Math question that is a bit complicated

[RiverDave]

So there is a jar in particular this jar..

129 oz. Clear PVC Hexagon Jar with 110mm Snap-on Neck (Lid Sold Separately) | U.S. Plastic Corp.

These hexagon jars are reusable and recyclable. Stands on side for angled opening or on bottom for top opening. These jars are perfect fo...

www.usplastic.com

It is a pentagon shaped jar that is filled with erasers in particular these erases

Whaline 80Pcs Smile Mini Bulk Erasers Welcome Back to School Yellow Smile Face Round Erasers Happy Face Collection Erasers for Students Classroom Rewards Gift Filler Supplies https://a.co/d/d12hT3L

There is a “little room” at the top..

Need to know as close as possible how many erasers are in the jar.. daddy’s little girl wants to win the contest.

RD

 

[RiverDave]

Jar volume Dimensions: 8.375" L x 5.185" W x 7.5" Hgt.
Erasers 1x1x.3.

You can’t just calculate the volume and the volume of the erasers and divide because there’s gonna be a lot of air gap due to the shape so what factor would you use to get pretty close to that?

 

[JL95]

Dumb it down and assume they probably ordered +-6 orders of the 80 quantity erasers lol. What are the odds the school would want to spend over $100 on a guessing game filler. I’ll just throw in 480 count for shits

Looking at the jar now rather than while driving I’ll say add two orders to my guess. 720 like mandelon said sounds closer and looks closer.

 

[RiverDave]

Jar volume 5.33698509 liters
Eraser volume .2356. Inches or .00072 in liters.

Total amount of erasers if they were liquid = 7412.47929167

There ain’t no way it’s that.. so what am I doing wrong here.. even if I scaled it back 50% that seems way high

 

[RiverDave]

Dumb it down and assume they probably ordered +-6 orders of the 80 quantity erasers lol. What are the odds the school would want to spend over $100 on a guessing game filler. I’ll just throw in 480 count for shits

We wanna win.. someone in here can do the math on a rough volumes of the space in between.. @Racey

 

[Mandelon]

Volume Calculator

This free volume calculator computes the volumes of common shapes, including sphere, cone, cube, cylinder, capsule, cap, conical frustum, ellipsoid, and more.

www.calculator.net

 

[RiverDave]

Volume Calculator

This free volume calculator computes the volumes of common shapes, including sphere, cone, cube, cylinder, capsule, cap, conical frustum, ellipsoid, and more.

www.calculator.net

I can calculate the volumes I am unsure how to calculate shapes inside of shapes

 

[Racey]

Jar volume Dimensions: 8.375" L x 5.185" W x 7.5" Hgt.
Erasers 1x1x.3.

You can’t just calculate the volume and the volume of the erasers and divide because there’s gonna be a lot of air gap due to the shape so what factor would you use to get pretty close to that?

Buy some of those erasers and do a sample test in a known volume of liquid, weigh the, glass and the glass liquid brimmed full, the stuff all the erasers in, then weigh it. Then dry the erases off and weigh just them.

Now you can determine how much water as a percentage of volume remains in a sample size, then extrapolate.

 

[Mandelon]

I'm thinking 30% is gonna be air. So that leaves 3.5 liters of solid erasers. Each eraser is 1 inch by .3 inches. This gives you 3 erasers per cubic inch of remaining space. 3.5 liters is about 213 cubic inches so the internet says.


So 213 x 3 = 639 erasers.

10 orders of erasers on Amazon would be 720 erasers. I would estimate in that range.

 

[Runs2rch]

Buy some of those erasers and do a sample test in a known volume of liquid, weigh the, glass and the glass liquid brimmed full, the stuff all the erasers in, then weigh it. Then dry the erases off and weigh just them.

Now you can determine how much water as a percentage of volume remains in a sample size, then extrapolate.

Might be easier just to duplicate the "prize"

 

[Racey]

Might be easier just to duplicate the "prize"

Hey, if you wanna win ... Winning aint free

 

[RiverDave]

So I asked chat gpt and it came back with

To find out how many round objects can fit into a pentagon, we need to calculate the volume of the pentagon and the volume of the round objects.

The volume of a round object with a diameter of 1" and a depth of 0.3" can be calculated using the formula for the volume of a cylinder:

Volume = π * (radius)^2 * height

First, we need to calculate the radius of the round object. The diameter is 1", so the radius is 0.5".

Volume = π * (0.5)^2 * 0.3
Volume = π * 0.25 * 0.3
Volume = 0.2356 cubic inches

Now, we need to calculate the volume of the pentagon. The volume of a pentagon can be calculated using the formula:

Volume = base area * height

The base area of the pentagon can be calculated as:

Base area = 8.375 * 5.185
Base area = 43.48 square inches

Now, we can calculate the volume of the pentagon:

Volume = 43.48 * 7.5
Volume = 326.1 cubic inches

Now, we can calculate how many round objects can fit into the pentagon by dividing the volume of the pentagon by the volume of the round object:

326.1 / 0.2356 = 1384

Therefore, approximately 1384 round objects can fit into the pentagon.



————————————


So I said “you forgot the air gap between the circles” and it came back with


You are correct, I apologize for missing that detail. In order to account for the gap between the round objects, we need to adjust our calculation.

Since the round objects have a diameter of 1" and are being placed in a pentagon that measures 8.375" x 5.185", we can calculate the maximum number of round objects that can fit in one layer of the pentagon.

If we arrange the round objects in a hexagonal pattern (similar to how balls are packed in a box), the centers of the round objects will be touching each other. The diameter of each round object is 1", so the distance from the center of one round object to the center of another adjacent round object will also be 1".

To calculate the number of round objects that can fit in one layer of the pentagon, we can divide the length of the pentagon by the diameter of the round object:

Number of round objects along the length = 8.375" / 1" = 8.375

Similarly, for the width of the pentagon:

Number of round objects along the width = 5.185" / 1" = 5.185

Therefore, the maximum number of round objects that can fit in one layer of the pentagon is the product of these two quantities:

Total number of round objects in one layer = 8.375 * 5.185 ≈ 43.46

Since the height of the pentagon is 7.5", we need to divide the volume of the pentagon by the volume of one round object to find out how many layers of round objects can fit in the pentagon:

326.1 / 0.2356 ≈ 1384

Therefore, considering the gap between the round objects, approximately 1384 round objects can fit into the pentagon.



————————-


So I said how can they be the same answer when you forgot the air gap? And it came back..


I apologize for the oversight in my responses. Let's correct the calculations:

First, let's calculate the number of round objects that can fit in one layer of the pentagon. We will arrange the round objects in a hexagonal pattern to account for the gap between them.

Number of round objects along the length = 8.375" / 1" = 8.375
Number of round objects along the width = 5.185" / 1" = 5.185

Total number of round objects in one layer = 8.375 * 5.185 ≈ 43.46

Now, let's calculate the volume of one round object:
Volume = π * (0.5)^2 * 0.3
Volume = π * 0.25 * 0.3
Volume = 0.2356 cubic inches

Next, we calculate the volume of the pentagon:
Volume = 8.375 * 5.185 * 7.5
Volume = 324.34 cubic inches

Now, we divide the volume of the pentagon by the volume of one round object to find out how many layers of round objects can fit in the pentagon:
324.34 / 0.2356 ≈ 1376

Therefore, considering the gap between the round objects, approximately 1376 round objects can fit into the pentagon. Thank you for pointing out the error, and I appreciate your understanding.







That still seems crazy high.. maybe it’s because it’s stacking them in an organized fashion in “theory” and not in the “random” or chaos theory that they would be dumped in the jar

So I am guessing maybe a 1/4 of that in reality would put is in the 344 range and now we gotta figure that “little bit at the top”. So I’m thinking like 315?

Total random numbers at this point?

 

[RiverDave]

Buy some of those erasers and do a sample test in a known volume of liquid, weigh the, glass and the glass liquid brimmed full, the stuff all the erasers in, then weigh it. Then dry the erases off and weigh just them.

Now you can determine how much water as a percentage of volume remains in a sample size, then extrapolate.

But we gotta assume the teacher just dumped them in there and didn’t stack them for efficiency.. that would take forever..

 

[Javajoe]

Have her take a pic of the jar and just ask AI

 

[Racey]

But we gotta assume the teacher just dumped them in there and didn’t stack them for efficiency.. that would take forever..

It would take like 5 minutes, dump the erases in a bowl and fill it up

 

[RiverDave]

Have her take a pic of the jar and just ask AI

Had I known I woulda done just that.. lol. She came to me right now with the

“There’s a jar full of smiley face erasers and we need to guess the closest to win something big”

So we looked online until I found the jar, and u til we found the erasers..

With that and the “little bit at the top” is the information we have

 

[RiverDave]

It would take like 5 minutes, dump the erases in a bowl and fill it up

I can’t get the jar and erasers in two hours lol

 

[oldboatsrule]

Cut a cardboard 7.5" hexagon.
Cut one inch circles.
Have your daughter fill the hex with the circles.
Count and multiply by 17.28 ( the amount per stack to fill 5.185 wide container.

Thus will give you the amount IF they were all neatly stacked.

Deduct (my guess 18%) for lack of efficiency in random piling the erasers inside.

Or draw it up in cad as described above

 

[2Driver]

But we gotta assume the teacher just dumped them in there and didn’t stack them for efficiency.. that would take forever..

Yours or hers?

 

[RiverDave]

Yours or hers?

?

 

[Angler]

Yours or hers?

YES

 

[Sandlord]

Hey, if you wanna win ... Winning aint free

Or duplicate the question.
fab up a cardboard jar, and some fake erasers…

 

[monkeyswrench]

Are the erasers "stacked" or loose? Pure volume of the little coin type erasers stacked isn't too tough, but the randomness of how they fill the void is a bitch...

Ask ChatGPT maybe?

 

[RiverDave]

Are the erasers "stacked" or loose? Pure volume of the little coin type erasers stacked isn't too tough, but the randomness of how they fill the void is a bitch...

Ask ChatGPT maybe?

Loose to my knowledge

 

[Badchoices03]

1085

 

[monkeyswrench]

There are no less than 100 absolute geniuses on here. I'm sure someone knows some magic formula from some calculus or statistics class.

I'd find a tub like that at a candy store maybe, and then a bunch of rubber washers. I have to do things a more practical way.

 

[RiverDave]

There are no less than 100 absolute geniuses on here. I'm sure someone knows some magic formula from some calculus or statistics class.

I'd find a tub like that at a candy store maybe, and then a bunch of rubber washers. I have to do things a more practical way.

Same here but they are leaving for the open house in like 20 minutes and the teacher sprung it on them at the end of the school day today.. lol

 

[Runs2rch]

Same here but they are leaving for the open house in like 20 minutes and the teacher sprung it on them at the end of the school day today.. lol

Accidently knock it over, then rough count as you put them all back.

Solved haha

 

[wallnutz]

Dumb it down and assume they probably ordered +-6 orders of the 80 quantity erasers lol. What are the odds the school would want to spend over $100 on a guessing game filler. I’ll just throw in 480 count for shits

Im sticking with this. Don't over contemplate this. By the way, I am a math genius. Also I stayed at a Holiday Inn last night.

 

[lbhsbz]

It’s a hexagon…start there

 

[gottaminute?]

3d model the eraser. then use the modeling software to recreate the jars inner dimensions. then use the nesting option to fill it up.
subtract a few % for a non optimised fill ??????

anybody asked AI to get an answer?

 

[Captain Dan]

Accidently knock it over, then rough count as you put them all back.

Solved haha

 

[Racey]

Have her take a pic of the jar and just ask AI

This isn't a bad idea tbh

 

[FUN4ME]

https://www.tiktok.com/video/7363693291361127722
try the mark rober method

 

[stokerwhore]

Jar volume 5.33698509 liters
Eraser volume .2356. Inches or .00072 in liters.

Total amount of erasers if they were liquid = 7412.47929167

There ain’t no way it’s that.. so what am I doing wrong here.. even if I scaled it back 50% that seems way high

that is one small ass eraser. doesn't seem correct, have you tried google ing "estimating fill" for irregular objects? edit- in your first post you give eraser dims as 1x1x3. that's three cubic inches not .2365"

 

[rightytighty]

Put the erasers in a food chopper dealio and grind them small. Then you’d get pretty close to 1 to 1 jar volume. No air gaps.

If you aint cheatin, you aint tryin’!

 

[rightytighty]

Your first post tell you how to figure the volume of an eraser.

My money is on my idea.

 

[4Waters]

Above my pay grade

 

[Mandelon]

that is one small ass eraser. doesn't seem correct, have you tried google ing "estimating fill" for irregular objects? edit- in your first post you give eraser dims as 1x1x3. that's three cubic inches not .2365"

1" round by .3" tall

 

[Icky]

Did we confirm that said teacher isn't member here yet

 

[TonyFanelli]

So I asked chat gpt and it came back with

To find out how many round objects can fit into a pentagon, we need to calculate the volume of the pentagon and the volume of the round objects.

The volume of a round object with a diameter of 1" and a depth of 0.3" can be calculated using the formula for the volume of a cylinder:

Volume = π * (radius)^2 * height

First, we need to calculate the radius of the round object. The diameter is 1", so the radius is 0.5".

Volume = π * (0.5)^2 * 0.3
Volume = π * 0.25 * 0.3
Volume = 0.2356 cubic inches

Now, we need to calculate the volume of the pentagon. The volume of a pentagon can be calculated using the formula:

Volume = base area * height

The base area of the pentagon can be calculated as:

Base area = 8.375 * 5.185
Base area = 43.48 square inches

Now, we can calculate the volume of the pentagon:

Volume = 43.48 * 7.5
Volume = 326.1 cubic inches

Now, we can calculate how many round objects can fit into the pentagon by dividing the volume of the pentagon by the volume of the round object:

326.1 / 0.2356 = 1384

Therefore, approximately 1384 round objects can fit into the pentagon.



————————————


So I said “you forgot the air gap between the circles” and it came back with


You are correct, I apologize for missing that detail. In order to account for the gap between the round objects, we need to adjust our calculation.

Since the round objects have a diameter of 1" and are being placed in a pentagon that measures 8.375" x 5.185", we can calculate the maximum number of round objects that can fit in one layer of the pentagon.

If we arrange the round objects in a hexagonal pattern (similar to how balls are packed in a box), the centers of the round objects will be touching each other. The diameter of each round object is 1", so the distance from the center of one round object to the center of another adjacent round object will also be 1".

To calculate the number of round objects that can fit in one layer of the pentagon, we can divide the length of the pentagon by the diameter of the round object:

Number of round objects along the length = 8.375" / 1" = 8.375

Similarly, for the width of the pentagon:

Number of round objects along the width = 5.185" / 1" = 5.185

Therefore, the maximum number of round objects that can fit in one layer of the pentagon is the product of these two quantities:

Total number of round objects in one layer = 8.375 * 5.185 ≈ 43.46

Since the height of the pentagon is 7.5", we need to divide the volume of the pentagon by the volume of one round object to find out how many layers of round objects can fit in the pentagon:

326.1 / 0.2356 ≈ 1384

Therefore, considering the gap between the round objects, approximately 1384 round objects can fit into the pentagon.



————————-


So I said how can they be the same answer when you forgot the air gap? And it came back..


I apologize for the oversight in my responses. Let's correct the calculations:

First, let's calculate the number of round objects that can fit in one layer of the pentagon. We will arrange the round objects in a hexagonal pattern to account for the gap between them.

Number of round objects along the length = 8.375" / 1" = 8.375
Number of round objects along the width = 5.185" / 1" = 5.185

Total number of round objects in one layer = 8.375 * 5.185 ≈ 43.46

Now, let's calculate the volume of one round object:
Volume = π * (0.5)^2 * 0.3
Volume = π * 0.25 * 0.3
Volume = 0.2356 cubic inches

Next, we calculate the volume of the pentagon:
Volume = 8.375 * 5.185 * 7.5
Volume = 324.34 cubic inches

Now, we divide the volume of the pentagon by the volume of one round object to find out how many layers of round objects can fit in the pentagon:
324.34 / 0.2356 ≈ 1376

Therefore, considering the gap between the round objects, approximately 1376 round objects can fit into the pentagon. Thank you for pointing out the error, and I appreciate your understanding.







That still seems crazy high.. maybe it’s because it’s stacking them in an organized fashion in “theory” and not in the “random” or chaos theory that they would be dumped in the jar

So I am guessing maybe a 1/4 of that in reality would put is in the 344 range and now we gotta figure that “little bit at the top”. So I’m thinking like 315?

Total random numbers at this point?

 

[JJ McClure]

You need to solve for the area of a hexagon. Not a rectangle. Then multiply by height for volume.

 

[Vamodsquad]

We're not supposed to do our kids homework around here. I think they call it cheating !

 

[stokerwhore]

Jar volume 5.33698509 liters
Eraser volume .2356. Inches or .00072 in liters.

Total amount of erasers if they were liquid = 7412.47929167

There ain’t no way it’s that.. so what am I doing wrong here.. even if I scaled it back 50% that seems way high

.2356 cubic inches equals 0.0038607923 liters though.
5.33698509 liters divided by .0038607923 is 1382.355
which still seems high.
i'm going with 1000 lol

 

[Javajoe]

I didn’t read thru all of this but it tells you how big the package is for 80 erasers. Do the math on the package dimensions vs the school container

 

[monkeyswrench]

Is the hexagonal prism loaded with said erasers on a treadmill?

 

[Javajoe]

Im guess about 700 due to my shitty math. 714 actually if you do the math.
But that is nicely stacked so let’s say if ya toss them in there you may be looking at 600

 

[BigAl96]

Based on the container volume at 100% fill with no gaps it'd hold 818 erasers. Assuming 30-40% of the container can't be filled due to air gaps and irregular stacking you'd be between 491 and 573 erasers (a cylinder will fill a cube to appx 78% based off the same thickness and width). Being that the erasers are sold in packs of 80, I'd work off multiples of that and say either 480 or 560 erasers depending on how well packed the container looks.

 

[Terminal Velocity]

Got any 1” bar stock at the shop? Start cutting get a similar candy container and strat dropping then in.

 

[Heylam]

Tell the teacher she’ll let him know in two weeks. Lol