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lesolee

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Posts 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. 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.

  3. 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.

  4. 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.

  5. 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.

  6. given nothing interacting with a magnet, will it never lose its magnetism? i thought it would as energy has been put into the magnet to align the particles to make them magnetic in the first place, and over time these will revert back to random alignments and thus the field will diminish, sort of like radioactive decay.

     

    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.

  7. I want to be able to calculate the force required by a hydraulic cylinder to lift a 3500kg load vertically to a height of 1.2 meters in a time of approximately 3-4 seconds. Can someone help. All comments will be hugely appreciated.

     

    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 :doh:

    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!

  8. My question is this:

     

    If a current running through a wire produces a magnetic field, why does current in computer hardware not strip data from the HDD?

    Is the magnetic field produced simply not strong enough? I'm not too sure of standard amperage in computer parts.

     

    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.)

  9. Hi, Ive been running these particle simulations the last few months, and I dont know what exactly these fields are. If you know, or you know someone that could know... thatd be great! It would be nice to know the mathematical formulas for these things :)

     

    ... Basicallz I have a reference point and particles which have a direction and velocity. the distance the particle travels is proportional to the distance to the reference point.

     

    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.

  10. can we store static electricity in any form like with the help of rechargeable batteries, if possible pl let me know how?

     

    "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.:eek:

  11. Hello! Since I haven't solved any examples using functions in a while, it's hard for me to solve this example. It would be really nice if someone could help me in some way and explain how i works.

     

    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

  12. Convert 7 and -7.25 to twos complement with 6 bits and add them.So , -7.25 is smth like 1001,01. When I add 000111 to it,the result is 1000,01...and if i convert this to twos complement the result is not -0.25...whats wrong?

     

    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.:mad:

     

    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.

  13. Convert -128 ( decimal) to binary,representing the number with Two's complement with 8 bits...when I convert it to binary it is 10000000..so it is already 8 bits, if I put the bit of sign then it is 9 bits...110000000

    I also have to convert 115,375 (octal) to binary representing the number with 16 bits,but when I convert it I have 001001101,011111101..which is far more than 16 bits..where am I wrong?

     

    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.

  14. The frequency is zero, and the virtual photons are the field, whose force you feel when holding two magnets.

     

    Given E=h.f

    a zero frequency photon should have no energy :unsure:

    This is clearly a difficult area.

  15. what is the energy of a photon coming out of a standard everyday bar magnet?

    what end of the electromagnetic spectrum will it be in?

     

    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.

  16. Well it is a bit unusual - as we're using OLEDs, devices from different areas of the glass substrate have slightly different thicknesses and operate at different voltages, however the current which flows at a given brightness is always the same. Hence if we have a fixed voltage supply of say 3V, then all the devices will be a different brightness, but if we have a fixed current supply of 0.2mA, then they will all have the same brightness. That's the reason that I am using the circuit above. The problem is that sometimes I want to run a load of devices at 0.2mA and other times 0.3mA and so on, and trimming every device manually would be an utter pain.

    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.

     

    post-81967-0-21242800-1354131055_thumb.jpg

    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.

  17. 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.

  18. I think it has something to do with focusing the image at infinity, so the viewer doesn't have to constantly break his or her focus on the things around them to focus on the image.

     

    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

  19. I am not too strong when it comes to optics / lenses, but I am trying to figure out a way to get a person to be able to focus on something substantially closer than his or her point of focus. I understand this technology has been around for centuries (eyeglasses), but I am looking for a way to be able to focus on an image at a distance of about 1-2 inches from the eye. I think it has something to do with focusing the image at infinity, so the viewer doesn't have to constantly break his or her focus on the things around them to focus on the image. I think I may be able to do this by collimating the light using a fresnel lens, but I am still not sure.

     

    Can anybody offer advice? I'd love to discuss this.

     

    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 :(

     

    post-81967-0-94027600-1354043719_thumb.jpg

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