# True Position on CMM result vs Equation

• 03-28-2014, 08:12 AM
canteri
True Position on CMM result vs Equation
Hello Guys,

I'm new on this forum. If someone can help me, I'd appreciate very much :-D

Team I need your support to understand if I'm correctly measuring a hole (representative drawing attached).
Result on my CMM Faro Arm doesnt reflect the equation 2*((X)^2+(y)^2)^0,5.
So I'm not sure if I should consider Z on this calc as the surface which I'm measuring is inclined 12.5º.

My CMM result is the image below.

[ATTACH=CONFIG]994[/ATTACH]

[ATTACH=CONFIG]993[/ATTACH]

Somehow my drawing would be something similar to the link below. But we don't have the M condition material.

• 03-28-2014, 10:04 AM
Kelly Bramble
Yes, you need to include the Z axis in your calculations.

Try [COLOR=#333333]2*[(X)^2+(y)^2+(z)^2]^0,5

The 12.5 deg inclined surface will result in an offset to the as-measured location of each measurement point.

See [URL="http://www.engineersedge.com/calculators/spherical-true-position.htm"]Spherical True Position Calculator[/URL] [/COLOR]
• 03-28-2014, 11:45 AM
canteri
Hello Kelly, thanks for the feedback!!!

I did 2*((0,166)^2+(0,288)^2+(0,228)^2)^0,5 and the result is 0,770 however on my CMM report shows 0,637...
So, is this difference because of the 12.5 deg? Do you know how I could make the calc to actually get on the same result of my CMM Report?
• 03-29-2014, 10:49 AM
Kelly Bramble
Yes, using a spherical position equation will not work when the measured circle is 3D and at some angle not normal to the datum coordinate system.

What you will need to do is know the orientation of the circle to the datum's and the relative coordinate assignments (x,y,z). I tried to determine what the coordinate assignments where but have not been successful yet.

Next you will need to project the deviations from those angled coordinate axis to get a deviation that is normal to the axis.

The equation will likely look some thing like

Projection x = (deviation x) X sin(12.5)
Projection z = (deviation z) X cos(12.5)

Add these together to get a single deviation from the angled measurement (Projection x + Projection z) = Projection T

Then you will need to combine the deviations into a 2D calculation

[COLOR=#333333]TP = 2*((Projection y)^2+(Projection T)^2)^0,5.
[/COLOR]
If you could post more information of the angle relative to the perspective axis we may be to reproduce the calculations.

The following attachment is an example calculation (by Ray Xing).

[ATTACH]998[/ATTACH]
• 03-31-2014, 10:26 AM
canteri
Kelly,
Thanks!
I did the calc again and now I'm using the correct angle (12.2) gives me the same result as my CMM machine.
But I had to change the 'Total dev.' calc. Instead of doing "Total Dev. = ProjX(sen) - ProjY(cos)" {because this gives me 0,449} I have to do "Total Dev. = ProjX(sen) + ProjY(cos)" so I got the same result as my CMM Arm.
If you know why I had this issue, please let me know.

• 03-31-2014, 01:12 PM
Kelly Bramble

• 03-31-2014, 01:38 PM
canteri
Sorry... My explanations was not that clear... now I can see.
Thanks!

[ATTACH=CONFIG]1001[/ATTACH]
• 03-31-2014, 02:41 PM
Kelly Bramble

Therefore:

Projection X = (-0,615)(sin(12.2)) = -0,22031
Projection Y = (0,350)(cos(12.2)) = 0,326772

Projection XY = 0,326772 - (-0,22031) = 0,547082

or

Projection XY = 0,326772 + 0,22031 = 0,547082

TP = 2((0,547082)^2 +(0,220)^2)^0,5 = 1,17955
• 03-31-2014, 03:37 PM
canteri
Kelly, once more thanks a lot for the feedback.
So, at least I'm not crazy and we could get on the same result (1,179)... :-)
On this case my Faro CMM Arm shows me TP = 1,534. Is this a normal difference? Any final recommendation?

Thanks again!!!
• 03-31-2014, 04:33 PM
Kelly Bramble
Well, no the math should work.. Let's review your datum setup.

Datum A-B give you an Axis and two perpendicular datum planes (that can rotate). Primary Datum

Next, Datum C which is a plane perpendicular to Datum axis/two planes A-B. Secondary Datum

You now have three axis and three planes two of the planes can rotate about axis A-B.

Datum D locks the rotation of the two planes rotating about Datum axis A-B. This creates the Tertiary Datum about one of the two planes on axis A-B now locked down.

So,

Which datum planes are "X", "Y" and "Z"?

Also, you need to verify and define the angle to one or more of the Datum planes.
• 03-31-2014, 04:38 PM
Kelly Bramble
You might want to create a drawing that illustrates the X,Y, and Z planes to then show the angle relative to the specific datum plane.

This will simplify determining the equations to verify the CMM.
• 11-06-2015, 12:37 AM
nikky
why 3D-CMM True Position result not match with Equation ?