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Thread: Rests help! Urgent

  1. #1

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    Default Rests help! Urgent

    Ok, first time I asked for help like this, I need to know while playing 16th notes, how many notes does an 8th note rest go till? Does it end on the "&" of 1? (1 e &) a 2 e & a 3 e & a 4 e & a To clarify, I think that the duration of an 8th note rest is in the parentheses of the diagram I just put out above, please tell me if I am wrong, and how long the rest goes!
    This is really confusing, and makes a lot more sense when I say it verbally.
    ( I need the answer by Friday morning!)
    -DrumRookie

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  2. #2

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    Default Re: Rests help! Urgent

    DR, think of it this way. There will be two 16th notes to an 8th note. So if your 16th notes are counted 1 e & a, 2 e & a, 3 e & a, 4 e & a in 4/4 time, then the 8th notes are counted 1 & 2 & 3 & 4 &. Therefore, your 8th note rest in 16th note counting, using your parentheses would take two counts and look like this:

    (1 e) & a, 2 e (& a), 3 (e &) a, 4 e & a - each parentheses covering two counts.
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  3. #3

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    Default Re: Rests help! Urgent

    If you're counting 16th notes, an 1/8 note rest will equal two of those 1/16th note counts....examples as follows:

    1(e)+(a) is 2 1/8th notes played on "1" & "+".

    1e+(a) is 3 1/16th notes played on "1e+", followed by a 1/16th note rest. However, you can also say (and therefore write) it as 2 1/16th's and an 1/8th, as the last 1/16th note in the group ("+") can be joined to the 1/16th note rest. So "+" 's note value can be changed to an 1/8th, as mathematically as we all know, 1/16 +1/16 = 1/8.

    1e(+a) is 2 1/16th notes played on "1e", followed by an 1/8th note rest....this is because in this example, two 1/16th note rests can be added together. BUT if the very last rest was changed to a 1/6th note, it'd be very different, as you will then have 1e(+)a.
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