For my GCSE Engineering project, I am building a device which attaches to the frame of a bike, and uses a magnet (Attached to the wheel) to detect 1 turn of the wheel (I have also compensated for contact bounce etc.) The only problem I face is that I have never used reed switches before, and I don't know what sensitivity of reed switch to use.

Assuming the bike wheel is 2150mm circumference, the max speed is 13 m/s, the magnet diameter is 50mm and it will be 180mm from the centre of the wheel, I calculated that (hopefully) the smallest amount of time the switch will be directly passing the magnet is 0.0037 seconds.

I thought this seemed like a very small amount of time for a switch to close and open again before 1 rotation is over. A 1AT switch apparently closes at the north pole from the earth itself, so just to be safe i want to avoid disaster from it being too sensitive. Does anyone know what type/sensitivity of reed switch to use?

  • 1
    \$\begingroup\$ I think you should consider an optical encoder. Less susceptible to vibration. \$\endgroup\$ – vicatcu Sep 18 '19 at 16:15
  • 2
    \$\begingroup\$ Use a hall effect sensor \$\endgroup\$ – Voltage Spike Sep 18 '19 at 16:22
  • \$\begingroup\$ Reed switches is used in some commercial bicycle computers. But I agree with @VoltageSpike suggestion about a hall sensor, it's not mechanical! \$\endgroup\$ – MatsK Sep 18 '19 at 16:52

Somewhere around 25 AT will be more than enough to prevent accidental triggering from the earth's field, although you could go higher depending on your magnet and the proximity to the switch.

The trick is to get the alignment right. You want to align the magnet and reed switch so that the reed switch is magnetized along its length. See https://standexelectronics.com/wp-content/uploads/Reed-Switch-And-Magnet-Interaction.pdf .

If your magnet is 180 mm from the center (360 mm diameter), the magnet will pass in .0073 seconds, and a reed switch should be fine. They have operation times on the order of .0005 seconds.

Good luck!


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.