Building an expression for the alignment

Do not judge less you be judged - and if it should come to that my son, judge them by the company they keep. And if there is any doubt, throw them OUT!!

Do not judge less you be judged - and if it should come to that my son, judge them by the company they keep. And if there is any doubt, throw them OUT!!

Every day I thank God that I don't believe in him.


Seriously though. Where are the holes? I'm genuinely interested. Are they true diameters i.e. round/cylindrical. Are they square to the faces? Or are they coaxial or co-linear with the 'centreline'.

'UKCMM' nice to make your acquaintance! Finally, someone who speaks the language, thanks for chiming-in 'UK'. I've written off 'DAN_M'. I thought he had a little more sense, but it's hard to tell when you have people hide behind words rather than a face to face conversation. It's immature and it fails every attempt at humility. So, I thank you for your satisfying delivery 'UK'.
Your questions are exactly the type of questions I was expecting to receive from 'louisd', but that was all a waste of time and on that note, it's what started this whole bratty immature exchange of trying to get down to solving the problem. But for some it's all about pride, and that can be dangerous if in the wrong hands like I have discovered with a few names on here.
Getting down to answering your questions:
1) The flats on the ends do exist on the CAD model ----------- Ok
2) The end faces do have a flat; it's small but there is a slight flat to work with and to use as a leveling device ----------- Debatable, but Ok we'll go with that assumption for now
3) The end faces with the larger one on the right side is used for leveling and I also translated as my 'X' origin ----------- Ok
4) Step 3) is also taking place on the left end ----------- Hmmm, if you level on the second face and origin on that you override the leveling on the other end but we'll go with that for now
5) The difficult part is getting the center as I am producing a circle Ø on the inside as close as I can possibly can be to the end at the top. This is where you're losing us - we have one end and another end - where's this diameter at the 'top'?!?!? The same thing with a Ø on the inside bottom. What do you mean by bottom?!?!?
...now here is where 'louisd' and some of his other crew dropped the ball, and started down a path they shouldn't have gone down. It only told me there are limits to some abilities. That's just the way it is and that's ok. Whatever... Lovely dig at people who were trying to help you - ever thought you've maybe not explained the issue properly?
6) I take the Ø's and project them out onto the face of the 'X' flat(s). Here's the critical part ARE THE BORES SQUARE TO THE FACE, OR COAXIAL TO THE C/L? Now I have two centroids that I can work with out on the very flat to produce my midpoint origin What?
7) The same is done on the other end thus creating a centerline of the part to which I need a distance from center to center on the ends
8) The centroids on the ends do share the centerline OMFG - The centroids of what?!?! UK gave an example with three different centroids yet you don't clarify what effing centroid you are dealing with?

Now, if you were to read in the very beginning of my post you should see that all I was looking for was or code or some syntax I could use to enter into my program to fix (straighten) my first angle to a right angle whether it was 15° or 14°.437 or 16°.352 etc. Put a code in using the 'Expression Builder' to accomplish the feat of having the angles read the code to straighten to a right angle.

This is the issue - we could tel you how to do that (in fact we have) BUT I think we're all thinking that your methodology is flawed.

Look at it this way - if you have the part perfectly aligned to the CMM you could grab the angle (using an assignment expression) of the plane and adjust it by the measured value, BUT if the part isn't perfectly aligned to the CMM then you are going to get an error - let's take an extreme example, first lets assume the part CL is meant to be along the CMM x axis okay?

If the part is aligned perfectly you could get the angle and rotate around the Y axis by the measured angle to get you square to the Xaxis yes? If the part was 90° (like I said, an extreme example) out, you'd need to rotate around the Zaxis to get square to the Xaxis. In reality you're going to be close to the first case, but not perfectly! I think this is what Louis was trying to explain but you just didn't get it.

But that just went south with these guys 'UK', they didn't know if it could be done or they knew, but just didn't elaborate or they didn't know and were just too foolish to admit it. Some kids got to have it their way or they will go crying, and I just can't tolerate that so I left. Then, you showed-up 'UK' and that s a great sign, but I thought DAN_M was a good advisor until he went to no man's land himself joining the rest of that crew.
There you have it in a nutshell 'UK'...good to make your acquaintance. You have the outline of the beginning of my program with this part I've been dealing with. To tell you the truth, I think you have the pizzazz and the finesse to go after this and solve what I've been trying to do from the start