What strategy would you use to estimate the number of hazelnuts
from otter@lemmy.ca to nostupidquestions@lemmy.world on 29 Aug 21:33
https://lemmy.ca/post/50664380
from otter@lemmy.ca to nostupidquestions@lemmy.world on 29 Aug 21:33
https://lemmy.ca/post/50664380
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: 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!
#nostupidquestions
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All of them are in the cup.
1, all is not a numeric value 2, not all of them are in the cup. Some of them are resting on top of others that are in the cup, but they themselves are outside of the cup’s boundaries.
No, I have some in a jar at my house, so it can’t be all of them.
I saw some at the store, but I need to go back and confirm that they are still there
For a tapered section like that, you could estimate bottom and top layers and then average them. Then estimate height and multiply. You’d want to include an overlap factor as the roughly spherical nuts would settle in between each other somewhat. I’d imagine there’s some accepted value out there for that.
This is what I did. Roughly 5 wide at the bottom and 7 wide at the top or roughly 12 nuts per layer average. Then the stack appears to be about 12 nuts high so 144 total and maybe round up to 150 since it’s heaping at the top.
Edit: I didn’t initially see OP’s link to the site and I call complete bullshit on that ‘correct’ answer without seeing them poured out and counted on video. According to it, my answer is off by several hundred.
A = pi(r^2) is a hell of a drug.
Only when it jives with a sensible guesstimation.
Without outright spoiling the answer, according to the site, roughly every visible hazelnut in the image makes up just 16% of the total. If my guess of 12 layers is roughly accurate, every visible hazelnut only makes up two layers of the total which doesnt seem correct.
Considering this is a user submitted puzzle with zero verification (as far as I know), the simplest answer is that they gave a fake/incorrect number.
Pie are round. Not square.
Not sure where you’d land at 12 nuts per layer on average. If you go off 5 nuts “wide” as your diameter you’d end up with at least 20 nuts at the bottom layer (area = π(5/2)² = 19.6).
I did 5 wide at the bottom and 7 wide at the top. 5x2 and 7x2 =24/2 (average) = 12 per layer. I now see how this isnt exactly correct as it’s a circle but the ‘proper’ answer puts it at 36 per layer which doesnt seem correct either.
Ah, but you’re calculating the area, so it would be length * width, not just multiplying by 2.
So you’re looking at 25 at the bottom and 49 at the top, making your average 37 per layer.
For me, it would be more about figuring out the rough volume. So, like, you look at a hazelnut, it appears to be about a half-inch spherical value. There seem to be about 6 wide across the bottom, so you could assume that the bottom is somewhere around three inches wide. The top is probably closer to about 5 inches wide. And the height is going to be something like 6 inches.
This is also using rough guesstimation from my own personal knowledge of cups.
So with that information I would use what I remember of the cylinder formula, which is pi times diameter times height I think.
And average the two diameters for a four, so you would go four times six times pi is about 75 cubic inches of volume. Each hazelnut uses about 3/4 of an inch of volume so I would guess there are about 100 hazelnuts in the cup.
That being said, the question is not, what is the correct way to guess it, just how would I do it, and this is how I would do it.
And after actually putting my quotes into it, I was horribly wrong, but still, that’s how I would have done it.
I think that this is more or less the approach I would take, but you shouldn't worry about the actual diameter of anything. It's not important, after all - if everything was scaled up twice as big, the answer would be the same. Just call the diameter of the cup a nice round number and then see how the hazelnuts compare to it. In this case I think there's about five hazelnut widths to the glass, so I'm gonna call the glass diameter 50, the nuts 10, and the glass height 80.
You'll need to change your formulae, though.
pi*dis the circumference of a circle, but we need the area here, sopi*r*r(and then multiply by height for volume). That gives me 157,050 whateverunits cubed for the volume of the cup. For a sphere it's(4/3)*pi*r*r*r, so 524 for the hazelnuts. Now, I know that spheres don't pack perfectly into a volume, but I don't remember the factor even for optimal packing, so I'm just gonna take a wild guess and say that 70% of the internal volume of the cup is actually occupied by hazelnuts. That gives me... 209 hazelnuts in the cup. Which seems worse than your answer on a gut level, but I can count 86 visible ones so it's maybe actually not badChecking my results
Hah, I was way off too
Weigh 10 hazelnuts, weigh all hazelnuts, do basic math.
For this one I would count half of the from, half of the top, multiply them, the multiply by 4, the subtract the 10%.
28 front 8 top (28x8x4)×(0.9) = 691.2 ~ 691
I was 8 off, notbad
Used cylinder volume formula using hazelnuts as units
I don’t math so good, so I roughly counted layers and used the formula for area of a circle, reduced slightly for gaps. First guess was a good 10%-15% off but got within 12 on the second. Then I seriously overcorrected on my third.
Nice work. Cylinder volume is just area of circle multiplied by height, nothing too fancy. Imagine a circle being extruded by one dimension (height)
Can't math so would do it entirely visually and I would get it wrong but feel happy that I'd picked a number (in this case 215) and been able to move on 🤣
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.
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.
Empty the cup and then count them all
I would look at them and pull a number out of my ass.
Count the visible ones and multiply by pi. Not that accurate, considering the shape of the glass and the top of the nut stack, but it’s about as close as I can be arsed calculating now. I refuse to be nerd sniped today, so a ballpark figure will have to suffice.
None, that’s a jar.
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
licks screen
210
You don’t have to actually lick it, you can just imagine
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.
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.
Is there a name for this equation ands can you explain the pix part of the equation?
It’s the volume of a cylinder, pi times height times radius squared
Oh. Thanks. :)
Its not “pix” its “pi * r² * h”. It’s a basic equation for the volume (V) of a cylinder.
Gotcha, thanks.
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.
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
<img alt="" src="https://discuss.tchncs.de/pictrs/image/188f155c-6d89-4af6-bca5-b5188d827b33.mp4">
Context:
<img alt="" src="https://discuss.tchncs.de/pictrs/image/dced0e60-93dc-4c49-8695-761d4a9281ab.mp4">
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.
Avarage all the guesses of others. (Doesn’t work if they also use this method)
Someone say a God Damn number. _
480
Thank you. I feel better about this whole thing now.
Was I right?
Off by 38, not bad, I got similar
x
27???
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.
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 got it laughably wrong
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.
None, those are pretzel bites
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
<img alt="" src="https://sh.itjust.works/pictrs/image/8d9357b5-efe0-4cae-9df8-dd6a259a359c.png">
i’ll eat one every day and tell you how many years are inside the jar
this was fun. thanks
My method was that I waited for @m0darn@lemmy.ca to do the work and used it as a baseline. I got 8 away.
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.
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
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
Totally unrelated: Now I have the urge for a doner kebab.
<img alt="" src="https://lemmy.world/pictrs/image/8b091088-5077-432c-8f5b-582a8543f1a6.gif">
Zero.
Ceci ne sont pas noisettes.
If you put them in the trash there’s 0