update pid loop

[MichaelMohn]

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[MichaelMohn]

Change output to uint8_t if it only returns values from 0-8. I'm pretty sure your pid code works but instead of doing (-ErrorSum)/INTEGRAL*1000 do (-ErrorSum)*1000/INTEGRAL. The otherway kind of gets rid of precison I'm p sure.

Nah I might not understand this part of C perfectly... but at the moment ErrorSum is capped at 500,000 and integral is 250. So the max output is 2. If I place the 1000 in the numerator I would get a massively incorrect value right?

[ray4477]

Change output to uint8_t if it only returns values from 0-8. I'm pretty sure your pid code works but instead of doing (-ErrorSum)/INTEGRAL*1000 do (-ErrorSum)*1000/INTEGRAL. The otherway kind of gets rid of precison I'm p sure.

Nah I might not understand this part of C perfectly... but at the moment ErrorSum is capped at 500,000 and integral is 250. So the max output is 2. If I place the 1000 in the numerator I would get a massively incorrect value right?

It should be equivalent mathematical statements. The problem with the before is if ErrorSum is say less than 250 than if it is divided by INTEGRAL (250) it will become zero because int division rounds down. If you do the other way then it will return fine. Here is an example below

[MichaelMohn]

Change output to uint8_t if it only returns values from 0-8. I'm pretty sure your pid code works but instead of doing (-ErrorSum)/INTEGRAL*1000 do (-ErrorSum)*1000/INTEGRAL. The otherway kind of gets rid of precison I'm p sure.

Nah I might not understand this part of C perfectly... but at the moment ErrorSum is capped at 500,000 and integral is 250. So the max output is 2. If I place the 1000 in the numerator I would get a massively incorrect value right?

It should be equivalent mathematical statements. The problem with the before is if ErrorSum is say less than 250 than if it is divided by INTEGRAL (250) it will become zero because int division rounds down. If you do the other way then it will return fine. Here is an example below

This has been fixed I believe.