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. The calculation in the original post fail to take into account the latitude of the couple. It’s fairly easily done, however, if you’ve got a GPS that gives your coordinates…

    The (theta) == 2*pi*T/day becomes 2*pi*cos(latitude)*T/day.

    At 60 degrees latitude (Scandinavia) it adds a factor of two to the duration of the sunset!

    On a related subject, the sun moves at a constant rate of 15 degrees per hour, which for most people corresponds to the distance between thumb and little finger on an outspread hand at arms length. (I have been known to frequently tell people that – sure, we’re in the sun now, but wait 20 minutes, and we’ll be in the shade, after waving my hand around in the air.)

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  3. Why not just do it with something that allows you to go lower than its local ground level. Start up the beach a little bit, lower it to sea level(10m or so down), and then raise it up 30m

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  4. I did this exact same calculation for a high altitude balloon launch that was carrying a camera. I wanted to get some shots of sun set at 80k-90k feet. Given the rate of ascent of the balloon and that height, we needed to calculate a launch time that would put the balloon in the right spot at the right time, it worked out pretty well. Check it out at http://nearspace.0x58.com

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  5. I know this is terribly un-mathematical, but… Couldn’t he rent the cherry picker before the date, park it in his desired location, wait till sunset, and try it out?

    Maybe, I’m speaking to the wrong crowd. Maybe, when I ask you if you could hammer a nail into a piece of wood; you need to know the type and age of wood, weight and drag coefficient of the hammer, and detailed notes on the forging process that formed the nail.

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  6. I love how the google calculator link displays this blog entry right below the calculation.

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  7. One time on a ship in the Indian Ocean I saw the green flash twice because of a fortuitous combination of swell and tilt of the deck. So based on practical experience I think you can achieve at least part of this if you’re moving smartly enough.

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  9. You’re ignoring the lensing effect of the atmosphere. I realize this would complicate things quite a bit, but at this level of granularity, it could have a significant impact on the difference between sunset times at different heights.

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  12. If it’s not been suggested, San Diego would be an interesting test of this phenomena. The glide port near UCSD sits on a bluff that drops (nearly) straight down to the beach. Have an observer stand on the cliff, another on the beach, and have each report the sunset time.

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  16. I’ve done this, without the mechanical help. Near SCSU (New Haven,CT), you can watch the sun go down over West Rock, then quickly climb West Rock (takes about 10 minutes if you run) and watch the sun go down again. The first sunset isn’t all the way to the horizon, but the second one is pretty long.

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  24. I suppose taking the refractive properties of the atmosphere into consideration, rummaging through calculations, and determing the effects of latitude would be prudent, but would miss the most important point:

    It’s a “cherry” picker! Hohoho! Double entendres…
    Or triple, get the 11th Hussars involved and make it that much stranger.

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  26. …and to think I thought it was just going to turn out to be some dirty pun about cherries. OK phew apparently I wasn’t alone in that!

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  27. Im physically 5 minutes from Blacks Beach in San Diego at work right now…I’d go tonight but uh, the NCAA’s are on tonight.

    And also tomorrow I can’t because I’m lazy.

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  28. The sun forms an appreciable disc in the sky. Does this affect things noticeably? Also, you will continue to see the glare for a while after sunset. You’ll probably want to speed up your cherrypicker quite a bit so that you can let it get dark (a little) before you watch the sunset again.

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  29. I’ve done this in a small airplane — touch and go’s at sunset are perfect. You watch the sun set, then rise, then set again. On a sufficiently small field, you can do this twice. Very cool.

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  30. Trebuchet built for two would be faster, and that way you can see if
    a) she’s a strong swimmer
    b) she has a sense of adventure/humour
    c) how she looks in wet date clothes

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  31. Some have alluded to the effect of latitude. This is very important – the Sun does not, in general, go straight down.

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  32. “Were you so sad, then?” I asked, “on the day of the forty-four sunsets?” But the little prince made no reply.

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  34. How to correct for atmospheric distortion!!

    Here’s a site that deals with the angular deflection of the sun due to atmospheric distortion:

    calgary.rasc.ca/horizon.htm

    Some rough calcs based on their graph (for celestial bodies from ground level) plus their slidery thing (for horizon viewed from a height) give the following:

    Sunset is about 2min 15s later than predicted by a straight line calculation, wheras at 100m up this increases by a further 13 seconds. The effect is that the sunset at 100m won’t be 76s later than at ground level as predicted by straight line calcs, but 89s later. Which, if you follow the calcs through, means you don’t need to go to 100m, but 100m*(76/89)^2 = 73m.

    In fact all heights predicted by straight line calcs are overestimated by roughly the same fraction, so a correction of -27% on all calculated values should be enough to take into account atmospheric refraction.

    This will make things a bit easier on your cherry picker – it only needs to lift you to 73% of the height given by a straight-line calculation.

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  35. Cool idea … and very doable. And amazing that Norm saw *TWO* Green Flashes … an unusual phenomena that few people see once!

    Speaking of which, you might enjoy a related problem is can you see the (complete) moon set (or rise) twice. Now you have to factor in the apparent “width” of the moon since you have to wait for it to completely set, and then go high enough to see the entire moon again. And yes, if you do any type of photography with a long lens, it moves right along – a nifty sequence over the Colorado Rockies here.
    http://www.komar.org/faq/lunar-eclipse/moonset/

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  36. Or, you could go to this one town in Canada almost directly on the arctic circle on June 21, watch the sun dip below the mountains for about a minute and then come back up.
    Or use a hot air balloon, like John said.
    Additionally, if you study the wind charts enough, you can ideally have your balloon not land for a few hours… all alone in the balloon gondola with her :3

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  37. To follow the sunset up, keeping it on the horizon for as long as possible, all you need do is accelerate upwards at a constant rate of 0.0246m/s/s.

    This is 73% of (earth radius)*(2pi/86400)^2

    The hot air balloon idea would work. Just maintain uplift at one quarter percent greater than the mass, and off you go. Everlasting sunset!

    (works as long as the McLaurin expansion holds out – maybe an hour – then it’ll slip below the horizon unless you start turning up the uplift…)

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  39. I would take her to a beach with a tall building close-by, and then once the sun has set, go to the roof of that building to get another view. Will I make it on time given the average elevator speed? If not, do I need a special high-speed elevator?

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  43. #!/usr/bin/php5
    <?php
    $last = 0;
    function getans($query){
    global $last;
    if (in_array($query[0],array(“+”,”*”,”x”,”/”,”^”))){
    $query = “(@)” . $query;
    }
    $ans = file_get_contents(“http://www.google.com/search?q=” . urlencode(str_replace(“@”,$last,str_replace(“x”,”*”,$query))));
    if (strpos($ans,”/images/calc_img.gif”)){
    $ans = substr($ans,strpos($ans,”<h2″));
    $ans = substr($ans,strpos($ans,”“)+3);
    $ans = substr($ans,0,strpos($ans,”    “));
    $last = str_replace(array(” × 10″,””,””),array(“E”,”^”,””),trim(substr($ans,strpos($ans,”=”)+1)));
    return str_replace(
    array(” × 10″,”2″,”^2 “,”^2)”,”3″,”^3 “,”^3)”,””,””),
    array(“E”,”²”,”² “,”²)”,”³”,”³ “,”³)”,”^”,””),
    “$ansn”);
    }else{
    return “This does not look like a calculation.n”;
    }
    }
    unset($argv[0]);
    if(isset($argv[1])){
    echo getans(implode(” “, $argv));
    }else{
    $noexit = true;
    while($noexit){
    echo “GCalc> “;
    $in = fgets(STDIN);
    if (substr($in,0,4)==”exit”){
    $noexit = false;
    }elseif(substr($in,0,4)==”help”){
    echo “Enter an equation to be evaluated using Google Calculator. If you have alreadyndone a computation, the @ sign may be used to designate the last result.nAdditionally, begining with a +, *, /, or ^ operator will continue computationnfrom the last result. This does not apply to – to allow for negative numbers.nTo subtract a value from the last result use @-stuff or +-stuff.n”;
    }else{
    echo getans($in);
    }
    }
    }
    ?>

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  45. Do people actually watch the sunset all for the one instant when the sun disappears?

    I think this would be better used to simply make a sunset last longer – raising 6 metres results in 20 seconds of extra sunset. Now get in a huge, slow, elevator.

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  46. $ apt-cache search qalc
    –> qalc, qalculate, etc. (has the same functionality as google but with suggestions.)

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  48. Bob D — your answer agrees with my order of magnitude calculation (I’m not going to bother to type it up) which says about one km in height needs to instantaneously gained to see the entire disk of the sun again, using the radius of the Earth as 6*10^3 km and the angle subtended by the sun as about 0.5 degrees.

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