Why gas balloons keep rising with the increase of altitude?

Something was bothering me. The whole concept of the law of buoyancy is based of the fact that in order that the balloon rises it is necessary that the gas inside the balloon (in our case Helium) is lighter than the air. And the difference between the weight of the air and the weight of the gas is the weight that can be lifted by the balloon. The bigger the balloon, the bigger this difference, and the bigger weight can be lifted. The difference between the total weight that can be lifted and the weight of what is lifted (payload, balloon skin, strings) will define the rate of ascent. But we know that as the altitude rises, the weight of the air decreases, and even if the weight of Helium also decreases, it is intuitive thatthe total weight that could be lifted should decrease. If this was true it would mean that somewhere the balloon would cease rising and would just float. But is this so?

Well there's nothing like going back to the maths to clarify this issue and better understand how the variables relate to each other. To do this let's consider the weight of 1 litre of air and helium at 0km and 20km of altitude, or put in another way, the air density and helium density at 0km and 20km.

We have seen before that at Standard conditions for temperature and pressure (STP) - we can consider those as being the conditions at 0km altitude - we have:

Gas / Mixture	Weight of 1 litre (g) at 0km  <=> density at 0km	Weight of 1 litre (g) at 20km <=> density at 20km
Helium	0.176	?
Air	1,275	?

Now let's determine the density of Helium and Air at 20km of altitude. To do that we shall used our old friend, the ideal gas law, but first let's see what is the pressure and temperature at 20km, that like we have seen before, and according to the International Standard Atmosphere (ISA), is:

• Absolute pressure at 20 km: 5474.9 Pa= 0.054 atm
• Temperature at 20km: -56.5 ºC= 216.65 K

So equipped with these values let's do the maths.

Step 1. Determine the volume (in litres) occupied by 1 mole of the Helium or Air (at 20 km):

At 20 km we have:

T (Temperature)= 216.65 K
P (Absolute pressure)= 0.054 atm

Additionally we have the ideal gas constant R= 0.08205746(14) L atm K^-1 mol^-1 which we can approximate to R= 0.082 L atm K^-1 mol^-1

And n= 1 since we want to determine the volume of 1 mole of helium or air.

Now, by using the ideal gas law we just have to replace the variables by their correspondent values:

PV = nRT
V= (nRT) / P
V= (1 x 0.082 x 216.65) / 0.054
V= 17.765 / 0.054 = 328.98 litres

Notice that above in the ideal gas law we have user as units for n, T and P, the same units in which the R constant is expressed: mol, K, and atm respectively.

So we have just determined that both 1 mole of Helium as well as 1 mole of Air occupy 328.98 litres at 20 km of altitude.

Step 2. With the result in 1. determine to how much moles correspond 1 litre of Helium or Air (at 20 km):

Now we have:

1 mol of Helium/Air ____(corresponds)_____ 328.98 litres
X mol of Helium/Air____(corresponds)_____ 1 litres

And by applying the rule of three, which is a particular form of cross-multiplication:

X = (1 * 1) / 328.98 = 0.003 mol

So we have just determined that 1 litre of Helium or 1 litre of air contain 0.003 moles.

Step 3. Using the molar mass of Helium and of the Air ,which is expressed in g/mol, and the result in 2, get the weight in grams (at 20 km):

And now we have reached to the last step. As seen in a previous post, the molar mass (molecular weight) of Helium is equal to its atomic weight (since Helium has only 1 atom). In this way, and knowing that the atomic weight of Helium A_r(Helium)= 4.003 g/mol, it is easy to calculate the molar mass of Helium, M(Helium):

M(Helium)= A_r(Helium)= 4.003 g/mol

In this way, we can now use once again the rule of three:

4.003 g         ____(corresponds)_____ 1 mol
X g of Helium____(corresponds)_____ 0.003 mol

X = (0.003 * 4.003) / 1 = 0.012g

Which means that the weight of 1 litre of Helium at 20 km of altitude is 0.012g, which is equivalent to:
Helium density (ρ) = 0.012g/l = 0.012 Kg/m³

Regarding the Air, we know by a previous post, that the molar mass (molecular weight) of Air is equal to:

M(Air)= 28.972 g/mol

In this way, we can now use once again the rule of three:

28.972 g         ____(corresponds)_____ 1 mol
X g of Air ____(corresponds)_____ 0.003 mol

X = (0.003 * 28.972) / 1 = 0.087g

Which means that the weight of 1 litre of Air at 20 km of altitude is 0.087g, which is equivalent to:
Air density (ρ) = 0.087g/l = 0.087 Kg/m³

And now we can complete the previous table containing the Helium and Air density at 0km and 20km as follows:

Gas / Mixture	Weight of 1 litre (g) at 0km  <=> density at 0km	Weight of 1 litre (g) at 20km <=> density at 20km
Helium	0.176	0,012
Air	1,275	0.087

Now, and after revising the post on how much weight can a Helium balloon lift, we know that:

• At 0 Km of altitude, 1 litre of Air weights 1,275g, whereas 1 litre of Helium weights 0,176g. This means that a 1 litre Helium balloon (plus its payload) shall rise if the weight of the balloon (without gas) plus the payload is less than 1,275-0,176g= 1,099g (if the weight is the same the balloon will float and not rise);
• At 20 Km of altitude, 1 litre of Air weights 0,087g, whereas 1 litre of Helium weights 0,012g. This means that a 1 litre Helium balloon (plus its payload) shall rise if the weight of the balloon (without gas) plus the payload is less than 0,087-0,012g= 0,075g (if the weight is the same the balloon will float and not rise).

This result is surprising, or not...

At 0 Km if you fill-in a balloon with, for instance, 500 litres of Helium,  the balloon (plus its payload) shall rise if the weight of the balloon (without gas) plus the payload that it may carry is less than 1,099 * 500 = 549,5g. This means that if we had a balloon that without gas weights 300g, we could carry a payload of a bit less than 549,5-300= 249,5g, for instance 240g.

But applying the same maths, the same balloon at 20km of altitude,  would only rise if the weight of the balloon (without gas) plus the payload that it carries is less than 0,075 * 500 = 37,5g. This means that with the same balloon weighting 300g, the balloon would already been descending much time before, meaning that would never reach the 20km of altitude. But we know that at this altitude the ballon is still rising. So what is failing in all this reasoning?

Well the problem is that we were assuming that the balloon filled with a volume of 500 litres of Helium at 0 km, would still have 500 litres of volume at 20 km of altitude. And that is wrong as we already have seen in our previous Post. So, let's determine the volume of the balloon at 20Km of altitude. To this let's use the general formula bellow:

$$V_{2}=V_{1}\frac{P_{1}T_{2}}{P_{2}T_{1}}$$
So the above formula is a general formula that allow us to determine the volume of the balloon in state 2 (S₂) given the balloon conditions in state 1 (S₁) and given the temperature and pressure of the balloon in state 2 (S₂).
Regarding the state 1, let's consider the that balloon at 0 km of altitude is filled in with 500 litres of Helium and that the pressure and temperature follow the STP values that we have indicated in this previous Post.

State 1 (S₁)

• Altitude= 0 meters (sea-level)
• Pressure= P1= 100000 Pa (at STP)
• Temperature= T1= 273.15 k (at STP)
• Volume= V1=  500 L=  0.5 m³

Then for state 2 (S₂) of the balloon, we would like to know the volume of the same balloon when at an altitude of 20000 meters. So  let's consider that for the absolute pressure at 20 km we have  5474.9 Pa  and the temperature is  -56.5 ºC, as indicated by the International Standard Atmosphere (ISA):

State 2 (S₂)

• Altitude=  20000 meters (20km)
• Pressure= P2= 5474.9 Pa
• Temperature= T2= -56.5 ºC= 216.65 k
• Volume= V2=  ?

Now it is just a question of using the general formula presented above as follows:

$$V_{2}=V_{1}\frac{P_{1}T_{2}}{P_{2}T_{1}} <=> V_{2}=500 \times \frac{100000 \times 216.65}{5474.9 \times 273.15} <=> V_{2}=500 \times \frac{21665000}{1495468.935} <=>$$

$$<=> V_{2}=500 \times 14.49 <=> V_{2}=7245L= 7.25m^{3}$$.

So now that we know that the balloon at 20km occupies a volume of 7245 litres (instead of the 500 litres at 0 Km of altitude), we also know that the balloon at 20km of altitude would rise if the weight of the balloon (without gas) plus the payload that it carries is less than 0,075 * 7245 = 543.38g. This is a value very similar to the 549.5g that the balloon is able to carry at 0 Km of altitude, corresponding to a drop of approximately 1 % in the weight that can be lifted.

This is just fantastic. As the balloon rises from 0km to 20km of altitude, the difference between the weight of the Helium and the Air around it drops substantially (14.65 times) what would makes us imagine that the lifting capability of the balloon would also drop 14.65 times. But in fact that is not the case because the volume occupied by the Helium rises also substantially (14.49 times) as the balloon rises up to 20km of altitude. This seems like magic, and it is... it is the magic of Nature and of the Physical world.

Just by curiosity let's see what would be the situation for the balloon in a third state at 30 km of altitude.

Step 1. Determine the volume (in litres) occupied by 1 mole of the Helium or Air (at 30 km):

At 30 km we have:

T (Temperature)=  226.5 K
P (Absolute pressure)= 1197 Pa= 0.012 atm

Additionally we have the ideal gas constant R= 0.08205746(14) L atm K^-1 mol^-1 which we can approximate to R= 0.082 L atm K^-1 mol^-1

And n= 1 since we want to determine the volume of 1 mole of helium or air.

Now, by using the ideal gas law we just have to replace the variables by their correspondent values:

PV = nRT
V= (nRT) / P
V= (1 x 0.082 x 226.5) / 0.012
V= 18.573 / 0.012 = 1547.75 litres

Notice that above in the ideal gas law we have user as units for n, T and P, the same units in which the R constant is expressed: mol, K, and atm respectively.

So we have just determined that both 1 mole of Helium as well as 1 mole of Air occupy 1547.75 litres at 30 km of altitude.

Step 2. With the result in 1. determine to how much moles correspond 1 litre of Helium or Air (at 20 km):

Now we have:

1 mol of Helium/Air ____(corresponds)_____ 1547.75 litres
X mol of Helium/Air____(corresponds)_____ 1 litres

And by applying the rule of three, which is a particular form of cross-multiplication:

X = (1 * 1) / 1547.75 = 0.00065 mol

So we have just determined that 1 litre of Helium or 1 litre of air contain 0.00065 moles.

Step 3. Using the molar mass of Helium and of the Air ,which is expressed in g/mol, and the result in 2, get the weight in grams (at 30 km):

And now we have reached to the last step. As seen in a previous post, the molar mass (molecular weight) of Helium is equal to its atomic weight (since Helium has only 1 atom). In this way, and knowing that the atomic weight of Helium A_r(Helium)= 4.003 g/mol, it is easy to calculate the molar mass of Helium, M(Helium):

M(Helium)= A_r(Helium)= 4.003 g/mol

In this way, we can now use once again the rule of three:

4.003 g         ____(corresponds)_____ 1 mol
X g of Helium____(corresponds)_____ 0.00065 mol

X = (0.00065 * 4.003) / 1 = 0.0026g

Which means that the weight of 1 litre of Helium at 30 km of altitude is 0.0026g, which is equivalent to:
Helium density (ρ) = 0.0026g/l = 0.0026 Kg/m³

Regarding the Air, we know by a previous post, that the molar mass (molecular weight) of Air is equal to:

M(Air)= 28.972 g/mol

In this way, we can now use once again the rule of three:

28.972 g         ____(corresponds)_____ 1 mol
X g of Air ____(corresponds)_____ 0.00065 mol

X = (0.00065 * 28.972) / 1 = 0.019g

Which means that the weight of 1 litre of Air at 30 km of altitude is 0.019g, which is equivalent to:
Air density (ρ) = 0.019g/l = 0.019 Kg/m³

So, at 30 Km of altitude, 1 litre of Air weights 0,019g, whereas 1 litre of Helium weights 0,0026g. This means that a 1 litre Helium balloon (plus its payload) shall rise if the weight of the balloon (without gas) plus the payload is less than 0,019-0,0026g= 0,0164g (if the weight is the same the balloon will float and not rise).

Additionally, like we have seen before at 30 Km:


State 3 (S₃)

• Altitude=  30000 meters (30km)
• Pressure= P3= 1197 Pa
• Temperature= T3= 226.5 k
• Volume= V3=  ?

Using again the above formula:
$$V_{3}=V_{1}\frac{P_{1}T_{3}}{P_{3}T_{1}} <=> V_{3}=500 \times \frac{100000 \times 226.5}{1197 \times 273.15} <=> V_{3}=500 \times \frac{22650000}{326960.55} <=>$$

$$<=> V_{3}=500 \times 69.27 <=> V_{3}=34635L= 34.64m^{3}$$.

So now that we know that the balloon at 30km occupies a volume of 34635 litres (instead of the 500 litres at 0 Km of altitude), we also know that the balloon at 30km of altitude would rise if the weight of the balloon (without gas) plus the payload that it carries is less than 0,0164 * 34635 = 568.014g. This is a value even slightly bigger similar to the 549.5g that the balloon is able to carry at 0 Km of altitude, corresponding to an increase of 3% in the weight that can be lifted.

This also shows that the factor that will determine the return of the balloon to the Earth surface is not its capability to lift the weight that it carries, but the extensibility of the balloon material, meaning its capacity to be able to increase in volume without bursting - and in this post we have seen that the balloon occupies a volume of 500 litres at the Earth's surface and then 7245 at 20km and 34635 at 30km.

Increase of the size of an Helium balloon as it raises into the atmosphere

In this Post we shall determine the increase of size of a balloon as it raises up into the atmosphere. For this we shall set a practical scenario and start using some of the concepts and terminology that we have seen in previous Posts.

So let's consider a scenario where we want to send a balloon carrying a payload weighting 800g to an altitude of 30000 meters (30Km). From the Kaymont balloons manufacturer site lets choose the HAB-600 balloon. This is a 600 grams balloon recommended for individuals or groups launching a photographic mission for the first time. This balloon is recommended for payloads with a weight between 500 and 1000 grams.

Then lets use the CUSF balloon burst calculator and set the following (be sure that the selected gas is Helium):

• Payload mass (g): 800g
• Balloon mass (g): Kaymont - 600
• Target burst altitude (m): 30000

The result from the calculator shall be:

• Burst altitude: 30000m (as we have selected in the inputs for the calculator)
• Time to burst: 134m
• Launch volume: 1.81m3 (1811 L)
• Ascent rate: 3.72 m/s (223.2 m/min)
• Neck lift: 1258g

So now we know that by filling the 600-grams balloon with a volume of 1811 litres of Helium it is possible to send the balloon to an altitude of 30 Km carrying a payload weighting 800 grams. We discover also that this is possible with an ascent rate of 223.2 meters per minute - which is a bit far from the recommend ascent rate which is normally set to 320m/min - the consequence is that the flight will take longer, in this case, 134 minutes, which in my opinion is still reasonable. As we have seen in a previous post the neck lift is the difference between the gross lift (the total lift generated by the Helium gas) and the weight of the balloon (in this case 600 grams), meaning that is the lift available to the payload (if you imagine the balloon as the head, the nose will be in the beginning of the payload) and to generate the lift of the system. In this case the nose lift is 1258g, which means that if you subtract from that the value of the weight of the payload (800g) you get the amount of lift available to generate the lift of the system (458 grams).

To summarise we now have a 600-grams balloon at the Earth surface (let's consider at the sea level - 0 meters) filled in with 1811 litres of Helium. How can we then know what will be the size of the balloon at the surface and then at different altitudes?

Well for the size of the balloon let's apply the formula for the volume of a sphere (we are here considering the balloon to be a sphere):
$$Volume= V= \frac{4}{3}\pi r^{3}$$
Where:

• π= 3.14
• r= radius= ^diameter⁄₂

The units that we use for the radius will then have an impact on the resulting units for the volume. If we use meters for the radius, then we get the volume in m³. So, for the case above we have a volume of 1811 litres, which considering that 1 m³ corresponds to 1000 litres, gives a volume of 1.811 m³. In this way the sphere formula can be written in the following way:

$$1.811=\frac{4}{3}3.14r^{3}=4.19r^{3}\Leftrightarrow r^{3}=0.43\Leftrightarrow r=0.75m\Leftrightarrow diameter=d=r*2=1.5m$$
To summarise once again (but with more information), we have a 600-grams balloon at the Earth surface (let's consider at the sea level - 0 meters) filled in with 1811 litres of Helium and with a diameter of 1.5m.

Now let's determine the volume of this same balloon at a different altitude, for instance at 20km. As we have seen before, as the balloon rises, its volume increases and consequently the diameter of the balloon also increases. So let's use the ideal gas law to determine what will be the volume and diameter of the balloon at this altitude of 20km, but let's first deduce a general formula that allows to determine that. As we have seen before:

PV = nRT

Imagine that we have a balloon and that we have two different states, S₁ and S₂, associated to the balloon. In state 1 (S₁) the balloon is at the Earth's surface and in state 2 (S₂) the balloon will be at a different altitude, for instance, 20 km.
So now let's draw the ideal gas law for each one of the states S₁ and S₂:

For S₁:
P₁V₁ = n₁RT₁
^(P₁V₁)⁄_(n1RT1)= 1                        (eq. 1)


For S₂:
P₂V₂ = n₂RT₂
^(P₂V₂)⁄_(n2RT2)= 1                         (eq. 2)

As the first term of eq. 1 equals 1 and the first term of eq. 2 also equals 1, we can say that those terms are equal to each other:
^(P₁V₁)⁄_(n1RT1) = ^(P₂V₂)⁄_(n2RT2)

But as the amount of gas do not change between  S₁ and S₂, we can say that n =  n₁ = n₂, and consequently:
^(P₁V₁)⁄_(nRT1) = ^(P₂V₂)⁄_(nRT2) <=> ^nR(P₁V₁)⁄_(nRT1) = ^nR(P₂V₂)⁄_(nRT2) <=> ^(P₁V₁)⁄_(T1) = ^(P₂V₂)⁄_(T2) <=>
$$V_{2}=V_{1}\frac{P_{1}T_{2}}{P_{2}T_{1}}                       (eq. 3)$$
So the above formula is a general formula that allow us to determine the volume of the balloon in state 2 (S₂) given the balloon conditions in state 1 (S₁) and given the temperature and pressure of the balloon in state 2 (S₂). To make this more concrete let's apply to our example balloon. As we have seen before, in state 1 (S₁), we have the 600-grams balloon at sea level (0 meters) filled in with 1811 litres of Helium and with a diameter of 1.5m. Additionally let's consider that for the absolute pressure at sea level we have 101325 Pa and the temperature is 15ºC, as indicated by the International Standard Atmosphere (ISA):

State 1 (S₁)

• 600-grams balloon
• Altitude= 0 meters (sea-level)
• Pressure= P1= 101325 Pa
• Temperature= T1= 15 ºC= 288.15 k
• Volume= V1=  1811L=  1.811 m³
• Diameter= 1.5 m 

Then for state 2 (S₂) of the balloon, we would like to know the volume and consequently the diameter of the same balloon when at an altitude of 20000 meters. So  let's consider that for the absolute pressure at 20 km we have  5474.9 Pa  and the temperature is  -56.5 ºC, as indicated by the International Standard Atmosphere (ISA):

State 2 (S₂)

• 600-grams balloon
• Altitude=  20000 meters (20km)
• Pressure= P2= 5474.9 Pa
• Temperature= T2= -56.5 ºC= 216.65 k
• Volume= V2=  ?
• Diameter= ? 

Now it is just a question of using the eq.3 as follows:

$$V_{2}=V_{1}\frac{P_{1}T_{2}}{P_{2}T_{1}} <=> V_{2}=1811 \times \frac{101325 \times 216.65}{5474.9 \times 288.15} <=> V_{2}=1811 \times \frac{21952061.25}{1577592.435} <=>$$
$$<=> V_{2}=1811 \times 13.91 <=> V_{2}=25191.01L= 25.2m^{3}$$
And now that we have determined the volume of the balloon at state 2 (S₂), it is just a question of determining the new diameter of the balloon:
$$25.2=\frac{4}{3}3.14r^{3}=4.19r^{3}\Leftrightarrow r^{3}=6.01\Leftrightarrow r=1.82m\Leftrightarrow diameter=d=r*2=3.64m$$
Finally let's model a state 3 (S₃) of the balloon, where we would like to know the volume and consequently the diameter of the same balloon when at an altitude of 30000 meters. For the pressure and temperature at that altitude let's use again the ISA values but this time retrieved from the Engineering ToolBox site. So let's consider that for the absolute pressure at 30 km we have 0.01197 bar, which corresponds to 1197 Pa, and the temperature is 226.5k:

State 3 (S₃)

• 600-grams balloon
• Altitude=  30000 meters (30km)
• Pressure= P3= 1197 Pa
• Temperature= T3= 226.5 k
• Volume= V3=  ?
• Diameter= ? 

Now it is just a question of using the eq.3 as follows:

$$V_{3}=V_{1}\frac{P_{1}T_{3}}{P_{3}T_{1}} <=> V_{3}=1811 \times \frac{101325 \times 226.5}{1197 \times 288.15} <=> V_{3}=1811 \times \frac{22950112.5}{344915.55} <=>$$
$$<=> V_{3}=1811 \times 66.54 <=> V_{3}=120503.94L= 120.5m^{3}$$
And now that we have determined the volume of the balloon at state 3 (S₃), it is just a question of determining the new diameter of the balloon:
$$120.5=\frac{4}{3}3.14r^{3}=4.19r^{3}\Leftrightarrow r^{3}=28.76\Leftrightarrow r=3.06m\Leftrightarrow diameter=d=r*2=6.12m$$
The table bellow summarises the results for the three states, S₁, S₂, S₃:

	S1	S2	S3
Type	600-grams	600-grams	600-grams
Altitude (m)	0	20000	30000
Pressure (Pa)	101325	5474.9	1197
Temperature (K)	288.15	216.65	226.5
Volume (m3)	1.811	25.2	120.5
Diameter (m)	1.5	3.64	6.12

If you remember from the beginning of this post, we have used the CUSF balloon burst calculator and set the balloon parameters in such a way that the target burst altitude would be 30000m. Well, from the above calculations we have just determined that at 30000m the balloon will have a diameter of 6.12m. Consequently this means that the diameter of the balloon when bursting will be approximately 6.12m.
These results are very similar to the results presented in the Kaymont balloons manufacturer site regarding the weather forecasting (sounding) 600-grams balloon (with a 250g payload) - it is just a question of converting the units there in order to see that the resulting values are very similar to the ones we have just reached above. This means that we are now equipped to better understand the balloon specifications (and nomenclature) and the different possible configurations of the balloon and its consequences.

How much weight can a Helium balloon lift?

In this post I will describe how we can determine the lifting power of a Helium balloon. In other words, how much weight can a Helium balloon lift?
In a previous post we have already seen what is the intuition behind the reason why gas balloons rise: "...the air that was displaced to make room for the gas balloon, as it is heavier than the balloon, will want to claim its place back, and when it does, the balloon will have to occupy the space where that air was...". From this intuition it is clear that the Helium balloon will rise as long as the total weight of the balloon plus the gas inside, is lighter than the air that the balloon displaces.

Back in a previous post we have determined that the air density (weight in grams of 1 litre of gas ) of the Air and Helium is the following:

ρAir= 1,275g

ρHelium= 0,176g

This is equivalent to say that 1 litre of Air weights 1,275g, whereas 1 litre of Helium weights 0,176g. This means that a 1 litre Helium balloon (plus its payload) shall rise if the weight of the balloon (without gas) plus the payload is less than 1,275-0,176g= 1,099g (if the weight is the same the balloon will float and not rise). As you can imagine the balloon itself (without gas and payload) will weight quite more than 1,099g which means that this balloon will not rise. But if you fill-in a balloon with, for instance, 500 litres of Helium,  the balloon (plus its payload) shall rise if the weight of the balloon (without gas) plus the payload that it may carry is less than 1,099 * 500 = 549,5g. This means that if we had a balloon that without gas weights 300g, we could carry a payload of a bit less than 549,5-300= 249,5g, for instance 240g. This is the reason why the gas balloons are so big - as the air density difference between Helium and Air is only approximately 1g (more exactly 1,099g), we must have a balloon with much more Helium than just 1 litre in order that this 1g of difference between Air and Helium is much bigger allowing to compensate the weight of the balloon itself and some payload.

In the International System of Units (SI) the volume is expressed in cubic meters (m³). In this way it is quite convenient to know how much weight 1m³ of Helium can lift. Since 1m³  is exactly 1000 litres, and since, as we have seen before, 1 litre of Helium can lift 1,099g, we have:

1m³ of Helium can lift 1000*1,099g= 1099g

But in fact if our calculations are 100% correct (which aren't because we made the calculations are based on many assumption - for instance this applies to STP conditions) we should remove some grams from the value above. This is because if 1 litre of Helium carries 1,099g of weight, the balloon will, as we have seen above, only float and not rise because for this case the Helium balloon weight would be exactly the same than the weight of the Air the balloon has to displace. Is this way and in order to remain in the safe side let's say that:

1m³ of Helium can lift 1000g = 1Kg

For some countries is more convenient to know the above but for 1 cubic foot of Helium. By searching the Internet the commonly agree value is - see this introductory very well written article in the How Stuff Works site:

1 cubic foot of Helium can lift 28.2g

This value makes sense because people usually consider that 1 litre of Helium is able to carry 1g of weight (it is is a good approximation to our value of 1,099g) and then we know that 1 cubic foot = 28.3 litres. Well but this would mean that 1 cubic foot of Helium could lift 28.3g and not only 28.2g. Here is like we have seen above, a weight of 28.3g would mean that the balloon would be just floating and not rising, and so the value of 28.2 guarantees that the balloon will lift.

If we wanted to perform this calculation for a Hydrogen balloon it would just be a question of replacing the density of Helium by the Hydrogen density and repeat the calculations. If we did that we would conclude that the additional buoyancy of Hydrogen is comparison with Helium is approximately 8.0%, which many people argue that is not enough to compensate the fact that Hydrogen is potentially explosive in comparison with Helium which is inert.

Before concluding let's just introduce some notation to denominate the concepts just introduced:

• Gross lift (gr) - The total weight that the considered amount of gas (Helium, Hydrogen, other) can lift;
• Free lift (gr) - The difference between the gross lift and the total weight of the system (excluding the gas itself). It is the free lift that sets the ascent rate. If we add a weight equal to the free lift to the payload weight, we will achieve neutral buoyancy - the balloon will neither rise nor descend - this is a good technique to use before launching the balloon when doing the last preparations - when all is ready it is just a question of removing this extra weight for the balloon to rise up into the atmosphere;
• Neck lift (gr) - is the difference between the gross lift (the total lift generated by the Helium gas) and the weight of the balloon (in this case 600 grams), meaning that is the lift available to the payload (if you imagine the balloon as the head, the nose will be in the beginning of the payload) and to generate the lift of the system.
• Rate of ascent (m/min) - velocity with which the whole system rises up into the atmosphere  
• Recommended free lift (gr) - some balloon manufacturers like Totex refer to the recommended free lift which is the free lift that a system must have in order to reach to a given ascent rate. 

And finally to conclude and wrap it up, let's say that:

• The gross lift of 1m³ of Helium is approximately 1000g (= 1Kg) of total weigh - this means that when sizing the system you must decrease from the available 1000g, the weight of the balloon skin itself, the payload, the cords, the radar reflector (if any), any other items carried by the balloon, and the remaining value (free lift) must be at least bigger than or equal to 0;
• Bigger values of the free lift imply faster ascent rates - balloon manufacturers like Totex recommend which should be the free lift for each one of their balloons in order that the rate of ascent is always 320 meters per minute;
• The system should be sized to have as goal reaching a given altitude - see bellow.

And of course let's always remember that as the balloon gets higher, it expands due to the pressure outside decreasing while the pressure inside remains the same. So while it is true that at release we can have a smaller balloon filled in with more gas carrying the same weight than a bigger balloon with less gas, and both rising at the same ascent rate, it is also true that the smaller balloon will burst first than the bigger balloon (considering that both are made from the same type of material). So the whole system must be sized in order to reach to a given goal altitude.

Additionally it may also be important to consider that faster rates of ascent imply less chance of something going wrong - equipment failure, dead batteries, remote landing site, collision, among many, many, others.

From UKHAS you can find this excellent guide containing information and pointing to simulators where you can set different scenarios and ultimately decide on how will be your balloon mission in terms of type of balloon, quantity of gas, size of the balloon, among many other parameters. And with the information provided in this and previous posts hopefully you are much better equipped to dimension correctly your balloon mission.

What is the weight of 1 litre of Helium at STP (or density)? And 1 litre of Hydrogen? And 1 litre of Air?

To being able to calculate how much weight a given balloon can lift  (the lifting power) based on the volume of the balloon,  it is first necessary to know what is the weight of the gas inside a balloon with a given volume. Or more technically speaking, what is the density of the gas. This post aims to answer to this last question, by considering the gas to be Helium, Hydrogen and Air, and by considering the balloon as being subject to Standard conditions for temperature and pressure (STP).

We shall now apply what we have seen in my previous post to calculate the weight of X litres of Gas.

1. Determine the volume (in litres) occupied by 1 mole of the Gas;
2. With the result in 1. determine to how much moles correspond X litres of Gas;
3. Using the molar mass of the Gas ,which is expressed in g/mol , and the result in 2, get the weight in grams.

Realizing what we have just said, lets determine the weight of 1 litre of Helium:

Step 1. Determine the volume (in litres) occupied by 1 mole of the Helium:

At STP we have:

T (Temperature)= 273.15 K
P (Absolute pressure)= 0.986 atm

Additionally we have the ideal gas constant R= 0.08205746(14) L atm K^-1 mol^-1 which we can approximate to R= 0.082 L atm K^-1 mol^-1

And n= 1 since we want to determine the volume of 1 mole of helium.

Now, by using the ideal gas law we just have to replace the variables by their correspondent values:

PV = nRT
V= (nRT) / P
V= (1 x 0.082 x 273.15) / 0.986
V= 22.398 / 0.986 = 22.7 litres

Notice that above in the ideal gas law we have user as units for n, T and P, the same units in which the R constant is expressed: mol, K, and atm respectively.

So we have just determined that 1 mole of Helium occupies 22.7 litres.

Step 2. With the result in 1. determine to how much moles correspond 1 litres of Helium:

Now we have:

1 mol of Helium ____(corresponds)_____ 22.7 litres
X mol of Helium____(corresponds)_____ 1 litres

And by applying the rule of three, which is a particular form of cross-multiplication:

X = (1 * 1) / 22.7 = 0.044 mol

So we have just determined that 1 litre of Helium contains 0.044 moles.

Step 3. Using the molar mass of Helium ,which is expressed in g/mol, and the result in 2, get the weight in grams:

And now we have reached to the last step. As seen in a previous post, the molar mass (molecular weight) of Helium is equal to its atomic weight (since Helium has only 1 atom). In this way, and knowing that the atomic weight of Helium A_r(Helium)= 4.003 g/mol, it is easy to calculate the molar mass of Helium, M(Helium):

M(Helium)= A_r(Helium)= 4.003 g/mol

In this way, we can now use once again the rule of three:

4.003 g         ____(corresponds)_____ 1 mol
X g of Helium____(corresponds)_____ 0.044 mol

X = (0.044 * 4.003) / 1 = 0.176g

Which means that the weight of 1 litre of Helium is 0.176g, which is equivalent to:
Helium density (ρ) = 0.176g/l = 0.176 Kg/m³

If we wanted to determine the weight of 1 litre of hydrogen instead of Helium, we would just have to replace the molar mass of Helium by the molar mass of Hydrogen - please be aware that hydrogen in diatomic (H2). See bellow the calculations:

Step 1:

It is the exact same calculations that were made above for Helium, meaning that:

1 mole of Hydrogen occupies 22.7 litres.

Step 2:

It is the exact same calculations that were made above for Helium, meaning that:

1 litre of Hydrogen contains 0.044 moles.

Step 3:

The molar mass (molecular weight) of Hydrogen is equal to twice its atomic weight (since  Hydrogen  has 2 atoms - diatomic). In this way, and knowing that the atomic weight of Helium A_r(Helium)= 1.008 g/mol, it is easy to calculate the molar mass of  Hydrogen , M( Hydrogen ):

M( Hydrogen )= A_r( Hydrogen ) * 2= 2.016 g/mol

In this way, we can now use once the rule of three:

2.016 g                ____(corresponds)_____ 1 mol
X g of  Hydrogen ____(corresponds)_____ 0.044 mol

X = (0.044 * 2.016) / 1 = 0.089 g

Which means that the weight of 1 litre of Hydrogen is 0.089g, which is almost half the weight of Helium, and which is equivalent to:
Hydrogen density (ρ) = 0.089g/l = 0.089 Kg/m³

If we wanted to determine the weight of 1 litre of Air, we would just have to use the molar mass of Air. See bellow the calculations:

Step 1:

It is the exact same calculations that were made before, meaning that:

1 mole of Air occupies 22.7 litres.

Step 2:

It is the exact same calculations that were made before, meaning that:

1 litre of Air contains 0.044 moles.

Step 3:

The determination of the molar mass (molecular weight) of Air is complicated  because the chemical composition of air varies. The most notable variation is related with the presence of water vapour (H₂O), which is the gas phase of water. Following what is stated in Wikipedia, the water vapour represents 0.40% of the atmosphere and at the Earth's surface may represent 1% to 4%. Here, you can read a very interesting article on the air compositions and properties, including information regarding the water vapour and air density. But since the water vapour in the Air varies, let's simplify things and do the calculation for dry Air.
And as stated in the IUPAC Gold Book, "the composition of the major components in dry air is relatively constant (percent by volume given): nitrogen, 78.084; oxygen, 20.946; argon, 0.934; carbon dioxide, 0.033; neon, 0.0018; helium, 0.000524; methane, 0.00016; krypton, 0.000114; hydrogen 0.00005; nitrous oxide, 0.00003; xenon, 0.0000087". See bellow the table that presents and overview of these values and after the calculations to determine the molar mass of dry air.

Gas	Chemical symbol	% by volume	Atomic Weight	Molecular Weight
Nitrogen	N2	78.084	N: 14.007	28.014
Oxygen	O2	20.946	O: 15.999	31.998
Argon	Ar	0.934	Ar: 39.95	39.95
Carbon dioxide	CO2	0.033	C: 12.011 O: 15.999	63.996
Neon	Ne	0.0018	Ne: 20.18	20.18
Helium	He	0.000524	He: 4.003	4.003
Methane	CH4	0.00016	C: 12.011 H: 1.008	16.043
Krypton	Kr	0.000114	Kr: 83.80	83.80
Hydrogen	H2	0.00005	H: 1.008	2.016
Nitrus Oxide	N2O	0.00003	N: 14.007 O: 15.999	44.013
Xenon	Xe	0.0000087	Xe: 131.3	131.3

The atomic weights of the table above were retrieved from table 4 of the IUPAC Recommended Atomic Weights of 2009, where the atomic weights are abridged to four significant digits, and which apply to elements of natural terrestrial origin (as the gases in the air do). The values that are provided in an interval, like hydrogen, were retrieved from table 6 instead.

So now to determine the molecular weight of Air it is just a question of performing a weighted multiplication as shown bellow:

M(Air)= 0.78084 * 28.014 + 0.20946 * 31.998 + 0.00934 * 39.95 + 0.00033 * 63.996 + 0.000018 * 20.18 + 0.00000524 * 4.003 + 0.0000016 * 16.043 + 0.00000114 * 83.80 + 0.0000005 * 2.016 + 0.0000003 * 44.013 + 0.000000087 * 131.3 = 28,9715355715 = 28.972 g/mol

In this way, we can now use once the rule of three:

28.972 g             ____(corresponds)_____ 1 mol
X g of  Air           ____(corresponds)_____ 0.044 mol

X = (0.044 * 28.9715) / 1 = 1.275 g

Which means that the weight of 1 litre of Air is 1.275g, which is equivalent to:
Air density (ρ) = 1.275g/l = 1.275 Kg/m³


Quicker way to reach a value almost as accurate as the above of 1.275g:


If for the determination of the molecular weight of Air we used only the three elements most existing in Air, Nitrogen, Oxygen, and Argon, the final result would be very similar: 

M(Air)= 0.78084 * 28.014 + 0.20946 * 31.998 + 0.00934 * 39.95 = 28.950 g/mol

In this way, we can now use once again the rule of three:

28.950 g              ____(corresponds)_____ 1 mol
X g of  Air           ____(corresponds)_____ 0.044 mol

X = (0.044 * 28.950) / 1 = 1.274 g

A difference of just 0.001g

Of course, and before concluding, there is an even easier way to know the densities of helium and hydrogen - by consulting the many existing tables and charts, like this one in Wikipedia. For air, less sources exist.

To provide with a quick overview, the table bellow shows some of the values we have just determined for Helium, Hydrogen and Air.

Gas / Mixture	Molecular Weight / Molar Mass (g/mol)	Weight of 1 litre (g) / Density(ρ)
Helium	4.003	0.176
Hydrogen	2.016	0,089
Air	28,972	1,275

Please notice that for all the above calculations we have the following approximation:

- We are considering standard conditions for temperature and pressure (STP).

Fundamental (thermodynamics) concepts of gas (helium) balloons

This post details some aspects of the physical, chemical and mathematical behavior around high-altitude ballooning which after understood will allow us to derive important characteristics of gas balloons important to correctly dimension the system. We will focus on helium balloons but whenever applicable we shall explain the differences to hydrogen balloons.

In this post we shall introduce some important concepts that we will use in later posts.

Amount of substance

The amount of substance, also known as chemical amount, is a quantity that measures the size of an ensemble of elementary entities, such as atoms, molecules, electrons, and other particles. The SI unit for amount of substance is the mole, and its symbol is: mol. 

Atomic weight

Let's first clarify that the weight of the elements has different names which are all synonymous: standard atomic weight (as it appears in the entry for each chemical element in Wikipedia, for example for helium), relative atomic mass and atomic weight (of the element) - read here, and here, the controversy around these names. Hereafter I will use mostly the name atomic weight because its is the name used by IUPAC, which is the institution responsible for publishing the atomic weights of the elements. The atomic weights are published by the IUPAC and revised each 2-years. Here we have their revision of 2009, and here their periodic table. Other source could be the Wikipedia periodic table with atomic weights, but from there it is not clear which revision of IUPAC they are using. Other source, from the Stanford university, presents not the atomic weights but the molecular weight (relative molar mass), using the deprecated IUPAC Recommended Atomic Weights of 1997. The molecular weight for elements like Helium is the same than the atomic weight because Helium has only 1 atom. But for elements like Atmospheric Nitrogen (N2) that is diatomic, the molecular weight is 2x the molecular weight (in this case = to atomic weight) of the monatomic N.

The atomic weight of element E is represented by the symbol: A_r(E).

The atomic weight is presented as a dimensionless physical quantity, and even in the existing tables no units are provided. See also the IUPAC article "ATOMIC WEIGHT -THE NAME, ITS HISTORY, DEFINITION, AND UNITS" where the dimensionless nature of th eatomic weight is referred. But then in every place where the atomic weight is used for some calculation, it is expressed as a quantity of grams per mole. For instance in this well explained course of Carnegie Mellon University. So for all the calculation hereafter, and until we get some better explanation of the above, we shall consider the atomic weight has being a quantity of grams per mole.

To give an an example, by consulting table 4 of the IUPAC Recommended Atomic Weights of 2009, where the atomic weights are abridged to four significant digits, and which apply to elements of natural terrestrial origin (as our balloon helium does), we can see that the Helium has an atomic weight of approximately 4.003, which as explained above, we will interpret as Helium weighting approximately 4.003 grams per mole.

Standard conditions for temperature and pressure

The Standard conditions for temperature and pressure, better known as STP, are "standard sets of conditions for experimental measurements established to allow comparisons to be made between different sets of data". Different institutions may follow different values for the STP, but we shall follow the values recommend by the IUPAC for chemistry:

• Temperature of 273.15 K (0 °C, 32 °F) and
• an absolute pressure of 100 kPa (14.504 psi, 0.986 atm, 1 bar)

Latter on well shall use the STP for making some calculations that depend on the temperature and pressure: as the temperature and pressure depend on the altitude, and as a balloon may be in different altitudes, sometimes is good to make a simplification and just consider the STP.

Gas constant

The gas constant (also known as ideal, universal or molar gas constant) is a "physical constant which is featured in many fundamental equations in the physical sciences, like bellow in the ideal gas law. It is denoted by the symbol R. Physically, the gas constant is the constant of proportionality that happens to relate the energy scale to the temperature scale, when a mole of particles at the stated temperature is being considered. The value for this constant is recommended by the Committee on Data for Science and Technology (CODATA) for international use. The value we shall use comes from the "CODATA Recommended Values of the Fundamental Physical Constants: 2010" which are in-line with the values presented in the gas constant entry in Wikipedia:

R= 8.3144621(75) J mol^-1 K^-1

The gas constant can be presented in a multitude of different units. In the gas constant entry in Wikipedia we can count the constant R as being available in 23 different units. We should choose the unit of the R constant in accordance to the unit(s) that we then want then the result of our calculations. In our case, and for the calculations that shall be presented in future posts, we shall use the R constant as:

R= 0.08205746(14) L atm K^-1 mol^-1

Ideal gas law

The ideal gas law "is the equation of state of a hypothetical ideal gas" - it approximates the behaviour of many gases under many circumstances. But be aware that is just an approximation - the reality shows that different gases behave differently. This equation relates pressure, volume and temperature to determine the state of a gas. We shall use this equation in its common form:

PV = nRT

where:

P = pressure of the gas;

V = volume of the gas;

n = amount of substance of gas (see above);

R = gas constant (see above);

T = temperature

Intuition behind the reason why gas balloons rise

In this post I will try to explain the intuition that allows us to understand why gas balloons rise into the atmosphere. This post will prepare the readers to better understand the physics and mathematics of gas balloons that I will be presenting in my next post.

Water tank analogy
Imagine that you have a tank full of water. Imagine also that you have a one litter plastic bottle filled-in with air and with a string attached to the neck of the bottle. Now imagine that there is a magical hand (occupying no volume) that picks up the wire and dives the bottle in the water tank. The bottle will only stay suspended under the water if the magical hand keeps holding the string. If the hand let the string go, the plastic bottle will emerge. This is common knowledge, but lets try to gain an intuition on what is happening. When the one litter bottle is submerged in the tank, it has to displace one litter of water so that there is room for the bottle to occupy its space in the tank. This means also that the water level in the tank will increase in one litter (if the magical hand had volume we would have to consider it also to know the increase of the water level in the tank). Now lets think the following. The air is lighter than water, this means that the one litter air bottle is lighter than one litter of water. This means that, when the magical hand releases the bottle string, the one litter of water that was displaced to make room for the bottle, as it is heavier than the bottle, will want to claim its place back, and when it does, the bottle will have to occupy the space where that water was, and this process will be repeated as long has the bottle is surrounded by water. When the bottle emerges it will have water bellow and air above. As the bottle is lighter than the water it will not submerge again. And as it is heavier than the air (because of the weight of the bottle itself) it will not raise up into the atmosphere. This is the law of buoyancy. 

Fig. firstly the water tank is empty; then a magical hand dives an air plastic bottle; afterwards the magical hand releases the air bottle string and the water surrounding the bottle, which is heavier, claims it space where the bottle is located; finally the bottle is pushed up into the surface because whenever the bottle is surrounded by the heavier water it will claim the space it occupies and as the contents of the tank can only expand to above, it is to the above that the bottle moves until is not surrounded any more by water but also by the lighter air (@Paulo Carmo)

So, to summarise:

- Lets have two things. Thing 'A' and thing 'B'; A and B can be objects, gas, etc.;
- 'A' is lighter than 'B';
- If 'A' is surrounded by 'B', the space occupied by A will claimed by B;
-  Now we have 'A' occupying the space of 'B' and vice versa;
- If 'B' continues being surrounded by heavier things, then the space it occupies shall be claimed again;

- 'B' moves in the direction of where it can claim space

With the water tank analogy clear on our mind let's now mode deep into the atmosphere.

Gas balloons in the atmosphere

While the behaviour of objects in the water is very well know since we practice with it on a daily basis, the interaction of objects with the atmosphere is a bit more esoteric to most people. So, equipped with the previous water tank analogy, lets now imagine that all this air above our heads, the atmosphere, is in contained inside a air tank (only open on the top at the end of the atmosphere, 10000Km above the earth's sea level). Now imagine that we fill in a balloon with an element lighter than air like hydrogen (the lightest existing element) or with helium (the second lightest existing element) and string it to an heavy rock. If we do that we get an object which is lighter than the air that surrounds it (when released from the rock), and if we make it big enough we can even attach objects to the balloon that its total weight is still less than the air surrounding it. Now the mechanics of the raise of the balloon is exactly the same of when we had a water tank. When the gas balloon string is released, the air that was displaced to make room for the gas balloon, as it is heavier than the balloon, will want to claim its place back, and when it does, the balloon will have to occupy the space where that air was, and this process will be repeated as long has the balloon is surrounded by air and no other event (explained bellow) occur, which will make the balloon continuously rise deep into the atmosphere. But now the question is, and does it stop rising? Well, theoretically, as long as the gas balloon is surrounded by elements lighter that the balloon itself, it could keep rising until the "end of days", but here a different phenomena occurs - the atmospheric pressure - that does not allow the balloon to rise forever. So in the atmosphere the same law of buoyancy applies.

Atmospheric pressure and its impact in the gas balloons

The atmospheric pressure, or air pressure, is the force applied into a surface by the weight of air existing in the atmosphere above that surface. This is applicable to the Planet Earth or any other planet with an atmosphere. The referred force is applied to any object in the planet including yourself, or in what concerns this post, to a balloon.
The air in the atmosphere is constituted by small particles called air molecules, and depending on the altitude, the concentration of air molecules in the atmosphere is bigger (at sea level) or smaller (in the higher altitudes of the atmosphere). The figure bellow, which was built on data taken from the following site, shows the force which is applied by the atmospheric pressure to a man at different altitudes.  

Fig. Atmospheric pressure at different altitudes - not in scale (@Paulo Carmo)

For instance, at sea level, 0 meters of altitude, the atmospheric pressure applies a force of 1.03 Kg/cm2 to a man. This is due to the weight of all air existing in the atmosphere above the man. If the man were at an altitude of 12192 meters, the atmospheric pressure would be almost 9 times less, particularly of 0.191 Kg/cm2. This is due to the fact that the man at 12192 meters has much less air (molecules) above is head. Other related phenomena is that the air density (in the right side of the figure) decreases with the altitude (as does the air pressure). The air density is the mass per unit volume of Earth's atmosphere, and it changes also with changes in temperature or humidity. The air density is related with the fact that the air can be compressed to fit in a smaller volume (volume is the space that contains an object. In the case of a gas, or air, the volume of the container will tell you the volume of the gas. Volume is typically measured in litres or milliliters. For example, an open 1-litre bottle at sea-level contains 1-liter of air).When the air is compressed it is said to be under high pressure, which is what happens, for instance, at sea level.
A question that can be formulated is the following: why don't we fell, at sea level, the force of 1.03 Kg/cm2, that is applied to us by the air existing in the atmosphere above us? Well the reason is that we have air inside our body too, and that air balances out the pressure outside so that we do not feel the air pressure.

But all this information is to explain what is the impact of the air pressure in a gas balloon as it rises into the atmosphere. Well, as the balloon rises, and as we have seen, the atmospheric pressure drops, and so the force which is applied to the exterior of the balloon decreases. Since the gas inside the balloon stays the same (the balloon is sealed), also the force that it exerts outward stays the same while the balloon rises.
So, when you are at the Earth surface and fill-in the balloon with gas, the equilibrium of internal and external forces tin the balloon will result in a given size of the balloon. As the balloon rises, the air pressure external to the balloon will drop and since the internal gas pressure is maintained, what happens is that the balloon will increase in size - because the pressure that the gas inside the balloon applies encounter less resistance from the external air pressure. When the internal gas pressure is bigger than the sum of the external air pressure and the pressure (resistance) of the balloon fabric, the balloon bursts. And this is the event that prevents the balloon to rise forever - at a given altitude the balloon bursts.

High-altitude balloon stack

In this post it will be described what is the typical high-altitude balloon stack. In this and in the next posts the terms high-altitude balloon and near space photography will be used interchangeably.

Fig. High-altitude balloon stack (@Paulo Carmo)

The figure on the left shows a typical stack for high-ballooning. With stack we mean a pile or heap of sub-systems that together compose the high-balloon system:

Balloon (sometimes called envelope): the balloon is the device that provides the lifting force that allows the whole system to reach an high-altitude in the Earth's atmosphere. As the system rises the balloon expands until eventually it bursts and the whole system starts falling back into Earth's surface. It is filled-in with a gas lighter than the air, usually helium or hydrogen. Today helium is more used because it is less explosive than the hydrogen even though is more expensive.

Parachute: the parachute is activated after the balloon bursts and the whole system starts falling back into Earth's surface. Before this, the parachute is in tension between the balloon above and the radar reflector and payload bellow. When the balloon bursts, it stops providing the tension force for the parachute bellow and so the parachute naturally opens providing a drag force that decreases the falling velocity of the whole system.

Radar reflector: the radar reflector is a safety apparatus that is placed between the parachute and the payload and it serves the purpose of presenting an high radar reflectivity and in this way increasing the chances of the whole system being detected by a flying asset (air plane, ultra-light, etc) and consequently reducing the probability of an air collision.

Payload: the payload contains the whole purpose of the system, which is the set of equipment that will accomplish the mission. An example payload could be constituted by the following modules:

• Communications: this module is responsible for the communication with ground. Its major task is to communicate back the position of the system so that it can be tracked and found after landing; usually radio-frequency communication is used, but GSM communications is also employed;
• Sensors: this module contains any device that detects a physical condition in the world. It contains several sensors depending on the mission. Its primary sensor is the GPS receiver, but other sensors can be aboard like camera, temperature, pressure and others;
• Actuators: this module contains any device such as switches, that perform actions such as turning things on or off or making adjustments in the system; one practical example is an actuator that is able to move a camera to control the angle in which a photograph is taken;
• Computing: computing power is necessary to all sorts of things like:
• acquiring the sensor readings and storing and/or communicating them (through the communications module);
• controlling the electronics and other equipment;
• other.
• Power: this module will be responsible for the power to feed the equipments in the payload; it is wise to have separate and independent power lines in order that problems in one line does not affect other line.

It must be noticed that all components of the high-altitude stack must be tied up very securely. Depending on the components your are tying, different types of strings are used, like nylon string and others.

The balloon, the parachute and the radar reflector are sometimes named flight system. This means that the high-altitude balloon stack is in fact constituted by two major subsystems, the flight system (with the balloon, the parachute and the radar reflector) and the payload (with all its subsystems).