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March 22nd, 2017, 03:57 PM   #1
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3 legged stool

I have a mechanism that resembles a 3 legged stool. One of the legs has a fixed length. The other 2 can change length. I need to calculate d and e that will produce a desired pitch and roll angle. When (d) and (e) are 0, all 3 legs are the same length (c) and the pitch and roll angles are 0. The pitch and roll angles are limited to +/- 45 deg.
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March 23rd, 2017, 12:18 PM   #2
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ok, i'll give you an outline on how to do this

set up your coordinate system. I chose the front point of the triangle to be $(0,0,0)$.

The default rear points of the triangle are at

$pleft=\left(-\dfrac b 2,-\dfrac 1 2\sqrt{4a^2-b^2},0\right)$

$pright=\left(\dfrac b 2,-\dfrac 1 2\sqrt{4a^2-b^2},0\right)$

Now you can tweak these positions by $d$ on the left and $e$ on the right so after tweaking these become

$pleft=\left(-\dfrac b 2,-\dfrac 1 2\sqrt{4a^2-b^2},d\right)$

$pright=\left(\dfrac b 2,-\dfrac 1 2\sqrt{4a^2-b^2},e\right)$

Now form the vector from each of these points to the front point. How fortunate the front point is $(0,0,0)$ so we can use use these positions as the vectors.

Find $\hat{n}=\vec{pleft} \times \vec{pright}$

This is the (unnormalized) normal vector.

let $\hat{n} = (n_x, n_y, n_z)$

To find the roll find the angle between the z axis and $(n_x, 0, n_z)$

To find the pitch find the angle between the z axis and $(0, n_y, n_z)$
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March 24th, 2017, 06:12 PM   #3
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Looks like an elegant solution. I'm rusty on vector math, can you explain...

To find the roll find the angle between the z axis and (nx,0,nz)

...further?

Thanks.
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March 24th, 2017, 06:30 PM   #4
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Quote:
Originally Posted by mdbennett View Post
Looks like an elegant solution. I'm rusty on vector math, can you explain...

To find the roll find the angle between the z axis and (nx,0,nz)

...further?

Thanks.
the z axis points along the vector $(0,0,1)$

the angle between 2 vectors is $roll = \arccos\left(\dfrac{\vec{v}\cdot \vec{w}}{\|v\|\|w\|}\right)$

so letting $\vec{v}=(nx, 0, nz),~w=(0,0,1)$ we have

$v \cdot w = nz$

$\| v \| = \sqrt{n_x^2 + n_z^2}$

$\|w\|=1$

so

$roll = \arccos\left(\dfrac{n_z}{\sqrt{n_x^2 + n_z^2}}\right)$

Depending on the way you are measuring roll this may end up being $\arcsin$ instead of $\arccos$

it should be apparent when you start playing with it.
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