Vector Maps (Trig): Solving Flight Path from Lincoln to Manhattan

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In summary: In summary, the conversation discusses a training flight for a student pilot and the use of the method of components to find the distance and direction she must fly to get back to her starting point. The steps for finding the distance and direction are provided, but the speaker expresses confusion about how the information was cut up into right triangles. The solution involves thinking of the cities as points on a Cartesian x,y plane and using the differences in x and y co-ordinates, along with the angle, to find the components.
  • #1
PascalPanther
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Map: http://xs206.xs.to/xs206/06362/map.gif

On a training flight, a student pilot flies from Lincoln, NE to Clarinda, IA, then to St. Joseph, MO, and then to Manhattan, KS. The directions are shown relative to north: 0 degrees is north, 90 degrees is east, 180 degrees is south, and 270 degrees is west. Use the method of components to find:
1. the distance she has to fly from Manhattan to get back to Lincoln
2. the direction (relative to north) she must fly to get there.

I have the answer, and the steps to do it, but I don't understand it.

Here are the steps:

East displacement from Manhattan to Lincoln:
(147km)sin85 + (106km)sin167 + (166km)sin235 = 34.3 km

North displacement:
(147km)cos85 + (106km)cos167 + (166km)cos235 = -185.7

SqRt(34.3^2 + -185.7 ^2) = 189km

Direction relative to north, arctan (34.3/-185.7) = -10.46 = 349.5 degrees
----------------------------------------------------

I think the main thing I don't get is, how did they cut up that information into right triangles so that they could use sin and cos. I don't see it at all. What I would have thought for example the first length, would be was sin(5 degrees ) = x/147km since it is at 85 degrees or 5 degrees above east which is 90 degrees. Anybody have any insight or tips on seeing this?
 
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  • #2
PascalPanther said:
Map: http://xs206.xs.to/xs206/06362/map.gif

I think the main thing I don't get is, how did they cut up that information into right triangles so that they could use sin and cos. I don't see it at all. What I would have thought for example the first length, would be was sin(5 degrees ) = x/147km since it is at 85 degrees or 5 degrees above east which is 90 degrees. Anybody have any insight or tips on seeing this?
Just think of the cities as points on a Cartesian x,y plane. The components of each vector showing the displacement from one point to the other are simply the difference in x and y co-ordinates. The angle is simply arctan y/x.

AM
 
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  • #3


I can understand your confusion with the steps provided for solving the flight path from Lincoln to Manhattan using vector maps. Let me break it down for you.

Firstly, it is important to understand that vector maps use a mathematical approach called the method of components to solve problems involving direction and distance. This method involves breaking down a vector (in this case, the flight path) into its horizontal and vertical components, which can then be solved using trigonometric functions.

Now, looking at the given map, we can see that the flight path from Lincoln to Manhattan involves three different segments (Lincoln to Clarinda, Clarinda to St. Joseph, and St. Joseph to Manhattan). Each of these segments has a different distance and direction, which is why the problem asks for the total distance and direction from Manhattan to Lincoln.

To solve for the east and north displacement, we use the formula:

East displacement = (distance) x sin(direction)
North displacement = (distance) x cos(direction)

In this case, the east displacement is calculated by taking the distance (147km) and multiplying it by the sine of the direction. The direction for the first segment (Lincoln to Clarinda) is given as 85 degrees, which means that the angle between the direction and east is 5 degrees (90-85=5). This is why the formula is written as sin(85) = x/147km, where x is the unknown east displacement. Solving for x, we get 34.3km.

Similarly, the north displacement is calculated by taking the distance (147km) and multiplying it by the cosine of the direction. The direction for the first segment is still 85 degrees, but now we need to find the angle between the direction and north. This can be done by subtracting the direction from 90 degrees (90-85=5), which gives us 5 degrees. So the formula for the north displacement becomes cos(5) = x/147km, where x is the unknown north displacement. Solving for x, we get -185.7km.

Now, to find the total distance from Manhattan to Lincoln, we use the Pythagorean theorem, which states that the square of the hypotenuse (in this case, the total displacement) is equal to the sum of the squares of the other two sides (east and north displacement). So we have:

Total distance = √(34.3^2
 

1. How do you determine the flight path from Lincoln to Manhattan using vector maps?

In order to determine the flight path from Lincoln to Manhattan using vector maps, you will need to first plot the starting point (Lincoln) and the destination (Manhattan) on a vector map. Then, using the principles of trigonometry, you can calculate the direction and magnitude of the flight path from Lincoln to Manhattan.

2. What is a vector map and how does it differ from other types of maps?

A vector map is a type of map that uses mathematical equations and vectors to represent geographic features. This is different from other types of maps, such as topographic maps, which use contour lines to represent elevation, or political maps, which show boundaries between countries or states.

3. How does trigonometry play a role in solving flight paths using vector maps?

Trigonometry is used to calculate the angles and distances between different points on a vector map. This is essential in determining the flight path, as the direction and magnitude of the flight must be accurately calculated in order to reach the destination.

4. Are there any limitations to using vector maps for determining flight paths?

While vector maps are a useful tool for determining flight paths, they do have some limitations. For example, they may not accurately represent changes in elevation or terrain, which can affect the actual flight path. Additionally, vector maps may not be as detailed or up-to-date as other types of maps.

5. Can vector maps be used for other types of navigation besides flight paths?

Yes, vector maps can be used for various types of navigation, such as driving, boating, and hiking. They are useful for determining the direction and distance between two points, making them a valuable tool for any type of navigation.

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