Count triads?

Hello,

You must observe the tribal tones.

Each triad on C, for example, has the tribal tones C, E and G.

C-E-G is a thirsty-three sound (first the big, then the small terz). You don’t need any sign at C as a basic tone. But if you want to make the whale sound, you need the little terz first. It must be something with E.

If C-E is a big terz, C-It is a small one. Since Es-G is again a big one, everything fits.

For the reduced triad, You still lower the G next to the E, so C-Es-Ges. It would be two big ancestors over each other.

So C-E-Gis. Since it must always be something with C, E and G, you can’t do it at all.

With A as a basic tone, you still need the stem tones C and E:

A major (first big, then small terz): A-Cis-E; a-moll (first small, then big terz):

A-C-E; A reduced (two small terces): A-C-Es; A excessive (two large terzens):
A-Cis ice.

So you always write down the tribal tones first, so Grundton as first, the third and the fifth tone from the basic tone, calculated afterwards. Then you can consider which signs you have to give this trinket for the four possible trinities. Never deviate from the trunk tones!

D-Ges-A would not be a correct three-tone, even if it sounds like the thirsty sound D-Fis-A and is also beaten to the same piano keys, because G is the tribal tone of D from the (if D is counted) fourth tone (D-E-F-G). But you need the third tone, i.e. the tribal tone F, which is then provided with the necessary signs, depending on the type of trinity.

Thus, in the basic position always: tribal tone 1, tribal tone 3, tribal tone 5 in the order of the tone conductors with the corresponding basic tone.

Of course, it is also possible to reverse three-length: instead of the order 1-3-5 as the first reversal 3-5-1 (basic tone has evolved from bottom to top), as the second reversal 5-1-3 (the tones 1 and 3 have evolved upwards). But here too, the tribal tones 1, 3 and 5 are always in play (from the basic tone).

Best regards,

Willy