Page 164 - The ROV Manual - A User Guide for Remotely Operated Vehicles 2nd edition

152 CHAPTER 7 Power and Telemetry
Max tether length!
Max power needed Power needed
Tether length
FIGURE 7.13
Graphical depiction of maximum tether length.
Table 7.2 Standard Copper Wire Gauge Resistance over Nominal Lengths
Wire Gauge (Approximately) Ohms/1000 ft
20 10 18 6 16 4 14 2.5 12 1.5
Source: Example and Table Courtesy of DeepSea Power and Light.
Thus, based on Ohm’s law, the voltage drop over a length of cable can be calculated by using the formula, V 5 IR, where V is the voltage drop, I is the current draw of the vehicle in amps, and R is the total electrical resistance of the power conductor within the tether in ohms. The current draw of a particular component (light, thruster, camera, etc.) can be calculated if the wattage and voltage of the component are known. The current draw is equal to the component wattage divided by the component voltage (or amps 5 watts/volts).
For example, referring to the table of electrical resistances for various wire gauges (Table 7.2), the voltage required to operate a 24-V/300-W light at 24 V over 250 ft of 16-gauge cable can be calculated as follows: The current draw, I, of a 24-V/300-W lamp operating at 24 V is 300 W/ 24 V 5 12.5 A. The resistance of 16-gauge wire is approximately 4 Ω/1000 ft (Table 7.2). Since the total path of the circuit is from the power supply to the light and back to the power supply, the total resistance of the cable is twice the length of the cable times the linear resistance, or for this exam- ple, R5(23250 ft)3(4 Ω/1000 ft)52.0 Ω. Since V5IR, the voltage drop, V, is equal to 12.5 A 3 2.0 Ω 5 25 V. This means that 25 V is lost due to resistance, so the power supply will need to provide at least 49 V (the 24 V necessary to operate the light plus the additional voltage loss of 25 V) to power this 24-V/300-W light over a 250-ft cable.
Total power
Power available