How do you know what probe "Overshoot amount" should be?

I cannot find anything to tell me what this should be or what it actually does. Is it used in any of the hidden probe calibrations or calculations? Or, is it just saying to travel this much further after the signal is tripped? Why would you need to travel past the point of the probe tripping? Why is this not set to 0?

Hi,
unless you are probing at high speed or your machine has very low acceleration overrun should be very small.

The distance of over run, d, is given by:

d=1/2. a.t²

where a is the acceleration (or de-acceleration to be precise) and t is solved by:

t=v_p / a

where v_p is the probing speed.

Using numbers from my mill where I have standardised on a probing speed of 100mm/min (1.66mm/sec) and the machine has an acceleration of 375mm/s².

t=1.666 /375
  =4.44ms
d=1/2 x 375 x (0.0044)²
  =3.7 um

3.7um is about the limit of the accuracy of my machine so I don't worry about it, its near enough to zero for my purposes.
Substitute the numbers from your machine and see if the overun is significant.

Craig

How would anyone know they are supposed to calculate all of this out with the given documentation?
My machine runs in inches, so this is not really working out for me.

My Acceleration Units/(Sec^2) is 50.00 (from the motors setup screen).

so,
d=1/2*a*t^2
d=1/2*50*t^2

My "Slow measure feedrate" is 5.0 and I am not positive what units these are?? inches per minute? inches/Sec^2 ??
So,
t=5/50
t=0.1

So,
d=1/2*50*.1^2
d=0.25

This cannot be right? Quarter inch of over travel would break my probe likely.

This all makes no sense to me. Something is not converting correctly for me here. Mismatch in units or something. 

I understand why this is Mach4 hobby. Your entire hobby becomes figuring out how to work it instead of using your machine

If I convert my "Slow measure feedrate" of 5.0 and assume it is inches/minute to inches/second (0.0833333), the math comes out like this...

t=0.0833333/50
t=0.0016666

d=1/2*50*0.0016666^2
d=0.000069438889

This is basically setting my overshoot to zero.....does this sound correct??


 

 

Oh, to throw another wrench in this. My Z axis has a brake on it that is controlled by the servo drive.
How do you account for that? The X and Y do not have a brake, but the bed mill is heavy. Will handle a 300 lb work piece according to the old manual.

So, this is interesting.
Before I had my Overshoot amount set to .005" (for no other reason than it sounded reasonable and wouldn't hurt the probe).

After I would run the calibration I would always get something like this:
"Auto tool setting complete, Offset = 1.0019"

The 1" seems to be programmed in as a default (which makes no sense to me), but I was never sure why I was getting some number just higher than 1"

I set my probe overshoot to 0.000 and ran it again.
Now I get "Auto tool setting complete, Offset = 1.0000"

Is this what I should be getting and possibly that zero calculation above was correct?

Slightly off topic, but related, why is my TLO for Tool 1 (the probe) set to 1" anyway??

Hi,

Oh, to throw another wrench in this. My Z axis has a brake on it that is controlled by the servo drive.

I would not expect the brake to come on UNLESS the servo is disabled. Even then, its common for the brake to take 10ms or so to come
on and so unless you have long  deceleration times programmed into your machine then it will have decelerated and stopped
BEFORE the brake has come on.

If you probe sufficently slowly relative to the machine acceleration overrun should be vanishingly small. That overun has been included
in the calculation is for those circumstances where it is significant or for nit-picking perfectionists who demand to chase every last micrometer.

Craig