I would like to scale a raster image inside Illustrator to a precise dimension - for example, a mechanical drawing with dimensions.  I can draw a precise length line over the dimension, but can I then scale the line and the image precisely to the desired length.  I realize I can do the math and use a scale percent, but I was wondering if there's a faster way.
In Illustrator is there a way to precisely scale to a pivot point or "reference point" that is off the screen? I mean when you want to zoom in somewhere far away from where you set that point and then scale precisely to align something.   Here's :     To scale relative to a different reference point , click where you   want the reference point to be in the document window, move the   pointer away from the reference point, and then drag until the object   is the desired size.   Well, I tried using this click-drag scaling method as well, but the level of precision is too low to be usable. It overshoots or undershoots the guide I'm aiming for.  The Scale tool's pop-up dialog isn't useful either. It only offers percentages. And moving/zooming are disabled while the dialog window is up, meaning I cannot navigate to the guide I need to see.  For these reasons, I need to be zoomed in on the guide, tweaking the measurements from the top Control Panel.  Some images to help illustrate what I mean:    ▲ The circle marks where I would like the center of my scale transformation to be.    ▲ Now zoomed in somewhere else, I am trying to get the black border of this bitmap image to align perfectly with the vector guide by scaling it.    ▲ I nudge the numbers in the Control Panel until the image is perfectly aligned.  One would think all would be perfect, but then I look and my reference point has vanished!   It seems the scaling has been executed relative to one of these limited nine reference point choices  instead of the reference point I specified.  There must be a way to scale precisely to a specific reference point in illustrator. Can somebody please enlighten me?
1st example:  \starttypescript [\s!serif] [paratype]   \definefontsynonym [ParatypeSerif-Regular] [\s!name:PTSerif-Regular] [\s!features=\s!default]   \definefontsynonym [ParatypeSerif-Italic] [\s!name:PTSerif-Italic] [\s!features=\s!default]   \definefontsynonym [ParatypeSerif-Bold] [\s!name:PTSerif-Bold] [\s!features=\s!default]   \definefontsynonym [ParatypeSerif-BoldItalic] [\s!name:PTSerif-BoldItalic] [\s!features=\s!default] \stoptypescript   2nd example:  \definefont[SerifS][\s!file:GenBkBasR at \smallfontsize]   Is \s! only related to fonts? Should I use it in my typescript files? Bootstrap dropdown sub menu missing Spacetime Torsion, the Spin tensor, and intrinsic spin in Einstein-Cartan theory In  gravity, the action is the usual  but now the Torsion tensor is allowed to vary as well (in usual GR, it is just set to zero).  Variation with respect to the metric gives:  $$R_{ab}-\frac{1}{2}R g_{ab}=\kappa P_{ab} \quad (1)$$  where $P_{ab}$ is the canonical stress energy tensor.  Variation with respect to the torsion tensor ${T^{ab}}_c$ gives:  $${T_{ab}}^c + {g_a}^c{T_{bd}}^d - {g_b}^c {T_{ad}}^d = \kappa {\sigma_{ab}}^c \quad (2)$$  where ${\sigma_{ab}}^c$ is the .  By contracting that equation, I can see that if the spin tensor is zero, the Torsion tensor is identically zero:  $${T_{ab}}^c + {g_a}^c{T_{bd}}^d - {g_b}^c {T_{ad}}^d = \kappa {\sigma_{ab}}^c = 0$$ $$ {g^b}_c({T_{ab}}^c + {g_a}^c{T_{bd}}^d - {g_b}^c {T_{ad}}^d) = 0$$ $$ {T_{ab}}^b + {g_a}^b{T_{bd}}^d - {g_b}^b {T_{ad}}^d = 0$$ $$ {T_{ab}}^b + {T_{ad}}^d - 4 {T_{ad}}^d = 0$$ $$ {T_{ad}}^d = 0$$ $${T_{ab}}^c + {g_a}^c{T_{bd}}^d - {g_b}^c {T_{ad}}^d = 0 \quad \Rightarrow \quad {T_{ab}}^c =0$$  My understanding is that:  The spin tensor and the stress energy tensor, are defined entirely in terms of whatever matter Lagrangian we add to the theory.  Therefore, from above, the equations in vacuum are exactly the same as normal GR (so solving for outside matter, only the boundary conditions with the matter could be different).  Assuming my understanding up to here is correct, my line of questioning is:   How is the Spin tensor (and hence Torsion) related to the concept of material with intrinsic spin? Hopefully answered by 1, but if the matter has zero intrinsic spin, yet we have an extended "spinning" body, is the spin tensor still zero (as I'd consider that orbital angular momentum then)? Does this mean Einstein-Cartan predictions are identical to normal Einstein GR if (and only if) the instrinsic spin of any particles and fields in the theory are zero? One of the first things ever taught in a differential calculus class:   The derivative of $\sin x$ is $\cos x$. The derivative of $\cos x$ is $-\sin x$.   This leads to a rather neat (and convenient?) chain of derivatives:   sin(x) cos(x) -sin(x) -cos(x) sin(x) ...   An analysis of the shape of their graphs confirms some points; for example, when $\sin x$ is at a maximum, $\cos x$ is zero and moving downwards; when $\cos x$  is at a maximum, $\sin x$ is zero and moving upwards.  But these "matching points" only work for multiples of $\pi/4$.  Let us move back towards the original definition(s) of sine and cosine:  At the most basic level, $\sin x$ is defined as -- for a right triangle with internal angle $x$ -- the length of the side opposite of the angle divided by the hypotenuse of the triangle.  To generalize this to the domain of all real numbers, $\sin x$ was then defined as the Y-coordinate of a point on the unit circle that is an angle $x$ from the positive X-axis.  The definition of $\cos x$ was then made the same way, but with adj/hyp and the X-coordinate, as we all know.  Is there anything about this basic definition that allows someone to look at these definitions, alone, and think, "Hey, the derivative of the sine function with respect to angle is the cosine function!"  That is, from the unit circle definition alone.  Or, even more amazingly, the right triangle definition alone.  Ignoring graphical analysis of their plot.  In essence, I am asking, essentially, "Intuitively why is the derivative of the sine the cosine?" If a function can only be defined implicitly does it have to be multivalued? Active Web Site as Gnome Background Is there a way to put an active website, such as Gmail, as your Desktop background in Gnome, running Ubuntu 10.10 64-bit? MS Win calls it "Active Desktop". Time ordering operator if commutator is $c$-number function Finding CPU usage of a service Because my laptop fan was spinning up for no reason, I had a look at the CPU usage and I see  svchost with 12.5% CPU, which is basically a full core.  Now, I'm aware that  svchost runs services and I know how to display the services it runs in Task Manager or Process Explorer ().  Now I know which services are run, but I don't know which service eats the CPU time. I could disable one service after another, but I don't really like doing this (it could break things that are currently in use and I could forget turning the service back on later).  Therefore I'd like to know whether it's possible to instruct Windows to run only one service per svchost the next time I boot my PC. While I might have the slight disadvantage of several processes, I could immediately find out which one it is and it would be a much easier decision of whether or not to stop or restart the service. I would like to create a half-turn staircase without a landing. Is there an easy way to accomplish this without having to manually calculate and create each step individually?  Here’s an image similar to what I would like to achive:    I have found the Archimesh add-on including . While I can bend the created stairs, I haven’t found any way to get anywhere close to the image above. My stairs just always look out of shape and I don’t know how to get the outline of the staircase into a square.  I should mention that I only have very little experience with Blender modeling, so it’s well possible that I’m just missing some curve modification basics. Perhaps Archimesh is also the wrong approach for this in the first place? How to salt PBKDF2, when generating both an AES key and a HMAC key for Encrypt then MAC? When using Encrypt-then-MAC with AES and HMAC by password, and given 128 bits of payload with the ciphertext to store a random salt, which would be more secure:   Using PBKDF2 with then entire 128 bit random salt and generating 512 bits of derived data to split into a 256 bit AES Key and a 256 bit HMAC key.   or   Using PBKDF2 separately for each key using two 64 bits salts to generate a 256 bit AES Key and a 256 bit HMAC key.    or    Better option? Why can't this be done with Master Theorem Apparently recurrences like this cannot be solved with the Master Theorem:  $T(n) = 2T\left(\frac{n}{2}\right) + \frac{n}{\log(n)}$  Because $n^{\log_b(a)} = n^1$ is not a polynomial multiple of $f(n) = \frac{n}{\log(n)}$ since $\frac{n}{\frac{n}{\log(n)}} = \log(n)$, which is less than $n^\epsilon$ for any positive $\epsilon$.  But when is this really even an issue?   For example why wouldn't the time complexity be $n \log n$ using case 2 of the Master Theorem? Unrolling the recursion:  $$T(n) = n\sum_{k=1}^{\log_2(n)} \frac{1}{\log_2(2^k)}$$  $$T(n) = n\sum_{k=1}^{\log_2(n)} \frac{1}{k}$$  Doesn't this suggest $\Theta(n \log n)$ even though it supposedly cannot be done with Master Theorem? QUESTION Find the number of triples $$(x,y,z)$$ of real numbers satisfying the equation $$x^4+y^4+z^4+1=4xyz$$  I have tried to solve this problem but can't quite understand how to manipulate the given data to reach a clear result. Could someone please explain how I should approach and solve this problem.Thanks :) biblatex: How to omit "p." at \cite? Submitting a URL to Google using PHP/cUrl Install suspension forks on non-suspension bike Are there substitutes for chain lubricant? With long journeys and particularly during winter I have noticed that the salt erodes the chain very quickly, even when I have lubricated the chain beforehand. When you are missing proper chain lubricants, what would you use as a substitute? Could you use cooking oil if you are missing the right stuff, or should I keep a small bottle of oil with me on my rides?  Related Are smaller p-values more convincing? What does Redis do when it runs out of memory? Check if macro is fully expandable Question  Is there a way to check whether a macro is fully expandable (or rather "safe in an expansion-only context" )?    Consider this code:  \def\a{Just a string} \def\b{\a} \def\c{\def\unsafe} \def\d{\c}   How could I check which of the macros (a-d) are safe in an expansion-only context? By looking at them I know that a and b are whereas c and d are not but if I wanted to know the same for a macro I haven't written myself this could get quite useful.      Background  I am working on a way to detect whether some input is a valid number in PGF. For this I developed  which makes use of passing the input into \pgfmathfloatparsenumber. The problem I have run into is that said macro appears to somehow manages to expand the input until there is an error (if the input is in fact not safe in an expansion-only-context). I tried using protected, noexpand and similar but somehow PGF manages to circumvent those.    So the idea is to check whether the input is safe before actually passing it to PGF. The problem is: I don't know how I'd go about that... XMLHttpRequest Origin null is not allowed Access-Control-Allow-Origin for file:/// to file:/// (Serverless)