The discussion centers on calculating the velocities required to generate an Archimedean spiral trajectory. The key takeaway is that the angular velocity (ω) is chosen as the input, while the linear velocity approaches ωr as the radius increases. The velocity vector is expressed as v = (dr/dt) a_r + (r dθ/dt) a_θ, simplifying to v ≈ (r ω) a_θ for large r. This formulation is critical for accurately modeling the spiral's motion.
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Thaks for the reply, It was really interesting but what about the linear velocities on x, y and z and the angular velocities on x y and z.Charles Link said:The following Insights article that I authored might be of interest to you. https://www.physicsforums.com/insights/frenet-equations-2-d-result-cornu-spiral/
For the OP @Benyoucef Rayane A google shows the Archimedes spiral has ## r=\theta^a ## with ## a=1 ##. This means ## r=\theta ## for this spiral. We can write the velocity ## \vec{v}=(\frac{dr}{dt}) \hat{a}_r+(r \dot{\theta}) \hat{a}_{\theta} ##. We have for ## r=\theta ##, that ## \frac{dr}{dt}=\dot{\theta}=\omega ##. This gives ## \vec{v}=\omega \hat{a}_r+(r \omega) \hat{a}_{\theta} ##. As ## r ## gets large, ## \vec{v} \approx (r \omega ) \hat{a}_{\theta} ## as @rumborak pointed out.rumborak said:Since the usual definition of the Archimedes Spiral takes the angle as the input, it means you choose the angular velocity ω. Regarding linear velocity, I don't have an exact formula at hand, but since it gets closer to a circle with increasing angle, the linear velocity will asymptomatically approach ωr.
Thanks guys, I got it now.Charles Link said:For the OP @Benyoucef Rayane A google shows the Archimedes spiral has ## r=\theta^a ## with ## a=1 ##. This means ## r=\theta ## for this spiral. We can write the velocity ## \vec{v}=(\frac{dr}{dt}) \hat{a}_r+(r \dot{\theta}) \hat{a}_{\theta} ##. We have for ## r=\theta ##, that ## \frac{dr}{dt}=\dot{\theta}=\omega ##. This gives ## \vec{v}=\omega \hat{a}_r+(r \omega) \hat{a}_{\theta} ##. As ## r ## gets large, ## \vec{v} \approx (r \omega ) \hat{a}_{\theta} ## as @rumborak pointed out.