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, the manufacturer recommends selecting an instrument
capacity as close as possible to the maximum measured load.
