How to create RGB-colors with equal vector norm?

I am trying to do the following:

• Interprete RGB color values as a 3-D vektor

• Transform the RGB color into polar coordinates L, teta and phi

• Create colors with constant L

The math behind this is the standard transformation into spherical coordinates:

b = L * Sin(teta) * Cos(phi)
r = L * Sin(teta) * Sin(phi) 
g = L * Cos(teta)

With L beeing the norm of the RGB vektor I want to create.

L=(r^2+b^2+g^2)^0.5 'between 0 and 441.67 = sqrt(3)*255

So far so good. Onto the first issue: r,g and b need to be between 0 and 255

This means that this holds true (I calculated that myself, it's possible that it has errors in it):

If L>255 Then
   arccos(255/L) < teta < pi/2 - arccos(255/L)
   arccos(255/L) < phi < pi/2 - arccos(255/L)
End If

Here my calculations:

The idea is that teta goes from (pi/4 - delta) to (pi/4 + delta) where
Which basically means that teta and phi are restricted in what they can be according to the ratio of 255/L. This restriction gets upto a point were L=441.67 the maximum length. Here phi and teta can only be pi/4.

I think upto this points my conclusions are valid, but there is a problem I cannot figure out.

If phi = teta = pi/4 and L = 441.67 the RGB value should be (255,255,255) and the forumlas above yield: RGB= (221,255,221)

Which is what the forumlas for r,g and b above yield if phi = teta = pi/4 and L = 441.67 are inserted.

So how would one generate RGB vektors with a constant norm that are still valid RGB values? (0 < r,g,b < 255)

So how would one generate RGB vectors with a constant norm that are still valid RGB values? ($0 \le r,g,b \le 255$)

There is a simpler algorithm. Let $L$ be the given norm.

• Verify that $0\le L\le 255\sqrt3$.

• Let r be a random number between $\sqrt{L^2-\min(2\cdot255^2, L^2)}$ and $\min(255, L)$.

• Let $M=\sqrt{L^2-r^2}$. Let g be a random number between $\sqrt{M^2-\min(255^2, M^2)}$ and $\min(255,M)$.

• Let $b =\sqrt{L^2-r^2-g^2}$.

Some minor adjustment is needed if $r,g,b$ should be integers, in which case only approximate solutions can be found in general.

Here is the right computation if we want to find the range of $\theta$ and $\phi$.

• If $0\le L\le 255$, then $0\le\theta,\phi\le\pi/2$

• If $255\lt L\le 255\sqrt3$, let $M=\sqrt{\max(0, L^2-2\cdot255^2})$.

we have
$$\begin{align}
&\quad\quad\begin{cases}b = L \sin(\theta) \cos(\phi)\\
r = L \sin(\theta) \sin(\phi)\\
g = L \cos(\theta)\\
0\le b,r,g\le 255\\
0\le\theta,\phi\le\pi/2
\end{cases}\\
&\Longleftrightarrow
\begin{cases}b = L \sin(\theta) \cos(\phi)\\
r = L \sin(\theta) \sin(\phi)\\
g = L \cos(\theta)\\
\sqrt{\max\left(0,(L\sin(\theta))^2-255^2\right)}\le b,r\le \min(255,L\sin(\theta)) \\
M\le g\le 255\\
0\le\theta,\phi\le\pi/2,\, \theta\not=0
\end{cases}\\
&\Longleftrightarrow
\begin{cases}
\sqrt{\max\left(0,1-\left(\dfrac{255}{L\sin(\theta}\right)^2\right)}\le \cos(\phi)\le \min\left(1, \dfrac{255}{L\sin(\theta)}\right)\\
\sqrt{\max\left(0,1-\left(\dfrac{255}{L\sin(\theta}\right)^2\right)}\le \sin(\phi)\le \min\left(1, \dfrac{255}{L\sin(\theta)}\right)\\
\dfrac{M}{L}\le \cos(\theta)\le \dfrac{255}L\\
0\le\theta,\phi\le\pi/2,\, \theta\not=0
\end{cases}\\
&(\text{the inequalities on}\cos\phi\text{ put on the same restrictions on }\phi\\
&\text { as the inequalities on }\sin\phi)\\
&\Longleftrightarrow
\begin{cases}
\arccos\left(\min\left(1, \dfrac{255}{L\sin(\theta)}\right)\right)\le \phi\\
\phi\le\arccos\left(\sqrt{\max\left(0,1-\left(\dfrac{255}{L\sin(\theta)}\right)^2\right)}\right)\\
\arccos(\dfrac{255}L)\le\theta\le\arccos(\dfrac{M}{L})
\end{cases}
\end{align}
$$

The formula above tells a way to generate RGB vectors with a given norm $L$. Select $\theta$ and $\phi$ in the range given by the inequalities. Compute $r,g,b$ accordingly.

In particular, if you take $L=255\sqrt3=441.67$, then $\dfrac{255}L=\dfrac{M}L=\dfrac3{\sqrt3}$. $\theta=\arccos(255/L)=0.9553$. Since $\dfrac{255}{L\sin(\theta)}=0.7071$, we will get $\arccos(0.7071)\le \phi\le\arccos(0.7171)$. So $\phi=\pi/4$ and we will arrive at $b=255$, $g=255$ and $r=255$, which is the only possible combination of RGB vector for that value of $L$, as expected.

There have been several computational mistakes in the question. There has been a few in the previous versions of this answer as well. The comments below this answer might have become irrelevant.

StackExchange Computer Science Q#104911, answer score: 5