Aggregate weight to be lifted will be the deciding factor. As far as buoyancy is concerned, one liter helium gas is sufficient to lift one gram weight in the air. A child’s average balloon is filled with about five liters helium, hence its lifting capacity is not more them 5 grams. Based on this rate of lift, if the person to be lifted by the balloon weighs 65 kilograms (65,000 grams) helium sufficient to inflate 13,000 balloons will be required.
Since it is the question of the personal experience of ballooning, the amount of helium required will have to be arrived at by taking into account the factors such as the weight of the gondola, essential supplies, ballast etc. Weight of balloon’s fabric and cordage will also have to be carefully taken into account. However, let us ignore these complex aspects of practical ballooning in order to keep the calculation simple. Let us also assume for the sake of simplicity that the balloon is perfectly globular. The following formula will help us calculate how many liters gas will be required to inflate a globular balloon:
4/3 X π X r X r X r, which means: 4/3 X pi X radius X radius X radius.
The value of π is 3.14159, whereas radius/r for calculating the cubic area of balloon is taken in centimeters. First, let us see the example of balloon having 30 centimeter diameter (i.e. 15 centimeter radius). Such a balloon will have 4/3 X 3.14159 X 15 X 15 X 15 = 14,137 cubic centimeters or little over 14 liters of helium and its lifting capacity will be 14 grams. We have already assumed that the person who wants to go up on the sky in a balloon weighs 65 kilograms (65,000 grams). Hence, we will require a balloon having a diameter of 5 meters (500 centimeters) or radius of 250 centimeters as follows: 4/3 X 3.14159 X 250 X 250 X 250 = 6,55,00,000 cubic centimeters = 65,000 liters helium having the buoyancy of 65.5 kilograms which is 500 grams more than the balloonist’s weight. He can utilize this spare capacity to take with him a small bottle of water!