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 sunset^{1}.

But we can also get the answer by just trying a few different heights. We plug it in to Google Calculator^{2}:

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 credit^{3} 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 level^{4}.

^{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.

another interesting stunt would be bungeeshooting right after the sun sets, that way you might get serial sunsets.

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Much as I hate to burn holes In a beautiful idea, I can see one thing wrong in this logic.

I would consider a sunset not to be a fixed moment in time, but to be the whole duration In which the sun sinks over the horizon. The shortest period of time this could be defined as is from when the lowest part of the sun appears to just touch the horizon (so the horizon appears like a tangent), to when the sun can no longer be seen. A definition of a sunset which would give a longer time would be the period of time when the suns light is refracted so It appears red.

The calculations shown here give the rate of climb that would have to be maintained for the sun to appear in the same position relative to the horizon. For the same sunset to be experienced twice you would have to wait until the sun has sunk, and then elevate yourself at a rate faster than the one given here, until the sun appears right above the horizon again.

This would take a different period of time depending on how long the sun took to set, which is different at all times of the year, (unless you happen to live at the equator).

This is not to say that the answer given by this calculation is useless. When Space Elevators are developed, this is the rate at which they will have to climb for people in them to enjoy a (semi) perpetual sunset. Now I for one think that that is a great date idea.

Yours In Maths

Hungry_Joe

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While on vacation a few weeks ago, I had a thought that led me down a similar line of reasoning:

How “curved” is lake tahoe? I.e., if you set a perfectly flat plane down at lake level on one shore, how far below the plane would the lake be at the opposite shore, 22 miles away, due to the curvature of the earth?

Using similar math, you wind up with somewhat surprising answer: 300 feet. I had to double check that a few times.

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Long time reader of XKCD, first time poster.

I get the math, and I have done this very thing; but with a Cessna airplane.

flew straight and level at slow cruse. A second after the sunset from view, pushed the throttle in and pulled up. Second sunrise and sunset of the day. Very cool and I got the girl!!

🙂

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Airplane? GAH! I JUST THOUGHT OF THAT!

You could also use a car and a hill.

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Instead of digging a hole, couldn’t you just pick a day that where sunset coincides with low tide?

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Rate of ascension for the cherry picker bucket would be doubled as anticipated load would be cut roughly in half.

This due to the fact hat no chick is going ANYWHERE in a cherry picker with a dork.

C’mon, people…think.

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I am a woman that occasionally gets asked out. Now, much as I find the idea of the double-sunset appealing, the cherry picker…not so much, those things are jerky and bouncy.

Therefore, I submit…the glass elevator. Guys, here’s what you do: 1. Find a many-storied beach hotel with a glass elevator facing the beach. (Google might help you…go to it!)

2. Arrange for the elevator to be empty with hotel management for roughly half an hour.

3. Get your date to the hotel…a vacation in which you are already booked works, or simply visiting someone nearby, be creative!

And there you go…sadly, for all you math geeks out there, you probably don’t need calculations of any sort…if you get the floor wrong…just go up. But I’m sure you can find something to derive if your mouth isn’t too busy getting hella kissed.

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The illustrations made the post much better and entertaining. =)

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wow, did I just read an article that was about dating bit included math and physics? Very clever 🙂 Never thought I would understand that but your illustrations helped quite a lot.

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Lol I like the illustration and the calculation! Honnestly who would of thought of dating a girl with a cherry picker ? some people have weird ideas 🙂

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berrywing, what kind of plane were you using? I tried to do this with a plane once, and I couldn’t outclimb the sunset. But, it was only a Piper Warrior, so the climb rate kinda sucked.

But, of course, since I’m a pilot I got the girl anyway. 😀

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Ouch, you smashed my head ! i left school long ago so forgot most of te trigonometry went back to the teacher 🙂

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So with this info can you Run Your Own Dating Site?

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Thank you For Post

My Name is EmeL .)

Thank you For admin

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Many people over analyze on how to date. They take every advice word for word and then later figure out why the date didn’t go well. The best thing to do is to be yourself. Many women will pick up little cues of men trying too hard and in the end it makes them look bad.

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ple over analyze on how to date. They take every advice word for word and then later figure out why the date didn’t go well. The best thing to do is to berder nomery

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ple over analyze on how to date. They take every advice word for word and then later figure out why the date didn’t go well. The best thing to do is to merdinan comple saha

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ple over analyze on how to date. They take every advice word for word and then later figure out why the date didn’t go well. The best thing to do is to dadaes return momphe shaped of liser

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ple over analyze on how to date. They take every advice word for word and then later figure out why the date didn’t go well. The best thing to do is to double

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very nice this blog thanks admin

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very nice this blog thanks admin

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To berrywing: Was flying back to base at sunset, just happened to be at 6,500 feet when the sun set. The sun looked squished and weird at that altitude, not sure why. I’ve always wished I’d had my camera that day.

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So inspiring post.Really great sharing.

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Nice post.I’m enjoy read that.

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Funny ending I need to brush up on my math 😦

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Math is a lot more interesting now then it was in high school actually I will prob go back and study it later in life

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Well, as far as the math goes, it’s really a simple transformation: instead of assuming theta starts at zero, simply calculate the angle at which they are initially viewing the sunset. I took the average heights of males and females in the US of age 20+ (available online), and averaged them, giving 1.6925m and I took the mean radius of the earth to be 6371.0km, or 6371000m, as those are the units of the calculation, and with some simple trigonometry, you come up with a modified formula for theta which can simply be plugged into the equations, and barely changes the data:

theta = 2*pi*t/day – arccos(6371000/6371001.6925)

arccos(6371000/6371001.6925) is approx. .0007289rad for those of you who like decimals

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Wolfram|Alpha: http://www.wolframalpha.com/input/?i=0.3+m%2Fs+%3D+%282*pi*x%29%2F%28day*arcsecant%28x%2Fradius+of+earth+%2B+1%29%29

I doubt I entered that right, but it comes up with h=6.283 m, which is close to 6, so I think that’s your number.

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good post very interesting to gras the contribution to knowledge

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Thanks very interesting post i like

http://www.de-pelicula.com

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Maths let me laugh in these cases. 🙂

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Interesante post muy bueno el amor con las matematicas jejej

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bueno is right!

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Another informative post. This is a very nice blog that I will definitively come back to several more times this year.

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berrywing, what kind of plane were you using? I tried to do this with a plane once, and I couldn’t outclimb the sunset. But, it was only a Piper Warrior, so the climb rate kinda sucked.

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Question: why don’t high school math books use word problem scenarios like this? The catty teens would be much less cruel to those actually interested in calculating the perfect date, and this could teach several types of equations at once, including social skills – that way even the nerdiest of nerds wouldn’t be lonely… like I was.

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While this is a fairly old post, you might think about updating your link to ConBust. The link presently directs you to a 404 page, though just navigating to the root of that directory presents you with the appropriate page.

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@hungry_joe:

What about climbing up the Willis Tower with the express lift?

Start from floor 10, then go all the way up to the top?

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ConBust sounds fun. Just looking at the schedule makes me sad I missed it.

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ConBust sounds fun. Looking at the schedule makes me sad I missed it. Hope they do it again next year.

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Refraction!!

You forgot the refraction effect of atmosphere, which changes the sunset time.

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I’ve done something similar to this in a general aviation aircraft. You take off before the sun rise, watch from altitude, and then dive down to a lower altitude to see a second or even third sunrise. It doesn’t work for sunset, however, because you can’t get the nessecary climb rate to beat the shadow of the earth as it climbs up (probably because of the shadow moving faster higher up, but it’s late and I skipped over the math).

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Yes, but what about the bending of the sunlight as it enters the atmosphere- Refraction is important!

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Great idea for a post.Thank you!

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It takes many thousands of feet of climb to watch a sun that has set, then set again. I have done it a few times in my career. It is very cool but rare to get all the requirements to line up just right.

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