Very Computer

To calculate the storage requirements for a text/image column the BOL
states:

The text chains that store text and image data have 112 bytes of overhead
per page.

Use the following formula to calculate the number of text chain pages that a
particular entry will use:

Data length / 1800 = Number of 2K pages

The result should be rounded up in all cases; for example, a data length of
1800 bytes requires two 2K pages.

Considering a page is 2048 bytes where do the other 248 bytes (or 136 bytes)
go?

Richard.
Polymorphic.
UK.

Richard,

It has to do with the way text/image data is stored.  The "text" data type
is a linked list of 2k pages.  It would appear to me that you would have a
lot of wasted space (1664 bytes).  You would need two 2k pages for each
record with the second page being almost empty.

Grant

Richard,

I was confused with the statement:

Quote:> Data length / 1800 = Number of 2K pages
>The result should be rounded up in all cases; for example, a data
>length of 1800 bytes requires two 2K pages.

You round up if it is over 1800.  For example if you had a record of 1900
bytes you would have:

1900/1800 = (1.06) two 2k pages wasting almost the whole page.

I am not sure what the 136 bytes is used for.  Maybe pointers for the linked
list.  Sorry for my confusion!

Grant