• anyhow2503@lemmy.world
    59·
    7 days ago
    1. Set lower bound by counting how many are visible in the photo
    2. Grow disinterested
    3. Guess some number above the result from step 1
  • ornery_chemist@mander.xyz
    12·
    6 days ago

    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.

  • rowinxavier@lemmy.world
    15·
    7 days ago

    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

  • Gustephan@lemmy.world
    5·
    6 days ago

    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

  • m0darn@lemmy.ca
    8·
    7 days ago

    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.

    • Corn@lemmy.mlEnglish
      3·
      6 days ago

      Too lazy, measure mass of 4 hazelnuts to get an average, then weigh them all.

  • Siegfried@lemmy.world
    6·
    6 days ago

    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

    • bizarroland@lemmy.worldEnglish
      3·
      7 days ago

      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.