<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN"> <HTML><HEAD> <META http-equiv=Content-Type content="text/html; = charset=iso-8859-1"> <META content="MSHTML 6.00.2715.400" name=GENERATOR> <STYLE></STYLE> </HEAD> <BODY> Yes, I would agree = with you about the oddity of this temperament, but I am pretty sure that I know what = happened. To me it sounds like this is a kind of Kirnberger II temperament, with = some modifications and some misunderstandings all mixed = together:<?xml:namespace prefix = o ns = "urn:schemas-microsoft-com:office:office" = /><o:p></o:p> <o:p> </o:p> The temperament = should have started on C instead of A. Unfortunately, the only way to make the = directions you were given make sense is to use A as a starting note for the = temperament. This is not such a huge deal, but it may have caused some confusion in = the transposition. <o:p> </o:p> With a temperament = based on a cycle of fifths, there should be alterations that total a -23.46 cent = comma. Now in theory, this amount can be distributed, either equally or unequally, = among any number of intervals (1-12). Kirnberger II has two unequal tempered intervals. The following is Kirnberger II transposed with A as the = starting note:<o:p></o:p> <o:p> </o:p> A         440.00<o:p></o:p> D         586.66 [P5 down]<o:p></o:p> G         782.22 [P5 down]<o:p></o:p> C         521.48 [P5 down]<o:p></o:p> F          695.31 [P5 down]<o:p></o:p> Bflt       = 463.54 [P5 down]<o:p></o:p> <o:p> </o:p> (A        440.00)<o:p></o:p> E          660.00 [P5 up]<o:p></o:p> B          495.00 [P5 up]<o:p></o:p> <o:p> </o:p> (A        440.00)<o:p></o:p> C#       = 550.00  [M3 up]<o:p></o:p> G#       = 825.00  [P5 up]<o:p></o:p> D#       = 618.75  [P5 up]<o:p></o:p> <o:p> </o:p> F# is = tempered with an equal beat rate between D-F# [M3] and F#-B [P4]<o:p></o:p> <o:p> </o:p> D-F# as a Just M3 = would make F# 733.32<o:p></o:p> F#-B as a Just P4 = would make F#  742.50<o:p></o:p> To get an equal beat = rate add the two frequencies and divide by two: <o:p> </o:p> F#        737.91  [Equal Beating]<o:p></o:p> <o:p> </o:p> <o:p> </o:p> <o:p> </o:p> The “Other” temperament<o:p></o:p> Tune by Just = 5ths<o:p></o:p> E should be tempered = A and B [mean tone?]<o:p></o:p> F# should be a Just = M3 with D<o:p></o:p> <o:p> </o:p> (First tune the cycle = without the tempered E)<o:p></o:p> A         440<o:p></o:p> E          660<o:p></o:p> B          495<o:p></o:p> F#        742.5<o:p></o:p> C#       556.87<o:p></o:p> G#       835.32<o:p></o:p> D#       626.48<o:p></o:p> A#       469.86<o:p></o:p> F          352.40<o:p></o:p> C         528.60<o:p></o:p> G         792.89<o:p></o:p> <o:p> </o:p> Next tune D as a Just M3 to F#: <o:p> </o:p> D         594<o:p></o:p> <o:p> </o:p> I am not sure I = understand exactly how to make E a meantone between A and B. Generally a meantone temperament uses narrow 4ths and wide 5ths. The only thing that I can = think of is that that you tune A - E - B, and then lower E to make a widened 5th = between E-B (a widened 5th between A-E would be reverse temperament, so that = will not work). But how much should it be lowered? How do you tell? The only way = to make this work [with only two tempered intervals] is to let E absorb the rest = of the comma (-21.576): <o:p> </o:p> E          651.83       (-8 bps) <o:p> </o:p> <o:p> </o:p> <o:p></o:p>  <o:p></o:p>  <o:p></o:p>  After you follow = through with the rest of the adjustments, the frequencies should be as listed below = (hopefully I haven’t made too many calculation errors): <o:p> </o:p> Kirnberger II           &nbs= p; "The Other"<o:p></o:p> A      = 440           &nbs= p;         440<o:p></o:p> A#    469.9           &nbs= p;      469.9<o:p></o:p> B       = 495           &nbs= p;         495<o:p></o:p> C      = 528.6           &nbs= p;      528.6<o:p></o:p> C#    556.9           &nbs= p;      556.9<o:p></o:p> D      = 594.67           &nbs= p;    594<o:p></o:p> D#    618.7           &nbs= p;      626.5<o:p></o:p> E      = 660           &nbs= p;         651.83<o:p></o:p> F       = 704.8           &nbs= p;      704.8<o:p></o:p> F#    737.9           &nbs= p;      742.5<o:p></o:p> G      = 792.9           &nbs= p;      792.9<o:p></o:p> G#    825           &nbs= p;         835.32<o:p></o:p> <o:p> </o:p> <o:p> </o:p> <o:p> </o:p> I told him that there weren't = going to be any thirds tuned "pure" in the Kirnberger.<o:p></o:p> Well, Kirnberger II = does have one M3, which normally falls on C-E. The M3 is used to lower the amount of = the comma by -21.5 cents. Remember the total comma is only -23.462 cents. <o:p> </o:p> And if any of the historical = temperaments would have a pure third, it would be C-E. Was I = wrong?<o:p></o:p> Not at all . . . well = a little bit. The general understanding is that well temperaments will start on C = and then move away. The result is that the closer you remain in relation to = C, the more harmonious the key area will sound. But there is nothing stopping a = tuner from starting on C# instead! <o:p> </o:p> <o:p> </o:p> <o:p> </o:p> One last interesting = note: in working through the calculations of Kirnberger II, I found that the = comma never really gets properly absorbed. The M3 accounts for -21.5 cents, but the = interval tempered with equal beating rates between M3 and P4 add an additional = -10.7 cents; this means that this last interval was tempered too much (+8.7 = cents). To officially close the loop, set the intervals to beat equally, and then = raise the tempered interval slightly. <o:p> </o:p> <o:p> </o:p> Bradley M. Snook</BODY></HTML>