"Logging Precision"
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dimondjack
- Posts: 101
- Joined: Tue Jan 15, 2013 1:37 pm
"Logging Precision"
What does logging precision stand for? Are these the number of bits recorded for a 0-5v channel?
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dimondjack
- Posts: 101
- Joined: Tue Jan 15, 2013 1:37 pm
Thanks for the clarification. Correct me if i'm wrong about the next bit
To potentially help a future user, the precision value does not affect the number of bits that the RCP microcontroller "sees". It will always see 8 bits of precision, or 1024 values (0-1023). If you log the raw value, you should always use a precision value of 0. The calibration you choose will change the precision value you choose. A simple example:
linear curve of y=x/10 (divide raw value by 10). Values now go from 0-102.3 in increments of 0.1, so you would want a precision of 1. Adding additional precision will only give you more zeros. In a more complex example, say y=x/8+10 The minimum value will be 10, but then will increment in values of .125. To get all potential values you would need to have a precision of 3. However, incrementing rounding to two decimal points (.13) would like be pretty darn close and you would save the logging space (if you need it). You will retain all of the precision (you'll get values of .13, .25, .38, .5) even though your points will be minutely skewed.
To figure out how many precision points you need, put "1" through your math formula and see how many decimal points it requires to divide out. Then decide at what decimal point you will see a significant difference to your next raw value (2) and make this value your precision.
Hope that helps.
To potentially help a future user, the precision value does not affect the number of bits that the RCP microcontroller "sees". It will always see 8 bits of precision, or 1024 values (0-1023). If you log the raw value, you should always use a precision value of 0. The calibration you choose will change the precision value you choose. A simple example:
linear curve of y=x/10 (divide raw value by 10). Values now go from 0-102.3 in increments of 0.1, so you would want a precision of 1. Adding additional precision will only give you more zeros. In a more complex example, say y=x/8+10 The minimum value will be 10, but then will increment in values of .125. To get all potential values you would need to have a precision of 3. However, incrementing rounding to two decimal points (.13) would like be pretty darn close and you would save the logging space (if you need it). You will retain all of the precision (you'll get values of .13, .25, .38, .5) even though your points will be minutely skewed.
To figure out how many precision points you need, put "1" through your math formula and see how many decimal points it requires to divide out. Then decide at what decimal point you will see a significant difference to your next raw value (2) and make this value your precision.
Hope that helps.
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dimondjack
- Posts: 101
- Joined: Tue Jan 15, 2013 1:37 pm