clarity in understanding the moment of inertia matrix D and Coriolisis and centripetal matrix C as per attached equations

Question

Would appreciate if anyone could help me understand the how if equation 6 and 7 are derived. Helping with any one D22 and C32 would suffice.

Answer 1

Where did this example come from? It looks like maybe a class handout or similar. I don't agree with all of the formulas given, but I'll try to explain where I think they should be coming from.

The $\bf{D}$ matrix is pretty straightforward - the key here is to understand the parallel axis theorem, which is a formula for finding the moment of inertia at one point given the moment of inertia at the center of mass and the mass:

$$ I_{\mbox{new}} = I_{\mbox{center of mass}} + mr^2 \\ $$

The linkages form a kinematic tree, where each joint "feels" the mass of its "downstream" link and the effective masses of all links attached after it as well.

So, start with the last term in the matrix - $\bf{D}_{44}$:

$$ D_{44} = I_{bu} + M_{bu}r_4^2 \\ $$

This is exactly the parallel axis theorem. You transfer the moment of inertia from the center of mass of the bucket to the stick-bucket joint, because that's where you're rotating. The next step is a little tricky because everything has already been reduced. I'll walk through the derivation of $\bf{D}_{33}$ and then hand-wave the rest saying the method is similar.

The distance from $O_2$ to $O_3$ is $a_3$. The distance from $O_3$ to $G_4$ is $r_4$.

If you wanted to get the distance from $O_2$ to $G_4$, then you could use the Pythagorean theorem, but that's only applicable for right triangles, so what you have to use instead is the law of cosines, which is a generalization of the Pythagorean theorem that applies to all triangles:

$$ c^2 = a^2 + b^2 - 2ab\cos{\gamma} \\ $$ Where $\gamma$ is the angle between $a$ and $b$. Side note: for right triangles, they're "right" because the angle between $a$ and $b$ is a right angle (90 degrees), so the cosine of 90 is zero, so that term drops off and you're left with the Pythagorean theorem.

So, for your example, you need not just the angle of the stick-bucket joint, but also the angle from that joint to the center of mass.

The angle of the stick-bucket joint is $\pi + \theta_4$, per the drawing, and now you need the angle from the $O_3 - O_4$ joint to the center of mass of the bucket; this is given as $\alpha_4$.

Now, going back to the parallel axis theorem, if you wanted the moment of inertia for the bucket about the stick-boom joint, you need the moment of inertia for the bucket at the bucket's center of mass, and then you need the straight-line distance from the bucket-stick joint out to the center of mass of the boom.

Call that distance $c$, call the distance from the boom-stick (joke) joint to the stick-bucket joint $a$, and the distance from the stick-bucket joint to the bucket center of mass $b$, then what you get is:

$$ I_{\mbox{bucket about boom-stick joint}} = I_bu + M_{bu}c^2 \\ $$

$$ c^2 = a^2 + b^2 - 2ab\cos{\left(\mbox{joint angle} + \mbox{center of mass angle}\right)} \\ $$

$$ c^2 = a_3^2 + r_4^2 - 2a_3r_4\cos{\left(\left(\pi + \theta_4\right) + \left(\alpha_4\right)\right)} \\ $$

Now a little regrouping in the cosine term:

$$ c^2 = a_3^2 + r_4^2 - 2a_3r_4\cos{\left(\pi +\left(\theta_4 + \alpha_4\right)\right)} \\ $$

And finally a period shift identity:

$$ \cos(\pi + \theta) = -\cos{\theta} \\ $$

And you get:

$$ c^2 = a_3^2 + r_4^2 - 2a_3r_4\left(-\cos{\left(\theta_4 + \alpha_4\right)}\right) \\ $$

Distribute the negative:

$$ c^2 = a_3^2 + r_4^2 + 2a_3r_4\cos{\left(\theta_4 + \alpha_4\right)} \\ $$

NOW FINALLY plug this $c^2$ back into the shifted moment of inertia equation:

$$ \begin{array}{ccl} I_{\mbox{bucket about boom-stick joint}} & = & I_{bu} + M_{bu}c^2 \\ & = & I_{bu} + M_{bu}(a_3^2 + r_4^2 + 2a_3r_4\cos{\left(\theta_4 + \alpha_4\right)}) \\ & = & I_{bu} + M_{bu}r_4^2 + M_{bu}(a_3^2 + 2a_3r_4\cos{\left(\theta_4 + \alpha_4\right)}) \\ \end{array} $$

Now recognize that $D_{44} = I_{bu} + M_{bu}r_4^2$, meaning that:

$$ I_{\mbox{bucket about boom-stick joint}} = D_{44} + M_{bu}(a_3^2 + 2a_3r_4\cos{\left(\theta_4 + \alpha_4\right)}) \\ $$

So that's the moment of inertia of the bucket about the boom-stick joint, now you also need to add the moment of inertia of the stick about the boom-stick joint,

$$ I_{\mbox{stick about boom-stick joint}} = I_{st} + M_{st}r_3^2 \\ $$

Put it all together and get: $$ \begin{array}{ccl} I_{33} = D_{33}& = & I_{\mbox{stick about boom-stick joint}} + I_{\mbox{bucket about boom-stick joint}} \\ & = & I_{st} + M_{st}r_3^2 + D_{44} + M_{bu}(a_3^2 + 2a_3r_4\cos{\left(\theta_4 + \alpha_4\right)}) \\ & = & \boxed{D_{44} + I_{st} + M_{st}r_3^2 + M_{bu}(a_3^2 + 2a_3r_4\cos{\left(\theta_4 + \alpha_4\right)})} \\ \end{array} $$

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Short P.S. -

The trouble with calculating the Coriolis force (why I'm not willing to do it at the moment) is that it's a cross-product. You're looking at a singular plane, but it looks like the axes are flipping for some reason; $O_3$ and $O_4$ seem to be rotated about the y-axis? Or maybe the y- and x-axes? I can't tell if that's a typo or not.

Anyways, you should take the cross-product and then keep the terms about the axes you care about; the forces about the axes in the plane don't affect the solution.

But, there are a lot of typos in the document;

• $D_{44} = I_{bu} + M_{bu}R_4^2$ should be $D_{44} = I_{bu} + M_{bu}r_4^2$,
• $D_{24} = D_{42} + D_{34} + (...)$ should probably be $D_{24} = D_{42} = D_{34} + (...)$,
• I think that the signs are wrong on the off-diagonal $D$ terms, but I'm not willing to study the problem hard enough to verify,
• there are $\dot{\theta}_{234}$ terms that aren't defined, etc.

I would definitely question the source/author of the paper you've acquired. Generally speaking, the method is as I've described above. Start where the only force is the local one, then work your way up (or down) the kinematic chain. Subsequent forces are the new local forces plus all of the previous forces expressed relative to the local frame.