Mark-10 defines accuracy as a percentage of full scale of the
instrument. To determine the measurement error as an actual
load value, multiply the accuracy percentage by the instrument’s capacity.
Example 1 – M5-50 force gauge:
The accuracy is ±0.1% of full scale (FS). Multiply ±0.1% by 50 lbF,
which equals ±0.05 lbF. This means that any displayed reading may be
higher or lower by up to 0.05 lbF. For example, if the displayed
value is 30.00 lbF, the true reading will be ≥29.95 lbF and ≤30.05 lbF.
Example 2 – Plug & Test® indicators and sensors:
The accuracies of the sensor and the indicator must be added together.
Models 7i and 5i indicators have accuracy values of ±0.1% FS,
while the Model 3i is rated at ±0.2% FS. Using the example of a
Series R50 torque sensor with Model 3i indicator, add ±0.35% to ±0.2%,
which equals ±0.55%. In a specific example for the Model MR50-12,
the accuracy becomes ±0.55% x 135 Ncm = ±0.7425 Ncm.
Percentage of Reading:
Because of these fixed errors, lower measured values will be more
inaccurate as a percentage of reading.
Further using the example of an M5-50 force gauge, a fixed error of
±0.05 lbF represents a higher error as a percentage of reading for
a load of 1.00 lbF than 30.00 lbF.
To calculate the error as a percentage of reading, divide the fixed
error by the measured value. For a 1.00 lbF load, the fixed error
equals ±0.05 ÷ 1.00 lbF = ±5% of reading. For a 30.00 lbF load,
the fixed error equals ±0.05 ÷ 30.00 lbF = ±0.17% of reading.
Conclusion:
Because of the relationship between load and accuracy, we recommend
selecting an instrument capacity as close as possible to the
maximum measured load.
Medical Device & Pharmaceutical
Packaging
Automotive
Consumer Products & Sports Equipment
Springs
Ergonomics Testing & Job Task Analysis
Aerospace
Food & Beverage
Electronics
Other Industries
