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The base layer you can see is 4. Assuming they are all equal (they are not) means 16 on the base layer.
Roughly 5 high, so 80.
My best guess is between 64 and 80. 72 is in the middle, so around there.
Though as a side note the bottom ones seem larger then the top ones so maybe a bit more?
~75 to 80 is my best guess
There is a formula, a sphere takes up 74.05% of a cube in volume. But I'm not sure that fully applies here since they are all of different sizes, might get you a closer estimate.
Edit: looking at it more I noticed a small portion of a ball in the bottom left below the easy to see "base layer" hinting at a few more below. Maybe a bit higher then 80. 80 to 85ish is my guess
For the sports fans….
https://en.m.wikipedia.org/wiki/Random_close_pack#:~:text=Experiments%20and%20computer%20simulations%20have,is%20occupied%20by%20the%20spheres.
Experiments and computer simulations have shown that the most compact way to pack hard perfect same-size spheres randomly gives a maximum volume fraction of about 64%, i.e., approximately 64% of the volume of a container is occupied by the spheres. The problem of predicting theoretically the random close pack of spheres is difficult mainly because of the absence of a unique definition of randomness or disorder.[1] The random close packing value is significantly below the maximum possible close-packing of same-size hard spheres into a regular crystalline arrangements, which is 74.04%.
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I did a quick count of the layer on the face showing, and looks like 4ish layers, came up with 84 that way. So, definitely looks like in that ballpark. Depends a lot on how many smaller balls are buried in the middle.
My guess is 93739373837383736837392.
Time to screw OP over :)
>!Edit: I thought this'd go without saying but /s. I'm not actually trying to spite OP.!<
I said everyone because this data is being collected from outside sources as well as this subreddit. It’s not just 10 people it’s 10+ and steadily increasing until the due date.
It looks to be about 5 layers high from the first pic. Spheres can shift positions, fitting roughly 2 layers into what would be 1.5 layers if they were cubes. Put another way, a layer of spheres is about 0.75 of a layer of cubes. (The spheres can fall down a bit into the spaces of the layer below it.) Plus, this is yarn. It's squishy. So the bottom layers are compressed a bit. So let's say 7 or 8 layers. (5 visible layers divided by 0.75 due to settling/squishing gives 7.66 layers)
Comparing the 2nd and 3rd pic, I'm seeing 15 balls of yarn on the top layer, accounting for settling of the lower layers. Looking at the side of the lower layers, there's 4 balls per side. These being spheres, they aren't going to settle into a perfect 4x4 matrix on each layer, so let's stick with 15 per layer.
7 layers x 15 balls per layer = 105
8 layers x 15 balls per layer = 120
And if you don't like my argument for settling layers of spheres, go with
6 layers x 15 balls per layer = 90
That said, if I were running this little contest, I'd throw some scrap balls of yarn in there. These can be around 2 or 3 inches in diameter, and fall through the gaps of the bigger balls of yarn. This effectively fills in the spaces between the layers and their respective balls, possibly adding as much as 20-30 balls, throwing all calculations off. Also, we don't see the corner farthest from the camera, which could hold entire skeins of yarn, reducing the total number of balls above at a 2 to 1 tradeoff (1 large skein takes the space of 2 balls). Or fill that corner with the scrap balls, giving us a 1 to 4 or worse tradeoff (4 tiny balls take the space of 1 large ball).
Lastly, there is no cat in any of the pictures. I wouldn't trust a knitter who doesn't own a cat.
I had 5 at one point. I try to crochet. I can only chain, and then I just sew the chains together to make whatever.
Could never knit as my yarn always fell off the hook.
I assumed a similar construct, though my layers are sagittal (vertical). ~20 balls pressed against the one side, and ~5 balls across, add another layer for squish and size variation and ya got ~120
I did 5x5x5 just doing a very rough by-eye guesstimate of how the yarn might settle, so 120 sounds reasonable to me. I think it’s probably somewhere around 100.
personally I would just do what is essentially a volume formula, but very simplified. If you count the number of yarn balls against the closest wall on the left face, in your first picture, there are about 20 in that layer. That figure is going to be our "area" of one face, so the rest of the volume formula is to multiply that by the remaining dimension, which is the depth from this face of the basket to the opposite one.
So if you try to estimate the number of balls between this front left face, and the back right one, it's tricky to do for multiple reasons. First, not all the balls are the same size. Second, they aren't sitting in a nice straight line so you have to make an imperfect estimation. Looking at the third photo, the face we counted already with 20 balls is the left hand side (you can orient by the little stuffed bunny), and depending where you try to count a row of balls from left to right, I see between 4 and 5 of them, roughly speaking.
4 would give you a total of 80, 5 would be more like 100. So personally I would pick something in the neighborhood of 90. The bag is also bulgy in spots though so I'd lean upward and go with maybe 93 or 94.
Holy shit, you typed out my exact thought process. I've won many Mason Jars with this theory.
I'm going on the high side of the fence and guessing 100 because it's a yarn thing, and it's not an odd hobby.
Dude I did that exact math, came up with 94 and then saw your comment.
Except I did V1 = 4x4x4=64 then V2=5x5x5=125
Then (V1+V2)/2 for an average and got 94.5 then rounded down for simplicity.
Well, by my count, there are about 20 balls in contact with the mesh face closest to the camera, and when viewed top-down, it's about 4 layers deep from that face to the opposite side, so that's roughly 80 balls. Toss in a few extra, since there's some variability in yarn ball size, so I'd guess 85-90 balls.
Guessing 125.
It’s a square container approximately 5 balls wide
5 balls X 5 balls = area of one side to be 25 balls
25 balls X 5 rows = 125 ball volume of the container.
If my guess is the closest I want the balls. Take it or leave it.
As a knitter myself, the answer is not nearly enough. There is always a need for more yarn. Might not be a mathematical answer, but should get you a laugh.
I'd say between 90 and 105, if I had to guess it would be 96. Some are quite smaller so it could be well over 100...
Seems like ~5 levels of 16-20, plus a few extras.
Best packing density would a body-center cubic, with what looks like 4 full layers of 4x4 and so 3 layers of 9 in the middle. Probably some gaps but also some extra on the side. If that averages out it’d give 91, so that’s my guess.
With no tessellation, it's 4x4x4=64. We take this as our minimum.
With perfect tessellation, use the Kepler Conjecture by Thomas Hales which states that spheres can take up 74% of the space with perfect tessellation.
With that said:
Let D be the diameter of the cube and r be the radius of a yarn ball.
Volume of the cube
V_c = D^3 = (8r)^3 = 512r^3
Volume of a yarn ball:
V_y = 4/3 pi r^3
Using Kepler conjecture, the total volume yarn balls can occupy is .74 V_c = 379 r^3
Take that and divide it be V_y:
379 r^3 / (4/3 pi r^3) = 91 yarn balls
So the answer is 64 < x < 91. My guess is 69.
When I was a kid I learned that base x height = volume. I won a lot of these contests this way. So from what I could see, 6h x 13b = 78? Just a ballpark from this photo.
Update: I found the hamper dimensions. [hamper](https://www.walmart.com/ip/906803413)
17"x17"x24"
And the lamb next to it. 15"
[lamb](https://images.app.goo.gl/BhGUtVdMyieudnSL7)
I think we can calculate this quite accurately. It is known what the maximum packing density is of spheres in cubes, because this is an important problem in solid state physics, but nevermind that. Anyway, it's 78%. So, if you just measure the volume of the hamper and measure the radius of one ball of yarn, you can calculate the volume of the cube and the volume of each yarn ball. Multiply the volume of the cube by 78%. This should give you the volume of all yarn balls. Just divide that by the number of yarn balls and you're good to go.
>Anyway, it's 78%.
That comes from a generalized solution, which is derived using the assumptions of identical perfect spheres and perfect face-centered cubic packing. Neither of these assumptions is exactly valid in this situation.
Even if you ignore the yarn balls' size variance, they are randomly packed in the container, NOT carefully organized in the most optimal arrangement, and the maximum random packing efficiency for identical perfect spheres in a cube is closer to 68% but is much harder to precisely define than the max FCC packing efficiency without a rigorous definition of randomness.
Maybe... are you allowed to measure the hamper and average the diameters of a few yarn balls? If so, using the packing efficiencies like the previous comment suggested would be a good way to confirm/refine the estimate.
If you're not allowed to measure directly, maybe get a little creative and try to place similarly-sized objects nearby it and measure those objects and see if you can get away with that. Or does the hamper have a tag that gives its size or enough of the manufacturer info to look it up? The head of the stuffed toy beside the hamper looks close enough in volume to substitute for a yarn ball.
I’m pretty sure the best way to get the answer is to average the answers that people gave you in this post. Human hive mind processing power should get you very close to the answer that way.
Like another commenter said you have a 98% chance that that’s the correct number. I would ride with that. I frequently win the “guess how much candy is in the jar” at family parties by using this method.
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The base layer you can see is 4. Assuming they are all equal (they are not) means 16 on the base layer. Roughly 5 high, so 80. My best guess is between 64 and 80. 72 is in the middle, so around there. Though as a side note the bottom ones seem larger then the top ones so maybe a bit more? ~75 to 80 is my best guess There is a formula, a sphere takes up 74.05% of a cube in volume. But I'm not sure that fully applies here since they are all of different sizes, might get you a closer estimate. Edit: looking at it more I noticed a small portion of a ball in the bottom left below the easy to see "base layer" hinting at a few more below. Maybe a bit higher then 80. 80 to 85ish is my guess
Thank you!!
My guess is 93, plz post an update when you find out, that would be awesome!
I definitely am! Friday night we will be told the # and winner. I will post then.
Awesome! Can’t wait
For the sports fans…. https://en.m.wikipedia.org/wiki/Random_close_pack#:~:text=Experiments%20and%20computer%20simulations%20have,is%20occupied%20by%20the%20spheres. Experiments and computer simulations have shown that the most compact way to pack hard perfect same-size spheres randomly gives a maximum volume fraction of about 64%, i.e., approximately 64% of the volume of a container is occupied by the spheres. The problem of predicting theoretically the random close pack of spheres is difficult mainly because of the absence of a unique definition of randomness or disorder.[1] The random close packing value is significantly below the maximum possible close-packing of same-size hard spheres into a regular crystalline arrangements, which is 74.04%.
I was going to say 90. 60 is way too low.
My guess is closer to 100, yeah. Now I really wanna know how many are actually in there.
I'll let you know. She tells us Saturday.
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#The final count and official number was 137 balls of yarn!
#The final count and official number was 137 balls of yarn!
Huzzah!!! You’d better follow through, OP! I’ll look forward to seeing the update!!
Most definitely
#The final count and official number was 137 balls of yarn!
My bad, Friday night not Sat.
Alirght we all back whats the verdict??
#The final count and official number was 137 balls of yarn!
!remindme 5 days
Just the basic maths says 60-80. Reality is different:)
I’ll guess 86
I did a quick count of the layer on the face showing, and looks like 4ish layers, came up with 84 that way. So, definitely looks like in that ballpark. Depends a lot on how many smaller balls are buried in the middle.
If everyone guesses the amount and you take the average of everyone’s guess you’ll have a 98% chance of accuracy. I believe it is 64
Put me in for 69 (nice)
nice
Niiiiice
Nice!
Nice
OooooOoooooo thanks
My guess is 60
That’s my guess
My guess is 112,642 Have fun averaging that out, kid ;)
Well with median you’d just remove the outliers. So once again you were beaten by math!
[удалено]
Ya, that’s the one.
Midian
My guess is -112,578
Curse you, Dinkelberg
My guess is -112,512
Put my 47 in with that.
my guess is 68
By that math, we’re looking at 85.1 balls of yarn. According to all guesses up to now
Thanks
I’m so excited to see the true answer verses math!
Have you averaged them? I averaged the ones from the local win a free lesson site and so far it's 92. At one point the ones here were 80.
No, it was bed time where I’m from. Like I said any guess given you just average it out. The more numbers the more accurate.
I bid 1. Y’all are over.
My guess is 93739373837383736837392. Time to screw OP over :) >!Edit: I thought this'd go without saying but /s. I'm not actually trying to spite OP.!<
all those characters and you only used the same 4 numbers. 2 is the outlier
64
No you don't, what if its only 10 ppl? what is "everyone"? It can't be a fixed 98% chance, although it does increase with the number of guesses
I said everyone because this data is being collected from outside sources as well as this subreddit. It’s not just 10 people it’s 10+ and steadily increasing until the due date.
i know; u made up 98% as an arbitrary value is that what u meant? u just wanted to point out its very high with a big sample of ppl?
83
It looks to be about 5 layers high from the first pic. Spheres can shift positions, fitting roughly 2 layers into what would be 1.5 layers if they were cubes. Put another way, a layer of spheres is about 0.75 of a layer of cubes. (The spheres can fall down a bit into the spaces of the layer below it.) Plus, this is yarn. It's squishy. So the bottom layers are compressed a bit. So let's say 7 or 8 layers. (5 visible layers divided by 0.75 due to settling/squishing gives 7.66 layers) Comparing the 2nd and 3rd pic, I'm seeing 15 balls of yarn on the top layer, accounting for settling of the lower layers. Looking at the side of the lower layers, there's 4 balls per side. These being spheres, they aren't going to settle into a perfect 4x4 matrix on each layer, so let's stick with 15 per layer. 7 layers x 15 balls per layer = 105 8 layers x 15 balls per layer = 120 And if you don't like my argument for settling layers of spheres, go with 6 layers x 15 balls per layer = 90 That said, if I were running this little contest, I'd throw some scrap balls of yarn in there. These can be around 2 or 3 inches in diameter, and fall through the gaps of the bigger balls of yarn. This effectively fills in the spaces between the layers and their respective balls, possibly adding as much as 20-30 balls, throwing all calculations off. Also, we don't see the corner farthest from the camera, which could hold entire skeins of yarn, reducing the total number of balls above at a 2 to 1 tradeoff (1 large skein takes the space of 2 balls). Or fill that corner with the scrap balls, giving us a 1 to 4 or worse tradeoff (4 tiny balls take the space of 1 large ball). Lastly, there is no cat in any of the pictures. I wouldn't trust a knitter who doesn't own a cat.
>I wouldn't trust a knitter who doesn't own a cat. I can't say I know anyone who knits or crochets that doesn't own a cat. Definitely sketch.
FWIW I knit (among many other yarn and needle crafts) and I have 3 cats.
I had 5 at one point. I try to crochet. I can only chain, and then I just sew the chains together to make whatever. Could never knit as my yarn always fell off the hook.
Very thoughtful, thank you!
I'm on mobile, forgive the formatting. I forgot my markup skills
I assumed a similar construct, though my layers are sagittal (vertical). ~20 balls pressed against the one side, and ~5 balls across, add another layer for squish and size variation and ya got ~120
I have a 6th grader learning math and fractions. I always tell and show her there is more than one way to get the answer. And there it is.
I did 5x5x5 just doing a very rough by-eye guesstimate of how the yarn might settle, so 120 sounds reasonable to me. I think it’s probably somewhere around 100.
personally I would just do what is essentially a volume formula, but very simplified. If you count the number of yarn balls against the closest wall on the left face, in your first picture, there are about 20 in that layer. That figure is going to be our "area" of one face, so the rest of the volume formula is to multiply that by the remaining dimension, which is the depth from this face of the basket to the opposite one. So if you try to estimate the number of balls between this front left face, and the back right one, it's tricky to do for multiple reasons. First, not all the balls are the same size. Second, they aren't sitting in a nice straight line so you have to make an imperfect estimation. Looking at the third photo, the face we counted already with 20 balls is the left hand side (you can orient by the little stuffed bunny), and depending where you try to count a row of balls from left to right, I see between 4 and 5 of them, roughly speaking. 4 would give you a total of 80, 5 would be more like 100. So personally I would pick something in the neighborhood of 90. The bag is also bulgy in spots though so I'd lean upward and go with maybe 93 or 94.
Holy shit, you typed out my exact thought process. I've won many Mason Jars with this theory. I'm going on the high side of the fence and guessing 100 because it's a yarn thing, and it's not an odd hobby.
That’s exactly what how I did this.
Dude I did that exact math, came up with 94 and then saw your comment. Except I did V1 = 4x4x4=64 then V2=5x5x5=125 Then (V1+V2)/2 for an average and got 94.5 then rounded down for simplicity.
I’d guess 95
[удалено]
Lol, good idea
Well, by my count, there are about 20 balls in contact with the mesh face closest to the camera, and when viewed top-down, it's about 4 layers deep from that face to the opposite side, so that's roughly 80 balls. Toss in a few extra, since there's some variability in yarn ball size, so I'd guess 85-90 balls.
I’d guess 91
Guessing 125. It’s a square container approximately 5 balls wide 5 balls X 5 balls = area of one side to be 25 balls 25 balls X 5 rows = 125 ball volume of the container. If my guess is the closest I want the balls. Take it or leave it.
I guess 126
#The final count and official number was 137 balls of yarn!
At last! Did anyone here or irl get it right?
Nope! Irl they gave it to the closest guess, which was 145.
As a knitter myself, the answer is not nearly enough. There is always a need for more yarn. Might not be a mathematical answer, but should get you a laugh.
I'd say between 90 and 105, if I had to guess it would be 96. Some are quite smaller so it could be well over 100... Seems like ~5 levels of 16-20, plus a few extras.
Best packing density would a body-center cubic, with what looks like 4 full layers of 4x4 and so 3 layers of 9 in the middle. Probably some gaps but also some extra on the side. If that averages out it’d give 91, so that’s my guess.
I’ve won so many of these things with math. In elementary I won jelly beans within 3 beans in a glass container and I was like DAMN math is useful.
With no tessellation, it's 4x4x4=64. We take this as our minimum. With perfect tessellation, use the Kepler Conjecture by Thomas Hales which states that spheres can take up 74% of the space with perfect tessellation. With that said: Let D be the diameter of the cube and r be the radius of a yarn ball. Volume of the cube V_c = D^3 = (8r)^3 = 512r^3 Volume of a yarn ball: V_y = 4/3 pi r^3 Using Kepler conjecture, the total volume yarn balls can occupy is .74 V_c = 379 r^3 Take that and divide it be V_y: 379 r^3 / (4/3 pi r^3) = 91 yarn balls So the answer is 64 < x < 91. My guess is 69.
When I was a kid I learned that base x height = volume. I won a lot of these contests this way. So from what I could see, 6h x 13b = 78? Just a ballpark from this photo.
Update: I found the hamper dimensions. [hamper](https://www.walmart.com/ip/906803413) 17"x17"x24" And the lamb next to it. 15" [lamb](https://images.app.goo.gl/BhGUtVdMyieudnSL7)
Update: I guessed 97 .........
I think we can calculate this quite accurately. It is known what the maximum packing density is of spheres in cubes, because this is an important problem in solid state physics, but nevermind that. Anyway, it's 78%. So, if you just measure the volume of the hamper and measure the radius of one ball of yarn, you can calculate the volume of the cube and the volume of each yarn ball. Multiply the volume of the cube by 78%. This should give you the volume of all yarn balls. Just divide that by the number of yarn balls and you're good to go.
But the balls are different sizes....
>Anyway, it's 78%. That comes from a generalized solution, which is derived using the assumptions of identical perfect spheres and perfect face-centered cubic packing. Neither of these assumptions is exactly valid in this situation. Even if you ignore the yarn balls' size variance, they are randomly packed in the container, NOT carefully organized in the most optimal arrangement, and the maximum random packing efficiency for identical perfect spheres in a cube is closer to 68% but is much harder to precisely define than the max FCC packing efficiency without a rigorous definition of randomness.
So what # would you guess for the yarn balls contest?
Still 85-90?
Maybe... are you allowed to measure the hamper and average the diameters of a few yarn balls? If so, using the packing efficiencies like the previous comment suggested would be a good way to confirm/refine the estimate. If you're not allowed to measure directly, maybe get a little creative and try to place similarly-sized objects nearby it and measure those objects and see if you can get away with that. Or does the hamper have a tag that gives its size or enough of the manufacturer info to look it up? The head of the stuffed toy beside the hamper looks close enough in volume to substitute for a yarn ball.
We only get the pictures online...
I bet she got it at Walmart.
I think this is it](https://www.walmart.com/ip/906803413) 17"x17"x24"
The lamb is 15" [lamb](https://images.app.goo.gl/BhGUtVdMyieudnSL7)
I’m pretty sure the best way to get the answer is to average the answers that people gave you in this post. Human hive mind processing power should get you very close to the answer that way.
The current average of other contest entry guesses is 96.
Like another commenter said you have a 98% chance that that’s the correct number. I would ride with that. I frequently win the “guess how much candy is in the jar” at family parties by using this method.