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8

This is a standard dynamics problem. Let's use this figure I drew: Some definitions: $$ \begin{align} m & \mbox{, the mass of the vehicle in kg.} \\ \mu_{\mbox{rolling}} & \mbox{, the rolling friction coefficient of your tires.} \\ \theta & \mbox{, the incline of the plane in radians.}\\ g & \mbox{, the gravitational constant, 9.81 } m/s^2 \...


7

How quickly do you want to go from stopped to 10rpm? This will define your angular acceleration. Regarding calculations, first you should convert to standard units, so meters instead of centimeters and radians per second instead of revolutions per minute: $$ \omega_{\mbox{rad/s}} = N_{\mbox{rpm}}*\frac{2\pi}{60} \\ \omega_{\mbox{rad/s}} = N_{\mbox{rpm}}*0....


5

There are 2 (or 3, depending on how the planned lifetime of the robot is) thing thats you have to consider. Static load: The motors stall (zero revolution) torques have to be able to hold the weight in the robots most unfavorable pose (usually the arm stretched out). You can determine this by static modelling, that involves only an equilibrium of forces for ...


5

Series elastic actuators tend to have more stable force control because the spring filters out the high-frequency motion of the mechanism. A low frequency in the system dynamics means that you can use slower control techniques, which is important when using digital controllers with naive control implementations, and sensors with significant abbe error and ...


5

Model Based: Low cost solution is always software: Develop a dynamic model which computes the load on the motors based on the motions (e.g. with Recursive Newton Euler Algorithm or Lagrange-Euler Method) Make as simple model to handle losses in driver/motor Make the difference between real torque and computed torque It will be not accurate. At least 10% ...


5

Write the forward kinematic equations $$\vec(x) = F\vec(\theta)$$ Taking the partial derivatives of each $\vec (x)$ term with respect to each joint variable $\vec(\theta)$ will give you $J$.


4

If you assume that the wind force you calculated is distributed evenly about the arm (reasonable assumption) then the torque about the shaft is the distance to the center of the arm (20 in) multiplied by the force in oz. So Torque = Force in ounces * Distance = (6.4 lb) * (16 oz / 1 lb) * (20 in) = 2048 oz. in.


4

You need enough torque to overcome friction, and to accelerate the load. If you know the friction torque ($\tau_f$), and the mass moment of inertia along the motor axis ($I$), then the minimum motor torque required is $$ \tau = \tau_f + I\alpha $$ where $\alpha$ is the required rotational acceleration.


4

Your expectations are rather aggressive for a DC motor. First - 40Nm (350in-lbs) is A LOT of torque! Ex: A max rated torque for 1/4-20 bolt is only 75 in-lb (8.5Nm). Second - The mechanical power of a motor results from RPM*Torque. 40Nm*100rpm is 0.56hp (420W at 100% efficiency). That is A LOT, about 1/2 of what today's high end cordless drills can do. ...


4

The Boston Dynamics’ Atlas robot doesn't actually use servo motors. It uses electrically-powered hydraulic actuation. I saw a presentation by Boston Dynamics' Marc Raibert where (if I remember correctly) he said they used to use Moog actuators designed for the aerospace industry, but that they were now using a motor that they build in-house that is ...


3

I've done a couple worm drive designs and I threw the numbers you gave (Assuming sd-spi nylatron gears) into my calculations. I'm using AGMA 6022-C93 Worm gear design guidelines for tooth width and thickness to get the shear area. I'm assuming 3 teeth in contact and mesh efficiency around 30%. The tooth stresses are a little more than one order of ...


3

What's the point of the middle gear? It's not doing anything for you reduction-wise because it's on a different shaft than the larger gear. There's an equation here that should give you the maximum allowable force for a nylon worm gear (Equation 18-4). A torque of 5 kg-cm caused by a force at (45/2) = 22.5mm (2.25cm) means the force is 2.2 kgf. While I ...


3

If your computed torques are out of the range for your robot, then you are no longer modelling your robot but a different one. You can ignore the torque limits as good as the joint angle limits of your robot (i.e. not at all). If the required torque is to high, your robot will either be much slower (if you are lucky) or won't be able to reach its goal ...


3

First I would like to caution your choice of motor torque. I don't know how you generated the drawing of your wrist, but CAD packages like Inventor or Solidworks can easily generate masses and moments of inertia if you select the correct material. For looking solely at holding torque, the torque required to not move, you need a torque that supports the ...


3

OK. as drawn, ignoring mass and accelerations, the force $F_p$ will appear as a torque on your ball screw. However, the total force on the ball screw, and hence the torque, depends on the mass of the thing you're moving with the ball screw interacting with gravity (if it's being moved in anything other than a horizontal plane), and on whether or not the ...


3

I think your Pmotor calculation is correct. The reason for the difference with the CIRC link is in the P=F*v equation they are referring to the linear velocity. Your conversion is converting the rotational velocity back into linear velocity. I don't think the Ftotal calculation is correct because your estimated weight isn't correct. Kg is a unit of mass, ...


3

From wikipedia http://en.wikipedia.org/wiki/Stepper_motor Because windings are better utilized, they are more powerful than a unipolar motor of the same weight. This is due to the physical space occupied by the windings. A unipolar motor has twice the amount of wire in the same space, but only half used at any point in time, hence is 50% efficient ...


3

Since the arm is in a vertical plane, certainly, motor torque is required to counteract the gravitational torque. Since it is a simple 1DOF arm with a rotary joint, the gravitational torque is $ \tau_{g} = mglcos(\theta) $ where m - mass of the link; l - perpendicular distance of the center of mass from the joint axis; $\theta$ - angle of rotation ...


3

One of the most common controllers is a computed-torque controller, also known as the inverse dynamics. The preceding controller is based on the feedback linearization principle which is an approach that maps a nonlinear model into a linear one and treats it as such as we will show momentarily. Consequently, one can utilize linear controllers such as PD and ...


3

The dynamics of robotic arms are fairly complex, especially when there are more than three joints to consider. The problem is that the movement of each joint moves all the links beyond it, which can induce torques at other joints. You have to consider how the movement of all links affects each individual joint. There is software that can automatically ...


3

I agree with @Andy 's answer, that chain efficiency is over 90%. I'll point out that your terminology is not quite right - using bicycle chain (or any other transmission) affects power transmission efficiency. It's a subtle but significant difference between what you said (torque transmission efficiency). First, efficiency is defined as the ratio of output ...


3

A robot like the one you refer to probably has a mix of transmission types that increase in power-to-weight ratio and decrease in size and stiffness as you move outwards on the kinematic chain. The base joints likely have planetary or cycloidal gearing, with maybe some spur gears in the middle, and maybe a harmonic drive reducer at the end effector. I don't ...


3

Actuation is a complex subject that cannot be reduces to 1 component alone. It is the interaction of software electronics and mechanics. "Toy" servos tend to incorporate everything in one small package. Industrial robots have 3 or 4 separate components (each costing several hundred or a few thousand euros) instead. To answer your question, worm gears are ...


3

Every action has an equal and opposite reaction. Each motor/joint in a linear chain of actuators (snake) needs to be capable of supplying the appropriate reaction forces. This mean that, if you have a 100cm long snake robot with a motor every 10cm, the first/"neck" joint (at 10cm) needs to support 8 other motors and 90cm of snake body. The second joint ...


3

One relevant equation is the lorentz force equation. which suggests that if you want to reduce current then you need to increase the magnetic field. The magnetic circuit is complex, with different materials moving relative to each other. As the rotor of the motor rotates, the magnetic field within the stator changes direction and there are losses. Choosing ...


3

When looking for motors, you usually are trying to figure out specific things with specific requirements. Holding Torque Torque at Speed Resolution (for steppers) NEMA size Positional Accuracy vs speed vs torque Movement Profile Budget So we need a little more info about your system. You're dealing with a 10kg load and your linear vel. is very slow at 0....


2

It is most likely kgf-cm, as kg/cm is not a valid unit for torque. Apparently writing kg/cm when kgf-cm is meant is a common mistake made on datasheets. According to this conversion of kgf-cm to Nm, 28 kgf-cm of torque is 2.7 Nm, and 150 kgf-cm of torque is 14.7 Nm.


2

The reason for this is that when attempting to hold a position between positions, you are not only fighting the torque from your load but that generated by the magnets in your stepper more info here


2

Only rotate 180º, right? Not continuous rotation? How about wrap a string around the pipe and then hang a weight off the string? As you rotate the pipe it will lift the weight and the torque calculation is straightforward. If you need more torque you can artificial increase the diameter where the weight is attached. Hmm, looking at your title again you say ...


2

Probably I'm missing your point, but the fundamental dynamics equation of a manipulator describes physical relations among quantities irrespective of the particular electronics/mechanics running underneath. If there is a rotating shaft with a joint mounted on it then there is always an acceleration. In a stepper, velocity and acceleration might be ...


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