[PSA] Hexagird, BPM manipulations, Center of triangle

Hello.
Since I'm continuously getting the same questions in #modhelp, I decided to make this mini-guide.
Contents:
• 1. Editing the 3rd+ digit after the dot of BPM multiplier
2. Easy way to make a hexagrid
3. Finding the center of triangle
1. Editing the 3rd+ digit after the dot of BPM multiplier

You can manipulate the 3rd+ (0.00x) digit of the BPM multiplier in timing menu without opening the .osu file.

Why?

Usually if you're using hexagrid or a very slow slider (<0.5x) the 2nd digit of multiplier after the dot may affect the slider length way too drastically, like this: if you add 0.01 more, it can move too much: tho if you add 0.005 then it might fit exactly on the hexagrid: The problem

The editor doesnt show the change and visually rounds it to the closest digit. So after editing the SV to, say, 0.806: and pressing OK you will see only rounded number in the menu: However, in actuality the SV is 0.806, not 0.81.

2. Easy way to make a hexagrid

Why?

Hexagrid is way of structured object placement, which allows the mapper to keep the consistency in visual spacing. Also it looks really nice: How to?

Step 1: Create 2 circles not so far from each other - depends on what hexagrid you're aiming for, but dont make them overlapping each other, the closest range is when the fading out approach circles are barely touching each other: Step 2: Copy paste these two circles to somewhere and put them in next timing interval: Step 3: Rotate (ctrl+shift+R) the copy-pasted pair of circles by -120 degrees: Step 4: Stack the rotated pair of copy pasted circles to the original pair of circles: Step 5: Remove the stacked circle of w/e pair so you can have a triangle now: Step 6: Copy paste and move that triangle to have it partially stacked with original triangle: Step 7: Remove the stacked on top of each other circles and continue doing the Step 6 till you get the hexagrid as big as you need: 3. Finding the center of triangle

Why?

Welll, usually I'm using this method to find the 3rd circle of some pair of cirls to keep the visual spacing, also its nice to make a proper triangle without using rotations and readjusting the triangle polygon (ctrl+shift+D).

How to?

Step 1: Get a triangle (Im showing it as a centered one, but this works for any equilateral triangle): Step 2: Copy paste 2 circles of that triangle to somewhere: Step 3: Use the scaling (ctrl+shift+S) tool on that copied pair and set the value to 0.5: Step 4: Stack the rescaled pair on the initial place and remove the stacked circle: Step 4: Cop paste and rescale that middle circle and the opposite circle to somewhere and rescale it to 0.66 with scaling tool: Step 4: Stack the final rescaled pair and remove the stacked circle: That's it for now, maybe I'll add something in future. Thanks for reading.
Hm this is really useful i was just going to use a hexagrid just now by coincidence thank you very much I'm pretty sure that any output of Ctrl+Shift+D is always centered around the 320,240 middlepoint and as such all the steps done in "Finding the center of a triangle" are unnecessary if you know beforehand that you want a note centered. Just place it in the middle of the playfield and move it along with the triangle.

Endaris wrote:

I'm pretty sure that any output of Ctrl+Shift+D is always centered around the 320,240 middlepoint and as such all the steps done in "Finding the center of a triangle" are unnecessary if you know beforehand that you want a note centered. Just place it in the middle of the playfield and move it along with the triangle.
UHHH yea i mean u can use that method for any equilateral triangle, so lets say u have 2 circles somewhere and you want to make a structured thing, so you find the 3rd one (to make a equilateral triangle) and then you find the center of it, without moving or rotating it.

fixed btw
Shouldnt tbis be under Mapping Techniques?
Cool guide.

Short-cut for finding the center of any triangle:

Create the triangle. Take any two of the circles, copy/paste. Rotate them by 30 degrees. Scale by 0.57735. And you'll get the center. The premise is that the ratio between the length from the mid-point of the triangle to one of its vertices and the ratio of its side length is 1/2(2 divided by root 3) or 0.577350269.