Dear Roger,

I am puzzled when I read the section"Data Storage Capacity-Readings" on page 78 of DT800 user's manual.

It says when the schedule contains one channel, the readings per MB of DT800 storage is about 14500, while 2, the readings is 25800. Could I induce that DT800 storage capacity is improved while the number of channels per schedule increase? If so, Why?

Could you do me a favor to explain my puzzle?

Thank you!

Good afternoon kangle2miao,

Overhead.

Let me explain each line of data has a header then the data and a tail. The length of the header and the tail are fixed in length.

This is the first line of data from the example in the DT800 manual (Page 22) D,080015,"Zz",2000/05/29,10:08:24,0.007411,1;F,0,13.94;0053;60CC

The header is D,080015,"Zz",2000/05/29,10:08:24,0.007411,1;F,0, and is 49 characters long. Then there is the data 13.94 In this case 1 reading

And the tail is ;0053;60CC The first number is the line length and the seconds is a 16 bit CRC. The entire line is 64 characters long with only 5 being actual data.

As the number of channels increase the ration of overhead to actual data reduces. Hence the more channels of data the more efficient it is in storage.

Cheers,
Roger

Oh, I understand now. Thank you very much!!

Regards!