{"id":1281,"date":"2016-08-02T20:56:36","date_gmt":"2016-08-03T01:56:36","guid":{"rendered":"http:\/\/android-originals.com\/tulsasound\/?page_id=1281"},"modified":"2016-08-05T15:31:32","modified_gmt":"2016-08-05T20:31:32","slug":"provisional-patent-application-filed-for-humbucking-quads","status":"publish","type":"page","link":"http:\/\/android-originals.com\/tulsasound\/tutorials\/humbucking-pairs\/provisional-patent-application-filed-for-humbucking-quads\/","title":{"rendered":"Provisional Patent Application Filed for Humbucking Quads"},"content":{"rendered":"<p>\u00a9 2016 Don Baker dba android originals LC<\/p>\n<p>Today, Aug 2, 2016, I filed a Provisional Patent Application for switching circuits to connect dual-coil humbucking pickups and matched single-coil pickups into humbucking quads, including digital controls.\u00a0 I\u2019ve read that no one can patent a mathematical equation, so that is what this document discloses \u2013 the number of humbucking quads that can be made from KK number of dual-coil humbuckers and K number of matched single-coil electromagnetic pickups. \u00a0The math shows that there are a surprising number of possible combinations.<\/p>\n<p><strong>Dual-coil humbuckers<\/strong><\/p>\n<p>We will assume that the individual coils in each dual-coil humbuckers can be connected together in either series or parallel.\u00a0 But always in-phase, unless the humbuckers is specially constructed with the same magnetic poles upwards towards the strings in each coil.\u00a0 And then they might as well be two single-coil pickups.\u00a0 So, series and parallel makes two different tones for each humbucker.\u00a0 From KK number of humbuckers, you can get 2*KK number of tones.<\/p>\n<p>A pair of humbuckers makes a quad \u2013 four individual coils connected together.\u00a0 A pair of humbuckers can be connected together in series and parallel, in-phase and contra-phase (out of phase).\u00a0 That\u2019s 4 different connections.\u00a0 The coils of each humbucker can be connected together in series or parallel.\u00a0 Over two humbuckers, that is 4 more choices, for a total of 16.\u00a0 Any number of KK humbuckers equal to 2 or greater can have KK*(KK-1)\/2 different pairs of humbuckers (by combinatorial analysis).\u00a0 That makes a total of 16*KK*(KK-1)\/2 number of tones.<\/p>\n<p>So KK number of humbuckers can have:<\/p>\n<p>KK = 2, 3, 4, 5\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 humbucking pairs<\/p>\n<p>2*KK = 4, 6, 8, 10\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 humbucking pair tones<\/p>\n<p>KK = 1, 3, 6, 10\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 humbucking quads<\/p>\n<p>16*KK*(KK-1)\/2 =16, 48, 96, 160 \u00a0\u00a0\u00a0 humbucking quad tones<\/p>\n<p>Total Tones = 20, 54, 104, 170<\/p>\n<p>&nbsp;<\/p>\n<p><strong>Matched single-coil pickups<\/strong><\/p>\n<p>We already know that K number of single-coil pickups can have K*(K-1)\/2 different humbucking pairs.\u00a0 Since the pairs can be connected in parallel or series, that makes K*(K-1) humbucking pair tones.\u00a0 But humbucking quads needs more discussion.<\/p>\n<p>Start from the fact that no single can appear twice or more in a quad.\u00a0 It just wouldn\u2019t be a humbucking quad.\u00a0 It is then easy to construct and see how many humbucking pairs from K = 4 singles, A, B, C &amp; D, can be made into unique quads.\u00a0 They are (AB,CD), (AC,BD) and (AD, BC).\u00a0 There are no more than K-1.<\/p>\n<p>Now add another coil, E, to get A, B, C, D, and E, for K = 5.\u00a0 All the combinations of A, B, C, and D are the same, plus those added by E.\u00a0 E can pair with the K-1 other coils, making K-1 pairs.\u00a0 To get the next pair, we take K-2 singles 2 at a time with combinatorial analysis, to get (K-2)*(K-3)\/2 pairs for each of the K-1 pairs with E.\u00a0 So with coil E we add (K-1)*(K-2)*(K-3)\/2 quads to the existing K-1 quads.\u00a0 The same analysis holds for K = 6 or more.\u00a0 The same factor of 16 tones per quad made of two humbuckers applies to these quads.<\/p>\n<p>The number of single coil quads from K single coils can be expressed by:<\/p>\n<p>Nqk = Summation of j = 4 to K: (K-1)*(K-2)*(K-3)\/2, \u00a0K &gt; 3<\/p>\n<p>And the number of quad tones is: 16*Nqk.\u00a0 This means that:<\/p>\n<p>K = 2, 3, 4, 5, 6, 7, 8\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 singles<\/p>\n<p>K*(K-1)\/2 = 1, 3, 6, 10, 15, 21, 28\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 humbucking pairs<\/p>\n<p>K*(K-1) = 2, 6, 12, 20, 30, 42, 56\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 humbucking pair tones<\/p>\n<p>Nqk = 0, 0, 3, 15, 45, 105, 210 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 humbucking quads<\/p>\n<p>16*Nqk = 0, 0, 48, 240, 720, 1680, 3360\u00a0\u00a0\u00a0 humbucking quad tones<\/p>\n<p>Total Tones = 2, 6, 60, 260, 750, 1722, 3416<\/p>\n<p><strong>Note<\/strong>: Some of <em>these<\/em> numbers were found to be inaccurate (too large) and were corrected 5 Aug 2016<\/p>\n<p>&nbsp;<\/p>\n<p><strong>Is this real?\u00a0<\/strong><\/p>\n<p>Could you possibly get that many different tones out of that many pickups?\u00a0 Not exactly.\u00a0 First, as the pickups get closer together the tonal differences get more subtle.\u00a0 Eventually, the differences may disappear.\u00a0 Second, in-phase quad tones, by the very separation of the pickups, can be spread and averaged over longer distances of under the strings than pairs.\u00a0 Thus they can be expected to be warmer, losing the higher harmonics, and getting closer to the fundamental harmonic of the string.\u00a0 The closer to the fundamental they get, they less distinct they become.\u00a0 Contra-phase quad tones will likely be something else.<\/p>\n<p>But still, most dual humbucker guitars, with a potential of 20 different tones, only have a 3-way switch.\u00a0 Three humbucker guitars have a potential of 54 different tones.\u00a0 Compared to a guitar with two humbuckers and a single, a guitar with five matched single-coil pickups, more spread out, has a potential of 260 tones.<\/p>\n<p>How much have you been missing?<\/p>\n<p>But check my math, please.\u00a0 I could have made a mistake.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>\u00a9 2016 Don Baker dba android originals LC Today, Aug 2, 2016, I filed a Provisional Patent Application for switching circuits to connect dual-coil humbucking pickups and matched single-coil pickups into humbucking quads, including digital controls.\u00a0 I\u2019ve read that no &hellip; 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