Problem 56 - College Trigonometry

Question

In calm air, a plane flying from the Pedimaxus International Airport can reach Cliffs of Insanity Point in two hours by following a bearing of \(\mathrm{N} 8.2^{\circ} \mathrm{E}\) at 96 miles an hour. (The distance between the airport and the cliffs is 192 miles.) If the wind is blowing from the southeast at 25 miles per hour, what speed and bearing should the pilot take so that she makes the trip in two hours along the original heading? Round the speed to the nearest hundredth of a mile per hour and your angle to the nearest tenth of a degree.

Step-by-Step Solution

Answer

The pilot should fly at 112.72 mph on a bearing of 358°.

Step 1: Calculate the Relative Wind Effect

The wind is blowing from the southeast at 25 miles per hour. The direction of the wind is northeast (45° from the north). We represent the wind as a vector with magnitude 25 mph in this direction. This wind vector can be broken down into its components: eastward and northward, using trigonometry: - Eastward component: \( 25 \times \cos(45°) \)- Northward component: \( 25 \times \sin(45°) \)Evaluating these components yields approximately 17.68 mph for both the eastward and northward components.

Step 2: Determine Required Airspeed Components

The original course is \( 8.2° \) east of north at a speed of 96 mph (plane's airspeed in calm air). This can be resolved into components:- Eastward component: \( 96 \times \sin(8.2°) \)- Northward component: \( 96 \times \cos(8.2°) \)These calculations give approximately 13.73 mph for the eastward and 94.98 mph for the northward components.

Step 3: Compensate for Wind

To maintain the original course to the destination despite the wind, calculate the required true airspeed:- Eastward: The pilot needs to negate the eastward component of the wind by adjusting her direction. Thus, the required eastward component should equal the eastward course component minus the eastward wind component: \[ 13.73 - 17.68 = -3.95 \]- Northward: Sum the northward course component and the northward wind component: \[ 94.98 + 17.68 \approx 112.66 \]

Step 4: Calculate Resultant Airspeed and Bearing

Compute the resultant airspeed and bearing using these components. Use the Pythagorean theorem to find the magnitude:\[ \text{Airspeed} = \sqrt{(-3.95)^2 + (112.66)^2} \approx 112.72 \]Determine the new bearing using inverse tangent:\[ \text{Bearing} = \arctan\left(\frac{-3.95}{112.66}\right) \approx -2.0° \text{ (east of north, adjusted)} \]Convert to correct bearing notation (\(360° - 2.0°= 358° \)), resulting in a very slight adjustment west from directly north.

Step 5: Final Adjustments and Corrections

Verify and round the calculations:- Speed is rounded to 112.72 mph.- The angle is adapted to the northward equivalent bearing of the intended course, resulting in \( 358° \) or simply north \( (0°) \) due west by approximately \( 2° \).

Key Concepts

Bearing Calculation

When flying, a **bearing** is the direction in which a plane is supposed to travel, measured in degrees. In aviation, bearings are usually given based on the cardinal directions, with North as 0° and angles increasing clockwise. For example, North 8.2° East means that the direction is slightly east of directly north. To understand and calculate bearings, both the desired path and external factors like wind must be considered. In this exercise, the original bearing, N 8.2° E, and the impact of the wind on this path was critical. The pilot must adjust to keep on course by calculating the bearing that neutralizes the wind's impact and maintains their desired path.

Vector Components

Vectors help break down complex quantities, like wind speed and direction, into simpler, usable components. These components are usually the eastward and northward parts of a direction vector. In this scenario, both the wind and plane's motion can be seen as vector sums: one traveling northeast and one towards the destination. - **Eastward Component**: Calculate using cosine of the wind's angle (45°): - **Northward Component**: Using sine for calculation, as it's the vertical direction. Trigonometry aids in breaking the vectors down:- for wind: \( 25 \times \cos(45°) \approx 17.68 \) mph east and north.- for airspeed: \( 96 \times \sin(8.2°) \approx 13.73 \) mph east and \( 96 \times \cos(8.2°) \approx 94.98 \) mph north.

Airspeed and Wind Compensation

To stay on course when wind alters your path, pilots must compensate by adjusting their airspeed. The plane's speed and direction naturally change when flying into a wind. This compensation involves: - **Headwind and Tailwind**: The extent to which wind is counteracted or aided by flight path. - **Side (Crosswind) Components**: These must be nullified to avoid drift. In the exercise, pilots compensate by adjusting the plane's vector components relative to the wind: - Eastward: Calculated to counteract the wind influence; here it becomes a negative value indicating a slight in-fight eastward adjustment is needed. - Northward: Needs boosting to keep the forward thrust, resulting in 112.66 mph as required.

Bearing Adjustments

Once the vectors are calculated, pilots adjust bearings to maintain the original path. This includes slight changes in direction to balance the wind factor. In the exercise, change involves computing new airspeed and recalculating the heading:- **Resultant Airspeed**: Use the Pythagorean theorem, resulting in 112.72 mph.- **Recomputed Bearing**: The arctangent function helps find the new angle:\[\text{Bearing} = \arctan\left(\frac{-3.95}{112.66}\right) \approx -2.0°\]approximating a correction west of north.Overall, these calculations ensure the plane compensates for wind to stay on course, with just a minor heading tweak, in this case, interpreted correctly as 358° due to the slight westerly correction.