Learning science up in the air?

Our micro-searching consists in an estimation about the number of helium balloons we would need if we wanted to have a science lesson up in the air. For that, we have estimated the weight of all the components inside the classroom (people and materials) based on approximated data. Then, we will convert all the weight according to the quantity of balloons we will need (taking into account that the balloons are 46cm diameter and in each balloon it fits 56L of helium).

Different data. We have…

    -25 tables of 15kg each one. 25x15 = 375kg

                -1 teacher’s table (with a computer below). 50kg

                -51 chairs of 3,5kg each one. 51x3,5 = 178,5kg

                -50 people of 70kg each one. 50x70 = 3500kg

                -50 bags of 5kg each one. 50x5 = 250kg

                -1 digital blackboard of 16,5kg

                -1 blackboard of 8kg

                -3 hangers of 1kg (all three)

Floor

We have searched which is the weight of a tile that measures 50x50 (5kg). Then, we counted how many tiles were inside the classroom (17x18 = 306) and finally the total weight of the tiles (306x5 = 1530kg). Now, we have to add the weight of the bricks which shape the inferior layer of the floor. Using a table of a brick that measured 20x20x40 and heighted 8kg. Changing those measures for our measures the result was 6800kg. Finally, we add the tiles’ weight with the bricks’ weight (1530+6800 = 8330kg).

Roof

Calculating the weight of the roof was easier because was the same surface as the floor. However, the roof didn’t have tiles, so we substrate those kg. That means, the roof weight is 6800kg.

Walls

Using the same table of bricks as in the floor, we changed those data by ours. The height of all the walls was 3m. Two of the walls were 8,5m long and the other two were 8m long. So, using the scale, the first two walls weighted 5100kg each one (in total 5100x2 = 10200kg) and the second two walls weighted 4800kg each one (in total 4800x2 = 9600kg). The total weight of the 4 walls is 19800kg.

The total weight is:

19800+16,5+8+1+8330+6800+375+50+178,5+3500+250= 39319kg.

We know that 1L of helium corresponds to 0,17 grams, so we need the total weight in grams. 39139kg = 39319000g. Right now, we have to transform those grams into L of helium, which are 231288235,3L

Finally, we only have to distribute the balloons (remember that each balloon contains 56L helium) according to the total weight of all the classroom. 231288235,3 : 56 = 4130147,059 balloons. We can’t break a balloon, so we will use 1 balloon extra. For that, the quantity of balloons we will need will be 4130148.

PS. We are sad to inform you that, even with that huge quantity of balloons, we “only” could be in class during 36 hours. Then, balloons will start to deflate.