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enter image description here

Here's the link to the entire answer: Designing a 5 bar linkage robot: Plot Clock

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It looks (to me) like there's a typo in that answer. The approach is to recognize that you have to wind up at the same point, whether you get there via the left arm (d->e->blue dot) or the right arm (a->b->c->blue dot).

This means that you can go:

$$ \Sigma_{x_{\mbox{left arm}}} = \Sigma_{x_{\mbox{right arm}}} \\ \Sigma_{y_{\mbox{left arm}}} = \Sigma_{y_{\mbox{right arm}}} \\ $$

The "middle linkage" $a$ is horizontal and there's no angle given for it, but if we suppose the angle of $a$ is $\theta_0$, and that $\theta_0 = 0$, then you can do:

$$ \Sigma_{x_{\mbox{left arm}}} = d\cos{\theta_1} + e\cos{\alpha}\\ \Sigma_{x_{\mbox{right arm}}} = a\cos{\theta_0} - b\cos{\theta_2} -c\cos{\beta}\\ $$

I'll point out that the minus signs in there are because of the angle definitions - $\theta_1$ is defined "conventionally," starting at 0=right and getting positive counter-clockwise, where $\theta_2$ is defined opposite - starting at 0=left and getting positive clockwise.

You can define the points for vertical closures:

$$ \Sigma_{y_{\mbox{left arm}}} = d\sin{\theta_1} + e\sin{\alpha}\\ \Sigma_{y_{\mbox{right arm}}} = a\sin{\theta_0} + b\sin{\theta_2} + c\sin{\beta}\\ $$

Since we're assuming $\theta_0 = 0$, you can simplify $a\cos{\theta_0} = a$ and $a\sin{\theta_0} = 0$, you you're left with:

$$ \left[\begin{array}{ccc} d\cos{\theta_1} + e\cos{\alpha} \\ d\sin{\theta_1} + e\sin{\alpha} \\ \end{array}\right] = \left[\begin{array}{ccc} a - b\cos{\theta_2} -c\cos{\beta} \\ b\sin{\theta_2} + c\sin{\beta} \\ \end{array}\right] $$

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  • $\begingroup$ How is ''a'' considered as an arm? In the video it's just the gap between two motors (main linkages). $\endgroup$ – Aman Agarwal Oct 22 '18 at 19:09
  • $\begingroup$ @AmanAgarwal - $a$ is a linkage that connects the left "shoulder" to the right "shoulder." $\endgroup$ – Chuck Oct 22 '18 at 19:12
  • $\begingroup$ I'm trying to create a robotic leg with this skeletal design. Here let θ1,θ2 be the motor angles. imgur.com/a/mvJMAC6 $\endgroup$ – Aman Agarwal Oct 22 '18 at 19:22
  • $\begingroup$ I am a first year engineering student and I do not have much knowledge about this. I could not understand much after [the "middle linkage"] line. Are there any special concepts that I need to learn first to thoroughly understand this? $\endgroup$ – Aman Agarwal Oct 22 '18 at 19:30
  • $\begingroup$ @AmanAgarwal - It's a geometry problem. The left arm and the right arm both terminate at the same location - this is the key to the solution. The sum of all of the horizontal components of the left arm must equal the sum of all of the horizontal components of the right arm, or else the left and right arms don't terminate at the same $x$ location. Similarly, the sum of the vertical components of the left arm must equal the sum of the vertical components of the right arm, or else the left and right arms don't terminate at the same $y$ location. $\endgroup$ – Chuck Oct 22 '18 at 20:47

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