Axonometry

The choice of the images of the axes and the foreshortenings

Notation:

• ${\displaystyle \alpha :}$ angle between ${\displaystyle \overline{z}}$-axis and ${\displaystyle \overline{x}}$-axis
• ${\displaystyle \beta :}$ angle between ${\displaystyle \overline{z}}$-axis and ${\displaystyle \overline{y}}$-axis
• ${\displaystyle \gamma :}$ angle between ${\displaystyle \overline{x}}$-axis and ${\displaystyle \overline{y}}$-axis.

The angles can be chosen so that ${\displaystyle 0^{\circ }<\alpha +\beta <{360}^{\circ }.}$
The foreshortenings: ${\displaystyle 0<v_x,v_y,v_z.}$

Only for suitable choices of angles and foreshortenings does one get undistorted images. The next diagram shows the images of the unit cube for various angles and foreshortenings and gives some hints for how to make these personal choices.

In order to keep the drawing simple, one should choose simple foreshortenings, for example ${\displaystyle 1.0}$ or ${\displaystyle 0.5}$.

If two foreshortenings are equal, the projection is called dimetric.
If the three foreshortenings are equal, the projection is called isometric.
If all foreshortenings are different, the projection is called trimetric.

The parameters in the diagram at right (e.g. of the house drawn on graph paper) are: ${\displaystyle \alpha ={135}^{\circ },\beta ={90}^{\circ },v_y=v_z=1,v_x=1/\sqrt{2}.}$ Hence it is a dimetric axonometry. The image plane is parallel to the y-z-plane and any planar figure parallel to the y-z-plane appears in its true shape.

Special axonometries

Engineer projection

In this case

• the foreshortenings are: ${\displaystyle v_x=0.5,v_y=v_z=1}$ (dimetric axonometry) and
• the angles between the axes are: ${\displaystyle \alpha ={132}^{\circ },\beta ={97}^{\circ }.}$

These angles are marked on many German set squares.

Advantages of an engineer projection:

• simple foreshortenings,
• a uniformly scaled orthographic projection with scaling factor 1.06,
• the contour of a sphere is a circle (in general, an ellipse) .

For more details: see de:Axonometrie.

Cavalier perspective, cabinet perspective

• image plane parallel to y-z-plane.

In the literature the terms "cavalier perspective" and "cabinet perspective" are not uniformly defined. The above definition is the most general one. Often, further restrictions are applied. For example:

cabinet perspective: additionally choose ${\displaystyle \alpha ={135}^{\circ }}$ (oblique) and ${\displaystyle v_x=0.5}$ (dimetric),

cavalier perspective: additionally choose ${\displaystyle \alpha ={135}^{\circ }}$ (oblique) and ${\displaystyle v_x=1}$ (isometric).

Birds eye view, military projection

• image plane parallel to x-y-plane.

military projection: additionally choose ${\displaystyle v_z=1}$ (isometric).

Such axonometries are often used for city maps, in order to keep horizontal figures undistorted.

Isometric axonometry

(Not to be confused with an isometry between metric spaces.)

For an isometric axonometry all foreshortenings are equal. The angles can be chosen arbitrarily, but a common choice is ${\displaystyle \alpha =\beta =\gamma ={120}^{\circ }}$.

For the standard isometry or just isometry one chooses:

• ${\displaystyle v_x=v_y=v_z=1}$ (all axes undistorted)
• ${\displaystyle \alpha =\beta =\gamma ={120}^{\circ }.}$

The advantage of a standard isometry:

• the coordinates can be taken unchanged,
• the image is a scaled orthographic projection with scale factor ${\displaystyle \sqrt{1.5}=1.225}$. Hence the image has a good impression and the contour of a sphere is a circle.
• Some computer graphic systems (for example, xfig) provide a suitable raster (see diagram) as support.

In order to prevent scaling, one can choose the unhandy foreshortenings

• ${\displaystyle v_x=v_y=v_z=\sqrt{2/3}}$ (instead of 1)

and the image is an (unscaled) orthographic projection.

Circles in axonometry

A parallel projection of a circle is in general an ellipse. An important special case occurs, if the circle's plane is parallel to the image plane–the image of the circle is then a congruent circle. In the diagram, the circle contained in the front face is undistorted. If the image of a circle is an ellipse, one can map four points on orthogonal diameters and the surrounding square of tangents and in the image parallelogram fill-in an ellipse by hand. A better, but more time consuming method consists of drawing the images of two perpendicular diameters of the circle, which are conjugate diameters of the image ellipse, determining the axes of the ellipse with Rytz's construction and drawing the ellipse.

Spheres in axonometry

In a general axonometry of a sphere the image contour is an ellipse. The contour of a sphere is a circle only in an orthogonal axonometry. But, as the engineer projection and the standard isometry are scaled orthographic projections, the contour of a sphere is a circle in these cases, as well. As the diagram shows, an ellipse as the contour of a sphere might be confusing, so, if a sphere is part of an object to be mapped, one should choose an orthogonal axonometry or an engineer projection or a standard isometry.

Coordinate calculation

The Cartesian ${\displaystyle x\text{-}y\text{-}z}$ coordinate system with origin ${\displaystyle O}$ is still used in space. Let the drawing plane be equipped with a Cartesian ${\displaystyle x^{\prime }\text{-}{y}^{\prime }}$ coordinate system with origin ${\displaystyle \overline{O}.}$ An axonometry where ${\displaystyle \overline{O}}$ is the image point of ${\displaystyle O}$ can be represented by the map ${\displaystyle f:{\mathbb{R}}^3\to {\mathbb{R}}^2,}$ $${\displaystyle p=\left(\begin{array}{c}x\\ y\\ z\end{array}\right)\mapsto f(p)=\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)}$$ which maps the coordinate vector ${\displaystyle p\in {\mathbb{R}}^3}$ of a general point ${\displaystyle P}$ in space to the coordinate vector in ${\displaystyle {\mathbb{R}}^2}$ of the image point ${\displaystyle \overline{P}.}$ The map ${\displaystyle f}$ is linear and can therefore be represented by its ${\displaystyle 2\times 3}$ transformation matrix ${\displaystyle A:}$ $${\displaystyle f(p)=Ap}$$ (matrix times vector). The three columns of ${\displaystyle A}$ are the image vectors ${\displaystyle f(e_x),f(e_y),f(e_z)\in {\mathbb{R}}^2}$ of the standard basis vectors $${\displaystyle e_x=\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right),e_y=\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right),e_z=\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)}$$ and thus describe the ${\displaystyle \overline{x},}$ ${\displaystyle \overline{y}}$ und ${\displaystyle \overline{z}}$ image axes. They do not lie on a common straight line through the origin, i.e., the real ${\displaystyle 2\times 3}$ matrix ${\displaystyle A}$ has rank ${\displaystyle 2.}$ Conversely, each such matrix generates an axonometry.

Examples

1. The factors ${\displaystyle v_x,v_y,v_z}$ and angles ${\displaystyle \alpha ,\beta }$ introduced above immediately lead to the transformation matrix $${\displaystyle A=\left(\begin{array}{ccc}-v_x\sin \alpha & v_y\sin \beta & 0\\ v_x\cos \alpha & v_y\cos \beta & v_z\end{array}\right).}$$ It is assumed that the ${\displaystyle y^{\prime }\text{-}}$axis and the ${\displaystyle \overline{z}}$ image axis point upwards and the ${\displaystyle x^{\prime }\text{-}}$axis points to the right.

2. In particular, ${\displaystyle v_x=v_y=v_z=1}$ and ${\displaystyle \alpha =\beta ={120}^{\circ }}$ give the standard isometry with the matrix $${\displaystyle A=\left(\begin{array}{ccc}-\frac{\sqrt{3}}{2}& \frac{\sqrt{3}}{2}& 0\\ -\frac{1}{2}& -\frac{1}{2}& 1\end{array}\right).}$$ For instance, the vertex ${\displaystyle p=(1,1,1{)}^{\mathsf{T}}}$ of the unit cube is mapped to ${\displaystyle f(p)=Ap=(0,0{)}^{\mathsf{T}},}$ as is the opposite vertex ${\displaystyle (0,0,0{)}^{\mathsf{T}},}$ where ${\displaystyle {}^{\mathsf{T}}}$ denotes transposition. As mentioned above, standard isometry is an orthographic projection scaled by the factor ${\displaystyle \sqrt{3/2}.}$

3. Multiplying the last matrix by the reciprocal ${\displaystyle \sqrt{2/3}}$ thus yields the transformation matrix $${\displaystyle A=\left(\begin{array}{ccc}-\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}& 0\\ -\frac{1}{\sqrt{6}}& -\frac{1}{\sqrt{6}}& \sqrt{\frac{2}{3}}\end{array}\right)}$$ of an (unscaled) orthographic projection. Incidentally, in its transpose ${\displaystyle A^{\mathsf{T}},}$ both columns $${\displaystyle e_x^{\prime }=\left(\begin{array}{c}-\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}\\ 0\end{array}\right),e_y^{\prime }=\left(\begin{array}{c}-\frac{1}{\sqrt{6}}\\ -\frac{1}{\sqrt{6}}\\ \sqrt{\frac{2}{3}}\end{array}\right)}$$ are orthonormal, i.e., ${\displaystyle e_x^{\prime }\cdot e_y^{\prime }=0}$ and ${\displaystyle e_x^{\prime }\cdot e_x^{\prime }=e_y^{\prime }\cdot e_y^{\prime }=1.}$ This characterizes orthographic projections, as is shown exemplarily hereinafter.

4. Another orthographic projection is constructed, now conversely, starting with the orthonormal vectors $${\displaystyle e_x^{\prime }=\frac{1}{7}\left(\begin{array}{c}6\\ -3\\ -2\end{array}\right),e_y^{\prime }=\frac{1}{7}\left(\begin{array}{c}3\\ 2\\ 6\end{array}\right).}$$ These two together with ${\displaystyle e_z^{\prime }=e_x^{\prime }\times e_y^{\prime }}$ and ${\displaystyle O}$ define another Cartesian coordinate system in space. A general point ${\displaystyle P}$ in space now has another coordinate vector ${\displaystyle p^{\prime }=(x^{\prime },y^{\prime },z^{\prime }{)}^{\mathsf{T}}}$ besides ${\displaystyle p=(x,y,z{)}^{\mathsf{T}}.}$ For these, always ${\displaystyle E{p}^{\prime }=p}$ with the ${\displaystyle 3\times 3}$ change-of-basis matrix ${\displaystyle E=(e_x^{\prime }|e_y^{\prime }|e_z^{\prime }).}$ As ${\displaystyle E}$ is orthogonal, ${\displaystyle E^{-1}=E^{\mathsf{T}}}$ and consequently ${\displaystyle p^{\prime }=E^{\mathsf{T}}p.}$ The orthographic projection, which projects perpendicularly onto the ${\displaystyle x^{\prime }\text{-}{y}^{\prime }\text{-}}$plane as the drawing plane, corresponds to omitting the bottom ${\displaystyle z^{\prime }\text{-}}$coordinate of ${\displaystyle p^{\prime }.}$ Its transformation matrix is obtained accordingly from ${\displaystyle E^{\mathsf{T}}}$ by omitting the bottom row: $${\displaystyle A=\frac{1}{7}\left(\begin{array}{ccc}6& -3& -2\\ 3& 2& 6\end{array}\right).}$$ This can be used to calculate points in the adjacent drawing of the unit cube. For instance, the vertex ${\displaystyle p=(1,1,1{)}^{\mathsf{T}}}$ is mapped to ${\displaystyle f(p)=Ap=(\frac{1}{7},\frac{11}{7}){}^{\mathsf{T}}.}$