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5

When selecting an actuator, you need to consider not just the force but also the power. Power = force x speed The faster you need to lift the lever, the more power you'll need. Companies like Maxon provide detailed graphs showing exactly how much torque you can expect at various speeds for their motors. Other companies may just state the motor's wattage. ...


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

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

The loop frequency is a parameter that needs to be tuned just like your proportional, integral, and/or derivative terms. Varying it has a similar affect on your output as varying your other parameters. Too low a frequency and you'll never reach your desired steady state. Too high and the output will oscillate. To determine the optimal loop frequency, you ...


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.


3

As a general remark, a PID controller can be designed to take into account only constraints occurring at the commanded variable - the end-effector velocity in your case - through the so-called antiwindup mechanism that is responsible for detaching the integral whenever those constraints are reached. This way, the PID is somehow informed that something "bad" ...


3

But my understanding is that all calculations happen relative to the base frame This is incorrect. Take a look at this two-link arm manipulator. The point ($x_2,y_2$) is obviously expressed in a moving frame $y_1,x_1$. Another way to look at the problem is to use Lagrangian approach which free you from constraint forces.


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

I always add 100-120% overhead (normally this just means the next bigger unit than exactly what I need) but also this comes with experience, a motor that can just barely lift your load will do so very slowly.


3

When the string isn't under tension you have a non-linear system (i.e. you're pushing on a rope) which may also make this harder to control. The stiffness of your string is going to limit your bandwidth. (The string acts as a low-pass filter, at least when it's under tension). I've actually worked a little on a similar setup and it was really hard to ...


3

Attach a metal square-shaped frame to the first robot, fill the edges with magnets, take your second robot (should have non actuated free moving wheels, like the ones on chairs) and fill it with opposite magnets on all sides, put it inside that metal square, drive your first robot.


3

There are very limited options for effectively transmitting force across a distance, other than with projectiles of some kind which won't achieve your goals or be allowed under the rules. Based on the rules as given (which mentions having an arm on the hybrid robot), I would assume that the hybrid robot is expected to extend an arm above the eco robot's ...


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, ...


2

As I understand, you do not actually need the force, you just want that your simulation to behave somewhat realistically. Instead of complicating everything with dynamics, I suggest you remain at kinematic models (will be much much faster, considering you will evaluate your models probably millions of times, if I undertand you goals corretly, in the ...


2

The wrench characterizes the forces and torques acting on the respective linkages. Part of this is motor torque, the other forces and torques are loading the mechanical structure (and motors down the chain). If you have followed the Denavit Hartenberg convention in defining the coordinate systems and you have a rotational joint the joint torque is the z ...


2

Yes, it is the same, $F$ should be the same as $F_{ext}$. And $F$ comes from a sensor or an estimation, in general from a force sensor. The diagram seems correct and you may have $F$ and $F_{ext}$ coming from the same source. The impedance controller is an indirect force control, you directly do not control the force exerted to the environment but rather ...


1

Short answer: you can use it. Long answer: Depend on your case, but I have seen many papers where they use the pseudoinverse of the non-square Jacobian matrix. The rectangular Jacobian matrix opens the window to optimize in terms of other variables such as space constraints or force minimization. The pseudo inverse by definition (in a redundant manipulator) ...


1

In the simplest case where targets and obstacles are just points in your space, it's not that difficult to come up with good representations of attractive and repulsive forces. For the sake of simplicity, let's stick to one-dimensional space and deal with massless points. Then, knowing the complete state of your robot in terms of position and velocity $\...


1

You can try the following. Set an admittance you want as $ F_e = m_d \ddot q + c_d \dot q$ then, you can solve for $q$ as the velocity reference $q_r = \frac{F_e - m_d \ddot q_r}{c_d}$ and approximate the acceleration through a forward euler with the velocity commands as $\ddot q _r = \frac{q_{r_{k}}- q_{r_{k-1}}}{dt}$ and insert that in the ...


1

By taking the time derivative of the forward kinematics equation, you get a Jacobian equation, as @steveo said in his answer. What is interesting is that by using some properties of rotation matrices, we can derive a rather impressive formula for computing a Jacobian. In short, a Jacobian can be computed as $$J = \begin{bmatrix}J_1 & J_2 & \cdots ...


1

I found a new paper titled "Quadrotor Parameters Identification and Control System Design" (https://ieeexplore.ieee.org/stamp/stamp.jsp?tp=&arnumber=7910684) by the same authors published on the same year, presumably using the same quadcopter platform. They provided formulas relating torque in Newton-meters and PWM percentage from 0 to 100. The results ...


1

If you only use Newton's second law, you'll get pretty close in your analysis. $\sum F = m a$ $\sum M = I \alpha$ Using only the first equation and the idea that a torque is a force at a distance ($\tau = F \times l$), and point masses for the payload and link masses, you will get a quick first estimate. So choose what acceleration is acceptable to you ...


1

I have tried to analyze the system, and the process is shown in the picture. The mechanism is called as Slider Crank Mechanisms, Please open the website (kinematics) for some other analysis. $(1)~\tau = F_1 . l_{DC} . sin \alpha \implies F_1 = \dfrac{\tau}{l_{DC} . sin \alpha}$ $(2)~F_1 = F_2$ $(3)~F_2 . cos \beta = F \implies F = F_2 . cos \beta = ...


1

I would suggest that your robot activate a relay which controls power to the heater. This is simpler than solving the original problem of developing the mechanical force to plug it in. You would use a "power relay", rated for at least 120Vac and 2.5A, something like this. A mobile robot will communicate via internet or local network to a controller which ...


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