What's new

How many apples can you fit in a Big Apple

Jordanovichy

Credit Whore 2016
So I've been engaging in discussion with Nealbie, Stone Cold, Mysterious Sue and Ploddish and we've discovered a problem. We dont know how many apples you can fit inside the big apple on a Big Apple.

Can you help us work it out? We have worked out that 16.42 Big Apples make up the same length of The Ultimate at Lightwater Valley but this in no way helps.
 

GuyWithAStick

Captain Basic
Rough estimate of 1,573 apples.

No calculations, or anything. Just a random guess.

Sent from my VS840 4G using Tapatalk
 

ATTACKHAMMER

Strata Poster
I guess it depends on the size of the apples. If you had giant apples, then perhaps two. However that is unlikely. It must be a lot! My guess is 950.
 

Hixee

Flojector
Staff member
Administrator
Moderator
Social Media Team
MATHS!

If the Big Apple apple has a volume of x and each small apple has a volume y, then the upper limit of the number of apples (assuming you can change their shape) would be x/y. However, intuition tells us that the actual number will be less than that, as there will be voids in the space.

This sort of lattice packing is actually incredibly important for many applications in materials science (it's why diamond is so strong, but graphite sides apart easily, even though they're both pure carbon - it's all to do with their lattice structure and bonds). As it's so important, a lot of people spend a lot of time (me included for a handful of modules at university), studying them and understanding how they work.

A close packed structure (either face centred cubic [FCC], or hexagonal close packed [HCP]) are the densest way to pack spheres. There's an upper bound, which can be proved with some maths, that says of a given volume the most space that you can fill up with spheres is about 74%. The rest of that space is unoccupied.

Let's assume the Big Apple apple, and all of the real apples, are spheres and plug the numbers in.

Let's assume the average apple is 8 cm in diameter. It's volume, therefore, is (4/3)*pi*0.04^3 = 0.000268 m^3
Let's also assume the average Big Apple apple is 3.5 m in diameter. It's volume, therefore, is (4/3)*pi*1.75^3 = 22.45 m^3.
However only 74% of that volume is occupied by apples (the rest is air), so we need to fill 22.45*0.74 = 16.61 m^3 with apples.
16.61/0.000268 = 61977. I'll round that to about 62000 apples.

Sounds like a hell of a lot - which it is. In reality, you'd probably get a packing density of more like 50%, which is about 42000 apples. And it's worth remembering that neither the Big Apple apple, nor the real apples are likely to be spherical. It still seems like a number much higher than you'd guess, but the maths is there (unless I've made a blunder)!

EDIT: For clarity.
I'm talking about this kind of Big Apple:
bot_ba.jpg


Not the ones that have just a smaller one at the top of the lift hill like this:
artikel5.jpg


But you can just repeat the process for any size Big Apple (or apple) you like.
 

Ian

From CoasterForce
Staff member
Administrator
Moderator
Social Media Team
Hahaha, that is amazing, Hixee. *goes off to the CF Twitter and Facebook page*
 

ThomVD

Giga Poster
That was absolutely brilliant Hixee! Thanks for making me laugh during days of non-stop studying :)
 

Pokemaniac

Mountain monkey
Staff member
Administrator
Moderator
Also keep in mind that the apples at the bottom might be a little squeezed together by the weight of the apples above. You'd be looking at a slightly higher packing density down there - plus a fair amount of apple juice.

By the way, Hixee, if you ever come across a definite answer for the random packing density of coins, please let me know. I've been trying to do some calculations on a money bin (à la Scrooge McDuck), but all I've ever got for the packing density of the coins in the vault - which are kind of vital for calculating the live load of the building - are very rough estimates.
 

Hixee

Flojector
Staff member
Administrator
Moderator
Social Media Team
If I get the time over the next week or so, I'll revise my calculations to take into account the compressibility of the apples. :p

Coins wise, I haven't come across that problem yet, but if for some reason I do then I'll let you know. :lol:
 

gavin

Moderator
Staff member
Moderator
Social Media Team
You also haven't taken into account the volume that the track takes up passing through the apple.

God, you're crap at Maths
 
Top