Balloon Flight Analysis

Weight

Meteorological Balloons

Balloon Type

1

2

3

Scientific Sales model #

-

8237

8244

Edmund Scientific order #

30665-68

-

-

Cost

$19

?

$72

Nominal Inflation diameter

4 ft

5 ft

6.3 ft

            Volume

33 ft3

65 ft3

130 ft3

Bursting Inflation diameter

8 ft

13 ft

30 ft

            Volume

270 ft3

1100 ft3

14,000 ft3

Mass of Balloon

110 g

300 g

1200 g

Weight of Balloon

1 N

3 N

12 N

Lift of Helium in std. air = 32 g/ft3 = 0.31 N/ft3 = 11 N/m3 of vol.

 

A common 4 ft cylinder of helium contains 244 ft3 at 2490 psig and 70° F.  This implies that the cylinder has a volume of 1.43 ft3. 

Assuming constant temperature: PV = constant

Every 10 ft3 will use up about 103 psi for this size cylinder.

 

Text Box: Flight String
150 #-test braided nylon twine
weighs 350g / 1000ft:   
m = 0.350 g/ft (9.8N/1000g) = 0.0034 N/ft
Fw = mL 

Gondola & Rig 
Gondola	500 g
Picavet rigging & hang-up	50 g
LabPro (w/batteries)	400 g
Barometer (or other sensor)	100 g
Temp probe (SS)	50 g
Camera & Intervalometer	350 g    
Wg 	=	1450 g


Forces

Text Box: Drag
Fd = ½ Cd r V2 A 
Cd = coefficient of drag » 0.5
r = density of air = 1.23 kg/m3
V = air speed  (1 knot = 0.514 m/s)
A = cross sectional area = pr2 
Fd = ½ Cd r p V2 (0.514m/s/knot)2 r2 (1m/3.28ft)2
     = k1V2r2 
k1 = 0.23 N/(knot ft)2

 

Free Lift

Fu = Lift due to Helium less weight of balloon & gondola

Fu = 31.7 g/ ft3 ·V – W = k2 r3 – W

k2 = 133 g/ft3 = 1.3 N/ft3

 

W1 = Wb + Wg =   110g + 1450g = 1560 g = 15 N

W2 = Wb + Wg =   300g + 1450g = 1750 g = 17 N

W3 = Wb + Wg = 1200g + 1450g = 2650 g = 26 N

In order to have a free lift of 500 g = 5 N at the ground:

r1 = 2.45 ft            C=15.4 ft        V=46 ft3          DP=473 psi

r2 = 2.53 ft            C=15.9 ft        V=51 ft3          DP=523 psi

r3 = 2.84 ft            C=17.8 ft        V=72 ft3          DP=739 psi

 

For every extra 100 g of payload, we will need an extra 3 ft3 of helium.  For a std 4 ft cylinder this will require that DP=31 psi

 

 

Tension

T = tension in the flight string as a function of flight string length from the balloon. 

Vertical component:        Tv = Fu – Ws

Horizontal component:    Th = Fd

 


Height

 

For a particular balloon, gondola, and 1000 ft of flight string, the altitude achieved will be determined by the ratio of the drag force and the net upward force.  For a particular situation with a required weight to lift and a given wind speed, the only adjustable factor is the Free Lift which requires that we change the inflation radius.

 

Assumptions and simplifications

Wind is moving uniformly with constant speed and direction and no turbulence.

Temperature is staying relatively constant and heating by sun is negligible.

The density of air remains constant over the entire range of altitude.

The string and gondola experience negligible drag.

The coefficient of drag is 0.5.

 

 


Flight Analysis

Using the above parameters, a mathematical model has been created (using Mathcad).  From it the following guidelines have been estimated.

Balloon type 1

Minimum requirements for various Wind Speeds

Given:   W = 15 N            L = 1000 ft      m = 350 g/1000 ft

with wind speeds:  v = 1, 2, 5,  &10 knots

In order the achieve at least 90% (900 ft) altitude we need the following:

 

Wind Speed

Desc

Free Lift at ground

Inflation Radius

Diameter

Circumference

1 knot

Calm

350 g

2.43 ft

4.86 ft

15.3 ft

2 knot

Light Air

375 g

2.44 ft

4.88 ft

15.3 ft

5 knots

Light Breeze

1.1 kg

2.72 ft

5.44 ft

17.1 ft

10 knots

Gentle Breeze

10 kg

4.43 ft

8.86 ft

27.8 ft

 

Maximum Altitudes and String Angles for various Wind Speeds

Given:       Fu = 500 gm   (Free Lift on the ground, ® r = 2.5 ft)

W = 15 N              L = 1000 ft      m = 350 g/1000 ft

with wind speeds:  v = 1, 2, 5,  &10 knots

 

Wind Speed

Desc

Maximum Altitude

String Angle from Horiz.

1 knot

Calm

999 ft

85 °

2 knot

Light Air

979 ft

70 °

5 knots

Light Breeze

647 ft

24 °

10 knots

Gentle Breeze

218 ft

6 °

 

 

General Guidelines

The balloon must be filled with helium until the balloon plus gondola has a net upward lift that is at least equal to the weight of 1000 ft of string (375 g).

 

In order to keep the maximum altitude within 10% (i.e. > 900 ft); for every knot of windspeed above 2 knots the balloon should have at least another 4 N (400 g) of net lift.

 

Increasing the radius by 10% will increase the free lift by about 60% of the total weight.

 


Appendix A:  Wind Speeds

Knots

Beaufort

Number

Name

Effects observed

Under 1

0

Calm

Calm; smoke rises vertically

1-3

1

Light air

Smoke drift indicates wind direction; vanes do not move.

4-6

2

Light breeze

Wind felt on face; leaves rustle; vanes begin to move.

7-10

3

Gentle breeze

Leaves, small twigs in constant motion; light flags extended.

11-16

4

Moderate breeze

Dust, leaves, and loose paper raised up; small branches move.

17-21

5

Fresh breeze

Small trees in leaf begin to sway.

22-27

6

Strong breeze

Larger branches of trees in motion; whistling heard in wires.

28-33

7

Near Gale

Whole trees in motion; resistance felt in walking against wind.

34-40

8

Gale

Twigs and small branches broken off trees; progress generally impeded.

Bowditch, American Practical Navigator, U.S. Naval Oceanographic Office, 1966