TeX, LaTeX and Friends

1:00 AM

@samcarter Depending on your preferences for the weather, you may or may not implement this for a `cast shadow` option (requires `shadows.blur` but I could redraw without that library if that's better)
\fill[gray,opacity=0,blur shadow] (1.5,0.2) arc(60:405:8mm and 2mm)
to[out=50,in=-130] ++ (0.4,0.2) to[out=50,in=50,looseness=0.5] ++ (-0.2,0)
to[out=-130,in=0,looseness=0.5] cycle;

I forgot to mention that people may need it at groundhog day.

CarLaTeX

5:33 AM

@UlrikeFischer Grazie! Is it Mondello? Maybe @Sebastiano is there.

Joseph Wright

6:12 AM

22

A: What does constructive criticism of a design change look like?

I really like this guidance, Jon; giving constructive feedback on design changes is something I've struggled quite a bit with over the years, both as a user and as an employee of the company making the changes. It's easy to say "I don't like it", much more work to articulate why... And I've often...

As suspected, the three column business comes down to needing to make money

2

Ulrike Fischer

7:40 AM

@CarLaTeX you are good ;-). And you can't read it at this resolution but one of the ducks is reading a book playing in Sicily.

CarLaTeX

@UlrikeFischer :)

Sebastiano

8:52 AM

@UlrikeFischer @CarLaTeX @egreg @marmot @Bernard @Mico ecc. Buon ferragosto with all my heart. I'm in Eastern Sicily :-) and I prefer to stay at home. There are a lot of people at sea and the weather is not the best. There are clouds, sun and even wind. Unfortunately on our beaches and in the clear sea there are many who do not respect cleanliness. I prefer to wait a few days to go to sea. A greeting and a hug to everyone.

@DavidZ Good morning also to you.

@egreg Halo'. :-)

David Z

@Sebastiano Thanks! It's late at night where I am, though.

(same to you)

Sebastiano

@DavidZ So I wish you a peaceful night and a good night's sleep when you go to sleep.

boycott.se - yo'

10:08 AM

@JosephWright with an important little detail: the useless empty left pane is to-become an advertisement to money-making thingies.

samcarter

11:37 AM

@marmot Oh, your marmot can fly :)

@marmot Try \marmot[ask phil] with github.com/samcarter8/tikzmarmots/blob/master/tikzmarmots.sty

(for now it is just a simple ellipse, need to figure out something to bring your marmot back to the ground :)

@marmot Can you ask your cousin Phil what the probability is that he sees his shadow?

Found a statistic: "According to the Groundhog Club's records, the various incarnations of Punxsutawney Phil have predicted 103 forecasts of more winter and 17 early springs. "

marmot

1:15 PM

@samcarter Jump, not fly
\documentclass[tikz,border={1cm 3.14mm 1cm 3.14mm}]{standalone}
\usepackage{tikzmarmots3d}
\usetikzlibrary{shadows.blur}
\newsavebox\Marmot
\sbox\Marmot{\tikz{\marmot[whiskers,teeth]}}
\begin{document}
\foreach \X in {0,10,...,350}
{\pgfmathsetmacro{\Jump}{0.15*(0.7+sin(\X))}
\begin{tikzpicture}
\path[use as bounding box] (-0.5,-1) rectangle (3,3);
\begin{scope}[xshift={cos(30)*\Jump},yshift={sin(30)*\Jump}]
\fill[gray,opacity=0,blur shadow] (1.5,0.2) arc(60:405:8mm and 2mm)

@samcarter Sooo nice!!! (I am wondering if one could make the shadow angle depend on the local time.... just kidding ;-)

samcarter

@marmot :) Unfortunately the azimuth angle of the users screen is not known

@marmot From the jumping speed I assume it is a Lunar marmot - Is this the explanation for the many craters? Are they entrances to marmot burrows?

boycott.se - yo'

@egreg ... stress $\sigma\textsubscript{a} [MPa]$

egreg

@boycott.se-yo' I keep seeing this. :-(

marmot

@samcarter Hmmh, if you adjust -delay in your convert (s)he will become a terrestic marmot ;-)

@samcarter Good point. The LaTeX gurus have to work on that. My iPad also knows if I am holding it upside-down ;-)

samcarter

@marmot I already reduced the -delay to 0.

barbara beeton

1:29 PM

@marmot -- looks like you've accomplished levitation. (even the jumping image looks more like you're levitating. or maybe being hauled up by an unseen string. no evidence of a gravitational effect. apologies for nitpicking.)

marmot

@samcarter Here is -delay 2. Perhaps -delay 0 doesn't work?

samcarter

@marmot Oh, thanks!

marmot

2

@samcarter I realized I screwed up. Needs to be \begin{scope}[xshift={cos(30)*\Jump*1cm},yshift={sin(30)*\Jump*1cm}], i.e. one has to multiply by 1cm since otherwise TikZ uses points

@barbarabeeton You were watching the slow motion version. ;-)

barbara beeton

@marmot -- yes. this one's much more compelling. good job!

samcarter

@marmot This is so great! I could watch him for ages!

samcarter

1:47 PM

3

Paulo Cereda

@samcarter ooh a balloon

marmot

@samcarter How do you know the marmot is a he? ;-)

TeXnician

@marmot You mean it should be it, because of "das Murmeltier"?

marmot

@TeXnician No, marmots are either she or he.

Paulo Cereda

@marmot \usepackage{heshe}

samcarter

2:05 PM

@PauloCereda Here is also one for you:

6

Paulo Cereda

@samcarter yay

samcarter

@marmot It was faster to type and I was under time pressure to edit the message in the allowed time :)

Ulrike Fischer

@samcarter this looks as if the ballon is coming out of the navel.

Skillmon

@marmot shouldn't it be a square function to calculate the height, not a sine?

marmot

@Skillmon What do you mean? A square function does not oscillate. But if you ask whether I solved the equations of motion, the answer is no.

Harald Hanche-Olsen

2:32 PM

@marmot While airborne, it should be a parabola, which I suppose is what @Skillmon meant by square function. While feet are on the ground, who knows what is an appropriate function? Perhaps another parabola, if the jumper is pushing against the ground with a constant force. Or perhaps like a piece of a sine function, if legs act more like springs. I imagine cangaroo legs are rather like that.

Skillmon

@marmot basically what @HaraldHanche-Olsen wrote :) square function is the wrong term, I guess, stupid English maths.

marmot

@HaraldHanche-Olsen I agree with the parabola part. However, if you also add the acceleration in form of another parabola, then the result agrees with a sine to second order in a Taylor expansion.

Harald Hanche-Olsen

@marmot Sure, just be careful what point you expand the Taylor series around. After all, at t=0, the quadratic term in the Taylor series of sin(t) is zero. If you expand around t=π/2, the quadratic approximation will be better, and probably hard to distinguish from the sine in the first half period (0 to π). A different issue, though, is how energetic the jumper is. The acceleration profile while on the ground could be greatly different from when airborne.

marmot

2:49 PM

@HaraldHanche-Olsen Absolutely. It also becomes a biological question: how do marmots jump? ;-)

3

3:07 PM

@UlrikeFischer That only gives me an obscure error message (An error occurred.)

@marmot Well, given relatively short legs, if they are to jump to a reaonable height, the acceleration on the ground must surely peak at more than 1g.

Phelype Oleinik

3:23 PM

Google translate tells me that he's looking for other type of latex... tex.stackexchange.com/questions/446161/latex-anzug-unter-18

marmot

@HaraldHanche-Olsen Well, that's independent of the length of the legs. On the surface of the earth you will always have to overcome the gravitational pull.

Harald Hanche-Olsen

3:40 PM

@marmot My point is this: If you are to jump higher than your legs are long, then you must accelerate upwards at more than 1g, simply because you have less distance available in which to reach the required velocity. And since you already need to overcome gravity, as you say, the conclusion is that at that point, your legs must provide a force greater than what is needed to support twice your weight.

marmot

@HaraldHanche-Olsen I agree with almost everything except for "twice". 1+\varepsilon is enough. Then you have a net acceleration of \varepsilon

Harald Hanche-Olsen

@marmot If your leg length is s, and you provide an acceleration a, starting from a standstill, then at the time t satisfying s=½at^2, you legs will have left the ground, and you cannot accelerate further. You are now accelerating down at a rate g. (Let us put g=1, as you seem to have done, to simplify the argument. This can always be done by appropriate scaling. [to be continued]

[…] At this point you have an upwards velocity v=at. The two equations give s=½v^2/a. This will propel you to a height h given by h=½v^2/g=½v^2 (g=1, remember?). So h=as, and you need a>1 to get h>s.

marmot

3:58 PM

@HaraldHanche-Olsen The pull to earth is F=-m g. To overcompensate it, we need F_\mathrm{jump} = m a with a > g.

samcarter

2

^^^ A post-honey-liquor-marmot

marmot

@samcarter Oh my. (S)he has drunk a lot, it seems. ;-)

boycott.se - yo'

@egreg Please, do you know the reason why Springer math articles look so ugly?

Ulrike Fischer

@samcarter One could add some ding-dong sound ;-).

2

samcarter

4:13 PM

@UlrikeFischer Indeed :)

Harald Hanche-Olsen

@marmot I totally agree. What I seem to be unable to communicate, is that my calculation is about trying to jump higher than your legs are long. Or to be more precise, to raise your center of gravity higher than the point where your feet leave the ground because your legs don't reach anymore, plus the length of said legs. Sorry if I did not make that clear.

marmot

4:30 PM

@HaraldHanche-Olsen I see. But even then it is more tricky. Let's assume that the legs excert a constant force over some time \Delta t. Then the velocity at the end of the "marmot is using her or his legs" period is v_0=(a-g) \cdot \Delta t. Then (s)he will feel the gravitational pull only and the velocity is v=v_0-g t. Obviously, the marmot will start falling down when v=0, i.e. t_1=v_0/g. This is when the initial kinetic energy equals what is usually called potential energy, [cont'd]

[cont's] i.e. m g h= \frac{1}{2}m v_0^2. Can you tell me now how h is related to the length of the legs? (Sorry if I am doing something stupid.)

egreg

@boycott.se-yo' No. But books used to be well printed.

boycott.se - yo'

4:48 PM

@egreg well, it's a strange mixture of fonts with ridiculous design decisions...

Harald Hanche-Olsen

@marmot I'll be a bit busy now; will read what you wrote and follow up later.

marmot

@samcarter It looks as if there was a large interest in \usebeamertheme{Marmots} ;-)

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TeX, LaTeX and Friends

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