Position of a point as a function of angular position

[Andrea Vironda]

Hi,
I have this scheme, in which there are 3 segments:
- I is coaxial to c axis and free to rotate in the origin. Length d1
- II is coaxial with a axis and free to rotate around c axis. There a fixed angle θ between a and c axis. Length d2
- III is welded to II, it's the PM segment. α is fixed between a axis and segment III. Length d2

Both a and c axis are free rotate of about 10" per step, so they have not continuous movement. About 130.000 possible position for a 360 deg rotation.
I'd like to find a relation between position of point M in function of angles: so $M(x,y,z)=f(\alpha, \gamma)$ where $\alpha = a_{axis}$ angular position and $\gamma = c_{axis}$ angular position, $0 \leq \gamma, \alpha \lt 360 °$

How could i proceed?
I can easily achieve the result fixing c axis and only rotating a axis, but i don't know how to combine them together.
For example, if a axis is fixed i will obtain $x^2 + z^2 = (d_3+d_2 \cos\theta)^2$

 

[anuttarasammyak]

Hi. Laying intermediate $C_2$ rod on X axis and $d_1$ rod on XY plane would be helpful. See attached figure.

 

[Andrea Vironda]

Hi,
I noticed I used $\alpha$ twice. It's the angle of rotation of d2 and d3 around a axis. It's not the angle between d2 and d3 (that's a constant angle). The 2 degree of freedom are the rotations around a and c axis.

 

[anuttarasammyak]

Hi. I am not sure I catch your setting, so I would say about general case.

Rod 1 with one end is fixed with Origin by free joint:
The (x,y,z) coordinate of the other end, using the polar coordinate :
(r_1 cos\theta_1, r_1\sin\theta_1\cos\phi_1,r_1\sin\theta_1\cos\phi_1)=\mathbf{r_1}

Rod 2 with one end is fixed with free end of rod1 by free joint:
The coordinate of the other end:
(r_1 cos\theta_1, r_1\sin\theta_1\cos\phi_1,r_1\sin\theta_1\sin\phi_1)+<br /> (r_2 cos\theta_2, r_2\sin\theta_2\cos\phi_2,r_2\sin\theta_2\sin\phi_2) =\mathbf{r_1}+\mathbf{r_2}

rod 3 with one end is fixed with free end of rod2 by free joint:
The coordinate of the other end:
(r_1 cos\theta_1, r_1\sin\theta_1\cos\phi_1,r_1\sin\theta_1\sin\phi_1)+<br /> (r_2 cos\theta_2, r_2\sin\theta_2\cos\phi_2,r_2\sin\theta_2\sin\phi_2)<br /> +(r_3 cos\theta_3, r_3\sin\theta_3\cos\phi_3,r_3\sin\theta_3\sin\phi_3) =\mathbf{r_1}+\mathbf{r_2}+\mathbf{r_3}

Constraints of the angle between rod1 and rod2,
\mathbf{r_1}\cdot\mathbf{r_2}=r_1 r_2 \ cos \alpha_{12}
\cos\theta_1\cos\theta_2+sin\theta_1\sin\theta_2(cos\phi_1cos\phi_2+\sin\phi_1\sin\phi_2)=\cos\alpha_{12}
\cos\theta_1\cos\theta_2+sin\theta_1\sin\theta_2 cos(\phi_2-\phi_1)=\cos\alpha_{23}
,rod 2 and rod3
\mathbf{r_2}\cdot\mathbf{r_3}=r_2 r_3 \ cos \alpha_{23}
\cos\theta_2\cos\theta_3+sin\theta_2\sin\theta_3(cos\phi_2cos\phi_3+\sin\phi_2\sin\phi_3)=\cos\alpha_{23}
\cos\theta_2\cos\theta_3+sin\theta_2\sin\theta_3 cos(\phi_3-\phi_2)=\cos\alpha_{23}

 

[Andrea Vironda]

Hi,
sorry but I can't understand how you set up angles, etc..
I've found a better image. How to outline the possible movements on an excel file?

 

[anuttarasammyak]

@Andrea Vironda Thanks for your sketch. Now I think I understand the problem. I consider the process of rotation a-a by $\alpha$ : M to M' then rotation c-c by $\gamma$ : M' to M" as attached.

 

[Andrea Vironda]

Consider that d1 and d3 are parallel at $\alpha, \beta = 0$. In my attachment there's the correct position in M, but when i have $\alpha = 90$ I can't find the z position.
I can't imagine in my mind how much d3 is raised by $\alpha$ rotation

 

[anuttarasammyak]

Hi. From your sketch attached, I see the problem says $\phi_2=\phi_1=\phi$ in my previous post.M[\alpha=0,\gamma=0]=(d_1+d_2 \cos\phi+d_3 cos 2\phi,\ d_2 sin\phi+d_3 sin 2\phi,\ 0)

As an initial position I take upper-full bent position and you take downer-parallel position, but they show no difference in essence. I will follow your way so as you said

M[\alpha=0,\gamma=0]=(d_1+d_2 \cos\phi+d_3, \ -d_2 sin\phi,\ 0)

Rotation with a-a axis by $\alpha$. M flies up above XY plane and draw a circle of radius $d_3 sin\phi$. P and the circle make a cone. Height of M (Z) is $d_3 \ sin\phi \ \sin\alpha$. You should put plus minus signs in front depending on your definition of $\alpha$ turning direction.
You observe M draws a "standing" circle so X and Y coordinates also change with $\alpha$. Careful consideration is needed as written in my sketch.