Rwanda has cultivated herself as a country of interesting surprises keeping her as a destination of interest from media, researchers, first time visitors and various fans.

From aligning with Arsenal football club, Paris Saint-Germain Football Club to introducing drones to facilitate the country’s health sector and many more others, this week Rwanda introduced hot air balloons onto offers for tourists at Akagera National Park.

“We are pleased to partner with Royal Balloon Rwanda to add yet another exciting product to Rwanda’s adventure tourism experiences,” said Clare Akamanzi, CEO of Rwanda Development Board.

For most Rwandans and other visitors, the hot air balloon may trigger so many questions on how they can really embrace it while visiting Akagera National Park.

**Phyisics of hot air balloon**

In simplistic terms, a hot air balloon is a lighter-than-air aircraft consisting of a bag, called an envelope, which contains heated air. The heated air inside the envelope makes it buoyant, since it has a lower density than the colder air outside the envelope. As with all aircraft, hot air balloons cannot fly beyond the atmosphere.

The principle behind hot air balloon physics is the **Archimedes Principle** which states that the buoyant force on a submerged object is equal to the weight of the fluid that is displaced by the object.

For a hot air balloon, the upward buoyant force acting on it is equal to the weight of air displaced. The density of air is 1.2 kg per cubic meter; therefore to lift a balloon is going to be necessary to displace a great volume of air so its weight exceeds the weight of the envelope, plus the weight of passengers, and the upward force is greater than the downward force of gravity.

Also an object floating in water stays buoyant using the same principle as a hot air balloon.

As shown in the figure above, the center of buoyancy acts through point *C*, which is the centroid of the volume *V* of the object. This volume is equal to the displaced volume of the fluid. The upward buoyant force *F _{B}* is equal to the weight of the displaced volume of fluid

*V*.

For the object to remain in an unconditionally stable orientation (i.e. not rotate) the center of mass of the object *G* must be directly below point *C*.

This means that if the object were to be rotated by any amount, it will automatically rotate back to the original position where point *G* lies directly below point *C*. This is what is meant by unconditional stability.

For a hot air balloon, the upward buoyant force acting on it is equal to the weight (or mass) of the cooler surrounding air displaced by the hot air balloon.

Since the air inside the envelope is heated it is less dense than the surrounding air, which means that the buoyant force due to the cooler surrounding air is greater than the weight of the heated air inside the envelope.

And for lift to be generated, this buoyant force must exceed the weight of the heated air, plus the weight of the envelope, plus the weight of the gondola, plus the weight of passengers and equipment on board.

As a result, the hot air balloon will experience sufficient buoyant force to completely lift off the ground.

As shown in the figure below, the weight of the hot air balloon is more concentrated near the bottom of the balloon (at the location of passengers and equipment), so the center of mass *G* of the hot air balloon is always below the center of buoyancy *C*.

Therefore, the balloon is always stable during flight (i.e. it will always remain in the upright position).

**Hot Air Balloon Physics – Operation**

If the balloon operator wishes to lower the hot air balloon, he can either stop firing the burner, which causes the hot air in the envelope to cool (decreasing the buoyant force), or he opens a small vent at the top of the balloon envelope (via a control line).

This releases some of the hot air, which decreases the buoyant force, which also causes the balloon to descend.

To maintain a steady altitude, the balloon operator intermittently fires and turns off the burner once he reaches the approximate altitude he wants. This causes the balloon to ascend and descend (respectively).

This is the only way he can maintain an approximately constant altitude, since maintaining a strictly constant altitude by way of maintaining a net zero buoyant force (on the balloon) is practically impossible.

If the balloon operator wishes to move the balloon sideways (in a horizontal direction) he must know, ahead of time, the wind direction, which varies with altitude. So he simply raises or lowers the hot air balloon to the altitude corresponding to the wind direction he wants, which is the direction he wants the balloon to go.

The balloon stays inflated because the heated air inside the envelope creates a pressure greater than the surrounding air.

However, since the envelope has an opening at the bottom (above the location of the burner), the expanding hot air is allowed to escape, preventing a large pressure differential from developing.

This means that the pressure of the heated air inside the balloon ends up being only slightly greater than the cooler surrounding air pressure.

An efficient hot air balloon is one that minimizes the weight of the balloon components, such as the envelope, and on board equipment (such as the burner and propane fuel tanks).

This in turn minimizes the required temperature of the air inside the envelope needed to generate sufficient buoyant force to generate lift. Minimizing the required air temperature means that you minimize the burner energy needed, thereby reducing fuel use.

**Hot Air Balloon Physics – Analysis**

Let’s examine the physics of a hot air balloon using a sample calculation.

The heated air inside the envelope is at roughly the same pressure as the outside air. With this in mind we can calculate the density of the heated air at a given temperature, using the Ideal gas law, as follows:

*P* = *ρ**R**T*

Where:

*P* is the absolute pressure of the gas, in Pa

*ρ* is the density of the gas, in kg/m^{3}

*R* is the gas constant, in Joules/kg.K

*T* is the absolute temperature of the gas, in Kelvins (K)

Now,

Normal atmospheric pressure is approximately 101,300 Pa

The gas constant for dry air is 287 Joules/kg.K

The air inside the envelope is typically heated to an average temperature of about 100 degrees Celsius, which is 373 K

Substituting the above three values into the Ideal gas law equation and solving for *ρ* we get *ρ* = 0.946 kg/m^{3}. This is the density of the heated air inside the envelope. Compare this to normal (ambient) air density which is approximately 1.2 kg/m^{3}.

Next, for an average size balloon with an envelope volume of 2800 m^{3} we wish to determine the net upward buoyant force generated by the envelope.

The net buoyant force is defined here as the difference in density between the surrounding air and the heated air, multiplied by the envelope volume. Thus,

*F _{B,net}* = (1.2−0.946)×2800 = 711 kg (1565 lb)

This is the net buoyant force pushing upwards on the heated air inside the envelope. The hot air balloon components (such as envelope, gondola, burner, fuel tanks, and passengers) can at most weigh 711 kg in order for the buoyant force to be able to completely lift the hot air balloon off the ground.

*compiled by Taarifa *