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 
Mathematica cannot determine h in terms of v exactly – it gives the error
Solve::tdep: The equations appear to involve the variables to be solved for in an essentially non-algebraic way. >>
However, it is possible to do a Taylor approximation.
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Read the latest, but not all, the comments: Key features missing in the equations I saw:
They tend to assume you’re at the equator, on a one of the solstices. Depending on latitude, the sun could be as much as 112 degrees off of vertical. (At either pole, during the winter (for that hemisphere) solstice).
The tendency has also been to assume that the oceans form a sphere; The variance, however, is within the precision of our other measurements.
Finally, they also neglect the value of any horizontal movement the cherry-picker might be capable of. I will continue in this, since I think the effect will be negative when compared to the lower vertical movement.
I figure it’s only worth it if I can watch the entire sun go down, twice.
What we need is the delta theta/delta time for the sun, as a function of latitude and day of the year and the delta theta/delta height, which should only be a function of the Earth’s diameter and current height.
Consider the situation where two observers on the surface of an unmoving spherical sun observe a rotating spherical Earth with pins in the surface, sticking straight out. First, how long would the pins have to be for all observers to see the tip, but none see the base, at any time?
1 391 000 kilometers: D(sun)
12 756.2 kilometers: D(Earth)
149 598 000 kilometers: Distance between sun and earth.
Just on the estimate, I don’t think there’s a cherry-picker with the height capability to do it at any speed, at sea level.
Different idea: Instead of sunset over the ocean, sunset under a mountain. Given a cherry-picker of known capabilities, what geometries of mountain provide suitable conditions, and where are those conditions exemplified?
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Yeah, I’ve seen this with rock face plateaus.
Sunset #1: behind the rock face.
10 minute drive to the other side.
Sunset #2.
(Now if only the cherry picker was already set up on the other side… hmmm)
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So, uh, why go through the trouble and expense of a cherry picker which is probably dirty and beat up since it is a rental. Why not watch the sunset. Then find a tall building and take the elevator up to the observation deck. May I suggest the John Hancock Building in Boston. You could watch the sunset from the plaza then race in and head to the top. I suggest getting your tickets first.
Then take your date to some swanky Beacon Street resturant, have lobster and then later make love like sea otters. If you are lucky you’ll get to relive that day over and over again if a weather predicting squirrel sees it’s shadow.
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your equation assumes he wants to be at H after the earth has gone through rotation theta, and he starts the cherry picker at t=0.
He doesn’t.
Sunset starts at t=0
He starts the cherry picker at t1 (when sunset is over)
He wants to be at h when t2
N(0) = Re
N(t1) = Re
N(t2) = Re + h
You need to know the duration of the first sunset to figure out how high he needs to be to see the start of the second sunset.
I believe what you calculated is the height at which he would be seeing the same part of the sunset that he was seeing at n=re.
You could use the apparent angular height of the sun Gamma = 2* arctan (sundiam/distance to sun) and the distance to the horizon at sea level for your location. Then, as the height of the cherry picker changes, the distance to the horizon changes. As long as the angular change of the view of horizon (based on cherry picker velocity) is greater than the rotation speed of the earth, you will be able to “rewind” the sunset.
Unfortunately, the distortion of the atmosphere is another factor which is hard to account.
Simplest way to check if this works is to get on a West facing ferris wheel at sunset and see it that works.
If it doesn’t, then a cherry picker isn’t going to be any better.
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Maybe, I should try this out first with a fellow hopeless romantic ?
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Never have I seen mathematics be more romantic . . . I’m seriously moved . . . .
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If you have a fast enougth plane its actually possable to chase the sunset. I’ve had the unique experince of doing this once in a RAF Jaguar and I have to tell you it’s truely exceptional, the lighting, the way you can make the sun settle as a huge firey orb on the horizon and either make it rise by speeding up or set by slowing down. This is the reason why many military transatlantic flights don’t come in on their expected ETA as the pilots are having just to much fun in that entierly magical zone, having the darkness behind you and the sun in front and the knowledge that you can outrun the dark is just, well there arn’t words, you feel like a god.
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best idea ever! speaking of which, i found a link of your chess-board-on-a-rollercoaster idea. it may or may not be based off of your comic, but i sent it to ya nonetheless. http://pics.livejournal.com/athelind/pic/00005bc1
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I think some sort of harness-and-pulley system could shoot you up faster than a cherry-picker.
Also, the ideal combination is to watch the sun set behind a rock face, drive past it, watch it again at the adjacent ocean, go up the cherry picker, watch for a third time, then zip back down and hop in a plane.
Alternately, make a sweet deal with Ra/Helios to manipulate the sun at your whim.
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They have that “Slingshot” ride that might do the trick. It’ll launch a person about 30 meters in three seconds or so.
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o.k. i thought i was a geek, while you guys have been debating how to get in a second sunset w/o getting a second date. i picked up that girl, partied a little, showed her my tattoo and told her that some people i knew were working out a way to see two sunsets in a day. ‘sunsets?’ she said, ‘i’d rather see your place,’ and she did. the moral of the story: it’s romance, don’t over-think it. if you’re building a bridge, dropping mortar rounds on an enemy position or putting a manned craft into orbit break out the trig and calc. but on a date leave that stuff at home. sure she’s glad you can do it, that’s not the same as her being interested in you doing it.
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you also missed the diffraction of the sun rays, when you see the sun go down, it is actually after the sun has went down, if that makes any sense. The light is “bent” (not really the right word) and it makes it so you see light even after the sun is below the equator
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I tried to do the shell script thing, but I only got this far:
#!/bin/bash
str=”http://www.google.bg/search?q=”
curl -A “Mozilla/4.0” $str$1 | grep “=”
the $1 param is given with gcalc “2+2”.
the problem is that the entire search page seems to be in 10 lines, so we need some way to cut out the part of it that contains the “2 + 2 = 4”. AFAIK this cannot be done with grep. With my limited knowledge of shell scripts, I can’t find a way to cut out parts of a string. it should be possible, since the part we want is surrounded by a . Any suggestions are welcome 🙂
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Meanwhile all you nerdgeeks are sitting here debating all the math and variables for this instance not getting laid at all, so its a moot point to begin with…..and im scorin’ with the chiks your only dreaming about…HAHAhahaha
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Thom Howe sounds suspiciously like somehow.
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I’m a bit of a novice with Mathematica, but here’s when solving for “h”, here’s what I got:
h = 5.354115518442182
Does that look legit to you guys?
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Lots of good comments, very analytical… But the key part that everyone is missing is that the conversation happened in Massachusetts. The sun sets over LAND here, not the ocean!!!
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@Thom: Inland, sure. But you certainly can see an ocean sunset on the Vineyard, Nantucket, and various parts of the cape.
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I think romantics already tried this way to often. The real challenge is trying to get as much sunsets in a given timeframe.
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If people want to be picky about things not accounted for here, I think the most important unconsidered effect is that of refraction through the atmosphere. It is this effect that makes the sun look massive and red as it sets.
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As much as the sentiment is thoughtful, and a truly moving way to do something romantic, it is the short length of the beauty of the sunset that makes it so magical. That being said, lengthening the time of the phenomenon reduces the rate of impressiveness/second enjoyed. Taking the law of diminishing returns into consideration we have to appreciate that the idea, though unique, has a finite gain and temps to hinder the pantie dropping effect of the sunset on the female in question.
Although, if the ultimate purpose for such exercises is to just to prove how awesome you are, whether by fork lift, elevator, airplane, hoverboard…. or cannon… (think Pilot Wings 64) then I say to hell with all that macho skirt chasing! GO SCIENCE!
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The proper equation to work out here is how quickly she’d run from your ass when she sees you pulling such a dorky stunt.
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I just put together a quick perl script for a command line google calculator(get it at http://retupmoca.com/google-calc.pl):
% ./google-calc.pl ‘2*pi*6 meters/(day*arcsec(6 meters/(radius of earth)+1))’
doing google search for 2*pi*6 meters/(day*arcsec(6 meters/(radius of earth)+1))
calculator result: (2 * pi * (6 meters)) / (day * arcsec(((6 meters) / radius of Earth) + 1)) = 0.318106531 m / s
%
Should work as long as you have perl and HTTP::Lite installed.
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holy shit, you’re so cool.
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Hey, if she can’t appreciate the thought and effort that went into making sure the cherry picker would do what I wanted it to, I don’t want a third date with her, anyway. If I wanted casual sex, I’d be hanging out in a casual sex bar.
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My wife and I actually did this in Aruba on our honeymoon, but did it by running up the stairs of our hotel. We were on the second floor, watched the sunset from our balcony, then sprinted to the top floor to watch it again from the roof. It was SPECTACULAR!
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I did the Google Calculator thing in Perl. Kind of a hack, but whatever, it’ll get it done.
#!/usr/bin/perl -w
# gcalc.pl
# Jon Purdy
# Commandline interface for Google Calculator.
# Copyleft (L) 2009 Jon Purdy. All rights reversed.
# Distribute as freely as the wind in the trees.
# Credit me if you’re not a meanie.
use LWP::UserAgent;
use URI::Escape;
$expr = uri_escape($ARGV[0]);
$agent = LWP::UserAgent->new;
$agent->agent(“Google Calculator Commandline Frontend/$0 “);
$request = HTTP::Request->new(GET => “http://www.google.com/search?hl=en&q=$expr”);
$result = $agent->request($req);
$page = $result->content;
if ($page =~ /]+>([^<]+)/i) {
print “$1n”;
} else {
print “No result.n”;
}
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Failure, of course. Here, have a link: http://evincarofautumn.infogami.com/blog/030409
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Your probably going to waste more time trying to work it out, rather than actually trying it out! Any one with a spare cherry picker want to try this out?
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Check this one out:
http://hypftier.de/en/gc-%E2%80%93-a-command-line-interface-to-parts-of-google-calculator
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OK – I like the idea – being an old lady geekie person – but my first thought was – why the math? Lets do it the visual artsy way – I am suggesting a very old romantic notion – I will make time stand still for you honey – THAT is romantic in anyones book right? So you have your Cherry Picker and you and your guy are in it and the sun is going down – as the sun sinks, you raise the cherry picker up a notch – sink – raise- you get the idea… Then – you go back to the beach and whisper sweet nothings under a beach blanket about arcs and tangents
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Mi clase de japones, esta comentando la cancion “chuchu ua” de piñon fijo…. Cualquiera que desee dejar su opinion al respecto, haga el favor de hacerse violar por una horda de vikingos hambrientos.
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Sigh. If only he would leave his dumb cherry picking toy at home, we could have had a great naked time on the beach watching the sunset.
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This is the sweetest thing.
I opt to be naked both on the beach, and in the cherry picker, watching the sunset twice…
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Would it not be better to have a silent cherry picker that rose at the same rate as the sun fell in order to give the illusion of one longer sunset? The sun’s position would appear not to change until the maximum range of the cherry picker was reached. Is that possible?
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There was (maybe still is) a cruise ship that stopped in Tonga on Christmas day, then zipped up to Samoa (on the other side of the International Date Line) for a second Christmas. Don’t know if this is relevant. No, it’s not. Forget I mentioned it.
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I’m a girl, and I would be HEAD OVER HEELS for a guy capable of doing all this 😀
Seriously, I’d marry him!
(But only after discussing with him a way to see two sunrises)
Stupid girls wouldn’t care nor understand all the thought put into this. Plus, they’d complain about the cherrypicker o_O’
(Yes, I’m saying that I’m smart HAHAHA)
Did you guys sent this to Mythbusters? XD
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Can a cherry-picker even drive on sand?
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How do u get that on the beach? Solve that one
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I think some things are special because they are rare. You don’t really need to watch the sunset twice in that short span of a time. It defeats the whole point, I think. Besides, I’d rather be on a blanket on the sand or cliff watching this romantic sunset on a date, than on a cherrypicker bucket!!
But my heart probably would skip a beat if a guy tried to show me the sunset twice, hoping that it might make me smile. 🙂 Kinda like catching the moon, I guess ;>
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I have watched at least a third of a sunset over again by running upstairs, in case anyone needs an anecdote to support the math.
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Anyone wondering about the off-road capabilities of cherry-pickers:
Do you really believe that nobody with money needs an off-road cherry-picker? You can get anything useful in construction work in an off-road version.
Hrm… a boom truck with a manlift cage would have a greater maximum height, and probably a better lift rate, but would require modifications to be controlled from inside the lift.
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This is great, but the date might have to span over a few days for the guy to take the girl to the beach in a cherry-picker! Unless the beach was close, but if it was that close he would probably have tried to find somewhere else for the date…
lol v. clever though.
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I see a problem with this idea of using a standard height above “sea level” since the point at sea level, when refered to as a distance from the earth’s center of mass, changes depending on where you are on the earth.
When dealing with meter level accuracy on the earth surface, you run into the problem where the earth is closes to the shape of an ellipsoid (a perfectly defined mathematical surface) than a sphere. Also, the geoid, which is the equipotential surface on the earth that used people define as sea level varies by several meters relative to the ellipsoid.
Going deeper into the concept of “sea level”, what is commonly used for definining the “sea level” in a physical point on the coast is more technically a mean water level. By this definition, it’s an average (mean) of all hourly tidal heights observations available for a given location. This means that the very definition of a more technically correct mean water level changes over time. Another problem with using mean water level is that different locations will have very different mean water levels due prevailing wind, evapouration, precipitation, freezing, melting, heating, cooling and sea slopes caused by the Coriolis effect.
What if you used a mean sea level for you calculations? This height does not change as much, at least not yet since the oceans haven’t been rising at an enormous rate at the moment. However, even so, this mean of all ocean surface heights will cause problems in that you still won’t be at water’s edge for every single location on the earth for which you may want to watch the sunrise twice. In such cases, you would still need check the tidal charts for your big day and make the appropriate height corrections which will influence on how fast and high you need to raise your cherry picker.
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use a trampoline !! then you could see it a few times ….
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Billy Chan makes a lot of good points about the shape of the earth. None of them is especially pertinent to the geometry of the problem, though. The eccentricity of the ellipsoid is small enough not to matter for the local area of the sunset, as it is the same for each of the lower and higher positions of the cherry picker. That is, comparing the two cases with each other cancels out most of the variations due to Earth’s shape.
Sea level in this case is only relevant within a dozen or so miles, which is small enough that it is relatively constant; the horizon and the shoreline are at about the same height. Even if they weren’t, the particular local conditions are being compared with themselves, slightly modified for the raised cherry picker, so any differences would cancel each other.
It seems the only two parameters affecting the required cherry picker movement needed are the distance to the horizon and the rotation speed of the Earth. The first is related to height above sea level and the second is constant. The distance to the horizon determines how far one needs to ascend to cross an angle defined by the width of the sun. The rotation speed of the earth determines how long one has to cross the angle.
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Google calendar is great. I use it just about every day. The only thing I wish it had were a random number generator. It would be very useful to be able to query random(a,b) and get a (pseudo) random number. I mostly want this so I can text a message to Google to help me make my life decisions for me.
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Ok, One biggg problem, you turn up with your cherry picker, Girl says yes ill be down in a minute
. ……. Time Passes …..
Turn up at beach in time to see the stars from a cherrypicker….
Not quite so cool.
Great math, but Girls cant be factored into this equation..
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You could always tow the cherrypicker to the beach using, say, a car. Then get out of the car, climb into the cherrypicker and watch both the sunsets.
Not quite as odd as cherry-pickering your way to her front door, but you still get the main effect.
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