Words Per Minute Test
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02-06-2013, 07:09 PM
RE: Words Per Minute Test
I got 90.2, but didn't take notice of how many errors. I'm assuming there were a good many. I think I could have topped that had I not been using my laptop. Sad Will have to try with my good keyboard at work!
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03-06-2013, 09:06 AM
RE: Words Per Minute Test
I forgot (and I was a bit preoccupied) to try this last night while drunk. Based on my drunken ramblings last night, and the complete lack of proper English, I assume I would have made the computer explode Drinking Beverage

"It was life, often unsatisfying, frequently cruel, usually boring, sometimes beautiful, once in awhile exhilarating." -Stephen King
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06-06-2013, 04:47 PM
RE: Words Per Minute Test
28.6
Apparently index fingers only aint the way to do it Undecided

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10-06-2013, 03:35 PM
RE: Words Per Minute Test
97 wpm with 1 error. I used to be able to type around 112 back in my youth when I typed all the time. Smile
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10-06-2013, 07:22 PM
RE: Words Per Minute Test
110, but it really messed with me not to be able to correct the words I messed up on (7 of 104, the scorecard handily informs me). So I lost a couple seconds on each typo.

Of course, I almost never actually type that much plain English, at work...
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11-06-2013, 07:21 AM
RE: Words Per Minute Test
I got 70 on my second try, but I can do a little better than that when my brain and hands are working together.

Give me your argument in the form of a published paper, and then we can start to talk.
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11-06-2013, 09:37 AM
RE: Words Per Minute Test
It'd be neat if there was such a test where you could use your own input text - anyone know of such a thing?

'Cause the crap I usually have to type looks more like
cjlr Wrote:In order to diagonalize this Hamiltonian, we attempt the following substitution.
\begin{equation} \begin{aligned}
\hat{c}_k^A &= v d_k + u f_k \\
\hat{c}_k^B &= -u^* d_k + v^* f_k \ \text{,}
\end{aligned} \end{equation}
where $d_k$ and $f_k$ are Fermions which, as linear combinations of $\hat{c}_k^A$ and $\hat{c}_k^B$, diagonalize $\mathbf{H}$. That is,
\begin{equation}
\begin{pmatrix}\hat{c}_k^{A \dagger} & \hat{c}_k^{B \dagger} \end{pmatrix} \mathbf{H} \begin{pmatrix}\hat{c}_k^A \\ \hat{c}_k^B \end{pmatrix} \equiv \begin{pmatrix}d_k^\dagger & f_k^\dagger \end{pmatrix} \begin{pmatrix}\lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix} \begin{pmatrix}d_k \\ f_k \end{pmatrix} \ \text{.}
\end{equation}
which is just tedious.
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