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Mostly Permanent Magnet Motor with minimal Input Power

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  • #16
    Originally posted by Peter Lindemann View Post
    Luc,

    Very interesting idea. With all of the motor ideas I have seen, I have never seen anything like this, so I'd say it looks like a Luc Original! I'd like to have more specific data on its performance, including volts, amps, and duty-cycle of the input, as well as the actual weight and stroke of the coil. Here's why:

    One "horse-power" is generally defined as 550 foot-pounds of work per second, which when you think of it, is quite a lot. If your coil weighs one pound and is moving back and forth one foot, once every second and a half, you are producing about 1/825th of a horse-power, which is about 0.90 watts in electrical equivalent. Your claim that the system is drawing about 0.25 watts is "in the ballpark" for what we are seeing in the movement.

    You can see why getting accurate information will help us determine its operational efficiency.

    Keep up the great work!!

    Peter
    Hi Peter,

    I have an update on the weight of the coil. I used my coil program to figure it out.

    The coil is .9 pounds or 408 grams. Less then I thought but very close to one pound.

    I'll try to get a scale to double check it also but I feel confident of that results.

    I believe I have now answered all your questions. Please let me know if you can answer mine or if you have any more questions.

    Thanks

    Luc

    Comment


    • #17
      Nice idea .

      You mention that it is north opposing north configuration. Would north opposing south would have bigger power or may not work at all?

      Comment


      • #18
        Originally posted by gotoluc View Post
        Hi Gyula,

        thanks for your interest

        The 1" inch square bar is just cheap steel bar.

        Is there less eddy currents in silicon steel laminations? I'm still learning some of this stuff.

        Luc
        Hi Luc,

        The answer to your question is yes, that is why transformers are made from laminated core. See here: File:Lamination eddy currents.svg - Wikimedia Commons

        Quote from another site:

        "Eddy currents are closed loops of induced current circulating in planes perpendicular to the magnetic flux. They normally travel parallel to the coil's winding and flow is limited to the area of the inducing magnetic field. "

        This is why using laminations, to chop up the bulk of the core, so reducing the thickness of the individual core layers.

        The quote is from here:
        Depth of Penetration & Current Density

        I guess however that your friction loss is higher than the eddy loss at this 1.5Hz frequency (this is why I mentioned linear bearing).

        EDIT: here is formula for eddy current loss, it turns out the loss depends on material thickness and frequency, both raised to the second power in the nominator. So at 1.5Hz it may be considered as negligible.

        Eddy current: Definition from Answers.com see formula under 'Strength of eddy currents'.

        regards, Gyula
        Last edited by gyula; 12-08-2009, 09:42 AM. Reason: addition

        Comment


        • #19
          Originally posted by sucahyo View Post
          Nice idea .

          You mention that it is north opposing north configuration. Would north opposing south would have bigger power or may not work at all?
          Hi sucahyo,

          If it was in north attracting to south the coil would stay in the center and not move at all.

          Luc

          Comment


          • #20
            Originally posted by gyula View Post
            Hi Luc,

            The answer to your question is yes, that is why transformers are made from laminated core. See here: File:Lamination eddy currents.svg - Wikimedia Commons

            Quote from another site:

            "Eddy currents are closed loops of induced current circulating in planes perpendicular to the magnetic flux. They normally travel parallel to the coil's winding and flow is limited to the area of the inducing magnetic field. "

            This is why using laminations, to chop up the bulk of the core, so reducing the thickness of the individual core layers.

            The quote is from here:
            Depth of Penetration & Current Density

            I guess however that your friction loss is higher than the eddy loss at this 1.5Hz frequency (this is why I mentioned linear bearing).

            EDIT: here is formula for eddy current loss, it turns out the loss depends on material thickness and frequency, both raised to the second power in the nominator. So at 1.5Hz it may be considered as negligible.

            Eddy current: Definition from Answers.com see formula under 'Strength of eddy currents'.

            regards, Gyula
            Thanks for all the information Gyula.

            I must say that I had thought of eddy currents but just like you concluded it may not have much of an effect at this low of a frequency.

            However, if we raise the frequency then we would need to consider using laminations.

            I would guess if laminations are used they would need to be going across the coil winding and not parallel to the windings? I'm not positive on this one.

            Thanks for sharing.

            Luc

            Comment


            • #21
              Hi All
              I like it,and I tried a simple copy,but with the a moving core,same thing I think,just using a sig.gen.not much energy used I would think,but it went like the clappers! Maybe it's a case of 1 mag. is 1 mag.2 mags. is 4 mags.I read somewhere.
              Dr Peter L. refered to 550lb per foot sec.but do's that not refere to vertical movement,and not hoeizontal? or have I got it wrong?
              peter

              Comment


              • #22
                Sorry, But I don't Understand....

                Originally posted by gotoluc View Post
                Hi Peter,

                In the video I'm using my SG to power the motor and it is set at 1.5Hz on square wave which is at 50% duty cycle. So there is no break in the power supplied, only a polarity shift. The Peak to Peak of the SG is 20v so we have a potential of 10v RMS. However power is limited.

                To calculate the possible power output of my SG, I connected a 1% 10 Ohm 50 watt resistor to it with scope probes across and it gave a 1.5 volts RMS reading. Just now re-tested with the same 1% 10 Ohm 50 watt resistor but using my Schlumberger 7150 meter set to AC volts and I get an average of 1.5 volts reading. So I'm quite sure about the power .25 watt available power at this frequency.

                Peter, at this time the motor is far from being efficient.

                While I have your attention , I have a question.... I'm thinking that surface area of magnets and coil (combination) will also boost the mechanical output as long as the coil geometry follows the Brooks coil for Maximum Inductance of wire used. I'll tell you why I'm asking... I'm thinking of making a coil that uses the full 2" square magnets for the center of the Coil opening which is a total of 8" of surface (circumference). I have a coil design program that calculates by wire diameter to get maximum Inductance and minimum resistance.

                Do you think my thinking of more coil & magnet surface = more mechanical output?

                Thanks for your time.

                Luc
                Luc,

                Honestly, your method of "calculating the possible power output of your SG" is not the same as "measuring the actual power consumption of your moving coil". I'm sorry, but I still can not agree with you on the INPUT, because you have supplied me with no real data.

                The weight of the coil, calculated using the wire charts is also a ballpark guess, but close enough for now.

                As for efficiency, my main concern is the "back EMF" in the system, which is always the limiting factor. To illustrate this, I would like you to run a little test. If you simply take the wires from the coil and disconnect them from your supply. Then, attach them to your scope probe, and manually move the coil back and forth between the magnets. What you will see on the scope is how well the unit works as a GENERATOR. The faster you move the coil back and forth, the higher the voltage will be. The wave-form will not be a sine-wave, but it will be a complex AC wave.

                Once you see this you will also understand that whatever you do to make the unit produce more mechanical energy, the faster the coil will move. The faster the coil moves, the more the coil will generate back EMF to counter your input. THIS is the process that limits the overall efficiency. In this sense, your set-up will behave "classically" and the probability of COP>1 operation is low.

                Your idea of how to produce more mechanical energy may be correct, but it will not increase the EFFICIENCY of the system above the limits set by "reverse generator effects".

                A motor can only produce a COP>1 IF the forward EMF to back EMF ratio is greater than one, AND/OR any electrical energy recovered is recycled for re-use.

                Peter
                Peter Lindemann, D.Sc.

                Open System Thermodynamics Perpetual Motion Reality Electric Motor Secrets
                Battery Secrets Magnet Secrets Tesla's Radiant Energy Real Rain Making
                Bedini SG: The Complete Handbook Series Magnetic Energy Secrets

                Comment


                • #23
                  Wesley Gary.

                  Take a look at Wesley Gary's Electro-magnetic motor of 1876. You can find his patent at Rex Research. He uses a mechanical DPDT. His point is that the power of the motor can be increased not only by increasing the power to the armature coil, but by increasing the number, or gauss strength of the magnets.

                  Comment


                  • #24
                    Originally posted by Peter Lindemann View Post
                    Luc,

                    Honestly, your method of "calculating the possible power output of your SG" is not the same as "measuring the actual power consumption of your moving coil". I'm sorry, but I still can not agree with you on the INPUT, because you have supplied me with no real data.

                    The weight of the coil, calculated using the wire charts is also a ballpark guess, but close enough for now.

                    As for efficiency, my main concern is the "back EMF" in the system, which is always the limiting factor. To illustrate this, I would like you to run a little test. If you simply take the wires from the coil and disconnect them from your supply. Then, attach them to your scope probe, and manually move the coil back and forth between the magnets. What you will see on the scope is how well the unit works as a GENERATOR. The faster you move the coil back and forth, the higher the voltage will be. The wave-form will not be a sine-wave, but it will be a complex AC wave.

                    Once you see this you will also understand that whatever you do to make the unit produce more mechanical energy, the faster the coil will move. The faster the coil moves, the more the coil will generate back EMF to counter your input. THIS is the process that limits the overall efficiency. In this sense, your set-up will behave "classically" and the probability of COP>1 operation is low.

                    Your idea of how to produce more mechanical energy may be correct, but it will not increase the EFFICIENCY of the system above the limits set by "reverse generator effects".

                    A motor can only produce a COP>1 IF the forward EMF to back EMF ratio is greater than one, AND/OR any electrical energy recovered is recycled for re-use.

                    Peter
                    Thank you Peter for taking the time to reply.

                    You are right! I only supplied you with the maximum power output of my SG using a 10 Ohm load. The motor maybe using less as you are correct about the generator effect, which I had noticed.

                    I have another design idea that is similar (strong magnet as core to coil) but this design can work with switched DC! which give the ability to easily collect the flyback of the coil since this design the coil will be turned off since it has 35% of a dead stroke area (no mechanical power output). Maybe this design would work best as a 2 phase motor. Just use the strongest 50% stroke area of each coil. This may give a full 360 degrees of mechanical power at the crankshaft and lots of time for each coil to recover the flyback.

                    Anyways, I guess it needs to be tested.

                    Luc
                    Last edited by gotoluc; 12-08-2009, 10:28 PM.

                    Comment


                    • #25
                      Originally posted by gotoluc View Post
                      ...
                      I would guess if laminations are used they would need to be going across the coil winding and not parallel to the windings? I'm not positive on this one.
                      ...
                      Hi Luc,

                      I think also for the laminations going across the coil winding, not parallel. This is just like in normal mains transformers, the actual winding encloses lengthwise the laminations that are layered above each other.

                      On some calculation of the input power. Your signal generator has a 50 Ohm output resistance. You mentioned in the video your moving coil has a 30mH inductance and 7.5 Ohm DC resistance.
                      I assume you measured the coil inductance with the steel rod inside it. If not, please measure it when the coil is in the middle and at the ends, with the magnets attached.

                      Supposing the 30mH inductance, it has a very low inductive reactance at 1.5Hz (XL = 2πfL), this formula gives XL=0.282 Ohm, this is quasy negligible to the 7.5 Ohm copper wire resistance.

                      So in your working setup shown in the video, you load the SG 50 Ohm output with a 7.5 Ohm "resistor" and there is the XL=0.282 Ohn coil reactance in series with this, your load is a series RL circuit. The SG tries to output the 10V rms voltage level from its "inside source" but its inner 50 Ohm resistance and your 7.5 Ohm coil resistance constitues a voltage divider. This is why you measured about 1.5V across your 10 Ohm precision resistor when you loaded the SG output with it: the "missing" 8.5V out of the unloaded 10V dissipates in the 50 Ohm output resistance, cannot come out.

                      With your coil loading the situation is a bit more severe: the 7.5 Ohm copper resistance loads the SG output, allowing even less useful output voltage (and current) reach the coil.

                      So what is the power that feeds your coil? Neglecting the inductive reactance of your coil, using a simple voltage divider where the 10V feeds a 50 and a 7.5 Ohm resistors in series, we get 10*7.5/(50+7.5)=1.3V across your coil, this gives 1.3*1.3/7.5= 0.225W power that feeds the coil.

                      Notice that this simple calculation is valid for this small inductance at this small frequency so that the phase angle (which is also very small) between the current and voltage can also be neglected. If this is not case, you have to apply normal AC theory calculations like is shown here:
                      Series resistor-inductor circuits : REACTANCE AND IMPEDANCE -- INDUCTIVE

                      You may wish to check with the scope or with your Fluke meter the voltage level while the setup is working. If your 30mH is not with the steel bar included in it, then values should differ from the calculations above.

                      rgds, Gyula

                      Comment


                      • #26
                        Originally posted by gyula View Post
                        Hi Luc,

                        I think also for the laminations going across the coil winding, not parallel. This is just like in normal mains transformers, the actual winding encloses lengthwise the laminations that are layered above each other.

                        On some calculation of the input power. Your signal generator has a 50 Ohm output resistance. You mentioned in the video your moving coil has a 30mH inductance and 7.5 Ohm DC resistance.
                        I assume you measured the coil inductance with the steel rod inside it. If not, please measure it when the coil is in the middle and at the ends, with the magnets attached.

                        Supposing the 30mH inductance, it has a very low inductive reactance at 1.5Hz (XL = 2πfL), this formula gives XL=0.282 Ohm, this is quasy negligible to the 7.5 Ohm copper wire resistance.

                        So in your working setup shown in the video, you load the SG 50 Ohm output with a 7.5 Ohm "resistor" and there is the XL=0.282 Ohn coil reactance in series with this, your load is a series RL circuit. The SG tries to output the 10V rms voltage level from its "inside source" but its inner 50 Ohm resistance and your 7.5 Ohm coil resistance constitues a voltage divider. This is why you measured about 1.5V across your 10 Ohm precision resistor when you loaded the SG output with it: the "missing" 8.5V out of the unloaded 10V dissipates in the 50 Ohm output resistance, cannot come out.

                        With your coil loading the situation is a bit more severe: the 7.5 Ohm copper resistance loads the SG output, allowing even less useful output voltage (and current) reach the coil.

                        So what is the power that feeds your coil? Neglecting the inductive reactance of your coil, using a simple voltage divider where the 10V feeds a 50 and a 7.5 Ohm resistors in series, we get 10*7.5/(50+7.5)=1.3V across your coil, this gives 1.3*1.3/7.5= 0.225W power that feeds the coil.

                        Notice that this simple calculation is valid for this small inductance at this small frequency so that the phase angle (which is also very small) between the current and voltage can also be neglected. If this is not case, you have to apply normal AC theory calculations like is shown here:
                        Series resistor-inductor circuits : REACTANCE AND IMPEDANCE -- INDUCTIVE

                        You may wish to check with the scope or with your Fluke meter the voltage level while the setup is working. If your 30mH is not with the steel bar included in it, then values should differ from the calculations above.

                        rgds, Gyula
                        Hi Gyula,

                        thanks for taking the time to calculate all this and posting it.

                        You're right! the coils inductance was taken without the steel bar in it. I'm so use to only using magnets in the coil (which has next to no change on coil Inductance) I forgot about the steel. So it's much more now. I measured 103mH when the coil is in the center of the bar and works its way down to 66mH when it's at the magnet (end of bar). Same measurement on each side.

                        Added:

                        Gyula,

                        I just did a new test. I attached my voltmeter to the coil to read the voltage when the coil is moving (attraction) from one end of the bar to the other. I used my variac through a FWBR and 3900uf cap and set the voltage to 1.50vdc with coil attached and held in place. When I let go of the coil the voltage rises to 3.03vdc during its travel across. I then used a D cell 1.5vdc battery and attached it to the coil and also attached my quality amp meter and repeated the experiment above. I connected the battery and held the coil in place and the readings are 1.51vdc @ 189ma. This seems to be the maximum power the coil uses at this voltage. I then disconnected the coil to relieve it for a few seconds. I let go of the coil and reconnected to take the readings while in movement. They are 1.53vdc @ 55ma (average ma after many tries) which = to 0.08415 Watts.

                        To me it looks like the coil has an advantage while it's in movement, so I'm not sure what Peter was trying to say (The faster the coil moves, the more the coil will generate back EMF to counter your input. THIS is the process that limits the overall efficiency)

                        It doesn't look like anything is countering the input it looks quite the opposite. The faster the coil is moving the less power it consumes I do agree that under load it will consume more power which is understandable but I don't understand the above statement.

                        Do you know what Peter is trying to explain?

                        Thanks for your time

                        Luc
                        Last edited by gotoluc; 12-10-2009, 05:13 AM.

                        Comment


                        • #27
                          hi luc,
                          Ill try to explain it as i see it. Your motor involves a coil moving by strong magnets. Anytime you move a coil by strong magnets, a current is generated in the coil, this is how all generators work. So while what you built is a motor, its also generating current too. This current that is being generated is in the opposite direction to the current that you are putting into the motor to make it run. The faster the motor speed, the more its generating and the more your fighting against. This is what Peters no bemf motor takes advantage of by simply removing the magnets. You could use the same theory with a solenoid motor like yours by removing your magnets and just letting the coil attract iron, like Bob Teal's motor i believe.

                          I think the results you are getting with the current decreasing with increased motor speed are more related to the shortened pulse duration. If you havnt read Peters electric motor secrets thread yet, i would highly recommend it. Its a very long read but it is full of priceless info on this subject. Someone please correct me if i got something wrong.

                          Comment


                          • #28
                            Originally posted by cody View Post
                            hi luc,
                            Ill try to explain it as i see it. Your motor involves a coil moving by strong magnets. Anytime you move a coil by strong magnets, a current is generated in the coil, this is how all generators work. So while what you built is a motor, its also generating current too. This current that is being generated is in the opposite direction to the current that you are putting into the motor to make it run. The faster the motor speed, the more its generating and the more your fighting against. This is what Peters no bemf motor takes advantage of by simply removing the magnets. You could use the same theory with a solenoid motor like yours by removing your magnets and just letting the coil attract iron, like Bob Teal's motor i believe.

                            I think the results you are getting with the current decreasing with increased motor speed are more related to the shortened pulse duration. If you havnt read Peters electric motor secrets thread yet, i would highly recommend it. Its a very long read but it is full of priceless info on this subject. Someone please correct me if i got something wrong.
                            Hi Cody,

                            thanks for your post

                            I do understand what you have explained and that was how I understood it as Peter explained it. So I wanted to verify if it is the case with this motor design. However, from my test results this is not happening or I'm missing something. This is why I'm questioning this.

                            Thanks

                            Luc

                            Comment


                            • #29
                              Hi folks, Hi gotoluc,

                              What
                              To me it looks like the coil has an advantage while it's in movement, so I'm not sure what Peter was trying to say (The faster the coil moves, the more the coil will generate back EMF to counter your input. THIS is the process that limits the overall efficiency)

                              It doesn't look like anything is countering the input it looks quite the opposite. The faster the coil is moving the less power it consumes I do agree that under load it will consume more power which is understandable but I don't understand the above statement.

                              Do you know what Peter is trying to explain?
                              What this means is that when at higher rpm your magnetic field will be less and so less shaft power due to the reverse generator built in, though at lower rpm's you can have higher efficiency, but still may be limited since the reverse gen effect is still there. So if we could use magnets and not have this reverse effect then at max rpm the torque and shaft power would be large, whereas in most motors now its feable until you lower rpm which allows more current to flow because less reverse voltage. hope that helps.
                              peace love light
                              Tyson

                              Comment


                              • #30
                                Bloch wall

                                Gotoluc - is your engine generating current unpowered in the same direction as you would use to move the coil by powering it? Ie; the wires are the same polarity?

                                If that is the case (and I'm no expert) then the counter-emf people are talking about would in this case be supporting-emf???

                                I reiterate, I have no idea what I'm talking about
                                Atoms move for free. It's all about resonance and phase. Make the circuit open and build a generator.

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