Physics of Giant Balloons

By Evan Angelus

As I was watching the Macy’s Thanksgiving Day Parade and all of the giant balloons go by, I wondered just how hard it actually is to hold those balloons down.

·      Buoyant Force:

·      The floats (filled with helium) are submerged in the air surrounding them.

o   F_B = PA = ρ_airghA = ρ_airgV = (ρV)g = m_airg

o   Vol_balloon = 16,780 ft³ = 475 m³

o   ρ_air = 1.225 kg/m³

o   (1.225 kg/m³ * 475 m³)(9.8 m/s²) = 5,704 N

·      Force of Gravity:

·      We will assume that the nylon fabric the balloon is made of does not contribute to the volume of the balloon, but for such a large balloon, its mass is important.

o   F = m_balloong = (m_uninflated + m_He)g

o   m_uninflated = 450 lbs = 204 kg

o   ρ_He = 0.1786 kg/m³

o   m_He = ρV = 0.1786 kg/m³ * 475 m³ = 85 kg

o   m_tot = 204 kg + 85 kg = 289 kg

o   F = 289 kg * 9.8 m/s² = 2,832 N

·      Net Force of Balloon = F_B – F_grav

o   5,704 N – 2,832 N = 2,872 N à 2,800 N

·      Will they be able to hold the Rugrats down?

·      Each float is held down by 2 vehicles weighing 800 pounds, and a number of volunteers. The only requirement to be a balloon handler is to weigh at least 120 pounds. The rugrats float requires a minimum of 30 people to maneuver it.

o   2 vehicles (800 lbs) = 362.874 kg * 2 = 726 kg

o   30 People (at least 120 lbs) = 54.4311 kg * 30 = 1,633 kg

o   Total = 726 kg + 1,633 kg = 2,359 kg

o   2359 kg * 9.8 m/s² = 23,115 N à ­– 23,000 N

·      This will be more than enough force to keep the balloon from floating away.

o   2,800 N – 23,000 N = – 20,200 N

o   Why is all of this force needed?

o   To maintain the normal force so that the balloon handlers have enough friction to direct the balloon.

o   20,200 N / 23,000 N = 88% à the normal force on each person on average is 88% what it would normally be. This means that they only have 88% as much friction.