Jump to content

darkenlighten

Senior Members
  • Posts

    158
  • Joined

  • Last visited

Posts posted by darkenlighten

  1. 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.

  2. 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.

  3. 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.

  4. 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.

    The wave function is a description of the state, the QM equivalent of "particle is at x and has a momentum p". It is the Schroedinger equation that determines the time evolution of this state (and thus the time evolution of the wave 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.

  5. 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?

  6. 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).

  7. 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.

  8. 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]

  9. what type of software are you using?

     

    If you are not using software, this still might be good:

     

    http://host.nigde.edu.tr/sayasun/documents/dc-motor-published.pdf

     

    But depending what this is for, what you are trying to do doesn't seem broad, but you make it sound like it is. Unless you are just trying to model a system that experience some type of input then goes through a damping, and has a feedback and output? Thats the most general I can think of, is feedback control.

     

    Which for feedback control, this looks good: http://www.google.com/url?sa=t&source=web&cd=1&ved=0CBwQFjAA&url=http%3A%2F%2Fciteseerx.ist.psu.edu%2Fviewdoc%2Fdownload%3Fdoi%3D10.1.1.129.1850%26rep%3Drep1%26type%3Dpdf&rct=j&q=feedback%20control%20theory&ei=Hdq8TJeCF8-aOr2u8c8M&usg=AFQjCNGttvlPjx1pUNhkHIHAVDGeZGSwvA&sig2=jbgjae6IxnDFoRWXjlsukQ&cad=rja, it opens straight to a PDF, but is legit.

  10. There are multiple ways in doing this. If you are trying to make a robot separate from it being connected to a computer, you would need to use a micro-controller of some type. Essentially you would write code for that specific controller (that has computational limits) and the controller would be connected to your actuators and sensors, which are the majority of what you would be controlling.

     

    It's not an easy task, imo. At least for doing a cheap route. You could use prototyping tools (most likely expensive), such as Simulink from Matlab (which could be a school license) and a controller from Motohawk or something like that. Or you could use lower level micro-controller, from TI that essentially uses assembler to do the functions you would require.

     

    For example, we use a Motohawk controller for our engine in our hybrid vehicle for the OSU EcoCAR team and a dSpace micro-autobox for our supervisory controller (the one the controls everything the vehicle does).

     

    The reason I mention this, is because you could essentially use this type of system to make a robot, same idea, different application.

     

    Or once again use a micro-controller that is more of a micro-processor, depending on what you want to do.

  11. Both of you are probably correct, so you wanna check your units. I converted all length units to meters, which gave me: [math] 1.353\times10^{-7} [/math].

     

    I think it is easier to go with this logic (since you have pretty much came to the conclusion). First realize what L is: [math] 1cm - 1 \mu m [/math] and then find the area. Since it's a hollow sphere, its area changes with distance, r, but r is just the length so [math] R = \rho*(\frac{L}{4*\pi*r^2}) [/math] and then since [math] r = L [/math] [math] \Rightarrow R = \rho*(\frac{1}{4*\pi*L}) [/math]

  12. I think saying the density of photons increases is misleading or just incorrect. The intensity increases, which is possibly what you guys mean. In a particle model the energy of each photon stays the same, but the amount coming in increases. In a wave model, the amplitude increases (constructive interference) and the frequency stays the same. That's the way I see it.

  13. my advice. learn it, it's easy. Scientific notation is just a short hand for showing how many zero's or places are in the entire number. For example [math] 9.0 \times 10^{16} [/math] is just [math] 9 000 000 000 000 000 0. 0 [/math]. So as you can see you just "move" the decimal point over 16 positions to the right. Or if it's negative it would be to the left. Another example for clarity [math] 10.5440 \times 10^3 [/math] which is [math] 10544. 0 [/math]

     

     

    Note: Google does display as many numbers as needed, displaying 16 zeros is redundant and who wants to count out 16 zeros anyways.

  14. Yea my point was that the flow was going to be minimal, especially with just a straight tube like that. The over all idea might do something if you use solar heating, but I doubt it'll be great. Though this might be the point of your project, try something, go through the design and analysis and see what happens.

  15.  

    This is a cheesy website but the idea is the same everywhere and it explains it in a simple fashion.

     

    Einstein believed that to give a single electron this energy to move, a single photon hit the metal surface (destroying itself), and transferred its energy to the one electron.

     

    Since the electron is attracted to the surface of the metal, some minimum amount of energy must be needed just to snap it off. Otherwise, electrons would just be dropping off of atoms all the time.

     

    Einstein called this the work function of the material, since you needed to do work on the electron to break it off.

     

    http://www.studyphysics.ca/30/Unit4/Light/Photoelectric/note.htm

     

    OK, we now know Einstein proposed a work function. (You can find other sources)

     

    Once we have introduced a work function for a photon, we must assume it operates this was across frequencies.

     

    I think this is where you start to mix things up. The work function is not a property of a photon, but a property of the material.

     

    And I guess you can say work is being done.

     

     

    Yea, it is a law.

     

    It's a property. It would be like saying Energy is a law, it itself is not a "law". But say the Conservation of Energy is a law.

     

    Because, one way or the other, infrared photons are delivered to wayter to turn it into steam. By the conservation of energy, we must assume the photons are doing work or transfering their energy to the water. But, energy transfer causes the stream or water molecules to move more rapidly. Now, why do they move more rapidly without work being done?

     

    No that's not necessarily how it always works. Once again, infrared does not equal heat, it can cause things to heat up by adding energy to the system. Thermal radiation can be caused by the whole EM spectrum, just not infrared.

     

    Wikipedia Article: Infrared: Heat

     

     

    This does not offend me and I appreciate your input.

     

    That's good.

  16. It is not because the current theory is wrong, it is because you current understanding is wrong.

     

    This is not how Einstein propsed the idesa. He did not make it materially based, though clearly it is.

     

    Please explain what you mean, because this does not make any sense. I'm not familiar what you are referring to, what did Einstein propose then?

     

     

    OK, let's assume, if the material is photo electric (perfect) then it obeys Einstein's work function.

     

    This does not make sense either, the work function is not a law or something like that, it's just the minimal ENERGY required to kick an electron out of a material.

     

     

    Otherwise, based on the material, if will obey it in a lesser way.

     

    I think this is what you are saying.

     

    Either way though, you must agree the work performed to eject an electron is less than that performed by the infrared photon otherwise, the steam engine would not exist.

     

    I'm confused on why you are referring to a steam engine, infrared is not directly heat, it can cause things to heat up, but is not itself heat. Heat is simply how fast atoms or molecules are moving or say how much energy it has.

     

     

    Yes, I will tell you my agenda.

     

    Something is wrong and I am trying to get ideas about the behavior of light.

     

    I do not have a theory as I do not have near enough information.

     

    But, there is something wrong with the current theory.

     

     

     

    Can you explain this in terms of the steam engine?

     

    This is the clearest experiment where the infrared photon does more work that the UV.

     

    Once again, your view is not correct, but I want to help you to understand why and put you on the right path.

×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.