Length budget: a container is not a count of pallet positions

Model version

1.0

The formula, in words

• Spend the length once. Take the internal length, take 30 mm off it for the doors, and you have a budget in millimetres. Every column spends its own footprint out of it, and when the budget is gone nothing follows, however much floor is left over.
• The count is a result, not a property. How many pallets a container holds depends on which pallet, so the same 40 foot box is a twenty position container for one footprint and a twenty four position container for another, and mixing the two gives you neither answer.
• Five long columns are six short ones. Both come to 6,000 mm, so swapping one long for one short returns 200 mm and changes no count at all, and the leftover length repeats every fifth long column.

A worked example

The 40 foot container. Internal length 12,032 mm, less the 30 mm the doors need, gives 12,002 mm to spend. Ten columns of 1,200 mm come to 12,000 and leave 2 mm. Twelve of 1,000 mm come to 12,000 and leave 2 mm. So the same box is either a twenty position container or a twenty four position container, and which one it is was decided by the pallet, not by the container.

The shipped plan. $8 \times 1,200 + 3 \times 1,000 = 9,600 + 3,000 = 12,600$ mm against 12,002. Over by 598 mm, on eleven columns and twenty two positions, when the box takes twelve columns and twenty four positions of the short kind. It is smaller by the count and too big by the tape at the same time.

The repair, one swap at a time. Each swap of a long column for a short one holds the column count at eleven and the positions at twenty two, and returns exactly 200 mm.

So three swaps, and the third is the one that lands. Nothing about the plan's size changed across those four rows, which is why no amount of counting could have found the fault.

The identity underneath it. $5 \times 1,200 = 6,000 = 6 \times 1,000$. Five long columns are exactly six short ones, so the waste returns to 2 mm every fifth long column and rises in 200 mm steps in between: 2, 802, 602, 402, 202, 2, 802, 602, 402, 202, 2 as A runs 0 to 10. Read the other way, five long columns occupy the length of six short ones while giving you one column fewer, and that is what a long footprint costs in a fixed length.

The 20 foot container, which makes the same point harder. Usable 5,868 mm. Best B for A of 0 to 4 is 5, 4, 3, 2, 1, so every one of those five plans is five columns and ten positions, without exception. The count is not merely misleading here, it is completely flat while the waste falls 868, 668, 468, 268, 68, a range of 800 mm across plans the count cannot tell apart at all. There is no exact fit and no sawtooth, because 5,868 mm is not long enough to complete one period of five long columns. Same rule, different remainder, which is why the rule is worth carrying and these particular numbers are not.

Where the numbers come from

Internal lengths and the two rows across are the calculator's container table, and everything else here is arithmetic.

The gross mass declaration is described from the international convention that requires it and from Incoterms 2020, both named and neither reproduced, and the point taken from them is only that the second does not allocate the first.

What this does not model

Pinwheeled or turned pallets, overhang, part columns, mixed products on one pallet, the height and weight limits that usually decide a load before length does, load restraint and dunnage, the order columns are loaded in, and the fact that a loader with a tape and a spare hour will often beat any of this by one column.