Vector Components
Circle with Points
Details
The given components \((v_x, v_y)\) define 5 points on a circle with radius \(r=\sqrt{v_{x}^2 + v_{y}^2}\):
Point |
Position |
\(\left( v_{\bar x} , v_{\bar y} \right)\) |
1 |
given |
\((v_x, v_y)\) |
2 |
3 o’clock |
\((r, 0)\) |
3 |
9 o’clock |
\((-r, 0)\) |
4 |
12 o’clock |
\((0, r)\) |
5 |
6 o’clock |
\((0, -r)\) |
Rules
Find point 1 at the end of the red line. Read the vector components \((v_x, v_y)\) as the point coordinates of point 1.
Drag the grey boxes to where they belong.
Check your result by checking the 5 points.
Example = First image: \((v_x, v_y) = (3, 4)\).
The circle radius is \(r=\sqrt{v_{x}^2 + v_{y}^2}=5\).
Point |
Position |
\(\left( v_{\bar x} , v_{\bar y} \right)\) |
1 |
given |
\((3, 4)\) |
2 |
3 o’clock |
\((5, 0)\) |
3 |
9 o’clock |
\((-5, 0)\) |
4 |
12 o’clock |
\((0, 5)\) |
5 |
6 o’clock |
\((0, -5)\) |
\(\varphi = \varphi_1 \Leftrightarrow v_{\bar x}\) maximal
Details
Find the angle \(\varphi=\varphi_1\), for which \(v_{\bar x}\) is maximal.
Angle |
counted |
from |
to |
\(\varphi_1\) |
\(\circlearrowright\) |
\(\color{red}{|}\) |
\(\color{red}{¦}\) |
Rules
Read \((v_x, v_y)\) as the point coordinates of point 1 at the end of the solid red line.
Find \(r=\sqrt{v_{x}^2 + v_{y}^2}\).
Find \(\varphi_1\) using 1 :
\[\begin{split}\tan \frac{\varphi_1}{2}&=\frac{v_y}{v_x + r }\\ \varphi_1 &=2 \arctan \frac{v_y}{v_x + r }\end{split}\]And find this angle in the diagram.
Drag the grey boxes to where they belong.
Example: \((v_x, v_y) = (3, 4)\) leads to \(r=5\) and \(\varphi_1\approx 53^\circ\).
Passive Transformation
Details
Use:
Winkel |
Zählrichtung |
von |
bis zu |
\(\varphi\) |
\(\circlearrowright\) |
\(\color{red}{|}\) |
\(\color{red}{¦}\) |
\(\varphi\) |
\(\circlearrowleft\) |
\(\color{blue}{¦}\) |
\(\color{blue}{|}\) |
Rules
Read \((v_x, v_y)\) as the point coordinates of point 1 at the end of the solid red line.
Read \(\varphi\).
Calculated \((v_{\bar x},v_{\bar y})\) for this angle.
Drag the grey boxes to where they belong.
Example: First image:
\((v_x, v_y) = (3, 4)\)
\(\varphi=30^\circ\).
- \[\begin{split}\begin{bmatrix} v_{\bar x} \\ v_{\bar y} \end{bmatrix} &= \begin{bmatrix} \cos 30^\circ & \sin 30^\circ \\ -\sin 30^\circ & \cos 30^\circ \end{bmatrix} \begin{bmatrix} v_x \\ v_y \end{bmatrix} \\ &\approx \begin{bmatrix} 4.60 \\ 1.96 \end{bmatrix}\end{split}\]
Active Transformation
Details
Use:
Angle |
Counting |
from |
to |
\(-\alpha\) |
\(\circlearrowright\) |
\(\color{red}{|}\) |
\(\color{red}{¦}\) |
\(\alpha\) |
\(\circlearrowleft\) |
\(\color{red}{|}\) |
\(\color{red}{¦}\) |
Rules
Read \((v_x, v_y)\) as the point coordinates of point 1 at the end of the solid red line.
Read \(\alpha\).
Calculate \((v'_x, v'_y)\).
Drag the grey boxes to where they belong.
Example: First image:
\((v_x, v_y) = (3, 4)\)
\(-\alpha=30^\circ\) so that \(\alpha=-30^\circ\)
- \[\begin{split}\begin{bmatrix} v'_x \\ v'_y \end{bmatrix} &= \begin{bmatrix} \cos( -30^\circ) & -\sin (-30^\circ) \\ \sin( -30^\circ) & \cos (-30^\circ) \end{bmatrix} \begin{bmatrix} v_x \\ v_y \end{bmatrix} \\ &= \begin{bmatrix} \cos 30^\circ & \sin 30^\circ \\ -\sin 30^\circ & \cos 30^\circ \end{bmatrix} \begin{bmatrix} v_x \\ v_y \end{bmatrix} \\ &\approx \begin{bmatrix} 4.60 \\ 1.96 \end{bmatrix}\end{split}\]
Test
Footnotes:
- 1
Note the periodicity: The domain of the angles in the grey boxes is \(-180^\circ < \varphi_1 \le 180^\circ\). Also your calculated will most likely return angles in this interval. However: The angles in the diagrams are angles in the domain \(0^\circ \le \varphi_1 \le 360^\circ\) so that all angles are positive.