• RE: "If we replaced the wire with foil that had 1/4th the mass and 4X the resistance per inch" ... - josh358 09:22:29 06/27/11 (9)
  In Reply to: RE: "If we replaced the wire with foil that had 1/4th the mass and 4X the resistance per inch" ... posted by tom_gootee on June 26, 2011 at 14:31:29

  What we need is for someone to measure the width of the Magnepan foil, or the resistance of a known length, because per Wendell we should use the same foil width Magnepan found worked best in their R&D effort. This will require some kind of series/parallel arrangement to get the impedance right. But we won't know what that will be until we have some measurements.
  

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  • RE: "If we replaced the wire with foil that had 1/4th the mass and 4X the resistance per inch" ... - tom_gootee 18:42:42 06/30/11 (8)
    In Reply to: RE: "If we replaced the wire with foil that had 1/4th the mass and 4X the resistance per inch" ... posted by josh358 on June 27, 2011 at 09:22:29

    Josh,

    I guess it depends on what you mean by "the Magnepan foil". All I have is a pair of MG-12/QRs. But in case it might be helpful, I measured their tweeter foil:

    On my MG12s, the tweeter foil is exactly 0.1-inch wide (2.54 mm).

    There are eight vertical runs of foil, with the outer two going to the connectors. There is a space of exactly 0.125 inch (1/8 inch) between each pair of vertical foil segments (so, 0.225" between their centers), except that the last vertical foil segment on the right is 0.250 inch from the one to its left, instead of 0.125 inch like the others.

    I did measure the total foil length (8.146 meters), and the resistance (3.60 to 3.66 Ohms), as precisely as I could. The calculated thickness, with measurement variations, came out to be in the range of .0009 inch to .0011 inch. So I'd have to assume that it is actually a .0010-inch thickness.

    I measured the length of the MG12 tweeter foil from connector to connector, as 8.146 meters (320.7075 inches). [No, my measurement precision wasn't .0001 inch. I just kept the four digits from the conversions of 16ths etc to decimal fractions, until after I added everything up.] (The lead-ins from the connectors totaled 16.8125 inches. The length of each of the eight vertical sections is 37.6875 inches. There are six horizontal runs of 0.325 inch each and one of 0.450 inch.)

    I unplugged the tweeter wires and measured the resistance, with my Tektronix DMM916 multimeter. Of course I also measured the resistance of the meter's probes and wires by shorting the probe tips, and subtracted that from the tweeter resistance measurement, each time. I repeated the entire process until the measurements stopped changing (five or six times). I got 3.6 Ohms, for the one tweeter that I measured (only one sock is off).

    I also tried measuring the resistance of the tweeter foil using my TH2821A LCR meter, and got 3.66 to 3.65 Ohms, using frequencies from 100 Hz to 10 kHz.

    To calculate the cross-sectional area, I used the equation A (in sq mm) = [0.026 (for Aluminum) x Length (in meters)]/Resistance (in Ohms). I then divided the Area by 2.54 mm to get the thickness in mm. Then I divided that by 25.4 mm/inch to get the thickness in inches.

    Cheers,

    Tom
    
    

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    Edits: 06/30/11

    • Wendell's answer - josh358 11:28:06 07/03/11 (4)
      In Reply to: RE: "If we replaced the wire with foil that had 1/4th the mass and 4X the resistance per inch" ... posted by tom_gootee on June 30, 2011 at 18:42:42

      Turns out the MMG and 12 use the 1 mil foil. Despite the fact that his replies are in red so they won't be missed by idiots like me, I managed to read right past it. Probably paying too much attention to the dancing girls.
      
      But, hallelujah, this gives us exactly what we need! With your measurements of the 12 it remains only for me and JBen to measure our new and old MMG's respectively. I've been unusually busy for the last week, but I'll make it a priority to do mine within the next few days.
      

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      • RE: Wendell's answer - JBen 23:22:13 07/11/11 (3)
        In Reply to: RE: Wendell's answer posted by josh358 on July 03, 2011 at 11:28:06

        No sooner had I begun to uncover one MMG than my wife warned we had visitors on their way Sunday afternoon. So, it all got moved up, again. However, I kept thinking that I must have saved my early measurements, at least the ones that I needed for the PLLXO-based bi-amp calculations.

        TaDahhh! I just found them and I also rechecked the impedances from outside each MMG. They match what I measured long ago. At the time, years ago:
        a. Both tweeters measured 2.61 ohms. This was 6 runs, 38 inches long each + 7 inches in bottom traces to the terminals. Aprox total=235+ inches.
        b. The mid/bass panels measured 4.1 & 4.14 ohms, respectively. There are 24 runs, of which 14 are double. Therefore, the equivalent of 38 runs, each 37.75 inches long (extending 38" when as twin loops). There are about 16.25 inches of additional connecting traces at the bottom. In total, roughly 1451+ inches.

        The actual lengths are probably just slightly longer because I don't think I added the short horizontals at each end. I do remember measuring at the looping (U) center between two mid/bass panel traces, that should be close enough.

        In fact, I did a quick proportional extrapolation with the numbers that Tom_G supplied earlier about the 12s. Tom is not clear if he included the traces to the terminals but there's enough thoroughness to suggest he would not have missed it. Electrically, the mid-panel numbers are right on. Using his figures backwards, mine should be 4.1~4.16 ohms, which is a virtual match.

        The tweeters are also close enough. Backing up from Tom's numbers I get that my tweeters would have been 2.64 ohms, I measure 2.61 ohms...so yeah, ballpark.

        Therefore, it seems that my old MMGs and Tom's 12s are made of the very same type of wire and foil.

        The newer MMGs probably are, as well. If so, it then means that their longer tweeter runs (vs old MMGs) would have around 3.5 ohms impedance, close to the 12s. OTOH, the newer MMGs mid/bass would be able to match the old MMGs impedance by using 22 rows of wire, 16 of them doubled for a total of 38. Is this the row count? If it is, you should see around 4.1 ohms there.

        In other words:
        - New MMGs with 8 tweeter runs and 22(38) mid/bass runs?
        - Old MMGs with 6 tweeter runs and 24(38) mid/bass runs

        Curiously, it appears that the 12s use less wire for mid/bass than the MMGs.

        Anyway, for QR on my mid/bass panels, it looks like I'd have enough "run length" for 24 runs of 38 inches (plus about 16" for horizontal & terminal runs). Using the same tweeter QR material, I would get a total impedance of about 10.5 ohms. Should I just thank God that I bi-amp at line-level or are we getting more suitable QR strips from Magnepan?

        
        
        

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        Edits: 07/12/11

        • RE: Wendell's answer - josh358 12:17:29 07/12/11 (2)
          In Reply to: RE: Wendell's answer posted by JBen on July 11, 2011 at 23:22:13

          Excellent, I'll measure my own MMG's tonight or tomorrow at the latest and we'll see how they compare.
          
          I think, with a DCR of 10.5 ohms, we're talking about paralleling the two halves of the woofer panel. That way, DCR will be 5.25 ohms, which is a bit high but not horribly so. The newer MMG's should be very slightly lower, owing to the missing woofer loop.
          

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          • RE: Wendell's answer - JBen 12:51:52 07/12/11 (1)
            In Reply to: RE: Wendell's answer posted by josh358 on July 12, 2011 at 12:17:29

            Huh! In parallel, how did you wind up with 5.25? If I split the runs in 2 equal, my finger calculator says I'll wind up with less than 3 -- ohms not fingers : - ))
            

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            • RE: Wendell's answer - josh358 06:27:23 07/21/11 (0)
              In Reply to: RE: Wendell's answer posted by JBen on July 12, 2011 at 12:51:52

              LOL sorry I must have been half asleep that night. Or maybe distracted by the dancing girls?
              

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    • RE: "If we replaced the wire with foil that had 1/4th the mass and 4X the resistance per inch" ... - josh358 09:49:12 07/01/11 (2)
      In Reply to: RE: "If we replaced the wire with foil that had 1/4th the mass and 4X the resistance per inch" ... posted by tom_gootee on June 30, 2011 at 18:42:42

      Cool, thanks, some figures at last! The only question is, does the 12 use the same foil Magnepan is offering us? Once we have that, we can figure out what kind of series/parallel arrangement we need to get the resistance right with the new foil.
      
      Wendell also wrote to say that while 1 mil foil would sound great on the woofer, it would be too inefficient, and that he's guessing that 2 - 2.5 mil foil would be right for the woofer. I counted 22 (doubled, he reminds me) wire runs on my MMG and 8 foil traces. So I get 2.75 times the thickness of the tweeter to keep the resistance/unit length and mass changes proportional. If we used .5 mil foil on the tweeter, that would come out to a single 1.5 mil trace on the woofer. But since there may be other factors that I'm missing, I'd say Wendell's guess is a better starting point.
      

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      • RE: "If we replaced the wire with foil that had 1/4th the mass and 4X the resistance per inch" ... - tom_gootee 13:51:32 07/02/11 (1)
        In Reply to: RE: "If we replaced the wire with foil that had 1/4th the mass and 4X the resistance per inch" ... posted by josh358 on July 01, 2011 at 09:49:12

        Since the total available foil length should be known (based on the panel where the wires are being replaced), I derived an equation that can be used to find, from a desired total length of foil, the cross-sectional area needed to get two identical parallel foil runs that result in a 4 Ohm load:

        Area = thickness x width = length x .001625

        where length is in meters, Area is in square millimeters, and thickness and width are in millimeters.

        If the desired width or thickness were known, then that could be rearranged to one of the following:

        thickness = length x .001625 / width

        or

        width = length x .001625 / thickness

        Or, if you had foil with a known thickness and width, you could calculate what total length would need to be divided in half to get 4 Ohms when the two halves were run in parallel, with:

        length = thickness x width / .001625

        The following conversion formulas can be used:
        mm / 25.4 = inches
        inches x 25.4 = mm
        meters x 39.37 = inches
        inches / 39.37 = meters

        Example:

        If the optimal width that Wendell mentioned turns out to be 0.1 inch, i.e. 2.54 mm, we can calculate what total lengths would give 4 Ohms when divided into two parallel runs, for each of several different available thicknesses:

        .0005" (.0127 mm) x 0.1" foil: .0127x2.54/.001625 = 19.85 m (65.13 ft)

        .001" (.0254 mm) x 0.1" foil: .0254 x 2.54 / .001625 = 39.7 m (130.25 ft)

        I'll stop there, since .0015-inch by 0.1-inch foil would need 59.55 meters split into two parallel runs, to get 4 Ohms.

        -------------------------

        With only a tiny bit more work, I was able to derive the equations for n aluminum foil runs in parallel:

        (1) Rn = .026 x Ltot / nnTW

        where Rn is the resistance in Ohms of n parallel foil runs, each with length Ltot/n (in meters), and Ltot is the total foil length in meters, T is the foil thickness in millimeters, and W is the foil width in millimeters. Note that nn is n x n which is "n squared".

        The derivation for n parallel foil runs was easy, if we can assume that we want the same current through all runs, and therefore we have to assume that their resistances are equal. For equal resistances, the parallel resistances equation [Rpar = 1 / (1/R1 + 1/R2 + 1/R3 + ... 1/Rn)] boils down to Rpar = R/n.

        (We could also use shorter lengths of thicker foil, for example, to get the same current, which would give the same force. But then we'd be accelerating a different mass, so the acceleration would be different, even though the current was the same. We could calculate the difference in mass needed for a different current, to get the same acceleration from a different length, but then other aspects of the behavior of the membrane would probably still be changed. Maybe someone should figure that out, though, because it would open up an infinite number of options, if it could be done that way while still retaining satisfactory characteristics.)

        If we assume that we want to end up with 4 Ohms, that gives us the following different rearranged versions of the same equation:

        (2) Area = T x W = .026 L / 4nn

        (3) T = .026 L / 4nnW

        (4) W = .026 L / 4nnT

        (5) L = 4nnTW / .026

        Examples:

        1) What total lengths of .0005 x 0.1 inch foil would give 4 Ohms when divided into different numbers of parallel runs?

        TW = .0127 mm x 2.54 mm = .032258 sq mm

        n = 2: L = 4 x 2 x 2 x .032258 / .026 = 19.85 m (whew, same as before)
        n = 3: L = 4 x 3 x 3 x .032258 / .026 = 44.665 m

        2) What thicknesses would work to get 4 Ohms from n parallel runs of 0.1-inch-wide foil that totaled 28.4 meters in length?

        T = .026 L / 4nnW

        n = 2: T = (.026 x 28.4) / (4 x 2 x 2 x 2.54) = .01817 mm (.0007153 inch)
        n = 3: T = (.026 x 28.4) / (4 x 3 x 3 x 2.54) = .008075 mm (.000318 inch)

        It doesn't look too good for the 28.4 meters needed for the MG12s, SO far.

        3) What widths of foil would work to get 4 Ohms from n parallel runs totaling 28.4 meters, for a couple of different thicknesses?

        W = .026 L / 4nnT

        T = .001 inch (.0254 mm):

        n = 2: W = .026 x 28.4 / 4 x 2 x 2 x .0254 = 1.817 mm (.0715 inch)
        n = 3: W = .026 x 28.4 / 4 x 3 x 3 x .0254 = .8075 mm (.0318 inch)

        T = .0005 inch (.0127 mm):

        n = 2: W = .026 x 28.4 / 4 x 2 x 2 x .0127 = 3.634 mm (.143 inch)
        n = 3: W = .026 x 28.4 / 4 x 3 x 3 x .0127 = 1.615 mm (.0636 inch)

        ---------------------------

        It's not looking overly-convenient for the MG12s, yet. But I guess it depends on what optimal foil width would be.

        We want the current through all runs to be equal, I assume. So if we can't get what we need with n identical parallel runs of foil, we're out of luck? It looks that way, to me, at the moment. The next-best thing would probably be to change the desired resistance to something slightly different than 4 Ohms. And I guess another option would be to sacrifice some power and add a resistance in series or parallel with the whole shebang.

        Cheers,

        Tom
        
        

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        Edits: 07/02/11   07/02/11   07/02/11

        • RE: "If we replaced the wire with foil that had 1/4th the mass and 4X the resistance per inch" ... - josh358 06:52:04 07/03/11 (0)
          In Reply to: RE: "If we replaced the wire with foil that had 1/4th the mass and 4X the resistance per inch" ... posted by tom_gootee on July 02, 2011 at 13:51:32

          That's an interesting question, actually. I'm still assuming we can't parallel foil as you can the wire. So you can't forex double just the central section, as in some of the early models. But that doesn't mean you can't run more current through say the central area of the diaphragm to produce the same effect.
          
          Anyway, as I see it, we'll be using foil with half the mass in the tweeter, so if it's the same width we just run two sections in parallel to restore the nominal 4 ohm DCR. And if we do the same in the woofer, that is, half the mass, that's a simple parallel run as well. The foil's been made available to us, so whey doesn't someone who's doing the mod just get some? Then we can measure it and find a source of the bass foil.
          

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• RE: "If we replaced the wire with foil that had 1/4th the mass and 4X the resistance per inch" ... - IEaudiodude 15:06:38 06/26/11 (1)
  In Reply to: RE: "If we replaced the wire with foil that had 1/4th the mass and 4X the resistance per inch" ... posted by tom_gootee on June 26, 2011 at 14:31:29

  OMG after all the work you did figuring that out you put a Discalimer at the end ? hahaha "Do Not Attempt Professional Driver Closed Course" I promise not to sue you if It's wrong. :)
  
  
  

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  • RE: "If we replaced the wire with foil that had 1/4th the mass and 4X the resistance per inch" ... - tom_gootee 19:28:41 06/26/11 (0)
    In Reply to: RE: "If we replaced the wire with foil that had 1/4th the mass and 4X the resistance per inch" ... posted by IEaudiodude on June 26, 2011 at 15:06:38

    Actually, I ran out of time and was pulled away but I suspected that there was an error, because when I calculated what the replacement of the MG12 bass/mid wire would be, with the same length and resistance, I think that I should have gotten foil dimensions that gave the same cross-sectional area as the original wire. But I didn't get anywhere even close. I figured I'd go back and fix it so I put the disclaimer in.
    
    I have since discovered that I measured the MG12's mid/bass wire diameter incorrectly. I guess I happened to place my micrometer on one of the double-run wire pairs with a little exta glue, instead of on a single wire, and got .05 inch. I went back and measured the single wire stub that was protruding from the bottom connector and got .022 inch, which is probably .0226 inch, i.e. 23 AWG wire.
    
    I corrected everything that referenced that wire size and then extended the post to include the calculations needed for changing an MG-12/QR's mid/bass panel into two wired-in-parallel foil panels of 8 Ohms each, per my previous post about that. It ended up having only 17.8% as much conductor mass as the original.
    
    I'll have to give a little more thought to what effect that might have on the power requirements. The same voltage would now give only half the current. But half as much current would still accelerate the membrane up to 2.8X as much (neglecting the mass of the mylar) as the full current did with the original wire.
    
    Cheers,
    
    Tom
    
    

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