Optimal Angle for Westward Flight with Crosswind | Vector Problem Solution

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Homework Help Overview

The discussion revolves around a vector problem involving a pilot's flight path in the presence of a crosswind. The pilot aims to fly directly west while dealing with a southward wind, and participants are attempting to determine the correct angle to head in order to maintain this course.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are sharing their calculations and methods for determining the angle, with some using trigonometric functions and others referencing vector analysis. Questions arise regarding the correct application of tangent and the interpretation of the angle in relation to the wind direction.

Discussion Status

The discussion is ongoing, with various calculations presented and some participants questioning the methods used by others. There is no explicit consensus on the correct angle, but multiple approaches are being explored, including the use of trigonometry and vector analysis.

Contextual Notes

Participants are discussing the implications of the crosswind on the flight path and the assumptions made in their calculations. There is a focus on the relationship between the wind speed and the plane's speed, as well as the need to account for the wind's direction when determining the heading angle.

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Homework Statement


A pilot wants to fly directly to the west. The engine pushes the plane at 100 m/s, and there is a crosswind blowing to the south at 30 m/s. Determine the exact angle at which the pilot should head.



Homework Equations





The Attempt at a Solution


Here's what I did:

100^2+30^2= 10900

I then did SqRT of 10900 and got 104

then I did tan negative 1 to find the angle and got 89 degreess

right?
 
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Hmm, I got 287 degrees. Check your work again.
 
my work shows how I got 89

how did you get your answer?
 
Well I did it using vectors and then by using a "whiz wheel" and both answers came out the same. You got 104 which is correct, but you can't just take the tan-1 of that. You have to take tan-1 of .3 because 30/100 = .3. That gives you 17 degrees but now if you are heading west and the wind is out of the north you have to point the aircraft into the wind to crab, so you will still track on a westerly heading. Therefore 270 + 17 = 287 degrees.
 
why 30/100 and not 30/100000, 30/89, 30/76, get what I am saying?
 
Simple trigonometry..

Opposite side = 30
Adjacent side = 100

Tan theta = opp divided by adjacent

To find the angle, you'd take the tangent inverse of it.

Tan-1 = 30/100

Which produces 17
 

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