In your first case:
   o. 4 %~ i.5
0 0.785398 1.5708 2.35619 3.14159
   o. (i.5)%4   NB. Same as above ...
0 0.785398 1.5708 2.35619 3.14159
   (o.i.5)%4    NB. Same again ...
0 0.785398 1.5708 2.35619 3.14159

So the numerator is (o. * i. 5) and the denominator is 4

In your last case:
   4p1 %~ (i.5)
0 0.0795775 0.159155 0.238732 0.31831
   (i.5) % 4p1  NB. Same as above ...
0 0.0795775 0.159155 0.238732 0.31831
   (i.5) % o.4   NB. Same again ...
0 0.0795775 0.159155 0.238732 0.31831

So the numerator is (i.5) and the denominator is o.4

i.e. you have inadvertently moved the o. (or 4p1) to the denominator, likely 
without realising

Remember in your first example that o. is a verb, whose argument is the result 
of  (4 %~ i.5).

HTH …/Rob

> On 21 Apr 2016, at 6:28 PM, Martin Kreuzer <[email protected]> wrote:
> 
> Hi all -
> 
> Somewhere along the way I picked up this small expression:
>   o.4 %~ i.5
> 0 0.785398 1.5708 2.35619 3.14159
> 
> Just for the fun of it, I tried building it from the ground up like this:
>   (i.5) % 4
> 0 0.25 0.5 0.75 1
>   1p1 * (i.5) % 4
> 0 0.785398 1.5708 2.35619 3.14159
>   1p1 * 4 %~ (i.5)
> 0 0.785398 1.5708 2.35619 3.14159
> 
> I then bravely thought this would work as well
>   4p1 %~ (i.5)
> 0 0.0795775 0.159155 0.238732 0.31831
>   (4p1) %~ (i.5)
> 0 0.0795775 0.159155 0.238732 0.31831
> but -unfortunately- it didn't ...
> 
> The "odot" page of NuVoc has given me the impression that the two should be 
> interchangeable (despite maybe performance issues).
> 
> Spoilers: A newbie at work.
> -M
> 
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