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  • COP test.

    There's a photograph of a DPDT blade switch in comment 160 above. Doubling the electrodes on the bifilar coil would allow 2 DMM's to attach to one side of the blade switch; One in series for amperage and the other in parallel for voltage. This would permit a quick and dirty COP measurement on the flip flop. I also received several timer switchs in the mail recently that will enable me to precisely channel exactly one second of pulse power to the neutralization coil.

    An accurate pre-calculation would help confirm the test measurements. These results will finish my contribution to this thread. There's a solid chance we may be able to confirm with a high degree of accuracy the existence of an OU COP coupled with a solid theory.
    Last edited by Allen Burgess; 10-17-2017, 10:30 PM.

    Comment


    • Electromagnetic equation

      Quote from smOky2,

      "The way you would need to go about this, is to use the electromagnetic equation, with the dimensions of your coil and by the current (Amps) you can determine how many Gauss the coil would produce.

      looking at it like that, ( in Teslas, which is 10^4 Gauss)

      T = 4(Pi)(10^-7)*(number of turns ^2)*(Amps)/ (Length)(Area^2)"

      This equation is for an air core coil.

      Look how simple this is; The coil dimensions factor into units of inductance, so the answer can be solved for simply by multiplying Henries times Amps: 75.5 mH x 1500 mA = 113250 or 113.25 Gauss.
      Last edited by Allen Burgess; 10-18-2017, 01:19 PM.

      Comment


      • Help verify?

        Originally posted by Allen Burgess View Post
        ...
        We may need member bistander to help verify any results!
        ...
        Originally posted by Allen Burgess View Post
        ...
        Look how simple this is; The coil dimensions factor into units of inductance, so the answer can be solved for simply by multiplying Henries times Amps: 75.5 mH x 1500 mA = 113250 or 113.25 Gauss.
        Wrong. The product of inductance (Henries) times current (Amperes) results in flux (Webers), not flux density (Webers per square meter or Tesla or Gauss). Neither are units of force (Newtons) or power (Watts).

        bi

        Comment


        • Inductance formula.

          @bistander,

          Thanks for responding to my inquiry:

          I initially stated that an Inductor of 1 Henry of inductance would generate 1 Tesla of magnetic force with an electrical input of 1 Watt/hour.

          1 Watt hour is equal to 3600 Joules.

          1 Ampere is equal to 1 coloumb per second.

          1 Coloumb is equal to 1 Joule/ volt.

          Therefore a magnetic force of 1 Tesla would require a current of 3600 Amperes in an Inductor of 1 Henry of Inductance, right?

          a Joule is equal to .7377 foot pounds, so 1 Tesla of magnetic force should lift an SUV of 4880 pounds. Correct me if I'm wrong.

          What do you think the magnetic force in Gauss is in the bifilar coil of 75.5 mH
          at 1500 mA?

          How do you feel we should go about converting Webers of flux to Webers of flux density?

          1 Weber of Flux is equal to 100,000,000 gauss per square centimeter right? What fraction of a Weber of Flux is the product 113250 of the Inductance times the Amps?

          Let's say I attach a spring scale to a piece of iron and see how much weight it measures before the coil pulls away from it; Do you feel this would produce a more accurate measure of magnetic attraction strength?
          Last edited by Allen Burgess; 10-18-2017, 08:39 PM.

          Comment


          • Ferrite core.

            Nevertheless, magnetic strength formula aside; I believe the bifilar coil is understrength and have ordered Iron powder and Magnetite to cold mold a high perm ferrite core for the bifilar air core coil.

            @bistander,

            How come smOky2's formula in comment #167 above has a T instead of a W at the beginning of his equation?
            Last edited by Allen Burgess; 10-19-2017, 12:46 AM.

            Comment


            • Electromagnet formula.

              It appears member bistander's correct:

              I see the Magnetic Constant (4 x PI x 10-7.) in this formula I failed to include in my calculation:

              "Calculate the force by writing the equation:

              F = (n x i)2 x magnetic constant x a / (2 x g2)

              Where, F = force, i = current, g = length of the gap between the solenoid and a piece of metal, a = Area, n = number of turns in the solenoid, and the magnetic constant = 4 x PI x 10-7.

              Analyze your electromagnet to determine its dimensions and the amount of current you will be running through it. For example, imagine you have a magnet with 1,000 turns and a cross-sectional area of 0.5 meters that you will operate with 10 amperes of current, 1.5 meters from a piece of metal. Therefore:

              N = 1,000, I = 10, A = 0.5 meters, g = 1.5 m

              Plug the numbers into the equation to compute the force that will act on the piece of metal.

              Force = ((1,000 x 10)2 x 4 x pi x 10-7 x 0.5) / (2 x 1.52) = 14 Newtons (N)".
              Last edited by Allen Burgess; 10-19-2017, 12:53 AM.

              Comment


              • Nonsense

                Originally posted by Allen Burgess View Post
                @bistander,

                Thanks for responding to my inquiry:

                I initially stated that an Inductor of 1 Henry of inductance would generate 1 Tesla of magnetic force with an electrical input of 1 Watt/hour.

                1 Watt hour is equal to 3600 Joules.

                1 Ampere is equal to 1 coloumb per second.

                1 Coloumb is equal to 1 Joule/ volt.

                Therefore a magnetic force of 1 Tesla would require a current of 3600 Amperes in an Inductor of 1 Henry of Inductance, right?

                a Joule is equal to .7377 foot pounds, so 1 Tesla of magnetic force should lift an SUV of 4880 pounds. Correct me if I'm wrong.

                What do you think the magnetic force in Gauss is in the bifilar coil of 75.5 mH
                at 1500 mA?

                How do you feel we should go about converting Webers of flux to Webers of flux density?

                1 Weber of Flux is equal to 100,000,000 gauss per square centimeter right? What fraction of a Weber of Flux is the product 113250 of the Inductance times the Amps?

                Let's say I attach a spring scale to a piece of iron and see how much weight it measures before the coil pulls away from it; Do you feel this would produce a more accurate measure of magnetic attraction strength?
                1 Watt hour is equal to 3600 Joules.

                1 Ampere is equal to 1 coloumb per second.

                1 Coloumb is equal to 1 Joule/ volt.
                These 3 equivalencies are correct except Coulomb is misspelled. The rest of your post is rubbish. And why ask me those questions when you don't believe my answers anyway and treat me with disrespect?

                Comment


                • Respect

                  @bistander,

                  You failed to calculate the correct answer!

                  Watch this:

                  https://www.youtube.com/watch?v=wTxh6zHXxIw
                  Last edited by Allen Burgess; 10-19-2017, 01:30 AM.

                  Comment


                  • What?

                    Originally posted by Allen Burgess View Post
                    @bistander,

                    You failed to calculate the correct answer!

                    Watch this:

                    https://www.youtube.com/watch?v=wTxh6zHXxIw
                    I wasn't calculating anything. I don't know what I'm doing here. Good bye.

                    Comment


                    • @bistander,

                      Henries X Amperes X Mu Naught = Magnetic Force in Newtons?

                      Did I supply enough data for you to calculate the correct answer for us, or are you going to ridicule me for another incorrect answer?

                      The missing factor is an approximation of the permeability of an air core. The Mu naught factor is practically infinitesimal, so the answer of 113 Gauss has to be very close to correct, right?

                      Stay out of here you turd.
                      Last edited by Allen Burgess; 10-19-2017, 02:10 AM.

                      Comment


                      • Bistander's wrong.

                        I realized that bistander is wrong as usual. The inductance measurement includes the permeability of the air core, so the Mu naught factor is irrelevant.

                        3600 Joules (1 Watt/Hour) of input, to a coil of 1 Henry of Inductance, generates 1 Tesla of Magnetic Force; As I maintained from the outset. It's that simple!
                        Last edited by Allen Burgess; 10-19-2017, 02:37 AM.

                        Comment


                        • Test confirmation.

                          I measured my 12 volt electro-magnet for Inductance, and was very surprised to find that it was 73.5 Henrys; very close to the Inductance of the air core bifilar. Armed with this coincidental equality, I moved forward with a comparison test:

                          I have a number of ceramic grade "1" block magnets from Harbor Freight, 3/8" X 7/8" x 1-7/8". I calculated the Gauss on the online "Centerline Calculator for a Rectangle Ceramic Magnet" and determined that the Gauss of the ceramic block at the centerline, at a distance of 2" was 97.56.

                          I attached the ceramic block magnet 2" behind the electro-magnet and it stuck to the refrigerator. I then applied 1500 mA of D.C. current to the electro-magnet and it neutralized the attraction strength completely, but not to the inside.

                          This test confirms the Gauss calculation I performed for the air core bifilar as perfectly correct at 113. Gauss is determined very simply by multiplying Amps times Henrys of Inductance; Now that's final; Case closed!
                          Last edited by Allen Burgess; 10-20-2017, 04:25 PM.

                          Comment


                          • Simplicity of neutralization.

                            Amperage can be measured by placing the DMM meter electrodes in series between the positive of the power source and the positive lead of the inductor. Make sure the input does not exceed the capacity of the meter fuse. My VICI has a 20 Amp fuse. The other factor is the inductance of the inductor in Henrys.

                            These two factors multiplied together will produce the magnetic force in Gauss of the electro-magnetic coil.

                            Next, consult an online calculator and determine the Gauss of the permanent magnet on the centerline. The calculator will give you the exact size of the spacer you need to equal the Gauss of the inductor. These simple measurements can save anyone a lot of time and pinched fingers over the trial and error approach.

                            Once the magnet neutralization coil is assembled, your'e next step would be to fashion an overhead oscillator portion. This is much simpler. The contact would consist of two wires to short the power source through the pulse coil.

                            Working with a battery, this kind of oscillator can reverse charge it's own source by "Super Position". The regauging of the permanent magnets generates a higher voltage then the power source, and the power flows backwards. Try it, it's a lot of fun and very easy to do do with the correct math.
                            Last edited by Allen Burgess; 10-22-2017, 03:45 PM.

                            Comment


                            • Ferrite core.

                              Good news! The 100g of "Atomised Iron Powder" and 100g of "Natural Magnetite Powder" arrived way ahead of schedule. I purchased a twin injector pack of metallic epoxy bonder and am preparing to mix everything in an old Tupperware dish to pack my bifilar coil with a high perm ferrite core.

                              I plan to try the solid state configuration with the latching Reed switch commutator once the core sets up.

                              Comment


                              • Bifilar coil and High Perm Ferrite core inductance

                                The finished coil and high perm ferrite core measures .121 Henrys. The air core was 73 mH.

                                I emptied the two bags of equal weight Iron powder and Magnatite powder in the large section of a three section Tupperware tray and blended them; I then emptied the entire two tubes of metal epoxy (Locktite) in a second tray and mixed them in a third, then packed the dry slurry into the coil core. Everything came out perfect. I then threw everything in the trash including two metal dining spoons.

                                The increase in inductance equals 48 mH. That's a 66% or a 2/3 increase in inductance. That equals a lot of wire!

                                My next step is to calculate the Gauss of my Neo ring disc magnets and determine how much amperage I need to neutralize the magnetic strength with my bifilar inductor of .121 H.
                                Last edited by Allen Burgess; 10-27-2017, 11:15 PM.

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