# Wheel encoder Tick calculation

Sorry for being a newbie. Can somebody explain me how many ticks a wheel encoder would deliver (based on one complete wheel rotation), if I have a slice with 500 windows? Is the assumption correct, that 1 tick is counted per window?

Thanks

• If one of the answers worked for you, don't forget to click the check mark. May 29, 2021 at 22:50

That depends on what type of digital encoder you have. A full quadrature encoder has 2 light (or magnetic) sensors detecting up and down pulses. A single light source encoder can only detect speed, but not velocity (speed and direction). A full quadrature encoder counts 2 up ticks per window and 2 down ticks for the same window. This is also how direction is calculated - which of the 2 sensors ticks up/down first determines which direction the encoder is rotating. This means there are 4 ticks for 1 window on a full-quadrature encoder.

• Oh sry, I missed the information "quadrature encoder". So it should be four times the number of windows, which means in this example 4*500=2000 ? Thanks for your answer.
– Alex
May 28, 2021 at 15:04
• If your encoder is a quadrature encoder, yes, you would have 4*500=2000 tick counts for 500 windows. I wasn't sure you understood what kind of sensor you had given the wording of the question, hence my answer. If the answer helped you, that green check mark would be appreciated. May 28, 2021 at 18:09

This depends on your hardware and the software stack you are using. A digital pulse will have two edges; rising and falling. A simple slotted optical interrupter sensor has a single LED emitter/detector pair that can generates the signal (and so the rising/falling edges). It does this as a disk passes by it that has slots in it. So each slot will generate a rising and falling edge. So the number of edges is 2 * number of slots.

A quadrature encoder has two sensors. The sensors my be optical, like the example above, in which case there is a disk with slots. The sensors my be hall effect sensors, in which case the 'disk' has magnets that cause the hall effect sensors to generate rising and falling edges. So if we think of those magnets as the same as optical disk 'slots' then we can say that for a quadrature encoder that the number of edges is 4 * number of slots per rotation.

So the number of signal edges is determined by the number of sensors in the encoder and the number of 'slots' in the 'disk'. However, if you are using a software stack to read the encoder, it may impose it's own semantics on 'ticks'; for instance, it may only count rising edges, or it may count a complete transition between rising/falling for all sensors as a single 'tick'. So you should consult the documentation to determine what it is reporting. Also look at your encoder specification of PPR, pulses per rotation.

See DroneBot Workshop for a good explanation of encoders.