Aerostatics and the High Altitude Balloon
Zak Weaver
Darrell Randall
Tyler Burke
9/21/2012
Dr. Jim Gregory
Friday 1:50-3:40
Introduction
The purpose of this experiment is to become familiar with the amount of lift that a helium
balloon can generate based on its volume as well as become familiar with the flight environment
through theoretical calculations as well as experimental data collected from a high altitude
balloon.
The hands on part of the lab examines how a lighter than air vehicle achieves lift and the
different forces that act on the balloon. With the measured diameter of the balloon, one can find
the weight of the air that the balloon displaces and thus find the amount of weight that the
balloon should be able to lift.
Theory
The buoyant force of an object is the weight of the fluid that it displaces. For air,
𝑊𝑎𝑖𝑟 = 𝜌𝑎𝑖𝑟 𝑔𝑉
This equation states that the weight of air in a certain region is equal to the density of the air
times the acceleration of gravity times the volume of the region. For a balloon to achieve lift, the
weight of the balloon must be less than the weight of the air that it displaces due to Archimedes’
principle. For a lighter than air gas, such as helium, the weight can also be given using this
equation.
𝑊𝐻𝑒 = 𝜌ℎ𝑒 𝑔𝑉
As helium is less dense than air, the weight of the helium is less than that of the air it displaces,
thus, the balloon can achieve lift.
However, the weight of the helium isn’t the only force that is acting downward on the balloon.
The weight of the balloon material and the weight of the payload must also be considered when
determining if the balloon can achieve lift. This gives an equation of lift
𝐿 = 𝑉𝑔(𝜌𝑎𝑖𝑟 − 𝜌𝐻𝑒 ) − 𝑊𝑏𝑎𝑙 − 𝑊𝑃𝐿
Where Wbal is the weight of the balloon material and WPL is the weight of the payload. This
equation should allow for the successful calculation of lift produced by the balloon to a close
degree of accuracy.
Procedure
The first step of the hands on portion of the lab was to record the ambient temperature and
pressure as these values are used to calculate the density of the air. Then, the balloon material
was weighed using a scale, and then the balloon was inflated using helium. The volume of the
inflated balloon was then recorded using the measured circumference around the balloon at
multiple points. Then the circumferences were averaged and an average volume was calculated
from this measurement. A paper cup and string was then attached to the balloon and weight was
added to the cup until the balloon achieved static equilibrium. Finally, the payload was weighed
and the volume of the balloon, the weight of the balloon material, and the weight of the payload
was reported to the TA.
Results and Discussion
Post Lab Work
Table1. Values of volume for each balloon, forces acting on the balloon, theoretical lift values,
actual lift values, and % error
Vbal
(in3)
Wair
(lb)
WHe
(lb)
Wbal
(lb)
625.439
503.077
610.318
0.027008
0.021724
0.026355
0.003732
0.003020
0.003641
0.005740
0.006200
0.006316
Theoretical
Lift
(lb)
0.017536
0.012504
0.016398
Actual Lift
(lb)
Percent
Error
0.01730
0.01636
0.01852
1.345
30.830
12.940
The values of actual lift for the balloons are very close compared to the values of theoretical lift
that were devised using the lift equation.
The error in the lab could be a result of a couple of different factors.
1. The balloons are not perfect spheres. The equation that was used to determine the
volume of the balloons holds true for the volume of a sphere, but as the balloons are not
spheres, there will be error in the volume.
2. It was rather difficult to attain a true circumference of the balloon, as the circumference is
different depending on where one measures it.
3. The temperature and pressure vary depending on one’s exact location. The pressure and
temperature used to calculate density were recorded in the room at an exact time and an
exact location. Both of these variable can change over time so they may not be the exact
same.
Theoretical Lift
0.06
0.05
y = 3E-05x
Lift (lb)
0.04
0.03
Payload Weight (lb)
0.02
0.01
0
0
500
1000
1500
Balloon Volume (in^3)
Plot 1. Theoretical Lift vs. Balloon Volume, both theoretical and experimental data from the
class.
Plot 1 shows the theoretical relationship of lift to balloon volume as the solid line and the
experimentally obtained values of lift as individual data points. One can determine that as
volume increases lift also increases in a linear manner.
Pre-Launch Work:
Ascent Rate for the given balloon.














At time of class the temperature was 79 °F and the pressure was 29.86 Hg in.
Converting temperature to °R yield 79+459.7 = 538.7 °R.
Converting pressure to lb/ft2 yields 29.86/29.92 = p/2116.2, P = 2111.96 lb/ft2
Density (ρ) = 𝑃/𝑅𝑇 = 2111.96/(1716 * 538.7) = .002285 slug/ft3
Weight of the payload (WPL) = 3.7 lb
Weight of the balloon (Wbal) = 2.7 lb
Circumference of the balloon = 250 in. or 20.833 ft
𝑐𝑖𝑟𝑐𝑢𝑚𝑓𝑒𝑟𝑒𝑛𝑐𝑒
Radius (r) =
= 20.833/2*п = 3.32 ft
2𝜋
Volume of the balloon = 4⁄3 𝜋𝑟 3 = 4⁄3 𝜋(3.32)3 = 152.69 ft3
Frontal area (s) = 𝐴 = 𝜋𝑟 2 = п(3.32)2 = 34.63 ft2
Acceleration due to gravity (g) = 32.2 ft/s2
Drag coefficient (CD) = 1.0
Molecular Weight of He (MWHe) = 4
Molecular Weight of air (MWair) = 28.8
Forces acting on the balloon: Buoyancy Force upwards, Weight of helium, Weight of payload,
Weight of the balloon and the Force of drag downwards.
Drag = FB – WPL – WHe - Wbal
1/2 * ρ * V2 * s * CD = ρair * g * Vbal * ρair * (1-(MWHe/MWair)) – WPL - Wbal
𝑉=√
𝑀𝑊
2 ∗ 𝑔 ∗ 𝑉𝑏𝑎𝑙 ∗ (1 − 𝑀𝑊𝐻𝑒 )
𝑎𝑖𝑟
(𝑠 ∗ 𝐶𝐷 )
−
2 ∗ 𝑊𝑃𝐿
2 ∗ 𝑊𝑏𝑎𝑙
−
𝜌𝑎𝑖𝑟 ∗ 𝑆 ∗ 𝐶𝐷 𝜌𝑎𝑖𝑟 ∗ 𝑆 ∗ 𝐶𝐷
V = 9.10 ft/s or 546 ft/min
Descent rate of the parachute / payload combination at an altitude of 50,000 ft







Density (ρair) at 50,000 ft = .00036391 slug/ft3
Weight of payload WPL = 3.7 lb
Drag coefficient (CD) = 2.5
Parachute diameter = 5 ft
Assume parachute is a half sphere
Surface Area (s) = ½(4*п*r2) = 39.27 ft2
Only 2 forces acting on the balloon. Upward force of drag and downward force of the
payload.
½ * ρair * V2 * s * CD = WPL
𝑉= √
𝜌
2∗𝑊𝑃𝐿
𝑎𝑖𝑟 ∗𝑠∗𝐶𝐷
V = 14.39 ft/s or 863.51 ft/min
Post Launch Work
2.5
4
x 10Experimental and Theoretical Variation of Temperature with Altitude
Ascent Temperature
Descent Temperature
Theoretical Variation of Temperature
Altitude(m)
2
1.5
1
0.5
0
220
230
240
250
260
270
280
Temperature(K)
290
300
310
320
Plot 2. Shows variation of temperature with an increasing/decreasing altitude with both
experimental and theoretical values.
The plot shows a general agreement in the shape of the experimental values with the shape of the
theoretical values, however, the values of the experimental temperatures vary from the
theoretical temperature by a noticeable amount.
This difference can be explained as the theoretical relationship with temperature and altitude
assumes standard atmosphere conditions. These conditions are an idealized set of conditions and
are often not found in the real world. These variations in atmospheric conditions explain the
difference between the theoretical and experimental values. Also, the descent temperatures are
much lower than the ascent temperatures because as the balloon gets higher, it gets cooler and
doesn’t warm as fast.
The shape of the plot stays consistent, which allows us to determine where the regions of the
atmosphere are located. The troposphere is the first layer and that is the region where the
temperature decreases linearly with altitude. After the troposphere, comes the stratosphere
where the temperature is constant. This is represented by the vertical regions in the plot.
12
4
x 10 Experimental and Theoretical Variation of Pressure with Altitude
Ascent Pressure
Descent Pressure
Theoretical Variation of Pressure
Experimental Pressure (Pa)
10
8
6
4
2
0
0
0.5
1
1.5
Altitude(m)
2
2.5
4
x 10
Plot 3. Shows variation of pressure with an increasing/decreasing altitude with both
experimental and theoretical values.
The plot shows a very good relationship between the theoretical values and altitude. The
theoretical pressure is always a bit lower than the experimental pressure, but it is such a little
difference that one cannot see a separation in the plot except at about 11000 meters. That
altitude is about the altitude that the troposphere and the stratosphere meet.
There is error in the experimental data of the ascent of the balloon. The transducer
malfunctioned, but righted itself so one can ignore the red diamonds that appear at the top of the
plot as they are erroneous data points.
Bonus Post Launch Work
See Appendix A for anticipated trajectory.
The balloon actually did not travel as far as anticipated and traveled mostly north and east. The
burst altitude was not 100000 feet due to overfilling the balloon slightly with helium. This
caused the flight time to be shorter and the balloon not traveling as far.
The reasons for some of the discrepancies of trajectory mostly come from varying ascent rate
and weather conditions. The weather conditions are constantly changing. The most recent
weather conditions and predictions were used to generate the anticipated trajectory, but those
change from second to second and can only act as an overall guideline, not an exact
interpretation of the weather.
Also, the rate of ascent significantly affects the balloons trajectory as it will reach different
weather conditions at different times than those that were used to generate the anticipated
trajectory. This ascent rate and descent rate are also constantly changing while the anticipated
trajectory was generated with constant ascent and descent rates.
Questions
1. A balloon of weight W will have a higher buoyant force in water. This is because water is
much denser than air. The density of water is 1000 kg/m3 and density of air at sea level is
1.2250 kg/m3. Water is almost 1000 times denser than air so the buoyancy force in water
will be almost 1000 times greater than the buoyancy force in air, assuming all other
variables are held constant.
𝐹𝐵 = 𝜌𝑤𝑎𝑡𝑒𝑟 ∗ 𝑔 ∗ 𝑉𝑏𝑎𝑙 ∗ (1 −
𝑀𝑊𝑔𝑎𝑠
) − 𝑊𝑃𝐿
𝑀𝑊𝑤𝑎𝑡𝑒𝑟
2. If the volume of the balloon were to remain fixed at its initial value at ground level the
maximum altitude to which the balloon would rise to can be determined by finding the
density of air at which the weight of air displaced is equal to the total payload weight.
Then using the standard atmosphere tables, altitude can be determined with the calculated
density.
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

Radius of the balloon = 4.33 ft
Vbal = 4/3 * п * r3 = 340.06 ft3
g = 32.2 ft/s2
ρair = 0.0003463 slug/ft3
WPL = 3.56 lb
ρair * Vbal * g = WPL * ρHe * Vbal * g
𝜌𝑎𝑖𝑟 =
𝑊𝑃𝐿 + 𝜌𝐻𝑒 ∗ 𝑉𝑏𝑎𝑙 ∗ 𝑔
𝑉𝑏𝑎𝑙 ∗𝑔
ρair = .0006714 slug/ft3
Using linear interpolation the altitude associated with a ρair = .0006714 slug/ft3 is
37,056.51 ft.
3. The volume of a spherical helium balloon that can lift a 175-lb person and 25 pounds of
equipment, assuming the balloon material weight .02 lbs/ft2 can be found by solving for
the radius in the following equation and then computing the volume of the sphere.




Ρair = .0023769 slug/ft3
g = 32.2 ft/s2
MWHe = 4
MWair = 28.8
4
𝑀𝑊𝐻𝑒
200 = 𝜌𝑎𝑖𝑟 ∗ 𝑔 ∗ ( ∗ 𝜋 ∗ 𝑟 3 ) ∗ (1 − (
)) − (4 ∗ 𝜋 ∗ 𝑟 2 ) ∗ (.02)
3
𝑀𝑊𝑎𝑖𝑟
Solving for r yields a value of 9.29 ft.
4
Plugging into the volume equation for a sphere,𝑣 = 3 ∗ 𝜋 ∗ 𝑟 3, the volume needed to
support the 175-lb person, 25 pounds of equipment and the weight of the balloon is
3,358.43 ft3.
Conclusion
The theoretical lift of a balloon was shown to increase linearly with volume of the balloon,
however it was hard to determine this experimentally because of difficulty accurately measuring
the volume of a balloon. These factors include difficulty of measuring a true circumference as
well as the assumption that a balloon is a sphere which is not the case.
Using equations of force acting on the balloon, it is possible to derive the velocity of ascent
incorporating the forces of the balloon at static equilibrium and adding the force of drag into the
mix to show that the balloon is now in motion. When the balloon bursts, the buoyant force
disappears and it is possible to find the descent rate using this data.
The temperature’s variation with altitude was indeed shown to be linear and the pressure’s
variation was shown to be a decreasing exponential relationship. These relationships match up
with theory very well.
The theoretical values were found to agree relatively well with the experimental values in most
cases. There were portions of the experiment where the theory and practice did not necessarily
match up, but these portions could be explained by variations in the atmosphere that could not be
accounted for completely.
Appendix A (Anticipated Trajectory)
Appendix B (MATLAB Script File)
close all
clear
clc
%Input values from excel spreadsheet.
Texpout= xlsread('calibrated_data.xlsx','e2:e683');
Hg=xlsread('calibrated_data.xlsx','h2:h683');
Paexp= xlsread('calibrated_data.xlsx','d2:d683')
%constants
g0=9.81;
T1=309.9;
P1=99036;
H1=324;
R=287;
Hf=max(Hg);
a1=-.0065
a2=0
%ascent/descent temperature
for g=1:length(Hg)
if(Hg(g)==max(Hg))
Top=g;
end
end
Tup=[]
Tdown=[]
for u=1:length(Hg)
if(u<=Top)
Tup=[Tup Texpout(u)];
else
Tdown=[Tdown Texpout(u)];
end
end
Lup=[1:Top]
Ldown=[Top:681]
%ascent/descent pressure
Pup=[]
Pdown=[]
for u=1:length(Hg)
if(u<=Top)
Pup=[Pup Paexp(u)];
else
Pdown=[Pdown Paexp(u)];
end
end
%theoretical temperature
Ttheo=[];
for r=1:length(Hg)
if(Hg(r)<=11000)
Ttheo(r)=T1+a1.*(Hg(r)-H1);
elseif(Hg(r)>11000)
Ttheo(r)=min(Ttheo);
end
end
%theoretical pressure
for r=1:length(Hg)
Ptheo(r)=P1.*exp(-((g0)./(R.*Ttheo(r))).*(Hg(r)-H1));
end
%plotting fun!
figure(1)
plot(Tup,Hg(Lup),'rs',Tdown,Hg(Ldown),'bs',Ttheo,Hg,'k-')
xlabel('Temperature(K)')
ylabel('Altitude(m)')
legend('Ascent Temperature','Descent Temperature','Theoretical Variation of
Temperature')
title('Experimental and Theoretical Variation of Temperature with Altitude')
figure(2)
plot(Hg(Lup),Pup,'rd',Hg(Ldown),Pdown,'bd',Hg,Ptheo,'k-')
xlabel('Altitude(m)')
ylabel('Experimental Pressure (Pa)')
legend('Ascent Pressure','Descent Pressure','Theoretical Variation of
Pressure')
title('Experimental and Theoretical Variation of Pressure with Altitude')