A quick google search converting 1.9 Nm to pounds equals 1.4 foot-pounds or 16 inch- pounds. For your motor to handle 30 pounds maximum, the arm must be 0.5 inches long or less. A gearbox will reduce the speed and increase the torque available.
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Having both speed and weight capacity will probably require a larger motor.
I believe that choosing between direct or indirect drive is more about where you want to install the motor in relation to the arm. Do you have the space and strength required to install the stepper motor and gearbox at the pivot point of the arm or must it go somewhere else?
Those are worse than your current stepper motor. The add shows 170 Newton-centimeter which means 1.7 Newton-meter compared to your current 1.9 Nm.
I believe sharing how I did my math will help you. Torque is defined as twisting or motor shaft rotation force. It is measured in compound units of force times distance. Picture your ratchets and sockets in your toolbox. The smallest ratchet has a handle length about 6 inches. When you press on the handle with some force (say 10 pounds), you have 10 pounds of force and 6 inches of length. This means 60 pound inches of torque or 5 pound feet torque. When you grab the large ratchet (12 inches), the same 10 pounds has 12 inches now so 120 inch pounds or 10 foot pounds.
It appears that your design has not been adequately thought out and you need to figure this out before buying more parts. Start with drawing a stick-figure of your arm and label the length of each arm segment. Next, decide the maximum weight capacity of the arm in the hand. This will allow you to calculate the minimum hold torque at each joint; stronger components will help things stay cooler and last longer.
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Another critical design specification is the required speed. If you decide the maximum rotation angle of the joint (shoulder or elbow) and how long you can wait for it to move from one stop to the other, then you can calculate the required rpm (rotation speed) of the joint. So, if the rotation is 180° and you want it completed in one second, then multiply the number of rotations (1/2 rotation) by the time (1 second) for the rotation speed 0.5 rotations per second).
Motor speed is measured in RPM (revolutions per minute). To make the above speed in minutes, multiply (0.5 rotation / second) by (60 seconds / 1 minute [this is like multiplying by one, but changes the unit of measure]). This equals 30 RPM.
Your electric stepper motor spins much faster than this (30 RPM)! The degree angle of your stepper motor and the maximum drive frequency of it determines the maximum speed. The second stepper motor link has 1.8° steps so 200 steps per 1 rotation. At 600 Herz (cycles per second), you get 3 rotation per second or 180 RPM.Through a gearbox, pulley system, or other mechanical advantage type device, you can use a mechanical advantage of 6 to match the motor speed to your desired arm speed. Slowing the speed by 6X increases the torque approximately (friction and other factors reduce the actual torque output) 6X. With your 1.9Nm motor, the 1.4 ft-lb goes to 8.4 maximum (~8 ft-lb likely).
If you want a 3 foot arm to hold 30 pounds, the hold torque is 90 ft*lbs or 11 times more than your current motor with 6:1 gearbox.
There is the possibility of using an electric linear actuator on the arm for movement and hold strength similar to how an excavator boom (arm) functions. Much easier to get inexpensive arm strength. The drawback is that it requires a separate circuit to measure the movement and position of the arm.
Good luck with your project.
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