On Mon, 03 Oct 2005 11:21:42 -0500, don holly
wrote:

Greetings Don,
I don't understand exactly. Oh, yes I do. I drew a picture and it's
all clear. Duh. With the mirror at 45 degrees the included angle is
90. And the mirror only has to turn 45 degrees to make the light shine
directly back on itself.
Thanks,
Eric

Eric,

This should work fine. Note that as the mirror moves through angle A,
the reflected beam moves through angle 2A, so that if the target is 159
inches from the mirror and the mirror moves 360 degrees/10,000 =
0.000628 radians, the laser spot moves 2* 0.000628 * 159 inches = 0.2
inches. Should be easy to see.


Don Holly

Eric R Snow wrote:
This may seem off topic but it's not really. Measuring is important to
machining. And I need to make a measurement without putting pressure
on the item being measured. Now, for the purists, I know that light
will put a little pressure on what I'm measuring, but this will have
way less effect than any kind thing I can measure. I have an encoder
that outputs 10,000 pulses per revolution. With a shaft measuring
.19515" (radius) each pulse equals .0001 on the circumference. So, if
the radius keeps multiplying by 10 then the amount traveled by the
circumference will increase by 10. Here's the plan:
Put a mirror on the shaft.
Shine a laser at the mirror.
Adjust the mirror until the laser is visible on a target 1591.5"
(132.625 feet) away.
This will increase the amount traveled to .1" for every pulse.
So, measuring the distance the spot travels will show how many pulses
should be generated.
The reason for this measurement is to rule out backlash in the
encoder. It appears that there is a 9 pulse error. In other words, if
the encoder shaft is turned one way and the count is noted, and then
turned the other way until the count changes, it appears that the
shaft turns an amount equal to 9 pulses. I need to rule this out
because this is the error I'm getting is a mechanical system and it
appears that all the mechanical lash has been reduced to less than
.0001". The last thing seems to be the encoder itself.
Thank You,
Eric R Snow,
E T Precision Machine