The vector diagram for the observation site south (180°) will mirror the one going north.
Base camp
θ
200 km h-1 Resultant velocity
Observation site
50 km h-1
θ = Sine-1 (1/4)
θ = 11.25°
Bearing = 360° – 11.25°
Bearing = 348.75°
Speed of aircraft = resultant velocity
2002 = V2 + 502
2002 – 502 = v2
37500 = v2
v = √37500
Resultant speed of aircraft = 193.6km h-1 (3s.f)
V = S/T (speed = distance/time)
T = S/V
T = 50/193.6
T = 0.258 hours (3s.f)
Time = 15 minutes 28 seconds
As with the time going north, the total time will be 2T, as the vector diagram will be the same in both directions. Total time is 2T = 2 × 0.258 = 0.516 = 31 minutes 36 seconds.
The next observation site will be east of the base camp, at 90°.
Base camp Observation site
Resultant velocity
Aircraft speed 200 kmh-1
Wind speed 50 km h-1
Speed of aircraft = 200 kmh-1 + 50 kmh-1
= 250 kmh-1
Time taken to reach observation site – T = S/V
T = 50 / 250
T = 0.2 hours
T = 12 minutes
Resultant velocity returning from observation site to base camp –
Base camp Observation site
Resultant velocity
Aircraft speed 200kmh-1
Wind speed 50 kmh-1
Aircraft speed = resultant velocity = 200 – 50 = 150 kmh-1
Time taken to return to base camp – T = S/V
T = 50 / 150
T = 0.333 hours
T = 20 minutes
Total time taken to travel to observation site and return = 20 minutes + 12 minutes = 32 minutes
The time taken to reach the observation site west (270°) of the base camp will be the same as the time taken to reach the observation site east of the base camp. This is because the vector diagram travelling from the base camp to the east observation site will be the same the vector diagram from the west observation site to the base camp, and the vector diagram from the base camp site to the east observation site will be the same as the vector diagram travelling from the west observation site to the base camp. So total time taken to reach went observation site and return to base camp = 32 minutes.
The results taken would not be very realistic, as the journey times are quite short take off and landing times will be significant. In a place like the arctic take off and landing times would be very significant, as it would take variable and mostly long periods of time to get the aircraft prepared. The times would be variable because the weather is so unpredictable in the Arctic. The journey may only be, say 32 minutes, but the preparation time could be much longer. For a model of this simplicity, we don’t take into account this extra time for take offs and landings; we just treat the aircraft as a particle, assuming instantaneous and constant speed.
The speed of the wind is also unlikely to be constant in direction and speed.
It is clear from the readings taken that a graph can be drawn that repeats itself after 180°. The time taken for 0° is the same as the time taken for 180°, and the total time for the site 90° is the same as the degree 180° later, 270°.
I now need to derive a formula to calculate the time taken for any wind speed, aircraft speed and wind direction.
W Observation site
β θ+90 θ-90
V R
Base camp
R = resultant velocity
W = wind speed
V = aircraft speed
R =?
Using sin rule
Sin A / a = sin B / b
Sin (90+θ) / V = sin α / W
Therefore, sin-1 (V sin 90 + θ / V) = α
Sin 90 + θ = cos θ
Therefore, sin-1 (W cosθ / V) = α
There are 180° in a triangle, therefore 180 = (90 + θ) + α + β
So 90 - θ - α = β
As α = sin-1 (W cos θ / V), we can substitute this for α in 90 - θ - α = β
Therefore, 90 - θ - [sin-1 (W cos θ / V)] = β
We can then substitute this back into the sin rule
Sin β / R = sin (90 + θ) / V
We can use this to work out the value of R, by making it the subject of the formula
V sin β / cos θ = R
Therefore,
V sin [90 - θ - sin-1 (w cos θ / V)] all divided by cos θ = R
We can then use this to work out the time it will take to make the journey, using v = s/t – t = s/v
Where s = radius of circle = R, and v = the resultant velocity
R / V sin [[90 - θ - sin-1 (w cos θ / V)] / cos θ] = t
This formula will give the time taken to travel to the observation site.
To calculate the time taken to return to base camp, 180° needs to be added to θ. This is because the bearing at θ + 180° = the bearing for the vector going in the opposite direction to the original vector, which equals the return journey time.
R / V sin [[90 – (θ +180) - sin-1 (w cos (θ + 180) / V)] / cos (θ + 180)] = t
These formulas will give me the time taken for any aircraft speed, and wind speed or direction, where –
S = radius of circle
V = velocity of the aircraft
W = wind speed
We can derive another formula to calculate the speed of the resultant vector using the quadratic equation. This will only give the speed from the centre of the circle.
b = wind speed
θ
a = aircraft velocity
c = relative velocity
θ
Both the plane and the wind have constant speed.
a2 = b2 + c2 – 2bc cos θ
Therefore c2 + (-2 b cos θ) c + (b2 – a2) = 0
Using the quadratic formula, we can take this as a quadratic in C2 –
C = 2b cos θ = √(-2 b cos θ)2 – 4 (b2 – a2)
Therefore, C = 2b cos θ ± √4b2 cos2θ - 4(b2 + a2)
Where b = wind speed, a = aircraft speed, c = resultant velocity, θ = angle of resultant relative to the wind.
To work out the distance the plane needs to travel from any base to any point on the circumference of the circle, we need to know where the base and the observation site is.
Observation site: measured in bearing from the centre of the circle, and will be the radius away from the centre of the circle.
Base camp: measured in bearing from the centre of the circle, with the distance in km.
Base camp
e
Observation site
d
r
θ1
θ
θ2
Centre of circle
Where d = distance, e = distance to observation site from base camp, r = distance to observation from centre of circle (the radius), θ = observation site bearing minus base bearing (θ = θ2 - θ1).
Using cosine rule – a2 = b2 + c2 – 2bc cos A, where a = e, b = d, c = r, A =θ.
Therefore, e2 = d2 + r2 – 2 × r × d cosθ
This formula can be used to calculate the length of the resultant velocity from any given point.
Conclusion - the formula to find time taken to travel to the observation site:
R / V sin [[90 - θ - sin-1 (w cos θ / V)] / cos θ]
To calculate the time taken to return to base camp;
R / V sin [[90 – (θ +180) - sin-1 (w cos (θ + 180) / V)] / cos (θ + 180)]
Where R = radius of the circle, V = velocity of the aircraft, W = wind speed.
So total time to and from base camp:
[R / V sin [[90 - θ - sin-1 (w cos θ / V)] / cos θ]] + [R / V sin [[90 – (θ +180) - sin-1 (w cos (θ + 180) / V)] / cos (θ + 180)]
Formula to calculate the speed of the resultant vector;
C = 2b cos θ ± √4b2 cos2θ - 4(b2 + a2)
Where b = wind speed, a = aircraft speed, C = resultant velocity, θ = angle of resultant relative to the wind.
Formula to calculate the length of the resultant velocity from any given point;
e2 = d2 + r2 – 2 × r × d cosθ
Where d = distance from centre of circle to observation site, e = distance to observation site from base camp, r = distance to observation from centre of circle (the radius), θ = observation site bearing minus base bearing (θ = θ2 - θ1).
