Position of a point as a function of angular position

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
7 replies · 2K views
Andrea Vironda
Messages
69
Reaction score
3
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##
 

Attachments

  • IMG_20201204_205730.jpg
    IMG_20201204_205730.jpg
    31.5 KB · Views: 240
Last edited:
Physics news on Phys.org
Hi. Laying intermediate ##C_2## rod on X axis and ##d_1## rod on XY plane would be helpful. See attached figure.
 

Attachments

  • img20201206_08145602.jpg
    img20201206_08145602.jpg
    77 KB · Views: 225
Last edited:
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.
 
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 :
[tex](r_1 cos\theta_1, r_1\sin\theta_1\cos\phi_1,r_1\sin\theta_1\cos\phi_1)=\mathbf{r_1}[/tex]

Rod 2 with one end is fixed with free end of rod1 by free joint:
The coordinate of the other end:
[tex](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}[/tex]

rod 3 with one end is fixed with free end of rod2 by free joint:
The coordinate of the other end:
[tex](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}[/tex]

Constraints of the angle between rod1 and rod2,
[tex]\mathbf{r_1}\cdot\mathbf{r_2}=r_1 r_2 \ cos \alpha_{12}[/tex]
[tex]\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}[/tex]
[tex]\cos\theta_1\cos\theta_2+sin\theta_1\sin\theta_2 cos(\phi_2-\phi_1)=\cos\alpha_{23}[/tex]
,rod 2 and rod3
[tex]\mathbf{r_2}\cdot\mathbf{r_3}=r_2 r_3 \ cos \alpha_{23}[/tex]
[tex]\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}[/tex]
[tex]\cos\theta_2\cos\theta_3+sin\theta_2\sin\theta_3 cos(\phi_3-\phi_2)=\cos\alpha_{23}[/tex]
 
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?
1607339423281.png
 
@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.
 

Attachments

  • img20201209_15223462.jpg
    img20201209_15223462.jpg
    77 KB · Views: 187
Last edited:
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
 

Attachments

  • IMG_20201209_213352.jpg
    IMG_20201209_213352.jpg
    40.5 KB · Views: 194
Hi. From your sketch attached, I see the problem says ##\phi_2=\phi_1=\phi## in my previous post.[tex]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)[/tex]

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

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

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.
 
Last edited: