I don’t do conventions very often, but I recently went to ConBust out in Northampton, MA, while visiting some friends. While I was there, I had a guy propose something fascinating to me. I can’t remember the guy’s name, so if he or one of his friends sees this, post your info in the comments. (Edit: it was a dude by name of Thom Howe.)
The guy Thom had an idea for a date. He wanted to rent a cherry picker, drive it to her door, and pick her up in it.
Then, he’d drive to the beach, and get there at just the right time to watch the sun set.
Once the sun had set, he’d activate the cherry picker, they’d be lifted up above the beach …
… and they’d watch the sun set again.
Clearly, this is an excellent idea, and any girl would be lucky to see this guy Thom at her door. But is it plausible? How fast and how high does the cherry picker have to go?
I tried to work out the answer for him there at the table, but there was a line of people and there wasn’t time. But when I got home, I remembered it again, and I’ve worked out the solution.
Here’s the situation:
By the time the earth has rotated through angle theta, the cherry picker will have to have climbed to height h.
After t seconds, theta in radians is:
The height of the lift above the center of the earth is:
So the height above the surface (sea level) is:
Substituting everything so far we get this expression for the height the lift needs to reach t seconds after sunset to stay even with the sun.
Now, an actual cherry picker has a maximum lift rate (I Googled some random cherry picker specs, and 0.3 m/s is a normal enough top lift rate.) We’ll call that rate v, so the actual height of the lift will be this:
Substituting that in and solving for v, we get this:
(That’s arcsecant, not arcsecond). This equation tells us how fast the lift has to go to get from the ground to height h in time for the sunset1.
But we can also get the answer by just trying a few different heights. We plug it in to Google Calculator2:
2*pi*6 meters/(day*arcsec(6 meters/(radius of earth)+1))
and find that h=6 meters gives about the right speed. So, given a standard cherry picker, he’ll get his second sunset when they’re about six meters up, 20 seconds later.
You might notice that I’m ignoring the fact that he’s not starting at sea level — he’s a couple meters above it. This is actually pretty significant, since the sunset line accelerates upward, and it brings down his second-sunset height quite a bit. If he got a faster lift, or used an elevator, the correction would become less necessary. Extra credit3 for anyone who wants to derive the expression for the height of the second sunset given the lift speed and height of first sunset. For now, I recommend he dig a hole in the sand and park the lift in it, so their eyes are about at sea level4.
1 Ideally, we’d solve for h, but it’s inside the arcsec and that looks like it’s probably hard. Do one of you wizards with Maple or Mathematica wanna find the result?
2 If you work in one of the physical sciences and don’t use Google Calculator for all your evaluatin’, you’re missing out. I wish there were a command-line version so I could more easily look/scroll through my history. I know Google Calculator is largely a frontend to the unix tool units, but it’s better than units and available everywhere.
3 Redeemable for regular credit, which is not redeemable for anything.
4 I suggest a day when there aren’t many waves.
xkcd 
R is a lucky lady, man. ; )
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I want to do that, so very very badly now. Off to google local rental places!
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Give that man a free sexual favor for his brilliant idea.
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Frink is an offline calculator, which actually has a command line interface (which is available offline). I find that Frink is almost as good as google calculator, though it has a few annoying properties.
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Whats about the flattening of the earth? If you want to imply the flattening (lets say 1:298), you might want to establish geographic latitude as a new parameter 🙂
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No computer algebra system is necessary. Theta (and h/r) is small, so just use Taylor expansions.
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Excellent. I shall make use of math and my cherry-picker permit. I knew that was a handy thing to have.
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You don’t need an elevator to see the sunset twice. Just watch the first one while lying on the beach, then stand up. (Tested.)
Also, I don’t recommend digging a hole. It will be wet.
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This idea is so charming it just might get you out of getting a ticket for driving a cherry picker onto the beach.
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Can somebody please write an extension for LaTeX that generates equations in Randall’s handwriting?
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Also, the sequence of button-commands to make a motorized lift go down is typically more complex than those required to make it go up. You usually hold one button to go up, but have to hold/turn two buttons to go down (as some sort of precaution)
Getting stuck in the dark might actually be a plausible scenario.
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Command line google calculator? http://johnbokma.com/perl/google-calculator.html or http://search.cpan.org/~typester/WWW-Google-Calculator-0.05/lib/WWW/Google/Calculator.pm
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Very mathematical. 🙂
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I once landed in a plane at exactly sunset, and was amazed at how fast the sun seemed to go down as the plane landed.. so yes scenario passes the smell test for me.
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I wish someone would make google calculator into a separate application, I hate when I try to use it, but it interprets my query as a search rather than a calculation.
Using mathematica’s Series function to approximate, I got h = (day)^2 v^2 / (2 pi^2 r)
For v = .3, you get h = 5.33641
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Fantastic idea. Loved your analysis, too.
Graham, the following site:
http://www.soople.com/soople_intcalchome.php
is apparently some sort of a frontend to google calendar, and in the generated link appends “&sa=Calculate!” to the query. Maybe you should try it.
(Also this site: http://www.xnumber.com/google_calc.htm)
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http://www.google.com/search?hl=en&q=2*pi*5.33641+meters%2F(day*arcsec(5.33641+meters%2F(radius+of+earth)%2B1))&btnG=Search
However its obvious that a higher v will give a higher h, which doesnt quite make sense as far as the real situation goes. That is because you are starting moving up as soon as the line of the sun passes you and continue moving until the line accelerates and meets you again, so really if you are that good with your reflexes, any height would give you a second sunset. The problem is more interesting if you wait, then try to meet the sun again.
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It seems that your formula does not take into account that you have to actually not only overcome the relative sun’s movement due to the turning of the earth, but also to move even faster so that the sun rises over the horizon again in order to see the sunset for a second time.
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I remember back in 2005, laying on a beach with my new girlfriend on Île d’Oléron, a small island off the west coast of France. The sun set. I said, “Quick, let’s stand up and we’ll see the top of the sun set again!”
We stood up quickly.
We didn’t see anything so I presumed I was talking bollocks, and have never tried it again. But the experience stayed with me, and I make a slight reference to it in a song I wrote about that first trip around France with my girlfriend (now together 4 years): Leave Montpelier
PS. I think the further north or south you are from the equator the slower you’d need to stand up. I base this only on anecdotal evidence on my various experiences of sunsets crossing the equator and seeing two sunsets in the same day in Antarctica.
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Well, seeing as h/r is really small, the expression simplifies to h = v*day, which is about as simple as a quick calculation carried out in my head can get.
It works out too: v*day = 0.3 * 24 ~ 7 meters.
The primary motivator for approximation is laziness, not complexity!
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Man, this is so awesome. I am saving this page.
Of course, there’s still the problem of how a cherry picker drives on sand if there isn’t any concrete (as is the case with most beaches I know of).
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Ah. That would be wrong. arcsec has a slope of near infinity around 1, so the correction to 1+h/r is crucial.
Neglecting the correction actually gives v*day = 0.3*24*3600 ~ 26 km.
Man! I’ll have to reach for pen and paper now.
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There exists a command-line Google, alas, the calculator function doesn’t actually query Google and therefore rarely works.
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He could drive forward while raising to reduce the required lift rate. Note that this is probably against the safety operating manual.
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Are you taking into account the refraction of the light by the Earth’s atmosphere? Refraction causes the sun to appear flattened when it gets close to the horizon, because there’s significantly more atmosphere for the light from the bottom of the sun to pass through in comparision to the light from the top. If you don’t think it’s significant, watch a film of a sunset sometime, such as a nature documentary. You’ll notice that the sun is far from a perfect circle.
I grew up in Seattle, so I’m used to seeing the sun set with the Olympic mountains in the background.
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This idea is hardly new my roommate (a pilot) has been taking dates to watch the sunset and then surprises them with a little flight to 10,000 feet for sunset round 2 for years
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Finally, my CERN safety certification for operating a platform at height will come in handy. Now if I could only get one out of the cavern…
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MathCAD will handle all your unit calculating needs – you can even save programs to use them again later. Engineers love it because half of the mistakes made in engineering are not putting the number in to your equation in correct units. With MathCAD you simply write your equation and just plug in the variables (in whatever units you were given the variables in) and the program will spit out the simplest unit, or one of your choosing.
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A few items:
-Several years ago I derived a similar calculation to determine the distance to the horizon as a function of one’s height above sea level. As has been suggested, the equation is extremely sensitive for small values of h; just a few feet above sea level theoretically allows one to see for miles.
-A friend of mine went skydiving once, a tandem jump. They left the plane at 10,000 feet, and he got to watch the sun set rather rapidly as he freefell to about 4,000 feet before popping the chute.
-Another suggestion for math software, a program called EES:
http://fchart.com/ees/ees.shtml
One of the really nice things is that it’s got built-in physical/thermodynamic properties for a whole array of substances. Very handy.
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At my grandparents’ house we could see the sun set from the porch. Close to that moment, someone would run upstairs. We’d say “going, going, gone!” on the porch and then the upstairs person would call out a few seconds later… Thanks for reminding me. (My grandfather also had the location of the sunset for different dates marked off in angles on the porch railing.)
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A wee McLaurin expansion gives height h (in metres) needed after t minutes:
h = R (pi/720)^2 t^2
in other words,
h = 60.9 t^2 (accurate to 3 sig fig up to at least t=10 minutes)
The sun takes about 4 minutes to actually set (at 50 degrees latitude – see http://bit.ly/LeGjE ). You’d want the whole show, wouldn’t you? After 4 minutes, you’d need to jump instantanously to an altitude of 974m to catch the start of the show again.
Assuming you don’t want to subject your date to g-forces in excess of 0.5, you could still whizz up to arrive at rest at 1251m above sea-level in 32 seconds (catching a very quick sunrise on the way) in time for a second show.
So we’re looking at a cherry-picker with a three-quarter mile arm and vertical acceleration of 0 to 175mph in 16 seconds.
It’d certainly be a cool ride…
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This reminds me of the story of professor Minnaert, who once observed the green flash at sun set for 20 seconds, by running up a 6 meter high dike.
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re my calc above –
There’s also a half missing from the first formula for h, sorry.
R is the radius of the earth, which averages about 6370km. The 60.9 should really be 60.6, with the proper figures. It only affects the conclusions by a percent or so.
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The screen scraping feels gross, but urlencoding input, passing it to google vial curl and sucking out the result that follows the image of the calculator shouldn’t be too hard – depending on how often they update. Anybody want to write the script?
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The other option here is to find a restaurant on the beach that has a rooftop deck from which to drink margaritas and simply make a quick dash upstairs, rather than going to all the trouble with the cherry picker.
Also, are you forgetting the sun is not a point light source? The time from when the first part of the sun touches the horizon to when the last part disappears below it is significant (read, “more than 20 seconds”).
So all this would allow Romeo to do is to “replay” the last few seconds of the sunset, or extend it for 20 seconds more.
(However, it’s all romantic as hell and the real problem being solved here is all about hormones, not physics… and I’m pretty sure the cherry picker option is the optimal solution there.)
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I got 23.4 m, and I took a eye-height of 1.8 m and the diameter of the sun in account. But maybe I just screwed up…
Here is my octave/Matlab programm:
y = sin(asin(R/ (R + 1.8))+2 * pi * 0.53 / 360- 2 * pi * h / (24 *60*60*0.3)) – R/(R + h + 1.8);
Where 0.53 is the diameter of the sun (32′).
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the sea level problem could be solved by renting a barge and driving the cherry picker onto a barge and floating out somewhat (about 400 miles should do it) into the atlantic.
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(reposting, as last one didn’t appear, probably because I linked to a sunset-duration website, and you know how dodgy they are)
A quick McLaurin expansion gives height h (in metres) needed after t minutes:
h = (1/2) R (pi/720)^2 t^2
where R is the radius of the earth, about 6370km (on average).
This gives
h = 60.6 t^2
(this is ok to 3 sig fig up to about t=10 minutes)
The sun takes about 4 minutes to actually set (at 50 degrees latitude). You’d want the whole show, wouldn’t you? After 4 minutes, you’d need to jump instantanously to an altitude of 970m to catch the start of the show again.
Assuming you don’t want to subject your date to g-forces in excess of 0.5, you could still whizz up to arrive at rest at 1245m above sea-level in 32 seconds (catching a very quick sunrise on the way) in time for a second show.
So we’re looking at a cherry-picker with a three-quarter mile arm and vertical acceleration of 0 to 175mph in 16 seconds.
It’d certainly be a cool ride…
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Less romantic, but more hardcore – Take cherry picker to artic circle around the start of august and watch all the sunsets you like until the battery goes dead.
Alternatively, giant trampoline on beach. I reckon that would even beat the cherry picker – trampolines are great. And also… bouncy.
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Just noticed there’s “extra credit” for a derivation involving lift speed. Top banana!
I thought a constant acceleration would be more appropriate – and I used g/2 rather than leave it general – so maybe I don’t qualify for extra credit after all.
If you want a general speed or a general acceleration, the formula will be one of those awful quadratic solutions with the speed or acceleration nested within the square root. I’m not sure it’d be very illuminating.
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It’s a really cool idea, though I’m just lazy and cheap enough to do the same thing by watching the first sunset and then rushing to the elevator of a really tall building nearby. I wonder if we could make it in time…
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I reckon that if you wait for 24 hours or so…
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Doing this in Death Valley, where the mountains around you make the apparent horizon occur much higher in the sky than at sea level without mountains would reduce the needed height and speed to see the second or even third sunsets. The sky itself would still be bright, too, after the last apparent sunset, so you can still get down and hop back on your brontosaurus.
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Daniel Snyder is correct – you should probably take into the account the refractive index of the atmosphere, and how it changes the apparent rate of sunset. The good news is that it will allow you a little more time for the cherry picker to get up to h. So, within your approximation, your probability of scoring remains the same. As a side note, If Mr. Cherry Picker could derive the full calculation on the way up, I’m sure that would also garner extra points with any woman impressed by this demonstration in the first place.
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You’re right. I’m going to build a CLI front end to Google Units.
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I grew up in a house in the mountains, facing west. It was a three story house, and when the sun set you could run up the stairs to each level and watch it set again – twice.
It’s a really cool feeling.
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LaTeX is for wussies. It’s all about scanning in your own hand written expressions.
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Ok – for the extra credit –
If you solve, for h2, the quadratic
h2 = 60.6 ( sqrt(h1/60.6) + T + 60(h2-h1)/v )^2
which a mathy program would do quite happily, that would give you the second height, h2, in metres, at which to view the sunset the second time after spending T minutes watching the first one at height h1 and then ascending at v metres per second.
The 60.6 is 1/2(earth radius)*(pi/720)^2, and the 60 converts between minutes and seconds.
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This site tells you sunset durations, including corrections for atmospheric distortion:
individual.utoronto.ca/kalendis/twilight/sun-rise-set-durations.pdf
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Maple says:
h = csc(RootOf(2 _Z v day sin(_Z) – Pi v day sin(_Z) + 4 Pi r[e] – 4 Pi r sin(_Z))) r – r
Sooo, gross.
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