jaclaz

Good. There is NO 'chainloader' command involved and definitely not a MBR of another partition. That is an entry in BOOT.INI. It doesn't exist such a thing as "the MBR of another partition", it may exist the "PBR of another partition" (i.e. "bootsector of another partition") or "the MBR of another disk" (that I doubt will work). It seems like BOOTMGR will read a BOOT.INI contents and add it's entry to the BCD entries (or you can add manually the "legacy" entry to the BCD): http://www.boot-land.net/forums/index.php?showtopic=12557&st=7 What is not clear is whether it needs a NTLDR also in order to boot an entry (exception made for "arcpath ones"). The safer/easiest would be to add an entry in BCD for grub4dos' grldr.mbr and add it, together with grldr and menu.lst to the drive. Once in grub4dos you can boot any Linux, using the old (and nowadays deprecated) chainloading of bootsector method or the more elegant and less prone to error direct loading of the kernel and initrd. Here: http://diddy.boot-land.net/grub4dos/Grub4dos.htm http://diddy.boot-land.net/grub4dos/files/install_windows.htm#windows3 http://www.boot-land.net/forums/index.php?showforum=66 jaclaz

Here there is no problem with numbers, it must be still some form of communication problem. You DO NOT want to change the (head) 254 to 255 You DO NOT want to change the (head) 254 to 256 You want to change the (cylinder) 4079 to 255, etc. The first line, that is now: 0/1/1/4079/254/63 should become: 0/1/1/255/254/63 Second line, that is now: 4080/0/1/8159/254/63 should become: 256/0/1/511/254/63 Third line, that is now: 8160/0/1/12239/254/63 should become: 512/0/1/767/254/63 Fourth line, that is now: 12240/0/1/16319/254/63 should become: 768/0/1/1023/254/63 I have no other way to explain it. You DO NOT change Heads. You DO NOT change Sectors. You ONLY change Cylinders. You already have (for the first time) all values "white" (which is good ), BUT currently you created in Partition Logic 4 (four) partitions around 2 Gb (4080x16x63x512) which resulted in RPM in 4 (four) partitions around 32 Gb (4080x255x63x512). Since you are experimenting with partition sizes around 2 Gb, you want to obtain them in RPM too, before attempting using bigger sizes. You can simulate the behaviour using the spreadsheet "Automatic", you try the sizes you want, by inputting them in cells U24:U27, in the lower half you see how Partition Logic will deal with them, in the upper part you see the corresponding Cylinder (and NOT Head, and NOT Sector) values you have to change in RPM. You don't need to be good with numbers, nor even to understand the mathematics arithmetics behind the set of instructions, you simply have to follow a simple set of instructions EXACTLY. If you don't understand this simple set of instructions, say so, try to communicate which parts you don't understand, and I may be able to further explain those parts in more detail. jaclaz

Lenovo's are tricky. They often use a non-standard geometry in BIOS, that can lead, EXPECIALLY if using the HP USB utility or petousb to NOT booting. Typically a nx240x63 geometry instead of the "normal" mx255x63 First thing to try is to use RMPREPUSB to do the partitioning/formatting of the stick. http://www.boot-land.net/forums/index.php?showtopic=7739 Next, once and if the above failed, you should make a copy of the MBR (use HDhacker to copy first sector of Physicaldrive), compress it in a .zip archive and post it. http://dimio.altervista.org/eng/ You should try simply making the USB stick bootable with RMPREPUSB and try booting from it, and post EXACTLY what happens if it fails booting, like: a completely blank (black) screen a flashing cursor in top left a "j"or g" an error of some kind (post the EXACT error message) jaclaz

Let's make a deal. If you post how you use the 'chainloader' command in XP to pass control to the mbr of another partition, I'll show you how to do it under Vista :ph34r:/2008/7. jaclaz

WHAT kind of question is this? OF COURSE they are THERE.(with that method) OF COURSE they are NOT HERE(in the current version of WinsetupfromUSBwithGUI) jaclaz

That's exactly the reason for the small spreadsheet. That is to simulate the behaviour and let you fine tune the sizes. It's not that difficult, you enter numbers in the "Automatic" sheet until you have filled the disk as you like (making sure that values are multiple of 4080). You see how the result should look in Partition Logic. Then you enter those values in Partition Logic. Then you run RPM and change the cylinders number as seen in the top part of the spreadsheet. Do the test with the base "4080/4080/4080/4080" set. The values you should change in RPM are: 4079->255 4080->256 8159->511 8160->512 12239->767 12240->768 16319->1023 jaclaz

I'll try again. In the "top" line you always enter the MINIMUM value supplied to you by Partition Logic. In the "bottom" line you always enter 4080c (i.e. the size of the partition) that will be: top-bottom 0-4080c 4080-4080c 8160-4080c 12240-4080c OR you enter in the Top the Minimum value supplied to you by Partition Logic and in the bottom the progressive End sector: top-bottom 0-4079 4080-8159 8160-12239 12240-16319 In other words, since cylinders are numbered from 0: (4079+1)-0=4080c (8159+1)-4080=4080c (12239+1)-8160=4080c (16319+1)-12240=4080c I know it is not easy to make calculation with 0 based things, as we are used to use 1 based objects in everyday life, but this is how it works with CHS, think about this: you cannot have anything less than 0/0/1 (first sector of first head on first cylinder, i.e. the MBR, i.e. sector 0 LBA). You cannot have more than the max number of heads (let's say 255). Last head (since they are numbered from 0) is then 254. You cannot have more than the max number of cylinders (let's say you have a small hard disk with just 1 cylinder). This small hard disk will start at 0/0/1 and end at 0/254/63. Since we know that it is exactly 1 cylinder in size, you have to use the formula: (0+1)-0=1 to calculate it's size in cylinders. Get SPREAD32 to try using the spreadsheet: http://www.byedesign.co.uk/ jaclaz

If it's not a trade secret , can you name it (including it's EXACT version)? And also post some details of the actual problem you are having? It is possible that there is a "specific" solution for the "specific" problem you have (completely UNconnected with an "universal" uninstaller for Win9x programs). Or, if you prefer, it is possible that you are "slipping" on a chocolate covered banana : http://homepages.tesco.net/J.deBoynePollard/FGA/put-down-the-chocolate-covered-banana.html jaclaz

YES. http://homepages.tesco.net/J.deBoynePollard/FGA/questions-with-yes-or-no-answers.html or NO? Can you explain WHAT you would like to do? jaclaz

No. There is no opening password, the protection is only about locked cells. You have to make up your mind, you can EITHER: open the spreadsheet OR: NOT open it if the latter you can't do ANYTHING with the "Manual" sheet in it (nor with the "Automatic" one, for that matters since you cannot open the spreadsheet at all). If you can open it, please read again my previous post, I guess I explained the idea: the "manual" is ONLY to "play" the "automatic" is to simulate what happens with varying partition sizes (number of Cylinders in a nx16x63 geometry) as seen BOTH by Partition Logic and RPM. You have only a very limited number of cells in which you can change the values, it shouldn't be difficult to get the hang of it. Simplified version: FORGET about the "Manual" sheet Open the "Automatic one" Input in cells U24÷U27 the value 4080. The "main part" of the sheet should represent the partitioning screen as "seen" in Partition Logic (bottom, with nx16x63 geometry) and in RPM (top, with mx255x63) Please READ the text in Partition Logic: (where y=x+1) Can you please put in the: 4080c,4080c,4080c,4080c and post the result in RPM? (both CHS and LBA) Then, try : 4079,8159,12239,16319 and post the result in RPM? (both CHS and LBA) jaclaz

Naaah , we don't have a lot of crocodiles and swamps in the middle of Europe and not even that many borders (corrupt or not ), real problem is elephant stampedes . jaclaz

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: 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: 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: 4080 4080 4080 4080 Input these values in partition logic, then check how RPM reads them. Report. jaclaz CURRENT_PL_RPM.zip

Still : jaclaz

Yes , it is. The info may not be seemingly there, but it is . http://mirror.href.com/thestarman/asm/mbr/MSWIN41.htm http://mirror.href.com/thestarman/asm/mbr/ntFAT32BR.htm And : http://mirror.href.com/thestarman/asm/mbr/FAT32brcomp.htm Hey, that's MS , you remember. Compare with the neat cmdcons trick : http://www.boot-land.net/forums/index.php?showtopic=2362&st=6 but in this case they probably did it to keep compatibility with a number of utilities that only "understood" first sector. The ReactOS guys were probably "forced" to use yet another unused sector for the same compatibility reasons. jaclaz

No problem , though there is still a proplem jaclaz

What normally Seagate does is: get drives to be refurbished refurbish drives Since they have NO obligation whatsoever about your DATA they will (99% probabilities): put your drive on a stack get ANY drive from the stack, check it, test it, refurbish it, REALLY low level format it, install to it the new firmware, apply to it a "refurbished label", send it to you If you obtain that your drive is treated (Free of charge) by their Data recovery division i365 they will (99% probabilities): get your drive on a stack get your drive from the stack, check it, test it non-destructively, make sure your data is readable, apply to it the new firmware, make sure your data is readable, send it to you When you send a drive there are the ADDITIONAL risks of it: being damaged during shipping being lost during shipping being stolen during shipping ( x 2 times as the drive goes to Seagate and back) You choose , but now you know the risks involved and a not-so-random estimation of probabilities of what may happen. jaclaz

You should post this under "News": http://www.boot-land.net/forums/index.php?showtopic=7739&st=92 http://www.boot-land.net/forums/index.php?showtopic=7739&st=100 http://mirror.href.com/thestarman/asm/mbr/index.html#ntfat32 http://mirror.href.com/thestarman/asm/mbr/index.html#mswin41 Well, NO. Giving up is NOT an option. I know I am grumpy, but if I ask you to use 485520c, why don't you use it? DO just the first partition and use for it 485520c (and not ANY other number). Then, post TWO screenshots of the drive in RPM, one like the ones you already posted (CHS) and one showing the LBA (press F4). Then, try creating the other three partitions ALL as primary (it is likely that the "shift" is caused by the Extended) Then, post TWO screenshots of the drive in RPM, one like the ones you already posted (CHS) and one showing the LBA (press F4). jaclaz

...and proplem to problem? Possibly the original source : http://www.geekstogo.com/forum/topic/223632-cant-launch-any-excel-files/ jaclaz

Exactly. Point is that in many places (wrongly) the deal is not (as it should): There is still the (wrong) conception linked to "factory worker": This also doesn't help much in distinguishing "good" IT admin from "bad" ones (in the sense of "smart/capable/efficient" vs. "unknowledgeable but hard working"). jaclaz

Well well: using IE on Win 3.x using IE to access the "ShieldsUp!" page you are lucky that we live in the 21st century, if you did the same in say the 14th you would have probably been processed (and condemned) for blasphemy! Things to try: Opera! (3.62 should be the "right" one - though NOT free as recent versions) http://www.oldapps.com/opera.php?old_opera=12 stay clear of grc.com Let's say that the consensus on Steve Gibson's work is NOT unanymous , expecially when it comes to internet security http://attrition.org/errata/charlatan/ http://attrition.org/errata/charlatan/steve_gibson/ jaclaz

Also, there are other ways: http://www.boot-land.net/forums/index.php?showtopic=9830 jaclaz

The problem lies in this. Obviously the data is shown in cylinders (noone ever talked of bytes). In a "normal" VM with attached a "normal" hard disk image (8 Gb) in size, partitionlogic behaves correctly, proposing "127 cylinders before" on second iteration. Some simple CHS math: (127)x(254+1)x63=2,040,255 Sectors x 512 bytes = 1,044,610,560 bytes i.e. roughly 1 Gb or 1,004 Mbytes, i.e. the 1000m BUT from your previous screenshot, partition logic detects your hard disk (INcorrectly) as having a geometry of nx16x63 INSTEAD of the "right one" of mx255x63 So, numbers change: (2031)x(15+1)x63=2,047,248 Sectors x 512 bytes = 1,048,190,976 bytes i.e. roughly 1 Gb or 1,048 Mbytes, i.e. the 1000m RPM detects the drive as having mx255x63 geometry thus all the values set by partition logic are "crazy" from a CHS point of view. To have a "robust" (meaning both CHS and LBA correct) partitioning, using this approach we have some limitations. Since 16x255=255x16 we need to find Cylinder numbers on a 16 head device that "fall" on the same boundary on a 255 head one. Since 255 and 16 have no "common divider" we have a "minimal step" of 255x16=4080. In other words: 4080x16x63= 4,112,640 x 512 = 2,105,671,680 i.e. roughly 2 Gb and: 256x255x63= 4,112,640 x 512 = 2,105,671,680 i.e. roughly 2 Gb This is the "minimum" partition size you can have and all partition sizes must be a multiple of this. Following the idea of 4 LBA FAT 32 partitions, let's see which can be the "right" number(s) for each in partition logic. Let's see: RPM "allows" for: 121,600x255x63= 1,953,504,000 sectors the screenshot from partition logic is not too readable and some numbers are "strange", but it looks like reasonable: 1,938,000x16x63= 1,953,504,000 sectors But we need 63 sectors for MBR+hidden sectors (Track 0) and 63 sectors for the Extended partition. So you could use: 63<-MBR+hidden sectors 485520<-first primary partition 63<-Extended 485520<-first logical volume inside extended 481440<-first logical volume inside extended 481440<-first logical volume inside extended 3954<-unused You should get the primary partition and one of the logical volumes: 485,520x16x63=489,404,160 x 512 = 250,574,929,920 bytes or 30,464x255x63=489,404,160 x 512 = 250,574,929,920 bytes and the other two logical volumes: 481,440x16x63=485,291,520 x 512 = 248,469,258,240 bytes or 30,208x255x63=485,291,520 x 512 = 248,469,258,240 bytes 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 from #2 always using the "sectors before supplied by partition logic. Once done that, in RPM you should have all partitions going to the right boundaries, i.e. for each partition starting head only 0 or 1 and Ending Head only 254. I hope that the above is clear. jaclaz

Actually doorknobs can be useful, a few examples : hxxp://www.youtube.com/watch?v=GpRSEJY0cTI hxxp://www.youtube.com/watch?v=2TvhG_B5d20 hxxp://vids.myspace.com/index.cfm?fuseaction=vids.individual&videoid=34036139 jaclaz

In my experience "doorknobs" tend to have problems in managing a knife/cutter, a soldering gun, a multimeter, etc. Not wanting in any way to put down anyone , only trying to point out the "easiest" ways.... Sure , I may be cheap (please read as "am") , but I know where my towel is. Yep, that's point #14 of the read-me-first in a nutshell. jaclaz

Sure , it's called "ask to a professional" BUT requires materials that are hard to find (banknotes, in a rather huge amount). Come on, use your own little gray cells, if there was an easier way, why do you think we are hanging to this one? For the record, this lacks totally of logic: This is logical: Anyway, it's not really difficult, you just need some patience and time, and be calm. If you have a friend with a minimal experience in electronics/electricity that can help you there won't be ANY problem at all. @Bloubul Obviously the idea of suggesting to a "doorknob" using a DKU-5/CA-42 cable is making things really difficult. As written n times by now, a technically knowledgeable peep won't have ANY problem using, in order of difficulty: a "proper" converter with KNOWN connections, voltage requirements, etc. a "proper" converter with UNKNOWN connections, voltage requirements, etc. ANY converter with KNOWN connections, voltage requirements, etc. ANY converter with UNKNOWN connections, voltage requirements, etc. a "known to be not fake" DKU-5/CA-42 cable with KNOWN connections, voltage requirements, etc. a "known to be not fake" DKU-5/CA-42 cable with UNKNOWN connections, voltage requirements, etc. ANY DKU-5/CA-42 cable with UNKNOWN connections, voltage requirements, etc. WITHOUT ANY converter and just build one from a handful of scrap parts he has in his basement What you suggested to Coffecup is #7 in the above list. Difference between the cost of #1 and #7: absolute maximum 30 US$ usually much less, around 5÷8 US$ EXAMPLES of #1: USB to TTL 3V: This is the most expensive I have ever seen in my life (NOT recommended): http://cgi.ebay.it/CAVO-USB-3V-TTL-per-HD-Seagate-7200-11-VER-PRO-rs232-/190426361136?pt=Accessori_per_Radioamatori&hash=item2c564bad30 This is the "average": http://cgi.ebay.it/TTL-INTERFACCIA-Seagate-7200-11-RS232-BSY-FIX-FIRMWARE-/170541063840?pt=Accessori_per_Radioamatori&hash=item27b50a1ea0 This is cheap: http://cgi.ebay.it/2-x-PL2303-USB-TTL-RS232-Serial-Port-Converter-Adapter-/220663946707?pt=LH_DefaultDomain_0&hash=item3360989dd3 jaclaz