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Arctic Research (Maths Coursework)

Extracts from this document...

Introduction

MATHS COURSEWORK

ARCTIC RESEARCH

Introduction:

In my investigation, a group of researchers are to set up a series of observation sites at equal distances from each other in the Arctic on the circumference of a circle of radius 50 km. The aim of my investigation is to figure out how long the researchers should be prepared to travel for on each journey. I am also to find where in the circle that the base camp should be located in order to minimize these journey times as short as possible. Using some calculations I will determine all of this taking into account that there is a strong westerly wind present constantly at all times in the area. I found out, from research, that average wind velocities in the Arctic are from 25 – 35 km/h. I will use this range in my investigation. The bearing at which the planes must depart is affected by this westerly wind therefore, this angle must be calculated.

If this investigation was to be physically carried out there would be some factors that needs to be considered. In the arctic there are many mountains and obstacles a plane would have to take into account during flight. To take into account these factors in my calculations would be impractical therefore, I must establish variables that could be controlled and make some assumptions in order to make it practical.

These are the variables that I must establish;

  • Strength of the westerly wind.
  • Speed that the researcher’s plane would be travelling at. I will investigate into how both of these affect the overall flight time by varying the velocities for both the plane and the wind.

These are the assumptions that I must establish;

  • I will be treating the people in the plane and the plane itself as particles. This will mean that they have no weight at all and therefore, this will not affect the flight of the plane. On the other hand, if it was considered the flight of the plane would be affected.
  • The flight velocity of the plane is constant and the speed at which the plane travels will be achieved instantaneously. This will prevent complications concerned with takeoff and landing times.
  • Wind velocity is also constant and always blowing in the exact, same direction of the west, as the flight times for a constantly changing wind velocity would be almost impossible to calculate.
  • Mountains and obstacles in the arctic that a plane would have to take into account during flight causes complications that are unnecessary for this investigation, so I will assume that the path of the flight is perfectly straight.
  • The plane will fly parallel to the ground, therefore, keeping the distance that the plane travels equal to the horizontal distance from base camp to observation site.
  • It takes no time for the plane to take of or land.
  • Time is flowing forward at a constant rate.
  • There is no air resistance or friction except for the wind. In real life there are other forces that could affect the flight of the plane, however, to make this investigation practical I will not be considering any of these.
  • The height at which the plane will be during flight will not be considered.
...read more.

Middle

The top half of the circle of radius 50 km is equal to the bottom half; therefore, opposite observation will have similar calculations. Observation site A is opposite to observation site E in this case they have similar calculations.

Departure Journey from Base Camp to observation site E

image09.png

Pythagoras’ Theorem to find the speed of the flight

a2 = b2 + c2

3402 = 302 + R.V. (resultant velocity) 2

R.V.2 = 115600 – 900

R.V.2 = 114700

R.V. = √114700

R.V. = 338.67 km/h

Time = distance ÷ speed

         = 50 ÷ 338.67 km/h

         = 0.148 hrs

         = 8.86 minutes

Trigonometry to find the angle θ and direction

Sin θ = opposite

            Hypotenuse

Sin θ = 30

            340

Sin θ = 0.088

θ = Sin – 0.088 = 5.06o

In this case the pilot will have to fly on a bearing of 180o slightly southeast to arrive at observation site E.

Return Journey

image10.png

a2 = b2 + c2

3402 = 302 + R.V. (resultant velocity) 2

R.V.2 = 115600 – 900

R.V.2 = 114700

R.V. = √114700

R.V. = 338.67 km/h

Time = distance ÷ speed

         = 50 ÷ 338.67 km/h

         = 0.148 hrs

         = 8.86 minutes

Trigonometry to find the angle θ and direction

Sin θ = opposite

            Hypotenuse

Sin θ = 30

            340

Sin θ = 0.088

θ = Sin – 0.088 = 5.06o

The pilot will have to fly on a bearing of 0 to arrive back at base camp.

Non-right angled triangle

The calculations used to find the resultant velocity and bearing of non - right angled triangles are different to right angled triangles. To find the resultant velocity and bearing for non - right angled triangles I am going to use the sine or cosine rule because these are used to find an unknown length or angle.    

Here are both rules:

Cosine rule; a2 = b2 + c2 – 2bc cosA

Sine rule;     sin A = sin B = Sin C

                         a            b          c

Departure Journey from base camp to observation site B

image11.png

Return Journey from observation site B to Base camp

image12.png

Observation site B is opposite to observation site D therefore they will have similar calculations.

...read more.

Conclusion

image05.png

For this, I will use trigonometry to find the angle x

Calculating the resultant velocity using the sine rule

image06.png

Calculating the distance using Pythagoras.

EVALUATION

My investigation was very unrealistic because there were some limitations that I left out e.g. the fact that there are a lot of mountains obstacles that the plane will have to take into account during flight. Things like these will affect the journey times and therefore I came up with a list of assumptions to prevent causing complications that are unnecessary for this investigation. In this case it makes my investigation unrealistic.                                          

OVERALL CONCLUSION

The first part of my investigation was to find an appropriate overall journey time for the researchers, which comprised of a return flight to each observation site.

I have come to the conclusion that the centre is the best place to put the base camp as all the sites would be reached at a faster time. I came to this conclusion because I figured out that if I was to relocate my base camp, and find all the total times for my four relocated base camps it would make no difference. This is because if 1 of my relocated base camp was situated slightly west of from my original base camp. All the sites on the west would be reached at a quicker time whereas all the sites on the east side would be reached at a much longer time. The overall time would be more or less than the original base camp, this applies to all the relocated base camp, and therefore it makes no difference.

...read more.

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