How does load force impact load inertia?

Question

I am trying to simulate a winch as a speed-regulated motor that works through a gearbox to lift a mass. The output of the gearbox is a drum, which rotates to accumulate cable.

I feel comfortable converting the mass to a moment of inertia and I also feel comfortable with converting that moment of inertia (output-side) to the moment of inertia "seen" by the motor (input-side) with the gearbox ratio. With a simple simulation, I have no problem writing the equations of motion.

My complication comes when I want to model "stretch" in the cable. I thought I could do this by simply putting a spring of arbitrary stiffness between the winch drum and the mass, as pictured below.

With this model, for the sake of simulation, I'm assuming I know the "drum height", which would be how far the drum has turned multiplied by the drum radius, and the height of the load. The spring force would be $k(\phi r - y)$, but how do I apply this to the motor?

I have a motor model:

$$ \frac{\Theta}{V} = \frac{K_T}{R_a Js+K_T K_b} $$ and a PI controller model:

$$ \frac{V}{\Theta_\mbox{error}} = \frac{k_p \left(s + \frac{k_i}{k_p} \right)}{s}\\ $$ where $\Theta$ is the motor speed, $V$ is terminal voltage, $J$ is the inertia of the load and machinery, and $R_a$, $K_T$, and $K_b$ are the motor armature resistance, torque constant, and back EMF constant, respectively.

The interaction I'm interested in studying occurs when the PI controller is tuned to the anticipated load inertia $J$, which would be found with the motor, gearbox, drum, and load mass, but the system actually "sees" the springy-mass.

Simplification is done by setting the $k_i/k_p$ ratio equal to $K_TK_b/R_aJ$, giving:

$$ \frac{\Theta}{\Theta_{\mbox{error}}} = \frac{V}{\Theta_{\mbox{error}}}\frac{\Theta}{V} = \left( \frac{k_p \left( s + \frac{K_TK_b}{R_aJ} \right) }{s} \right) \left( \frac{\frac{K_T}{R_aJ}}{s+\frac{K_T K_b}{R_aJ}} \right) $$

(Note I can leave $k_p$ as variable because the ratio $k_i/k_p$ can be set to whatever I want via $k_i$ as long as $k_p$ isn't zero.)

So, in an ideal world, where the value of "total" inertia $J$ is known in advance, the pole cancels, and the entire system reduces to:

$$ \frac{\Theta}{\Theta_{\mbox{error}}} = \left( \frac{k_p}{s} \right) \left( \frac{\frac{K_T}{R_aJ}}{1} \right) \\ $$ $$ \frac{\Theta}{\Theta_{\mbox{error}}} = \frac{1}{\frac{R_aJ}{k_pK_T}s} $$

Finally, $\Theta_{\mbox{error}} = \Theta_{\mbox{ref}} - \Theta_{\mbox{out}}$, so, with algebra:

$$ \boxed{\frac{\Theta_{\mbox{out}}}{\Theta_{\mbox{ref}}} = \frac{1}{\frac{R_aJ}{k_pK_T}s + 1}}$$

So, kind of sorry to shotgun so much detail, but I wanted to impress on anyone reading that I feel confident with all of my steps so far and that I have spent considerable effort working on this problem. Now, again to my question - I want to simulate stretch in the cable between the drum and the load, but I'm not sure how to use the spring force to modulate load inertia.

One thought I had was to try to fake an "equivalent mass", by assuming:

$$ F = m_{\mbox{equivalent}}a \\ m_{\mbox{equivalent}} = \frac{F_{\mbox{spring}}}{a} \\ $$

but this doesn't feel right, and I'm not sure what I would use for acceleration $a$.

I'm frustrated to be this far along on the problem and getting stumped by what seems like should be an easy issue, but I really can't think of a way to approach this problem. I think if I could frame it correctly I could work out the mechanics, but it's the force-to-inertia conversion I feel like needs to be made that has me stumped.

Finally, for the record, I have also tried back-tracking my motor model to include the load torque. This gives seemingly-reasonable results, but in the end I subtract the load torque from the motor torque to get net torque, then apply that net torque to the total inertia to get motor acceleration. That feeds on down the line and, again, I'm not sure that I'm treating total inertia correctly.

Answer 1

Let's first compute the model. The control design is a separate effort.

The torque applied to the drum is $n T_M $, where n is the gear ratio and $T_M$ is the output produced by the motor. $T_M= K_T i(t)$, where $K_T$ is a proportionality constant and $i(t)$ is the motor current.

Now we can write the equations for the mechanical system: $$ m y''(t)+m g-k (y(t)-r \theta (t))=0$$ $$ J \theta ''(t)+k r (y(t)-r \theta (t))=n K_T i(t)$$

Here m is the mass and k is the spring constant.

To write the motor equation, we need to determine the back emf. The back emf is proportional to the motor speed and to write it in terms of the drum speed we multiply it also with the gear ratio n. $$L i'(t)+R i(t)+n K_b \theta '(t)=V(t) $$

Here $V(t)$ is the applied voltage, $L$ is the inductance, $R$ is the resistance, and $K_b$ is the proportionality constant.

These three equations have $V(t)$ as the input and $i(t)$,$\theta (t)$, and $y(t)$ as states/outputs. This can be used to obtain the state-space model or transfer-function model. (The following were obtained using Mathematica)

Now the control design can begin...

Update

Since there has been some confusion about the inertia to be used, let me clarify the answer. I am going to assume one set of gears in the gearbox - a gear with inertia $J_1$ on the drum side and a gear with inertia $J_2$ on the motor side.

In the answer above I neglected the inertia of the gears. The only change that needs to be done now is modify the second equation as follows.

$$ \left(J+J_1\right) \theta ''(t)+k r (y(t)-r \theta (t))=n i(t) K_T$$

If the equation to describe the transient dynamics of the motor shaft is also desired then it is an additional equation involving $\theta_M$(rotation of the motor shaft), the inertia $J_2$, etc. However, this is not necessary if the objective is to control the drum position.

Answer 2

Stretch in the spring delta $ Y = A.sin(\omega.t) = A.sin\sqrt(k/m) . t $ So the delta Y is not constant but if you are interested in delt Y_max

delta $Y max = m/k$, by Hooks law.
Because your system doesn't accelerate except at the beginning and end assuming the pulley starts and stops suddenly that's you maximum. Any gradual start/stop acceleration will have to be subtracted from the acceleration of spring which is
$ - \omega^2 . t$
$ \omega = \sqrt(k/m)$

looking at free body diagram of mass
As you noted force is $ K(\phi.r - y)$

$m.dx^2/dt^2 = -K(\phi.r-r)$
divide both sides by K we get:

$m/K.dx^2/dt^2 +\phi.r=y$

$\omega^2.dx^2/dt^2 +\phi.r =y$

I hope this will help.

Answer 3

I realize this is an old thread, and I am not sure how deep of a dive you finally took on this, but one thing I don't see accounted for in your equations is drum/cable friction. This will be small, and like the accumulated mass of the wound steel wire rope you did not include, it may not be on your list. The cable could be pre-stretched and pre-loaded, however any movement between the cable and drum due to cable stretch will also encounter friction. In my industry (theater rigging, stage machinery design), the groove contacts a greater area than a flat drum application, and we usually have additional friction along the redirect sheaves and mules in the lineset to account for especially in 2:1 or 4:1 mechanical advantage systems.

Answer 4

I think the approach of Suba Thomas gives a good model: start with the sum of forces at the load and the sum of moments at the drum. Then determine the needed motor model.

The initial motor model of chuck needs a stiff system where a single value for the moment of inertia can be calculated, while the goal of the model is:

The interaction I'm interested in studying occurs when the PI controller is tuned to the anticipated load inertia $J$, which would be found with the motor, gearbox, drum, and load mass, but the system actually "sees" the springy-mass.

One note about the inertia in the drum moment equation of Suba Thomas: Do not forget the inertia of the motor increased to the drum. Depending on the chosen motor, its influence can be significant. So I would choose $J = J_{motor} * i^2 + J_{drum}$