Find the optimal number of photos of infinite cylinder.

Question

I need to solve the following problem :

Suppose I have a perfectly cylindrical object with diameter D, and a infinite height, placed on a normal surface. I have a camera, with angular aperture alpha , and I position myself in a distance $d$, always pointing the camera normally to the center of the facing cylinder.

Considering the camera is always at a fixed height, and that the distance you will be from the cylinder will always be a fixed distance d, although you can walk radially around the cylinder, what is the maximum number of n photos you can take with the camera, before any photos overlap? Also $d < D$.

I have the following answer until now :

$$n= \frac{360}{2( \pi - \alpha - \arcsin ( (\frac{D+2d}{D}) \sin (\alpha) ))} $$

I modeled the problem to a Side Side Angle triangle problem and found the internal angle of the arc photographed by the camera.

But this answer doesn't seem to make sense - a bigger distance from the cylinder doesn't reduce the number of photos required.

What is the maximum number of $n$ photos you can take with the camera, at fixed distance $d$, of cylinder of diameter $D$, before any photos overlap?

Answer 1

With ref. to the sketch, applying the Law of Sines $$ \frac{{\sin \left( {\alpha \backslash 2} \right)}} {{D/2}} = \frac{{\sin \left( {\pi - \left( {\alpha + \beta } \right)\backslash 2} \right)}} {{D/2 + d}} = \frac{{\sin \left( {\left( {\alpha + \beta } \right)\backslash 2} \right)}} {{D/2 + d}} $$ hence: $$ \beta = 2\arcsin \left( {\left( {1 + \frac{{2d}} {D}} \right)\sin \left( {\alpha \backslash 2} \right)} \right) - \alpha \quad \left| \begin{gathered} \;\alpha _{\,\max } = 2\arcsin \left( {1/\left( {1 + \frac{{2d}} {D}} \right)} \right) \hfill \\ \;\beta _{\,\max } = \pi - \alpha _{\,\max } \hfill \\ \end{gathered} \right. $$ where $\alpha _{\,\max }$ and $\beta _{\,\max }$ are the angles corresponding to the tangent line (indicated in red in the picture): for a larger aperture the portion of the cylinder portrayed remains the same, with some background added.

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Check with the solution provided by @gowrath
The solution above coincides with that provided by gowrath, which in fact , changing $\theta$ with $\beta$, can be further developed as: $$ \begin{gathered} \left( {1 + \frac{{2d}} {D} - \cos \frac{\beta } {2}} \right)\tan \frac{\alpha } {2} = \sin \frac{\beta } {2} \hfill \\ \left( {1 + \frac{{2d}} {D} - \cos \frac{\beta } {2}} \right)\sin \frac{\alpha } {2} = \sin \frac{\beta } {2}\cos \frac{\alpha } {2} \hfill \\ \left( {1 + \frac{{2d}} {D}} \right)\sin \frac{\alpha } {2} = \sin \frac{\beta } {2}\cos \frac{\alpha } {2} + \cos \frac{\beta } {2}\sin \frac{\alpha } {2} = \sin \left( {\frac{{a + \beta }} {2}} \right) \hfill \\ \end{gathered} $$

Answer 2

Let $D$ be the diameter of the cylinder, $d$ the distance of the camera from the outer edge of the cylinder, $\alpha$ is the aperture of the camera, and $\theta$ is the angle of the sector of the circle that joins the maximum coverage of the aperture at one point. I know that's wordy but a diagram should explain:

So $\theta = ADC$.

If you can find $\theta$, you can find the length of the sector $ADC$ and divide the circumference by that to get the number of unique photos.

We know length

$$ AC = \left(d+\frac{D}{2}\left(1-\cos{\frac{\theta}{2}}\right)\right)\tan{\frac{\alpha}{2}} = \frac{D}{2}\left(\sin{\frac{\theta}{2}}\right) \\ \left(\frac{2d}{D} + 1-\cos{\frac{\theta}{2}}\right)\tan{\frac{\alpha}{2}} = \sin{\frac{\theta}{2}} \\ $$

You can rearrange the above to solve for $\theta$. Thank you @GCab for the suggestion on how to proceed with the simplification. If you want to see what a tired man's approach to maths can be, check out the edits :P

$$ \left(\frac{2d}{D} + 1-\cos{\frac{\theta}{2}}\right)\tan{\frac{\alpha}{2}} = \sin{\frac{\theta}{2}} \\ \left(\frac{2d}{D} + 1-\cos{\frac{\theta}{2}}\right)\sin{\frac{\alpha}{2}} = \sin{\frac{\theta}{2}}\cos{\frac{\alpha}{2}} \\ \left(\frac{2d}{D} + 1\right)\sin{\frac{\alpha}{2}} = \sin{\frac{\theta}{2}}\cos{\frac{\alpha}{2}} + \sin{\frac{\alpha}{2}}\cos{\frac{\theta}{2}}\\ \left(\frac{2d}{D} + 1\right)\sin{\frac{\alpha}{2}} = \sin{\left(\frac{\theta + \alpha}{2}\right)}\\ \theta = 2\arcsin{\left(\left(\frac{2d}{D} + 1\right)\sin{\frac{\alpha}{2}}\right)} - \alpha \\ $$

Once you have $\theta$, you ca get the number of unique pictures, $n$, by dividing the circumference by $\frac{D}{2}\theta$ and take the floor function of it:

$$ n = \left\lfloor\frac{2\pi}{\theta}\right\rfloor $$