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I have a kinematic model like the simplified following image. Assume that all points are fixed. I'm interested in how the force $\vec{m}$ distributes across $a_2$ and $b_2$ up to $a_1$ and $b_1$.

Finally I'm interested in the torques that result in $A_1$ and $B_1$.

enter image description here

I know that the kinematic as presented is not stable. The stabilizing part was omitted for simplicity. Simply assume the kinematik is in a stable state and I simply want the forces/torques.

Some notational stuff (edit if you need more):

$$ c := \overline{A_1B_1}, \quad d := \overline{A_3B_3}, \quad M := \frac12(A_3+B_3), \quad A_3B_3 || A_1B_1 $$ The mass of the kinematic itself is neglectable.

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1 Answer 1

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Your system is very underdetermined. By applying a torque at $A_1$ and $B_1$ you can hold bars $a_1$ and $b_1$ still, but even without those you are still left with a four bar linkage, which has one more degree of freedom.

If there is no torque applied at any of the four lower joints, then the bars can only transmit force along their axes. That means that the forces in $a_2$ and $b_2$ will be in the direction of those bars. Some basic trigonometry will get you the forces on each bar to balance $m$. The problem is that, in a general position, those forces do not compensate the torque of $m$.

But if they did, then the force that $b_2$ exerts on $B_3$ has the same magnitude, but opposite direction, than the one it exerts on $B_2$, and similarly with $a_2$, $A_3$ and $A_2$. Knowing the force applied at $B_2$ or $A_2$ and its direction, computing the torque at $B_1$ or $A_1$ should be trivial.

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The kinematic shall be a 2 dimensional delta robot delta robot on wikipedia. I couldn't find a suitable image but there are extra tubes and joints that always keep the lower part parallel to the top part. But I'll want to expand a solution to the Delta 3 case later on. – Onur Nov 8 '12 at 12:20
If I get you right I can neglect the length of $d = \overline{A_3B_3}$? Then I would guess I'll get the force along $a_2$ and $b_2$ like this: $ \alpha \vec{A_2A_3} + \beta \vec{B_2B_3} = m$ like in a parallelogram of force? – Onur Nov 8 '12 at 12:31
If there are missing bars, you cannot simply neglect them. But in the system in your picture, yes, $d$ plays no role in determining the forces on $a_2$ and $b_2$, if there is equilibrium. But again, in a general position, as is yours, the system is not in equilibrium. – Jaime Nov 8 '12 at 22:06

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