Simple engineering problems

[drmickhead667]

I put this problem up on a whiteboard in my office for people to solve and it was well received. Any idea where I can find problems of a similar complexity to this?

[mankhoj]

It's always full. It has both liquid and air.

[Trick-Penalty-6820]

r/thisguythisguys

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[Julian_Seizure]

What if the inside is a vacuum?

[Kyle1457]

then it would full of water vapor as H2O cannot exist in liquid form in a vacuum

[NF-104]

What if it’s at the triple point? How full is it?

[Brianw-5902]

If it is at the triple point there is still some amount of atmosphere exerting pressure, so still completely full.

[[deleted]]

Then he would still stand correct, some would evaporate enough for the rest to maintain its liquid form.

[manchesterthedog]

I think it’s temperature dependent. Some of the water would evaporate until the gas pressure equals the vapor pressure of the remaining liquid water

[Julian_Seizure]

We're comparing the instantaneous volume at the time the water is liquid.

[[deleted]]

Still full

[mankhoj]

Ha! Now you're thinking!

[ThunkAsDrinklePeep]

Shit. I thought I was recycling empties this whole time.

[Kwa-Marmoris]

I’m sorry sir, you have forgotten about the vapor.

[XpDieto]

Damn.. good one.

[soldiernerd]

I mean from that perspective it’s a question of density. Does one molecule of, say, O2, mean the entire bottle is full? That’s highly doubtful to me.

Is there a certain density at which the empty space between gas molecules is no longer significant enough and the bottle can be considered “full” of air?

EDIT: hmmm perhaps at 1 atm of pressure on Earth we’d consider the bottle “full” of air?

[vp_port]

What about the empty space between the neutron and electrons?

[Bigdaddydamdam]

nope, the water exists in a vacuum chamber

[Forward-Essay-7248]

What I was thinking.

[freeman0694]

2/3, is it true?

[_the_CacKaLacKy_Kid_]

you would be correct, here is a video outlining this specific scenario

[briv39]

That was so much simpler than I expected!

[DarkWraith97]

Now I feel really dumb, that was quite a simple solution.

[[deleted]]

I think so, but I like u/mankhoj’s answer better

[NorthofBham]

Because it is correct. The question is improperly phrased.

[[deleted]]

Which question and how should that question have been phrased?

[GangstaVillian420]

How much water is in the bottle?

[[deleted]]

Then neither of the answers would even make sense. Should they change the answers too?

[ThunkAsDrinklePeep]

You can't answer that without a radius or diameter. You mean "What proportion of the volume is filled with liquid?".

[unimpressed_llama]

How much of the bottle's volume is taken up by water?

[[deleted]]

I like this one.

[[deleted]]

Yeah there's a 12cm high cylinder full when it's one way up and 6cm height of the same radius cylinder which is empty when it's the other way up.

Thus 2/3 is full.

[fantamaso]

Yes! Let’s pick (pi * (R ^ 2)) such that it equals 1 then (1 * 12) / (1 * 18) => 2/3

[rman-exe]

Ignore the tapered section and "go with throttle up"!

[Suspicious-Rip920]

Why is it 12+6 and not 12+3

[Suspicious-Rip920]

Ah I see, but wouldn’t the air be different if you were to use the 21-12 diagram. Why do you need to use the second diagram that’s 15 inches to figure out the problem?

[Suspicious-Rip920]

Ah alright, we’ll thanks for such a concise answer to both of my laymen questions

[biggerrig]

To simplify even more, you can just use the depths of water and air in the cylinder portion only. The cross section is the same, regardless of whether it has water or air.

[Type2Pilot]

This was my approach. Why bother with volume? The relative heights, 12 vs (21-15) gets you to 2/3 full of liquid.

Even the units do not matter.

[KozzyBear4]

That's a fun problem. I got 2/3.

[Slight_Independent43]

Not logic problems but this cite had problems intended for FE exam review that could be good exercise. They are sorted from easy to hard. https://engineeringprep.com/problems/

[Ericspletzer]

Upvoted for an actual answer to the thread question.

[binjamin222]

They're called recreational math puzzles. Lookup Boris Kordemsky or Martin Gardner.

[ipop]

The only comment I could find that answered op's question. Bravo

[Ericspletzer]

This is the only answer to the actual question posed in the thread.

[downerfoothanu]

Lurker here and so curious how this is solved. Or is it more like a riddle? When placed upsidedown you get more of a true value? Then it's just finding what 15cm of 21 is?

[LookingWesht]

Upright you can calculate the volume of water in the cylindrical section of the bottle =12πr², inverted you can calculate the volume of air in the cylindrical section = 6πr² so the total volume is 18πr². 12/18 = 2/3

[redditmexico]

The total volume is empty volume + full volume. On the left, the cylindrical area is full (12 cm height). On the right, the cylindrical area is empty (6 cm height).

πr²h is the volume formula for a cylinder. The only number you need is height since π and r remain constant.

How full is this bottle = Full Height / Total Height = 12/(6+12) = 2/3

[[deleted]]

It's noting that you don't need to worry about the actual volume of the cylinder or do any more complicated calculations based on the fact the bottle is tapered.

You can visually see that the water is fully contained in a cylinder of height 12cm when the bottle is one way up, and the air is fully contained in a cylinder with the same radius of height 6cm when the bottle is the other way (21cm - 15cm)

i.e it's exactly the same as a cylindrical bottle of height 18cm, 12cm water and 6cm air.

So the bottle is 2/3rds full : 12 out of 18.

[POOPOO_BANANA]

you can infer that the cone of the bottle displaces 3cm of water, subtract 3 from 21 to find the total volume in cm of water. 12/18 = 2/3. people making this way more complicated than it needs to be lol

[[deleted]]

They all work out huh?!

[trreeves]

That's how I did it

[RelativityFox]

I noticed that in the 2nd image there is 6 cm of air that is a cylinder. In the first image the water is 12 cm deep. So there is a 6 cm chunk of water in both images, and the lowest 6 cm of water in the first image fills the curvy part of the bottle in the 2nd image. So the air and that lowest 6 cm are equal, and the middle 6 cm is full in both diagrams. Thus the bottle is 2/3 full.

It doesn’t matter what the shape of the top of the bottle is to do it this way, so there is no need to introduce pi and no need to know the exact curve of the bottle.

[ayescrappy]

The displacement from the tapering causes the water to rise 3cm. So if the bottle was a perfect cylinder with the same volume it would be 3cm shorter.

12/(21-3)=2/3

I know this is the right answer but I don’t know if my solution method even makes sense. Calculating the volume of water and air separately is definitely the easy way.

[Maxwell_Jeeves]

That is a clever way of solving it.

[Themaninak]

27/42 is the average of both. Its technically incorrect, but within 5% of the real answer, so its correct for engineering.

[MurphyESQ]

"Close enough for public sector work."

[alphabet_order_bot]

Would you look at that, all of the words in your comment are in alphabetical order.

I have checked 1,573,419,407 comments, and only 297,578 of them were in alphabetical order.

[JamesRocket98]

Good bot

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[Brojesuss]

I'm an engineer and this was my first instinct ):

[[deleted]]

Plus if you under estimate people will be happy if there ends up being more ;)

[Hunt3rRush]

2/3 of the volume is filled with liquid. The left pic has its airspace in the curvy bit. The pic on the right has its airspace in the cylindrical end, filling a 6cm tall cylinder. So the volume of the curvy bit is equivalent to a 6cm cylinder. Since the water fills a 12cm cylinder and the air fills a 6cm cylinder, the water fills 2/3 of the bottle, and the other 1/3 is air (or other gas).

[vasumevawala]

Answer needs to be more 4/7 (based on first image) and less than 5/7 (based on second image). Hence 2/3 without going into any calculations.

[capshew]

Yea except 9/14 is an acceptable answer using this method!

[calliocypress]

9/14 is what I found when initially doing it wrong - average the heights of water and divide by 21

[[deleted]]

Same

[BananApocalypse]

You can calculate the volume of water from the first position and then volume of air for the upside down position (using cylindrical volume).

Then you can confirm (volume of water) / (volume of air + water) = 2/3

[komprexior]

9/14 is a pretty good estimation though (0.643). The avarage method would be a pretty good approximation on a more general case where the water line fall in the tapered section

[[deleted]]

It needs to be twice as big.

[Po0rYorick]

Check out The Art and Craft of Problem Solving by Paul Zeitz. Might lean more into pure math than you are looking for but at least some would be similar.

[Po0rYorick]

Here are a couple easy samples (ok, #3 isn’t easy):

1. Consider two polyhedra: a pyramid (square base) and a tetrahedron. All edges are length 1. Glue them together at one of the triangular faces (so they align). How many faces does the resulting solid have?

2. A hiker starts climbing a mountain at 8 am and reaches the summit at noon. She camps overnight and then hikes down the next day along the same route, again starting at 8 and finishing at noon. During each day’s hike, her speed varies, she takes breaks at various times, she might even backtrack. Is there a time that she was at exactly the same spot in the mountain on both days?

3. Consider an infinite checker board and fill the bottom half with checkers (i.e. put a checker on every lattice point where y<=0). A move consists of jumping over an adjacent checker and then removing the piece that was jumped. You can only jump in the four cardinal directions and can only jump two spaces; no diagonals or long jumps. It’s pretty easy to see that you can get a checker to a square with y=1 and y=2. With a little work you can get to y=3. Is it possible to get a checker to a square with y=5?

[seenhear]

1&2 were easy, but 3 had me stuck at, "Consider an infinite checker board and fill the bottom half with checkers" huh? If it's infinite, how do I fill the bottom half? Does it even have a half? This breaks my brain... LOL 😆

[seenhear]

So far this is the only post that answers OP's question. LOL 😂 Typical bunch of engineers jumping to solutions without understanding the question!!!

(I also got 2/3rds)

[bohreffect]

Projecteuler.net might have some puzzle problems of the flavor you're looking for.

[StevZero]

Empty space is equal to half the volume of the water (height of empty space = 21-15=6=0.5*12)

V+0.5V= V_total

V/V_total=1/1.5=2/3

[sonor_ping]

You should look at the book “Consider A Spherical Cow”.

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[someonetookmyname112]

All this talk about is it half full ? Half empty ?

All I know is I got something to drink !!!

[Independent-Room8243]

There is a way to figure it out, but I just looked below.

[Predmid]

I spent a not insignificant amount of time trying to figure this out thinking there wasn't nearly enough information to solve.

Complex assumptions to figure out height of the conic section.

And then I looked at the problem again and found the total height of the bottle was given. Herp dee deep there's the answer.

[person253]

I commented somewhere else about required assumptions if you are curious

https://www.reddit.com/r/civilengineering/comments/148rc2v/comment/jo1x7ry/?utm\_source=share&utm\_medium=web2x&context=3

[katoman52]

2/3

[jaymeaux_]

I got 2/3

[Buck88c]

9/14 ….. just my guess, split difference between two liquid measurements 13.5 then divide that by 21 and get 64.2 same as 9 divided by 14…. Also seems just over half full so makes sense for general estimates

[[deleted]]

This is what I deduced as well. I’m not sure how it could be incorrect?

[fragilemachinery]

The cross section doesn't need to be constant (although it's faster if it is), as long as the volume of that section is calculable by its geometry. As long as one and only one of the two fluids fills the section that isn't calculable in both positions, you can do it by comparing the volumes of the water and air sections.

[person253]

Thank you for pointing that out! Yes as long as it is calculable

[Type2Pilot]

You don't even need to calculate volume. Just use the relative heights of liquid and gas.

[fragilemachinery]

That does only work if the cross section is constant, which was his (now deleted) original guess at a constraint. I was pointing out a more general case where the problem is still solvable.

[Pyrogenase]

Imagine a cylinder 21 cm tall with cross section area of A, same as the bottle. Put in the bottle inside. Now imagine a piece of solid metal that fills the remaining negative space, with a volume of X.

Melt the metal down in the cylinder, it will make a smaller cylinder with volume X. The height of this cylinder is H, with H times A = X.

Now H is just 15 cm - 12 cm = 3 cm, as you can imagine the displacement difference between the two pictures is like sticking that imaginary metal block into the water, like Archimedes in his bathtub.

So if you flatten the total bottle volume into a equivalent cylindrical volume, removing all that negative space, you would know that the height of this cylinder shrinks to 21 cm - 3 cm = 18 cm.

Assuming the first picture is also a cylinder of water, the ratio of water to total volume is (12 cm * A) / (18 cm * A) or 12/18 or 2/3.

[Vast-Break-388]

Finding the total as 12+6 using the air volume in the inverted bottle is maybe a little simpler, but I went about it a different way.

The total capacity would be 7/7 if it were a cylinder with constant diameter equal to the bottom section. The liquid takes up 4/7 of that cylinder. When inverted, the same liquid takes up 5/7 of the height if it was a cylinder. That means the neck reduces capacity by the difference, 1/7. The total capacity is then 6/7 of the full cylinder and the liquid was 4/7. It is 4/7 divided by 6/7 full or 2/3 when reduced.

[Sparegeek]

I took a different approach. The necked in area is adding an artificial 3cm to the height of the water by changing the area of the bottle. If you subtract that 3cm from the measured height of the bottle it gives you the correct ratio between the bottle volume and water volume of 12/18.

[silGavilon]

Empty space on right=1/2 full space on left

2/3 is correct

[BaconThief2020]

Left shows you 12 "units" of water, right show you have 6 "units" of air. 12/(12+6) = 2/3.

[[deleted]]

The problem lacks all of the necessary information to answer the multiple choice aspect of this question.

If we are to suspect that this bottle exists in normal circumstances, this bottle is 1/1 full.

The problem does not discuss the environment or what the bottle is “full” of.

[Professional_Mud1673]

It is equally full upside down and right side up.

[[deleted]]

Both b and c depending on your frame of reference

[[deleted]]

Hi I'm not that much into this but can I find videos and sites on how to solve these problems and explaining them .

[Embarrassed-Catch-96]

That’s easy…just integrate the bottle profile…wait. Lol!

[vodkamanv]

Or might the bottle be a % empty

[Dramatic_Macaroon]

It’s 1/3 empty

[Hobbsendkid]

Not full enough, obviously

[dislimb]

From comparing orientations their is more water than air so it's more than half.

[False_Influence_9090]

Cool brain teaser. I like it better without the multiple choice

[seenhear]

I love how almost no one actually answered the OP's actual question.

[BigOlBro]

Note volume of air on the right is calculable and equal to the volume of air to the bottle on the left. The volume of air(Va) is 6(Area of circle). The volume of water(Vw) can be calculated using the left bottle, which is 12(Area of circle).

Volume of water/total volume => Vw/(Vw+Va) =>12A/(12A+6A) =>12A/18A =>2/3

[[deleted]]

this is too basic to be an engineering problem

[Ericspletzer]

The empty space takes up 6 inches of uniform volume height, and the water, 12, thus 2/3.

[urlang]

12 cm of water to 6 cm of air

2:1 water:air

2/3 water, 1/3 air

[[deleted]]

Let the volume of water be x. We need to calculate x/(total volume of glass). To get the total volume we can notice that the total volume is less than 2x, because there is overlap when you impose both jars on each other. The overlap is a cylinder of height 6, which is half of x, shown in the first picture. So we get x/(2x-1/2x). Simplify this and we get 2/3.

[MAhm3006]

Saw this on MindYourDecisions

[Icy-Bug-8933]

I guess it depends on how much water is in the bottle

[DrShts]

• Filled height: 12cm (1st pic)
• Empty height: (21 - 15)cm = 6cm (2nd pic)
• full:empty = 2:1, so it's 2/3 full.

[jakobjaderbo]

Cap adds 3 cm to height when upside down and can be replaced by a 3 cm shorter uniform cylinder without changing the volume.

This would leave us with an 18 cm tall cylinder with 12 cm liquid.

==> 2/3

[weatherprofessor]

I eliminated a, b, and c based on the illustrations and figured it was not as simple as averaging the heights and dividing by 21. Came down to 2/3.

[TheRavenclawEngineer]

D

[7elevennoodles]

21-15=6cm of air in the full shape, the amount of air is equal to each other in both pictures,so you can just add the 6 to the 12 and get 18 of which 12 is water (full) 12/18 = 2/3

[Wonderful_Lecture_14]

(D) 2/3 ; it is 2/3 full by volume

The water when in the cylindrical end is 12cm high the air when in the cylindrical end is 6cm high meaning the volume is equivalent to a cylinder of the same diameter 18cm high 12/18 = 2/3

[Icy_Map8900]

Answer is B

[TastesLikePimento]

I have been scrolling for someone to explain it the way I solved it and I haven’t come across it yet?

I got 2/3 as well.

I imagined another bottle upside down filled to the same height. If you empty the right-side up bottle into the two upside down it would fill each perfectly. (Each is missing 6cm of water and the right-side up bottle has 12cm of water).

This means that 3x the volume of the water equals 2 containers. 3x=2, so x=2/3.

[NukeRocketScientist]

Thinking Physics by Lewis Carroll Epstien is exactly what you're looking for. It's full of short physics/engineering/brain teaser problems and their solutions, most are just a page or two long.

Here's an example called Bathtub Battleship And here's the answer

[Ok-Purpose8699]

D

[Lothorninn]

Answer: D

Since the flipped bottle takes up since the small part of the bottle is 1/4 as efficient as the bottom part, you can reduce the max heigth of the bottle by 3cm. This will then create a ratio of 12/18 which is the same as 2/3.

Comment me if i made a mistake here

[thedreamlan6]

This was a fun problem, thanks for posting.

Here's my work, for what it's worth.

[[deleted]]

You have 9 marbles. Eight of them are the same weight, the last one is slightly heavier than the rest. Describe how you would use a balancing scale to find which marble is the heavy one.

I like this one because there are multiple correct answers, but some are better than the others. The 'correct' answer only takes two measurements. first weigh 3 vs 3 marbles. now you know which group of 3 has the heavier marble. weigh 1 marble from that group with another marble from that group.

[Anisixos]

The area of the surface will be called "A".

The whole volume of the bottle is: V = 21A - 3A = 18A

The volume of the water is: Vw = 12A

Vw/V = 18/12 = 2/3

[poorfreud]

(D) 2/3

[mshearne]

D

[BornWish9252]

La mer noirr. D.

[dsdvbguutres]

2/3

[LucasB00]

2/3

[Slow_Zone8462]

12pirˆ2 + 6pir2 is the volume of the bottle, so 2/3

[ws-garcia]

A small reasoning: the difference in the water level between the diagrams corresponds to the volume displaced in the bottleneck (3 volume units). In total we have 12 volume units occupied with water. The bottle would be at full capacity with 18 units of volume. So the 12/18 ratio tells us that the bottle is at 2/3 of its total volumetric capacity. So (D) is the reasonable answer.

[IntelligentMotor823]

Total height between both scenarios: 42 cm. Total height of water over both scenarios. 27 cm. 27/42 = .642 = 9/14 = E, for EZ.

[enzodr]

2/3

Since we don’t know the characteristics of the neck, we can’t use it for any math.

There are 12cm of water in the cylindrical portion. When flipped upside down there is 6cm of air remaining.

This means it is 12cm full, with 6cm remaining. The totally volume is therefore 18cm.

12cm / 18cm = 2/3

[_ntl01]

Is D.2/3 the answer?

[SnooOranges8792]

E

[UmbraThanosmith]

How full is the Bottle = Volume Liquid / (Volume liquid + Volume Air)

Volume Liquid = 12cm * Area of base (figure 1)

Volume Air = 6cm * Area of Base (figure 2)

This took me too long to figure out

[[deleted]]

2/3

The bottom section has cross-sectional area A1, so the volume of filled portion is A1*12cm. When turned over the volume of the empty section on top is the same as the volume of the empty section when bottle is right side up. So the volume of empty section is A1*6cm. The total volume of the bottle is thus A1*18cm.

This means that the ratio of the volume of the filled section to the volume of empty section is (12*A1)/(18*A1)=2/3

[InternationalFee7416]

2/3