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[Cobbler89]

This is a hypothetical meant to figure out how to handle a certain part of a different problem for my own purposes... think of it as education.

Let's say I have two trianglular blocks, of the same size and shape but different densities -- one is half the mass of the other. They're back to back, mirrored as it were, and either they're acute or they've got their hypotenuses together. Two balls of equal size and density roll into each at similarly mirrored angles and the same speed, hitting at the same time. So the triangles should both be knocked back (and the balls likewise deflected), but also into each other--though of course they're already at each other's backs, so this force is just going to... either knock them apart in the end or put some kind of stress on their material or whatever.

Code: [Select]

  AB
 AABB
AAABBB
 AABB
  AB
 ^  ^
 C  D(Where A and B are the trianglular blocks and C and D are the balls.)

I could dig up enough physics to solve this for one triangle and one ball, or even two balls on one triangle, but this thing where there's a larger system but also smaller parts that will be separately affected is giving me trouble. My initial thought is, "Find out how much force is applied to each triangle. Then find out what it does with that force and as part of that how much force that triangle passes on to triangles behind it. But wait, if one of them is also one of the initial triangles, then that force affects its calculations too, doesn't it? What about dealing with it as one system? But then how would I get back to knowing the push and shove on the individual parts?"

Feel free to post pictures or ASCII diagrams if the right approach is helped by a visual depiction. Feel free to say, "Beats me."



[EDIT:]
Ok, so I just realized I can calculate the forces applied by the triangles on each other the same way I would if they'd been moving toward each other and collided (or at least very similarly). This ultimately results in forces applied to each triangle away from each other (though as in any collision the net force may merely slow one down and only completely bounce one of them around), which is why they'll be knocked apart (which is what I'd expect if I tried this in real life. I'm not sure if that resolves the entire larger problem or not (I'll have to think about how to handle it if there are, say, ten triangles in a pile and two of them are hit by balls at the same time, or better yet ten arbitrary shapes each having both momentum and torque). I'll keep you posted.

(Actually, a couple more details in case anyone actually finds this helpful to helping me [cue Aperture Science references]: 1: the triangles are taped together, and 2: in case it isn't obvious I want to know the forces applied to each and every one if there's a larger group.)

[Drakale]

If you assume them taped together you should really abstract the 2 triangles as one simple object and find it's center of mass from the respective densities of the triangles. Then find out the net force of the 2 impacts on this simplified object and from there you have an accurate and simple force vector problem that represent your situation.

[Cobbler89]

It does up to a point... I'm actually looking for how much the triangles get knocked into each other (or squished if they're squishy), and how much the tape gets pulled, though. I think I may need to rewrite the example problem now that I've thought it through more.

[Drakale]

Finding how much compressive force happens between the triangles at the moment of impact is easy. The sliding on each other is harder, you have to take into account the friction between the two, and if you want to be precise there are a lot of parameters to take into account in real world friction.

If the goal of the exercise is to figure out if the 2 triangles would slide of each other, assuming their densities are different, then yes they would due to the center of mass being offsetted , as long as the adhesive effect is of lesser strength than the differential force on the triangles.

Interestingly, if the 2 balls are of equal weight, they would bounce off with perfect symmetry despite the difference in mass between the triangles assuming they are inelastic.

[Eagle_eye]

Since he's dealing with macroscopic objects, it's probably safe to assume the collision won't be even close to elastic.

[Cobbler89]

The objects in question are more a generic example, but yeah, it's macroscopic... I'm actually trying to figure out a good way to deal with the inelasticness of a system of objects each connected to each other, assuming that the connections experience a sort of inelasticness in transferring the energy/force/momentum/whatever. That's interesting about the balls bouncing off at the same rate -- I thought their end momentum depended on the force, which depended on the momentum of the objects, which depended on their mass? Is it equal anyway because the triangles have no momentum if they're sitting still at the start, or because in regard to bouncing the balls they act as a single system like you mentioned earlier, or have I just forgotten some of the details about how you get from velocity to momentum to force? (It's been a while since I took physics. I understood it at the time but haven't used it much since, so it's lying around in my brain's equivalent of wherever Link puts the heavy metal boots so they don't weigh him down until he pulls them out, even though he must be carrying them with him somehow. If you know what I'm talking about.)

It might be a moot point if I can find a way to get the overall effect I want while glossing over this level of detail. I'm still thinking about that too. Bouncing the ideas around -- or at least thinking about how to explain what I want to know how to do -- is helping even if none of the ideas pan out, so thanks.

[Drakale]

Yeah I assume that since the impact is simultaneous the struck object act as a single entity. The only reason I can do that is because the impact energy is evenly distributed on both sides at he moment of impact and I assume no elasticity.

[SewingPin]

Well, if you want the simple result of simple objects, it would look something like this if the differences were significant/exagerrated, considering shape, angles, densities, laws of motion, friction, all the normal stuff. You just make the forces go through the objects and subtract the portions that are expended on the transfer. You could even probably do it step by step instead of it all happening at once.


If you wanted to dissect the problem for its worth in deeper underpinnings of many more laws applied, you'd have to use Finite Elements Method. And ho boy that is some cheese, this kind of thing should be only used if you realy care about the result it will have in practical application.

[TeleDwarf]

in this problem it will be hard to assume inelasticness(is there even a word like that?): if impact is not instantaneous- balls start to show the structure as they collide.  a lot of different variables comes into play and you will end up just simulating whole thing using some kind of abstract model(as it is done in any modern physics problem).

For a simple case let's assume that all parts are rigid, and impact is an instantaneous transfer of momentum.
From here we can assume without any loss that triangles are separated by a tini tiny distance of D. In this case the problem is separated into two simple problems. 2x a rigid ball impacts a stationary rigid triangle. 1x two rigid triangles moving as (solution from the first problem) impact.
Both can be easily solved using tools of theoretical mechanics.

[Cobbler89]

Hmm... good ideas; thanks.

I think I've found a way to fudge the actual problem so it avoids anything complicated in how the parts interact, thankfully.

[smakemupagus]

I'm actually trying to figure out a good way to deal with the inelasticness of a system of objects each connected to each other, assuming that the connections experience a sort of inelasticness in transferring the energy/force/momentum/whatever.

Just as a random piece of philosophical advice, feel free to take it or leave it.  A good place to begin is by thinking about being more precise in how you use technical language.  Inelastic collisions conserve momentum, but not kinetic energy (nor "force" nor "whatever").  Using words consistently and to mean the same thing that other people do, you can more easily translate your ideas into accurate equations, not to mention you'll have an easier time searching for information.

>> inelasticness(is there even a word like that?)

Coefficient of restitution

>> the triangles are taped together

This changes the problem pretty substantially.  Are they taped so securely that they cannot come apart at all?  then from a mechanical point of view it's hardly two separate triangles, but effectively one object with an irregular distribution of mass.

[Cobbler89]

Hmm, well, I was trying to figure out how to approach the problem (or a bigger problem this attempts to isolate a tricky part of) without necessarily being concerned with the exact units/formulas involved in the real physics until I get to actually resolving it... but I suppose how these factors interact may depend on which ones we're talking about...

...Speaking of which, and in answer to your last point, I am (or was) in part concerned with figuring out how much strain the tape has to endure, with the idea of loosely approximating the physics for finding out whether they hold together or not -- something like, "does the amount of force transferred exceed an arbitrary strength/stickiness constant for the tape," or better yet, "how much is the stickiness of the tape worn down by the pulling/pushing on it, given an arbitrary ratio of push/pull to wear and a constant, also arbitrary point at which the wear exceeds the total stickiness and thereby breaks the hold of the tape." Of course, to complicate things, the "tape" analogy is a substitute for these parts being stuck together in some other manner, and I figured -- and my initial description focused more on -- the loss of force on the assumption that if some of the force went into pulling the sticky off the tape (to follow that particular analogy) then that force would not be transferred to the other objects. My main problem in figuring out how to deal with all this is that, the way I am (or was) thinking about it, the amount of force lost transferring between A and B involves the amount of force experienced by A and B, which itself depends how much force A and B transferred to each other, which itself depends how much force is lost between A and B... then obviously we have a problem that involves itself in its answer, creating a scenario of infinite recursion; which can't be right, so I must be approaching the whole matter wrong...

...Though to be honest, I think I thought of a way to simplify the scenario to begin with so I don't need such detail -- something like, only consider it in the first place where a single piece of tape alone connects two sections of the system, and then first figure out the forces on each section besides that connection, and then consider it similarly to how you'd consider a collision between those two sections.

It's all a mess that comes from me not having taken physics in a while (nor taken it particularly far when I did) but wanting to at least loosely simulate a situation for which the actual real-world physics would involve some of the most advanced/complicated stuff around, as I believe SewingPin pointed out. I'd like objects made of solid, non-deformable parts that can be ripped apart from each other but tend to stay together until they're hit too hard; I'm probably overthinking the whole matter, though, and my attempts to simplify the problem and to describe where I'm getting stuck obviously haven't been much less messy than the original problem. I might have to start over with that situation I'd like to loosely simulate and look more at "how could I fudge a simulation of this so the right general effect is achieved without getting into physics between the parts?" (And I should probably update the original post... In any case it's clear I explain even worse than I do physics...)