A Date Idea Analyzed

Posted byRandall 487 Comments on A Date Idea Analyzed

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.

487 replies on “A Date Idea Analyzed”

  1. But doesn’t light bend around the surface? That is – when the sun has just gone over the horizon the light waves curve around the earth ever so slightly, but noticably, so the final light you see is actually after the sun has gone past a direct line. If that’s not a mistake – then surely this effect would be more pronounced close to the surface than away. Wouldn’t that make a difference to your calculations?

    More importantly, wouldn’t it be easier to just take the girl to the movies?

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  3. I dont understand a bit of that math, but I LLLLOOOOVVVVVEEEE the webcomic and anybody here who stumbled this and has never seen the webcomic should go to it immediately.the address is: xkcd.com

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  4. Sure, we could take our lensing effects and trig and do lots of math and get an answer that’s maybe in the right ballpark…

    Oooor we could go to the beach and watch the shadow climb the seawall. (It climbs pretty slow. This would work fine.)

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  5. I like the idea of the tall building and elevator. I’m sure there are some nice park benches near tall buildings with suites or rooftop verandas that would make for good viewing. I suspect an elevator would travel faster than a cherry picker.

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  6. I think this is a very nice idea, and while there are sections of the mathematics I don’t know shit about, I hope to rent a cherry picker and try this. I’m thinking running up and down Mt. Mansfield won’t work as well.

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  7. I hate to point this out, but you need a better base on your cherry picker or the sun will set a little faster than you thought.

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  10. I’ve gotten on skydiving loads that take off just after the sun passes down below the horizon – the plane then ascends to ~13,000 feet, where the sun is pretty definitively not-set, and then watch it set again in freefall (with a better view of a realer horizon, as opposed to being on the ground where you’re subject to whatever treeline you happen to be closest to).
    So my initial answer was about to be “a couple thousand feet, maybe” – which would be a relatively impressive cherrypicker 😉

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  11. This reminds me of the story “How to Derive the Maximum Enjoyment from Soda Crackers” by Mason Williams (the brains behind the Smothers Brothers). It is another example of someone who will clearly never get a date dreaming of what they would do on a ridiculous date. And it’s wonderful. It could be an XKCD cartoon, easily.
    Full text is here:
    http://ribald.livejournal.com/2022.html

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  12. I guess just watching the sun set at the top of a building, and then taking a step to the side to watch it set at the horizon, isn’t enough… you have to see it hit the true horizon in the far distance. It’s better for lazy people, though.

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  13. WAIT WAIT WAIT – all that is assuming the earth is a circle. The earth is a sphere (sorry if you didn’t already know) and the spin is in no way even. So – let’s assume the guy did this trick in summer and is close to a pole. This would mean that the sunset is slow, and would also mean that there would be much more time to do the cherry picker lift.

    Taken to the extreeme – far enough north and you could watch the sunset all night long. A bit south of that and you could see the sunset, raise the cherry picker and watch it some more.

    Anyway – my point is that the maths is completely flawed.

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  14. @Jeff: Wrong coast.

    @Philip: the uneveness of the spin probably has little effect over these times and distances, the fact the earth’s spin is not perpendicular to its orbit is probably more of a problem…

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  16. A second thought:

    For a command-line version of google calculator, try SURFRAW(*). It’s a command-line wrapper for a number of useful websites. Including google, wikipedia and many others.

    The only problem I’ve encountered trying to use it with google calculator is how to correctly escape everything. (Backslashes? huh?)

    (*)It is in fact an acronym: “Shell Users’ Revolutionary Front Rage Against the Web”

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  17. @snig: Love the trebuchet!

    I want to see a calculation that takes into account the ~ 32 arcminute width of the disc of the Sun. The time it takes THAT to set is the sunset. In fact, I would calculate for that BEFORE I calculated in climb rate of the cherry picker. That would serve as a base-line necessary height assuming instantaneous height-attainment (The Nathan Petrelli baseline). Then you could insert a variable for lift-rate.

    The “lensing effect” of the atmosphere is a fallacy — the sun is still 32 arcminutes wide, even on the horizon. it’s an optical illusion. There is some refraction of the light due to the atmosphere (you can see the sun when it’s technically below the horizon) but that should not matter as the whole disk has to pass through the same refractive angle. But you have to remember sunset is 54 arcminutes after it would be if the earth had no atmosphere. (see below)

    @Phillip: The Sunset DOES vary greatly depending on latitude and longitude, and time of year. I personally would use an astrology program (like Astrolog32 freeware) to get the time for the sun to rise and set and then adjust the calculations for height. (Some of the fancier programs out there might have height calcs built in).

    Some relevant data from http://www.answers.com/topic/atmospheric-refraction

    ” Values

    “The atmospheric refraction is zero in the zenith, is less than 1? (one arcminute) at 45° altitude, still only 5? at 10° altitude, but then quickly increases when the horizon is approached. On the horizon itself it is about 34? (according to FW Bessel), just a little bit larger than the apparent size of the sun. Therefore if it appears that the setting sun is just above the horizon, in reality it has already set. Formulae to calculate the times of sunrise and sunset do not calculate the moment that the sun reaches altitude zero, but when its altitude is ?50?: 16? for the radius of the sun (solar positions are for the centre of the sun-disc, but sunrise and sunset usually refer to the appearance and disappearance of the upperlimb) plus 34? for the refraction. In the case of the Moon one should apply additional corrections for the horizontal parallax of the moon, its apparent diameter and its phase, although the latter is seldom done.

    “The refraction is also a function of temperature and pressure. The values given above are for 10 °C and 100.3 kPa. Add 1 % to the refraction for every 3 °C colder, subtract if hotter (hot air is less dense, and will therefore have less refraction). Add 1 % for every 0.9 kPa higher pressure, subtract if lower. Evidently day to day variations in the weather will affect the exact times of sunrise and sunset as well as moonrise and moonset, and for that reason are never given more accurately than to the nearest whole minute in the almanacs.

    “As the atmospheric refraction is 34? on the horizon itself, but only 29? above it, the setting or rising sun seems to be flattened by about 5? (about 1/6 of its apparent diameter).”

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  18. If my data and calculations are right, tonight, (at sea level) the Sun should take 2 minutes 30 seconds to set at Mission Beach, San Diego, CA, (7:11:02 – 7 13:32) and 2 minutes 56 seconds to set at Richmond Beach, Edmonds, WA (north of Seattle) (7:45:14 – 7:48:10)

    Anyone want to verify? (I’m in Baltimore/D.C.)

    Anyone want to calculate height variance?

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  19. Maple! Aie! My eyes! (I just remembered math classes from college, a long time ago, where the math department has the oldest computer screens and it was near impossible to use maple…)
    Zak

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  20. You’re considering the sun as a point, which means that when you view the second sunset, you’ll just get a tiny glimpse of the sun before it sinks below the horizon again. That’s hardly romantic at all. You need to take into consideration the angular size of the sun in the sky (about half a degree). So, not only do you need to stay even with the sun, you need to beat the sun by a relatively large degree.

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  21. I would like to meet the guy who figured this out so I can tell him to give me something for the head ache!!!!!!!

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  22. I don’t think the two sunsets would be worth all the effort.

    HOWEVER, a cherry picker at the beach would be worth it if you parked right at the water’s edge and extended it out over the water, so it’s just you and your date floating above the water a few feet and watching the sunset.

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  23. “Or, you could go to this one town in Canada almost directly on the arctic circle on June 21″…

    That should be possible for any point within the Arctic or Antarctic Circles, though the dates would vary. I was in Barrow, Alaska (approx. 71N, 155W) on January 2005, and for most of my trip we had a few hours of twilight, but no actual sunrise. On my next-to-last day there (Jan 25), I finally got to see the sun skim the horizon for maybe twenty minutes, then drop back down to begin another long night. At one, possibly two points during the summer, you could see the opposite: the sun would tap the horizon, then come back up.

    Not exactly the same, but fun and easily accessible is Fairbanks, Alaska (approx. 65N, 147W) and the Midnight Sun Baseball Classic, played on the night of the Summer Solstice.
    http://www.goldpanners.com/MidnightSunGame/index.html
    We have a pretty active summer league up here, so it’s actual college players from around the country, not just some random group of scrubs. The game starts at 10:30pm, and runs straight through until 1:00 or 2:00am with no artificial lighting. It gets a little dark around the seventh inning, but by the end of the game it’s bright and sunshiny again. Anchorage (61N, 150W) has lots of parties and festivals to mark the Solstice as well, but it actually does get dark briefly. Still a good option, though, if your primary objective is to get drunk and sing off-key with your arm draped around a stranger.

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  25. Huh. What an excellent idea. *scribbles it down*

    Yet another date idea I’ll never be able to use.

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  26. the calculus involved on your part in this idea is why he didnt take you on the date

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  28. Is it more feasible in an office tower? Like, a tall one with an ocean view? Um, where you’re the security guard with access to all the floors? …and offices?

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  30. I work in physical sciences and I still use Google calculator for a lot of my evaluatin’. One of my professors once said, “If I can use Google *and* Mathematica I’ll be twice as smart!”

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  32. Forget to mention, if you’re using that script on Linux, then ‘chmod +x gcalc’ it to allow it to be executed. Then the syntax is very simple:

    gcalc “1 hour 5 minutes in seconds”
    1 hour 5 minutes = 3900 seconds

    gcalc “100 + 2 * 5”
    100 + (2 * 5) = 110

    gcalc “41000 yen in british pounds”
    41,000 Japanese yen = 277.129372 British pounds

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  36. Anyone wonder why in an article about a cherry picker, we’re looking at what appears to be a forklift?

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  37. I think a more interesting problem would be “chasing the sunset” in a plane. How many times would you be able to climb altitude in your average 2-person prop plane to watch the sunset again?
    Also, would increasing forward acceleration help you “chase” it faster, or would it have no noticeable effect?
    How long would you be able to chase the sunset in your average Piper Cub? Leer jet? Commercial airbus?
    Bonus points for anyone who can do the math: I failed my last class, and that was Algebra 3 in high school.

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  40. I don’t care how “natural” they are, your thumbs and feet are not real units. just give it up.

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  41. The closer you get to the poles the easier this would be… I would suggest going on a date on a mountain in interior AK around the 21st of June. Then you can watch the sun set and rise on the same date. Of course not everyone thinks of a 10000 mile round trip is a “date”…

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  42. If you want to make it even more complicated, you should take in account the speed of light, and the extra distance, light has to travel to reach you, which would add practicly nothing. But still, for the geek factor, it’s worth noting.

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  43. One small problem I noticed: wouldn’t the wheels of the cherry picker get stuck in the sand? I suppose you could get one of those huge industrial ones…

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  44. eq := v = 2*Pi*h / (86400 * arcsec(h/6378100 + 1));
    Pi h
    v = —————————
    / 1
    43200 arcsec|——- h + 1|
    6378100 /
    fsolve(subs(v=0.3, eq), h);
    5.336403083

    So according to Maple, if the lift runs at 0.3 m/s, the maximum height needed will be 5.33 meters.

    evalf(subs(h=6, eq));
    v = 0.3180590352

    and if the lift goes to a height of 6 meters, it must do so at 0.318 m/s.

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  45. Oh my. This blog doesn’t do ASCII art very well, does it. Here it is in TeX:
    documentclass{article}
    usepackage{amsmath}
    DeclareMathOperator{arcsec}{arcsec}
    pagestyle{empty}
    begin{document}
    [
    eq := v = frac{pi h}{43200 arcsec left(frac{h}{6378100} + 1 right)}
    ]
    end{document}

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  46. The light’s path isn’t a straight line. Over water, the temperature differences in the water will refract the light more or less depending on the ambient vs. water temp.

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  48. One problem, the cherry picker in the picture looks to be of the same kind my theater uses. Unfortunately, this type of cherry picker cannot move it’s self, and you have to put in additional legs for it to ascend. However, this type of cherry picker would be perfect if you wanted to play a prank on the happy couple and trap them up there (we do this to initiate new techies) as there is an easily accessible button that will do just that. So, you could volunteer to push them to the beach, put the legs in, let them get however high, then trap them there for fun and profit.

    Also, there is a warning that states you should not have multiple people in the cherry picker, but I have disregarded about half of the warnings and am still alive.

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  49. I bet someone has done this before me, but here goes:

    #!/bin/bash

    # calculate url encoding: s/+/%2b/g etc.
    CHARS=”$-\$ &-& +-+ ,-, /-/ :-: ;-; =-= ?-? @-@”
    for CH in $CHARS ” “; do
    REPLACEWITH=”$(echo $CH |sed ‘s/.*-(.*)/1/’)”
    CH=”$(echo $CH |sed ‘s/(.*)-.*/1/’)”
    URLENCODE=”$URLENCODE s/$REPLACEWITH/%$(printf ‘%X’ “‘$CH”)/g;”
    done
    USERAGENT=’Mozilla/5.0 (X11; U; Linux i686; en-US; rv:1.9.0.2) Gecko/2008092313 Ubuntu/8.04 (hardy) Firefox/3.1’
    ANS=”$(echo -n “$@” |
    sed “$URLENCODE” |
    xargs -IQUERY wget -q -O – -U”$USERAGENT” “http://www.google.com/search?q=QUERY” |
    sed ‘/(.*).*/1/’)”

    if [ -z “$ANS” ]; then
    exit 1
    fi
    echo $ANS

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