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Elsewhere I’ve briefly written about Leo’s obsession with Portal, a clever, (mostly) non-violent, Steam game. As always, I try to turn his obsessions into lessons (I should trademark that!) and so it is with Portal. In another post I talked about the game’s clever “AIs” (scare-quoted because, of course, they are merely scripted, not real AIs!) Here I talk about a cool little physics lesson that we recently did using Portal.

One of the really fun and clever things that you can do with portals is to put an In-Portal in a floor, and an Out-Portal in a wall, and then drop something into the In-Portal and it’ll come out of the wall. If you drop it from a distance into the In, it’ll blow out the Out with what we shall hypothesize is the same horizontal velocity at which is hit the In in the floor.(Watch this happen quite dramatically at about 4:40 in this video.)

This can be quickly turned into an interesting physics question: How high do I have to jump from such that I go a specified distance across the room? Here’s the setup (I’ve redrawn all this because we initially did it on a white board, and between erasures and scribbling, the white board is entirely incomprehensible!)

20180223-PortalPhysics1

A box is dropped (or a person jumps!) from the tube at the top left at a height of Hf, and accelerates (presumably at 9.8m/s^2), hitting the In portal (the circle at the bottom left) with Vf. It instantaneously blows out of the Out portal (the circle on the wall in the middle left of the picture, at a height of Hp) with the same velocity, Vf and immediately begins to fall, eventually hitting the floor and stopping at distance D. (We neglect a lots of the usual things like rolling along the floor, air resistance, and that the box isn’t a point mass.)

The question is: How high to I need to put the Out portal (Hp), if the drop point (Hf) is, say, 20 meters high, and I want the box to land (D) at, say, 20meters from the left wall?

Notice that there are three equations involved here [all equations from this wikipedia page, or any of the other zillion pages that have the same info]: For the drop we can compute Vf from Hf (20m) (given g=9.8m/s^2). But in order to figure out how far it’s going to go to the right after coming out of the wall need to compute for how LONG (i.e., t in secs) we want it to travel in order to hit the ground in D (20m), and then back-compute the high (Hp) given how long we want it to be in the air before hitting the ground.

Okay so V_f=V_0+{\sqrt{2gH_f}} [Since the drop is static, V_0=0.] We’ll round 9.8 to 10m/s^2, so V_f={\sqrt{20*10*20}}={\sqrt{20^2}}, which, coincidentally, is really easy to calculate: Vf=20m/s.

Okay, now at 20m/s going 20m upon exit from the Out portal, leads to the sort of trivial use of t=D/V_f=20/20, which is exactly 1 sec. (I actually didn’t make these numbers up to come out so nicely!)

Finally, we need to figure out how far something will fall in 1 second, which is just 1/2*g*t^2, where t=1, and g=10, as above, so H_p=1/2*10*1 = 5 meters. So, we have to put the (center of the) portal 5 meters up the left wall.

By a little algebra we can reduce these three equations:

V_f={\sqrt{2gH_f}}
t=D/V_f
H_p=1/2*g*t^2

into one:

H_p={gD}/({2\sqrt{2gH_f}})

I’ll update this post with a screen cap from Leo’s actually implementing this so check it out (spoiler alter: it worked out pretty well!)

[By the way, there are many quite clever videos where people work out aspects of the physics of Portal, as well as many funny thought experiments about what could happen if portals were real!]