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Chapter 11: Problem 102

A pilot is flying at 168 mph. She wants her flight path to be on a bearing of \(57^{\circ} 40^{\prime} .\) A wind is blowing from the south at \(27.1 \mathrm{mph} .\) Find the heading the pilot should fly, and find the plane's ground speed.

Short Answer

Expert verified
The pilot should head 54.89掳 and the ground speed is approximately 162.38 mph.

Step by step solution

01

Convert Bearing to Degrees

The bearing of \(57^{\circ} 40^{\prime}\) needs to be converted into decimal degrees for calculation. The degrees and minutes are converted using the formula: \(57^{\circ} + \frac{40}{60} = 57.67^{\circ}.\) This is the angle the pilot aims to maintain her flight path relative to true north.
02

Set Up the Wind Effect

Wind blowing from the south implies it acts towards the north at \(27.1\) mph. We need to consider this in our calculations, as it will push the plane off course.
03

Use Vector Addition to Find Course

The plane's intended velocity becomes a vector at an angle of \(57.67^{\circ}\). The wind adds a \(27.1\, \text{mph}\) northern component, so the plane must adjust direction to compensate.
04

Calculate Effective Velocity

Using vector addition, the effective velocity \(v\) of the plane can be found using the formula: \[ v = \sqrt{(168 \cos(57.67^{\circ}) + 0)^2 + (168 \sin(57.67^{\circ}) - 27.1)^2}.\]Solve for \(v\) to find the resultant ground speed.
05

Determine Heading Adjustment

Since the wind has a northern component, calculate the angle \(\theta\) the plane needs to counteract using trigonometry:\[ \theta = \arctan\left(\frac{168 \sin(57.67^{\circ}) - 27.1}{168 \cos(57.67^{\circ})}\right).\]Adjust the original heading to find the correct heading \(X\): \[ X = 57.67^{\circ} - \theta.\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Trigonometry in Aviation
In aviation, trigonometry plays a crucial role in navigation and flight path correction. Pilots take advantage of geometric relationships to calculate directions and distances.
Understanding how trigonometry applies to aviation can help pilots navigate complex scenarios like deviating around obstacles or correcting their course due to wind.
For example, when a pilot intends to fly along a certain bearing, trigonometric functions can be used to determine the exact heading required to stay on course, considering factors such as wind speed and direction.
Angle Conversion
When working with bearings in aviation, it's crucial to convert angles from degrees and minutes to decimal degrees.
This conversion facilitates smoother calculations, especially when using formulas for vector addition or trigonometric functions.
For instance, the bearing of \(57^{\circ} 40^{\prime}\) can be converted as follows: \(57^{\circ} + \frac{40}{60} = 57.67^{\circ}\). It's essential to accurately perform this conversion to maintain the precision needed in navigational computations.
Wind Effect on Flight Path
Wind can significantly affect a plane's navigation, altering its intended path. Wind blowing towards the north, for example, can push an aircraft off its intended bearing, resulting in a deviation from the plotted course.
In our exercise, the wind at \(27.1\text{ mph}\) from the south impacts the flight path to the north.
To counteract this, pilots must adjust their heading so that the aircraft maintains its intended course. This process involves calculating the wind's vector and altering the navigation vector accordingly to ensure the flight stays on track.
Ground Speed Calculation
Calculating ground speed is vital for pilots to determine how fast an aircraft is moving over ground.
This differs from the airspeed, which is the speed relative to the surrounding air mass.
To find the ground speed impacted by wind, pilots use vector addition to combine the aircraft's velocity vector with the wind velocity.
In the given problem, the plane's effective ground speed \(v\) is calculated through the formula:\[v = \sqrt{(168 \cos(57.67^{\circ}) + 0)^2 + (168 \sin(57.67^{\circ}) - 27.1)^2}.\]The parameters \(168\text{ mph}\) and \(27.1\text{ mph}\) represent the plane's speed and the wind's contribution, respectively.
By solving for \(v\), pilots get the resultant speed over ground, accommodating both the desired course and the wind effect.

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Most popular questions from this chapter

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