- #1
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Anyone know a good web site, or two, that will tell me about working out the instantaneous centre of rotation; for links only I imagine.
TIA.
TIA.
Instantaneous centres of rotation
- Thread starter Fermat
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SUMMARY
The discussion focuses on the concept of instantaneous centres of rotation in mechanics, specifically how to calculate them using velocity and angular speed. A key formula mentioned is \(\vec{v}_{B}=\vec{v}_{A}+\vec{\omega}\times \vec{r}_{BA}\), which relates the velocities of points A and B. Users shared resources, including a website from the Czech Technical University, and referenced Kennedy's theorem for determining instantaneous centres. The conversation highlights the efficiency of resolving velocities directly compared to using instantaneous centres.
PREREQUISITES- Understanding of angular velocity and linear velocity
- Familiarity with vector mathematics
- Knowledge of Kennedy's theorem in kinematics
- Basic principles of mechanical linkages and mechanisms
- Research the application of Kennedy's theorem in various mechanical systems
- Explore advanced topics in kinematics, such as instantaneous centres of velocity
- Learn about different methods for analyzing mechanisms, including graphical and analytical techniques
- Investigate online resources and tutorials on calculating instantaneous centres of rotation
Mechanical engineers, students studying kinematics, and anyone involved in the analysis of mechanical systems will benefit from this discussion.
- #2
radou
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The first thing that came up on google [
]: http://www.fsid.cvut.cz/en/U2052/node27.html". Not much, depends on how deep you need to get into it.
By the way, if I recalled it correctly, if you know the speed [tex]\vec{v}_{A}[/tex] of a point A, and the angular speed [tex]\vec{\omega}[/tex], then you can find the speed of any point B with [tex]\vec{v}_{B}=\vec{v}_{A}+\vec{\omega}\times \vec{r}_{BA}[/tex], where [tex]\vec{r}_{BA}[/tex] is the vector from A to B. The condition on the centre of velocity is [tex]\vec{v}_{B} = \vec{v}_{c} = \vec{0}[/tex], so you can find its position.
]: http://www.fsid.cvut.cz/en/U2052/node27.html". Not much, depends on how deep you need to get into it.By the way, if I recalled it correctly, if you know the speed [tex]\vec{v}_{A}[/tex] of a point A, and the angular speed [tex]\vec{\omega}[/tex], then you can find the speed of any point B with [tex]\vec{v}_{B}=\vec{v}_{A}+\vec{\omega}\times \vec{r}_{BA}[/tex], where [tex]\vec{r}_{BA}[/tex] is the vector from A to B. The condition on the centre of velocity is [tex]\vec{v}_{B} = \vec{v}_{c} = \vec{0}[/tex], so you can find its position.
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- #3
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Many thanks.
I found a site (eventually - it took me ages, even on google) that told me what I was supposed to do. I used Kennedy's theorem to find the IC's and worked things out from there.
It was faster the first way I did it though. I just resolved velocities at the link ends and worked my way through the mechanism. Instantaneous centres gave the same answer, but took longer
I found a site (eventually - it took me ages, even on google) that told me what I was supposed to do. I used Kennedy's theorem to find the IC's and worked things out from there.
It was faster the first way I did it though. I just resolved velocities at the link ends and worked my way through the mechanism. Instantaneous centres gave the same answer, but took longer
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