Is it possible to calculate how much fluid a crisp packet (potato chip packet for US readers) can hold without actually filling it?
Question #148458. Asked by gmackematix.
Last updated May 14 2021.
Originally posted May 13 2021 9:48 PM.
looney_tunes
Answer has 1 vote
looney_tunes Moderator 19 year member
3326 replies
Answer has 1 vote.
Since the volume of the packet cannot be determined by the mass of crisps (there is a lot of air in there), one could make a guess by measuring the dimensions of the packet. But then you need to make some assumptions about the shape it will take on when you start to add the liquid. It will develop a flattish bottom, and vaguely cylindrical sides (or maybe spherical, depending on exactly how it was constructed), but the exact shape would depend on the way the bottom sealing section interacts with the rest of the packet, and with how elastic the material of which it is made is. If the plastic stretches, that increases the volume.
May 14 2021, 3:26 AM
gmackematix
Answer has 0 votes
gmackematix 22 year member
3206 replies
Answer has 0 votes.
I would have thought the shape it makes when filled up is predetermined. The metallised plastic that a crisp packet is made of is flexible but not particularly elastic. The seam at the bottom, and also the top if it is unopened, is clearly a limiting factor which will stop it bulging into a sphere, but I can't find anything online about the sort of shape so formed.
May 14 2021, 6:35 AM
looney_tunes
Answer has 0 votes
looney_tunes Moderator 19 year member
3326 replies
Answer has 0 votes.
The shape is predetermined, but would depend on the exact crimping used - which would affect the shape of the base and the way the lower edge of the sides starts out. Foil packs are constructed slightly differently from purely plastic ones. Also the material of the packaging affects the way it flexes (and possibly stretches under pressure), which also affects the shape. I would assume an open top, so that the liquid can be poured in; a sealed top would mean the need for a hole through which to inject the liquid, and that could affect the structural integrity of the packet.
What I believe is that we need more information to make accurate calculations, as the shape formed is not a precisely-defined one. (Even if we had that information, it looks to me as if we would need to solve a differential equation of some complexity to take all the factors into account, as the different portions of the surface have different constraints.) Or you could cheat and use a Pringles container - that one is easy.