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.

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