Re: [SuSE Linux] Toshiba Libretto
Jim-- If it really is a Libretto, it is *not* a WinCE machine. The Libretto is just an incredibly small PC that runs Win 95, and yes, it can run Linux (or FreeBSD). To answer your questions... for a new screen, you can always call Toshiba and ask them. I do know that Toshiba replacement screens are NOT cheap. As for the password, I don't know. You might check on comp.sys.laptop and see if anyone knows the answer there. I'm no expert at portables, but it may end up that you might have to remove the internal battery and let the machine "forget" all of it's settings. Again, check on comp.sys.laptop or check the Toshiba Web site. Mark -----Original Message----- From: James (Jim) Hatridge <hatridge@straubing.baynet.de> To: suse-linux-e@suse.com <suse-linux-e@suse.com> Date: Saturday, February 06, 1999 11:40 AM Subject: [SuSE Linux] Toshiba Libretto
Hi All;
A friend was given a Toshiba Libretto with a broke screen. Someone dropped it :(. It's running Windows CE. As soon as we can fix it we want to put Liunx on it. (I hope)
There are two problems with it. First when he starts it up it demands a password. This password seems to be in the BIOS. Without it the whole computer locks up. Does anyone know how to get around this? It seems to be wanting the password before it even boots W$ CE.
Next, does anyone know where to get a replacement screen for a Libretto, and what it costs?
We got it from a guy who got it from the first owner, so the password is completely gone. :(
TIA!
J I M ----------------------------------------- Jim Hatridge Germany hatridge@straubing.baynet.de
Proud Linux User #88484 !!!!!!!!!!
Micro$oft -- Ghostdriver* on the road to the future! (*German Slang for the guy driving on the wrong side of the road!) ---------------------------------------------- "If a President of the United States ever lied to the American people he should resign." Bill Clinton, 1974
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Just an idea -- If you like to play with the hardware, all you need to do is upgrade the BIOS chip. Pick one you like and replace the existing BIOS chip. It will not have a password until you set one. Check your local computer shows. Here, in Las Vegas, there is a "computer show" at least once a month. It is actually a swap meet for for all kinds of computer parts, pieces, and paraphenalia. You might find a Libretto with a fried board and a good screen. HTH John
-----Original Message----- From: James (Jim) Hatridge <hatridge@straubing.baynet.de> To: suse-linux-e@suse.com <suse-linux-e@suse.com> Date: Saturday, February 06, 1999 11:40 AM Subject: [SuSE Linux] Toshiba Libretto
Hi All;
There are two problems with it. First when he starts it up it demands a password. This password seems to be in the BIOS. Without it the whole computer locks up. Does anyone know how to get around this? It seems to be wanting the password before it even boots W$ CE.
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I thought I had this information somewhere, but I can't find it. The 'funny' images that accompany SuSE Linux (in 5.2 it was blue, in 5.3 green), can they be downloaded from somewhere? - To get out of this list, please send email to majordomo@suse.com with this text in its body: unsubscribe suse-linux-e Check out the SuSE-FAQ at <A HREF="http://www.suse.com/Support/Doku/FAQ/"><A HREF="http://www.suse.com/Support/Doku/FAQ/</A">http://www.suse.com/Support/Doku/FAQ/</A</A>> and the archiv at <A HREF="http://www.suse.com/Mailinglists/suse-linux-e/index.html"><A HREF="http://www.suse.com/Mailinglists/suse-linux-e/index.html</A">http://www.suse.com/Mailinglists/suse-linux-e/index.html</A</A>>
<PRE> Hi, On Sun, Feb 07, 1999 at 09:48 +0100, Kaare Rasmussen wrote:
I thought I had this information somewhere, but I can't find it. The 'funny' images that accompany SuSE Linux (in 5.2 it was blue, in 5.3 green), can they be downloaded from somewhere?
Have a look at the attached message. Ciao, Stefan </PRE> -- BEGIN included message</EM></P> <BLOCKQUOTE> <UL> <LI>To</em>: <A HREF="mailto:suse-linux-e@suse.com">suse-linux-e@suse.com</A></LI> <LI>Subject</em>: Re: Mousepad</LI> <LI>From</em>: Hubert Mantel <<A HREF="mailto:mantel@suse.de">mantel@suse.de</A>></LI> <LI>Date</em>: Fri, 17 Apr 1998 13:00:08 +0200 (MEST)</LI> <LI>In-Reply-To</em>: <199804162138.XAA12989@bohr.webline.dk></LI> <LI>Reply-To</em>: <A HREF="mailto:suse-linux-e@suse.com">suse-linux-e@suse.com</A></LI> <LI>Sender</em>: <A HREF="mailto:owner-suse-linux-e@suse.com">owner-suse-linux-e@suse.com</A></LI> </UL> <PRE> Hi, On Thu, 16 Apr 1998, Kaare Rasmussen wrote:
It's one of our CD covers, I think the on of the 5.0. It's some kind of Enriques surface, but please don't ask me about details, the formula to compute this image it pretty complicated...
But perhaps you could tell me what an Enriques surface is?
It is a special family of algebraic surfaces studied by Federigo Enriques. A simple algebraic surface for example are all points that satisfy the equation x²+y²+z²-1=0, which is a sphere with radius 1 and center at origin. Enriques surfaces are something "similar", but the formula is somewhat more complex ;-) In fact, you cannot see an Enriques surface itself, but only some model of it. You can find details in the preface of our handbook if you are interested. It is nontrivial to compute these surfaces. The program that computes the surfaces is on the CD. It's called "surf". Hubert PS: Have a look at <A HREF="http://www.mi.uni-erlangen.de/~endrass/en/page.html"><A HREF="http://www.mi.uni-erlangen.de/~endrass/en/page.html</A">http://www.mi.uni-erlangen.de/~endrass/en/page.html</A</A>> Stephan Endrass, a good friend of mine, is the author of "surf". There are very beautiful pictures on this site (go to the "parameter map"). Also, the animation is worth watching. PPS: Want some challenge? Find the formula that gives you a torus ;-) -- To get out of this list, please send email to majordomo@suse.com with this text in its body: unsubscribe suse-linux-e </PRE> </BLOCKQUOTE> -- END included message</EM></P>
/ PPS: Want some challenge? Find the formula that gives you a torus ;-) It's pretty neat.The flat torus formula: [u,v] -> [cos(u + v), sin(u + v), cos(u - v), sin(u - v)]/sqrt(2) where u and v both run from zero to 2 pi. The sum of the squares of these four coordinates is 1 so the object is completely contained in the hypersphere of radius 1 centered at the origin in four-space. -- Subject: [SuSE Linux] Re: The SuSE images
Hi,
On Sun, Feb 07, 1999 at 09:48 +0100, Kaare Rasmussen wrote:
I thought I had this information somewhere, but I can't find it. The 'funny' images that accompany SuSE Linux (in 5.2 it was blue, in 5.3 green), can they be downloaded from somewhere?
Have a look at the attached message.
Ciao, Stefan
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Hi, On Mon, Feb 08, J-L Boers wrote:
/ PPS: Want some challenge? Find the formula that gives you a torus ;-)
It's pretty neat.The flat torus formula:
[u,v] -> [cos(u + v), sin(u + v), cos(u - v), sin(u - v)]/sqrt(2)
where u and v both run from zero to 2 pi. The sum of the squares of these four coordinates is 1 so the object is completely contained in the hypersphere of radius 1 centered at the origin in four-space.
Cute ;) But that's too easy. The question was meant to be: Find an equation f(x,y,z)=0 so that all solutions of the equation form the surface of a torus. To be honest: I don't know the solution. I even don't know if this equation exists ;) -o) Hubert Mantel Goodbye, dots... /\\ _\_v - To get out of this list, please send email to majordomo@suse.com with this text in its body: unsubscribe suse-linux-e Check out the SuSE-FAQ at <A HREF="http://www.suse.com/Support/Doku/FAQ/"><A HREF="http://www.suse.com/Support/Doku/FAQ/</A">http://www.suse.com/Support/Doku/FAQ/</A</A>> and the archiv at <A HREF="http://www.suse.com/Mailinglists/suse-linux-e/index.html"><A HREF="http://www.suse.com/Mailinglists/suse-linux-e/index.html</A">http://www.suse.com/Mailinglists/suse-linux-e/index.html</A</A>>
Hubert Mantel wrote:
On Mon, Feb 08, J-L Boers wrote:
/ PPS: Want some challenge? Find the formula that gives you a torus ;-)
It's pretty neat.The flat torus formula:
[u,v] -> [cos(u + v), sin(u + v), cos(u - v), sin(u - v)]/sqrt(2)
But that's too easy. The question was meant to be: Find an equation f(x,y,z)=0 so that all solutions of the equation form the surface of a torus. To be honest: I don't know the solution. I even don't know if this equation exists ;)
Well I havn't looked at my Vector Calculus books in about 20 years, but I found a set of formulas in "toroidal coordinates" for a torus. They are quite complicated: I will give the x equation only, but there are y, and z ones also. x = ((a)sinh(v)cos(w))/((cosh(v)-cos(u)) where "a" is a constant, and u,v,w are the toroidal coordinates, w is actually an angle theta. If you want, I could scan the page with the full formula set, including a cartesian coordinate graph , and email it to you. But it would probably delay work on Suse 6.0 . :-) - To get out of this list, please send email to majordomo@suse.com with this text in its body: unsubscribe suse-linux-e Check out the SuSE-FAQ at <A HREF="http://www.suse.com/Support/Doku/FAQ/"><A HREF="http://www.suse.com/Support/Doku/FAQ/</A">http://www.suse.com/Support/Doku/FAQ/</A</A>> and the archiv at <A HREF="http://www.suse.com/Mailinglists/suse-linux-e/index.html"><A HREF="http://www.suse.com/Mailinglists/suse-linux-e/index.html</A">http://www.suse.com/Mailinglists/suse-linux-e/index.html</A</A>>
On 08-Feb-99 Hubert Mantel wrote:
But that's too easy. The question was meant to be: Find an equation f(x,y,z)=0 so that all solutions of the equation form the surface of a torus. To be honest: I don't know the solution. I even don't know if this equation exists ;) -o) Hubert Mantel Goodbye, dots... /\\
Well, how about: Suppose the torus is swept out by a circle of radius r1 whse centre is carried round a circle of radius r0 (r1 < r0 for a proper torus). Let u = x/r0, v = y/r0, w = z/r1, C = r1/r0. Then u^2 + v^2 = (1 + C sqrt(1 - w^2) )^2 (Consider r = r0 + r1 cos q, z = r1 sin q, x = r cos p = r0 cos p + r1 cos p cos q y = r sin p = r0 sin p + r1 sin p cos q and eliminate the angles p, q. p is the angle in the x-y plane from a fixed direction to the centre of the sweeping circle; q is the "angle of elevation" from the centre of this circle to a point on the torus as seen along the line from the origin to the centre of this circle; r is the distance from the origin to the point in the x-y plane vertically beneath the point on the torus.) I think it's right ... Ted. -------------------------------------------------------------------- E-Mail: (Ted Harding) <Ted.Harding@nessie.mcc.ac.uk> Date: 08-Feb-99 Time: 20:34:37 ------------------------------ XFMail ------------------------------ - To get out of this list, please send email to majordomo@suse.com with this text in its body: unsubscribe suse-linux-e Check out the SuSE-FAQ at <A HREF="http://www.suse.com/Support/Doku/FAQ/"><A HREF="http://www.suse.com/Support/Doku/FAQ/</A">http://www.suse.com/Support/Doku/FAQ/</A</A>> and the archiv at <A HREF="http://www.suse.com/Mailinglists/suse-linux-e/index.html"><A HREF="http://www.suse.com/Mailinglists/suse-linux-e/index.html</A">http://www.suse.com/Mailinglists/suse-linux-e/index.html</A</A>>
On Mon, 08 Feb 1999 Ted Harding wrote:
On 08-Feb-99 Hubert Mantel wrote:
But that's too easy. The question was meant to be: Find an equation f(x,y,z)=0 so that all solutions of the equation form the surface of a torus. To be honest: I don't know the solution. I even don't know if this equation exists ;)
Well, how about:
Suppose the torus is swept out by a circle of radius r1 whse centre is carried round a circle of radius r0 (r1 < r0 for a proper torus).
Let u = x/r0, v = y/r0, w = z/r1, C = r1/r0. Then
u^2 + v^2 = (1 + C sqrt(1 - w^2) )^2
(Consider r = r0 + r1 cos q, z = r1 sin q, x = r cos p = r0 cos p + r1 cos p cos q y = r sin p = r0 sin p + r1 sin p cos q
and eliminate the angles p, q.
p is the angle in the x-y plane from a fixed direction to the centre of the sweeping circle; q is the "angle of elevation" from the centre of this circle to a point on the torus as seen along the line from the origin to the centre of this circle; r is the distance from the origin to the point in the x-y plane vertically beneath the point on the torus.)
I think you can express this equation in a slightly more simple form, namely: (sqrt(x^2 + y^2) - a)^2 + z^2 = b^2 where a>b Namely revolve a circle of radius b about an exterior line in its plane, that is distance a from the center. Gives a ring shaped torus. At last the mailing list is becoming interesting :-) Alexander ------------------------------------- Alexander Volovics Dept of Methodology & Statistics Maastricht University, Maastricht, NL ------------------------------------- - To get out of this list, please send email to majordomo@suse.com with this text in its body: unsubscribe suse-linux-e Check out the SuSE-FAQ at <A HREF="http://www.suse.com/Support/Doku/FAQ/"><A HREF="http://www.suse.com/Support/Doku/FAQ/</A">http://www.suse.com/Support/Doku/FAQ/</A</A>> and the archiv at <A HREF="http://www.suse.com/Mailinglists/suse-linux-e/index.html"><A HREF="http://www.suse.com/Mailinglists/suse-linux-e/index.html</A">http://www.suse.com/Mailinglists/suse-linux-e/index.html</A</A>>
participants (9)
-
awol＠STAT.UNIMAAS.NL -
bjgilger＠earthlink.net -
jlboers＠usa.net -
kar＠webline.dk -
mantel＠suse.de -
medavid＠ix.netcom.com -
stefan.troeger＠wirtschaft.tu-chemnitz.de -
Ted.Harding＠nessie.mcc.ac.uk -
zentara＠mindspring.com