that

(1.11) $A\text{=}\sqrt{A\text{·}A}$

The dot product operation is used to project one vector onto the line of action of another. We can imagine bringing the vectors tail to tail for this operation, as illustrated in Figure 1.6. If we drop a perpendicular line from the tip of B onto the direction of A, then the line segment B_A is the orthogonal projection of B onto line of action of A. B_A stands for the scalar projection of B onto A. From trigonometry, it is obvious from the figure that

Figure 1.6. Projecting the vector B onto the direction of A.

$B_A\text{=}B\cos \theta$

Let ${\widehat{u}}_A$ be the unit vector in the direction of A. Then

$B\text{·}{\widehat{u}}_A\text{=}\Vert B\Vert \stackrel{1}{\overbrace{\Vert {\widehat{u}}_A\Vert }}\cos \theta \text{=}B\cos \theta$

Comparing this expression with the preceding one leads to the conclusion that

(1.12) $B_A\text{=}B\text{·}{\widehat{u}}_A\text{=}B\text{·}\frac{A}{A}$

where ${\widehat{u}}_A$ is given by Equation 1.1. Likewise, the projection of A onto B is given by

$A_B\text{=}A\text{·}\frac{B}{B}$

Observe that A_B = B_A only if A and B have the same magnitude.

Example 1.2

Let $A\text{=}\widehat{i}\text{+}6\widehat{j}\text{+}18\widehat{k}$ and $B\text{=}42\widehat{i}\text{-}69\widehat{j}\text{+}98\widehat{k}$ . Calculate

(a)

The angle between A and B;

(b)

The projection of B in the direction of A;

(c)

The projection of A in the direction of B.

Solution

First we make the following individual calculations.

(a) $A\text{·}B\text{=}(1)(42)\text{+}(6)(\text{-}69)\text{+}(18)(98)\text{=}1392$

(b) $A\text{=}\sqrt{{(1)}^2\text{+}{(6)}^2\text{+}{(18)}^2}\text{=}19$

(c) $B\text{=}\sqrt{{(42)}^2\text{+}{(\text{-}69)}^2\text{+}{(98)}^2}\text{=}127$

(a)

According to Equation 1.7, the angle between A and B is

$\theta \text{=}{\cos }^{\text{-}1}\left(\frac{A\text{·}B}{AB}\right)$

Substituting (a), (b) and (c) yields

(b)

From Equation 1.12 we find the projection of B onto A:

$B_A\text{=}B\text{·}\frac{A}{A}\text{=}\frac{A\text{·}B}{A}$

Substituting (a) and (b) we get

(c)

The projection of A onto B is

$A_B\text{=}A\text{·}\frac{B}{B}\text{=}\frac{A\text{·}B}{B}$

Substituting (a) and (c) we obtain

The cross product of two vectors yields another vector, which is computed as follows,

(1.13) $A\text{×}B\text{=}\text{(}AB\text{sin}\theta \text{)}{\widehat{n}}_{AB}$

where θ is the angle between the heads of A and B, and ${\widehat{n}}_{AB}$ is the unit vector normal to the plane defined by the two vectors. The direction of ${\widehat{n}}_{AB}$ is determined by the right hand rule. That is, curl the fingers of the right hand from the first vector (A) towards the second vector (B), and the thumb shows the direction of ${\widehat{n}}_{AB}$ . See Figure 1.7. If we use Equation 1.13 to compute B × A, then ${\widehat{n}}_{AB}$ points in the opposite direction, which means

Figure 1.7. ${\widehat{n}}_{AB}$ is normal to both A and B and defines the direction of the cross product A × B.

(1.14) $B\text{×}A\text{=-}(A\text{×}B)$

Therefore, unlike the dot product, the cross product is not commutative.

The cross product is obtained analytically by resolving the vectors into Cartesian components.

(1.15) $A\text{×}B\text{=}\text{(}{A}_x\widehat{i}\text{+}{A}_y\widehat{j}\text{+}{A}_z\widehat{k}\text{)}\text{×}\text{(}{B}_x\widehat{i}\text{+}{B}_y\widehat{j}\text{+}{B}_z\widehat{k}\text{)}$

Since the set $\widehat{i}\widehat{j}\widehat{k}$ is a mutually perpendicular triad of unit vectors, Equation 1.13 implies that

(1.16) $\begin{array}{l}\widehat{i}\text{×}\widehat{i}\text{=}0\text{undefined}\text{undefined}\text{undefined}\text{undefined}\text{undefined}\widehat{j}\text{×}\widehat{j}\text{=}0\text{undefined}\text{undefined}\text{undefined}\text{undefined}\text{undefined}\text{undefined}\text{undefined}\widehat{k}\text{×}\widehat{k}\text{=}0\\ \widehat{i}\text{×}\widehat{j}\text{=}\widehat{k}\text{undefined}\text{undefined}\text{undefined}\text{undefined}\text{undefined}\text{undefined}\widehat{j}\text{×}\widehat{k}\text{=}\widehat{i}\text{undefined}\text{undefined}\text{undefined}\text{undefined}\text{undefined}\text{undefined}\widehat{k}\text{×}\widehat{i}\text{=}\widehat{j}\end{array}$

Expanding the right side of Equation 1.15, substituting Equation 1.16 and making use of Equation 1.14 leads to

(1.17) $A\text{×}B\text{=}\text{(}{A}_y{B}_z\text{-}{A}_z{B}_y\text{)}\widehat{i}\text{-}\text{(}{A}_x{B}_z\text{-}{A}_z{B}_x\text{)}\widehat{j}\text{+}\text{(}{A}_x{B}_y\text{-}{A}_y{B}_x\text{)}\widehat{k}$

It may be seen that the right-hand side is the determinant of the matrix

$\left[\begin{array}{ccc}\widehat{i}& \widehat{j}& \widehat{k}\\ A_x& A_y& A_z\\ B_x& B_y& B_z\end{array}\right]$

Thus, Equation 1.17 can be written

(1.18) $A\text{×}B\text{=}\left|\begin{array}{ccc}\widehat{i}& \widehat{j}& \widehat{k}\\ A_x& A_y& A_z\\ B_x& B_y& B_z\end{array}\right|$

where the two vertical bars stand for determinant. Obviously the rule for computing the cross product, though straightforward, is a bit lengthier than that for the dot product. Remember that the dot product yields a scalar whereas the cross product yields a vector.

The cross product provides an easy way to compute the normal to a plane. Let A and B be any two vectors lying in the plane, or, let any two vectors be brought tail-to-tail to define a plane, as shown in Figure 1.7. The vector C = A × B is normal to the plane of A and B. Therefore, ${\widehat{n}}_{AB}\text{=}C\text{/}C$ , or

(1.19) $n_{AB}\text{=}\frac{A\text{×}B}{\Vert A\text{×}B\Vert }$

Example 1.3

Let $A\text{=-}3\widehat{i}\text{+}7\widehat{j}\text{+}9\widehat{k}$ and $B\text{=}6\widehat{i}\text{-}5\widehat{j}\text{+}8\widehat{k}$ . Find a unit vector that lies in the plane of A and B and is perpendicular to A.

Solution

The plane of the vectors A and B is determined by parallel shifting the vectors so that they meet tail to tail. Calculate the vector D = A × B.

$D\text{=}\left|\begin{array}{ccc}\widehat{i}& \widehat{j}& \widehat{k}\\ \text{-}3& 7& 9\\ 6& \text{-}5& 8\end{array}\right|\text{=}101\widehat{i}\text{+}78\widehat{j}\text{-}27\widehat{k}$

Note that A and B are both normal to D. We next calculate the vector C = D × A.

$C\text{=}\left|\begin{array}{ccc}\widehat{i}& \widehat{j}& \widehat{k}\\ 101& 78& \text{-}27\\ \text{-}3& 7& 9\end{array}\right|\text{=}891\widehat{i}\text{-}828\widehat{j}\text{+}941\widehat{k}$

C is normal to D as well as to A. A, B and C are all perpendicular to D. Therefore they are coplanar. Thus C is not only perpendicular to A, it lies in the plane of A and B. Therefore, the unit vector we are seeking is the unit vector in the direction of C, namely,

${\widehat{u}}_C\text{=}\frac{C}{C}\text{=}\frac{891\widehat{i}\text{-}828\widehat{j}\text{+}941\widehat{k}}{\sqrt{{891}^2\text{+}{(\text{-}828)}^2\text{+}{941}^2}}$

In the chapters to follow we will often encounter the vector triple product, A × (B × C). By resolving A, B and C into their Cartesian components, it can easily be shown (see Problem 1.1c) that the vector triple product can be expressed in terms of just the dot products of these vectors as follows:

(1.20) $A\text{×}(B\text{×}C)\text{=}B(A\text{·}C)\text{-}C(A\text{·}B)$

Because of the appearance of the letters on the right-hand side, this is often referred to as the bac-cab rule .

Example 1.4

If F = E × {D × [A × (B × C)]}, use the bac-cab rule to reduce this expression to one involving only dot products.

Solution

First we invoke the bac-cab rule to obtain

$F\text{=}E\text{×}\left\{D\text{×}\stackrel{bac\text{-}cab\text{undefined}rule}{\overbrace{[B(A\text{·}C)\text{-}C(A·B)]}}\right\}$

Expanding and collecting terms leads to

$F\text{=}(A\text{·}C)[E\text{×}(D\text{×}B)]\text{-}(A\text{·}B)[E\text{×}(D\text{×}C)]$

We next apply the bac-cab rule twice on the right-hand side.

$F\text{=}(A\text{·}C)\left[\stackrel{bac\text{-}cab\text{ rule}}{\overbrace{D(E\text{·}B)\text{-}B(E\text{·}D)}}\right]\text{-}(A\text{·}B)\left[\stackrel{bac\text{-}cab\text{ rule}}{\overbrace{D(E\text{·}C)\text{-}C(E\text{·}D)}}\right]$

Expanding and collecting terms yields the sought-for result.

Another useful vector identity is the interchange of the dot and cross :

(1.21) $A\text{·}(B\text{×}C)\text{=}(A\text{×}B)\text{·}C$

It is so-named because interchanging the operations in the expression A · B × C yields A × B · C. The parentheses in Equation 1.21 are required to show which operation must be carried out first, according to the rules of vector algebra. (For example, (A · B) × C, the cross product of a scalar and a vector, is undefined.) It is easy to verify Equation 1.21 by substituting $A\text{=}{A}_x\widehat{i}\text{+}{A}_y\widehat{j}\text{+}{A}_z\widehat{k}$ , $B\text{=}{B}_x\widehat{i}\text{+}{B}_y\widehat{j}\text{+}{B}_z\widehat{k}$ and $C\text{=}{C}_x\widehat{i}\text{+}{C}_y\widehat{j}\text{+}{C}_z\widehat{k}$ and observing that both sides of the equal sign reduce to the same expression (Problem 1.1b).

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RL Circuit Analysis

John Clayton Rawlins M.S. , in Basic AC Circuits (Second Edition), 2000

Calculating Impedance

Impedance of an RC circuit can be calculated by adding the resistance and reactance vectorially on the phasor diagram. This is accomplished by simply applying the Pythagorean theorem. As illustrated in Figure 9.6 , the length of the reactance vector, X _L , is the same as the length between the tip of the resistance vector and the tip of the impedance vector Z. In fact, the phasor diagram can be drawn with the reactance vector placed as shown in Figure 9.7 . Now, the right triangle of the Pythagorean theorem becomes clearly evident. Applying the Pythagorean theorem, the total impedance is equal to the square root of the resistance squared plus the inductive reactance squared.

Figure 9.7. Right Triangle Relationship of Z, R, and X_L

For example, a typical series RL circuit is shown in Figure 9.8 . If the resistance in that circuit is 60 ohms and the inductive reactance is 80 ohms, then the total impedance of the circuit can be determined as follows:

Figure 9.8. Impedance Calculation Example

$\begin{array}{l}\text{Z=}\sqrt{{\text{R}}^{\text{2}}{\text{+X}}_{\text{L}}{}^{\text{2}}}\hfill \\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=}\sqrt{{\text{60}}^{\text{2}}{\text{+80}}^{\text{2}}}\hfill \\ \text{\hspace{0.17em}\hspace{0.17em}=}\sqrt{\text{3,600+6,400}}\hfill \\ \text{\hspace{0.17em}\hspace{0.17em}=}\sqrt{\text{10,000}}\hfill \\ \text{\hspace{0.17em}\hspace{0.17em}=100Ω}\hfill \end{array}$

The total impedance of that circuit is 100 ohms.

When drawing voltage and impedance diagrams for circuits like this, it is often convenient to draw them in the form of a right triangle as shown in Figure 9.9 .

Figure 9.9. Right-Triangle Relationship of Z, X_L, and R of the Example Circuit

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Mathematics

In Standard Handbook of Petroleum and Natural Gas Engineering (Third Edition), 2016

1.2.13 Lengths and Areas of Plane Figures

For definitions of trigonometric functions, see "Trigonometry."

•

Right triangle (Figure 1.2.1)

$\begin{array}{ll}\hfill {\text{c}}^2& ={\text{a}}^2+{\text{b}}^2()\hfill \\ \hfill \text{area}& =1/2•\text{ab}=1/2•{\text{a}}^{2}cot\text{A}\hfill \\ \hfill & =1/2•{\text{b}}^{2}tan\text{A}=1/4•{\text{c}}^{2}sin2\text{A}\hfill \end{array}$

•

Any triangle (Figure 1.2.2)

area=1/2 base • altitude=1/2 • ah=1/2 • ab sin C

= ± 1/2 • {(x₁y₂ −x₂y₁)

+ (x₂y₃− x₃y₂)

+ (x₃y₁− x₁y₃)}

where (x₁, y₁), (x₂, y₂), (x₃, y₃) are coordinates of vertices.

•

Rectangle (Figure 1.2.3)

area=ab=1/2 • D² sin u

where u=angle between diagonals D, D.

•

Parallelogram (Figure 1.2.4)

area=bh=ab sin c=1/2 • D₁D₂ sin u

where u=angle between diagonals D₁ and D₂.

•

Trapezoid (Figure 1.2.5)

area=1/2 • (a+b)h=1/2 • D₁D₂ sin u

where u=angle between diagonals D₁ and D₂

and where bases a and b are parallel.

•

Any quadrilateral (Figure 1.2.6)

area=1/2 • D₁D₂ sin u

Note: ${\text{a}}^2+{\text{b}}^2+{\text{c}}^2+{\text{d}}^2={\text{D}}_1^2+{\text{D}}_2^2+4{\text{m}}^2$

where m=distance between midpoints of D₁ and D₂.

•

Circles

area= πr² =1/2 • Cr=1/4 • Cd=1/4 • πd² =0.785398 d²

where r	=	radius
d	=	diameter
C	=	circumference=2πr= πd.

•

Annulus (Figure 1.2.7)

area= π(R² −r²)= π(D² −d²)/4=2πR′b

where R′ =mean radius=1/2 • (R+r)

b=R−r

•

Sector (Figure 1.2.8)

area=1/2 • rs= πr²A/360° =1/2 • r² rad A

where rad A $=$ radian measure of angle A

s=length of arc=r rad A

•

Ellipse (Figure 1.2.9)

area of ellipse= πab

area of shaded segment=xy+ab sin⁻¹ (x/a)

length of perimeter of ellipse= π(a+b)K,

where K=(1+1/4 • m² +1/64 • m⁴ +1/256 • m⁶  +   …)

m=(a−b)/(a+b)

•

Hyperbola (Figure 1.2.10)

For any hyperbola,

shaded area A=ab • ln[(x/a)+(y/b)]

For an equilateral hyperbola (a=b),

area A=a² sinh⁻¹ (y/a)=a² cosh⁻¹ (x/a)

where x and y are coordinates of point P.

•

Parabola (Figure 1.2.11)

shaded area A=2/3 • ch

In Figure 1.2.12,

length of arc OP=s=1/2 • PT+1/2 • p • ln [cot(1/2 • u) ]

Here c	= any chord
p	= semilatus rectum
PT	= tangent at P
Note: OT   =   OM=x

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Analytic Geometry: Part 1—The Basics in Two and Three Dimensions

Howard Mark , Jerry WorkmanJr., in Chemometrics in Spectroscopy (Second Edition), 2018

The Distance Formula

In two dimensions (x and y), the distance between two points (x ₁, y ₁) and (x ₂, y ₂) in two-dimensional space (as shown in Fig. 17-1 ) is given by the Pythagorean theorem as:

Fig. 17-1. The distance between two points in a two-dimensional coordinate space is determined using the Pythagorean theorem.

(17-1) $\begin{array}{c}{D}^2={\left|x_2-x_1\right|}^2+{\left|y_2-y_1\right|}^2\\ ={\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2\end{array}$

and

(17-2) $D=\surd {\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2$

Note: This relationship holds even when x ₁ or y ₁ or both are negative (also shown in Fig. 17-1). In three dimensions (x, y, z), we describe three lines at right angles to one another, designated as the x, y, z axes. Three planes are represented as xy, yz, and zx, and the distance between two points (x ₁, y ₁, z ₁) and (x ₂, y ₂, z ₂) is given by

(17-3) $\begin{array}{c}{D}^2={\left|x_2-x_1\right|}^2+{\left|y_2-y_1\right|}^2+{\left|z_2-z_1\right|}^2\\ ={\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2+{\left(z_2-z_1\right)}^2\end{array}$

and

(17-4) $D=\surd {\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2+{\left(z_2-z_1\right)}^2$

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Automobile Technology

Hans Joachim Förster , Hermann Gaus , in Encyclopedia of Physical Science and Technology (Third Edition), 2003

III.A.1 Road Grip

The vertical load F _N on each wheel is determined statically by the total weight of the vehicle (passenger car in the range 1–3   tons, commercial vehicles up to 50 tons), the number of wheels (car, 4; commercial vehicle, 4–10), and the position of the center of gravity of the vehicle. Superimposed on this static load distribution are three types of additional dynamic forces, which occur during acceleration or braking, during cornering, and through oscillations of the body. During acceleration, the rear axle is loaded further and the front axle is correspondingly relieved, while during braking the front wheels take up more load and the rear wheels are relieved. During cornering, the centrifugal forces acting at the center of gravity in addition to the weight are supported on the ground by an additional load on the wheels at the outside of the curve and a corresponding reduction on the wheels at the inside of the curve. Because in principle only one axle is necessary to support the centrifugal forces, the way the division between the two axles occurs provides additional freedom for influencing the self-steering performance during cornering. Both types of additional forces depend on the height of the center of gravity (Fig. 6), so that it is kept as low as possible, though this is limited by requirements for sufficient passenger or cargo space.

FIGURE 6. Effect of dynamic forces on the normal load of the wheels. S, center of gravity; R, Wheel base. (Top) Braking (deceleration): mg, weight; ma, forward force due to deceleration (both are effective in the center of gravity), (N _S )_v, static load, front wheel; (N _S)_h, static load, rear wheel; (N _d)_v, dynamic load, front wheel; (N _d)_h, dynamic load, rear wheel; N _v  =   (N _s  + N _d)_v, total load, front wheel; N _h  =   (N _s  − N _d)_h, total load, rear wheel. (Bottom) Cornering: mv ²/r, centrifugal force when cornering with speed v and bend radius r; (N _s)_vr, static load, front wheel right; (N _s)_vl, static load, front wheel left; (N _s)_hr, static load, rear wheel right (N _s)_hl, static load, rear wheel left; (N _d)_vr, dynamic load, front wheel right;(N _d)_vl,dynamic load, front wheel left; (N _d)_hr, dynamic load, rear wheel right; (N _d)_hl, dynamic load, rear wheel left; (N _v)_r  =   (N _s  − N _d)_vr, total load, front wheel right; (N _v)₁  =   (N _s  + N _d)_vl, total load, front wheel left; (N _h)_r  =   (N _s  − N _d)_hr, total load, rear wheel right; (N _h)₁  =   (N _s  + N _d)_hl, total load, rear wheel left.

Finally, vibrations of the body and unevenness of the road can increase or decrease the load on the wheels. Since the maximum ground adhesion is achieved with an even distribution of the total weight among the load-carrying wheels, the dynamic forces, which create additional load fluctuations, must be kept as low as possible.

The maintenance of contact between the road and the wheels involves the whole vibratory system, including the pneumatic tire, wheel or axle against the frame, and the frame against the body, with their frequencies and attenuations. The following are technical devices for achieving good road grip: low-pressure tires, small unsprung masses of the wheel-axle system (e.g., independent wheel suspension), long spring displacements even under load, low-friction springs, and adequate damping.

The traction coefficient μ_w, the second factor that determines road adhesion, depends on many parameters (Fig. 7). The following apply to the rolling tire:

FIGURE 7. (a) Circumferential force coefficient μ_B as a function of slip λ between tire and road (slip angle α   =   0). 1, radial-ply tire on dry asphalt; 2, radial-ply tire on wet asphalt; 3, radial-ply tire on gravel; 4, radial-ply tire on powder snow; 5, radial-ply tire on glaze. (b) Lateral force coefficient μ_S (–) and circumferential force coefficient μ_B (—) as function of slip λ and slip angle α between tire and road. L  =   lateral force; traction coefficient, $\sqrt{\mu \text{w}\approx {\mu }_{\text{B}}^2+{\mu }_{\text{S}}^2}$ .

1.

The total force that a tire transmits is divided into circumferential and lateral forces roughly according to the Pythagorean theorem.

2.

Circumferential force is transmitted only when there is slip between tire and road surface.

3.

Lateral force is generated only if the wheel (tire) is running at an angle to its plane (slip angle), and it tends toward a limiting value with an increase in slip angle.

4.

The circumferential force increases rapidly with increasing slip and reaches a maximum at ∼15% slip, after which it decreases.

5.

The lateral guidance force is at a maximum at zero circumferential slip and reduces roughly linearly to almost zero at 100% slip. A locked wheel has almost no lateral guidance.

Superimposed on these rules are the effect of the tires, such as their construction (height/width ratio and whether they are radial- or cross-ply), rubber mixture, type and depth of tread, and tire pressure, as well as the larger effects of the road surface, such as type and roughness, wetness and cleanliness, and degree of water removal with rain, snow, and ice. The grip-force limits can thus vary over a very wide range of 1–10, although in practice they are mainly utilized only over less than 20%. The surprise (and incorrect behavior) is then all the greater under special driving conditions with small values of the traction coefficient, such as icy roads and aquaplaning.

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Piping Isometrics

Roy A. Parisher , Robert A. Rhea , in Pipe Drafting and Design (Third Edition), 2012

Dimensioning Offsets

With isometric offsets changing a pipe's routing from one plane to another or from one geographic direction to another, coordinates and elevations no longer provide all the dimensions necessary to describe a pipe's total length. However, the use of 90° and 45° elbows to form the offsets results in a problem that can be easily solved with simple mathematical formulas. The 90° and 45° elbows form right triangles. By using the Pythagorean theorem which states that the sum of the squares of the two sides is equal to the square of the triangle's hypotenuse, this problem can be solved. Simply stated, A ²+B ²=C ². Figure 13.18 identifies the sides and angles of a right triangle and their resulting solution formulas.

Figure 13.18. Pythagorean theorem formulas.

These formulas can be used to solve the length of an unknown side when the other two sides are known. They work no matter the degree value of angle X. Some angles seem to be used repeatedly in pipe drafting. The chart in Figure 13.19 can significantly reduce the amount of time spent calculating unknown sides of right triangles. Use the appropriate decimal value when X is one of the provided angles.

Figure 13.19. Decimal equivalents of common angles.

As mentioned previously, 90° elbows can be rolled to form any degree of angular offset. To fabricate such a roll, a pipe fitter should be provided with the lengths of the three sides of the triangle and the degree value of angle X. Solving for an unknown value of X requires some additional trigonometric formulas. Use the formulas provided in Figure 13.20 to solve for the unknown value of angle X. Notice that, relative to X , side A is identified as the Side Adjacent (SA), side B is identified as the Side Opposite (SO), and side C is identified as the Hypotenuse (HYP).

Figure 13.20. Right-triangle formulas.

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Vector and Matrix Norms

William Ford , in Numerical Linear Algebra with Applications, 2015

Abstract

This chapter begins the study of numerical linear algebra. The length, or norm, of a vector is defined, and the 2-, 1-, and infinity vector norms are defined. There are inequalities that bound each norm in terms of the others. The chapter develops properties of the 2-norm, since it is most frequently used in applications. In particular, the Cauchy-Schwarz inequality and the Pythagorean theorem are satisfied by the 2-norm. The 2-norm is orthogonally invariant and thus is very important in computer graphics. If a set of k vectors in n-dimensional space are orthogonal, they form a basis for a k-dimensional subspace. Multiplication by an orthogonal matrix effects a change of coordinates. Spherical coordinates are introduced, and an orthogonal matrix is developed to change to and from Euclidean coordinates and spherical coordinates. The spherical coordinate basis vectors are the columns of the orthogonal matrix. The chapter defines the matrix norm and first presents the Frobenius norm. The induced matrix norm is defined, and the unit spheres for the infinity, 1- and 2- norm vector norms are drawn. The value of an induced matrix norm is the maximum value of Ax for x on the unit sphere. The infinity norm of a matrix is developed both experimentally and mathematically. The submultiplicative property for matrix norms is defined and it is shown that the Frobenius norm is submultiplicative. The 2-norm is the most complicated norm to compute. Begin with the symmetric matrix A ^T A whose eigenvalues are nonnegative. A singular value is the square root of an eigenvalue of A ^T A. Using the spectral theorem, it is shown that the matrix 2-norm is the largest singular value. Important properties of the 2-norm are developed, including orthogonal invariance, the fact that the 2-norm of a matrix equals the 2-norm of its transpose, and that the 2-norm of the inverse of A is the reciprocal of the smallest singular value of A.

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Analytic Geometry: Part 4—The Geometry of Vectors and Matrices

Howard Mark , Jerry WorkmanJr., in Chemometrics in Spectroscopy (Second Edition), 2018

Principal Components for Regression Vectors

Fig. 20-3A shows the projection of two column vectors— C1  =   (1,   3) and C2  =   (3,   1) onto their vector sum (or principal component, PC1). We note that the product [1,   3]   ×   [3,   1]   =   [1   ×   3,   3   ×   1]   =   [3,   3]. The vector sum of the two column vectors passes through the point (3, 3), but the projection of each column onto PC1 gives a vector with a length equal to line segments B  +  C as shown in Fig. 20-3B.

Fig. 20-3. (A) The representation of two columns of a matrix in row space. The vector sum of the two column vectors is the first principal component (PC1). (B) A detailed view of (A), illustrating the line segments, direction angles, and projection of columns 1 and 2 onto the first PC.

To determine the geometry for Fig. 20-3A and B, we begin by calculating the length of line segment E (column 1) by using the Pythagorean theorem as:

$E^2={\mathrm{Hyp}}^2={\left(3-0\right)}^2+{\left(1-0\right)}^2=3^2+1^2=10$

(20-3) $\text{Therefore},E=\sqrt{10}=3.162$

Then the angle C can be determined using

(20-4a) $tan\left(C\right)=\frac{\mathrm{opp}.}{\mathrm{adj}.}=\frac{1}{3}$

and

(20-4b) ${tan}^{-1}\frac{1}{3}=18.435\text{degrees}$

So ∠  C  =   18.435   degrees ∠  D  =   18.435   degrees and ∠  α  +   ∠ β  −   2   ×   18.435   degrees   =   90   degrees. Thus, both ∠  α and ∠  β are each equal to 26.565   degrees.

It follows that the projection of the vectors represented by columns 1 and 2 onto the vector PC1 yields a right triangle defined by the three line segments C  + B, D, and E. The length of PC1 (the hypotenuse) is equal to line segments C  + B and is given by

(20-5) $cos\alpha =\frac{\mathrm{adj}}{\mathrm{hyp}}\Rightarrow cos\alpha =\frac{E}{C+B}\Rightarrow 0.8944=\frac{3.162}{\mathrm{hyp}}=3.5353$

So the length of the hypotenuse (segments C + B) is 3.5353. We can check our work by calculating the opposite side (D) length as:

(20-6) $tan\alpha =\frac{\mathrm{opp}}{\mathrm{adj}}\Rightarrow tan\alpha =\frac{D}{E}\Rightarrow 0.500=\frac{\mathrm{opp}}{3.162}=1.5810$

And by using the Pythagorean theorem, we can calculate the length of the hypotenuse:

(20-7) ${\left(3.162\right)}^2+{\left(1.5810\right)}^2={\left(3.5352\right)}^2$

By representing a row vector in column space, or a column vector in row space, we can illustrate the geometry of regression. These concepts combined with matrix algebra will be useful for further discussions of regression. Readers may wish to study additional materials related to the subject of analytical geometry and regression. We recommend two sources of such information below.

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Draw Impedance Triangle And Show Various Quantities On It

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