Hi everyone, I'm a musician working on a developing computer program for control over expressive parameters of a sound in live performance. I’m trying to build an equation that uses real-time input data from an accelerometer to smoothly interpolate between two sets of variables between 0.0 and 1.0 at the same time a time-based envelope function runs. The effect is that both the accelerometer and the envelope are influencing the output for musical performance data.
I'm posting this in the algebra category because although my equation variables in exponents, I think the part of the problem I'm stuck on is one of algebra around getting the exponents on one side of the equation; after that, I know I can take a logarithm of both sides to find out the exponent.
I created the scenario in figure 1 to visualize this. This drawing is only a possible representation. The lowest and highest values are always 0 and 1 respectively but any of the values can be the lowest. The relationships between a, j, x; b, k, y; c, l, z; a, b, c; j, k, l; x, y, z are all to be the same:
j = (x - a) * (0.5)^p + a
k = (y - b) * (0.5)^p + b;
l = (z - c) * (0.5)^p + c
b = (c - a) * (0.5)^p + a
k = (l - j) * (0.5)^p + j
y = (z - x) * (0.5)^p + x.
Variables a, c, x, z and k are known when a performance begins. I want to find variables b, j, l and y. I think I can do this by finding p from the known variables a, c, k, x and z.
I think that I can find p by substituting the equations for j and l into the equation for k:
k = (l - j) * (0.5)^p + j; j = (x - a) * (0.5)^p + a; l = (z - c) * (0.5)^p + c
k = ( ( (x - a) * (0.5)^p + a ) - ( (x - a) * (0.5)^p + a ) ) * (0.5)^p + j
This feels like a big mess! I know I can substitute a variable for (1/2)^p and simplify it a little bit but I'm not getting very far:
Q = (0.5)^p
k = ( ( (x - a) * Q + a ) - ( (x - a) * Q + a ) ) * Q + j
k - j = ( ( (x - a) * Q + a ) - ( (x - a) * Q + a ) ) * Q
I know that if I can get Q on one side, then I can substitute (0.5)p back in and take the logarithm of both sides to find the power. Then, knowing p, I can easily find b and y using the earlier equations. However, I can’t seem to reduce this complex equation.
Can someone help me? If it's unclear, perhaps let me know how I can present it clearer. Thanks so much, Christian
I'm posting this in the algebra category because although my equation variables in exponents, I think the part of the problem I'm stuck on is one of algebra around getting the exponents on one side of the equation; after that, I know I can take a logarithm of both sides to find out the exponent.
I created the scenario in figure 1 to visualize this. This drawing is only a possible representation. The lowest and highest values are always 0 and 1 respectively but any of the values can be the lowest. The relationships between a, j, x; b, k, y; c, l, z; a, b, c; j, k, l; x, y, z are all to be the same:
j = (x - a) * (0.5)^p + a
k = (y - b) * (0.5)^p + b;
l = (z - c) * (0.5)^p + c
b = (c - a) * (0.5)^p + a
k = (l - j) * (0.5)^p + j
y = (z - x) * (0.5)^p + x.
Variables a, c, x, z and k are known when a performance begins. I want to find variables b, j, l and y. I think I can do this by finding p from the known variables a, c, k, x and z.
I think that I can find p by substituting the equations for j and l into the equation for k:
k = (l - j) * (0.5)^p + j; j = (x - a) * (0.5)^p + a; l = (z - c) * (0.5)^p + c
k = ( ( (x - a) * (0.5)^p + a ) - ( (x - a) * (0.5)^p + a ) ) * (0.5)^p + j
This feels like a big mess! I know I can substitute a variable for (1/2)^p and simplify it a little bit but I'm not getting very far:
Q = (0.5)^p
k = ( ( (x - a) * Q + a ) - ( (x - a) * Q + a ) ) * Q + j
k - j = ( ( (x - a) * Q + a ) - ( (x - a) * Q + a ) ) * Q
I know that if I can get Q on one side, then I can substitute (0.5)p back in and take the logarithm of both sides to find the power. Then, knowing p, I can easily find b and y using the earlier equations. However, I can’t seem to reduce this complex equation.
Can someone help me? If it's unclear, perhaps let me know how I can present it clearer. Thanks so much, Christian
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