# How to count polyrhythm 3 against 4 in common time?

Polyrhythm 4:3 in 3/4 meter can be "inverted" to 3:4 in 4/4 meter, like this...

...the notation is easy enough to write, because we can use tuplets to gloss over the division problem.

Converting the 3/4 meter notation pattern to a counting grid I get...

``````3 |x| | |x| | |x| | |x| | |
4 |x| | | |x| | | |x| | | |
1     a 2   &   3 e
``````

...the bottom line, under the x's, is how I count it out loud.

But if I convert the 4/4 meter notation to a grid, the uneven division problem is apparent, and I run into a counting problem...

``````4 |x| | | | |x| | | | |x| | | | | |
4 |x| | | |x| | | |x| | | |x| | | |
1       2 e     3   +   4
``````

...the upper part isn't evenly spaced. Obviously the problem is 16 boxes on the grid can't be divided evenly by 3.

1st question:

If the 4 tuplets in 3/4 are timed to the exact same speed as the straight quarter notes in 4/4 will the other parts with 3 notes have the exact same timing regardless of which meter is used? I think the answer is yes, but I don't know how to prove that.

2nd question:

More important to me is how to count the pattern in 4/4 meter. It seems important to me to be able to count this pattern with either the 3 or 4 note base meters. If the triplet group is encountered in a 4/4 meter setting, I need to know how to count it while keeping a solid 4/4 count. Even if the 3/4 and 4/4 patterns are mathematically equivalent, I need some practical way to count it in 4/4.

I suppose the counting in my 4/4 example is close enough to be acceptable, but I don't want to practice it if there is a better way to count it.

Also, I understand that 4:3 polyrhythm in African music is normally written with a compound meter so so in that sense trying to count the pattern in 4/4 meter is contrary to that tradition.

EDIT

Is this the better way to do it? Either use 12/8 meter and make the tempo of the quarter note equal to the dotted quarter note, or use triplet eighths with ties.

• I think you're going the wrong way the moment you try to fit any rhythm to a base-2 grid. Don't do that :-) Oct 24, 2018 at 13:37
• Hi @CarlWitthoft, don't put any rhythm in a base-2 grid? Plain quarter notes in a base-2 grid would be ok. Do you mean don't put any polyrhythm in a base-2 grid? Oct 24, 2018 at 14:23
• @CarlWitthoft, I made an edit to my question which I hope addresses your point about base-2 vs. base-3 grid while at the same time uses 12/8 or 4/4 meter. Oct 24, 2018 at 19:09

Your problem is that your 4/4 grid divides the quarters into four sixteenths, and, as you've noticed, sixteen is not divisible by three without a remainder. The subdivision of the quarter note must have three as one of its prime factors; if it does not, then the triplet half notes will not coincide with the subdivision.

The easiest factor to use, unless the tempo is very slow, is of course three (that is, subdivide the quarters into 8th note triplets). Equivalently, you can think of it as 12/8.

Then each of the triplet half notes in your example gets four of those subdivisions. Note that this preserves the normal relationship between half notes and eighth notes, that is, a factor of four, but within the parallel triplet universe.

When you do that, you just invert the first grid:

``````    _3_   _3_   _3_   _3_
/   \ /   \ /   \ /   \
4 |x| | | |x| | | |x| | | |
4 |x| | |x| | |x| | |x| | |
``````

Then you could count it

``````    _3_   _3_   _3_   _3_
/   \ /   \ /   \ /   \
4 |x| | | |x| | | |x| | | |
4 |x| | |x| | |x| | |x| | |
1     2 &   3   a 4
``````

I chose the offbeat syllables based on the assumption that you count the triplet quarter notes as "1 and-a 2 and-a 3 and-a 4 and-a"; you can of course change them to suit:

``````     _3_   _3_   _3_   _3_
/   \ /   \ /   \ /   \
4 | | | | | | | | | | | | |
4  1 & a 2 & a 3 & a 4 & a
``````

In my experience, polyrhythmic music involving a 3:4 polyrhythm against a 4-beat meter, as in your second example, is better conceived in terms of 12/8. A true juxtaposition of 4/4, one that is subdivided in a duple fashion into eights and sixteenths, against a true 3/4 interpretation of the same span isn't something I've ever encountered in musics that have idiomatic polyrhythmic structures, since all of the ones that I'm familiar with rely on an irreducible grid built on the smallest possible rhythmic duration that makes sense in both meters (sixteenth notes in your first example, and triplet-eighths in your second). I'm sure there are experimental musics that contrast the two directly, but I don't think that's what you're getting at here. Therefore I'd counsel using triplet counting in the second case: either (1-and-a 2-and-a...etc) or (1-trip-let 2-trip-let), whichever is more comfortable, which will both enable you to exactly place the beats belonging to the contrasting meters and to conceive of the way that they relate to each other in a nominal 4-beat meter.

Basically,your explanation is quite limited and does not get to the brass tacks of polyrhythms. To be able to fit any possible rhythmic value into any kind of beat, or beat division(eight note in 6/8 is a bit division) e.g 4:5 (5/4) 5:6 (6/8) one needs to create a grid equal to multiplication outcome of both digits displaying the actual ration of proposed division. In the first example, 4x5=20, which means the grid will have 20 units. One can easily see that there will be 5 beats each comprised of 4 units, or 4 beats each comprised of 5 units as depicted by the actual ratio of 4:5! Now that we know we can divide the grid both ways, we can go ahead and connect 5 units by use of a slur creating 4 new beats over the timeline of our grid. Every possible polyrhythm can be solved that way. As a bonus, I’ll add that each of these polyrhythms is a palindrome (reads the same forth and back).

• I wasn't explaining, I asked a question. But, I understand your point: the grid size will be the product of the two numbers. Thanks. May 17, 2022 at 16:44

My take on this.

To your first question: If the tuplets in the 3/4 are each exactly the same length as the crotchets in the 4/4 then each measure will take the same amount of time. Since you have three evenly spaced notes in each or the two bars (the tuplets in the 3/4 and the triplet in the 4/4) they must also be timed exactly the same.

To your second question: I think you have made an assumption here. When fitting four notes into three beats you have subdivided the beat into 4 and everything works. When fitting three notes into four beats you have also subdivided into 4 - but why? You need to subdivide the beat into three and then it works doesn't it?

The usual pattern mmnemonic here is "pass the goddamn butter". A counting grid needs a division of the smallest common multiple as factor, let's use the minimal 12:

``````|x| | |x| | |x| | |x| | |
|x| | | |x| | | |x| | | |
``````

You can count this out in twelveths but the final execution really is a regular 4-group and a regular 3-group occupying the same time and running independently. You start by trying to make the pattern match the mnemonic, then you listen to each group alternatively to check that it is self-consistent.

The smallest common subdivision is a theoretical tool, it won't really work well for stuff like 5 against 4 except when programming drum computers.

• Thanks, but you only restated the issue. I understand why it's easy in triple meter and how to count it. I'm looking for an approach in common time. Oct 24, 2018 at 12:51