I'm trying to figure out the image width generated by a projector. I can calculate the image width with the throw ratio and the distance if the projector is beaming light in a straight line but I would like to calculate the image width if the projector is rotated to the left or to the right.
The initial situation looks like this:
AD is the distance between the lens and the wall
BC is the width of the image that can be obtained by using the throw ratio
AB = AC, those ones are also easy to calculate thanks to AD and BC.
Now, if I move slightly the projector to get this:
A: same coordinates and same angle
D: moves slightly to the left (I should have adapted its position on the picture, sorry about that)
AB and AC are no longer equal, AB is now longer than AC
Let's assume that the rotation is a rotation of 20°
How could I calculate the new value of BC? I'm always missing one variable to be able to calculate this.
Thanks!
Laurent
Let's assume that $AD$ is known, and it's a constant value $d$. The $\angle BAC=2\theta$ is not changing. So originally $\angle BAD=\angle DAC=\theta$. After the rotation by an angle $\alpha$, $\angle BAD=\theta+\alpha$ and $\angle DAC=\theta-\alpha$. Then, using trigonometric functions in $\triangle BAD$ and $\triangle DAC$, you get $$\tan(\theta+\alpha)=\frac{BD}{AD}\\\tan(\theta-\alpha)=\frac{DC}{AD}$$ or $$BC=BD+DC=d\left[\tan(\theta+\alpha)+\tan(\theta-\alpha)\right]$$ If you have your original $BC_{original}$ instead of $\theta$, just plug in $\alpha=0$ in the above equation:$$\tan\theta=\frac{BC_{original}}{2d}$$