Projection of a cylinder from XYZ 3D space on to XY plane.

Question

A cylindrical body of length $ \ L \ $ lies in the XYZ 3D space. The base of the cylinder makes an angle $\psi$ with the Z axis, as shown in image link. I would like to know the projection of that cylinder in the XY plane in terms of length and azimuthal angle $\psi$.

Answer 1

Let $u = (u_x, u_y, u_z)$ be the unit vector along the axis of the cylinder, with $u_z \gt 0$. And let $C$ be the center of the lower base of the cylinder, and let $R$ be the radius of the bases.

Generate two vectors $v_1 $ and $v_2$ with $v_1$ horizontal, that are orthogonal to the axis unit vector $u$. Hence,

$ v_1 = (u_y, -u_x, 0 ) / \sqrt{u_x^2 + u_y^2} $

Then it follows that

$ v_2 = u \times v_1 = (u_x, u_y, u_z) \times (u_y, - u_x, 0) / \sqrt{u_x^2+ u_y^2} = (u_x u_z, u_y u_z, -(u_x^2 + u_y^2) ) / \sqrt{u_x^2 + u_y^2} $

The boundary of the lower base is the ellipse

$ P_1(\theta) =\text{Proj}(C) + R \cos(\theta) \text{Proj}(v_1) + R \sin(\theta) \text{Proj}(v_2) \hspace{50pt} 0 \le \theta \le \pi $

This simplifies to

$ P_1(\theta) = (C_x, C_y) + \dfrac{1}{\sqrt{u_x^2 + u_y^2}}\bigg( R \cos(\theta) (u_y, - u_x) + R \sin(\theta) (u_x u_z, u_y u_z) \bigg) \hspace{50pt} \theta \in [-\pi, 0] $

Similarly the projection of the upper base is

$P_2(\theta) = (C_x + L u_x, C_y + L u_y ) +\dfrac{1}{\sqrt{u_x^2 + u_y^2}}\bigg( R \cos(\theta) (u_y, - u_x) + R \sin(\theta) (u_x u_z, u_y u_z) \bigg) \hspace{50pt} \theta \in [0 , \pi ] $

Finally we have the boundary of the projection of lateral surface of the cylinder, and this is a couple of line segments, given as follows

$P_3(t) = (C_x , C_y) + \dfrac{R}{\sqrt{u_x^2 + u_y^2}}( u_y, - u_x ) + t L (u_x, u_y) \hspace{50pt} t \in [0, 1] $

and

$P_4(t) = (C_x, C_y) - \dfrac{ R}{\sqrt{u_x^2 + u_y^2}}( u_y, - u_x ) + t L (u_x, u_y) \hspace{50pt} t \in [0, 1] $

The segments $P_1, P_2, P_3, P_4 $ define the boundary of the projection of the cylinder.

The above is seen on this GeoGebra graphic which I prepared.

Answer 2

For better nomenclature in spherical coordinates, $(\theta,\phi, \psi)$ are azimuth or longitude , latitude or elevation and co-latitude respectively.

$$ \psi+\phi= 90^{\circ};\quad $$

Origin O, $X,Y,Z$ are Cartesian coordinate axes. $OP$ is simple direct projection of $L$ on horizontal plane $YOX,$ (treating the tube as a line through the origin)

$$ OP= L \cos \phi = L \sin \psi. $$