Name: AC1123 Spare Chuck Key
SKU: AC1123
Price: 10.00 USD
Availability: BackOrder

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.

This field is for validation purposes and should be left unchanged.

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