3
$\begingroup$

I'm trying to compute the shadow of 3D objects on the ground. To do this I'm assuming parallel rays from Sun origin.

Let's assume the Sun direction is given by azimuth $\varphi$ and elevation $\theta$ angles. So the direction unit vector is retrieved as:

$$ \hat{\mathbf{s}} = \mathbf{R}_y\left(\theta\right)\mathbf{R}_z\left(\varphi\right)\hat{\boldsymbol{\imath}}$$

where $\mathbf{R}_{x/y/z}$ are the rotation matrices along the $x, y, z$ axes given by, for example, Euler-Rodrigues equation, and $\hat{\boldsymbol{\imath}}$ is the unit vector along $x$-axis.

A point on a ray-line has the equation: \begin{equation} \mathbf{r} = \mathbf{r}_p + \lambda\hat{\mathbf{s}} \tag{1} \end{equation}

being $\mathbf{r}_p$ a point on that line.

A point on a general plane is given by the equation:

\begin{equation} \left(\mathbf{r} - \mathbf{r_0}\right)\cdot \hat{\mathbf{n}} = 0 \tag{2} \end{equation}

being $\mathbf{r}_0$ a point on the plane and $\hat{\mathbf{n}}$ the unit vector normal to that plane.

Ensuring the validity of both equations (1) and (2) allows to compute the $\lambda$ multiplier and retrieve the position vector of the projected point (writing $\tilde{\mathbf{r}} = \mathbf{r_p} - \mathbf{r_o}$)

\begin{gather} \lambda = - \left(\frac{\tilde{\mathbf{r}}\cdot \hat{\mathbf{n}}} {\hat{\mathbf{s}}\cdot\hat{\mathbf{n}}}\right) \\ \mathbf{r}_\text{proj} = \mathbf{r}_p - \left(\frac{\tilde{\mathbf{r}}\cdot \hat{\mathbf{n}}} {\hat{\mathbf{s}}\cdot\hat{\mathbf{n}}}\right)\hat{\mathbf{s}} \tag{3} \end{gather}

Applying eq. (3) to all the vertices of a mesh $\left\{\mathbf{r}_{p_i}\right\}$ gives the shadow.

A problem can be noted: $\hat{\mathbf{s}}\cdot \hat{\mathbf{n}}$ can be zero (e.g. Sun is horizontal, i.e. the ray is ortogonal to the normal of plane).

Are there some conceptual mistakes in my reasoning? Is there a way to rewrite the operation as a linear operator applied to the set of vertices position vectors?

$$ \left[\mathbf{A}\right]\left\{\mathbf{r}_{p_i}\right\} = \mathbf{r}_\text{proj}$$

  • 0
    The silhouette projection..surface normal perpendicular to ray. – 2017-02-25
  • 0
    Sorry. I said that $\hat{\mathbf{s}} \cdot \hat{\mathbf{n}}$ was zero when Sun is vertically over the object, that is not right. That happens when the sun is parallel to the object and intuitively the projected shape goes to infinity. I added a new question about rewriting the requested operations as a linear operator applied to the set of vertices coordinates. – 2017-02-25
  • 1
    WLOG I think you could consider a rigid object of arbitrary orientation in space when rays parallel to z-axis cast a shadow on the horizontal x-y plane. – 2017-02-26
  • 0
    [Here](https://gist.github.com/rubendibattista/8731dd1a82553bd3b692616405e349f3) I uploaded an example script in Python2. – 2017-02-26
  • 0
    OK, also works for opaque solids? – 2017-02-27
  • 0
    @Narasimham in theory what I would do is assigning to each vertex an opacity value. So the projected point will have an intrinsic light intensity value that is $I_\text{proj}=\epsilon_p I_p$ where $0<\epsilon_p<1$ is the opacity. Could this be reasonable? – 2017-02-27
  • 0
    Hope I understand clearly. Unless the body is opaque solid normalcy condition would not form a smooth continuous projection line of silhouette. – 2017-02-27

1 Answers 1

1

I address here your second question, but I think that it also answers a little the first one.

Consider oblique projection where unit vertical vector $\vec{k}$ has for its shadow $\left(\begin{array}{c}a\\b\end{array}\right)$ on the $xOy$ plane.

Its matrix is:

$$\left(\begin{array}{ccc}1 & 0 & a\\0 & 1 & b\\0 & 0 & 0\end{array}\right)$$

This is how I have realized the following figure representing the projection (in black) of an helix (in red):

enter image description here

The explanation relies plainly on the way a transformation matrix is built: its successive columns are the images of vectors $\vec{i},\vec{j},\vec{k}$ (directing $x$,$y$,$z$ axis resp.)

  • $\vec{i},\vec{j}$, being already on $x0y$ plane are projected on themselves.

  • $\vec{k}$ is projected on its shadow $\left(\begin{array}{c}a\\b\\0 \end{array}\right)=\left(\begin{array}{c}\cos(\theta)\sin(\lambda)\\ \sin(\theta)\sin(\lambda)\\ 0 \end{array}\right)$

($\theta$ being the azimuth and $\lambda$ the latitude of the sun's direction).

Edit: Another way to define the third vector using the sun's direction $\vec{s}$ is $\left(\begin{array}{c}a\\b\\0 \end{array}\right)=\left(\begin{array}{c}\left(\dfrac{\vec{s}.\vec{i}}{\vec{s}.\vec{k}}\right)\\\left(\dfrac{\vec{s}.\vec{j}}{\vec{s}.\vec{k}}\right)\\ 0 \end{array}\right).$

  • 0
    No comment ?... – 2017-02-26
  • 0
    Yes, here I am. Thanks for your comment, I really did not think about the simple bare definition of linear operators (that is they are the transformed $\hat{\boldsymbol{\imath}}, \hat{\boldsymbol{\jmath}}, \hat{\mathbf{k}}$). Could you please add the matrix form of the operator in terms of pure vectors (so just considering the sun direction as a the $\hat{\mathbf{s}}$ unit vector (without formal reference to the angles)? If my computation are right the projected $\hat{\mathbf{k}}$ vector should be: $\hat{\mathbf{k}} - \frac{1}{\hat{\mathbf{s}}\cdot \hat{\mathbf{n}}}\hat{\mathbf{s}}$ – 2017-02-26
  • 1
    See the little Edit I have added. – 2017-02-26