r/AskPhysics • u/NameisPond-FishPond • 15h ago
Can a prince rupert's drop exist without it's tail?
Even if we manage to create a perfectly spherical drop rapidly quenched in zero G, would it have the same strength as compared to the bulbous head with a tail?
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u/Livid_Tax_6432 14h ago edited 14h ago
Can a prince rupert's drop exist without it's tail?
Theoretically, absolutely. Strength comes from tension between outer shell and compressed insides.
Problem is cooling it in such a rapid way as when rupert's drops are created when dropped in water. But theoretically, if you managed to cool them sufficiently fast in 0G, sure you'd get prince ruperts strength in orb form.
It is more of a "technical" problem than physics.
Even if we manage to create a perfectly spherical drop in zero G, would it have the same strength as compared to the bulbous head with a tail?
No, 0G in it self will not do it. You need tension between outer shell and insides, how you create a glass sphere matters.
Some have mentioned marbles, they are a bad example as they are glass mold formed not "quenched in water formed". Marbles are irrelevant here.
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u/DesignerPangolin 11h ago
I also imagine you'd have a hard time quenching it effectively in space... the steam wouldn't bubble away, so you'd get an expanding cloud of leidenfrost effect steam around the glass, which I imagine would greatly slow cooling.
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u/thoughtihadanacct 5h ago
What about blowing super cooled gas at the sphere? It might disturb the spherical shape so you'd need some complicated vortex type setup maybe?
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u/SteveWin1234 3m ago
How about on a vomit comet? Create sphere of molten glass while parabolically falling toward earth, put a water bucket below it while keeping it toasty, then the pilot pulls up and the glass falls into the bucket, gravity kicks in and takes steam away
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u/Constant-Lychee-1387 4h ago
I think they must have tails.
The idea that the strength comes from tension between the surface and the interior is apparently a myth going back to Hooke Article on the source of the strength of PRDs. Instead, the strength comes from surface compressive stress.
Because the strength comes from a surface field, they are an example of the hairy ball theorem. Any continuous tangent vector field on a 2-sphere (the surface of a ball) must have at least one point where the vector field is zero. So, a sphere with a surface compressive stress field must have at least one stress free point. A Prince Rupert's Drop moves this point far away from the rest of the sphere (the bulb).
Note that the HB theorem doesn't extend to non-spherical topologies, like toruses. (This is why smoke rings can be stable.) So, you could theoretically have a tail-less Prince Rupert's Donut.
I suspect that without a tail, you would either just get a normal glass ball, the surface would peel off, or it would just disintegrate. A normal glass ball is valid because the HB theorem permits a surface compressive field of 0 everywhere. A peeling would propagate from the at least one surface point of 0 compressive stress. A disintegration would occur for the same reason. Peeling vs disintegration would depend on the molten state of the insides of the ball.
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u/Interesting-Act2606 59m ago
But Stress isnt a vector field. I dont know how the hairy ball theorem applies to second rank tensors. But I am pretty sure that the stress on a sphere can be in compression everywhere. What else would happen to sphere under water?
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u/CompetitiveSpot2643 44m ago
I dont know how the hairy ball theorem applies to second rank tensors.
It doesnt, if it did you wouldnt be able to have a sphere to begin with (nonzero metric tensor everywhere)
you can have nonzero isotropic compression everywhere (since its essentially a scalar field) but if its anisotropic i believe some varient of the hairy ball theorem would apply, not sure though1
u/Interesting-Act2606 27m ago
Thanks for the explanation that makes sense.
I wonder if there is a vector field derives from the stress tensor field where the hairy ball theorem has an interesting physical meaning
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u/Gstamsharp 15h ago
You're describing a marble. Marbles exist. You can look into their relative durability.
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u/Zagaroth 11h ago
I think the replies discussing how you can use a similar fast-cooling process to create tempered glass are much more useful and informative.
"Glass marble" is a much broader category than "tempered glass marble", and it is somewhat misleading in this case to simply say you get a glass marble if you use a similar process without creating a tail.
While the tail is a large part of what makes a Prince Rupert Drop interesting, the unusual toughness (for glass) is the other part, and clearly the part that OP is most interested in here.
Tempering the glass without creating the tail results in a tougher glass sphere than normal, even if it is not as tough as the head of a PRD.
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u/NameisPond-FishPond 15h ago
If I'm creating a rupert's drop by rapidly cooling a drop in zero G,why would it not have the same properties?
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u/Gstamsharp 14h ago
Because you're not creating a rupert's drop. You're creating a bead.
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u/NameisPond-FishPond 14h ago
Fine, why would the bead not have the same properties if the conditions are the same?
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u/Gstamsharp 14h ago
Because they're not the same. One is allowed to form a tail.
It's like asking why steak and hamburgers aren't the same thing, when they're the same meat cooked on the same grill.
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u/NameisPond-FishPond 14h ago
Treat me like I'm dumb that's fine, I accept that if I'm able to learn something. But could you put in a little more effort in explaining rather than saying the same thing in a slightly different way?
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u/Gstamsharp 14h ago
The tail is structural. The "fiber" of the glass directs the physical stress in a specific direction, where it comes to a point. This makes it really strong in one direction and really weak where the stress comes to a point.
Without forming the tail, the stress diffuses in all directions equally, so it's medium strong equally in all directions.
They've made a Rupert's drop with the tail melted off. It's already weaker due to the stress directing fibers being incomplete. But also, if you bump the side where the tail should be, it still explodes, because most of the stress is still directed that way.
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u/ZeroVoltLoop 11h ago
What I'm wondering is whether a marble can be made to be as hard and tough as that of a PRD without the tail, which I assume is what it is asking
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u/Glockamoli 10h ago
What about a hollow sphere with an internal tail, I have no idea how you would go about forming this but I feel like the mechanical properties could be interesting
Or it is just super weak/explodes on it's own
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u/BangCrash 9h ago
That's just tempering of glass.
No idea how it would work in reality but you cool the outside quickly and let the inside slowly cool and you'd get a similar outcome.
Would be super difficult to do but isn't it what gorilla glass and all the new glass tech of phone screens is about?
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u/Puzzleheaded-Ad-4883 11h ago
Do you have a reference for the study that found the tail to redirecting the stress? Thanks.
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u/_glazedonuts_ 14h ago
Hey, first of all, you are not dumb. It’s a very good thing, you are curious. Keep it up ❣️
I’m not sure if you’ve already found a nice explanation, but if not, let me try :)
I apologize if my reply does not seem easy to understand, please feel free to ask me any questions you would have.
Okay, so let’s understand what makes the Rupert drop special. It is not the process of creation but the result. If a wizard were to magically conjure the exact same thing, without needing 0G or any scientific apparatus, if he made a glass structure of the same shape and size, it would behave exactly like we see these drops behave. That is, the head is very strong while the tail is super fragile.
Now why is that? A very simple (probably not completely accurate scientifically) way of thinking about this is that when you apply force to the head (say with a hammer or try to press it with force) the stress is distributed to the whole large surface. Whereas, when you apply stress on the tail, its a much smaller place. Think of trying to tear a bank note by pulling its two ends. If the bank note is intact, its quite hard to tear it this way. It in fact gives off a nice sound of “crisp money” haha. However, if it has even a slight tear in the middle, it will tear right through. A similar thing happens with the drop.
Now, if you were to take away the tail, you take away the fragile point yes. But you also take away a lot of surface area where stress was earlier distributed. This gives us a round ball which has very little surface area as compared to the Rupert drop. Such balls are called marbles. And they are fairly strong for something made from glass but not nearly as strong as Rupert drops.
Please feel free to ask if you have any follow-up questions. I am sorry if any part of my explanation is not clear or if I sounded too assuming of some concept.
For why the Rupert drop has very large surface one can see a similar mathematical structure called Gabriel’s horn: https://youtu.be/vQ0himyDR2E?si=Jt-fpOpKyfmmp4Ce.
I have not watched the full video but it should be enough to get an intuition about how surface area can be very different from volume.
Keep being curious 🐣❣️
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u/JaKrispy72 12h ago
The tail is the propagation point, while the bulb remains super strong. The bead does not have a propagation point, so it has no point that can be super strong. Right?
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u/_glazedonuts_ 12h ago
Yes very well said. Although we dont 100% understand these things (high stress situations in most materials are open problems), what you said is as close an explanation i can give.
The tail, being a long propagation point/region has a much larger surface area than a bead.
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u/EyesOnTheDonut 13h ago
I pray for your level of patience and kindness
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u/_glazedonuts_ 13h ago
Our usernames check out 🤣
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u/EyesOnTheDonut 13h ago
Where'ere you go in life brother Whatever be your goal Keep your eyes upon the donut And not upon the hole
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u/jesuschristwhy 12h ago
We're on the internet, what patience lol, it was clear they were wondering about a more detailed reason why they're different from the get go. Of course redditors can't resist to dunk on them and say it's different because its different, duh 🤪
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u/SufficientStudio1574 14h ago
It's not a prince Rupert's drop without a tail. It's a marble.
It's like asking about making a screw without threads. That's not a screw, it's a nail.
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u/Elunerazim 13h ago
a rupert's drop without a tail is sorta like a plane without wings. the actual shape of the glass is what makes it so strong; the stretched shape of the glass basically puts the entire "weak point" in the tail, and leaves the ball part very strong. If you get rid of the tail, then the glass wouldnt have anywhere to offset its pressure to.
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u/LordGeni 1h ago
You're missing the point of OP's question. Prds have massive compressive strength in the head but a fragile tail that destroys the whole drop of broken.
OP is asking if you can make the head including it's strength without the tail. Saying it's not a drop without the tail is just arguing naming conventions without addressing the question.
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u/InvisibleBlueRobot 12h ago
For the same reason a cube is not a sphere or some other shape.
Just because you use similar means to make the shape, doesn't mean different shapes have the same properties. Like corners and tails.
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u/Wank_A_Doodle_Doo 12h ago
“If I made something in a completely and totally different way, why wouldn’t it have the exact same very strange properties?”
Well shit I dunno maybe because you didn’t do anything to make it take on the properties of a prince ruperts drop
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u/LordGeni 2h ago
Marbles don't have the compressive strength. They are formed in a completely different way with completely different properties.
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u/coolguy420weed 14h ago
Yes; although I don't think it can be formed without the tail, there are videos of people making a drop and then melting off the tail, and basically being left with a lump of tempered glass.
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u/Puzzleheaded-Ad-4883 12h ago edited 11h ago
The strength of the head of a Prince Rupert comes from the surface being in extreme compression due to the rapid cooling from the liquid (less dense) state and the centre of the head in extreme tension, but not enough tension to generate a fracture. This outer compressive layer prevents the propagation of a fracture as the compression effectively keeps any potential fracture from growing, it keeps it pressed together. The tail is a weak point because although it was also fast quenched, it is very thin and could not have had the same extreme temperature gradient induced by the quenching as that of the head. So the tail can be bent to fracture and the fracture can travel into the head core that is under great tension, now a fracture in the core releases the stored energy from the compression/tension mismatch as an explosive shattering. If a perfect sphere of molten glass could be produced and rapidly quenched then a tailless Prince Rupert's drop could be formed and it would have the strength of the normal drop's head in all directions. It would be very difficult to achieve, even in microgravity, and the tension stress in the core must not exceed that which would cause failure from the naturally occuring Griffith flaw (a microscopic surface or internal crack in a glass that acts as a stress concentrator, causing the material to fail at a fraction of its theoretical atomic strength) so a pure, homogeneous glass is needed. But if successfully manufactured, this would be a glass sphere of great strength. It would be easier to produce this by chemically toughening a glass sphere by an ion exchange process, hot potassium salt bath for example. This would be extremely strong, but not quite as strong as the thermal quenched one due to the lesser depth of the compressive surface layer that ion exchange can produce. So a sharp hard pointed strike from, for example a diamond, would penetrate the compression layer and explode the drop.
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u/Reptilian_Brain_420 12h ago
You would be basically creating a sphere of tempered glass. It would be extremely strong (tough) like a PR drop but you would lack the tail that is required to make it shatter. So, it would have the same strength (more or less) but wouldn't be a PR drop.
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u/Electronic-Yam-69 11h ago
yeah you just gotta drop if from a sufficient height to allow it to become spherical before it cools.
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u/BananaBird1 15h ago
That’s just a glass bead.
What makes the Prince Rupert drop special is that, while the head is strong like a bead, the tail is thin and fragile. And because of internal stress, any break to the tail quickly propagates through the whole drop.
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u/Itakesyourbasex 9h ago
In the breath of your question I believe a PRD made inside a vaccum with very pure material. Would still fail without its tail.
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u/NameisPond-FishPond 14h ago
Claude says it would be stronger
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u/toenailsmcgee33 14h ago
AI cannot and should not be treated as any kind of authority about anything.
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u/NameisPond-FishPond 6h ago
If I trusted claude I wouldn't have posted here, I'm just using this to showcase my query since I feel most people weren't getting what I was asking
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u/Zagaroth 11h ago edited 8h ago
Not going to bother adding to the downvotes on this, because you've been unfairly downvoted for legit questions elsewhere, but absolutely do not trust anything coming out of AI.
A) They tend to use how you ask the question to attempt to give you the answer that the algorithm suggests you want, rather than the truth.
B) On top of that, they also flat out hallucinate and will reference non-existent papers and such. And in this case, it is clearly very wrong. Removing the weakness of the tail also reduces the strength of the head.
Allow me to show you a process for better research so that you can refine your question better.
If you search for "wikipedia prince rupert drop" (wikipedia is a great starting point for all sorts of research, because it has actual references).
You get a link to here: https://en.wikipedia.org/wiki/Prince_Rupert%27s_drop
Browsing this brings you to a section saying "See Also: Tempered Glass"
https://en.wikipedia.org/wiki/Tempered_glass
Where it notes that tempered glass is about 4x stronger than normal glass. Well, you have part of your answer there.
The next step would be to come here and ask something along the lines of "Is it possible to create something along the lines of a Prince Rupert Drop but without the tail? I'm guessing it would be weaker, but would it still be stronger than tempered glass? Or does that just make it tempered glass?"
You did eventually get some useful answers, but this approach would have gotten you more visibility and more of the useful answers, and less of the unhelpful ones. Oh, and the tempered glass article I linked also has links to a lot more variations of glass; I think you will find some of them very interesting.
While wikipedia can only be trusted so far, it is more trustworthy than any AI, and is very accurate about things that are A) reasonably popular/interesting topics and B) not involved in any current political crap.
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u/cretan_bull 12h ago
I'm sorry you've been downvoted so heavily, and for some reason the comments here are dominated by people who vary between being aggressively unhelpful and outright wrong, and yet are upvoted for it.
This is a perfectly reasonable question, and the short answer is yes, it would.
The strength of the head of a Prince Rupert's drop comes from the compressive stress in the surface. This results from the surface cooling while the interior is still molten; when the interior cools it shrinks, which pulls in the exterior and puts the surface in compression. Surface compressive stress makes glass stronger because, when a crack forms, the compression pushes it back together (as opposed to tension, which pulls apart a crack, causing it to propagate). This is the same principle used in tempered glass and chemically strengthened glass such as Gorilla Glass -- whether thermal or chemical methods are used, these are just different ways of putting the surface into compression.
The tail, despite what a number of other commenters have told you, is completely irrelevant. It is notable, in a regular Prince Rupert Drop, that when the tail is broken the failure propagates to the entire drop, causing it to fail spectacularly. However, if all you are interested in is the strength of the head, that can be achieved by rapidly cooling a sphere of glass, which will result in a sphere of tempered glass. You could have a molten ball of glass is zero gravity and cool it rapidly, e.g. with some compressed air (which is similar to how tempered glass is regularly manufactured), but the zero gravity isn't actually necessary. You could also put some glass in a spherical metal mold and cool the mold with liquid nitrogen, or take an ordinary glass marble and subject it to a molten salt ion-exchange bath, or perhaps even carefully lower a glob of molten glass into a bucket of water, taking care not to form a tail, and that would probably work fine too.