0
$\begingroup$

I am running a 3D cad program. Z +up, Y +to left, X + into page.

If I want a new plane (A) that is 20 degrees from the YZ about Y, no problem. Simple rotation about Y axis and enter 20 degrees. If I want a new plane (A) that is 35 degrees from the YZ about Z, no problem. Simple rotation about Z axis and enter 35 degrees.

However if I want a new Plane (A) with a line of intersection between Plane (A) and the XY plane that has an angle with the Y axis of 35 degrees and a line of intersection between Plane (A) and the XZ plane with angle of 20 Z, my mind keep telling me its not as simple as a rotation of 20 about Y followed by a rotation of 35 degrees.

I there a simple enough trig equation to adjust on or both rotation to achieve a set angle about Y and a set angle about Z axis?

I have other ways of generating this plane within cad and drawing a lot of lines. I was just curious if I could bypass all that and just do two successive rotations.

$\endgroup$

2 Answers 2

Reset to default
1
$\begingroup$

So you know that the vectors $$v_1 = (\sin(\alpha_1), \cos(\alpha_2), 0)$$ and $$v_2 = (\sin(\alpha_2), 0, \cos(\alpha_2))$$ are both situated in the plane $A$. Then the cross product $v_1 \times v_2$ is a normal vector for $A$. Starting with a given plane (e.g. $XY$) take two axis rotations that map its normal (in this case $(0,0,1)$) to the target normal of $A$.

$\endgroup$
3
  • $\begingroup$ I cant enter a cross product into a cad program. I need an angle or rotation about each axis axis $\endgroup$ Commented Apr 26 at 14:27
  • $\begingroup$ First do a rotation about the $Y$-axis to match $(0,0,1)$ to the $Z$ component of the normal to $A$. Then rotate about the $Z$-axis to match the $X$ and $Y$ components. $\endgroup$ Commented Apr 26 at 18:41
  • $\begingroup$ I'm lost. how do I find the amount of angle rotation about each axis in degrees? Since all I have a a plane which I can rotate or translate. And in this case I see no need for translation. $\endgroup$ Commented Apr 26 at 18:53
0
$\begingroup$

It is that simple. Let $L_{XY}$ be the intersection of (A) with the XY planes, and $L_{XZ}$ with the XZ plane. Imagine (A) is initially the YZ plane. First rotate $20^\circ$ about Y. This gives us a $20^\circ$ angle between $L_{XZ}$ and Z. Now apply a $35^\circ$ rotation about Z. Notice that this fixes $L_{XY}$, which is equal to Y at this point. You can imagine this rotation as applying to $L_{XZ}$ and Z both (since it doesn't change Z); and rotations preserve angles, so the angle between $L_{XZ}$ and Z must still be $20^\circ$. At the same time, we now have an angle of $35^\circ$ between $L_{XY}$ and Y as desired.

$\endgroup$
6
  • $\begingroup$ Really? dang I will have to follow this and test it. My visualization with my phone as the plane and rotating it as plane did not think this was the case $\endgroup$ Commented Apr 26 at 19:42
  • $\begingroup$ my tests indicates differently. I am measuring 36.69 degrees and 20 degrees. $\endgroup$ Commented Apr 28 at 12:56
  • $\begingroup$ @ForwardEd Are you sure you're measuring the right angle? The angle between Y and (A) and the angle between Y and $L_{XY}$ are two different quantities, where $L_{XY}$ is the intersection of (A) with the XY plane. You asked for the latter in your question, not the former. I did not make this clear in my answer, so I have edited it. $\endgroup$ Commented Apr 29 at 1:56
  • $\begingroup$ The more I think about this to more I do not think it is right. Here is my reason. Take the YZ plane pane and rotate it some arbitrary angle say 45 degree about the X axis and call this new plane "A". A is now has a slope of 45 degrees in the X direction. If the plane "A" is now rotated to the extreme of 90 degrees about the Z axis, I would be left with a slope of 0 degrees in the in the X direction instead of the 45 I need. $\endgroup$ Commented Apr 30 at 11:17
  • $\begingroup$ Rotating YZ about X does nothing, I assume you mean about Y? In which case you operations give $45^\circ$ with Z and $90^\circ$ with Y, as I promised. I think you need to think very carefully about what angles between what things it is that you want. You never said you wanted an angle with the X axis, you said you wanted Y and Z, which is what my answer provides. Please also read my previously comment carefully and be sure that what you asked for in your question is what you truly want. $\endgroup$ Commented Apr 30 at 14:22

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.