Relative Velocity and The Point of No Return

1. Relative Velocity andThe Point of No Return By Colin Clarke

2. Relative Velocity When you are measuring velocity you need to measure it with relation to a point. Two cars on a motorway the red car is moving at 70mph and the green car at 60mph both relative to a fixed point on the motor way. The red car however is moving at 10mph relative to the green car. eg.

3. Planes Relative Velocity With a plane its speed can be measured relative to the ground, its speed will then vary according to wind speed and direction or relative to the air. For a fixed engine speed the air speed of a plane is always constant. Air speed 300 km/h

4. The Point of No Return You have recently became the owner of a powered aircraft and wish to go out for a flight, but you would like to know how far from your base you can fly and still return in safety.

5. Modelling Flight When modelling the flight we will start with a very simple model and gradually complicate it more until we have a more realistic model. I will work out the speed relative to the ground from the set speed that an aircraft flies at then use this to work out the distance the aircraft can fly.

6. Constant Level Flight First I will assume the plane is flying in no wind at the same level as where it takes off. Assume Air speed =300km/h, 4 hours of Fuel onboard. The plane has equal speeds on forward and return journey so ratio of times for forward and return journey is 1:1 Time journey to furthest point =4x½=2 hours Using “s=ut” s = 300 x 2 = 600km Maximum distance from Base = 600km

7. With 50km wind Head on Towards Plane Airspeed =300km/h On way out Ground speed = 250 km/h On way in Ground speed = 350 km/h Ratio of Speeds is 5:7 Time needed to return = 4 x 512 = 1⅔ hour Using “s=ut” s= 1⅔ x 350 = 583⅓ km Maximum distance from Base = 583⅓ km

8. With 50km wind Towards Plane From Behind The maximum distance from the base will be the same as the wind blowing from head on because the two trips will have the same relative velocities but the two journeys will be swapped around.

9. With 50km/h Wind at 30° angle Intended direction of motion The plane needs to fly slightly towards the wind to fly in the intended direction. 50km/h 30° θ Speed of plane V Resolving Wind To direction of intended motion sin θ = 112 θ=4.78° 50 sin 30= 300 sinθ 300 sinθ = 25 To direction of intended motion Wind speed = 50 cos30 = 25√3 For Plane Going Out Speed in direction of intended motion = 300 cosθ -50 cos30 = 25(√143 - √3) =255.7 km/h Coming in Speed in direction of intended motion = 25(√143 + √3) = 342.3 km/h Time Flying in 255.7 x4 = 1.71 hours 255.7 + 342.3 Using “s=ut” s = 342.3 x 1.71 = 585.5 km Maximum Distance from Base = 585.5 km

10. With 30° angle of ascent and descent to 6.1 km high Time to reach 6.1 km 300 km/h 300 km/h t= 6.1150 =0.0406 300 km/h 300 sin30 t = 6.1 30° 30° Time to Descend = 0.0406 hours Time spent climbing and descending = 2 x .0406 = 0.0813 hours Time spent in level flight = 3.9186 hours Distance travelled when climbing = 300 cos30 x 0.0406 = 10.6 km Time spent going out in level flight = ½ x 3.9186 = 1.9593 hours Distance travelled going out in level flight = 1.9593 x 300 = 587.79 km Total distance from base = 587.79 + 106 = 598.39 km Maximum distance from base = 598.4 km

11. General equations All of the previous results ca be generalised for a plane that can fly at v km/h, Has enough fuel for t hours of flight, Climbing to a height h km, With wind blowing at an angle of θ, at a speed of w km/h With an angle of ascent of α and an angle of descent of β.

12. General equation with no wind The time needed to fly out to the furthest point from the base is half of the time of fuel you have on board = ½ t Using “s=ut” s= v x ½t =vt2 Maximum Distance from base = vt 2

13. With wind at angle θ and speed w km/h Intended direction of motion θ = angle of wind with y axis v γ γ = angle of motion of plane with intended direction of motion θ w Resolvingto intended direction of motion w sinθ = v sinγ To direction of intended motion For plane going out For plane going in Ground speed = v cosγ – w cosθ v cosγ + w cosθ v cosγ + w cosθ tv cosγ + tw cosθ x t = Time flying out = (v cosγ – w cosθ) + (v cosγ + w cosθ) 2v cosγ (v cosγ – w cosθ)(v cosγ + cosθ)t Distance from base = 2v cosγ (v2 + w2)t (v² cos²γ - w² cos²θ)t = Maximum distance form base = 2v cosγ 2√(v2 – w2sinθ)

14. With angle α of ascent and angle β of descent to h km high h h Time to descend = Time to ascend = v sinβ v sinα ( ) h h Time spent in level flight = t - + v sinα v sinβ h h Distance travelled ascending = Distance travelled descending = tanα tanβ tv sinα sinβ – h sinβ – h sinα Distance travelled in level flight = sinα sinβ Distance from base =½ total distance tv sinα sinβ – h sinβ – h sinα h h = + + 2sinα sinβ 2tanβ 2tanα tv sinα sinβ – h [sinβ + sinα – sin(α + β)] Maximum distance from base = 2sinα sinβ

15. Summary From the equations that I have invented using relative velocities I can see that wind in any direction decreases the distance from the base you can fly. These results shown from relative velocity do have a use, similar equations to these were used in the second world war by navigators on bomber aircraft to reach their target. The wind speed changes at different heights however so these equations would be more complicated.

16. The End Note: These equations should not be used for people flying light aircraft