0
\$\begingroup\$

So I came across a Persistence of Vision globe based video where a YouTuber he makes DIY slip rings to power and control some LEDs as seen in this picture: He soldered the wires inside those rings

The contacts from a microcontroller, touching the rotating rings using a binder clip handle

End product

My question is, given the high rate of data transfer needed to turn on and off the right leds at the right time, would this set up have an effect of the project? If an LED has 400Hz refresh rate, will this set up manage to get that happening?

\$\endgroup\$
7
  • \$\begingroup\$ are you asking if the last picture is fake? \$\endgroup\$ – jsotola Mar 9 '20 at 7:01
  • \$\begingroup\$ Not at all, I was just questioning the limits of such a setup. The image shown is relatively simple compared to something like a moving graphic or a video (with very poor resolution). I was planning to make my own and wondered if buying a slip ring would prevent obstacles like that. \$\endgroup\$ – Jake Mar 9 '20 at 7:04
  • \$\begingroup\$ the device does not have to spin very fast ... probably less than 10 revolutions per second ... you can buy a POV display for a bicycle wheel ... a bicycle wheel spins at about 3 revolutions per second at 30 km/h \$\endgroup\$ – jsotola Mar 9 '20 at 7:11
  • 2
    \$\begingroup\$ You can place the controller in the rotating part. Only power and maybe a single (slow) data line is going via the slip rings. \$\endgroup\$ – filo Mar 9 '20 at 7:14
  • \$\begingroup\$ that looks like steel to me, not aluminium. \$\endgroup\$ – Jasen Mar 9 '20 at 8:30
1
\$\begingroup\$

You can send video frequencies through a slipring without trouble. Maybe more if you use LVDS and the arrangement is symmetrical enough. You can get commercial sliprings for a reasonable price, from China for example.

However the operating lifetime might only be a few hundred hours for that sort of thing.

I believe the heads in VCRs used a rotary transformer for the signals and did not use sliprings at all.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.