And is correct. For RPM measurement a count of 6 is still required. Here's my math for this. Perhaps you can explain what position interpolated means.
So we have a spindle turning 6000 RPM or 50 RPS. That means if we didn't have the longer slot we'd see an edge every 0.0033333333333333333333 seconds. Now assume our counter (parallel port for example) is running 20kHz. That's a tick every 50 uS so divide that into 0.00333333 seconds and you get 66.6666666666 counts before the next hole on the disk. In reality you would see 66 counts and then 67 counts and then 66 counts etc. and then 133.333333 counts for the index slot. Anyway, assume we test and can tell the difference between the slot and the holes we get 66. Multiply that by 0.000050 (50uS) and we get the time for one hole edge to hole edge which is 0.0033 seconds or inverted 303. Scale it by 10 to get RPM which is 3030 RPM; out by 30 RPM. Next time we count 67 which works out to 2985 RPM; out by 15 RPM. Or we could accumulate 5 readings of 66,67,66,67,66 which gives us 66.4 a bit closer at 3012 RPM but really not the 3000 we're actually turning. And if it's the other way around 67,66,67,66,67 it's 66.6 average and is 3003 RPM. Closer. Increasing to 60 holes improves things but there will still be a certain amount of jitter since we have repeating decimals in the math. With 20kHz we're still able to count 5 or 6 ticks but above that encoder count the 20kHz tick clock has to be much faster. Anyway, that's how I'd calculate RPM. What's the position interpolated. And is there a better way to get more accurate RPM? John > -----Original Message----- > From: Sam Sokolik [mailto:samco...@gmail.com] > Sent: September-19-22 10:20 AM > To: Enhanced Machine Controller (EMC) > Subject: Re: [Emc-users] missing tooth index questions.. > > That was my plan - use position-interpolated.. 100 counts is barely enough > for smooth motion. > > sam > > On Mon, Sep 19, 2022 at 11:51 AM andy pugh <bodge...@gmail.com> wrote: > > > On Mon, 19 Sept 2022 at 17:06, John Dammeyer <jo...@autoartisans.com> > > wrote: > > > > > No. > > > A slot is still a hole. So if you are counting, say falling edges, you > > > will get 5 of them. If you count rising edges you will get 5 of them. > > > > > > Actually, the missing-tooth scheme takes a different opinion. Each hole > > corresponds to 1/6 of a revolution, so the correct scale is 6. But it only > > responds to actual edges seen, so does not create a synthetic count in the > > middle of the gap, instead it recognises the approximately double period > > and adds two (or more) counts when it sees the gap. > > > > Using position-interpolated will smooth through this jitter. > > > > -- > > atp > > "A motorcycle is a bicycle with a pandemonium attachment and is designed > > for the especial use of mechanical geniuses, daredevils and lunatics." > > � George Fitch, Atlanta Constitution Newspaper, 1912 > > > > _______________________________________________ > > Emc-users mailing list > > Emc-users@lists.sourceforge.net > > https://lists.sourceforge.net/lists/listinfo/emc-users > > > > _______________________________________________ > Emc-users mailing list > Emc-users@lists.sourceforge.net > https://lists.sourceforge.net/lists/listinfo/emc-users _______________________________________________ Emc-users mailing list Emc-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/emc-users
