Geometric interpretation of orbital angular momentum eigenvalues The eigenvalues of the square of the orbital angular momentum are defined as:
$$ \hat{L}^2  |  \phi \rangle = \hbar^2 l(l+1) |  \phi \rangle $$
$$ \hat{L}_z  |  \phi \rangle = \hbar m |  \phi \rangle, \,\, \text{where}\,\,  m = -l, -l+1, ..., l$$
This derivation typically (for example in Griffiths) goes based on the corresponding operator form of $\hat{\vec{L}}$. On the other hand when explaining the concept of how does orbital angular momentum behave in QM a certain geometrical intuition based on the following diagrams that treat $\hat{\vec{L}}$ as a vector, is helpful (at least for me). Here is the example with $l = 2$:

My question is whether there is a simple geometrically based argument that connects such a diagram (including a corresponding uncertainty relation for components of $\hat{\vec{L}}$), and the showed above eigenvalues for $ \hat{\vec{L}}^2$  or is it only "spin can be viewed as rotating electron" type of analogy?
EDIT: After response of Johny, I want to reformulate the question.
The radius of the cone base can be derived as: $$ R^2_{cone} = \hbar^2 (l(l+1) - m^2 )$$
How to derive the same R using commutators or uncertainty relations of the components of $\vec{L}$?
 A: That diagrams (if I am not mistaken) represent the possible directions of the total angular momentum.
The angle between two vectors is given by (in this case the angular momentum $\vec{L}$ and its component in the $z$-direction):
$$\theta =\cos^{-1}\left(\dfrac{\left\langle j,m\left|\vec{L}\cdot\left(0,0,L_{z}\right)\right|j,m\right\rangle }{\sqrt{\left\langle j,m\left|\vec{L}\cdot\vec{L}\right|j,m\right\rangle }\sqrt{\left\langle j,m\left|L_{z}^{2}\right|j,m\right\rangle }}\right)
 =\cos^{-1}\left(\dfrac{\left\langle j,m\left|L_{z}^{2}\right|j,m\right\rangle }{\sqrt{j\left(j+1\right)}\sqrt{m^{2}}}\right)
 \\= \cos^{-1}\left(\dfrac{m}{\sqrt{j\left(j+1\right)}}\right)$$
If you replace $m$ and $j$ by the values you obtain the following angles:
$$(j=2,m=2) =\cos^{-1}\left(\sqrt{\dfrac{2}{3}}\right)\quad
(j=2,m=1) =\cos^{-1}\left(\sqrt{\dfrac{1}{6}}\right)\quad
(j=2,m=0) =0$$
These are the angles that the total angular momentum of the states $|j,m\rangle$ makes with the $z$-axis. The set of points that makes a fixed angle with the $z$-axis is given by the surface of a cone, like the ones represented in your picture.
The interpretation is that, even when you are in the state of maximum angular momentum in the $z$-direction, there is a projection into $x$ and the $y$ direction, which comes from the non-commutativity (uncertainty) of the angular momentum operators in each direction. Hence, the interpretation of the blue surfaces is the possible total angular momentum that you can have when you are in the state $|j,m\rangle$.
