Ditron D80 DRO linear compensation...help?

Winegrower

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Updating my experience per above, I found these issues with the D80 DRO:

1) DROpros replaced the magnetic X axis strip, not clear if that changed much, but it was a clean platform to restart with.

2) Finally, it became clear that the DRO needed a 0.001” per inch linear compensation. Trouble was, the manual looked nice but was wrong, or at least incomplete and essentially not understandable. Eventually I taught myself how to compensate the scale, and the accuracy is good now.

3) there seemed to be some problems with tiny ferrous chips along the edge of the scale. This is speculation on my part, based on changing readings as I blew chips away from the scale. I could be wrong.

4) relating to that, the mag strip is adhered to a frame with a slightly smaller stainless steel cover. There is a tiny gap where chips could work in between the strip and cover. I filled the gap with a very small amount of clear caulk. Will it help? Too soon to tell, too late to back out.

I am still disappointed with error magnitude when in “diameter” mode. It seems like larger errors than the 0.0002 resolution of the x axis scale would imply. Still pondering this.

It’s been an adventure!
 

Winegrower

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FYI I had a 5 micron scale on my cross slide and had similar issues with rounding and calculation errors when entering values on my Easson ES-12B, these went away when I switched to a 1 micron scale on the cross slide scale.

I spent a lot of time trying to understand error sources on my Takisawa lathe and D80 DRO, and how to read and understand the manual. I never achieved that completely, just enough to do a linear compensation. I finally decided mksj was correct that a 1 micron X axis resolution was needed to have satisfactory performance in Diameter mode, which doubles the error.

DROpros were very helpful in that 1) they warned me that this would happen, but only 5 micron was available at the time, 2) they swapped my 5 micron out when the 1 micron was available, and 3) they listened to me fumble through several issues before arriving at an acceptable answer. I’m happy I didn’t buy direct from China, I would have been even more confused.
 

A watkins

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Sorry for the old thread resurrection, but this thread touches on a problem I have with the Ditron DRO. and I haven't seen it discussed anywhere: I can't make any sense out of how the linear compensation feature works, and it is not documented in the manual.

Once you have zeroed your DTI against one edge of the gage block and then zeroed again at the other end, you have on the DRO what it thinks is the gage block length. At that point you go to the compensation feature and find three fields, L1, L and S. L1 is the displayed figure for the length of the gage block you just measured and is not editable. L is the actual length of the gage block and editable. S is the "correction coefficient" which according to their description of the arithmetic involved is the error expressed as a unitless ratio.

The odd thing is the DRO allows you to edit both L1 and S, which I don't get. I expected simply to edit to tell it the actual gage block size and for it to take the correction from there. However, when I enter L1 (typically 4.0000") it modifies what I enter somewhat, and then supplies yet another figure for S, neither of which make any sense to me, in part because the L1-L and S differentials don't add up to the error. So between the auto-modifying L and magically computed S, I have no idea what it wants.

Can someone who's figured this out clue me in?

Thanks,
Alan
 

Winegrower

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Alan, I agree that the compensation procedure is described absurdly. What finally worked for me is that everything must be in metric. Be sure to do that, then I think everything will begin to make sense.
 

ddickey

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I just got done monkeying around with mine for hours. My readings were always inconsistent. Ended up going to back to no compensation for all the slides. May revisit at a later time.
 

RJSakowski

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I have a different system so this may not apply. I found that the compensation entered is applied to the last compensation factor used. So if you are reading 2.997" over 3" and apply a correction factor (mine is in ppm) of 1000 ppm, the scale should be right on. If for some reason, it was now 3.001" over 3", I would have over-corrected and need to apply a new factor of - .001/3 x 1,000,000 or -333. If the new measurement was 3.0002", the last correction wouldn't be enough and the new factor would be -.0002/3 x 1,000,000 or -67. Continue until you are satisfied with the calibration or bored whichever comes first.

If using a 1-2-3 block for the step, I would measure the block with an accurate micrometer. Use that measurement rather than the nominal size.
 

RJSakowski

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A little prep work will help with consistency of measurement. On the mill, I clamped a 6" bar parallel to the axis being calibrated and swept the edge to verify. Then I clamped a1-2-3 block at one end for a stop. For my step, I used a 6" parallel measured with a micrometer. The test indicator was mounted in the spindle and the spindle fixed so it couldn't rotate. I now had a stable repeatable system. I zeroed the test indicator on the stop block and zeroed the DRO. Then I moved the table 6+ inches, inserted the parallel, and moved in until the test indicator zeroed.

A similar setup was used for the lathe.
 

Winegrower

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I say again, put the DRO in metric, convert your 3” block to mm, and do it just as it says in the manual, which is actually better than most import manuals. It will work.
 

A watkins

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do it just as it says in the manual, which is actually better than most import manuals. It will work.
Before you send me back to the manual, let me remind of what the manual says:
---------------------------------------------
2.9 Linear compensation
Function: Linear error compensation function is used to correct the system errors of the grating
ruler measurement system linearly.

Note: the calculation formula of correction coefficient is:

Correction coefficient S = ( L - L1) / C L / 1000) mm/m L: Stands for the actual measured
length (mm)
L1: Stands for the displayed value (mm) on the DRO
S: Stands for correction coefficient (mm/m) (+ indicating lengthening and - indicating
shortening)
Compensation range: - 1.9 mm/m to + 1.9 mm/m

Example: The actual length of the machine's X axis table is 1000.000mm and the
displayed value on the DRO is 999.880mm. The correction coefficient is calculated as
follows:
S = C 1000.000 - 999.880) / C 1000.000 / 1000.000 ) = 0.120
Set the linear compensation coefficient according to the following operation (Note: Set the
compensation method as linear compensation in the system parameter setting section
firstly. The detailed operations are described in system parameter setting section.)


2. Basic functions

Step 1:Entering the linear compensation Press C to enter the linear
compensation. Select the X axes by pressing (left/right arrow) .After selecting well, press Enter to enter the
linear compensation.

Note: When any axes need to be done the linear compensation. The corresponding compensation way in
system parameter setting need to set as the linear compensation(linear).

----------------------------------------------------------------

Note that it says absolutely nothing about two boxes L and S that are editable on the subsequent screen.

So, Winegrower, when you had your displayed length of X mm, and your known length of Y mm, what did you enter into L and S? Did you literally do their computation S from L and L1, and then enter the result along with L? If so, do you understand why they (apparently) require us to enter L and calculate S when the latter is derived from the former by simple arithmetic?
 
Last edited:

A watkins

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So, Winegrower, when you had your displayed length of X mm, and your known length of Y mm, what did you enter into L and S? Did you literally do their computation S from L and L1, and then enter the result along with L? If so, do you understand why they (apparently) require us to enter L and calculate S when the latter is derived from the former by simple arithmetic?
The person from Ditron who told me I was to do the calculation and enter both L and S has recanted her testimony and now says:
—————
1. Clearance Zero under ABS model
2.Press button C ,then enter compensation model
3.ensure a start position,clearance zero,then move linear scale to another position
4.you will see a date (she means “number”) in L1 ,then input a actual measuring length at L, DRO will calculate S
Please note: L means the length between the start position and the end position
—————-


So, has the above worked correctly for you other Ditron users? If not, how are you making linear compensation work?
 
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