11.9
Vectors in 2D: Problem Solving
A plane is flying due north at a constant airspeed of 180 kilometers per hour. After 30 minutes, it is found 80 kilometers away, but 5 degrees east of north. This means the wind has changed the plane’s intended path.
The goal is to find the wind’s velocity and the direction the plane should fly to stay on course.
The plane’s ground velocity comes from combining its intended velocity with the wind’s velocity, forming a vector triangle.
The plane covers 80 kilometers in 30 minutes, giving the magnitude of its ground velocity. Since this velocity points 5 degrees east of north, it can be resolved into two components: northward and eastward.
The intended velocity is due north at 180 kilometers per hour. Subtracting this from the ground velocity gives the wind velocity. Its magnitude is found using the Pythagorean Theorem.
Finally, to stay on course, the pilot must aim slightly west of north to cancel the wind’s effect.
This adjustment is found by resolving the adjusted velocity into components and matching the westward part to the wind’s eastward push. The result shows the plane must aim about 4.4 degrees west of north to stay on course.
A plane traveling due north at 180 km/h in still air was found to be 80 km off-course after 30 minutes, deviating approximately 5 degrees east of north. This deviation means the influence of a crosswind alters the plane’s intended trajectory. The actual ground path formed a diagonal, suggesting that the aircraft’s effective ground speed was reduced to 160 km/h and directed slightly to the east due to the wind.
By analyzing the displacement from the intended path, the velocity contributed by the wind was inferred. The wind introduced a noticeable eastward component, causing the aircraft to veer from its intended northward route. The magnitude of this wind-induced motion was estimated to be about 24.9 km/h, acting consistently throughout the observed duration.
To counteract this deviation and ensure a direct path to the intended destination, a directional correction is required. This involves adjusting the aircraft’s heading slightly west of north. Such an adjustment introduces a westward motion component that effectively cancels the wind’s eastward influence. A tilt of approximately 4.4 degrees west of north provides the necessary balance, allowing the resulting trajectory to align with the desired northward course. This technique exemplifies the application of vector principles in aerial navigation.
A plane is flying due north at a constant airspeed of 180 kilometers per hour. After 30 minutes, it is found 80 kilometers away, but 5 degrees east of north. This means the wind has changed the plane’s intended path.
The goal is to find the wind’s velocity and the direction the plane should fly to stay on course.
The plane’s ground velocity comes from combining its intended velocity with the wind’s velocity, forming a vector triangle.
The plane covers 80 kilometers in 30 minutes, giving the magnitude of its ground velocity. Since this velocity points 5 degrees east of north, it can be resolved into two components: northward and eastward.
The intended velocity is due north at 180 kilometers per hour. Subtracting this from the ground velocity gives the wind velocity. Its magnitude is found using the Pythagorean Theorem.
Finally, to stay on course, the pilot must aim slightly west of north to cancel the wind’s effect.
This adjustment is found by resolving the adjusted velocity into components and matching the westward part to the wind’s eastward push. The result shows the plane must aim about 4.4 degrees west of north to stay on course.
From Chapter 11:
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11.9 : Vectors in 2D: Problem Solving
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11.1 : Introduction to Vectors
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11.2 : Vectors in Engineering Applications
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11.3 : Dot Product of Two Vectors
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11.4 : Cross Product and Its Geometry
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11.5 : Lines in Space
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11.6 : Planes in Space
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11.7 : Cylinders in Three-Dimensional Space
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11.8 : Quadric Surfaces
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11.10 : Vectors in Space: Problem Solving
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