Don.......You're the physicist so correct me if I'm wrong.........There is something to be said about the length of time required to come up to speed based on the applied voltage versus time constants. Thus in a given amount of time, the 220 should bring it up to speed faster. It's not a linear thing so it won't be twice as fast but faster. 2ndly, the line loss based on any resistance in the line (loose or bad connections) times the current drawn results in less power loss in the lines ....thus more available for consumption by the motor. The total power available remains the same but less current drawn through a bad connection/resistance results in less loss there and more applied power to the motor.
That's absolutely correct first of all the lower percent in voltage drop results in increased torque. As the speed of the motor increases resulting in less amps drawn to start the motor and less voltage drop. Actually, I believe (need to review my calc) the amps drawn go down as a square of the increase in speed. Because of this the motor gets up top design slip speed faster and the amps drain drops off faster with the higher voltage supply. That why the table saw seems to jump, more torque in a shorter period of time.
You are also correct in that not only the line losses due to conductor size and length contribute to the resistance (Voltage losses). There are high resistance connections where the wires screw into the Circuit breakers, where the plugs plug into each other and where the line wire connect to the motor. These all add to the problem.
The proof is in the puddin’. The purpose of a saw is to cut wood well.
I understand starting current and voltage drop during the brief time it takes motors to come to running speed. Maybe this is why your saw jumps on the 220V start and not the 110V? OK. Fine.
But since your shop wiring has been of sufficient gauge for both voltages, I do wonder why the performance is better at 220V.
The nameplate on my Delta saw gives HP ratings of 1.5 and 2.0 at low and high voltage. When the voltage is doubled the amperage draw is not halved. Why?
The Service Factor at low voltage is 1.15, compared to 1.0 at high. Temp, I suppose. But this is contrary to so many arguments I have heard about motors lasting longer at high voltage.
Still confused,
Maybe hopelessly so,
Frank
First of all lets consider the ratings of your motor. If it is rated at 1.5 hp at 120 Volts then that means it can handle 1.15% full load on a continuous rating assuming STP (standard temp and Pressure) as defined by NEMA (the
National
Electrical
Manufacturers
Association). They define it this way because of cooling. If the ambient of a motor is hotter such as you will find in Tucson in the summer or the atmospheric pressure is lower then sea level such as you would find in Denver then you must create the motor. Standard temp assumes an ambient of 80 degrees F. Now your two HP rating at 220Volts is because of the lower I squared R losses in the motor. In any motor the internal heating is defined as the load amps squared times the internal resistance. Take your 2 Hp 220 Volt motor at full load it will be carrying 10 amps. Lets assume an internal resistance of 1 ohm then the watts loss or heating in the motor would be 100 watts. Think of touching a 100 watt light bulb.
Now lets take you 1.5 HP 1.15 110 volt rating. The would translate into a full load amps rating of 17.25 amps. Now because the motor is wired for half the voltage lets assume the internal resistance is dropped in half. The the IsquaredR losses are around 150 watts which would be equal to the heat generated by a 150 watt light bulb.
That should address the service factor issue.
Now to address the torque issue at various voltages. HP = Torque RPM/ 63025 (assuming you want the answer to be in in.lbs.) That means that a 1 HP motor that runs at 3450 Rpm will deliver 18 in. Lbs of torque. This is a law of physics. So at 2 hp and 3450 Rpm at 220 Volts can deliver 36 In. Lbs. of torque.. The same motor that delivers 1.5 HP at 110 volts can deliver 27 in.lbs. and the rest is given up in internal motor losses.
OK Frank any more questions...