How do I calculate the angle of the handrail transition if both need to be set at 33 degrees around a 90 degree corner?

Question

How do I calculate the angle of the handrail transition if both need to be set at 33 degrees around a 90 degree corner?

Answer 1

If you mean that the handrail comes in $33^\circ$ below the horizontal on one side and leaves $33^\circ$ above the horizontal on the other side, bending around the $90^\circ$ corner, and you want to find what angle to bend the handrail, then you want the vector dot product.

In general, position two vectors so they start at the origin (0,0,0), one pointing left (along the negative y-axis) and down by $\theta^\circ$, the other pointing into the page (along the negative x-axis) and up by $\alpha^\circ$. For convenience, write these vectors with magnitude 1, so they become:

• $\vec{u} = \langle0,-\cos(\theta),-\sin(\theta)\rangle$

• $\vec{v} = \langle-\cos(\alpha),0,\sin(\alpha)\rangle$

The angle between these vectors, say $\beta$ (and hence the one you'd like to bend into your handrail) can be found by the definition of the cosine for vectors:

$$\cos(\beta) = \frac{\vec{u}\bullet\vec{v}}{||\vec{u}||\cdot||\vec{v}||}$$

Solving for $\beta$ gives us the angle we want:

$$\begin{align}\beta &= \arccos\left(\frac{\vec{u}\bullet\vec{v}}{||\vec{u}||\cdot||\vec{v}||}\right) \\ &=\arccos\left(-\sin(\theta)\cdot\sin(\alpha)\right)\end{align}$$

Using $33^\circ$ for both of these angles gives: $\beta = \arccos\left(-\sin^2(33^\circ)\right) \approx 107.26^\circ$

Here's confirmation using 3DCalcPlotter: