I am making an arm and am designing the should rn, and was wondering your alls opinions, should i use a direct drive or and indirect drive to move the arm.The main goal of the of the indirect drive would be to get extra speed and torque. Is it worth it? or should I just use a direct drive?
With pullies/belts or gears: if you gain speed, you loose torque because they are opposites.
then direct drive?
For an initial design, these are all the wrong questions! The prime question is what is the minimum movement necessary for the end of the arm and does the arm need to hold that position indefinitely?
Take a look at your motor speed range, generally you will want to reduce it via a drive train. Doing so will slow the output but will increase the torque. It is easier to control a bunch of turns on a motor verses 1/2 turn. I am assuming this is not a stepper.
im sorry to inform y ou that this would be a stepper, so would all of this still apply?
Is your arm going to be used to pay ping-pong? Why the need for speed, but using a stepper motor?
Yes it would have to hold that position indefinitely while the thing is running. So this is a 1.9Nm stepper motor holding torque so... is this strong enough to hold like 30 lbs (the weight of the arm on the stepper while it is extended) if its not should I use two steppers in direct access? indirect access to make the arm faster?
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.
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?
Sounds good. Can you control the stepper to move the increments you want the arm to move?
What does extended mean for a rotating arm?
The hand is stretched out horizontally as far as can be. This would definitely increase torque neccesseties.
so the stepper Im using has a 1.9Nm torque thing right? so if the arm was about as long as a normal human arm... I mean you dont need anyy specs to relize that there would definitely not be enough torque, so should I get two motors involved in an indirect drive? should I use a gearbox? What would be the best solution?
Im not sure how you figured that out but I now see that there definetely is not enough torque, would two of these motors be better? Or should I just rethink my design. https://www.amazon.com/23HS8430-Phase-Stepper-Stepping-Diameter/dp/B08RB5W2GM Its crazy that a motor like this or even two for that matter would not be enough.
Perhaps it would help if you just showed us your design! How will you couple two motors so as to double the torque?
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.
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.
1 Like
Thank you so much I really appreciate it!!
luke16, look at 110mm (or longer) nema23 stepper motors in the 3Nm ( = 30kgcm) sort of range, they are big and need constant current driver modules to run them, but if you can keep them all in the arm base and use pulleys or gears and shafts to take the rotations to the arm joints then those sort will have plenty of torque, especially with a gearbox added. if you can make it fairly big a 3d printed gearbox can (provide you have purchased a metal input coupler to adapt from the 6mm/8mm/10mm motor shaft out to a scrw pattern for the first printed gear) take 30kg*cm input and give you several times more torque out.
Unfortunately I already got motors that only output 1.9 Nm of static torque that's around 1.25 lbs-foot. Let's say that the arm was 2 feet long and 5 pounds. That would mean that 10 lbs-foot would be minimally necessary to lift the arm. With two of the motors I have there would only be a torque of 2.5 lbs-foot vs the arm which would need 10lbs-foot. Even if I would get these motors that you said and used two to lift the arm, that would only be 4.4 lbs-foot of torque vs the arm which would need 10lbs-foot. This is kind of the problem that I am having. I didn't realize how much torque I would actually need.
A linear actuator like this would have the strength to move and hold the arm. This will only move when power is applied.
ECO-WORTHY Heavy Duty 330lbs/1500N Solar Tracker Linear Actuator Multi-Functions with Mounting Brackets (12V, 12") IP54 Waterproof 300mm Stroke Linear Motion Actuator https://a.co/d/1bpDRDo
Hydraulic digging machines like excavators and backhoes use strong linear motion attached to both sides of the arm pivot points to provide tons of strength to dig, lift, and move the attached bucket. The downward force of the arm will lift the digging side of the machine off the ground.
The key to using linear motion to pivot an arm joint is locating the actuator attachment point the correct distance away from the arm pivot point on the moving arm segment. This needs to be far enough out so the actuator has enough leverage to lift the desired weight. On the other hand, less leverage means shorter actuators and faster arm speed for a given actuator length.
So, you want a 2 foot long arm weighing 5 pounds able to lift 30 pounds. The linear actuator I shared has a strength of 330 pounds lift. Reducing the expected load down to 200 (picked as easy round number) pounds for actuator service life means 100 pounds strength at 1 foot from the pivot and 50 pounds at 6 inches from the pivot. To get 35 pounds, the point belongs 4.2 inches away.
The drawback to using linear actuators for directly rotating an arm is getting the desired strength and required rotation range.