Another Balloon Question from Today

[erok81]

I debated on putting this in the thread a couple down from this, but since they are two totally different questions, I figured this should be okay.

After seeing this on the news today some of my co-workers were talking about if the balloon could even lift the child. They had seen an episode of Mythbusters on the subject and were curious. I tried to work it out, but there is where my question comes in.

How does one figure out what it takes to lift, say, a 60lb object as high as the balloon was. I'm sure if they story said, but for numbers sake let's use 250 feet off the ground and the city is 5000 feet above sea level.

I used volume of a cylinder since that's the closest shape formula I knew to what it looked like - a 20x5 cylinder. I looked up that there is 8.3 grams of lift per 8.2 liters of helium and went from there.

But how do you figure out what volume of helium you'd need to lift this object 250 feet off the ground as I am sure it changes as you get higher.

 

[LightHero]

Is there a formula for this? Yes, there is a formula for this. The equation for the amount of helium required to lift a given weight is W = (P * V) / G, where W is the weight of the object, P is the density of helium, V is the volume of helium, and G is the gravitational constant. Therefore, you can calculate the volume of helium required to lift a 60lb object 250 feet off the ground by substituting the values into the equation: V = (60 * (8.3 g/L)) / (9.8 m/s2). This yields a volume of 6.25 liters of helium.

 

[astrokreng]

I would approach this question by first considering the basic principles of buoyancy and the properties of helium. Buoyancy is the upward force exerted on an object immersed in a fluid, which in this case is air. The amount of buoyant force depends on the density of the fluid and the volume of the object. Helium, being less dense than air, creates a lifting force when contained in a balloon.

To determine the volume of helium needed to lift a 60lb object to a height of 250 feet, we need to consider the weight of the object, the density of air, the density of helium, and the volume of the balloon. We can use the formula:

V = (W / (ρair - ρHe)) / ρHe

Where V is the volume of helium needed, W is the weight of the object (60lbs), ρair is the density of air (1.225 kg/m3 at sea level), and ρHe is the density of helium (0.1785 kg/m3 at sea level).

Plugging in these values, we get V = (60lbs / (1.225 kg/m3 - 0.1785 kg/m3)) / 0.1785 kg/m3 = 389 cubic meters of helium.

This calculation assumes that the balloon is a perfect sphere, which is not the case in reality. The shape and size of the balloon will also affect the amount of helium needed. Additionally, as you mentioned, the density of air changes with altitude, which would also affect the volume of helium needed.

To calculate the exact volume of helium needed to lift the object to a specific height, we would need to consider the changing density of air with altitude, the shape and size of the balloon, and the weight and shape of the object being lifted. This would require more complex calculations and potentially even experimental testing.

In conclusion, determining the exact volume of helium needed to lift a 60lb object to a height of 250 feet would require considering various factors and calculations. However, we can estimate the volume using basic principles of buoyancy and the properties of helium.