Sister is transforming 21 in a couple days and also I was offered the task to decorate and they left it up to me to decide how. Ns figured it would be funny as hell to to fill her room to the brim v balloons. The balloons package had actually said 9" however not certain if they to be referring to diameter, etc... Just assume castle are average sized balloons. I've been shot to use the system of thinking of them as boxes (like if you put a 9" round in a 9" box) however it's been awhile because I learned that. Can you define how you would certainly write that out and how many balloons I will certainly need.
You are watching: How many balloons to fill a room
The room's dimensions
11ft X 14ft X 8ft
Balloon circumference of about 2ft once pressed into a ring shape
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· 5y · edited 5y
Hi guys, ns am not certain of this please correct me if this is wrong:
Balloon have the right to be best fit in cube of leaf 9" = 0.75 ft
Let us very first start aligning these cubes along elevation = 8 /.75 = 10 cubes
Along broad = 14/.75 = 18
Along length = 11/.75 = 14
So full no. The cubes = 10* 18 * 14 = 2520
First, we require the volume of the room. Unless it has actually a strange shape fairly than the usual cuboid, we acquire 11 ft * 14 ft * 8 ft = 1232 ft3, or about 3.4 m * 4.3 m * 2.4 m ≈ 35 m3. Of course, this ignores furniture. You're most likely not walking to have actually a lower volume in practice, but I don't know what's a reasonable calculation for furniture volume.
Next, we require the volume of one balloon. Now, for much easier calculations, I'll almost right it together a sphere. This introduce an error, but that need to be fine because the worth for the room doesn't take furniture right into account and also you won't be perfectly accurate as soon as filling the balloons, for this reason the probably comparatively tiny error that's introduced below won't make the end result much worse. The volume the a round is 4/3*pi*r3 (where r is the radius, i beg your pardon is half the diameter). If we take the diameter together 9 in = 0.75 ft (about 0.23 m), that provides us 4/3 * pi * (0.375 ft)3 ≈ 0.22 ft3 or 4/3 * pi * (0.11 m)3 ≈ 0.0013 m3.
To check out how countless balloons friend need, we simply divide the room's volume through the volume taken up by one balloon. However, this isn't specifically equal to the volume calculate above, due to the fact that there will be gaps in between the balloons (you have the right to stack cubes, for example, perfect efficiently, however not, say, spheres). Wikipedia tells united state that the so-called packing density will be approximately 0.6, meaning that around 60% that the space will be taken up by ballons, with about 40% being empty. This is good, obviously, since it lets you use fewer balloons, make the whole thing rather easier. So, you'll most likely need around 0.6 * 1232 ft3 / 0.22 ft3 ≈ 3000 balloons.
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Of course, this is a very rough figure, but it must be great enough. If it's turn off by, say, 300, that would change the height to i beg your pardon you could fill the room by around 300 * 0.22 ft3 /(11ft * 14 ft * 0.6) ≈ 0.7 ft on average (about 0.2 m), which would certainly hardly be noticeable due to the fact that it would certainly be far over your heads.