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  • Matthew Jones
    replied
    Originally posted by garrypm View Post
    G'day Matt,

    Your left side electrolytic capacitor on the boost module is blown and needs replacing.

    Have a good one
    It came like that. About every other one does, its not blown its just bulged.

    Matt

    Leave a comment:


  • garrypm
    replied
    Boost module

    G'day Matt,

    Your left side electrolytic capacitor on the boost module is blown and needs replacing.

    Have a good one

    Leave a comment:


  • bistander
    replied
    Analogy

    Originally posted by ilandtan View Post
    ...
    One battery scenario may not be the best way illustrate the concepts? I maybe incorrect here, but aren't you going to always muddy the power numbers by measuring how much you have going in, versus what is being consumed at the load when you are looping the energy back? It's predictable contention.
    ...
    You have a tank of water with 5 gallons in it. You have a 4 gpm pump. It takes 1.25 minutes to empty the tank. Now you use a 5 gpm pump but split the output flow so 1 gallon per minute flows back into the tank. How long does it take to empty the tank?

    Same concept as putting energy back into the battery.

    bi

    Leave a comment:


  • Matthew Jones
    replied
    Testing

    Here is a test I did today running a modified motor. Ran it 2 hours with 4+ amp draw on a 12v 7ah battery. By adding the boost converter and the efficiency of my motor I effectively doubled the output of the battery, or got 2 times the work out of the motor for the same cost as running it as a standard load.

    Start.
    https://www.youtube.com/watch?v=X-tP...ature=youtu.be

    Finish.
    https://youtu.be/qBbQ7DMyTP0

    Sorry about the sound in the second one I had to move a box fan to help cool the motor and forgot to turn it off.


    Its just video its not proof. You have to prove it to yourself.

    Matt
    Last edited by Matthew Jones; 07-23-2018, 08:03 PM.

    Leave a comment:


  • Matthew Jones
    replied
    Originally posted by ilandtan View Post
    I don't care what the power numbers are --because it's intuitave trying to place the power in a neat box. I know that we want to measure and compare, but what if closed system analysis isn't descriptive? All I care are the results; that for the same battery that should only provide 1hr at 8A, gets significantly more run time. Specially if I am consuming 100W through a load, while charging a different battery.

    One battery scenario may not be the best way illustrate the concepts? I maybe incorrect here, but aren't you going to always muddy the power numbers by measuring how much you have going in, versus what is being consumed at the load when you are looping the energy back? It's predictable contention.

    If I follow the initial 3B concept, you always recharged battery 3, while the other two discharged (something has to be doing the work). But here you are losing the advantage of that third battery charge. Harder to prove one battery works, unless there is more secret sauce.

    I think it would be easier to to use the generator to charge a separate battery through a load (so you are doing work) while the destination charges the charge battery, and then swap 1 & 2(after a rest period) and see how long you can keep doing that, while not adding any additional power.
    If you can run the load side by side with 2 like batteries and the conventional runs less time, your saying the thing. As long the measurements to that load add up real similar. You gotta make sure the draw on the LOAD for either setup is identical within reason.

    Matt

    Leave a comment:


  • ilandtan
    replied
    Slippery Slope

    I don't care what the power numbers are --because it's intuitave trying to place the power in a neat box. I know that we want to measure and compare, but what if closed system analysis isn't descriptive? All I care are the results; that for the same battery that should only provide 1hr at 8A, gets significantly more run time. Specially if I am consuming 100W through a load, while charging a different battery.

    One battery scenario may not be the best way illustrate the concepts? I maybe incorrect here, but aren't you going to always muddy the power numbers by measuring how much you have going in, versus what is being consumed at the load when you are looping the energy back? It's predictable contention.

    If I follow the initial 3B concept, you always recharged battery 3, while the other two discharged (something has to be doing the work). But here you are losing the advantage of that third battery charge. Harder to prove one battery works, unless there is more secret sauce.

    I think it would be easier to to use the generator to charge a separate battery through a load (so you are doing work) while the destination charges the charge battery, and then swap 1 & 2(after a rest period) and see how long you can keep doing that, while not adding any additional power.

    Leave a comment:


  • Turion
    replied
    Cha cha cha changes

    Pulse motor REDUCES costs to run the system and contributes a spike. That spike is an ADDITION. You get two for one. Coils on the generator that speed up under load reduce amp draw. Elimination of magnetic cogging reduces amp draw AGAIN. It is not one BIG thing that makes this work, it is the combination of many little things. Any one of which, if disregarded, AFFECTS the results dramatically.

    Leave a comment:


  • Matthew Jones
    replied
    Originally posted by dragon View Post
    All the conversions add to the total losses... It appears that the biggest problem is the motor/generator in this scenario. Just a quick calculation showing around 25% efficiency over all with an even lower % return to the battery.

    Simply using an inverter with a 24volt transformer rectified through a load back to the battery would yield a higher return. Assuming 90% efficiency of both inverter/charger yielding 80% overall you could return 36 watts with a 45 watt input ( not including the energy dissipated by the load ).

    If there were some cheap "isolated" boost converters, that might be the only unit needed to achieve a 90% return. At that point it would seem logical to utilize a system that would use the input energy to the converter more than once bringing you closer to unity.

    So what would it take to drive your load while charging a capacitor and return a small portion to the battery, then using the energy from the capacitor to drive the load while the battery is being charged? You need to either find a source of energy to add to the system or make better use of the energy you have available... yes? no?

    That the point is we have small motor that is rewound and a couple of coils that accelerate reducing the load of the motor and delivering power.

    Someone back aways just used the modified motor with the boost convert and ended up with a higher charge then they started. I am going to look at that hear soon now the chores are done.

    Matt

    Leave a comment:


  • dragon
    replied
    All the conversions add to the total losses... It appears that the biggest problem is the motor/generator in this scenario. Just a quick calculation showing around 25% efficiency over all with an even lower % return to the battery.

    Simply using an inverter with a 24volt transformer rectified through a load back to the battery would yield a higher return. Assuming 90% efficiency of both inverter/charger yielding 80% overall you could return 36 watts with a 45 watt input ( not including the energy dissipated by the load ).

    If there were some cheap "isolated" boost converters, that might be the only unit needed to achieve a 90% return. At that point it would seem logical to utilize a system that would use the input energy to the converter more than once bringing you closer to unity.

    So what would it take to drive your load while charging a capacitor and return a small portion to the battery, then using the energy from the capacitor to drive the load while the battery is being charged? You need to either find a source of energy to add to the system or make better use of the energy you have available... yes? no?

    Leave a comment:


  • bistander
    replied
    Test 4

    Here are data from most recent test.

    Wattmeter on battery:
    12.05V, 3.4A, 40.97W, 9.09Ah, 111.21Wh

    Wattmeter on 50Ω load resistor:
    18.31V, 0.33A, 6.1W, 0.885Ah, 15.3Wh



    Regards,

    bi
    Attached Files

    Leave a comment:


  • altrez
    replied
    Originally posted by Matthew Jones View Post
    I am finding funny stuff happening, if you can measure everything you can think of, every kind of way you can think of, and give us some details.

    Appreciate it.
    Matt
    I will do that and if you need me to buy some different parts to test different configurations I will.



    -Altrez

    Leave a comment:


  • Matthew Jones
    replied
    Originally posted by altrez View Post
    This is very interesting! I know that the 3 battery system that John shared so long ago works if you check my posts in the past you will see I have spent lots of money on that. However the one battery system is new. I ordered some boost modules that Matt posted a few pages back and will build / test the circuit and post some results.



    -Altrez

    I am finding funny stuff happening, if you can measure everything you can think of, every kind of way you can think of, and give us some details.

    Appreciate it.
    Matt
    Last edited by Matthew Jones; 07-23-2018, 02:44 AM. Reason: I type faster than I think.

    Leave a comment:


  • altrez
    replied
    This is very interesting! I know that the 3 battery system that John shared so long ago works if you check my posts in the past you will see I have spent lots of money on that. However the one battery system is new. I ordered some boost modules that Matt posted a few pages back and will build / test the circuit and post some results.



    -Altrez

    Leave a comment:


  • bistander
    replied
    Single battery test

    Glad to see you into it Matt. I ended my test after 3 hours. It was still running strong but motor and generator were well done (very warm) and battery down to 12V. I took a round of data before turning it off. I'll put the data on the diagram like before and post it. I'll leave it together for a while. Just ask if you want additional test.

    Regards,

    bi

    Leave a comment:


  • Matthew Jones
    replied
    Originally posted by dragon View Post
    The boost circuit is separate from the motor circuit. I see the boost portion operating between the positive and negative of the battery - the motor circuit is operating between potentials of the boost coil with no connection to the battery negative.... basically two potentials in the same loop. This loop can't contribute anything to the battery's state of charge.

    It appears to me that the generator is the sole contributor to battery charge in this case - or in other words - The motor/generator is the critical area where most of the losses appear to be.

    Not being critical - just an observation...

    Edit: "Ground" is sort of a universal term for zero voltage or negative. We tend to use the term loosely when we should be using "potentials" in reference to the connection. One might be more negative than the other making the other "more" positive, such as connecting between the negatives.
    I appreciate your input Dragon never had a problem with it.

    I ran some test today to look at this. The excess current on the negative potential or ground of the battery only shows up under this certain setup. With 3 batteries it doesn't act like this and with conventional setup both the positive side and negative side reflect each other.
    I am going to isolate some stuff with diodes and see if the numbers change. From the ratios Bi's test showed.

    Also going to put the OUTLINED circuit to a test using the correct equipment. Just got to get it together.

    This is the sh*t I love, mystery!!! I want answers..

    Cheers
    Matt

    Leave a comment:

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