Question 3.15: CORRECTING FOR A CROSSWIND With wind and airspeed as in Exam...

CORRECTING FOR A CROSSWIND

With wind and airspeed as in Example 3.14, in what direction should the pilot head to travel due north? What will be her velocity relative to the earth?

The Blue Check Mark means that this solution has been answered and checked by an expert. This guarantees that the final answer is accurate.
Learn more on how we answer questions.

IDENTIFY and SET UP:

Like Example 3.14, this is a relative velocity problem with vectors. Figure 3.36 is a scale drawing of the situation. Again the vectors add in accordance with Eq. (3.35) (\overrightarrow{\boldsymbol{v}}_{P / A}=\overrightarrow{\boldsymbol{v}}_{P / B}+\overrightarrow{\boldsymbol{v}}_{B / A}) and form a right triangle:

\vec{v}_{\mathrm{P} / \mathrm{E}}=\overrightarrow{\boldsymbol{v}}_{\mathrm{P} / \mathrm{A}}+\overrightarrow{\boldsymbol{v}}_{\mathrm{A} / \mathrm{E}}

As Fig. 3.36 shows, the pilot points the nose of the airplane at an angle b into the wind to compensate for the crosswind. This angle, which tells us the direction of the vector \overrightarrow{\boldsymbol{v}}_{\mathrm{P} / \mathrm{A}} (the velocity of the airplane relative to the air), is one of our target variables. The other target variable is the speed of the airplane over the ground, which is the magnitude of the vector \overrightarrow{\boldsymbol{v}}_{\mathrm{P} / \mathrm{E}} (the velocity of the airplane relative to the earth). The known and unknown quantities are

\overrightarrow{\boldsymbol{v}}_{\mathrm{P} / \mathrm{E}}= magnitude unknown       due north

\overrightarrow{\boldsymbol{v}}_{\mathrm{P} / \mathrm{A}}= 240 km/h                          direction unknown

\overrightarrow{\boldsymbol{v}}_{\mathrm{A} / \mathrm{E}}= 100 km/h                          due east

We’ll solve for the target variables by using Fig. 3.36 and trigonometry.

 

EXECUTE:

From Fig. 3.36 the speed v_{\mathrm{P} / \mathrm{E}} and the angle β are

v_{\mathrm{P} / \mathrm{E}}=\sqrt{(240 \mathrm{~km} / \mathrm{h})^{2}-(100 \mathrm{~km} / \mathrm{h})^{2}}=218 \mathrm{~km} / \mathrm{h}

 

\beta=\arcsin \left(\frac{100 \mathrm{~km} / \mathrm{h}}{240 \mathrm{~km} / \mathrm{h}}\right)=25^{\circ}

The pilot should point the airplane 25° west of north, and her ground speed is then 218 km/h.

 

EVALUATE: There were two target variables—the magnitude of a vector and the direction of a vector—in both this example and Example 3.14.

In Example 3.14 the magnitude and direction referred to the same vector (\overrightarrow{\boldsymbol{v}}_{\mathrm{P} / \mathrm{E}} and \overrightarrow{\boldsymbol{v}}_{\mathrm{P} / \mathrm{A}}).

While we expect a headwind to reduce an airplane’s speed relative to the ground, this example shows that a crosswind does, too. That’s an unfortunate fact of aeronautical life.

3

Related Answered Questions