This is a diagram of my model. I have defined the direction of the north.
I did some research on the internet to find a reasonable estimate for the speed of the plane. From a website about planes I found that 340 km/h is the approximate cruising speed of the type of plane used for transport in the Arctic region. I will use this speed in my investigation. An example of such a plane is the Rockwell Aero Commander (AC – 500S) as seen in the diagram below.
Rockwell Aero Commander
Lockheed WP-3D Orion Aircraft
The average speed of the wind that I am going to use will be 30 km/h because in the arctic, from my research, the average wind speeds are mostly 30 km/h.
Simple Scenario
This is a sketch of all journeys the researchers would need to make to visit every observation site and return to base camp. There are 16 flights altogether. There are two for each observation site, an outward flight and return flight. The distance covered per journey is 50 km. If we forget about the wind, we can use speed (s) = distance (d) / time (t) to find the time it would take, ignoring the wind, for one journey. If we multiply this result by 16 it will give us a fair idea of time it should take for all the 16 flights. I am going to do the calculations below;
Time for one journey = 50 km / 340 km/h -1 = 0.147 hrs (0.147 × 60 = 8.82 minutes)
Time for all 16 flights = 0.147 × 16 = 2.352 (2.352 × 60 = 141.12 minutes)
In my investigation I will also calculate the resultant velocity, journey time and bearing of travel, for each observation site A – H. I will make separate calculations for the departing and returning journeys, as this will allow me to compare how they are affected by the wind. After doing this, I will then add them together to produce the final overall journey time for that observation site.
Since the positioning of half the observation sites is opposite to the other half, I will perform one of three similar, but slightly different methods of calculation on each of them. These three calculations are Pythagoras theorem, trigonometry of a right angled triangle and non-right angled triangle and the sine and cosine rules.
Right angled triangles of velocities
I am going to start with the observation site A, which makes a right angled triangle as the diagram shows below. For right angled triangle velocities only 2 calculations are needed to find the time it takes for the journey. I will use Pythagoras theorem and trigonometry of a right angled triangle.
Departure Journey from Base Camp to observation site A
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
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 0 to arrive at observation site A.
I can now use the flight of (V.R.) to work out the journey time.
Time = distance ÷ speed
= 50 ÷ 338.67 km/h
= 0.148 hrs
= 8.86 minutes
The return journey back from A to Base camp has similar processes as the magnitudes are the same, only the directions change.
Return Journey
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 back at base camp.
Overall flight time = 17.72 minutes
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
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
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
Return Journey from observation site B to Base camp
Observation site B is opposite to observation site D therefore they will have similar calculations. I will calculate D similarly to B and record my results into my results table on page 11.
Observation Site C
Overall Velocity (resultant velocity) = 340 km/h – 30 km/h
= 310 km/h
Speed = distance ÷ time
Rearrange to find time
time = distance ÷ speed
= 50 ÷ 310
= 0.16 hrs
For this particular journey I can state, after doing my calculations, that in order for the researchers to arrive at observation site C, their plane must leave the base camp at a bearing of 90o and must expect 0.16 hrs (0.16 hrs × 60 = 9.68 minutes) for the journey in which they will travel with a R.V. of 310 km/h.
Return Journey
Overall Velocity = 340 km/h + 30 km/h
= 370 km/h
time = distance ÷ speed
= 50 ÷ 370
= 0.135
= 8.1 minutes
I worked out the bearing and resultant velocity for all the other triangles and filled the results into a table.
TABLE
Details when plane velocity = 340 km/h and wind velocity = 30 km/h
Observing from the table, I can now inform that if the researchers are to visit each observation site once and return to base camp after each trip, whilst flying against 30 km/h westerly wind, they will have to expect to travel for a total of 2.384 hrs (2.384 × 60 = 143.04 minutes) at an average speed of 340 km/h. This is a reasonable total time as comparing this to my calculation earlier of the time it will take for all 16 flights when ignoring the wind was 2.352.
I will now alter the velocities of both the plane and the westerly wind in order to see how this affects the journey times.
I will use these results to establish which of the two, either the plane or the wind, has the dominant effect on the journey time, when each changes its velocity.
Firstly, I am going to vary the plane’s velocity and keep the wind speed constant at 30 km/h.
Plane Velocities; 320, 340, 360
After, I will keep the plane velocity constant at 340 km/h and vary the strength of the wind.
Wind Velocities; 25, 30, 35,
Total flight time summary
The difference between the two extremities of the wind velocities, which is the amount of change of the journey time, is 2.39 hrs – 2.373 hrs = 0.017 hrs
On the other hand, the difference between the two extremities of the plane velocities is 2.539 hrs – 2.249 hrs = 0.29 hrs.
These results show that a percentage increase in the plane velocity will have much more effect on the journey time than if the wind velocity changes by the same amount.
Base Camp Relocation
In order to get a fairly rough idea of how the movement of base camp (B.C) will affect the journey times, I will relocate it 4 times. Each by 10 kilometres as this is halfway between 50 km, which is the radius, and each in the direction of the four main points on a compass. I did this because I felt it will give me a fairly broad range of results that roughly cover relocation in any direction.
This diagram shows the model of how my relocation will be done.
For all four of my new base camps, I will keep wind velocity of 30 km/h constant and also the plane velocity of 340 km/h constant.
Base Camp (B.C) 1 to Observation site A
The angle at which the wind will act on the plane is the same as from the original base camp, which was in the middle, to observation site A, as the movement of the base camp only has a vertical component. Therefore the resultant velocity is the same as the original which is 338.67 km/h. I can now find the time for my new journey.
Time = distance ÷ speed
= 40 ÷ 338.67 km/h
= 0.118 hrs
The journey back to the new base camp will also have the same time and velocity. This is because the wind velocity is at 90o to the path of the resultant velocity. So the overall journey time, departing and return, is double the departing journey time which is 0.118 hrs × 2 = 0.236 hrs (0.236 × 60 = 14.16 minutes)
Base Camp (B.C) 2 to observation site A
Since B.C 2 is located to the east of observation site A, Pythagoras is required to calculate the distance between B.C. 2 and observation site A. Also the times for the two parts of the journey are different and must be calculated separately. Firstly the time for the departure journey must be calculated but the distance that the base camp lies from the observation site must be established.
Base Camp (B.C) 3 to observation site A
For this particular flight, the resultant velocity will be the same as that of B.C 1, 338.67 km/h, as its components are the same and are working in the same angles as well as directions. However due to the position the distance that must be flown has increased. 50 + 10 = 60 km
Departure journey time is therefore,
Time = distance ÷ speed
= 60 ÷ 338.67
= 0.177 hrs
The resultant velocity is the same for the returning part of the journey so the overall journey is 0.177 × 2 = 0.354
Base Camp (B.C) 4 to observation site A
Due to the similar distances by which they have been moved, the distance between this base camp and site A is the same as that between B.C 2 and site A.
Base Camp (B.C) 1 to observation site B
As this B.C.1 is 10 km from the original base it going to have different angles that will be slightly difficult as it is not going to be judgemental and therefore these angles has to be calculated in order for me to find the resultant velocity. Below is a diagram of this situation and my calculations.
For this, I will use trigonometry to find the angle x
Calculating the resultant velocity using the sine rule
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.
