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!
- Set lower bound by counting how many are visible in the photo
- Grow disinterested
- Guess some number above the result from step 1
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.
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.
funily enough I picked random numbers then just chose 420 as a funny number and got the same answer as you
Start with the bottom, count how many are visible on the lowest layer. This gives you half the circumference of the lower end.
Repeat for the highest part of the cup that is still cup, not above the line.
Multiply each of these by 2 and you have the circumferences of each end of the cut cone.
Rearrange the circle area formula to get the radius from circumference and you have the radius of each end of the cone.
Now use the cut cone formula to calculate the volume in terms of hazelnuts.
Next, take the radius of the top circle and estimate how far above that the highest nut is. Use whatever formula seems more appropriate, in this case maybe just a right angle triangle formula with a full rotation, to estimate the volume of the top.
Sum that together with the conic volume and you have a good estimate.
My estimate, at least 3 hazelnuts.
Good luck
My method was that I waited for @m0darn@lemmy.ca to do the work and used it as a baseline. I got 8 away.
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’ll eat one every day and tell you how many years are inside the jar
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.
licks screen
210
You don’t have to actually lick it, you can just imagine
Imagej
In arbitrary units, I need in pixels estimations for:
Height of the vase H
Top diameter of the vase Dt
Bottom diameter of the vase Db
Mean diameter of a hazelnut Dh (measuring a few of them)
Packaging factor F… something arround .6? I’m sure there is a better way of estimating it. Like counting air and hazelnut areas on the image but I’m not sure how to correlate 2D to 3D in this case.
The volume of the vase is Vv (H,Dt,Db)
The volume of a single hazelnut Vh (Dh)
N = Vv F / Vh
Totally unrelated: Now I have the urge for a doner kebab.
this was fun. thanks. I like the stats on that website
Zero.
Ceci ne sont pas noisettes.
, side view
I did
V=pi×r²×h
and estimated:
r=4 (from some at the top left)
h=12
This gave 603. The link said too high, so realized I’d neglected packing factor. Google said that spheres typically pack at 64% efficiency, so I guessed 386. Too low.
These aren’t spheres, though, so the packing efficiency could be better, like 72% or better. These look better packed than those vases filled with glass marbles. So it could even be in the 80s.
Yeah I went 80% next which was too high. 72% is very close to the answer in the link.
Is there a name for this equation ands can you explain the pix part of the equation?
Its not “pix” its “pi * r² * h”. It’s a basic equation for the volume (V) of a cylinder.
Gotcha, thanks.
It’s the volume of a cylinder, pi times height times radius squared
Oh. Thanks. :)
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.
