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View Full Version : Understanding calibration results on CMM reports


lee01
12th December 2002, 12:23 PM
I’m having a lot of trouble of late and one of the problem is understanding CMM calibration reports with especial consideration to the following reading:

E= 3.5 + L / 150? I understand this is the measurement of both repeatability and accuracy using a length, but can someone please go through it with me as you would type it into the calculator to gain a simple micron reading please!

Many thanks for your assistance!

Lee01

lee01
12th December 2002, 01:25 PM
I’m having a lot of trouble of late and one of the problem is understanding CMM calibration reports with especial consideration to the following reading:

E= 3.5 + L / 150? I understand this is the measurement of both repeatability and accuracy using a length, but can someone please go through it with me as you would type it into the calculator to gain a simple micron reading please!

Many thanks for your assistance!

Lee01

Ryan Wilde
12th December 2002, 05:00 PM
I would help, but I need to know units, such as:

L in meters
E in µmeters

Without that, the formula would basically be calculated:

(for this example, L=0.150 meter, or the length of CMM travel is 150 mm)

E = 3.5 + L/150
E = 3.5 + (0.150)/150
E = 3.5 + 0.001
E = 3.501

Now this example does not take into account the units, which is probably L in millimeters, E in µmeters. This would make the possible error 4.5 µm as follows:

E = 3.5 + 150/150
E = 3.5 + 1
E = 4.5

Hope this helps.

Ryan

IGORTS
13th December 2002, 09:00 AM
Hi,

Usually L is in milimeters and the result E is in µmeters.

Almost all the CMM calibration reports that I have ever seen are specified in these units.

Bye

Rick Goodson
30th December 2002, 11:56 AM
lee01,

There doesn't seem to be much help here. I am going to move a copy to the 17025 thread as that thread also includes MSA questions. Maybe someone there can help.

Regards,