User:BudjarnLambeth/The intervals I see as important: Difference between revisions
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! style="background:#fff0c8;" | 13/3<br/><small>lg tri neu 2nd^</small> | ! style="background:#fff0c8;" | 13/3<br/><small>lg tri neu 2nd^</small> | ||
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|37 | |||
| style="background:#90EE90;" | 0.0¢ ✓✓ | | style="background:#90EE90;" | 0.0¢ ✓✓ | ||
| style="background:#90EE90;" | 0.1¢ ✓✓ | | style="background:#90EE90;" | 0.1¢ ✓✓ | ||
| Line 1,131: | Line 1,132: | ||
Use the attached PDF to calculate the correct numbers. | Use the attached PDF to calculate the correct numbers. | ||
All the columns N/1 (eg 2/1, 3/1) are already correct. For the other columns, some cells are correct and some not. What you need to do is subtract the error (in cents (¢)) of the two harmonics involved, but don't disregard their direction. For example if harmonic 5 has +10¢ error and harmonic 3 has -9¢ error, then the total error of interval 5/3 is 19¢ (opposite errors add). If harmonic 5 has +10¢ error and harmonic 3 has +9¢ error, then the total error of interval 5/3 is 1¢ (same-direction errors subtract).</pre> | All the columns N/1 (eg 2/1, 3/1) are already correct. For the other columns, some cells are correct and some not. What you need to do is subtract the error (in cents (¢)) of the two harmonics involved, but don't disregard their direction. For example if harmonic 5 has +10¢ error and harmonic 3 has -9¢ error, then the total error of interval 5/3 is 19¢ (opposite errors add). If harmonic 5 has +10¢ error and harmonic 3 has +9¢ error, then the total error of interval 5/3 is 1¢ (same-direction errors subtract).</pre> | ||