Rewinding - bit more complex than most rewinds

Discussion in 'Electric Bicycles' started by abudabit, Oct 6, 2008.

  1. abudabit

    abudabit New Member

    If I want to increase a brushless motor's N-m/A by 3.44 times and reduce it's kV by equal amounts, I would rewind it with 3.44 times more windings using the same total mass of copper. Pretty easy to understand.


    I don't want to reduce current capacity that much. So let's say I want to double the copper mass as well as modify the N-m/A and kV ratings. Since I will be increasing total power of the motor I know I will have to increase the perm magnetism of the core. But what do I do with the windings? Is it inductance that determines the N-m/A and kV ratings? Or will 3.44x windings have the same effect regardless of what diameter of wire I use?

    Hope there's some electrical engineers on board. :grin:

  2. duivendyk

    duivendyk Guest

    What on earth is a kV rating?,kiloVolts? (1000V),from a battery?This post makes absolutely no sense to me at all,you want to increase the torque to current ratio by a factor of 3.44?,why?,Explain in comprehensible terms what you are trying to do,increase power? operating voltage?,what?.I am an electrical eng. but not a clairvoyant.
  3. abudabit

    abudabit New Member

    I'm sorry, I did the capitalization wrong. Kv is the term used to describe ideal non-loaded RPM / V for brushless motors. If you are putting 90 v into a motor with a Kv of 50 the closer it gets to 4500 rpm the less current it will draw. Likewise the further away it is from V * Kv the more current it will draw.

    N-m/A is inversely proportional to Kv.

    Obviously what I want to do is increase torque and decrease rpm. Typically people would just use a finer wire and use more windings. But since I would need to significantly increase windings (and hence wire length) 3.44 times that would mean a wire length of around (glibly speaking) 3.44 times as long, which would mean a wire with 3.44 times less volume per length. I know realisticaly I will have to use a finer wire than the original, but I was hoping maybe just half the volume per length. So what I need to know is how will only reducing the wire volume / length moderately instead of linearly effect how many windings I need to do the 3.44 ratio.
  4. duivendyk

    duivendyk Guest

    Your winding resistance will go up by 3.44 squared which close to 12 times the original,so will your copper losses,this will be intolerably inefficient,if you used say, twice as many turns instead the increase would be 4 to 1. Lets say your input power is 1 your losses 0.1 and you efficiency is 0,9 in the original the loss is 0.4 and the output 0.6 your back emf is 0.6, which is proportional to rotational speed times number of windings (assuming that the flux is constant)therefore .
    0.9/(Sxw)= 0.6/(S'x2w) or 0,9/S= 0.3/S' or S'= S/3 the rotational speed S' will be 1/3 of the original S, if the input voltage and current are the same.The same input power that is,your efficiency will be 60% instead of 90% due to increased copper losses.Your no load speed will be somewhat less than half the original speed, but drop off to 1/3 due to the increased winding voltage drop under load.You end up overloading the motor unless you reduce the current.I made a gues at the original winding voltage drop (0.1Vin),it may have been optimistic,so things could be even worse.The torque will have gone up by a factor 2.twice the torque at 1/3 of the speed means 60% of the original output power. Reducing the operating speed substantially ends up reducing the power output capability of the motor,or thermally overloading it.This is obviously not a good idea.
    Last edited by a moderator: Oct 8, 2008
  5. abudabit

    abudabit New Member

    The idea is to use less current and more voltage. I was going to use half the current of the original design. But you are right, it would be less efficient in theory. If I have 3.44 times the length but half the current that will be (3.44/2)^2 or 2.96 more heat dissipation from current through the windings. That's not 2.96 times less efficient though, because that is only one portion of the inefficiencies of the motor. Still, that's enough of a drain of output power to warrant looking elseware for the solution.

    What relation does rotor diameter have with N-m/A : RPM/V? Is it linear?
    Last edited: Oct 9, 2008
  6. duivendyk

    duivendyk Guest

    I can't make any sense out of this, are you prepared to to have half the power.EXPLAIN what you really want to accomplish with this rewinding, in general terms, then I might be able to figure out wether it makes any sense or not.Attempt to be clear!!.When you write something, try to think of what someone is to make of it.I am not a MINDREADER!
    Last edited by a moderator: Oct 9, 2008
  7. abudabit

    abudabit New Member

    Now that I think about it, resistance by adding more length is linear. 50% current through 3.44 times longer wire would be 3.44/4, so that would be 86% power consumption. It actually would be more efficient, not less.

    Obviously the increase in power comes from a dramatic increase in voltage. Running more voltage and less current is a more efficient way to get power.
  8. duivendyk

    duivendyk Guest

    You are dead wrong, it goes up 4 to 1, if you double the number of turns in the same winding space. I have designed electrical machinery and know this stuff .In order to get twice as many turns in the same crosssectional AREA, the area of each wire has to be TWICE as small,for instance,think of a block of resistive material of length lets say 10, crosssection 1x2. We cut it ,so that we have two wires still 10 long,but each with a crosssection of 1x1.The resistance of each of these 10x 1x1 wires is twice of that of the original wire. since the area of each is 1/2 of that of the original wire,but these wires are put in series so that the total resistance of the 20x1x1 wire is 4 times the original resistance, 2x2 = 4 .If you don't believe me try it,you will find out.You have still not explained what you are trying to accomplish.Frankly I have had enough, I've tried to help you but concluded that it's just a waste of my time.
  9. abudabit

    abudabit New Member

    First off, I mentioned that I'm not reducing the cross section of the wire significantly. This is enlarging the motor.

    You are being extremely negative. You never had any interest in giving advice. Your posts from the beginning have been rude and condencending and a bit ignorant. If you had read the initial post you would have known what I want to do. It is mentioned plain as day.

    The only thing I neglected to mention was I would be increasing voltage.

    Luckily I posted on a different board and got all the answers I needed in the first post.

    If I'm wrong about your intentions I apologize, but it's doubtful.
    Last edited: Oct 10, 2008
  10. duivendyk

    duivendyk Guest

    Good luck to you,you'll need it .If you don't outline your problem in comprehensible terms,don't expect useful advice,and what I said about the resistance increase was CORRECT as youwill have occasion to find out.