I have FLEXY 205 without extra SD storage. if there are 20 REAL, 20 BOOLEN, 20 INT tags, if they log every minute, how many days of history logging it should save up to?
thanks, just to have a roughly idea. currently I see the history logging only up to 3 days. is it correct?
I think that this topic might be a good reference for you:
hi Tim, I read them. kind of understand the concept, but don’t know how to estimate my logging data size. for test purpose, I will remove some tags, give longer cycle time. so in concept, the logging period should be extended. right?
The less often that you are polling your data the longer the logging period should be. You can also change your logging method from a time based one to a deadband one that will only record the value if it changes by a certain threshold. If you want to keep it time based you could take a look at the internal memory information. You should then be able to see how much memory is currently being stored and then you could log for 10 minutes or so and calculate roughly how long it would be before the memory is filled
for a test, I removed all tags just to have 4 tags left. 1 BOOL, 1 REAL, 1 INT and 1 DINT, history cycle time is 1 min. I let the PLC run for more than 10 days, please see my paste for the internal memory info. I don’t know how to calculate the memory size. as of today June/24, I can see the logging is back to June/11.
could you help me to know if it reaches max storage, or the logging volume will still get extended more. how much more?
thanks in advance.
see my screen shots:
Missing Historical Values
Hi Charles I found this doc that might be useful
It looks like if you have 4 tags that you’re polling at once per minute you’ll get about 24 days if you have your memory size set to 139264
(139264 points / 4 tags) * 60 seconds = 2,088,960 seconds.
if there’s 86400 seconds in a day, then that leaves you with about 24.17 days worth of data
thanks Tim, your reply is very help. I am doing some math.
here is my memo config in that flexy205:
my flexy model is 205, so its recording size is sitting on 6MB or 16384 points.
in case of 16384/4 tags * 60s = 245760 second ===>> 2.84 days;
in case of 73728/4 tags * 60s = 1105920 second ==>> 12.8 days (this is close to my actual).
am I doing right?
also if you use 3 recording memory size options(6MB, 14MB or 19MB) divided by their 3 different points (16384, 73728, 139264), the actual size (Byte) per point is not the same.
sorry for asking too much, but I really want to understand how the recording memory size works, so I can estimate my logging design.
Memory Allocation for Logging
I was just checking this with some colleagues in Belgium and it seems the doc I sent was a little bit dated and was based off of the CD devices. For all devices you can calculate points this way: Recording size * 1024 * 1024 / 20
|Recording Size (MB)||Points|
So for this one, by default you’ll be using the 6MB. This will give you the (314572 Points / 4 Tags) * 60 seconds * 1 day / 86400 seconds = about 54.61 days
great, all clear:
6MB x 1024 KB x 1024 Byte = 6291456/20 byte per point = 314572 points
thank you so much
in case if we need to extend the record storage, we need to insert the SD card. will its size all be counted as record storage? for example in case of a 16GB SD, that will give us a lot of points if 20 byte per points.
could you send me couple links of topics of how to use SD card to extend the storage and how to set it up?
I think this ticket should be a good reference for that