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#1 dense

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Posted 24 February 2013 - 03:13 AM

When you boot using grub4dos, there are two timeouts: the one in the menu.lst, which in the sample is set to 30;

but before that menu, there is the other coundown of 5 seconds. I can't figure out where that timeout is set.



#2 RoyM

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Posted 24 February 2013 - 08:06 AM

See here



#3 dense

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Posted 24 February 2013 - 07:40 PM

This makes no sense:

using a hex editor, I searced both grldr and grub.exe for the string: "pxe detect" , and soon saw: "timeout 1";

  Yet, when invoking grub4dos on bootup, the prelude to the regular menu.lst(the one on the root drive) there is clearly

a *five* second timeout. I am stumped.



#4 dense

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Posted 24 February 2013 - 07:45 PM

Let me clarify:

In the root-drive menu.lst, I am using 'timeout 30';

before that menu, the embedded menu.lst shows timeout 1, but it is counting down from 5.



#5 steve6375

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Posted 24 February 2013 - 08:26 PM

What you don't mention is how you installed grub4dos???

 

Try installing grub4dos using RMPrepUSB.

If you use grubinst then use t=0 in the command line

 

It is the timeout before grub will boot from the previous MBR rather than load grldr.



#6 dense

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Posted 25 February 2013 - 12:26 AM

Here is how I install grub4dos:

place grub.exe,grldr,grldr.mbr,and bootlace.com, as well as menu.lst on c:\ drive, which is fat32.

Execute: bootlace 0x80,

then execute grub, which brings up the usual grub4dos menu. This only has to be done once.



#7 dense

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Posted 25 February 2013 - 01:07 AM

Amazing; just reinstalled grub4dos using grubinst(gui);set (hd0),only checked don't search for floppy,set t=1; BINGO! tnx all.

Still don't see where it is hiding though, as per the above.



#8 steve6375

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Posted 25 February 2013 - 09:19 AM

The timeout value is a byte in one of the first 16 bytes of the MBR.




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