Archive for June, 2009

Greece

For the first 4 nights of the trip, we stayed in Volos, a major port city where I attended a workshop on Information Theory. The following day, we took a bus to Kalambaka.

Natalie and I spent two days in Kalambaka hiking on Meteora, which might be the most amazing natural landform I’ve ever seen. Kalambaka is small town nestled up against some gigantic rock protrusions. Several of the rocks are topped with ancient (but still active) monasteries.

It’s a striking view and in fact one of the monasteries was used in the James Bond film, “For Your Eyes Only.” Here is a picture Natalie took of that monastery with me in it

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I can’t overstate how awesome Meteora was. I love mountains, views and nature. And I got it, for example, here is a mountainside meadow
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and the city of Kalambaka seen from an outcropping
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This was one of the best hikes of my life. Other pictures can be found here.

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Building a Picnic Table

Someone writes

Jonathan and/or [redacted],

I’ve got a basic geometry question for either of you, that I need a little help with…

I am working on making a picnic table with [redacted] and we wanted to design our own legs for the table which are angled and cross. I have been able to work out all relationships between heights, lengths, widths, angles. etc. and now have two equations with two unknowns that will allow me to solve for the correct angle at which to cut the wood for the legs. Unfortunately I have forgotten how to solve basic cos, sin, and tan when variables are involved. The two equations are (all units are in inches, not that it matters):


$\displaystyle \textrm{sin} (\alpha)$ $\textstyle =$ $\displaystyle \frac{5.5}{x}$ (1)
$\displaystyle \textrm{tan} (\alpha)$ $\textstyle =$ $\displaystyle \frac{29}{35.5-x}$ (2)

How do I solve for $x$ and $\alpha$? Also, what is the answer :-)

Thanks,
[redacted]

Because

\begin{eqnarray*}\textrm{tan} (\alpha) & = & \frac{\textrm{sin} (\alpha)}{\text......rac{\textrm{sin} (\alpha)}{\sqrt{1 - \textrm{sin}^{2} (\alpha)}}\end{eqnarray*}

we have

$\displaystyle \frac{29}{35.5-x}$ $\textstyle =$ $\displaystyle \frac{\frac{5.5}{x}}{\sqrt{1 - \left(\frac{5.5}{x}\right)^{2}}}$ (3)

Simplifying and solving by computer algebra, we get

\begin{eqnarray*}\frac{3243}{841} x^2 + \frac{8591}{841} x - \frac{1017005}{3364} & = & 0\end{eqnarray*}

and

\begin{eqnarray*}x & = & \frac{ \pm 1276\sqrt{2071} - 8591}{6486}\end{eqnarray*}

Only the positive choice is a root of Eqns. (1) and (2), giving

\begin{eqnarray*}x & = &7.62835577107986... \\\alpha & = & 0.805235953662952... \\& = & 46.1366216570791... \textrm{(degrees)}\end{eqnarray*}

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