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Given a note duration (whole, half, quarter, etc.), a tempo (in beats per minute), and a time signature, how long does that note last in milliseconds?

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A slightly better question to ask would be, “how much time is there between this note and the next note?” In a typical situation a note isn’t sounded for the entire length of time allotted to it :^) –  bdesham Oct 19 '11 at 3:52
This is very closely related: physics.stackexchange.com/q/8133/124 –  David Z Oct 19 '11 at 6:19

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You can solve this by picking a simple, specific example and then generalizing it as necessary. Let’s start with a simple case, where the tempo is such that there are 120 quarter notes per minute. The frequency of quarter notes is then $$ \frac{\text{120 quarter notes}}{\text{1 minute}} \cdot \frac{\text{1 minute}}{\text{60 seconds}} \cdot \frac{\text{1 second}}{\text{1000 milliseconds}} = \frac{\text{0.002 quarter notes}}{\text{millisecond}} . $$ This is the number of quarter notes that occur each millisecond. For a more useful number, we can invert this fraction to see that the number of milliseconds taken by one quarter note is $$ \frac{1}{\frac{\text{0.002 quarter notes}}{\text{millisecond}}} = \frac{\text{1 millisecond}}{\text{0.002 quarter notes}} = \frac{\text{500 milliseconds}}{\text{1 quarter note}} . $$ You can do fancier things involving other notes, tempos, key signatures, etc. by first finding out how many of your note occur in one minute (120 quarter notes in this example), and then following this same process.

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Well, actually the key signature is irrelevant ;-) –  David Z Oct 19 '11 at 4:06
Heh, well spotted :-) –  bdesham Oct 19 '11 at 4:09
But what happens if you're in an obscure time signature? And you want to find the duration of an obscure note? Thanks for trying, but I think you picked a simple example. –  Zach Rattner Oct 19 '11 at 5:10
Let's say you want a 7/32nd note in a 5/4th time signature at 120 bpm. Then again, you have 120 quarter notes per minute = 2 quarter notes per second -> 500 ms per quarter note. Let's pass to milliseconds: 500 ms / 8 = 62.5 ms per 32nd note. Finally, you multiply by 7 to get the duration of the note: 62.5 ms * 7 = 437.5 ms per 7/32nd note, and you're done. You can remark that only the 'denominator' of a time signature counts, i.e. what counts as one beat - you don't need to know how many beats go in one measure. –  Gerben Oct 19 '11 at 5:21

protected by Qmechanic Apr 28 '13 at 23:18


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