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About darkenlighten

  • Birthday 03/06/1987

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  • Location
  • Interests
    ecocar.osu.edu and a list of other things.
  • College Major/Degree
    The Ohio State University: Major Engineering Physics
  • Favorite Area of Science
    Physics :)
  • Biography
    Graduated from OSU in 2011. Studied Engineering Physics, where I concentrated in Electrical Engineering and minored in Industrial/Visual Comm Design.
  • Occupation
    Hybrid Vehicle Battery Modeling and Testing


  • Baryon

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Baryon (4/13)



  1. I would look up spooler.exe fix because its most likely something wrong with the file. You should be able to find a legit download of it and replace it or if its missing, put it where it belongs.
  2. Okay so I'm gonna take a shot at this. I would like to critique/comment your design on just some initial thoughts: -I think the coils could be better wounded -Depending on how you are turning this "magnetic bearing", there are going to be rotational losses, if and when you get any induction into these solenoids, due to the back emf it will produce. -The magnetic field interaction from the bottom magnets, as you suggested, will effect the magnetic field the coils are experiencing. -With that you might want to get a very well wounded coil and try varying its orientation to see where it peaks, or if any current is flowing. Maybe try one with a core, though using a core will cause more rotational loss. Granted I'm no expert, but hopefully this will help.
  3. To complete the nomenclature, I think you are really comparing a raster image versus a graphic/vector image. So with that, the raster image is a pixel by pixel description, while a graphic/vector image is represented by mathematical vector equations of the paths used to create the object. Rasters: Great for producing complicated color changes, ie Photographs. Do not scale well and programs use interpolation methods to do so. Graphics/Vectors: Great for illustrations which consist of solid colors or gradients. Scale perfectly. Not good for photographs. This is more of a Visual Communication Design description, but is still valid in the computer science world. sorry for the week late response lol. hope this helps.
  4. Yea I messed that up didn't I haha. I'll admit its been a little bit since I studied quantum mechanics. My main motivation was the several threads previous to this one about the wave function. I made a mistake, but that is why I have guys. Nonetheless, the following of that is correct.
  5. In this case it means it is no longer a superposition of multiple possible outcomes, but now just the one known state.
  6. Okay I am here to help. It seems like a lot of people misunderstand what exactly the wave function is, so I want to give my educated understanding of this. The wave function itself is not physical, but it is mathematical, hence a function. So when someone says, the wave function collapsed, it does not mean something physically collapsed, but that the mathematical model collapsed. To say further, the wave function describes all possible states in which the system can be in. For example the photon traveling towards a double slit; now while the photon has a particle-wave duality, this is separate from its wave function, though the wave function might describe the physical property. The wave function mathematically describes what the possible states of the photon are, given the initial conditions. So when we observe the photon at say slit 1, we have collapsed the wave function, because now there is no longer a superposition (addition) of all possible states, but now just a single state, the state that we observed. Thus setting up another set of initial conditions to our wave function, changing the possible states from there on out.
  7. I'm actually creating a separate thread since this one is old: http://www.scienceforums.net/topic/55917-the-wave-function-what-exactly-is-it/
  8. I would like to stress, it is NOT the act of observation or awareness that an event is happening, but the interaction that is needed to observe. Fundamentally we cannot observe an event without interacting with it and changing its current state.
  9. I personally like the actual definition of productive: http://dictionary.reference.com/browse/productive Sorry it doesn't really allow you to make your point, but hey why not make up our own definitions to meet a specific agenda. As you know, many things exists today that would not be here without science, especially quantum theory and yes they are productive. Now if you would like to argue about your statement of what you think productive is, then go ahead, it won't lead anywhere. If you think scientists are doing it wrong, what is your solution?
  10. I have an idea...let's change definitions so that it seems like what we are stating are contradicting natural thought, but it is actually just redefining the word so that it seems to make sense, though it's just the same thing twisted around.
  11. The simple, not too exciting answer is that a magnetic field is produced when you have a moving charge, aka a current. I'm not sure it can be explained any deeper that that, at least in simplicity.
  12. This would not be practical. The magnetic field needed in order to pick something up from 5-10 feet away would be too powerful, most likely, in the area which you would be wanting to use this. Obviously you could not isolate these small objects and the smaller the object, the less force it will feel from the magnet (which also needs to counter its mass).
  13. An electric motor, is essentially a solenoid , but this would be trick to create an electrical input that give a mechanical out put with out wrapping wire in some fashion, which is solenoids are more expensive than other inductor type of components, hence why they are usually avoided if possible in applications, where they are not the only option. With that in mind, this is not a trivial task, finding something that can cause a mechanical output not using electromagnets will be very tricky.
  14. I dont think that integration is good logic. What I would do if you really want to put an integral in there would be to find the area using a double integral [math] A = \int_{0}^{2\pi} \int_{0}^{\pi} r^2 sin(\theta) d\theta d\phi [/math] And then realizing that [math] r = L [/math] and solving the rest of the equation as stated before. Though doing this integral is really trivial, since the area of a sphere is already derived. But I used: [math] R = (\frac{1.7\times 10^{-8}}{4\pi*[(1\times 10^{-2}) - (1 \times 10^{-6})]}) [/math]
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