Where is the optimum lift on the inflation graph?
The lift is directly proportional of the volume of helium in the balloon. The volume of helium is directly proportional to the length of the balloon. (A fully inflated 280 will have 33% more lift than a 260) The volume of helium is also directly proportional to the cross-sectional area of the 260. BUT, the cross-sectional area of the 260 is proportional to the square of the balloon diameter. So, the fastest way to increase the lift is to increase the diameter. That’s why I suggested soaking the balloon in boiling water before inflating. The water is absorbed into the latex and reduces its elastic modulus, allowing it to stretch more. It will stretch more in the direction of maximum stress, the circumferential or “hoop” direction, and you’ll see this as a diameter increase. (No, I don’t know how much water is absorbed, and yes, the weight of the absorbed water does make the balloon heavier, and yes, there is probably an optimum amount of time to soak the balloon which maximizes the diam
