Problem 39 - Trigonometry

Question

For Problems 37 through 42, use your knowledge of bearing, heading, and true course to sketch a diagram that will help you solve each problem. True Course and Speed A plane is flying with an airspeed of 160 miles per hour and heading of \(150^{\circ}\). The wind currents are running at 35 miles per hour at \(165^{\circ}\) clockwise from due north. Use vectors to find the true course and ground speed of the plane.

Step-by-Step Solution

Answer

The plane's true course is the angle of the resultant vector, and its ground speed is the magnitude of the resultant vector.

Step 1: Define the Vectors

First, we need to define the velocity vectors for both the airplane and the wind. The plane's velocity vector has a magnitude of 160 miles per hour and a direction of \(150^{\circ}\). Let's denote the plane's velocity vector as \(\mathbf{v_p}\). The wind's velocity vector, \(\mathbf{v_w}\), has a magnitude of 35 miles per hour and a direction of \(165^{\circ}\).

Step 2: Decompose the Vectors into Components

Next, decompose each vector into its components along the x-axis (eastward) and y-axis (northward). For \(\mathbf{v_p}\), the components are given by:- \(v_{px} = 160 \cos(150^{\circ})\)- \(v_{py} = 160 \sin(150^{\circ})\)For \(\mathbf{v_w}\), the components are:- \(v_{wx} = 35 \cos(165^{\circ})\)- \(v_{wy} = 35 \sin(165^{\circ})\)

Step 3: Calculate the Resultant Vector Components

Add the components of the plane's velocity and the wind's velocity to find the resultant ground velocity vector \(\mathbf{v_g}\):- \(v_{gx} = v_{px} + v_{wx}\)- \(v_{gy} = v_{py} + v_{wy}\)

Step 4: Magnitude of the Ground Velocity

The magnitude of the resultant ground velocity vector \(\mathbf{v_g}\) is the ground speed:\[ v_g = \sqrt{v_{gx}^2 + v_{gy}^2} \]

Step 5: Determine the True Course

Finally, the true course of the plane can be found using the arctangent function to get the direction of \(\mathbf{v_g}\):\[ \theta = \tan^{-1}\left(\frac{v_{gy}}{v_{gx}}\right) \]Ensure that you adjust \(\theta\) based on the quadrant in which \(\mathbf{v_g}\) lies.

Key Concepts

Bearing in Navigation

In navigation, bearing is the direction or path along which something moves or along which it lies. This is essential for pilots as it helps them determine the direction in which they should travel. Bearings are usually measured in degrees from the north direction.

• A bearing of \(0^{\circ}\) or \(360^{\circ}\) means directly north.
• A bearing of \(90^{\circ}\) is directly east.
• \(180^{\circ}\) is directly south, and
• \(270^{\circ}\) is directly west.

One important aspect to remember is that bearing can also be described as 'clockwise from north'. Therefore, a bearing of \(150^{\circ}\) indicates a direction that is east of south. As in the case of our exercise, the airplane had a heading of \(150^{\circ}\), which means it was flying southeast.

Vector Decomposition

Vector decomposition involves breaking down a vector into its component parts, which makes it easier to analyze and work with, especially in navigation problems.
Each vector is represented by horizontal (x-axis) and vertical (y-axis) components. This can be visualized as the lengths of the vector's shadow on the two axes, often found using trigonometric functions.The basic formulae used in vector decomposition are:

• To find the x-component (horizontal), multiply the vector's magnitude by the cosine of its angle: \(v_x = v \cos(\text{angle})\).
• For the y-component (vertical), multiply the vector's magnitude by the sine of its angle: \(v_y = v \sin(\text{angle})\).

Using these calculations, one can find the plane’s speed forming a vector represented by its airspeed (\(160\) mph) and direction (\(150^{\circ}\)), as well as the wind’s speed, a vector with wind speed (\(35\) mph) and direction (\(165^{\circ}\)). Breaking them into components helps compute the resultant vector, which informs us about the true motion.

Ground Speed Calculation

Ground speed is the speed at which an airplane moves relative to the ground. It differs from airspeed, which is the speed of the airplane relative to the air around it.Ground speed incorporates how both airspeed and wind speed affect the airplane's movement. To calculate ground speed, you need the vector sum of the airplane's velocity and the wind velocity.
The ground speed is essentially the magnitude of the resultant vector (v_g), which combines both the plane's velocity and the wind velocity.To find this, compute it using:\[v_g = \sqrt{v_{gx}^2 + v_{gy}^2}\]Here, \(v_{gx}\) is the sum of the x-components of the plane's and wind's velocities, and \(v_{gy}\) is the sum of their y-components.This way, the resultant vector gives a clear picture of how fast and in what direction the airplane truly travels over the ground.

Wind Correction Angle

In aviation, the wind correction angle is critical as it allows pilots to adjust their heading to compensate for wind drift.Without making this correction, the plane would end up traveling off its intended path.
The wind correction angle is the difference between the heading and course intended to fly. After finding the resultant vector component from vector decomposition, one uses the arctangent function to find the true course:\[\theta = \tan^{-1}\left(\frac{v_{gy}}{v_{gx}}\right)\]Ensure to adjust the angle based on which quadrant your resultant vector falls into. This adjustment determines the required correction to the plane's heading to ensure that it stays on its intended course despite the wind pushing it off path.
Understanding this angle helps maintain accurate navigation, ensuring that flights are efficient and arrive at their destination timely.