Relative Velocity Relative Velocity andand

The Point of No ReturnBy Colin Clarke

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.

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

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.

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.

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

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 5

12 = 1⅔ hour

Using “s=ut” s= 1⅔ x 350 = 583⅓ km

Maximum distance from Base = 583⅓ km

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.

With 50km/h Wind at 30° angle50km/h

Intended direction of motion

30°

The plane needs to fly slightly towards the wind to fly in the intended direction.

Resolving Wind To direction of intended motion

50 sin 30= 300 sinθ

Speed of plane V

θ

300 sinθ = 25 sin θ = 112

θ=4.78°

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 255.7255.7 + 342.3

x4 = 1.71 hours

Using “s=ut” s = 342.3 x 1.71 = 585.5 km

Maximum Distance from Base = 585.5 km

With 30° angle of ascent and descent to 6.1 km high

30° 30°

Time to reach 6.1 km

300 sin30 t = 6.1300 km/h t= 6.1150 =0.0406

Time to Descend = 0.0406 hours

Time spent climbing and descending = 2 x .0406 = 0.0813 hours

300 km/h

300 km/h

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

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.

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 =vt2

With wind at angle θ and speed w km/h

θ

γ

Intended direction of motion

w

vθ = angle of wind with y axis

γ = angle of motion of plane with intended direction of motion

Resolving to intended direction of motion

w sinθ = v sinγ

To direction of intended motion

Ground speed = v cosγ – w cosθ

For plane going out For plane going in

v cosγ + w cosθ

Time flying out =v cosγ + w cosθ

(v cosγ – w cosθ) + (v cosγ + w cosθ)x t = tv cosγ + tw cosθ

2v cosγ

Distance from base =(v cosγ – w cosθ)(v cosγ + cosθ)t

2v cosγ

Maximum distance form base =(v² cos²γ - w² cos²θ)t

2v cosγ=

(v2 + w2)t

2√(v2 – w2sinθ)

With angle α of ascent and angle β of descent to h km high

Time to ascend =h

v sinαTime to descend =

v sinβh

Time spent in level flight = t - hv sinα v sinβ

h+( )Distance travelled ascending = Distance travelled descending =

htanα

htanβ

Distance travelled in level flight = tv sinα sinβ – h sinβ – h sinαsinα sinβ

Distance from base = ½ total distance

= tv sinα sinβ – h sinβ – h sinα2sinα sinβ

h2tanβ

++ h2tanα

Maximum distance from base =tv sinα sinβ – h [sinβ + sinα – sin(α + β)]

2sinα sinβ

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.

The End

Note: These equations should not be used for people flying light aircraft