Submitted 7 months ago by otter@lemmy.ca to nostupidquestions@lemmy.world
https://media.piefed.social/posts/9Y/KR/9YKRuOSuZK5h3WP.jpg
For some of these, I estimate the number of items the base and then multiply by the height. Is there a better strategy?
Original post crossposted from !dailygames@lemmy.zip: piefed.social/post/1205620
Guess here: 🔗 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.
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’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).
This ends up being about 5% lower than the true answer. I’m surprised it’s that close. The underestimate makes sense given the wall effects I mentioned, but the error bars on my eyeballed measurements are surely higher.
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.
No. Take the mean, not the mode.
Someone say a God Damn number. _
27???
X
480
Thank you. I feel better about this whole thing now.
Avarage all the guesses of others. (Doesn’t work if they also use this method)
I love it 😄
Did you extract those clips for this post, and do you have a recommended method for doing that? I sometimes find clips on getyarn, but the site barely loads half the time
Just looked up “Gibby fat cakes” on YouTube.
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
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.
It’s the volume of a cylinder, pi times height times radius squared
Find volume of one of them (can probably be found online), calculate volume of container, see how many could fit if you didn’t have to worry about physics, take a bit less than that.
There are some outside the container but I still think the amount of empty space is greater than that.
licks screen
210
You don’t have to actually lick it, you can just imagine
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.
Good luck
None, that’s a jar.
Count the visible ones and multiply by pi
I would look at them and pull a number out of my ass.
Empty the cup and then count them all
Find a cup that is similarly shaped and sized, fill it with hazelnuts and then count it manually
Too lazy, measure mass of 4 hazelnuts to get an average, then weigh them all.
This is a 2D image of a cup of hazelnuts. The number is just the ones you can see. The ones you can't see aren't part of the image.
sopularity_fax@sopuli.xyz 7 months ago
If you put them in the trash there’s 0