DoomslugTD Posted August 27, 2024 Posted August 27, 2024 (edited) Math and physics heavy. Edit: I am putting a clear list for what all the equations mean before starting, this edit has also come with some changes to the body text as well, but they are mostly just to make things clearer. x(t) = distance as a function of time in the presence of gravity. (The motion we see in the book that I am basing this experiment on.) x'(t) = velocity as a function of time in the presence of gravity. x"(t) = accelerations as a function of time in the presence of gravity. x"(x(t)) = acceleration as a function of the specific distance x(t) in the presence of gravity. (It is there its just in the same line as x"(t).) d(t) = distance as a function of time with no gravity. (I have not solved for this (yet) because I am unsure how to solve the differential equation for d"(t).) d'(t) = velocity as a function of time with no gravity. (I have not solved for this (yet) because I am unsure how to solve the differential equation for d"(t).) d"(t) = accelerations as a function of time with no gravity. (I have not solved for this explicitly but we can write it in terms of d(t).) d"(d(t)) = acceleration as a function of general distance with no gravity. (I did not show this explicitly but its pretty easy to derive from the force equation. This is the differential equation you would need to solve in order to find d(t)) F(t) = total force as a function of time in the presence of gravity. F(x(t)) = total force as a function of the specific distance x(t) in the presence of gravity. Fₚ(t) = force provided by the steelpush as a function of time (this one depends on whether or not there is gravity or other external forces and requires us to solve for d(t) to figure out.) Fₚ(x(t)) = force provided by the steelpush as a function of the specific distance x(t) with no gravity. Fₚ(d(t)) = force provided by the steelpush as a function of general distance with no gravity. (This is the equation we are trying to solve for.) So I have repeatedly attempted to come up with a satisfying force equation for the steelpushing in mistrborn to no avail. At first it seemed like a simple question, something along the lines of F∝1/x, F∝1/x² or F={x≤a/b: a-xb, 0} (I am using x to represent distance) but upon closer examination of the book you will notice that coinshots will approach the crests of their jumps smoothly and wont overshoot when x(0)=0 and x'(0)=0, while the equations I provided prior both overshoot their max height. This lead me to figure out some other option. I came to the conclusion I would have to come up with the motion I was looking for then extrapolate the force equation from there. I can explain the process I went through and the equations I tried in the comments if people would like, but I settled on x(t) = h · Tanh(ax)² as the motion for coinshots when in the precense of gravity for 6 key reasons. 1. x(0)=0 // because the motion we have the most knoledge about starts ideally from beside the object, though if solved, the final solution would let us solve for any value of x(0). 2. x'(0)=0 // because the motion we have the most knoledge about starts with 0 velocity, though if solved, the final solution would let us solve for any value of x'(0). 3. Lim t->∞(x(t))=h // because we want it to settle to one value over time. 4. ∀t(0<t→x(t)≤h) // because we dont want it to overshoot the final distance. 5. x"(t) exists and is continuous on [0,∞] // because its a physics equation. 6. x"(t) must be expressible only in terms of x(t) and constants // because there is no indication that the strength of a steel push depends on time or any derivative of x(t) such as velocity or acceleration, as such I have constrained it to be only in terms of the distance from the metal and any constants. x(t) = h · tanh(at)² satisfies all these conditions (6. was definitely the hardest to figure out), and thus I will use it for the rest of my calculations. There probably are other equations that may also fit these paramaters, and may do so even better, I have just not found them yet. I will skip over the tons of math I did to get to this point (ask me for it in the comments, and I will get to posting it sometime) but I will give the equations I came up with. x'(t) = 2ha · tanh(at) · sech(at)² x"(t) = 2ha² · sech(at)² · (sech(at)² - 2 tanh(at)²) = 2a²(h - 4x(t) + (3/h)x(t)²) Using F=ma... F(x) = 2ma²(h - 4x(t) + (3/h)x(t)²) We do know however that there are two forces at play here gravity and the steelpush so F(x) = Fₚ(x) - mg Fₚ(x) = F(x) + mg Fₚ(x) = 2ma²(h - 4x(t) + (3/h)x(t)²) + mg This gives us for the force equation for the steelpush as a function of x, this is so close to what we want but there is one issue. x is the distance specifically when the equation for motion is x(t) = h · tanh(at)². This the area I am the least sure of, but as reason stands we should be able to switch out x(t) for a more general distance d(t), this is because no matter what the equation for motion looks like, the force provided by the steelpush at any given distance from the metal should remain the same. So... Fₚ(d(t)) = 2ma²(h - 4d(t) + (3/h)d(t)²) + mg This Fₚ(d(t)) is what we were looking for, the force of a steelpush as a function of distance. Now this raises two big questions, what is the gravitational constant of scadrial and the mass of the person doing in the equation for the force provided by a steelpush. Both these questions can be answered by just considering that I used h as one of the initial parameters, which I am using to represent height. You can explain g's precense with a little thought experiment. If you provide the same force on another planet with a different g you would also have to change the h and a constants appropriately (For example: if you had a planet with a higher g, h would be smaller because you would reach a shorter final height, and a would decrease as well because if it didnt the person would approach the final height much faster for no reason.) You can also explain m's precense. If someone has increased mass, h and a will decrease for the same reasons as I mentioned above. In conclusion, the reason both g and m are in the final force equation is because the force equation is defined in terms of the final motion it causes (paramatarized by the maximum height h and how quickly they get there a.) If we did not define the force in terms of the final motion it would cause those terms would dissapear. Some final notes. If this is indeed the equation for the force of a steelpush it has some interesting implications. Either people with more mass must not be able to achieve the same height as people with less mass, or the strength of a steelpush must scale off of mass. From my previous reads of mistborn I am swayed to believe the second statement as not only do I never see any final height differences mentioned when two coinshots or mistborn rise into the air (which knowing brandon, he would definitely make use of in his writing if it was the case), I also believe I have read of much more powerful pushes being done when wax increases his weight in era 2 (though people in the past have raised a valid point that it could be because people with more mass stay closer to the object they are pushing for longer, allowing for greater total force to be exerted, though that would mean people with more mass accelerate slower when steelpushing. As one more note I have not read mistborn in about half a yaer so I could be misremembering things. I just bring this up because I have seen many opinions on this matter and thought I might add my thoughts to it. If anyone knows of any way to solve the final equation of motion I presented above I would love to hear it. I have not gone to university yet, so my knoledge on how to solve non-linear systems of equations is very rough. I know some basics, but I also know that a grand majority are unsolvable, so I let this one be for now. If anyone can tell me how to solve it, or if its solvable at all though I would be very grateful! Feel free to comment and give criticism and feedback, I would love to engage in some discussion over this! Thanks to anyone who has read through to the end. If you would like here is the graph of the functions in this post, it was just for personal use, so sorry if it is a bit unintuitve. https://www.desmos.com/calculator/133h5d3hpo Edit: There is a very massive problem with my solution as presented right now. The lim x(t)->∞ of F(t)=∞ which is nonsensical as we can all logically assume that when not in the precense of gravity steelpushing does not suddenly increase in strenght with distance, and that at large distances F(t)≈0. This means we need to find a new solution that fulfills a 7th stipulation and 8th stipulation. (As a note F(d) uses d(t) as a function parameter so the derivative d/dd(t) F(d) will look weird, sorry!) 7. ∀t(0<t→d/dd(t)·F(d(t))≤0) // because F(d(t)) should always be decreasing as we move farther away from the object following the logic of the books. I used ≤ instead of < because there is technically nothing in the books that say F(d(t)) does not have a region where it does not change based on distance, though I do suspect it probably does not. 8. Lim d->∞(F(d))=0 // because F(d(t)) should approach 0 as we go an infinit distance away, stipulation 7. by itself could technically allow for a force function that eventually pulls you back in at a certain distance as F(d(t)) decreasing could mean it goes all the way into the negatives, so I implimented this as well. Edited October 6, 2024 by DoomslugTD 5
Quantus he/him Posted August 27, 2024 Posted August 27, 2024 Having the steelpush scale to the user's base mass makes a lot of sense to me, given how many physical effects in the cosmere scale Subjectively to the user's base value (feruchemy, Gravitation Surge, Heartbeats for time measurement, etc). I Like it! For the math, I recall the basic pathway but I havent touched that level of math in a couple decades. The short Answer is Matrix Math to solve for the unknowns, (I think the method was called a Hessian Matrix?) but there's some factoring to be done before you could plug the terms in. Devil's advocate possible Complications: Much like the ever-problematic Weight Storing, it's possible we dont yet understand where the Magic-t-physics hand-off is happening. It's possible that what's being exerted is not a Force at all, but is only believed to be that in their current era. It could be secretly a Pressure that scales based on something obscured like how blunt the mental image of the misting actually is (something they might not realize until more scientific testing is done and compared). Or it could be a magical Displacement that mimics force opposition when two casters are applying conflicting Intents. Or it varies a lot and comes down to an Energy/Investiture exchange equation that we dont have yet, and the only thing conserved is Investiture itself (via governing Laws we dont know yet). 1
ChillPenguin Posted August 27, 2024 Posted August 27, 2024 (edited) Correct me if I’m wrong but I think all allomancers aren’t created equal. Elend probably can produce a stronger steel push than Vin, and probably also Kelsier, because he (Elend) ate a Lerasium bead (same with Hoid I guess). We do know that a coinshot (or mistborn) needs to balance the forces so if they want to go into the air they have to push against something really heavy. If two coinshots got in a fight and are pushing against the same metal object between them, I don’t think the mass of the coinshot determines how hard they COULD push. If both coinshots are braced against a wall then the one who wins might not be the heavier one. If they are just standing without bracing the heavier one might win out because they have more inertia. You are right though that if a coinshot/mistborn wants to hover then heavier ones have to be able to push more than a lighter one. If you’re a weak coinshot though and you are heavy well…. you probably can’t hover. As to Wax being able to push harder I think it’s because he increases the mass he can push against so he has more capacity to push other objects. If he had his back to a wall, he would be able to allomantically push as hard as he possibly can (assuming his body can take it and the wall doesn’t break). Edited August 27, 2024 by ChillPenguin 3
CtrlAltDepressed Posted August 27, 2024 Posted August 27, 2024 25 minutes ago, ChillPenguin said: Kelsier because he ate a Lerasium bead Kelsier did not eat Lerasium. He was snapped in the Pits and was a 'natural' Mistborn. 1
ChillPenguin Posted August 27, 2024 Posted August 27, 2024 (edited) 2 hours ago, CtrlAltDepressed said: Kelsier did not eat Lerasium. He was snapped in the Pits and was a 'natural' Mistborn. Yes exactly. Vin and Kelsier did not eat Lerasium so they would be weaker steel pushers than Elend and Hoid who ate it. I edited my post to make it more clear. Thanks for pointing that out. Edited August 27, 2024 by ChillPenguin 3
DoomslugTD Posted August 28, 2024 Author Posted August 28, 2024 6 hours ago, Quantus said: Having the steelpush scale to the user's base mass makes a lot of sense to me, given how many physical effects in the cosmere scale Subjectively to the user's base value (feruchemy, Gravitation Surge, Heartbeats for time measurement, etc). I Like it! For the math, I recall the basic pathway but I havent touched that level of math in a couple decades. The short Answer is Matrix Math to solve for the unknowns, (I think the method was called a Hessian Matrix?) but there's some factoring to be done before you could plug the terms in. Devil's advocate possible Complications: Much like the ever-problematic Weight Storing, it's possible we dont yet understand where the Magic-t-physics hand-off is happening. It's possible that what's being exerted is not a Force at all, but is only believed to be that in their current era. It could be secretly a Pressure that scales based on something obscured like how blunt the mental image of the misting actually is (something they might not realize until more scientific testing is done and compared). Or it could be a magical Displacement that mimics force opposition when two casters are applying conflicting Intents. Or it varies a lot and comes down to an Energy/Investiture exchange equation that we dont have yet, and the only thing conserved is Investiture itself (via governing Laws we dont know yet). Thanks for the steps for solving! As for that last bit, it could just be something that appears as a force but itsn't, but that feels extremely complicated for something that is somewhat based on Newtons 3rd law as a core tenant (equal and opposite reaction). As far as I understand pressure is just force over a certain area, so that would ultimately just mean adding an area component to the equation which feels odd though possible, though I do agree that there could by something in the steelpushers mind that effects strength (though that again feels mostly unsuported for now.) As for the displacement, its entirely possible it just moves you in a certain way (displacing you a set amount), but like I said, that feels like a really overcomplicated way to go about producing a system that seems so inspired by forces and newtons 3rd law, not to mention thinking about how other forces would interact witht the user while being displaced, does the displacment ignore other forces? And if it doesn't it could logically also just be represented by a force in the end. I do agree we dont know for sure how it works, but everything so far seems to suggest some sort of force, and that is by far the simplest answer, especially with the languege being used in the book. As a final note, I dont really need to know how investiture turns itself into energy for this calculation as I am only interested in the motion here, how much investiture it consumes is not in the scope right now (maybe one day, it sounds interesting) though laws governing investiture and its transition to energy would definetly help solve this question. Thanks for the reply!
DoomslugTD Posted August 28, 2024 Author Posted August 28, 2024 5 hours ago, ChillPenguin said: Correct me if I’m wrong but I think all allomancers aren’t created equal. Elend probably can produce a stronger steel push than Vin, and probably also Kelsier, because he (Elend) ate a Lerasium bead (same with Hoid I guess). We do know that a coinshot (or mistborn) needs to balance the forces so if they want to go into the air they have to push against something really heavy. If two coinshots got in a fight and are pushing against the same metal object between them, I don’t think the mass of the coinshot determines how hard they COULD push. If both coinshots are braced against a wall then the one who wins might not be the heavier one. If they are just standing without bracing the heavier one might win out because they have more inertia. You are right though that if a coinshot/mistborn wants to hover then heavier ones have to be able to push more than a lighter one. If you’re a weak coinshot though and you are heavy well…. you probably can’t hover. As to Wax being able to push harder I think it’s because he increases the mass he can push against so he has more capacity to push other objects. If he had his back to a wall, he would be able to allomantically push as hard as he possibly can (assuming his body can take it and the wall doesn’t break). Responding paragraph by paragraph. That is correct, users dont have the same strength always, which can be represnted by changing the h and a values for them appropriately. Most users however do have mostly the same allomantic strength. Steelpushers do need to push against something heavy to go into the air, but as far as I understand that is just because the farther away the ancor is, the less force you would be able to get out of it, so if you ended up with a light anchor it would push itself away from you, leaving you with minimal force. Whether or not the mass of the metal effects the power of the push is a very interesting debtate but one I will have another time. As for the coinshot duels, you bring up a good point that if two coinshots are in the air one will win the push just by matter of intertia whether or not they have more pushing strength, so we are not able to use those as examples to indicate the strength is mass dependent. With two steelpushers pushing on a wall however, I dont believe we have much evidence of this in the books, so its hard to say whether one would push harder or not, if we have an example of this we could pretty easily prove one way or another if the strength is mass dependent. I personally believe that the more massive one would push harder, just due to how the rest of the system appears to work, but I cannot say for sure. For the hovering I agree, if the strength is not mass dependent then heavier people would not be able to push as high, but as far as we know that is not the case. We never got examples of Vin going significantly higher than Kelsier or anyone else thats larger then her, despite Brandon loving to use weight differences in steelpushing fights in other cases. This indicates to me that most average steelpushers can go to about the same height regardless of weight, which must mean they provide more force on their pushes. For Wax, because of inertia he would be able to stay much closer to the objects and push for longer because he would not be shot away from the building. But in the scene it defintely feels like he is producing more force at a given distance, which should not be effected by inertia (only the total force over the whole duration should be effected by intertia as someone will stay closer to the object for longer.) Either he is producing more force cause weight lets him push more powerfully, or allomantic pushes depend on the mass of the metal you are pushing on, in which case he decided to push on a lot more metal at once (the whole building's worth of metal) because he had more weight so he would not immediately be rocketed off, or both of those options. In either case it needs to depend on either the mass of the user or the mass of the metal, otherwise he would not have been able to produce the insane amount of force he did at that distance. That or I am reading the scene wrong and he produces a normal amount of force for a steelpusher at that distance, but due to the longer duration of the push he managed to collapse that building (in which case someone would be able to get an identifcal effect to Wax by putting their back against a wall and pushing, because their intertia would effectively become that of the wall and the earth its attached to, rendering them to massive to noticabely move, letting wax get an even greater amount of net force over the full duration of the push. Though they would probably get very squished on that wall if they didnt have some support. I dont believe this is the case as this would defintely be a tactic used by people if they could.)
DoomslugTD Posted August 28, 2024 Author Posted August 28, 2024 (edited) Hello everyone who is reading. I have made an edit to the original post that I deem fairly important, it is at the bottom of the original post now. Here it is in this reply as well though. Edit: There is a very massive problem with my solution as presented right now. The lim x(t)->∞ of F(t)=∞ which is nonsensical as we can all logically assume that when not in the precense of gravity steelpushing does not suddenly increase in strenght with distance, and that at large distances F(t)≈0. This means we need to find a new solution that fulfills a 7th stipulation and 8th stipulation. (As a note F(d) uses d(t) as a function parameter so the derivative d/dd(t) F(d) will look weird, sorry!) 7. ∀t(0<t→d/dd(t)·F(d(t))≤0) // because F(d(t)) should always be decreasing as we move farther away from the object following the logic of the books. I used ≤ instead of < because there is technically nothing in the books that say F(d(t)) does not have a region where it does not change based on distance, though I do suspect it probably does not. 8. Lim d->∞(F(d))=0 // because F(d(t)) should approach 0 as we go an infinit distance away, stipulation 7. by itself could technically allow for a force function that eventually pulls you back in at a certain distance as F(d(t)) decreasing could mean it goes all the way into the negatives, so I implimented this as well. Edited October 6, 2024 by DoomslugTD
DrPhysics he/him Posted September 23, 2024 Posted September 23, 2024 On 8/26/2024 at 10:52 PM, DoomslugTD said: because there is no indication that the strength of a steel push depends on time or any derivative of f(x) such as velocity or acceleration, as such I have constrained it to be only in terms of the distance from the metal and any constants. There is a lot of evidence that suggests the maximum force is also limited by the power that an allomancer can deliver. (See this discussion for that and other rules that I've seen) Another point you haven't considered is the cognitive aspect of pushing. We don't know what part of the rules are there because they are part of the limits of pushing, or if it is just how the allomancer expects it to be (example: Wax seeing the bullet as multiple pieces, rather than one object because he started thinking of it as more than one object). I think we have three hard limits: A Coulomb-like force that depends on the amount of metal and the connectedness/"allomantic mass" of the allomancer. For most objects, the distance dependence would be like 1/r^2 because of geometry (everything is a point mass when you are far away), but you could get other ones with different shapes of metal. A power limit (Fv) that is tied to how quickly the allomancer can tap investiture A belief limit - my brain expects this to happen, so my push is matched accordingly (We do this all the time with our muscles - it's a safety feature so we don't tear them). This is the only way I can think of that explains Vin stopping (instead of bobbing up and down) at her max push while hovering over a coin. The other important piece to consider for any physics model, is that you should have a "why" for a fit. 1/r^2 pops up so much because of how things expand in 3d space. What physical property implies a tanh? 1
DoomslugTD Posted September 25, 2024 Author Posted September 25, 2024 On 9/23/2024 at 12:04 PM, DrPhysics said: There is a lot of evidence that suggests the maximum force is also limited by the power that an allomancer can deliver. (See this discussion for that and other rules that I've seen) That would be included in the "constants" that I talked about as that should not depend on time or distance so is effecitively a constant according to this example. In general this is a highly simplified example where we are assuming they are doing a full push the whole time (though I will touch on what you mentioned about the cognitive aspect is a second) and not changing it with time as this gives us the simplist example and is reasonably what you would expect vin to be doing when she pushes to her apex. On 9/23/2024 at 12:04 PM, DrPhysics said: A Coulomb-like force that depends on the amount of metal and the connectedness/"allomantic mass" of the allomancer. For most objects, the distance dependence would be like 1/r^2 because of geometry (everything is a point mass when you are far away), but you could get other ones with different shapes of metal. We really dont have anything in the books that says it has to take the form of 1/r^2, though it is the obvious suggestion given most forces in real life work like this (though not all.) I detailed the biggest problem with this in my orignial post, but the biggest thing is that it does not produce the motion that has been indicated in the book such as the smooth stop at the highest point. (We will get to the cognitive aspect in a sec, but I was assuming a full push the whole time to get to this as an experiment.) Regardless of the equation you are correct that the formula will depend on geometry at close ish distances as long as we assume steelpushing on the continuous surface or volume of the metal, though in the example I am basing this off of she is pushing off a coin at a relatively large distance, so it should mostly act as a point mass, thus simplifing our thought experiment to finding the equation of force based on a point mass like I am doing. On 9/23/2024 at 12:04 PM, DrPhysics said: A power limit (Fv) that is tied to how quickly the allomancer can tap investiture This is actually a good point that I thought of, because investitiure and energy can be directly transformed into eachother there definitely could be a hard limit on the energy the allomancer can add to their system per second, but we do not really know enough about the conversion rate or how allomancy accesses investiture to make this concrete of a statment. Namely two things. One we dont know the conversion effeciency of allomancy, though that would be a fun thing to eventually experimently produce (we dont know if some investiture is dispersed in the process, or other small forms of energy or mass are created in the process, though it is not entirely unreasonable to make an ideal situation where the effeciency is 100% as this whole situation is highly idealized anyways. We still run into the conversion ratio for investiture to energy which we could only figure out experimently which seems like it might be a very difficult task) The second thing is that we dont actually know if allomancy has an investiture per second limit (whether through the person or the magic itself) because its magic. Because the investiture isnt being drawn from the metal (if it was we could just say allomancy converts whatever investiture / mass thats in the metal into investiture and thus must have a limit) and instead the metal is a catylist for drawing from invesititure, allomancy could easily just take as much investiture is required at any given moment to produce the effect we see without being limited in investiture / second. Reading the post you showed, they kind of a just assumed there was a power limit to allomancy without justifying it. I suspect that is because all real world systems have it because you are not just pulling energy from a 3rd magical source, but it actually has to come from somewhere in the system. This is not the case in the cosmere as there is nothing in the books that says someone must pull a constant amount of investiture at any given time. They are litteraly pulling power from god to do things, so they dont really need to limit how much investiture they turn into energy. Essentially we dont know that allomancers can only tap a certain amount of investiture per second, that is just one possible theory, albiet one that seems somewhat reasonable. Ideally we would come up with a force system that also coresponds to a constant addition of energy to the system, but it is not a requirment as far as I know. It would be intresting to add the constraint to see if we could narrow down the possible equations for motion / force but it is just one method we can use to get to a solution. On 9/23/2024 at 12:04 PM, DrPhysics said: A belief limit - my brain expects this to happen, so my push is matched accordingly (We do this all the time with our muscles - it's a safety feature so we don't tear them). This is the only way I can think of that explains Vin stopping (instead of bobbing up and down) at her max push while hovering over a coin. This is a very good point, and something I have though about briefly, but is also something that almost entirely makes this thought experiment useless and leaves us where we started, with no idea what equation governs the forces in steelpushing other than guesses based on real life phenomina. If we let someone subconciously change their push strength during their ascent (such as vin subconciously changing her push strength to not bob up and down) we introduce a variable that can be any function of time, and thus we end up with at least one (and probably more) family of infint equations that could describe the motion and corresponding forces in the books. This is entirely reasonable, but defeats the thought experiments purpose and we just dont have enough info to solve for the equation if we allow the push force to depend implicily on time, thus I have made the (potentialy wrong) assumption that the person is pushing at full force the whole time as they believe they do in the books. I would love to be able to solve for the equation allowing for a variable that implicilty depends on time, but it is impossible right now, so I wanted to see if I could solve for an equation that works for this simplified scenario (that also may be entirely true, we do not know yet if their subconcious is actually able to control the output of this particular power.) I should have specified this all in my initial post, but it is also only a "Potential physics equation for steelpushing". It is one possiblility given certain assumptions, and an entirely different conclusion (such as 1/r^2 or an infinit family of other equations) can be reached with other assumptions that I didnt make, such as allowing the equation to implicitly depend on time. On 9/23/2024 at 12:04 PM, DrPhysics said: The other important piece to consider for any physics model, is that you should have a "why" for a fit. 1/r^2 pops up so much because of how things expand in 3d space. What physical property implies a tanh? So there is a little error here, though I do understand your point. The tanh is the equation for motion taking time as an input that I proposed, wheras 1/r^2 is the equation for force that takes distance as an input. X(t) versus F(x) so they will understanbly take very differnt forms. F(x) according to the tanh model for motion is a quadratic which feels much closer to being reasonable than the tanh you were comparing it to. If you are confused you can check out the orignial post again where I derive that quadratic. I also made an edit to the post a while ago mentioning how tanh doesnt actually work which I think you will find interesting, which actually has somethign to do with that quadratic force equation being unreasonable. In the end I do actually need a F(x) that converges to 0 at infinity such as e^-r or 1/r^2 that also produces motion similar to the tanh for reasons I listed in the original post, which will produce a result that feels more reasonable. In total I do get your point, it does feel odd to have force or motion equations that are not similar to what we have in real life, but within the confines of the cosmere, an equation using magic does not really need to have any physical properites that imply its motion, because the laws of the magic system are the thing that determine the motion and forces and can be entirely arbitrary as they are accidental or puprosful results of a god messing with things. It is very nice if we get an equation that is similar to what we have in real life (and like a mentioned earlier, it will probably have to be cause of constraints I listed in the original post near the end), but it is not a requirment like in real systems as far as I can tell, despite how unintuitive that feels. I do really appreciate all the input though, you raise very good points, and we could definitley come to many different conclusions if we took it in all these exciting directions. I would be very happy to discuss the equations that would come out of these assumptions if you would like to, I would just like to acknoledge that they are essentially answer a "different question" if we do. 1
DrPhysics he/him Posted September 25, 2024 Posted September 25, 2024 (edited) On 8/26/2024 at 10:52 PM, DoomslugTD said: x(t) = h · Tanh(ax)² Edit: I'm an idiot. I just realized that i missed the squared on tanh. I still want to know if it should be x(t) = h · Tanh(at)². If that's the case, then I can second the math. What you have is true. You did make one mistake in your edits. The limit of x(t) as t goes to infinity is h, not infinity. And if you plug that into your force equation, you get that as t goes to infinity, p=g. If you take the limit of p as h goes to infinity (i.e. final height is infinite and time is infinite), you still get that p=g. So your equation handles the limits just fine. Original Comment: I interpreted it the first time that you were saying the distance dependence went like tanh, not time. I'm guessing that instead, the equation is supposed to be x(t) = h · Tanh(at)²? (if it is x, you have an unsolvable transcendental equation) If so, x'(t) = ah · Sech(at)² and x''(t) = 2ah · tanh(ax) · sech(ax)² Somehow you ended up ahead by a derivative. Also, when you simplified x''(t), you plugged in both x(t) and x'(t) as x(t). What you should get is x''(t) = 2/h x(t) x'(t), which you can't just separate out a g from and get an effective force. Making x(t) proportional to tanh means you will get particular solutions for F that will follow hyperbolic trig functions, not a polynomial. Tanh also violates the condition x'(0)=0, so you'll have to try a different function. I think you have a good start with the rules (so keep going), but your math still needs work. Edited September 26, 2024 by DrPhysics Don't post if you haven't double checked your math... 1
DrPhysics he/him Posted September 26, 2024 Posted September 26, 2024 4 hours ago, DoomslugTD said: This is a very good point, and something I have though about briefly, but is also something that almost entirely makes this thought experiment useless and leaves us where we started, with no idea what equation governs the forces in steelpushing other than guesses based on real life phenomina. For the sake of full disclosure, I'm in this camp. There just isn't enough data to come up with a solid model. But, it's always very exciting to be proven wrong, so please do. 1
DrPhysics he/him Posted September 26, 2024 Posted September 26, 2024 (edited) On 8/26/2024 at 10:52 PM, DoomslugTD said: Now this raises two big questions, what is the gravitational constant of scadrial and the mass of the person doing in the equation for the force provided by a steelpush It has the same g as earth. In arcanum unbounded, it lists Scadrial's gravity as one cosmere standard, which is set to Yolen's gravity, which is equal to earth's (though I couldn't find the WOB claiming that Yolen=Earth gravity, I know I've read it before on previous research). On 8/26/2024 at 10:52 PM, DoomslugTD said: though people in the past have raised a valid point that it could be because people with more mass stay closer to the object they are pushing for longer, allowing for greater total force to be exerted, though that would mean people with more mass accelerate slower when steelpushing. This is another fact that supports power as a limit. The slower velocity at the beginning lets you increase the force and lets you do more work (bigger force for the same displacement). If we were only force limited, you'd have the same max force and same displacement, so the same amount of work would be done. You'd have more momentum, but the same amount of energy. With a power limit, you'd end up with both more energy and momentum. On 8/26/2024 at 10:52 PM, DoomslugTD said: If anyone knows of any way to solve the final equation of motion I presented above I would love to hear it. Also, You've already solved it. x(t) = h · Tanh(at)² And I just noticed the part where you say that you haven't attended University yet. That's an impressive piece of calculus that I wouldn't expect one of my physics majors to be able to do until they were in their third or fourth year of university studies. Edited September 26, 2024 by DrPhysics 1
DoomslugTD Posted October 6, 2024 Author Posted October 6, 2024 @DrPhysics as a note before jumping into the rest of the responses, I believe you were double posting (making more than one post before someone responds) which I got in trouble for in this exact thread earlier. I was told that if you have more things to add you can edit your most recent post and add the things there, and if you make a new quote it will still alert the person just like if you created a new post. Just thought I would let you know I have made some edits to the original post to make things clearer and fix a couple mistakes you mentioned. To make sure we are talking about the same thing here is a list of the updated equation and what they mean. x(t) = distance as a function of time in the presence of gravity. (The motion we see in the book that I am basing this experiment on.) x'(t) = velocity as a function of time in the presence of gravity. x"(t) = accelerations as a function of time in the presence of gravity. x"(x(t)) = acceleration as a function of the specific distance x(t) in the presence of gravity. (It is there its just in the same line as x"(t).) d(t) = distance as a function of time with no gravity. (I have not solved for this (yet) because I am unsure how to solve the differential equation for d"(t).) d'(t) = velocity as a function of time with no gravity. (I have not solved for this (yet) because I am unsure how to solve the differential equation for d"(t).) d"(t) = accelerations as a function of time with no gravity. (I have not solved for this explicitly but we can write it in terms of d(t).) d"(d(t)) = acceleration as a function of general distance with no gravity. (I did not show this explicitly but its pretty easy to derive from the force equation. This is the differential equation you would need to solve in order to find d(t)) F(t) = total force as a function of time in the presence of gravity. F(x(t)) = total force as a function of the specific distance x(t) in the presence of gravity. Fₚ(t) = force provided by the steelpush as a function of time (this one depends on whether or not there is gravity or other external forces and requires us to solve for d(t) to figure out.) Fₚ(x(t)) = force provided by the steelpush as a function of the specific distance x(t) with no gravity. Fₚ(d(t)) = force provided by the steelpush as a function of general distance with no gravity. (This is the equation we are trying to solve for.) On 9/25/2024 at 4:27 PM, DrPhysics said: Edit: I'm an idiot. I just realized that i missed the squared on tanh. I still want to know if it should be x(t) = h · Tanh(at)². If that's the case, then I can second the math. What you have is true. YES, sorry! Thats a blunder on my part. It is meant to be a t. On 9/25/2024 at 4:27 PM, DrPhysics said: You did make one mistake in your edits. The limit of x(t) as t goes to infinity is h, not infinity. And if you plug that into your force equation, you get that as t goes to infinity, p=g. If you take the limit of p as h goes to infinity (i.e. final height is infinite and time is infinite), you still get that p=g. So your equation handles the limits just fine. In my edits I was taking the lim d->inf of f(d(t) (originally it was in terms of x but due to the edits some terminology was changed, but it was always in terms of distance) not lim t->t of x(t). This has left me a little confused about what you are talking about in that sentance. As for the rest of this paragraph I am a little unclear on what you are trying to say. We dont actually know what lim t->inf Fₚ(t) is because we only have Fₚ(d(t)) in terms of d(t) not explicitly in terms of t. We want that limit to also be 0 preferably, but like I said my edit was not focused on the lim t->inf it was focused on lim d->inf because we also know that the force function should tend to 0 as distance goes to infinity and we have the function explcility in terms of d(t). This is where we get the problem with the limits of my original htanh(at)^2 motion equation because the Lim d->∞( Fₚ(d(t))) where Fₚ(d(t)) = 2ma²(h - 4d(t) + (3/h)d(t)²) + mg is inf. This is obviously inccorect as the force from a steelpush should not blow up to infinity as the distance goes to infinity, but if we take htanh(at)^2 as our equation of motion in the presence of gravity that is what we end up with. (You can go through the original post again to see how I got the final form of Fₚ(d(t)) from htanh(at)^2. Essentially the equation of motion htanh(at)^2 satisfies requirments 1-6 but does not satisfy 7-8 which were the requirments I added in post after having this realization. I am sorry if I am not quite getting what you are saying, online forums are not the most condusive enviroment to well interpreted discussion sadly. On 9/25/2024 at 4:27 PM, DrPhysics said: I interpreted it the first time that you were saying the distance dependence went like tanh, not time. I'm guessing that instead, the equation is supposed to be x(t) = h · Tanh(at)²? (if it is x, you have an unsolvable transcendental equation) If so, x'(t) = ah · Sech(at)² and x''(t) = 2ah · tanh(ax) · sech(ax)² Somehow you ended up ahead by a derivative. Also, when you simplified x''(t), you plugged in both x(t) and x'(t) as x(t). What you should get is x''(t) = 2/h x(t) x'(t), which you can't just separate out a g from and get an effective force. Making x(t) proportional to tanh means you will get particular solutions for F that will follow hyperbolic trig functions, not a polynomial. Tanh also violates the condition x'(0)=0, so you'll have to try a different function. I am pretty sure x"(t) = 2ha² · sech(at)² · (sech(at)² - 2 tanh(at)²) I even just plugged the stuff into a derivative calculator and it confirmed it. Here is a link where you can look at it visually to intuitivily confirm its true https://www.desmos.com/calculator/lzrnge3drx. I then used the trig identity sech(x)² = 1 - tanh(x)² to get everything in terms of x(t) (specifically so I could avoid using x'(t) as that goes against one of the stipulations). I double checked it in the same calculator link I gave you earlier so if you want to test that x"(t) = 2ha² · sech(at)² · (sech(at)² - 2 tanh(at)²) = 2a²(h - 4x(t) + (3/h)x(t)²) it is all there, just turn on f₂(t) and j(t) and seeing that they have the same value for all values of t. If you want to see my work for how I got it in terms of x(t) just send me a message, but I believe everything I did in that section was correct including getting the final solution of x"(t) as a polynomial in terms of x(t). As for the last point. Tanh does violate x'(0) = 0 but I used Tanh^2 which does not violate that, I think this is what you were correcting in your edit, but I just wanted to make sure. On 9/25/2024 at 6:09 PM, DrPhysics said: For the sake of full disclosure, I'm in this camp. There just isn't enough data to come up with a solid model. But, it's always very exciting to be proven wrong, so please do. Honestly... so am I. I fully believe that there is more to be revealed, and there is a very high likelyhood someones perceptions effect how much force they output. The problem is I couldn't really do a fun math problem like this if I justified the weird motion like that. Basically this thought experiment goes under the assumption that perceptions do not effect the force because that is simply the only case for which we can currently do any math. We dont know for sure that it does or does not, so I will make what little progress I can in this niche assumption until more is revealed in hopes that this groundwork helps some people in the future. I acknoledge this wont be a definitive model for steelpushing as, like you said, we just dont have enough data, but a lot of real physics is based upon the idea of constraining ourselves to very niche and honeslty unrealistic assumptions in order to make progress towards a more general understanding. I am taking the same approach here and creating a model that fits a very narrow set of assumptions in order to progress towards a more general understaning. Ideally it will be revealed that everything wonderfully fits the crazy assumptions I made and my little project becomes a basis of understanding for steelpushing, but more likely I hope that it can be a stepping stone towards a fuller understanding by exploring a very narrow aspect of the greater whole if that makes any sense? On 9/25/2024 at 9:27 PM, DrPhysics said: It has the same g as earth. In arcanum unbounded, it lists Scadrial's gravity as one cosmere standard, which is set to Yolen's gravity, which is equal to earth's (though I couldn't find the WOB claiming that Yolen=Earth gravity, I know I've read it before on previous research). Not disputing that, I am just specifiying the gravitational constant of scadrial to point out its particular to the planet. The whole reason I pointed out g and m originally is because when I finished and saw them there I was very confused why the force of a steelpush would depend on the gravitational constant of a planet, until I realized that I initially paramatized the function in terms of height. I wanted to note that in the original post in case any others came away with the same confusion I did at first when seeing those seemingly unreleated constants in the final form of the equation. But thank you for confirming scadrials gravitation constant is the same as earths, I suspected as much but did not really want to go searching through WOBs. On 9/25/2024 at 9:27 PM, DrPhysics said: This is another fact that supports power as a limit. The slower velocity at the beginning lets you increase the force and lets you do more work (bigger force for the same displacement). If we were only force limited, you'd have the same max force and same displacement, so the same amount of work would be done. You'd have more momentum, but the same amount of energy. With a power limit, you'd end up with both more energy and momentum. I am a little confused at what you are saying, mostly because my mind has been jumping all over the place on this but I will reply with what i can. I think there is a really good argument to be made for a power limit, it makes some intuitive sense, but we just dont have any confirmation in the book that it exists as far as I know. I am completely fine though if a power limit emerges naturally from the force equation I provide, like for example if it means heavier people will stay near the source for longer thus doing more work I am fine with that, I just dont want to constrain my initial equation even more with things we dont have proof of in the book if that makes sense? I would be interested in seeing what equations might result from that constraint though. But I think I need a lot more clarity and detail about what you are saying here, because it does sound very interesting and promising, I just dont quite understand it. On 9/25/2024 at 9:27 PM, DrPhysics said: Also, You've already solved it. x(t) = h · Tanh(at)² That is the distance function in the presence of gravity, I am interested in solving for d(t) which would be the distance function without gravity which would require solving the system d"(t) = 2a²(h - 4d(t) + (3/h)d(t)²) + g. On 9/25/2024 at 9:27 PM, DrPhysics said: And I just noticed the part where you say that you haven't attended University yet. That's an impressive piece of calculus that I wouldn't expect one of my physics majors to be able to do until they were in their third or fourth year of university studies. Thank you! I have just started my first year as a physics major now. I was always interested in this kind of stuff so I just found what I could online, though that is getting harder and harder the more advanced it gets (hence finding it difficult to even know if there is a solution to the differential equation i put above.) I take it you are a proffesor? If so I am honored to have someone so knowledgable helping me with my little theory!
DrPhysics he/him Posted October 7, 2024 Posted October 7, 2024 (edited) 20 hours ago, DoomslugTD said: I believe you were double posting (making more than one post before someone responds) which I got in trouble for in this exact thread earlier. I was told that if you have more things to add you can edit your most recent post and add the things there, and if you make a new quote it will still alert the person just like if you created a new post. Just thought I would let you know Thanks for the heads up. I'll do that in the future. 20 hours ago, DoomslugTD said: I take it you are a proffesor? Yes. I've been a professor for about 10 years now. As you go through your program, never forget that the only big difference between you and me is 20 years of experience. We all remember being that brand-new student who was trying to figure out what was going on. 20 hours ago, DoomslugTD said: I am pretty sure x"(t) = 2ha² · sech(at)² · (sech(at)² - 2 tanh(at)²) I even just plugged the stuff into a derivative calculator and it confirmed it. Here is a link where you can look at it visually to intuitivily confirm its true https://www.desmos.com/calculator/lzrnge3drx. I then used the trig identity sech(x)² = 1 - tanh(x)² to get everything in terms of x(t) You're right. I made a mistake (I hadn't squared tanh when I copied it to paper) and put an edit in the top, but left my original post in case you had already seen it. That said, it looks like in doing so I caused more confusion than anything. Next time I'll just fix the whole comment. With that new x(t) notation, you have everything correct. 20 hours ago, DoomslugTD said: I am interested in solving for d(t) which would be the distance function without gravity which would require solving the system d"(t) = 2a²(h - 4d(t) + (3/h)d(t)²) + g. Ans here's where the teacher will come out a little bit. What you are looking for is called a "particular solution". Since you already know that the solution to the homogeneous (the equals zero case) is x(t) = h · Tanh(at)², you assume that d(t) = h · Tanh(at)²+<some function of t>, and just take guesses at that something until you get somethings that works. Maybe start with d(t) = h · Tanh(at)²+At² (since you are left with just a constant in the second derivative). Then, you plug in initial conditions (like d(0)=0) and see if you can solve for what "A" is. That said, some equations are unsolvable, so you'd have to use numerical methods instead, but it wouldn't give you an equation. 20 hours ago, DoomslugTD said: it was focused on lim d->inf because we also know that the force function should tend to 0 as distance goes to infinity and we have the function explcility in terms of d(t) Some things to add context to my answer: This might be me misunderstanding your notation. When I read this statement, it looks like you are trying to take the limit of your d(t) function as the function itself goes to infinity. I'm going to answer assuming this limit, but if that is now what you meant my answer might not apply. Since we don't know d(t), I'll use x(t), but once we have d(t) that satisfies the equation, the same limits should hold true If you want to see the behavior of an equation as you approach infinity, you can't just take the limit of the equation. Instead, you need to take the limit of individual parts of the equation and see how they behave. If you want to see how your equation behaves at really long distances, you need to be really careful about how you apply boundary conditions. If you simply take the limit of x''(t) as h goes to infinity, x''(t) does become infinite. However, that limit assumes I am starting on a mass and ending up and infinite distance away, so my acceleration should be infinite. Your constraint needs to be x''(t) is zero as both h and t go to infinity (I'm now really far from the planet and have been pushing a long time). In that case, x(infinity) = h, so: x''(t) = 2a²(h - 4x(t) + (3/h)x(t)²) = 2a²(h - 4h + (3/h)h²) = 2a²(h - 4h + 3h) = 0 Once you have a full solution to d(t), you should be able to let h go to infinity to get zero because you've eliminated gravity from the situation. In short, two things to remember: Don't panic about a limit not working when you don't have the full solution. Don't apply physical limits to an equation without considering the constraints you used when making the equation. Edited October 7, 2024 by DrPhysics
DoomslugTD Posted October 7, 2024 Author Posted October 7, 2024 6 hours ago, DrPhysics said: Yes. I've been a professor for about 10 years now. As you go through your program, never forget that the only big difference between you and me is 20 years of experience. We all remember being that brand-new student who was trying to figure out what was going on. Awesome! Well I am still super happy to be talking to someone who has considerable experience with physics so thanks for humoring my wild thoughts! 6 hours ago, DrPhysics said: Ans here's where the teacher will come out a little bit. What you are looking for is called a "particular solution". Since you already know that the solution to the homogeneous (the equals zero case) is x(t) = h · Tanh(at)², you assume that d(t) = h · Tanh(at)²+<some function of t>, and just take guesses at that something until you get somethings that works. Maybe start with d(t) = h · Tanh(at)²+At² (since you are left with just a constant in the second derivative). Then, you plug in initial conditions (like d(0)=0) and see if you can solve for what "A" is. That said, some equations are unsolvable, so you'd have to use numerical methods instead, but it wouldn't give you an equation. Ohhhh, I did not know that you could extrapolate out from the homogenous equation like that. I have not formally learnt how to solve differential equation as of yet, so my techniques are often very scattered. I usually end up just using the laplace transform for almost anything cause it applies to a lot of stuff, so I did not really notice you could do this. I will give it a shot, thank you! 6 hours ago, DrPhysics said: Some things to add context to my answer: This might be me misunderstanding your notation. When I read this statement, it looks like you are trying to take the limit of your d(t) function as the function itself goes to infinity. I'm going to answer assuming this limit, but if that is now what you meant my answer might not apply. Since we don't know d(t), I'll use x(t), but once we have d(t) that satisfies the equation, the same limits should hold true If you want to see the behavior of an equation as you approach infinity, you can't just take the limit of the equation. Instead, you need to take the limit of individual parts of the equation and see how they behave. If you want to see how your equation behaves at really long distances, you need to be really careful about how you apply boundary conditions. If you simply take the limit of x''(t) as h goes to infinity, x''(t) does become infinite. However, that limit assumes I am starting on a mass and ending up and infinite distance away, so my acceleration should be infinite. Your constraint needs to be x''(t) is zero as both h and t go to infinity (I'm now really far from the planet and have been pushing a long time). In that case, x(infinity) = h, so: x''(t) = 2a²(h - 4x(t) + (3/h)x(t)²) = 2a²(h - 4h + (3/h)h²) = 2a²(h - 4h + 3h) = 0 Once you have a full solution to d(t), you should be able to let h go to infinity to get zero because you've eliminated gravity from the situation. In short, two things to remember: Don't panic about a limit not working when you don't have the full solution. Don't apply physical limits to an equation without considering the constraints you used when making the equation. So, I end up with a final force equation Fₚ(d(t)) and I am wanting that to approach 0 as d(t) becomes larger because at large distances it should logically follow that the force is ≈ 0. With Fₚ(d(t)) = 2ma²(h - 4d(t) + (3/h)d(t)²) + mg I just noticed that with respect to d(t) Fₚ(d(t)) looks like a second degree polynomial and with that postivie d(t)² term it looks like that equation will blow up to infinity at extremly large distances which is a problem considering we want the force to decrease with distance. Am I misinterpreting something here, or is there an error in this work? Does it not matter that the force would be extremely large for high values of d(t)? Another thing, this might be bad notation on my part, but h is not a function of time or anything else, its not the current height of the allomancer it is defined as being the final height they would reach if they are pushing at full strength directly upwards in scadrials gravity. It is an initial paramter that comes from how strong the push is, just like how a is a paramter that defines how quickly the allomancer reaches that final height h. I used these values because they seemed like the easiest values to work out in readings and are fairly simple to fit into the skeleton structure of the original tanh(t)^2 with h multiplied on the outside and a being multiplied on the inside of the tanh(t)^2. So taking the limit as h approaches infinity is equivelent to asking how the function behaves as the allomancers push strength approaches infinity (though you would have to increase it in a very particular way to not also get a change in a as that is another paramter that depends on the strength of the push.) The current height of the allomancer in this example would be represented by x(t) or d(t) (depending on whether you are in gravity or not) as the allomancer is always pushing straight up away from the metal. I hope that clear some things up, though if you believe I am still approaching this wrong I am happy to be proven wrong. I would absoltely love it if my solution worked!
DrPhysics he/him Posted October 8, 2024 Posted October 8, 2024 1 hour ago, DoomslugTD said: I end up with a final force equation Fₚ(d(t)) and I am wanting that to approach 0 as d(t) becomes larger because at large distances it should logically follow that the force is ≈ 0. With Fₚ(d(t)) = 2ma²(h - 4d(t) + (3/h)d(t)²) + mg I just noticed that with respect to d(t) Fₚ(d(t)) looks like a second degree polynomial and with that postivie d(t)² term it looks like that equation will blow up to infinity at extremly large distances OK. I understand now. Don't forget that h is your max height, so d(t) must always by less than or equal to h (minus whatever the correcting factor is that includes gravity). If you let d(t) go to infinity, h also goes to infinity. And when they are both infinity, the force equation becomes mg, which is what we expect. 1
DoomslugTD Posted October 8, 2024 Author Posted October 8, 2024 (edited) 1 hour ago, DrPhysics said: the force equation becomes mg, which is what we expect. I think there was a misunderstanding, Fₚ(d(t)) is the force only of the steelpush. We specifically subtracted the force of gravity from the F(x(t)) (force equation we derived from x(t)) to get that. I will post the defininitiions of all the functions I am using right below this, I would especially focus on x(t), d(t), F(x(t)) and Fₚ(d(t)) so we can discuss this while having a common set of defined functions. On 10/6/2024 at 1:57 PM, DoomslugTD said: x(t) = distance as a function of time in the presence of gravity. (The motion we see in the book that I am basing this experiment on.) x'(t) = velocity as a function of time in the presence of gravity. x"(t) = accelerations as a function of time in the presence of gravity. x"(x(t)) = acceleration as a function of the specific distance x(t) in the presence of gravity. (It is there its just in the same line as x"(t).) d(t) = distance as a function of time with no gravity. (I have not solved for this (yet) because I am unsure how to solve the differential equation for d"(t).) d'(t) = velocity as a function of time with no gravity. (I have not solved for this (yet) because I am unsure how to solve the differential equation for d"(t).) d"(t) = accelerations as a function of time with no gravity. (I have not solved for this explicitly but we can write it in terms of d(t).) d"(d(t)) = acceleration as a function of general distance with no gravity. (I did not show this explicitly but its pretty easy to derive from the force equation. This is the differential equation you would need to solve in order to find d(t)) F(t) = total force as a function of time in the presence of gravity. F(x(t)) = total force as a function of the specific distance x(t) in the presence of gravity. Fₚ(t) = force provided by the steelpush as a function of time (this one depends on whether or not there is gravity or other external forces and requires us to solve for d(t) to figure out.) Fₚ(x(t)) = force provided by the steelpush as a function of the specific distance x(t) with no gravity. Fₚ(d(t)) = force provided by the steelpush as a function of general distance with no gravity. (This is the equation we are trying to solve for.) The final equation I am trying to solve for is Fₚ(d(t)) which is also the equation I am taking the limit of in this case. This equation should approach 0 as d(t) gets larger because this is purely the force of the steelpush as a function of distance (not in the presence of gravity) and that should approach 0 as d(t) gets infinitly large. 1 hour ago, DrPhysics said: OK. I understand now. Don't forget that h is your max height, so d(t) must always by less than or equal to h (minus whatever the correcting factor is that includes gravity). If you let d(t) go to infinity, h also goes to infinity. I dont believe d(t) is limited by h in the same way x(t) is either due to the fact that x(t) is in the presence of gravity and d(t) is not. There may be a way to prove this mathmatically, but at leasy logically the lim t->inf of d(t) should be inf because steelpushes can by definition only push and there are no other forces because by definition d(t) is the motion of the steelpusher only under the influence of the steelpush therefore there is only ever be a postive therefore the velocity will always be increasing with respect to time therefore the distance will also always be increasing with respect to time and therefore the lim t->inf d(t) = inf. I belive the majority of the confusion has come from unclear communication about what each function means as far as I can tell. If this is not the case and we were interprating the functions the same way then I would welcome some clarification on your statments! As an additional thing, before I solve for d(t) I do actually need to solve for x(t) which I am so close to (due to the fact that I started with x(t) = htanh(at)^2) but this is missing something very important being the initial conditions, because that is only true if x(0) and x'(0) = 0 so I dont have a general solution to the homogeneous equation as of yet let alone d(t). If you have any way to help me solve x"(t) = 2a²(h - 4x(t) + (3/h)x(t)²) or you have some technique for adding initial conditions when I have the specific solution when x(0) and x'(0) = 0 that would be super helpful! Edited October 8, 2024 by DoomslugTD
DrPhysics he/him Posted October 8, 2024 Posted October 8, 2024 1 hour ago, DoomslugTD said: I dont believe d(t) is limited by h in the same way x(t) is either due to the fact that x(t) is in the presence of gravity and d(t) is not. 1 hour ago, DoomslugTD said: due to the fact that I started with x(t) = htanh(at)^2) By.choosing to start with this, you limited the function (both x and d) such that their maximum value is h. Basing your differential equations on x forces this to be true. So either you constrain x and d to be less than h, or you throw it all out and start over. Or another way to look at it is this: you can't build a model on the assumption of how something works on near-surface gravity then hope that goes away by just subtracting off gravity. You can see that in your force differential equation. When you take the limit as d goes to h you are left with mg, which is how hard they would need to Push to hold themselves at a constant height.
DoomslugTD Posted October 8, 2024 Author Posted October 8, 2024 (edited) 1 hour ago, DrPhysics said: By.choosing to start with this, you limited the function (both x and d) such that their maximum value is h. Basing your differential equations on x forces this to be true. So either you constrain x and d to be less than h, or you throw it all out and start over. Or another way to look at it is this: you can't build a model on the assumption of how something works on near-surface gravity then hope that goes away by just subtracting off gravity. You can see that in your force differential equation. When you take the limit as d goes to h you are left with mg, which is how hard they would need to Push to hold themselves at a constant height. Why does limiting x do have a maximum of h limit d? We have real world examples of systems that dont tend to infinity in the presence of gravity but would if they were not. Like if you a negativly charged plate fixed to the ground and a postivly charged ball above it they charges would push away to a certain distance before gravity became stronger then the colobmn force and it settled at a final height, but if you removed gravity from this equation the ball would accelerate from the plate but never stop therefore tending towards infinity. It is a very similar system here where we have a pushing force that when combined with gravity (x(t)) results in a motion that maxes out at h, but when not in the presence of gravity there is no pulling force that could slow the ball down therefore it should tend to infinity (d(t)) I explained this in more detail in my previous post if you missed that. There is another force at play in x(t) that is not present in d(t) so as far as I know it does not make sense to say that because x(t) is limited to h d(t) should also be. I dont understand why I cannot build a model based on how the motion looks in the presence of gravity, determine a force function that takes distance as an input that is the combination of steelpushing and gravity, subtract off gravity and get a new equation of motion that would behave differently as a fundemental part of the forces on it have been changed. There is a part of the original post which explains how I switched the force equation from depending on x(t) (the distance specifically in the equation of motion I started with) to a more general d(t) that is not constrained to x(t). I would suggest re-reading that portion of the original post if that is where the misunderstanding is routed from. But essentially it stems from the fact that the part of the force equation that represents the steelpush should not depend on whether the system is in gravity or not, so I can convert x(t) from the distance specifically when the motion is x(t) to a more general for where I represent distance as d(t). As an example the strength of the steelpush should be the same at 1 meter regardless of weather the distance is following the motion of x(t) or any other arbitrary motion. A more thourough explenation, in case there is still confusion. Fₚ(x(t)) = F(x(t)) + mg because by definition F(x(t)) is the combination of the force of the steelpush Fₚ(x(t)) and the force of gravity -mg. We also have the fact that Fₚ(x(t)) = Fₚ(d(t)) because the force of the push at any given distance does not depend on overall motion of the particle, only on the distance at that given point in time, so we are able to replace x(t) (the motion constrained to htanh(at)^2) with d(t) (which is the path when the particle is not under the influnece of gravity) or any other path of our choosing. The same logic can be applied to get F(x(t)) = F(d(t)) because the only additional force is a constant one which should also not change based on the path of the particle therefore the total force still only depends on the distance from the metal regardless of the overall path of the particle. This leaves us with Fₚ(d(t)) = F(d(t)) + mg which is not constrained to htanh(at)^2 as d(t) can be a completely different path than x(t). This means that there is no reason to assume d(t) has a maximum of h. Finally, the last thing is just logic. I have iterated this in my last post, but it just doesnt make sense for d(t) to have a maximum of h because the only force in that system is the pushing force, which should only be able to push. This is where the x(t) that I chose runs into problems because not only can the force be negative at certain distances (which would result in pulling, and therefore does not fit with steelpushing in the books) it blows up to infinity at infinit distance which also does not fit the books. Edited October 8, 2024 by DoomslugTD
DrPhysics he/him Posted October 8, 2024 Posted October 8, 2024 11 hours ago, DoomslugTD said: Why does limiting x do have a maximum of h limit d? We have real world examples of systems that dont tend to infinity in the presence of gravity but would if they were not. Like if you a negativly charged plate fixed to the ground and a postivly charged ball above it they charges would push away to a certain distance before gravity became stronger then the colobmn force and it settled at a final height, but if you removed gravity from this equation the ball would accelerate from the plate but never stop therefore tending towards infinity. Every one of these systems is built starting from first principles using forces, and the description of motion comes second, so you can tweak it by adding/removing forces and seeing how that impacts motion. Your model is built on a description of motion, and you are adding in forces second. So, you can change motion and see how that impacts forces, but you can't change forces to see how that affects motion. What you are doing is equivalent to what happened with the Rayleigh-Jeans law and the Ultraviolet catastrophe. Short version: They came up with an equation that fit pretty well to the spectrum of a blackbody. It was an empirical formula designed to fit what we saw (like your motion equation). But, when we tried to use it to describe the thermodynamics/statistical mechanics of what was going on, it broke at infinity (specifically, every blackbody would emit an infinite amount of energy in Ultra-Violet and shorter wavelengths). It worked well to predict spectrum, but it couldn't be used on more fundamental ideas. Later on, Plank constructed what we use now from thermodynamic principles, and his solution became part of what inspired Einstein to describe the photon. But before then, the Rayleigh-Jeans law was still very useful in describing the radiation we see coming out of black bodies. Thermodynamics creates the radiation, so you can use thermodynamics to describe the radiation, but you can't use something that purely fits the radiation to describe thermodynamics. Similarly in your case, forces cause changes in motion, so you can use forces to describe that changing motion, but you can't use something that purely fits motion to describe forces. That doesn't mean what you have is useless, it just means it has limits. 12 hours ago, DoomslugTD said: Finally, the last thing is just logic. I have iterated this in my last post, but it just doesnt make sense for d(t) to have a maximum of h because the only force in that system is the pushing force, which should only be able to push. This is where the x(t) that I chose runs into problems because not only can the force be negative at certain distances (which would result in pulling, and therefore does not fit with steelpushing in the books) it blows up to infinity at infinit distance which also does not fit the books. Exactly this. x(t) can potentially describe a push while under gravitational pull, with a max of h. Anything you derive using your x(t) will have that same limitation. Even if you find a general and particular solution for your differential equation, it will still be bound by h because it was bounded by h when you created it. It only blows up at infinity if you don't let h go to infinity as well.
DoomslugTD Posted October 9, 2024 Author Posted October 9, 2024 10 hours ago, DrPhysics said: Every one of these systems is built starting from first principles using forces, and the description of motion comes second, so you can tweak it by adding/removing forces and seeing how that impacts motion. Your model is built on a description of motion, and you are adding in forces second. So, you can change motion and see how that impacts forces, but you can't change forces to see how that affects motion. What you are doing is equivalent to what happened with the Rayleigh-Jeans law and the Ultraviolet catastrophe. Short version: They came up with an equation that fit pretty well to the spectrum of a blackbody. It was an empirical formula designed to fit what we saw (like your motion equation). But, when we tried to use it to describe the thermodynamics/statistical mechanics of what was going on, it broke at infinity (specifically, every blackbody would emit an infinite amount of energy in Ultra-Violet and shorter wavelengths). It worked well to predict spectrum, but it couldn't be used on more fundamental ideas. Later on, Plank constructed what we use now from thermodynamic principles, and his solution became part of what inspired Einstein to describe the photon. But before then, the Rayleigh-Jeans law was still very useful in describing the radiation we see coming out of black bodies. Thermodynamics creates the radiation, so you can use thermodynamics to describe the radiation, but you can't use something that purely fits the radiation to describe thermodynamics. Similarly in your case, forces cause changes in motion, so you can use forces to describe that changing motion, but you can't use something that purely fits motion to describe forces. That doesn't mean what you have is useless, it just means it has limits. 23 hours ago, DoomslugTD said: Finally, the last thing is just logic. I have iterated this in my last post, but it just doesnt make sense for d(t) to have a maximum of h because the only force in that system is the pushing force, which should only be able to push. This is where the x(t) that I chose runs into problems because not only can the force be negative at certain distances (which would result in pulling, and therefore does not fit with steelpushing in the books) it blows up to infinity at infinit distance which also does not fit the books. Exactly this. x(t) can potentially describe a push while under gravitational pull, with a max of h. Anything you derive using your x(t) will have that same limitation. Even if you find a general and particular solution for your differential equation, it will still be bound by h because it was bounded by h when you created it. It only blows up at infinity if you don't let h go to infinity as well. I guess I must just have a fundemental misunderstanding of what is going on. But if you could point out where exactly my argument breaks down when switching between using the constrained x(t) to a non constrained distance I would love it, I have iterated it in multiple different ways in both my original post and other replies including my last one. I just know that in this model the force of gravity is constant regardless of the motion of the particle, and that the force of the steelpush is always the same at a certain distance regardless of the motion of the particle as well, I used this to justify why it should not matter if we replaced x(t) in the force equation with d(t) which is also represents distance but is not constrained to the original motion of the particle. Essentially the force of the push should not depend on the overall motion of the particle so we can generalize the force equation for a steelpush we find in terms of x(t) to a force equation for a steelpush in terms of a general distance d(t) that does not depend on path. Your example seems like it might have extra complexities that this simple idealized system does not have and might not be able to generalized because of the very specific requirments I layed out in order to get from x(t) to d(t) so I dont know if it proves what I am doing inccorect, though it does cast some doubt. Thanks!
DrPhysics he/him Posted October 9, 2024 Posted October 9, 2024 14 hours ago, DoomslugTD said: I just know that in this model the force of gravity is constant regardless of the motion of the particle, and that the force of the steelpush is always the same at a certain distance regardless of the motion of the particle as well This is where you went wrong in trying to solve for forces. Your model doesn't assume these things. It assumes that x(t) = h · tanh(at)². Could that be a result of those assumptions? Sure. But it isn't built on those assumptions. As I said before, that doesn't mean your model is useless or you should throw it out. It just means that it won't work if you try to generalize it to situations where motion might not be x(t) = h · tanh(at)².
DoomslugTD Posted October 9, 2024 Author Posted October 9, 2024 (edited) 2 hours ago, DrPhysics said: This is where you went wrong in trying to solve for forces. Your model doesn't assume these things. It assumes that x(t) = h · tanh(at)². Could that be a result of those assumptions? Sure. But it isn't built on those assumptions. As I said before, that doesn't mean your model is useless or you should throw it out. It just means that it won't work if you try to generalize it to situations where motion might not be x(t) = h · tanh(at)². The model for the motion does not, but the book itself suggests that it is that way which are where the assumptions are coming from. We know that the steelpush always gives the same force at the same distance and its a fact that the force of gravity is independent of path (when we assume its a constant) from there we can then determine what the force of the steelpush is by asking what forces must be present on an allomancer for them to follow a certain motion. I chose this motion to be htanh(at)^2 because it is one of many that fits the motion as described in the books, in this simplified example we assume there are only two forces acting on this allomancer, the push and gravity. We know what gravity is so there is only one unkown, the force of the push, we solve for it and put it in terms of the distance from the metal in this motion because we know the steelpush should only depend on distance and not velocity or time. We now have a force in terms of the distance of this very particular motion htanh(at)^2 but from the first assumption we made (that was not made from the motion but instead made from the material of the book which the motion was derived from) we can say that the force of the steelpush should not depend on what path the allomancer is following and should instead only depend on their distance regardless of what path it took to get there and what path it will continue to follow. We can now confidently say that the same force equation we came up for that took in the distance when distance(t) = htanh(at)^2 as an input can take in any distance(t). Essentially we know that Fₚ(x(t)) = Fₚ(distance(t)) where distance(t) is an arbitary function, not from any part of the equation of x(t) but from the underlying laws that come from the work this entire discussion is based off of. We know this because we know that the force of a push is the same at a given distance regarldess of the overall motion because the book has told us this not because of some underlying physical principle. Edited October 9, 2024 by DoomslugTD
DrPhysics he/him Posted October 10, 2024 Posted October 10, 2024 17 hours ago, DoomslugTD said: Essentially we know that Fₚ(x(t)) = Fₚ(distance(t)) where distance(t) is an arbitary function, not from any part of the equation of x(t) but from the underlying laws that come from the work this entire discussion is based off of. We know this because we know that the force of a push is the same at a given distance regarldess of the overall motion because the book has told us this not because of some underlying physical principle. Well, your next steps would be to test it out numerically and see if that gives you any insight into figure out the general solution. Mathematica has some good built-in numerical solvers and most universities have access to a free license for students (it's how they market - if we get you hooked as a student, we'll keep you hooked when you employed.) Or, you could write it up using Euler's method in python. These wouldn't give you general solutions, but they would let you run several test cases and see how they behave.
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