Position of a point as a function of angular position

In summary, the conversation discusses a scheme with three segments, each with different degrees of freedom and lengths. The goal is to find a relation between the position of point M in terms of the angles of rotation. The conversation also discusses how to outline the possible movements on an excel file. The conversation also includes a sketch outlining the problem and its different components. The final part of the conversation involves calculating the position of M for different angles of rotation and considering the changes in the X and Y coordinates.
  • #1
Andrea Vironda
69
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##
 

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  • #2
Hi. Laying intermediate ##C_2## rod on X axis and ##d_1## rod on XY plane would be helpful. See attached figure.
 

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  • #3
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.
 
  • #4
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)+
(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)+
(r_2 cos\theta_2, r_2\sin\theta_2\cos\phi_2,r_2\sin\theta_2\sin\phi_2)
+(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]
 
  • #5
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
 
  • #6
@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.
 

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  • #7
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
 

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  • #8
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.
 
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FAQ: Position of a point as a function of angular position

What is the position of a point as a function of angular position?

The position of a point as a function of angular position refers to the location of a point on a circle or curve, measured in terms of the angle formed between a fixed reference point and the point of interest. This position can be represented by coordinates or polar coordinates.

How is angular position measured?

Angular position is typically measured in radians, which is a unit of measurement for angles. One radian is equal to the angle formed when the arc length of a circle is equal to the radius of the circle. Alternatively, angular position can also be measured in degrees, with one full rotation around a circle equal to 360 degrees.

What is the relationship between angular position and linear position?

Angular position and linear position are related through the concept of arc length. The linear position of a point on a circle or curve can be determined by multiplying the radius of the circle by the angle formed between the reference point and the point of interest. This relationship is known as the arc length formula: s = rθ, where s is the linear position, r is the radius, and θ is the angular position in radians.

How does angular position change over time?

Angular position can change over time if the point is moving along a circular or curved path. In this case, the angular position will increase or decrease depending on the direction of motion. This change in angular position can be described by the angular velocity, which is the rate of change of angular position over time.

What is the significance of angular position in physics and engineering?

Angular position is a crucial concept in physics and engineering as it helps describe the motion of objects along circular or curved paths. It is used in various fields such as mechanics, robotics, and astronomy to understand and analyze the behavior of moving objects. Additionally, many physical quantities such as torque, angular momentum, and angular acceleration are defined in terms of angular position.

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