For a 30.00 lbF load, the fixed error equals ☐.05 ÷ 30.00 lbF = ☐.17% of reading. For a 1.00 lbF load, the fixed error equals ☐.05 ÷ 1.00 lbF = ±5% of reading. To calculate the error as a percentage of reading, divide the fixed error by the measured value. Percentage of Reading:īecause of these fixed errors, lower measured values will be more inaccurate as a percentage of reading.įurther using the example of an M5-50 force gauge, a fixed error of ☐.05 lbF represents a higher error as a percentage of reading for a load of 1.00 lbF than 30.00 lbF.
In a specific example for the Model MR50-12, the accuracy becomes ☐.55% x 135 Ncm = ☐.7425 Ncm.
Rs logix pro 500n free series#
Using the example of a Series R50 torque sensor with Model 3i indicator, add ☐.35% to ☐.2%, which equals ☐.55%. Models 7i and 5i indicators have accuracy values of ☐.1% FS, while the Model 3i is rated at ☐.2% FS. The accuracies of the sensor and the indicator must be added together. 500 N/mm2 D 550 N/mm2 Answer : B 90 As per IS:1139, the characteristic yield strength for cold twisted deformed bars is A 250 N/mm2 B 415 N/mm2 C 500 N/mm2 D 550 N/mm2 Answer : C 91 For reinforced concrete work, aggregates having a nominal size of - are generally used A 20 mm B 40 mm Civil Engineering Rocks Civil Engineering.
Example 2 – Plug & Test ® indicators and sensors: For example, if the displayed value is 30.00 lbF, the true reading will be ≥29.95 lbF and ≤30.05 lbF. This means that any displayed reading may be higher or lower by up to 0.05 lbF. Multiply ☐.1% by 50 lbF, which equals ☐.05 lbF. To determine the measurement error as an actual load value, multiply the accuracy percentage by the instrument’s capacity.
Rs logix pro 500n free full#
Mark-10 defines accuracy as a percentage of full scale of the instrument.
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