Jofwu Posted February 11, 2021 Report Share Posted February 11, 2021 Hey physics nerds. I've been toying with the equations of motion for paired gems (conjoiners and reversers). I'm happy with what I've got so far, but I'm interested in digging for thoughts on the decay factor that comes with distance.... But first let me back up and explain where I'm at. Using Pulleys I think the easiest way to think of paired fabrials is by comparing them to pulley systems. For a simple conjoiner, it's easiest to just think of them as rigidly connected, but the metaphor doesn't work for reversers. Pulleys are also a great way to account for the mechanical advantage discovery that Navani makes in RoW chapter 84. And I think friction is a good way to think of the decay they experience over greater distances. Two major caveats though. The metaphor only works for one dimension at a time, though that's not a big deal. And we have to pretend we can force our ropes in both directions. (can be pushed and pulled) Simple Equation Without the mechanical advantage discovery Navani makes and ignoring friction, we have the following equation: (m1 + m2) a = F1 + R F2 This is simply the equation of motion for a simple pulley, with one addition: that R factor. More on that in a second. The idea here is that F1 and F2 are the sum of all forces acting on the gems. (excluding the force being shared between the paired gems -- that is essentially the tension in our pulley system's rope) The masses m1 and m2 are the masses attached to each gem, plus the mass of the gem itself. If we've attached something fairly heavy to one and not the other, you can effectively say the mass of the lone gem is zero. So all this equation is saying is that the acceleration of the system is the sum of forces on the system divided by the total mass of the system. F=ma. We need to note that the accelerations of 1 and 2 aren't necessarily the same. I'm using a = a1 = R a2. In other words, the equation above is really for the acceleration of a1. If we want to know the acceleration of the other gem we need to substitute a = R a2. This R factor I've added is really just a convenience to account for the difference between conjoiners and reversers. Rather than deal with separate coordinate systems for each, I'm just using this one equation with R=1 for conjoiners and R=-1 for reversers. In other words, with a reverser the accelerations are in opposite diretions of one another and F2 acts opposite of F1 rather than with it. (F1 and F2 are in the same direction--whatever direction we want to label the positive direction.) Simple Examples Let's say we've got a pair of reversers with the same mass floating in the air. Their weights balance each other out just as if they were two masses hanging from a pulley. We have (m1 + m2) a = F1 - F2 where F1 and 2 are their equal weights, m1*g and m2*g. The net force of F1-F2=0 so we have a=0. Note that if we give one a push up or down (and ignore wind resistance) they will move at constant velocity in opposite directions. Point being, this isn't a scale that tries to balance back out. Just two weights balancing each other out. What if the second is twice the mass of the first? (m1 + 2*m1) a = m1*g - 2*m1*g using down as positive direction, which is 3*m1*a = -m1*g. Solving for a we get a = -g/3. So the first gem moves up into the air at 1/3 g. The second moves down at 1/3 g. What if the second is much heavier than the first, such that the mass/weight of the first is negligible? (m2) a = -m2*g so we are left with a = -g. The first gem, basically weightless, flies upward at 1g while the second gem falls at 1g as if it weren't paired. Makes sense, right? With a two equal masses conjoined, they just fall freely. (m1 + m2) a = m1*g + m2*g = (m1 + m2) g, which is just a=g and for conjoiners we have a=a1=a2. Mechanical Advantage In RoW chapter 84 Navani is able to use Raboniel's dagger to transfer the spren of a paired gem into a different, larger gem. When she moves the larger gem, the other moves three times as far. We see this same phenomenon with a pulley system that gives a mechanical advantage of 3. If Navani moves the smaller gem 30 centimeters, the larger gem will only move 10 centimeters, and vice versa. There's some force multiplication that happens here. Moving the smaller gem requires less force, though it doesn't move the other as far. Moving the larger gem is the opposite. It causes the smaller gem to move further, but it requires more force. When you work through the math, the equation above becomes this: (m1 + m2/G^2) a = F1 + R F2/G The G factor here is the mechnical advantage ratio. Usually we would use MA for this, but I felt that might be confusing alongside the m's and a's of mass and acceleration. I've been using gamma in my notes and don't have it on this keyboard, so you get G. Also note that this gives us a different acceleration relationship. Now we have a = a1 = R G a2. In other words, the smaller gem accelerates G times faster than the larger. Navani isn't clear on what ratio produces this observed ratio of 3, though I'm guessing the larger gem in her accidental experiment was 3 times larger by mass? Something like that. More Examples So let's say we have a reverser where m1 is basically negligible, a gem ratio of 2, and both only under the influence of their own weight. We have (m2/2^2) a = -m2*g/2. Which gives a = -2g. For the larger gem we have a2 = a/(RG) = g. In other words, the large gem/mass falls at a normal 1g, while the smaller, effectively weightless, gem moves upward at 2g. Note that if the small gem is half the weight of the larger gem they balance out in this situation. The force-half of the equation becomes (m2/2)*g - (m2*g)/2 = 0. You can use this discovery in two different ways. Let's talk about how they apply to the Fourth Bridge. Two options. One thing you can do with this discovery is reduce the number of chulls required. If you use the chulls to pull the small gems and attach the larger gems to the Fourth Bridge, they can get more force with their pulling. But the downside is they have to walk further to make the Fourth Bridge move the same distance. On the other hand, if you attach the large gems to the chulls and the small gems to the Fourth Bridge, your chulls don't have to walk as far... but they have to pull harder. I'm guessing the second will be what they do with the Fourth Bridge. Most of the time the ship should be moving at some constant cruising velocity. So they'll have a harder time getting the thing accelerated up to that speed... But once they're there, they just need to maintain that speed, which shouldn't take as much effort. And in the meantime, the Fourth Bridge will be moving 2 chull steps (or whatever the ratio is) for every chull step. It would be interesting to work out the math on all of this, but I haven't tried yet. Distance Decay So now we come to the distance decay. Navani notes that there's some amount of decay which occurs over greater distances. The further the paired gems are apart, the more resistance you experience. This is essentially like the friction you see in a pulley system. When you work out the math, this essentially just becomes an extra force on the system. In a real pulley system you have different friction amounts in each pulley, but I'm basically collapsing the sum of those into one value: Ff. (m1 + m2/G^2) a = F1 + R F2/G - Ff I'm... probably not handling the sign on this well. It's resisting movement so we're basically assuming here that a > 0. Otherwise, the sign on Ff flips. Now... we don't really have enough information from the books to figure a correct equation for this value... But I'm trying to think it through and figure some idea of what form it might take. We know that it's a function of the distance between the two gems, but it can't JUST be that because I think spanreeds wouldn't work that way. Think about it. If the Fourth Bridge experiences a notable amount of decay from Urithiru/Shattered Plains to Hearthstone and if we just have Ff as a function of distance, then a spanreed over the same distance would experience that same resisting force. That just doesn't really make sense. I don't think it makes sense for the masses to be involved. Thinking back to the pulley metaphor, the friction between a rope and pulley is going to depend on how much tension is in the rope. More force being transferred through the rope means it's pulling harder on the pulley, and that harder normal force leads to more friction. So perhaps it should be a function of F1 - R F2/G? Note the negative sign there. For a conjoiner, that gives us the difference between forces. So, for example, two reversers (G=1) floating in the air give m1*g + m2*g. If you're looking at a simple pulley that's the total forcing tugging on the pulley. The full weight of one is countering the other, so the full weight of each is carried by the rope. More friction. Change that to conjoiners and it's a little odd though. You get m1*g - m2*g = (m1 - m2)*g which... suggests the amount of friction depends on how comparable the masses are? I would think two falling conjoined gems just basically fall under their own weights, together, and so there's no transfer of force between them. No tension in the rope. So something doesn't seem right here. My brain is getting mushy. Feedback appreciated. If you have no ideas, hopefully you at least enjoyed the explanation of the rest! 8 Quote Link to comment Share on other sites More sharing options...

Lightdancer Posted February 12, 2021 Report Share Posted February 12, 2021 (edited) That was very enjoyable. 4 hours ago, Jofwu said: Change that to conjoiners and it's a little odd though. You get m1*g - m2*g = (m1 - m2)*g which... suggests the amount of friction depends on how comparable the masses are? I would think two falling conjoined gems just basically fall under their own weights, together, and so there's no transfer of force between them. No tension in the rope. So something doesn't seem right here. I think a more useful way would be to think of the conjoiners as lying on a frictionless surface, with the positive direction parallel to the ground. That negates the extra acceleration from gravity that's throwing that scenario out of wack. (The surface is frictionless so that our fictional friction can take real friction's place in the equation.) What I said above depends on a possibly-wrong assumption. In real physics, objects dropping like that would experience no friction (besides air resistance). However, objects moving along a surface would experience friction, since the normal force is pressing them into the ground. (The force of friction is equal to -F_{N}μ, with F_{N }normally equaling the force perpendicular to the main force. Thus, if F_{N}=0, there is no friction.) However, we've seen no reason for this to apply to conjoiners, since there would be no "normal force" affecting them. (Perhaps there's a Spiritual Realm equivalent?) When I visualize this situation, I leave out the ropes entirely and assume that whatever forces are exerted on the first mass are exerted on the second, and vice versa. I drew this diagram: Unfortunately that loaded sideways, but I basically drew a diagram of the force exerted on each mass. Since each individual mass had more force exerted on it than if it were falling on its own, it accelerates faster. For example, if F2=M2g, then M1 accelerates at A1=F2/M1+g, or A1=(M2/M1+1)g. In other words, each mass will accelerate faster than if it were falling on its own. How much faster depends on both masses. In the simplest language, both conjoiners would fall faster. That doesn't address how "decay" would affect this situation. I would assume that the greater the distance between the conjoiners, the smaller the additional acceleration would be. Hopefully that makes sense. Feel free to correct my math. Edited February 12, 2021 by Lightdancer 0 Quote Link to comment Share on other sites More sharing options...

Jofwu Posted February 12, 2021 Author Report Share Posted February 12, 2021 6 hours ago, Lightdancer said: I think a more useful way would be to think of the conjoiners as lying on a frictionless surface, with the positive direction parallel to the ground. That negates the extra acceleration from gravity that's throwing that scenario out of wack. (The surface is frictionless so that our fictional friction can take real friction's place in the equation.) Gravity isn't really the problem though; I just used it there as the force out of convenience.I wasn't particularly clear about this on my post I guess, but F1 and F2 are just a catch-all for forces acting on each gem, whether that be weight, an applied force, air drag, friction, whatever. And again, this is all constrained to one-dimension. So basically, what I did there (and most of the examples) was to use a case where we're interested in the vertical dimension and no forces except weight, so that F1=m1*g and F2=m2*g. Weight is just an easy force to use. We can take a horizontal case where weight isn't in the direction of interest if we want. Or just put the whole thing in deep space where there's no gravity to even consider. Anyways... You run into the same problem if we look at movement in a horizontal direction in any case where the ratio F1/F2 = m1/m2. Intuition says that F1/m1 = F2/m2 = a in such a case, so the conjoiners are accelerating together at the same rate and there's no differential that requires force to be transferred between the gems. Just like if we have two weights rigidly connected and in freefall, there's no stress in their connection. But that equation I ran into suggests there IS a "tension in the rope" (so to speak) when m1 isn't equal to m2. So I must be not thinking something through there... 6 hours ago, Lightdancer said: When I visualize this situation, I leave out the ropes entirely and assume that whatever forces are exerted on the first mass are exerted on the second, and vice versa. I drew this diagram: Yeah, no problem with that. Especially for a simple conjoiner. I just liked the pulley metaphor to help illustrate what's going on and to clarify how the situation with different gem ratios works out. And it helps avoid a mistake which I think you're making here, by giving a way to think through the problem carefully. You've saying the weight of each is applied to the other, which would indeed make them fall faster. I don't think there's any evidence in the books to suggest this happens though. If you work through it as a pulley problem you find that the force applied from 1 on 2 and vice versa should be zero. (which... is actually I think making me realize what I was doing wrong with that one issue...) So there's a difference in philosophy here. Consider a case without weight where the gems have the same mass m=m1+m2 and I apply a force F to m1 only. The way your math works there, you're applying F1 to both gems in full. So that means a1 = F/m and a2 = F/m. For a constant F applied for a duration of T the change in velocity of each (v1 and v2, respectively) is v1 = v2 = F/m*T. Change in kinetic energy (mv^2/2) then is KE1 = KE2 = F^2 * T^2 / (2m). So with a force F applied for a duration of T we have added an energy of F^2 * T^2 / m... But if we apply that same force to m1 when the gems aren't paired, we only add half of that. Same impulse. Twice as much energy. We just created energy. A more simple point to highlight the issue is that we've applied F to two masses but the person pushing only felt a reaction of F rather than 2F. We're violating Newton's second law. (equal, opposite reactions) To be fair, that's not outside the bounds of the rules for the cosmere. Investiture lets us break rules, so it COULD work that way. I don't think that's Brandon's intent however. I guess we would need something more explicit from the books on the matter to prove it. Quote If you work through it as a pulley problem you find that the force applied from 1 on 2 and vice versa should be zero. (which... is actually I think making me realize what I was doing wrong with that one issue...) Quoting myself and circling back to say some more on these thoughts, for my own sake... I guess the issue I had here is that I wanted to say the decay is a function of the "tension force" but then tried to take it as the difference between F1 and F2. The tension force also depends on the masses and accelerations... Doesn't help me get unstuck here though, because acceleration is the unknown we're really solving for. I feel like I must be headed into some kind of differential equation that I can't quite put my finger on... 1 Quote Link to comment Share on other sites More sharing options...

Lightdancer Posted February 12, 2021 Report Share Posted February 12, 2021 9 hours ago, Jofwu said: We just created energy. A more simple point to highlight the issue is that we've applied F to two masses but the person pushing only felt a reaction of F rather than 2F. We're violating Newton's second law. (equal, opposite reactions) To be fair, that's not outside the bounds of the rules for the cosmere. Investiture lets us break rules, so it COULD work that way. I don't think that's Brandon's intent however. I guess we would need something more explicit from the books on the matter to prove Oh! Good catch. I agree; that’s probably not what Brandon is trying to do. I’m going to have to think about this more. It sounds like you’re on the right track. 0 Quote Link to comment Share on other sites More sharing options...

Pagerunner Posted February 13, 2021 Report Share Posted February 13, 2021 Pulleys is where I looked to understand conjoiner and reverser physics, as well, and I came to basically the same conclusions. The math works out nearly identical, but then things get a whole lot nicer than they do for real pulleys. We can have negative tension force, and when the weights on either end move in multiple directions, the vector for the tension force doesn't have to change due to physical limitations. I've attached my current understanding of the system below. (Or tried to attach; it usually takes me a few tries to successfully put a picture in a post.) The equations can be derived using just gravity and the lifting force in one dimension; three equations (sum of forces on each half, and the physical constraint of equal or opposite accelerations) and three unknowns (both accelerations and the tension force). Moving in three dimensions, you just set up those same three equations for each dimension (or use simplified versions if you've isolated that particular plane of motion). I haven't looked too closely at the "inefficiency" situation, but I suspect it will be a property of both distance and the magnitude of the tension force. Like if the rope for the pulley went across a speed bump; farther you are, the bigger the speed bump (more surface area for friction); and the harder you pull, the harder the rope rubs against the speed bump. But I haven't compared that to the text, yet; if you are able to post the passages and find any specifics, that would be appreciated. 1 Quote Link to comment Share on other sites More sharing options...

Jofwu Posted February 15, 2021 Author Report Share Posted February 15, 2021 On 2/12/2021 at 9:09 PM, Pagerunner said: if you are able to post the passages and find any specifics, that would be appreciated. These are the two references I can recall off the top of my head. I searched "decay" and didn't find any others, but it's possible they talk about it in different terms somewhere and I'm forgetting. Quote Force was transferred: if the distant half was underneath something heavy, you’d have trouble lowering yours. Unfortunately, there was some additional decay; the farther apart the two halves were, the more resistance you felt in moving them. RoW ch 3 Quote Spanreeds had a certain decay to them. The farther apart they were, the heavier the pens became after activation. In most cases, this was a slight—almost imperceptible—difference. Today, the spanreed board, with pen attached, had been placed on Falilar’s most precise scale. The pen was hooked by strings to other instruments as well. [...] “Brightness,” Falilar said, “this can’t be correct. The decay is almost nonexistent.” “So they’re near to us,” Navani said. [...] You couldn’t tell the direction of a second spanreed, not directly, even from the decay. However, because you could measure the decay—and therefore judge distance—you could triangulate from multiple measurements and get a rough idea of location. RoW ch 28 That first is the main one I was going off of. "More resistance you felt in moving them" sounds very much like extra force is required. The second however... I'm not sure what to make of it. Pens become heavier? I tried to work it through just in the vertical direction. If the "receiving" pen is #1, the forces on it include its weight and the normal force from the table surface. (plus the "tension" from the other) The "transmitting" pen is the same, with an additional applied force from the person writing. (either up or down, depending on whether they're pressing it into the paper or lifting it. Of course, I'm sure the forces involved in writing with a pen are very dynamic, so I'm not sure how practical all this is. Anyways, all I end up with is that the sum of the weights equals sum of normal forces plus applied force from writer. (negative writing force being downward) Too many unknowns it feels like? What Navani is saying is that the normal force is greater when distance is greater, so I guess there ought to be some kind of additional force? Or else the tension force is greater than it normally would be (in some instances?)? It sounds like Brandon is saying that lifting a mass via paired gems takes more force at greater distance. So for example, if we have paired gems each attached to a 10 pound weight... In close proximity, lifting one weight requires lifting 20 pounds. At great distance, lifting one weight requires lifting more than 20 pounds, with the extra apparent weight being the "resistance you felt." And with the second quote it's saying that if we put the weights on scales, they would each read greater than 10 pounds. This doesn't jive with my notion of how the decay works like a friction force because two masses at rest have no tension in the pulley rope. It's almost like.... there's some extra force being applied to the rope. Like we have some person pulling the rope in the central portion of your "Conjoiner" diagram upward with constant force, making both gems heavier. But that makes no sense at all, because why would it pull that direction? If we turn the whole thing sideways, obviously our gems don't start sliding on one horizontal direction because there's a "resistance force" tugging them one way. Not sure what to make of the second passage. 0 Quote Link to comment Share on other sites More sharing options...

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