As shown in the illustrated digital gyrocompass functional block diagram and the associated communication protocols table, what would the rate of turn signal voltage be if the rate of turn is 30 degrees per minute to port? Assume that the rate of turn to port signal voltage is negative in polarity and that the rate of turn to starboard signal voltage is positive in polarity. Illustration EL-0194
• Check the Digital Gyrocompass Communication Protocols table for the "Rate of Turn" output specification, especially the mV per degree per minute value and the full-scale voltage vs. maximum rate. • Relate the given rate of turn (30°/min to port) to the stated scale factor to find the corresponding voltage, remembering that port is negative polarity and starboard is positive polarity. • Confirm how many degrees per minute correspond to the full‑scale voltage to help you sanity‑check your calculated value.
• What is the specified number of millivolts per degree per minute in the table, and how many volts would that be for 30°/min? • If the full‑scale rate of ±90°/min corresponds to ±4.5 VDC, what fraction of full scale is 30°/min, and what voltage would that fraction represent? • Given that port is negative, which answer choice matches both the magnitude you calculated and the correct polarity?
• Verify the scale factor in the table (mV/deg/min) and convert millivolts to volts correctly. • Double‑check the sign convention: port (negative) vs. starboard (positive). • Ensure your final voltage value is within the ±4.5 VDC full‑scale range stated for the rate‑of‑turn output.
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