Formula to create simple MIDI velocity curves - math

Question

I'm trying to figure out a simple math formula that will allow me to apply various velocity curves to an incoming MIDI value. In the picture below, the starting x,y is (0,0) and ending x,y is (127,127). I'm trying to get a formula with a single variable that will allow me to produce simple expanded or contracted curves always bumping in the middle (by a degree of the variable). My input will be a value between 0 and 127, and my output will always be between 0 and 127. This seems like it should be easy, but my college calculus is escaping me at the moment.

Answer 1

Let's start by marking the four corners as:
S = (0,0)
E = (127,127)
U = (0,127)
V = (127,0)
You are looking for a circle equation which passes through E, S and a third point Z on the line between U and V. Let's mark it Z:
Z(t) = t*U + (1-t)*V
I will use the fact the the perpendicular bisectors of two chords meet at the center.
The bisectors B1 and B2 are:
B1 = (Xz/2 , Yz/2)
B2 = ((127+Xz)/2 , (127+Yz)/2)
The slopes of the perpendiculars are the negative inverse of the slope of the chords so:
Slope1 = - (127-Xz)/(127-Yz)
Slope2 = - Xz/Yz
Using the straight line equation for a point and a slope
L1 = y = Slope1*(x - B1x) + B1y
L2 = y = Slope2*(x - B2x) + B2y
The center of the circle C is their intersection:
Cx = [B2y-B1y + Slope1*B1x - Slope2*B2x]/[Slope1 - Slope2]
Cy = [Slope2*(B2y-B1y + Slope1*B1x - Slope2*B2x)]/[Slope1-Slope2] - Slope2*B2x +B2y
So we have the left hand side of the circle equation:
(x-Cx)^2 + (y-Cy)^2 = R^2
What's missing is the radius. But it is just the distance between C and any of our initial points. Computing from S is the easiest because it is (0,0):
R = square-root( [Cx-Sx]^2 + [Cy-Sy]^2 ) = square-root( Cx^2 + Cy^2)
So finally if you substitute all the definitions (probably easier to do in a program than to type it here) then you will obtain a function of a single variable t:
(x-Cx)^2 + (y-Cy)^2 = Cx^2 + Cy^2
note: you will get a straight line for t = 0.5 , but you can easily substitute t' = t-0.5 and only play with t'

Answer 2

I wanted a simple quadratic bezier curve from p0 (0,0) to p2 (127,127) based on a movable control point p1 that would range between (0,127) and (127,0). I wanted the position of the control point to be determined by a deviation variable. This is how I solved it based on my deviation variable being a number between -100 and 100 (0 being a linear line, the negative side representing pic 2 and positive side representing pic 3).
/// <summary>
/// Converts a MIDI value based on a velocity curve
/// </summary>
/// <param name="value">The value to convert</param>
/// <param name="deviation">The amount of deviation from a linear line from -100 to 100</param>
private int ConvertMidiValue(int value, double deviation)
{
if (deviation < -100 || deviation > 100)
throw new ArgumentException("Value must be between -100 and 100", "deviation");
var minMidiValue = 0d;
var maxMidiValue = 127d;
var midMidiValue = 63.5d;
// This is our control point for the quadratic bezier curve
// We want this to be between 0 (min) and 63.5 (max)
var controlPointX = midMidiValue + ((deviation / 100) * midMidiValue);
// Get the percent position of the incoming value in relation to the max
var t = (double)value / maxMidiValue;
// The quadratic bezier curve formula
// B(t) = ((1 - t) * (1 - t) * p0) + (2 * (1 - t) * t * p1) + (t * t * p2)
// t = the position on the curve between (0 and 1)
// p0 = minMidiValue (0)
// p1 = controlPointX (the bezier control point)
// p2 = maxMidiValue (127)
// Formula can now be simplified as:
// B(t) = ((1 - t) * (1 - t) * minMidiValue) + (2 * (1 - t) * t * controlPointX) + (t * t * maxMidiValue)
// What is the deviation from our value?
var delta = (int)Math.Round((2 * (1 - t) * t * controlPointX) + (t * t * maxMidiValue));
return (value - delta) + value;
}
This results in value curves ranging between the three shown below (blue is -100, gray is 0 and red is 100):