So, analysis of the movie Up is pretty popular in the blogosphere. Figure I might as well surf the popularity wave. So, I have a couple more questions.
The most important thing to estimate is the mass of the house. I am going to completely ignore the buoyancy of the house. I figure this will be insignificant next to the buoyancy needed. Anyway, let me go ahead and recap what has already been done on this in the blogosphere.
Wired Science - How Pixar's Up House Could Really Fly - from that post:
- First, they calculated (seemingly correct) that the buoyancy of helium is 0.067 pounds per cubic foot of helium. This does not include the mass of a balloon holding this, nor any strings on the balloon.
- The weight of the house is 100,000 pounds. I like how they got this - they had a house mover estimate it. Seems legit.
- It would take 1,500,000 cubic feet of helium to lift the house. This would require 112,000 balloons (3 foot diameter balloons)
Slate Explainer: How Many Balloons Would it Take to Lift a House - from that post:
- If using party balloons, (11 inches in diameter) they would have a lift equivalent to 4.8 grams.
- The dude would need 9.4 million party balloons to lift his house.
- Pixar estimated it would take 23.5 million balloons to lift a 1,800 square foot house.
- Pixar used 20,622 balloons in the animation of the lift off and 10,297 balloons from the floating sequences
physics and physicists looked at up also. ZapperZ looked at how the balloons were deployed. He assumed that the balloons were originally attached to the house before being released (this would have the same buoyancy as if they were deployed). However, it seems like the balloons may have been attached outside the house to the ground or something.
Swans on Tea looked at the problem with precision in the calculations. I am guilty of this sin also. Basically, if you estimate that the house is 100,000 pounds and convert that to Newtons or something, you should keep the same level of precision. He (Tom) is correct.
Finally, here is stuff on buoyancy and floating things. That is from a post on the MythBusters making a lead balloon float. Oh, and here is my last post about Up (just for completeness). And now on to the calculations.
If the house were lifted with one balloon, how big would it be and what would it look like?
There is one important thing to estimate - the weight of the balloon material. I have no clue how much this would weigh. Let me just say that the balloon is a sphere with a thickness of t. If I had to off the wall guess t, I would say 1 mm. If this is latex, the density would be about ? = 950 kg/m3 (wikianswers.com). From this I can do some stuff. Here is my approach. I have the buoyancy force = weight of house plus weight of balloon plus the weight of the helium. Of course the weight of the balloon depends on the size of the balloon.
Here, r is the radius of the balloon. If I put this all together, I get:
I just need to solve this for r. Problem, this is a cubic equation. There are quite a few ways to solve a cubic equation, but I will use the plotting method. If I let f(r) be written as:
And then I can plot this to find the zeros. Here is a plot from zoho
Looking at the graph, r = 23 meters seems close to a solution. Just as a check, I calculated the size of the balloon if the mass of the balloon were negligible (it's in the zoho sheet). This is a much easier calculation, and I get a radius of 22 meters. Seems reasonable. Note, if I change the thickness of the balloon to 2 mm, the radius moves up to around 24 meters - but you can play with the spreadsheet yourself if you like. A couple more notes:
- I assumed that the balloon was a sphere - clearly, it would not be.
- I assumed that the thickness of the balloon was very small compared to the size. This way I can calculate the volume of rubber needed as surface area of sphere times thickness.
Ok, I want to draw this. I need the size of the house. The Slate article said that Pixar said the house was 1,800 square feet. Looks pretty square, so this would make it 42 feet by 42 feet or about 13 meters. Oh wait, it's a two story house. That means the bottom floor is maybe 900 square feet. This would make the side 30 feet or about 9 meters. Now I just measure the pixel size of the house and use the same pixel per meter for the spherical balloon. Here it is.
But wait, I am not finished. The next question:
If the house were lifted by standard party balloons, what would it look like?
The thing with party balloons is that they are not packed tightly, there is space between them. This makes it look like it takes up much more space. Let me just use Slate's calculation of 9.4 million party balloons. How big would this look? This could be tricky if I didn't know how to cheat. How tightly packed do party balloons fit? Who knows? Pixar knows. From that Slate post, Pixar said they used 20,600 balloons in the lift off sequence. From that and the picture I used above and the same pixel size trick, the volume of balloons is about the same as a sphere of radius 14 meters. This would make a volume of 12,000 m3. The effective volume (can't remember the technical term for this) of each balloon would be:
And then this would lead to an apparent volume of the giant cluster of 9.4 million balloons:
If this were a spherical cluster, the radius would be 110 meters. Here is what that would look like:
How long would it take this guy to blow up this many balloons?
You can see that there is no point stopping now. I have gone this far, why would I stop? That would be silly. The first thing to answer this question is, how long does it take to fill one balloon. I am no expert, I will estimate low. 10 seconds seems to be WAY too quick. But look, the guy is filling 9.4 million balloons, you might learn a few tricks to speed up the process. If that were the case, it would take 94 million seconds or 3 years. Well, you can see there is a problem because that time doesn't include union bathroom breaks. Also, a standard helium balloon will only stay inflated for a few days.
What if it was just 20,600 balloons like Pixar used in the animation? At 10 seconds a balloon, that would be 2.3 days (and I think that is a pretty fast time for a balloon fill). Remember that MythBusters episode where they filled balloons to lift a small boy? Took a while, didn't it?
How many tanks of helium would he need?
According this site, a large helium cylinder can fill 520 of the 11" party balloons and costs about $190. If he had to fill 9.4 million balloons, this would take (9.4 million balloons)(1 tank)/(520 balloons)= 18,000 tanks at a cost of 3.4 million dollars. You could buy an awesome plane for that much. Oh, maybe he got the helium at cost.
PS - I still haven't seen the movie, but I want to.
Not germane to the discussion, but Mythbusters lifted a small girl.
By the way, I seem to have graduated to the "physics blogging bigs." Uncle Al commented on and Tom Swans on Tea (listed in your blog roll) cited my post at http://hamiltonianfunction.blogspot.com/2009/06/cornucopia-of-cluelessn….
I couldn't be more proud! :)
Well, if Uncle Al commented on your blog, I guess I should add you to my blog roll. - oh wait. I already did.
Don't forget, though, that the "lift" of Helium depends upon the
atmospheric pressure. Do we know how "thick" the atmosphere is? What is
the composition of the atmosphere (Nitrogen/Oxygen mix? Sulfur
Hexafluoride?)? What is the value of gravity on the Up planet? What if
he was using Helium-3 instead of the much more common Helium-4? Are the
balloons elastomeric (e.g., latex) or non-elastic (e.g., mylar, perhaps
Aluminized mylar)? If latex, was a bit of Hi-Float added to the
balloons to slow the Helium leakage? At what altitude will the lift
I was thinking about the "balloons anchored outside" angle â you'd need an anchor that could hold > the weight of the house (if they were all anchored), which seems unlikely. I think the best you can hope for is that the balloons anchored were only the ones that provided the excess buoyancy, plus a few, so that the house was almost ready to float away.
@Rob. I had no idea a link from me carried so much weight ;)
Party helium is diluted with air to lessen virtual leak rate (sell air at helium prices) and restrict lift vs. altitude (jet engines). Air with an average MW =29 vs. He with MW = 4 obtains a 7:1 lift ratio. The obvious solution is to use two (one for backup) vacuum balloons with nearly infinite (no perfect vacua) lift ratios. Federal financial planning works exactly this way (including the backup - a vacuum economy plus a vacuum war). The whole of society must be diverted from cruel engineering to kind social solutions, mythos to CNC lathes. We grow closer every day.
In Babylon 5 the kid would prolongedly, horribly die along with the dog. Those who deny Grimm's fairy tales in the original are doomed to repeat them.
@Tom. I'm pretty sure I've never been quoted before, though I've occasionally been cited. Mostly, I talk to myself of course. I happened upon Rhett's site because of a post he made a long time ago regarding the most efficient speed to drive accounting for fuel efficiency and hourly wage. I'd done some similar calculations. In any case, I've really enjoyed following along with his blog, and now yours.
I'd forgotten about Leonard Pinth-Garnell.
@Rhett. Sorry to take up your bandwidth with what amounts to an email to Tom.
Yo must be a degree friki with a lot of free time. X-)
Yo must be a degree friki with a lot of free time. X-)
Its a cartoon movie my friend.