For some of these, I estimate the number of items the base and then multiply by the height. Is there a better strategy, especially for items that don't fit into distinct layers?

Original post crossposted from !dailygames@lemmy.zip: https://piefed.social/post/1205620

Guess here: ๐Ÿ”— https://estimate-me.aukspot.com/archive/2025-08-29

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[โ€“] 12 points 1 year ago* (last edited 1 year ago) (1 child)

Roughly a truncated cone with diameters ~7 nuts and ~9 nuts, and the cup is ~12 nuts high (loose guesses, it is hard to tell due perspective and nuts of different sizes). Throw in an extra layer to account for the heap at the top (which is a dome taller than 1 hazelnut, but treating it as a shorter but full layer should give some error cancellation) to give a height of 13. The volume of a truncated cone of those dimensions is ~657 cubic hazelnut diameters. Random sphere packing is 64% space-efficient (though wall effects should decrease this number) giving a total of 420 nuts (nice).

Multiple edits for clarity and typos.

AnswerThis ends up being about 5% lower than the true answer. I'm surprised it's that close. This is in the opposite direction from what I expected given wall effects (which would decrease the real number relative to my estimate). Perturbing one of the base diameters by 1 nut causes a swing of ~50, so measurement error is quite important.

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