help with mirror sizes

[Finale]

Can anyone help me with this math problem?

I have these two scanners I'm considering getting.
One has
X: 8x14mm, Y: 8x14mm
mirror sizes and the other one has
x: 8.6x10mm, Y: 8x14mm
mirror sizes.
I'd like to know what beam size can fit in these two for 40 degree and 50 degree scan angles before the beam bleeds out.
If there's a simple formula for calculating this please post that.
Thanks!

[rns0504]

Can you post a pic of the scanset?

It is important to see which axis of the X mirror is "oriented" with reference to the output shaft. On the scanners we design almost always the "wide" side of the X mirror is perpendicular to the output shaft, but this is not the case on many cheap chinese scanners. I assume on set 1 it will be the opposite...

Once I have this info I can do the math, the math for the X mirror is really simple. The Y mirror is a bit harder as it depends on the configuration of the XY Mount and whether or not a setback is used, but I can do some theoretical calculations for that as well...

[Finale]

Here it is

[rns0504]

From my post in another thread:

You can just use the equation

MINIMUM X MIRROR WIDTH = (Beam Diameter) * secant (45deg + (desired mechanical degrees past nominal))

So say you want to scan, say, 40 degrees optical (20 degrees optical in either direction from resting position of 45 degrees).

40 degrees optical is 20 degrees mechanical. So + and - 10 degrees mechanical from rest position of 45 degrees.

Right now we're interested in the PLUS direction as this is where the overspill will occur

So we use the equation MIN X MIRROR WIDTH = 6.5mm * secant (45deg + 10deg) = 11.33mm wide is the minimum width of the X mirror. Should be a bit bigger, say at least 11.5mm, as no mirror is perfect and there will be chipping around the edges.

So if you want 60 degrees optical, thats 30 mechanical, or + and - 15 degrees.

MIN X MIRROR WIDTH = 6.5mm * secant (45 + 15) = 13 mm wide


Alternately, to just calculate the minimum width of the X mirror at rest position of 45 degrees, just multiply the beam diameter by 1.41, to determine the minimum X mirror width, just to capture the beam

So solving for beam size, and knowing X mirror orientation...

SET 1 40 DEGREES:

8mm = (MAX BEAM DIAMETER) * sec (45+10)
8mm = (MAX BEAM DIAMETER)*1.74344679562
4.59mm = MAX BEAM DIAMETER

SET 1 50 DEGREES:

8mm = (MAX BEAM DIAMETER) * sec (45+12.5)
8mm = (MAX BEAM DIAMETER)*1.8611589967
4.30mm = MAX BEAM DIAMETER

SET 2 40 DEGREES:

8.6mm = (MAX BEAM DIAMETER) * sec (45+10)
8.6mm = (MAX BEAM DIAMETER)*1.74344679562
4.93mm = MAX BEAM DIAMETER

SET 2 50 DEGREES:

8.6mm = (MAX BEAM DIAMETER) * sec (45+12.5)
8.6mm = (MAX BEAM DIAMETER)*1.8611589967
4.62mm = MAX BEAM DIAMETER

__________________________________________________

This assumes a perfect world, where you place the beam exactly perfectly on the X mirror, and the mirrors are perfect. Most mirrors have a "chip margin" which is unusable around the edges, around 0.25 or 0.5mm per edge. So on an 8mm mirror, you can expect to only have a 7 to 7.5mm usuable width. You can redo the math for this if you wish.... this will compensate for not getting the beam placed perfectly (+/- 0.002" / 0.05mm) or chip margins...

[Finale]

Thank you. Thats all I needed.