dencorso, on 23 September 2010 - 11:56 PM, said:
2) Highlight partition 2, go to the ending head select it and type-in 255, in place of the current 254, then accept, save select format latter, and see whether it becomes white.
For the record (and FYI), on a
nx255x63 geometry:
- Last addressable Cylinder is n-1
- Last addressable head is 254
- Last addressable sector is 63
Cylinders are numbered strating from 0.
Heads are numbered starting from 0.
Sectors are numbered starting from 1.
The general idea was:
- do "logical" tests
- report
- loop to #1 until a suitable solution is found
NOT "change a value with a random (BTW wrong) one and see what happens.
The only thing we need now is a method, and being calm.
There may be a few tens reasons why RPM makes the partition "red", incuding rightful and wrongful ones.
This:
piikea, on 23 September 2010 - 02:45 PM, said:
Well dencorso mentioned 0 - 485519 in post #80. Besides it shows up as that when you do use 485520. Perhaps I should be putting a "1" in the "starting cylinder" box? Also didn't add "c" to the numbers - maybe that made some difference.
...
I put 485520 & I'm pretty sure it displayed as 0 - 485519.
is perfectly normal.
CHS sectors are numbered starting from 1.
LBA sectors are numbered from 0 (actually the meaning of a LBA values is "Sectors before" or "offset of the sector")
Now the issue at hand is the following:
first partition (notwithstanding the fact that RPM sees it in RED) seems OK:
- it starts at a correct CHS address (0/1/1)
- it ends at a correct CHS address (30,463/254/63)
- it's CHS values are correspondent to LBA ones:
(0+1)x(1+1)x63=126-63=63
(30,463+1)x(254+1)x63=489,404,160-63=489,404,097
following partitions don't.
We have now an added info.
The 1938020 end cylinder that was "needed".
However let's try again.
The original data you should have input was:
jaclaz, on 22 September 2010 - 04:46 AM, said:
So, now you have to use the "c" instead of the "m" and:
- Primary FAT32 LBA 485520c
- Logical FAT32 LBA 485520c
- Logical FAT32 LBA 481440c
- Logical FAT32 LBA 481440c
Then, I asked you to input the SAME values, only using ALL primaries.
What seemingly you did input is:
- Primary FAT32 LBA 485520c
- Logical FAT32 LBA 485521c
- Logical FAT32 LBA 485521c
or at least this is what partition logic wrote to the partition table.
Please find attached a small Excel spreadsheet.
It consists of TWO sheets "Manual" and "Automatic".
BOTH sheets are protected.
In sheet "Manual" you can ONLY change:
- cells I10÷I13 i.e. the End Cylinder in mx255x63 (RPM)
- cells I24÷I27 i.e. the End Cylinder in nx16x63 (Partition Logic)
The two sets of partition entries are independent, to allow playing a bit with values.
In sheet "Automatic" you can ONLY change:
- cells U24÷U27 i.e. the "c" value you can input in Partition Logic
this automatically calculates the data how it should be written by Partition Logic and how it should be seen in RPM.
If ANY of the cells in column J (end head) becomes
RED it means that the partition dos not end on cylinder boundary. (NOT "kosher").
Now, try some values in U24÷U27, remember that ANY value should be a multiple of
4080!
Start with small values, creating 4 Primary partitions like:
Input these values in partition logic, then check how RPM reads them.
Report.
jaclaz
This post has been edited by jaclaz: 24 September 2010 - 05:05 AM