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Everything posted by lesolee

  1. For 100 years we have been told that the reals form a larger infinity than the counting numbers, and that the reals are in fact "uncountable". Here is the pattern required to enumerate the reals (that is to "prove" that they are countable). continuum.pdf I don't for a moment suggest that Cantor's method is at all sensible or correct. In fact it is the most unmathematical hand-waving you could imagine. Commercial link removed by moderator
  2. I have quite a different take on counting the Cantor set. LINK DELETED
  3. Instead of using Cantor's (bogus) counting methods you could just count the rationals more directly. density.pdf
  4. Thank you. A very understandable answer. My concern is that the text says ""(it is possible to design a detector that can tell whether a photon went through it)". Now we know from classical experiments that it is necessary to "split the incoming wavefront" by using an earlier slit. This ensures good spatial coherence to both later slits. If we "regenerate" the photons by down conversion there seems no reason to suppose the resultant beams would be coherent, and if not coherent then no interference would be seen - as indeed is the case. The text's interpretation of this experiment then seems quite bogus ... unless there is a better "special detector" of course.
  5. I have been reading a book on QED by Richard Feynman (The Strange Theory of Light and Matter). In it he describes what appears to be a real dual slit experiment using a low intensity source into a single photon detector. I have found avalanche detectors and photomultipliers and that is all well and good. Where I have a problem is when he puts a detector near each slit to see if the photon went that way, saying that the interference effect goes away when you "look" like this. My problem is that he seems to be describing a real experiment but I can't understand how a detector can detect a photon as it passes by. The single photon detectors I have seen "consume" the photon in order to detect its presence. I am not looking for an explanation of the single photon dual slit experiment, what I would like to understand is what sort of apparatus is being used to perform the real world sensing of the photon path.
  6. Not much as it turns out, pre-supposing you mean AC RMS. The difference is that standard deviation divides by (N-1) and the RMS calculation divides by N, where N is number of samples. When N is above 100 the difference is going to be unimportant.
  7. I have heard of people writing "physics engines" so that they get realistic simulations of events such as dropping a bag of marbles or realistic games simulations. That is a lot different to simulating "particles". The macroscopic (big) world is very predictable using Newtonian physics. The microscopic world of particle physics is not so easy since the interactions are harder to formulate, although Prof Richard Feynmann said that QED is the most accurately tested and verified theory around.
  8. Old designs of magnets certainly did lose their magnetic properties over time. There was (and still is) something called a "keeper" to maintain the strength of the magnet. You don't "use up" the magnet by letting the magnetic flux increase, in fact the opposite is the case. http://www.coolmagnetman.com/magtypes.htm As for all the vitual photon interactions, that's all way beyond a humble electronics designer.
  9. Following on from swantsont's response, let's look at the acceleration. Remember your maths from when you were 16 years old? SUVAT S=ut +(1/2)a*t*t S= distance travelled t= time taken a= acceleration We start from stationary so u=0 S= (1/2)a*t*t a= 2S/(t*t) Accelerate 0.6m in 2 seconds. I get 0.3m/s/s The required acceleration is so much lower than the gravitational acceleration you can pretty much ignore it in this case. Of course re-reading the post title you want "constant velocity". Well you might want to moderate this and allow plenty of time to accelerate and decelerate, otherwise the inertial load will require infinite force!
  10. This is a very good question. The answer is that the computer is designed so that doesn't happen! In the first place what you do in a good design is to run the "go" and "return" wires next to each other. In a DC circuit the positive and negative conductors are run next to each other. The magnetic field at a distance from this pair is very small. The field would be even less if the two conductors are twisted together, a "twisted pair". I would expect the hard drive itself to have magnetic shielding. An iron, mu-metal or radiometal cover would give good magnetic shielding. (I don't know the specifics of what is actually done.)
  11. I don't see how anybody can help you. You just are not saying anything ... yet. "Running these simulations for months". Why? What are you trying to achieve? What is this simulation software? The attached picture was very pretty, but also pretty meaningless without a "key". In general terms a particle will carry on at its initial velocity (speed and direction) unless an external force is applied. So all the tracks should be straight lines. They are curved. A curved track implies a force, in this case possibly a magnetic force. But if you are running a simulation you should already know what you typed in to it, so why are you not saying? Notice that some curve one way and some curve the other. That means the particles have opposite charges. A different radius of curvature means a different mass or a different velocity and you can't tell which from the radius of curvature. If the radius of curvature gets bigger smoothly that suggests the particle is slowing down. The finite and increasing width of the tracks is strange. If this is a sensible program I can only guess that it is showing an uncertainty band around the particle's track.
  12. "Static Electricity" is not really any different to any other sort of electricity. The name is historic and relates to how the electricity was generated. These Van der graaf generators and Whimshurst machines are merely creating high voltage electricity. (Note that I have given you (bold) names that you can Google for pictures and more info.) A Van der Graaf generator, for example, generates high voltage (DC) by friction and stores it on a big globe or sphere. It is really just a big capacitor. You could, theoretically, store this electrical energy in a battery. The problem is one of scale. Consider a car battery. These are lead-acid rechargeable batteries. To get 12V you have 6 lots of 2V cells wired together. A Van der Graaf generator might be running at 30,000V. You would need 15,000 cells to hold that much voltage. Not very convenient.
  13. There is a fuller explanation here ... http://van.physics.i...ting.php?id=414
  14. So we have a parabola, Y= x*x The slope of the parabola at any point is what? (A) Having found the slope of the parabola you now want the equation of a line passing through that point but with a slope at 90 degrees to the parabola. (B) Do you remember the standard equation of a straight line? © Think about the above questions before peeking ... _________________ Peeking ......... A) if y = x*x then dy/dx = 2x (elementary calculus) B) The tangent slope is dy/dx and the normal to the tangent (the perpendicular) is -dx/dy. c) Y= mx + c
  15. I am not sure why you think it is ok to spam the forum like this. This is the third post on essentially the same subject, the last two posts being identical. Since you are now using a fractional representation we can consider the number as being scaled by x4. 7 x 4 = 28 --> 011100 using 6 bit signed binary -7.25 x 4 = -29 -> 100011 using 6 bit signed binary Add to give -1 -> 111111 using 6 bit signed binary It would be much easier to use standard word sizes, 8/16/32 etc, as then you can just use a binary calculator to do the working.
  16. What has gone wrong in your calculation is an overflow. You can't represent +128 in a signed 8 bit value. The number you wrote (10000000) is in fact -128 not +128. All the rest should make sense when you see this.
  17. Given E=h.f a zero frequency photon should have no energy This is clearly a difficult area.
  18. If a (permanent) bar magnet were continuously emitting photons it would be an energy source and would "run down" (be depleted) over time. Of course radioactive materials like glow-in-the-dark paint mixtures do run down eventually but that is a different mechanism. It would be an interesting question to ask what the frequency or wavelength of the virtual photons mentioned by swansont are. That is somewhat above my level. At my level there is simply a magnetic field storing the energy statically, not dynamically. I imagine the "virtual photons" are not able to be observed and have been postulated to exist to suit a theory.
  19. Well ordinarily that wouldn't be an issue, but as I don't know the details of these devices I have to assume that when you say "different voltages" you mean something significant like 2V across one and 3V across another. Obviously the more data I have the more I can help you. This next idea is at increased cost/complexity but improved current matching. Obviously you can get better matching accuracy by using a higher base drive voltage and you could add temperature compensation either using a diode or a transistor base-emitter junction or wrapping an op-amp around one of the drivers. If any of this needs more explanation then feel free to ask.
  20. The usual simple way of driving an LED at "constant current" is to just put a resistor in series with it to a power rail. So if your LED has a nominal volt drop of 2V at 1mA you might power it from a 5V rail and drop 3V across the resistor at 1mA (3000 ohm resistor). The volt drop across the LED is going to be pretty stable so the current will also be stable. There is no need to get complicated about this unless there is some unusual circumstance you haven't mentioned. Obviously the higher the power supply voltage you use, the more stable and known the current will be. One resistor per LED will give a pretty cheap solution and then you can use an adjustable power rail to change all the currents at once.
  21. In telescopes and microscopes you use an eyepiece. This ideally produces a virtual image at infinity so your eye is relaxed (unaccommodated). John's answer of a magnifying glass is in the right direction but will probably not have enough strength (short enough focal length). This wiki article explains it nicely. magnifying glass
  22. Well I'm not too strong on optics either, but am willing to try You use a Fresnel lens as a cheaper way of getting a large lens. Rather than have all the material in the middle of the lens you just use the curvature that you would get at various points on the lens. Wikipedia When you say you want to focus on an image at 2 inches from the eye I am reading that as wanting to see an object 2 inches from the eye. In this case it is too close to focus on as the rays are diverging too much. You therefore need a converging lens with a short focal length. Hopefully the attached ray diagram makes sense. (FP = focal point) My free-hand lens shape didn't quite work out as well as I might have hoped
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