Ghideon

Have you selected one of the options?

Thanks! Then guess the initial question regarding the vector math have been answered: "I need help with my math problem". The vectors I helped verify is correct. But due to a misunderstanding of the physics you end up with something physically impossible, the vectors does not describe a physical valid configuration. By the way, "fancy vector math", when applied to the correct set of forces, will show that internal forces cancel*. Unfortunately you have been unable to follow along in my attempts at explaining this. That means we have reached the end of this thread. From here I see two options, up to you. 1: continue to argue that Newton (and Einstein and others) were wrong, move to speculations and led the moderators close the thread**. You have no evidence that your physics works and seem to object to any attempts to show you how it works or where there are misunderstandings. 2: Continue asking about how physics works, how to analyse systems, add forces, implications of newtons laws etc (maybe a separate thread?) Since we seem to be able to communicate rather well using pictures such a discussion could be productive. But that is only if you are interested in learning, it's up to you. True. And used all the time in electric motors for instance. But have you any reference that this rule is allowed to break the rules of momentum conservation? You can't pick one rule from physics, apply the math in isolation, and expect a valid result. That is due to imaginations or misunderstandings. The laws of physics seems to disagree with your statement. By the way, maybe there is a third option: Maybe they could help you with understanding the basics about mechanics? *unless the stricture breaks apart of course, accelerating cables, iron cores and what not, all around. ** It would be cool to be proven wrong, that evidence supporting this lifter can be provided. But I doubt that.

I think you might want to slow down and address the more basic issues first. Note: you are not arguing against me, you are arguing against Newton. I'm just the messenger*. So I'll drop the details and go back to basics for a while until that is sorted out. Situation at the top. Newton's third again: "For every action, there is an equal and opposite reaction." The wall is pushed by two equal forces that cancels each other. Put person P1 and P2? They push against the wall and the wall pushes back with the same force. You have not drawn those forces. I will skip middle image since it is in principle the same as the bottom."For every action, there is an equal and opposite reaction." So the wall pushes back on the persons with force F1 and F2. And since you have nothing to balance the force FTotal F=ma tells us that the wall will accelerate. OR are there other forces you are not drawing? Are there a force from the ground, holding back the wall? And the individuals, are they standing firmly on the ground or are they sliding backwards while pushing? What you are drawing is a free body diagram of the wall, not a diagram for a system of two persons, wall, and ground. Note: the vector sum (F1 + F2 = Ftotal) seems ok. So yes FTotal becomes larger. As will the force from the ground holding back the wall, unless you accept the persons to push away the wall. You seem to constantly confuse components of a system and its internal forces with the sum of forces acting on the complete system. And each time we get close to the core you throw in a set of magnetic fields that seems to get the discussion of track. To further reduce the possibility for misunderstandings about the intention with the lifter, here is again a question about its basic workings: Let's put the lifter in a concealed box. The box has low mass to not interfere with the lifters capacity. "The lifter" means everything that is required for it to operate. Power sources, cables etc. There is no connection to the outside of the box; no electro magnetics, cables, noting. Note that I draw a simple 3d drawing. This is not a cross section of a larger structure. The lifter is started (via a timer since it is not readable from outside the box). If the lifter works as you intend it to do, will the box lift into the air, pushed by the lifter inside it? This is as simple and abstract as it can get I think. It does not matter how the lifter is constructed only that you managed to build it somehow. From my earlier questions and your responses I think you intend the lifter to work in this manner but I want to be sure. Standard disclaimer: There is always the possibility that I'm wrong. I'm trying to argue from a mainstream perspective but my knowledge is of course limited.

Thanks for the feedback! A quick check. Router A seems right. Step 2 gets a second route to B through C but since that is longer (=5) you have correctly rejected it. If a third step would be added the router A would get a route to D through B-C but more expensive (=7) so no changes will be made to A's routing table. Router B Seems OK, as argued above Router C has a typo on step 2, I think (0,A) should be (0,C). Also I think the route to A will not change on step 2. Unless there is some weird algorithm or additional information the router will likely keep (1,A) since that is cheaper than going through B at cost 6. Router D seems a little off: 0 | [ (/,/) (/,/) (/,/) (0,D) ] Seems OK 1 | [ (2,A) (0,B) (1,C) (0,D) ] This step could not contain information about A and B yet, how about this: | [(/,/) (/,/) (1,C) (0,D) ] 2 | [ (2,C) (4,C) (1,C) (0,D) ] (4,C) should likely be (5,C) since D->C->B is 1+4=5 3 | [ (6,C) (3,C) (1,C) (0,D) ] (3,C) should likely be (4,C) since D->C->A->B = 1+1+2=4 Important note: I have tried to stay very basic in the analysis above since the exam task does not provide much information. That means that if this is an exam for a specific implementation for instance a certification for a specific brand or manufacturer there could be other things that the exam expects to be taken into account. Additional, non important stuff: I note that this little example looks very simple but contains enough to be used as a basis for a discussion about a lots of additional questions for the interested. Example: if the routers keep the more expensive path instead of dropping them the routers would have backup paths in case of failure. But if they do, additional algorithms may be required. If A knows B can be reached through C, and C knows B can be reached through A, What happens if B goes off-line? Will a package intended for B bounce back and forth between A and C? Such questions is not likely something a network engineer will care much about on a daily bases in a modern environment. Designing the algorithms, improving them, and studying the math behind it is ongoing. The picture reference above is from a book for a course in distributed computing, 1993 edition. Much of it is obsolete now. I guess that newer sources could be based more on interest since the topic contains so much more nowadays.

Ok! No problem. Again: Adding to the above, since your drawing has a net force on the wall it is not a static situation. What will it look like at some time later? What does Newton say?

Yes. Do you? No. Idea: If you look at the wall and the individuals again, the last one on the page, what will it look like after a while? Do you know about the equation F=mA from Newtonian mechanics?

You have provided a drawing where two people pushing the wall and since there is a net force on the wall the wall will be pushed away, downwards. The distance between wall and individuals will increase. Now look at the sketches for your rig. You have drawn the force in the same way as in the wall example, forces are from the cables perspective. The cable is embedded the in the iron structure. Do you expect the cable to brake loose and accelerate upwards? Or do you expect the iron structure to push back at the cable with an equal and opposite force? These questions are quite crucial to the understanding of the issues. You can't calculate forces on an individual internal part of a structure and apply that force as an accelerating force on the whole structure.

Thanks. You have three bodies: the wall and two persons. Note that you have drawn the forces from the walls perspective, not the total sum of forces for the three bodies. The persons are external in your picture, not part of the wall, not attached to the wall. What do you predict regarding acceleration of the bodies in the bottom picture? What relative motion of the bodies do you expect?

The above contains many issues that I do not have time to adress right now. you have not yet managed to get to the core of the problem. maybe if you provide a picture of the two people and the wall you mentioned we can do some basic mechanics calculations? I might get some time later today to post a few other ideas.

if you provide a drawing we can calculate the outcome. This may be a good approach since it is more general and does not go into details about magnetic fields.

Sorry, you completely missed the point. Maybe you should check out Newtons laws first? When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body. Why do you think this law does not apply? When the cable at one of the corners is pushed with some force , it cant go anywhere because it is embedded in the iron triangle, and the iron triangle holds it back with the same force. Again, from what you have provided in the drawings and math and descriptions, this is a basic physics issue. You can't by any means, accelerate something by using internal forces. Nature just does not work that way. Either you push against something extarnal, like when you jump off the ground, or you eject mass, like a rocket engine. Don't get entangled in detailed geometry or magnetics. By what principle, compatible with basic laws of Newton, does your lifter operate?

Please use the qoute function. Please quote from the simplified analysis I provided. Please point out in detail where my simplifications does not apply to your rig.

Sorry, I do not follow you reasoning. Can you cite the exact assumptions or statements I made that you disagree with? Again, are you familiar with Newton's third law?

Vector addition will give you the answer. Do you want an example, see my earlier posts where the B-vectors are added, resulting in new vectors with a different direction and magnitude. Why does that matter? If I change one of my magnets in the simplified analysis to a smaller magnet it does not change the outcome.Forces cancel in the rig. The number of forces foes not matter. If they are all internal, acting on the structure, not pushing against anything external, they cancel. No calculation needed. But I could have been clearer: When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body. It is called Newton's third law*. Are you familiar with its implications and why it allows me to make generalized comments about the lifter? I've spent a considerable amount of time in this thread, sorry if that it does not count as caring. When did that happen? I dream all the time, just not so much about perpetuum mobiles and reactionless drives. I think the same goes for engineers and scientists at LIGO and CERN. Gravitational wave astronomy and the experimental evidence for Higgs came from dreaming about extending what is possible, not dreaming about what is known to be impossible. https://en.wikipedia.org/wiki/Newton%27s_laws_of_motion#Newton's_third_law or https://en.wikipedia.org/wiki/Reaction_(physics)

The picture above is from an era way before Home & Garden Television was launched! Easy to mix up

Thanks. I also note that this thread has stayed away from non-mainstream claims, so we can pursue this topic in a more collaboratory fashion than what is possible in a speculations thread. 😀 I will use the latest picture you posted, with the triangular core, and try some ideas before calculating. In a speculations this would have been a more blunt and short question but this time some more details will be provided, it will be easier to spot if my understanding of the setup is wrong. First we so a series of simplifications. -Let’s assume the the rig is large enough so that interference between the points a, B and D are neglectable. Hence we can look at one corner at a time. -It is not known yet how the magnetic field will look in detail at the corners. The magnetic flux will likely have some radius, there are no sharp turns so the vector calculations I did do not apply. But that does not matter. What matter is that the point A is located where the magnetic field B is horizontal. That is all we care about. Now we reverse the reasoning. We do not look at the suggested rig in drawings above, we look at an idealised rig, where we have managed to create the magnetic field we want and it affects the cable at point A pushing it upwards. How we managed to build it is not important. Here is corner A with force F pushing on the cable due to the magnetic field B. Now we simplify further. It does not matter what creates the magnetic field or what it pushes. AFAIK there is no fundamental differences between an electromagnet and a permanent magnet for the properties we want to investigate. We are interested in a force and a magnetic field and the interaction, nothing more at this point. So let’s replace the magnetic field with a permanent magnet. And since the force F is directed in positive y direction we can use another magnet instead of a wire. (since a piece of iron would be pulled towards the magnet) Now we have reduced the setup to its basics. The force F pointing up will have an equal force pointing down. If the magnets are attached to the same solid frame, as the cable and the iron core in your drawings, there is no way there can be an acceleration. The forces cancel internally in the rig. Does this show how any kind of magnetic fields, acting on internal parts of the rig, will be unable to generate accelerating force? It does not matter how the magnets are angled, if they have some unusual material or some unusual geometrical shapes. The forces cancel.

Thanks for the update, this allows for a more complete analysis. OK, that indicates the guess that the destination is implicit from the position the table is correct. Here is a modified attempt. I guess you are on the right path but I suspect the routing table for router A may only contain the name of the next router, the next-hop target. So the green (2,C) entry below means an outgoing packet from router A, intended for router D should be sent to C for further routing. The cost is 2. |Step | Router A | 0 | [ (0,A) (/,/) (/,/) (/,/) ] 1 | [ (0,A) (2,B) (1,C) (/,/) ] 2 | [ (0,A) (2,B) (1,C) (2,C) ] Router B is more interesting since I guess that there is some selection algorithm involved to make the task more interesting. Note that the table is not static for some tuples, the router B will calculate that the route to C is cheaper through A. In reality this is more complicated. A discussion about the history behind the algorithms, mathematic proofs and issues in real applications is off topic for this thread, but could be very interesting. |Step | Router B | 0 | [ (/,/) (0,B) (/,/) (/,/) ] 1 | [ (2,A) (0,B) (4,C) (/,/) ] 2 | [ (2,A) (0,B) (3,A) (5,C) ] 3 | [ (2,A) (0,B) (3,A) (4,A) ] C and D are similar to create. Attached below is a reference from a textbook on distributed computing. Note that the "B" router in the example only contains adjacent routers. Destination is explicit in the table. "kostnad" translates to "cost". The example is slightly more complicated than the task above since cost for the routes are asymmetric.

It looks like we are reasoning along the same lines. Just approaching it different. My point is that due to the deep principles of mechanics it has to cancel. All other options are incorrect or requires new physics. So no calculation is needed. That said, it is interesting and maybe educating, I may have time to do some calculations later today.

I do not calculate. I look at the rig and intuitively it tells me "reaction less drive" or "issues with conservation of momentum". I asked about it earlier. That leaves two* options as far as I can see: 1: I have misunderstood the setup completely; the lifter either ejects mass or it interacts with external things that is not in the pictures. 2: the lifter is not following basic laws of physics and cannot work. The forces that we calculated have equal and opposite forces in the rig. If a magnetic field pushing at something then the magnet will be pushed back. *) Number 3, new physics beyond current mainstream, is not probable.

At this point it is not so much of a math problem but physics. I tried to ask earlier regarding conservation of momentum and how magnetic fields are created to nudge you in that direction. Now that the sketch and expected outcome is more complete we can analyse it further. Again, slightly more detailed: If the rig is stationary on ground its momentum is zero. When started there will be internal stresses in the rig due to the currents and magnetic fields; there will be internal forces. So momentum is conserved. The rig's momentum is zero. The rig will not move.

Here is the corrected last step using the matrices. At point A we apply 270 ccw degree rotation of the vector by using matrix [math]\begin{bmatrix}0&1\\-1&0\end{bmatrix}[/math] A 90 degree ccw rotation, is applied to the vector at point B and D [math]\begin{bmatrix}0&-1\\1&0\end{bmatrix} [/math] The resulting vectors "F" at each point A,B and D is: [math]A: F_{1}=\begin{bmatrix}0&1\\-1&0\end{bmatrix} \cdot \begin{bmatrix}2B_{4x}\\0\end{bmatrix} = \begin{bmatrix}0\\-2B_{4x}\end{bmatrix} [/math] [math]B: F_{2}=\begin{bmatrix}0&-1\\1&0\end{bmatrix} \cdot \begin{bmatrix}B_{4x}+B_{6x}\\-B_{4y}\end{bmatrix} =\begin{bmatrix}B_{4y}\\B_{4x}+B_{6x}\end{bmatrix} [/math] D:[math] F_{3}=\begin{bmatrix}0&-1\\1&0\end{bmatrix} \cdot \begin{bmatrix}B_{6x}+B_{4x}\\B_{4y}\end{bmatrix} =\begin{bmatrix}-B_{4y}\\B_{6x}+B_{4x}\end{bmatrix} [/math] The result is vectors F pointing in directions predicted by calculations in OP Ok. Is the expected outcome that the rig will lift from the ground and accelerate upwards? As drawn that seems not possible.

I'm unable to see "0-3" in the post, should the end result be four lines so that the table is created alliteratively, filling more tuples on each line? If so I probably posted the final row for "A" without the intermediate steps. Maybe I misinterpreted the requirement. I think the table only may contain the adjacent routers so I guess that only router C will contain all of A,B,C,D. There is for instance no direct route B-D so B must know that packages intended for D should be sent to C. Maybe there is more context you can post?

Hello. Usually I have seen routing tables having a destination sub-net, cost and next-hop. Since destination is not included in the tuples I guess that the destination is implicit from the position in the tuples have in the table. Here is an attempt. It shows how a packets, at router A, will be routed next to reach routers A,B,C or D [ (0,A) (2,B) (1,C) (1,C) ] (0,A) = done, nothing to do (2,B) = B is reached directly by routing to B, cost is 2 (1,C) = C is reached directly by routing to C, cost is 1 (1,C) = D is reached indirectly by first routing to C (next-hop) , cost is 1. Apply the above reasoning to the other three tables. (Disclaimer: last time i did some serious work on routing tables was in the 90's my memory regarding this is rusty and maybe also obsolete at this time)

I think I made an error in the rotation, applying the wrong matrix at A and B,D. I'll post an update once i get some time. It does not affect my conclusion, but it affects the signs in the final matrices.

Ok! Here are some quick calculations regarding the original question. NOTE: I’ll handle this as a mathematics problem only. I do intentionally not mention any electromagnetics since there is not enough information provided to know if this setup have any physical meaning or is physically possible. I’ll write out conclusions made from the notes above to give the whole picture, maybe unnecessary detailed. It’s a little tricky to read all the indexes in the original image but the intention is to keep the numbering and letters from other pictures posted so far. Let’s start with an equilateral triangle ABD. ABD is placed in a cartesian coordinate system xy. Initially there is no need for a z-axis. The corners are associated with vectors and the vectors have x and y components identified by subscript. Since ABD is equilateral and the vectors having identical length that lead to some symmetries and one zero component. [math]B_{4x}=B_{5x}\\B_{4y}=-B_{5y}\\B_{6y}=0[/math] At each point A,B,D, add* the pair of vectors associated with that point. Use symmetries to simplify: [math]A: B_{1}=\begin{bmatrix}B_{4x}\\B_{4y}\end{bmatrix} +\begin{bmatrix}B_{5x}\\B_{5y}\end{bmatrix}=\begin{bmatrix}B_{4x}+B_{5x}\\B_{4y}+B_{5y}\end{bmatrix} =\begin{bmatrix}2B_{4x}\\0\end{bmatrix} [/math] [math]B: B_{2}=\begin{bmatrix}B_{5x}\\B_{5y}\end{bmatrix}+\begin{bmatrix}B_{6x}\\B_{6y}\end{bmatrix}=\begin{bmatrix}B_{5x}+B_{6x}\\B_{5y}+B_{6y}\end{bmatrix} =\begin{bmatrix}B_{4x}+B_{6x}\\-B_{4y}\end{bmatrix} [/math] D:[math] B_{3}=\begin{bmatrix}B_{6x}\\B_{6y}\end{bmatrix}+\begin{bmatrix}B_{4x}\\B_{4y}\end{bmatrix}=\begin{bmatrix}B_{6x}+B_{4x}\\B_{6y}+B_{4y}\end{bmatrix} =\begin{bmatrix}B_{6x}+B_{4x}\\B_{4y}\end{bmatrix} [/math] These directions seems to match the hand-drawn picture by OP. To check how “force” vectors “F” would look like, we apply rotation** by 90 degrees ccw by using matrix [math]\begin{bmatrix}0&-1\\1&0\end{bmatrix} [/math] at point B and D. At point A we apply 270 degrees of rotation** to match original picture. [math]\begin{bmatrix}0&1\\-1&0\end{bmatrix} [/math] [math]A: F_{1}=\begin{bmatrix}0&-1\\1&0\end{bmatrix} \cdot \begin{bmatrix}2B_{4x}\\0\end{bmatrix} = \begin{bmatrix}0\\2B_{4x}\end{bmatrix} [/math] [math]B: F_{2}=\begin{bmatrix}0&1\\-1&0\end{bmatrix} \cdot \begin{bmatrix}B_{4x}+B_{6x}\\-B_{4y}\end{bmatrix} =\begin{bmatrix}-B_{4y}\\-B_{4x}-B_{6x}\end{bmatrix} [/math] [math]D : F_{3}=\begin{bmatrix}0&1\\-1&0\end{bmatrix} \cdot \begin{bmatrix}B_{6x}+B_{4x}\\B_{4y}\end{bmatrix} =\begin{bmatrix}B_{4y}\\-B_{6x}-B_{4x}\end{bmatrix} [/math] Result is vector F1 in y direction at A and vectors F2 and F3 pointing inwards into the triangle at points B and D. This seems to match the directions of "F" vectors in OP's picture. Due to symmetry the x-components at B and D cancels. So the total sum of F1,F2 and F3 is in y direction. Note again: this is a set of vector calculations. It could possibly represent some setup of magnetic fields, currents and forces. But so far the thread lacks the responses to address that possibility. *) If this would have been magnetic fields, they add as vectors. **) If there would have been currents, right hand rule tells vector directions.