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: 🔗 https://estimate-me.aukspot.com/archive/2025-08-29
If you’d like to discuss your guesses, please use spoiler tags!
Zero.
Ceci ne sont pas noisettes.
Totally unrelated: Now I have the urge for a doner kebeb.
Volume of glass
Volume of hazelnuts (median)
Packing efficiency of spheres in a cylinder
Do some math
Be wrong by an order of magnitude because all the little hazelnuts are concealing one giant one
Vibes. I look at it and try to guess and submit a number within 5 or 10 seconds. It may not be as accurate but I feel like I get more benefit training my ability to estimate at a glance than I do training my ability to do math and spatial reasoning assisted by math. Im already really good at math and math assisted spatial reasoning
I counted 11 down and 7 across, so I assumed the radius is 3.5 hazelnuts and the height eleven, making pi r² h = 423 hazelnuts. that is, if they haven’t snuck in any walnuts.
My method was that I waited for @m0darn@lemmy.ca to do the work and used it as a baseline. I got 8 away.
this was fun. thanks. I like the stats on that website
i’ll eat one every day and tell you how many years are inside the jar
Roughly a truncated cone with diameters ~7 nuts and ~9 nuts, and the cup is ~12 nuts high (quite hard to tell due perspective and nits of different sizes). Throw in an extra layer to account for the heap at the top (which is a dome, but the dome is taller than 1 hazelnut, so treating it as a shorter layer should give some error cancellation) to give a height of 13. The volume using the formula for a truncated cone of those dimensions is ~657 cubic hazelnut diameters. Random sphere packing is 64% (though wall effects should decrease this number) giving a total of 420 nuts (nice).
Answer
This 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.
None, those are pretzel bites
I’m going to guess the cup is around 12 nuts high and 5 nuts in diameter at the center.
Let’s assume it’s a cylinder to get the volume (2/5)^2 * pi * 12 as the volume or around 236 nuts.
The packing density of a sphere is 74% but the edges do play a factor so nudge it down by a bit to get 70%
So I’m going to guess 165 nuts.
I got it laughably wrong
, side view
The cup is about 13 hazelnuts high. The top circumference looks like it could hold about 24, the bottom about 20, so I’ll average that to 22.
13*22=286
I’d ask a couple thousand people to guess in private. So the most popular answer would probably be either surprisingly close to correct or Cuppy McHazelnutface.
