How many candies are in a jar - packing efficiency and volume?

[hibernicus]

@Anonymous User

Also the air density will matter when there are mass problems.like I said if someone told you that the jar is 1200 grams then you can take 80 percent of that, and then divide if I the average density. Remembers mass divided by density meet equals volume. find the average density you will need to know the density of the candy and the density of the air. If you know that then you can take 70 percent of the density of the candy times and add 30 percent are the density of the air to it which will give you the average density. Now this example assumes that the packing efficiency is 70% but it can and will be different for others.

No, you should neglect the air density. The air in the pile of candy has neutral buoyancy. So when you weigh a pile of candy in air, you are weighing only the candy and not the air.

If you were weighing a jar of candy on the moon, or weighing an evacuated jar of candy on earth, then the air would have a small effect on the reading. However, bearing in mind that your estimate of the packing density has an uncertainty of +/- several %, it is pointless to consider effects that amount to less than 0.1%.

I note that Senegoid has asked you 3 or 4 times, and I once, for clarification of the context and rules of the contest (because it may help us to suggest an alternative approach), and you seem reluctant to give a straight answer. Why?

[si_blakely]

@hibernicus

I note that Senegoid has asked you 3 or 4 times, and I once, for clarification of the context and rules of the contest (because it may help us to suggest an alternative approach), and you seem reluctant to give a straight answer. Why?

Post #55 - the OP just wants to win open competitions using math. The rules are not fixed, but probably include no touching/measuring.
Si

[Leo_Bloom]

@Anonymous User

Also the air density will matter when there are mass problems.like I said if someone told you that the jar is 1200 grams then you can take 80 percent of that, and then divide if I the average density. Remembers mass divided by density meet equals volume. find the average density you will need to know the density of the candy and the density of the air. If you know that then you can take 70 percent of the density of the candy times and add 30 percent are the density of the air to it which will give you the average density. Now this example assumes that the packing efficiency is 70% but it can and will be different for others.

[nitpick, for now]
Anonymous, be sure to be attentive to the English grammar and style of your sentences when presenting your assumptions and results. English follows a logic and has the opportunities for descriptive style that should match as closely as possible–and not just in a blow-by-blow way–the clarity of what you are thinking about in other symbols. Sometimes it can fight or conceal the points you may think are obvious in the math.
[/nitpick, for now]

[R3d_Anonymous]

@si_blakely

Post #55 - the OP just wants to win open competitions using math. The rules are not fixed, but probably include no touching/measuring.
Si

Yes this is what I mean. Alright air density might not matter but candy density will.

[R3d_Anonymous]

@Leo_Bloom

[nitpick, for now]
Anonymous, be sure to be attentive to the English grammar and style of your sentences when presenting your assumptions and results. English follows a logic and has the opportunities for descriptive style that should match as closely as possible–and not just in a blow-by-blow way–the clarity of what you are thinking about in other symbols. Sometimes it can fight or conceal the points you may think are obvious in the math.
[/nitpick, for now]

I am sorry. I was typing on a smartphone using voice recognition.

[Leo_Bloom]

@Anonymous User

I am sorry. I was typing on a smartphone using voice recognition.

:eek:Holy shit.

[Senegoid]

@hibernicus

I note that Senegoid has asked you 3 or 4 times, and I once, for clarification of the context and rules of the contest (because it may help us to suggest an alternative approach), and you seem reluctant to give a straight answer. Why?

I think that Anonymous User doesn’t have a specific contest in mind, but instead is thinking of hypothetical contests he expects to enter. So I think he’s trying to figure out, in advance, as much of the problem as he can without knowing the details. I’ve asked several times about the geometry of the jar and what measuring rules will be allowed, and also about what raw data he will be given. He seems to think that the density of the candy is important (and even the density of the air!), and he is going to the lab to experiment with densities. (ETA: Makes me think of Sherlock Holmes stories.) (That sounds odd – I would think the LAST think he would know in advance is the exact kind of candy to be used. Candies could have all different kinds of densities. Maybe the jar will be full of marshmallows.)

He is doing good by researching such constants as packing densities, and studying formulas for deriving volumes given weight and densities (if the raw data will involve that). He might also develop formulas for various more-common jar shapes (perhaps those big urn-shaped glass jars you often see in store-window contests like this).

Here’s a Google image search for guess how many gum balls – shows several pictures of jars/urns/tubs of gum balls.

ETA: I think I’m learning something myself from this discussion. At first I thought, intuitively, that smaller spheres would fill a jar more completely than larger spheres. Apparently, from reading this (and some googling myself), that’s not true. It looks like sphere packing density isn’t affected by the size of the spheres. I would have thought, if you fill the jar with BB’s it would fill it much more fully than filling it with jawbreakers.

[Measure_for_Measure]

@fumster

I was in 1st grade in 1961 and we had the first generation of texts with set theory in them. What a colossal waste of time. The first three weeks of every year we spent learning unions and intersections and then never used them again. You don’t need to learn that “0 = the null set” to learn to count. Rant over.

@Senegoid

[Continuing fumster’s rant]

Hijack though this may be, it’s already come up in this thread, so I’ll run with fumster on this. This is right. Set Theory isn’t particularly useful until much more advanced math than this, at which time it becomes quite useful. Until then, fumster is right – you spend too much time studying it, too early on, with no particular use for it, and with no clear understanding of what it’s all good for.

I enjoyed set theory and Boolean logic because it was much easier than arithmetic. Years later Nexis Lexis and then google were a breeze to learn and use.

FWIW, my parents never quite got the handle of search engines. And some people here seem to have difficulty with them as well, for reasons that escape me. Would a natural grasp of Boolean logic help? If so, elementary school set theory may have been more helpful than they appear.

[Leo_Bloom]

@Measure_for_Measure

I enjoyed set theory and Boolean logic because it was much easier than arithmetic. Years later Nexis Lexis and then google were a breeze to learn and use.

FWIW, my parents never quite got the handle of search engines. And some people here seem to have difficulty with them as well, for reasons that escape me. Would a natural grasp of Boolean logic help? If so, elementary school set theory may have been more helpful than they appear.

My thoughts also. In addition, I can’t remember Venn diagrams being taught to me for arithmetic, but they were taught, and established for me an excellent mindset of how to parcel out a multiplicity.

[R3d_Anonymous]

@Leo_Bloom

[nitpick, for now]
Anonymous, be sure to be attentive to the English grammar and style of your sentences when presenting your assumptions and results. English follows a logic and has the opportunities for descriptive style that should match as closely as possible–and not just in a blow-by-blow way–the clarity of what you are thinking about in other symbols. Sometimes it can fight or conceal the points you may think are obvious in the math.
[/nitpick, for now]

@Senegoid

I think that Anonymous User doesn’t have a specific contest in mind, but instead is thinking of hypothetical contests he expects to enter. So I think he’s trying to figure out, in advance, as much of the problem as he can without knowing the details. I’ve asked several times about the geometry of the jar and what measuring rules will be allowed, and also about what raw data he will be given. He seems to think that the density of the candy is important (and even the density of the air!), and he is going to the lab to experiment with densities. (ETA: Makes me think of Sherlock Holmes stories.) (That sounds odd – I would think the LAST think he would know in advance is the exact kind of candy to be used. Candies could have all different kinds of densities. Maybe the jar will be full of marshmallows.)

He is doing good by researching such constants as packing densities, and studying formulas for deriving volumes given weight and densities (if the raw data will involve that). He might also develop formulas for various more-common jar shapes (perhaps those big urn-shaped glass jars you often see in store-window contests like this).

Here’s a Google image search for guess how many gum balls – shows several pictures of jars/urns/tubs of gum balls.

ETA: I think I’m learning something myself from this discussion. At first I thought, intuitively, that smaller spheres would fill a jar more completely than larger spheres. Apparently, from reading this (and some googling myself), that’s not true. It looks like sphere packing density isn’t affected by the size of the spheres. I would have thought, if you fill the jar with BB’s it would fill it much more fully than filling it with jawbreakers.

Density of air actually is not important but the candy one is. Why? Because like I said, it can become important especially when working with mass problems and conversions. Remember that mass divided by volume is density. So it also gives some other useful information. Also in the lab, I am going to find out the volumes of the candies as well. It is just handy information. However, if I know the volume of the jar then all I have to do is plug that in for x in that linear equation I have created. Also packing efficiency is not always going to be known but for spheres it is 64% (and this is crucial as it is part of that basic linear equation) and for oblate spheroids it is 66.5% (I.e. skittles and m&ms). This is not much of a problem because most are around 70% anyways so worst case scenerio is that you have to put in 70% if you do not know the actual which is okay. By the way, does anyone know the packing efficiency of aspherical elipsoids? That would be nice if someone knew because those are basically jelly beans and jelly beans are the most commonly used candies.

As for the contest type,yes you are right that I have no specific environment. Just no touching and measuring. Now like I said if the guy is foolish enough to tell me the mass then you know what I am going to do. But if he does not, that is what I am asking you guys about. How do I figure out the volume. I feel like who ever said creating a cylinder in your mind that makes two quantities equal and then find the radius of that (this guy was about the third post I think). He had a point I think. But if someone could tell me exactly how to do that accurately, that might be something to explore. As For the jar, think of the most irregular jar out there. As for hiding tennis balls in the middle, do not worry about that. We will figure something out later on, but for right now, just focus on the jar volume.

[Senegoid]

Well, here’s a thought: Call or write the Jelly Belly folks and ask them what the packing efficiency is for their jelly beans. Manufacturers of all sorts pay attention to the packaging of their product, so I think they’d be guaranteed to know. (No telling if your request would get forward to the right packaging engineers, or if they’d care to answer you – but you can try.)

Jelly Belly main page

Jelly Belly “Contact Us” page

There’s also a lengthy FAQ out there – 4 pages long (and that’s just the questions). Here’s one:
– How many jelly beans per pound? A: 25 per ounce, or approx. 400 per pound.
(Didn’t see a question about packing density, but I think there’s a page there where you could send in a question.)

ETA: And the M&M folks would probably know that statistic for their products too.

[R3d_Anonymous]

@Senegoid

Well, here’s a thought: Call or write the Jelly Belly folks and ask them what the packing efficiency is for their jelly beans. Manufacturers of all sorts pay attention to the packaging of their product, so I think they’d be guaranteed to know. (No telling if your request would get forward to the right packaging engineers, or if they’d care to answer you – but you can try.)

Jelly Belly main page

Jelly Belly “Contact Us” page

There’s also a lengthy FAQ out there – 4 pages long (and that’s just the questions). Here’s one:
– How many jelly beans per pound? A: 25 per ounce, or approx. 400 per pound.
(Didn’t see a question about packing density, but I think there’s a page there where you could send in a question.)

ETA: And the M&M folks would probably know that statistic for their products too.

What about the volume of the jar? How do you feel about creating a cylinder that makes two quantities equal? How should this be done?

[si_blakely]

I still think the photo method will work.

Take this photo of a jar of gumballs. It is basically just a cylinder for ease of demonstration.
I saved it to my computer, and opened it in a photo editor (the Gimp, which is free). I used a rectangle selection tool to isolate the gumballs to the edge of the internal volume - this gave me a image 253x124 pixels. I then selected a gumball (the white one in the middle). This was 16x16 pixels.

So, if a gumball is 1.5 cm (I don’t know how big they are - you need to do your research).

This is 1.77cm[sup]3[/sup]

pixels per cm = 17.33

So the diameter of the jar is 124/17.33 = 7.15cm.
The height of the jar is 253/17.33 = 14.60cm

The jars volume is pidiameterheight = pi7.1514.60 = 328 cm[sup]3[/sup]

You can now use your data on packing density to calculate how many gumballs are in the jar.

So 328 * 0.75 (~ packing density) / 1.77 = 139 gumballs

Si

[Senegoid]

@Anonymous User

What about the volume of the jar? How do you feel about creating a cylinder that makes two quantities equal? How should this be done?

I think this was ZenBeam’s idea. You seem to think it’s too vague and you don’t quite understand it.

Well, as best I can figure, that idea IS vague, and I don’t understand it either. The problem is still this: We don’t seem to have any idea what the jar is going to be shaped like, so what more can anyone say?

I guess the idea here is, if the jar is approximately cylindrical (like the one that si_blakely linked to, just above), then imagine the cylinder that just fits inside the jar, and figure the volume of that. Then try to make some adjustment for any irregularity in the shape – like adding a few to account for the gum balls up in the neck of the jar. Or something like that.

But suppose the jar is,say, rectangular in cross section. Then I don’t think any kind of cylindrical approximation is meaningful. Or, the picture that keeps coming to my mind is a volume of revolution but with a diameter that varies from top to bottom – like an amphora (photo). Here’s a picture of a jelly bean jar where I don’t think a cylinder makes a good approximation.

[R3d_Anonymous]

@Senegoid

I think this was ZenBeam’s idea. You seem to think it’s too vague and you don’t quite understand it.

Well, as best I can figure, that idea IS vague, and I don’t understand it either. The problem is still this: We don’t seem to have any idea what the jar is going to be shaped like, so what more can anyone say?

I guess the idea here is, if the jar is approximately cylindrical (like the one that si_blakely linked to, just above), then imagine the cylinder that just fits inside the jar, and figure the volume of that. Then try to make some adjustment for any irregularity in the shape – like adding a few to account for the gum balls up in the neck of the jar. Or something like that.

But suppose the jar is,say, rectangular in cross section. Then I don’t think any kind of cylindrical approximation is meaningful. Or, the picture that keeps coming to my mind is a volume of revolution but with a diameter that varies from top to bottom – like an amphora (photo). Here’s a picture of a jelly bean jar where I don’t think a cylinder makes a good approximation.

Well for rectangular prisms you would not do that. You could just do the area of the base times the height. But even for very irregular ones, such as the jelly bean one you showed, I think it still may be possible. Because still all you are doing is making those two quantities equal. You keep fine tuning the diameter of that cylinder until it makes sense, and then find the volume of that cylinder which is very easy from there. If that volume does not seem right, keep fine tuning the diameter. You would not have to do any measuring since you could use your two hands as the length of the diameter in a.sheet of paper and use a ruler to measure that distance. If the jar is toughly cylindrical, it is close enough to just take the diameter of that jar which is still possible without touching or picking up the jar. I do not know what do you think?

Also that pixel method cannot really be possible because how am I supposed to have a that high tech photo editor on my phone?

[si_blakely]

@Anonymous User

Also that pixel method cannot really be possible because how am I supposed to have a that high tech photo editor on my phone?

There are phone apps for doing measurements on photos - I found a couple of Andriod apps (non free), the same is likely for Apple phones.

But these competitions usually run for a reasonable time - plenty of time to snap the photos, do the math and get back to enter the competition. You didn’t think it would be easy, did you.

Si

[ZenBeam]

@Senegoid

I think this was ZenBeam’s idea. You seem to think it’s too vague and you don’t quite understand it.

Well, as best I can figure, that idea IS vague, and I don’t understand it either.

Maybe an example will help. I’ll use this gumball machine. Taking a gumball as having a diameter of 1, the maximum diameter of the globe is about 9. The minimum diameter is about 6. The height is also about 6. The volume is then somewhere between 6 * pi *3^2 = 170 and 6 * pi * 4.5^2 = 382, but you know 170 is too small, and 382 is too large. By looking at it, I’d guess a cylinder with a diameter of 8 would have about the same volume, so calculate using that volume: 6 * pi * 4^2 = 302. Using an efficiency of 64%, that would be 193 gumballs.

You could get the diameter at a bunch of different heights, and numerically integrate the volume to get a more accurate answer, but that’s a lot more work, and you’ve probably got enough inaccuracy elsewhere that it’s not worthwhile.

[R3d_Anonymous]

@ZenBeam

Maybe an example will help. I’ll use this gumball machine. Taking a gumball as having a diameter of 1, the maximum diameter of the globe is about 9. The minimum diameter is about 6. The height is also about 6. The volume is then somewhere between 6 * pi *3^2 = 170 and 6 * pi * 4.5^2 = 382, but you know 170 is too small, and 382 is too large. By looking at it, I’d guess a cylinder with a diameter of 8 would have about the same volume, so calculate using that volume: 6 * pi * 4^2 = 302. Using an efficiency of 64%, that would be 193 gumballs.

You could get the diameter at a bunch of different heights, and numerically integrate the volume to get a more accurate answer, but that’s a lot more work, and you’ve probably got enough inaccuracy elsewhere that it’s not worthwhile.

By the way, I don’t think the picture taking method is very practical and good. What if the people at the contest catch you? I know it may work, but it just doesn’t seem like a good way to do it. I don’t know why. I’d like Senegoid’s opinion on this.

I don’t think you really even have to do that much. You could just eyeball it right? Keep adjusting that imaginary cylinder until you get to the closest you can for the two quantities being equal?

Also, I’m going to the lab tomorrow!

[ZenBeam]

@Anonymous User

By the way, I don’t think the picture taking method is very practical and good. What if the people at the contest catch you? I know it may work, but it just doesn’t seem like a good way to do it. I don’t know why. I’d like Senegoid’s opinion on this.

I don’t think you really even have to do that much. You could just eyeball it right? Keep adjusting that imaginary cylinder until you get to the closest you can for the two quantities being equal?

Yes, exactly.

Also, I wasn’t assuming you’d be able to take a picture, but the picture sure made it easier to explain. If you can take a picture, I’d certainly do it. Just remember a picture isn’t a projection. Parts closer to the camera will appear bigger relative to parts farther away.

[Senegoid]

Do you know of any rules that say you can’t take a picture?

If you’re going to estimate that cylinder-matching technique simply by eyeballing the jar, and adjusting the cylinder until it “looks the right size”, doesn’t this assume you already have an estimate in mind of what “the right size” is? It sounds like circular reasoning – Start with an estimate of the right size, then adjust the cylinder until it agrees with that, then you will have an estimate the right size.

If you’re going to estimate just by eyeballing the jar, then just go ahead and do that. As far as I can tell (for jars like that gum ball machine and some of the other photos we’ve seen), that may just be what you have to do.

You might think of it this way: Picture a cylinder “overlaid” with the jar. It isn’t a perfect fit – part of the cylinder sticks out, outside the jar. Part of the cylinder falls inside the jar, with some extra space around it.

So there are some gum balls in the wider parts of the jar that are outside the cylinder. And for the narrower parts of the jar (and maybe the odd-shaped top or bottom), the cylinder will stick out. There, you will have room for gum balls in the cylinder that are outside the jar.

So you have some volume for balls in the cylinder outside the jar, and some volume for balls in the jar outside the cylinder. Try to picture the right size cylinder to make these volumes equal. Then compute the volume of the cylinder and you have your estimate. But I can’t think of a detailed formula for that, if that’s what you’re looking for.

( ZenBeam, is the above description what you had in mind? )

Now, back to the integrating technique: Let’s be clear: There’s a “formal” mathematical meaning to this, and a less formal meaning. The “formal” meaning involves having some formula for the profile of the jar, six months or so of calculus study, and integrating a volume of revolution (as I discussed above). You aren’t really going to be able to do this. I brought it up just to explain the idea of integration, which you’re going to enthusiatically study all on your own this year. (Right?)

Then there’s the “informal” integration method: This is the bread-slicing technique. I think you should seriously at least consider doing it this way, if you think you can:

Just picture the jar as a bunch of horizontal slices (maybe 2 inches thick or so?) stacked up one on top of another. For EACH slice: Measure or estimate the diameter, then use that to compute the cross-sectional area, then multiply by the thickness of the slice to get the volume of the slice. This way, you can estimate the volume of each slice independently of the other slices, which enables you to deal with the variable diameter of the jar. Then, add up the volumes of all the slices to get the volume of the whole jar. This method is called “numerical integration” and you’ll study that (a little more rigorously) in Calculus as well.

Tangential (heh) observation: Did read right, that you mentioned earlier that you haven’t learned the Pythagorean Theorem yet (or maybe that you’ve only just recently learned it)? And that you aren’t sure if you’ve learned any Trigonometry yet, since you think it might have been mixed in with your Geometry class?