# Raster scan with a rotating mirror

I'm trying to figure out a way to scan a 2D raster (like a TV or CRT monitor) with a beam using two rotating mirrors for the X and Y positions.

I'm looking for a shape that will, when spinning, deflect a beam in discrete steps on one axis only.

Combining two of these should let me scan a raster of theoretically very large size and a high speed.

I'm not sure this is the correct place for this kind of question as its more physics rather than math, but as they say, physics is just applied math right?

The problem i'm facing is that it seems that on a spinning disk any deflection would cause a 'wobble' or change the divergence of the beam (as it would be reflecting off a conical surface)

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The logarithmic spiral has the property that it deflects a beam of light directed at its centre always with the same angle. Maybe your shape could be a locally logarithmic spiral, one piece for each position. – WimC Dec 15 '12 at 8:07
More engineering than physics, I'd say... – Henning Makholm Dec 15 '12 at 11:10

I think it's a mistake to want discrete steps here.

Instead make the Y scan make a slow smooth scan down the scanning field once per frame, and then put the X scan mirror a slight angle from horizontal such that it rises during each scanline by exactly the same amount that the Y can drops.

Alternatively, simply adjust the timing of the signal such that each successive scanline starts a small amount earlier in the X mirror cycle, to compensate for the scanlines angling slightly downwards and still produce a rectangular image. Then tilt the entire deflection apparatus slightly (with the mirrors still perpendicular) so the scanlines become horizontal.

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This is genius, I didn't think correcting with a tilt would be this easy! I'm going to try model this in some CAD software to see what kind of angles and speeds I will need. Thank you – Yarek T Dec 16 '12 at 5:50

A prism will do that, spinning around its axis. If you have a collimated beam incident on a prism, the angle of the plane face is proportional to time, so the exit angle is proportional to (twice the) time. It recovers to the origin when the edge spins by. You can choose the number of sides by the range you want to scan and the rate. If you have two in perpendicular axes, one (probably the second) must spin much faster than the other.

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That will not make discrete steps as requested. – WimC Dec 15 '12 at 7:57
@WimC: I think it can, if you orient it correctly. The "horizontal scan", which I am imagining would be the fast one, needs to be on a slight upward angle to compensate for the vertical scan. – Ross Millikan Dec 15 '12 at 16:33