New angle formed after rotating pipe

Question

I am having a bit of an issue with a problem (home maintenance) and would need to figure out a new angle formed after rotating a pipe. I will try to be as descriptive as possible:

This diagram shows the top view and the side view, so the pipe is both going down AND to the side, forming an angle of $35$ degree when seen from above.

My question would be, if I were to rotate the STATIONARY horizontal pipe $11$ degrees, what would the NEW angle formed be from above (it is $35$ degrees now, what would it become after I rotate that horizontal component 11 degrees on itself, position stays the same).

if possible, I would like two answers: the new angle after it is rotated clockwise and counterclockwise

Thank you very much

Answer 1

[I may have some things off by $90^\circ$ here or have $\sin$ and $\cos$ swapped, but I think you’ll get the idea.]

If you rotate the stationary pipe through a full $360^\circ$, you would get this sort of funnel-like shape, where the angle between the horizontal pipe and the funnel is $180^\circ$ minus the angle by which your pipe has been bent. That’s the angle between the green and orange parts here, and they illustrate your pipe’s current position. [Mark McClure made a great animation of this in his answer!.]

So first, measure the angle by which the pipe is bent (about $20^\circ$ in this picture) and call it $B$.

Identify points in this picture with the origin at the bend, the positive $x$-axis going left, the $y$-axis going up, and the $z$-axis going towards you.

A circle on the cone is then the set of points $(x_0,x_0\sin \theta\tan B,x_0\cos \theta\tan B )$, where $\theta$ is the angle around the cone, with $\theta=0$ towards you and $\theta$ increasing counter-clockwise when looking into the funnel from the left.

So the bent part of the pipe lies along the ray with parametric equations $(x,x\sin \theta\tan B,x\cos \theta\tan B )$ for $x\ge0$ for some fixed $\theta$.

Now, figure out $\theta$ for your pipe. Here, it’s about $-30^\circ$.

If you got the angles right, the angle you see in the top view should match what it is mathematically. Since a top view basically ignores the $y$ coordinate, the tangent of the angle the pipe appears to make when viewed from the top is the slope of the line $z=x\cos \theta\tan B$. That slope is $\cos \theta\tan B$, so the angle you see is $\arctan(\cos \theta\tan B)$.

The angle from the side view is the angle seen when ignoring the $z$ coordinate, so it’s $\arctan(\sin \theta\tan B)$

The angles you’re looking for (the top-view angle after rotations) are then

$$\arctan(\cos (\theta+11^\circ)\tan B)\mbox{ and }\arctan(\cos (\theta-11^\circ)\tan B)$$.

Answer 2

This is not an answer but is a little long for a comment.

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I don't think there's quite enough information to answer the question. As I understand it, you've stated the apparent angle that a revolving pipe makes when viewed from one particular direction. You then ask for the apparent angle after rotation about an axis perpendicular to our line of sight. But that depends upon how much it has already been rotated from the spot where the apparent angle is at a maximum or (equivalently) where the axis and revolving pipe are in a plane that is perpendicular to our line of sight.

To illustrate, have a look at the figure below. The rotation of the horizontal axis occurs at a constant rate. Note, however, that the rate of change of the apparent angle appears to be faster when that apparent angle is near zero. As a result, the change of the apparent angle after a rotation about the axis of 11 degrees depends upon where you are in the rotation.

Answer 3

HINT:

$$ \tan \alpha = \tan \beta \cos \theta $$

where $ \alpha ,\beta ,\theta $ are projected angle, true apex angle of a cone generator and rotation from central plane respectively.Using the relation appropriately you can calculate all projected angles.