Imagine a frictionless and rigid hemispherical pit with a radius of 10m on the Earth. For a wizard in my world, it is easy to make such a pit, but hard to build a jail. The pit is the side effect of some space magic.

The prisoner is trapped at the bottom of the pit with no initial speed. Can the prisoner escape the pit? Assume that air resistance and lifespan are factors. Breathing or using the Magnus Effect might take too long.

Space Magic Description: This is short-distance teleportation magic. In order to avoid collision between the teleported and the molecules being teleported, the molecules are squeezed outward thus forming an irregular frictionless rigid mirror surface.

The Wizard was not carrying something such as a rope on their person. Their task is to capture the enemies again and again and is not willing to spend more time on one enemy. These enemies were normal humans who could not do magic.

Please let me know if you need any clarification or more details.

  • $\begingroup$ Does he get a skateboard? $\endgroup$ Commented May 19, 2023 at 16:38
  • $\begingroup$ @RobertRapplean For safety, the skateboard can be confiscated $\endgroup$
    – Voyager
    Commented May 19, 2023 at 16:39
  • $\begingroup$ Of interest: How would the world control an invulnerable immortal mass murderer?. $\endgroup$ Commented May 19, 2023 at 17:41
  • $\begingroup$ @Voyager: thank you for keeping your pronouns gender-neutral. I wish commenters (on the question and the answers) would similarly avoid gender bias. $\endgroup$ Commented May 20, 2023 at 17:32
  • $\begingroup$ if it was a cylinder instead of a bowl, it would be inescapable $\endgroup$
    – njzk2
    Commented May 21, 2023 at 11:57

7 Answers 7


They can escape.

It's basically the same principle as how one uses a swing without getting a push to start - a swing is essentially a 2D version of this setup, a semicircular pit with no frictional force in play, and an ability to exert net force only in the radial direction. Standing at the bottom of the pit, push your arms out in front of you. Your center of mass doesn't move, but your feet will slide backwards a little. Since there's no friction, your feet will oscillate underneath you, swinging back and forth over the center slightly. Now you just need to bend your knees and pump in time with the oscillations, which will make the swing larger and larger. Eventually, the swing is large enough to grab the lip of the pit. Even though the prisoner's center of mass is moving vertically and must remain above the pit when clearing the lip, the prisoner merely need extend his arms forward and legs backwards to get a handhold outside the pit without moving his center of mass outside the pit.

If you've ever used a swing set without someone pushing you to get started, you can get out of the pit. As another analogy, this is basically like riding a frictionless skateboard in a half-pipe. It is clearly possible to increase the amplitude of the oscillations, a competent skateboarder can jump much higher than the lip of the half-pipe where they enter, and they never need to push tangentially to the ground to do it.

  • 3
    $\begingroup$ @Voyager I don't think standing will be terribly difficult, it's not like it's impossible to stand up on a sheet of ice, you just need to do it carefully. But even if it is, this approach still works lying down, you just need to pump your legs back and forth to get started and then pump your arms up and down to increase the swing. $\endgroup$ Commented May 19, 2023 at 16:58
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    $\begingroup$ BTW, @Voyager, it may be less important for a yes-or-no type question like this, but it's usually recommended to wait 24 hours or more before accepting an answer. A check mark tends to discourage other answers, and might send away someone who's answer is actually better. $\endgroup$
    – Zeiss Ikon
    Commented May 19, 2023 at 19:16
  • 3
    $\begingroup$ I think there's a flaw in this reasoning: if the pit is actually hemispheric, even if the prisoner manages to pump up enough momentum to reach the rim, at that point his velocity will be vertically upward, and he won't be able to get a grip on anything outside the pit to avoid simply falling back in -- and if he falls, the constantly changing difference between gravity and centripetal force over a pretty short period (less than a minute) will make rising when not still at the bottom impossible (IMO) until air drag stops him. $\endgroup$
    – Zeiss Ikon
    Commented May 19, 2023 at 19:19
  • 1
    $\begingroup$ @LorenPechtel Ever seen kiiking? youtube.com/watch?v=xJkjyq_or70 $\endgroup$
    – biziclop
    Commented May 20, 2023 at 11:29
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    $\begingroup$ @Tristan: There's "can", and there's "think about". They may not think about the necessity of doing so, if no prisoner ever escaped their contraption before. $\endgroup$ Commented May 22, 2023 at 9:20

Escape Proof (with a tiny change)

Geometry is our friend and physics a fun companion when it comes to devising a gaol that is not only escape proof, but also educational and entertaining into the bargain!

There was once, in the land of Angera a particularly eager archon by the name of Crowell whose sole delight, it seems, was devising what some might consider cruel and unusual means of punishing the wicked. But never let it be said that Crowell was a mean sort! For he was, at heart, a jolly old soul, and so jolly was he that he was determined to bring judicial jollity to his subjects.

Thus, let us examine old king Crowell's merry old bowl of judicial jollity! In construction, it is a roughly hemispherical bowl made from thick and crystal clear glass some eighteen feet in diametre. It is suspended from a platform above, for the convenience of the Guard, and also a shit hole at the bottom. The colonnade upon which the Cauldron is suspended allows the public a grand view of those stuck within!

Penitents are brought up to the platform, stripped naked, oiled and have their charges read out to them. Upon public approbation of the proceedings, the penitents are shoved into the bowl, from which they might only be fished out at death.

The key to the success of keeping penitents in is two fold: geometry tells us that a rounded shape is best for keeping the penitents down. There is really no flat surface upon which one might stand, and the curve of the bowl itself prevents climbing. Even if one could climb, the lack of a rim upon which to hoist oneself out and the steep neck of adverse camber prevents the escape from continuing.

Secondly, physics, that old prankster, teaches us that motion tends to continue in its own direction unless otherwise directed. So, if one tries to slide oneself out of the bowl as if riding a swing by constant back-and-forth motions of the body, the circular motion of the body will simply be guided by the slight incurve of the bowl, sending the miserable miscreant out into empty space, where from he shall plummet back into the bowl, with much cheering and jollity among the crowd!

Having learned this valuable lesson of physics, the penitent might consider himself Smart, and says: "I shall beat this device by physics of mine own!" And so thrusting himself to and fro again, he shall attempt to twist himself at the last moment in an attempt to grasp at the rim! Had Crowell and his geometers not considered this clever means of escape, the penitent might be successful! However, the smooth curve of the cauldron's neck and the angle of its slope spell doom for the penitent, and with further jeers and guffaws from the crowd, he shall, his fingertips vainly grasping for the edge!, slide once again into the waiting bowl!

A diagramme: enter image description here

  • 5
    $\begingroup$ Well, this cauldron is not an exact hemisphere, but a tad more. So, a frictionless hemisphere can be escaped, yet modernized like this, indeed cannot. Unless the person inside could grow up nice strong claws and penetrate the surface, or plain break it up into oblivion :) $\endgroup$
    – Vesper
    Commented May 20, 2023 at 6:02
  • 2
    $\begingroup$ I think this might work even if the lower part were just a hemisphere. The key thing is to prevent the inmate from grasping the upper edge of the bowl. $\endgroup$
    – David K
    Commented May 21, 2023 at 13:45
  • $\begingroup$ @DavidK --- Exactly. Key is the upper edge. Whether it's an exact or super or infra hemisphere, if the rim is flat, it's graspable and thus escapable. Old King Crowell opted for slanted rim to give hope without the possibility of attaining said hope. The cauldron shape is purely for entertainment! $\endgroup$
    – elemtilas
    Commented May 21, 2023 at 15:39
  • $\begingroup$ @Vesper --- A pure hemisphere can only be escaped if there is a sufficient rim to grasp. Hence the cauldron shaped rim -- nothing to grasp! $\endgroup$
    – elemtilas
    Commented May 21, 2023 at 15:41
  • $\begingroup$ This might also work with a flat rim if the spherical part of the bowl continues far enough above the equator. At some angle I think there would be two possible outcomes: the prisoner does not have enough speed to stay in contact with the bowl, and falls without being able to reach the edge; or the prisoner has enough speed to reach the edge, but at that point has too much momentum in the direction away from the edge and will slide back into the bowl. $\endgroup$
    – David K
    Commented May 21, 2023 at 15:50

This is a straight-up Newtonian mechanics physics question, but it would probably get closed as homework-like on Physics SE, and it's fun, so I'll give you a Newtonian mechanics physics answer here. The answer is yes, as others have said, but it's not quite as others have described it.

Start with a more formal treatment of the mostly correct "oscillate to escape" answers.

Because there is no friction, the prisoner (whose name is Bob$^1$) can only accelerate his center of mass in the direction radially inward from his point of contact with the prison wall.

Because Bob can change his shape, he can temporarily move his point of contact with the prison wall in the tangent direction without accelerating his center of mass in the tangent direction.

Suppose Bob can change his shape such that the displacement between his center of mass and his point of contact, is described by the below triangle:

enter image description here

The force vector from a thrust at the blue point (the point of contact) will be in the direction of the green side of the triangle.

The proportion of thrust in the tangent direction (at the angle corresponding to the center of mass) is roughly the ratio of the blue segment to the green segment, but if one wants to be precise, it's the ratio of the green line segment to the yellow height of the purple triangle below.

enter image description here

For green line segment $r$, blue line segment $b$, note that the purple triangle is a similar triangle to the one above it, but with legs $b$, so its height $x$ is the height of the above triangle scaled by the ratio of $b$ to $r$:

$x = \frac{b}{r} \sqrt{r^2 - b^2/4} \approx b \text { for } b \ll r$

If Bob pushes such that his change of momentum is $\Delta \vec P_i$ in the green direction, then his change of velocity in the tangent direction is $\Delta v_i = \frac{x}{rm}|\Delta P_i|$.

If Bob times his changes of shape and his pushes such that his tangent velocity is zero whenever he pushes, each push adds

$\Delta T_i = 0.5m\Delta v_i^2$

until such time as Bob's total mechanical plus potential energy, $\sum \Delta T_i$, is greater than or equal to his potential energy at the height above which a displacement of $b$ puts Bob's point of contact outside of the hemisphere... which prevents him from pushing off, but should permit him to grab the rim and escape.

Bob, however, has a problem that the "oscillate to escape" answers have overlooked. He is not constrained to stay on the prison wall.

If Bob had an anchor point mounted at the center of the sphere, to which he had attached a rod that he could push against so as to constrain his motion to the prison wall, escaping would be easy and painless. Bob has no such anchor. While Bob's weight $m\vec g$ is roughly antiparallel to his direction of thrust, Bob is fine. All he needs to do is keep the force of his pushes low enough that gravity can pull his center of mass back down before his point of contact leaves the ground (just like you can stand up and sit down without ever leaping into the air or falling to a bone-jarring halt). However, as Bob reaches the more vertical portions of the wall, if Bob wants to continue adding significant $\Delta T_i$ per oscillation, Bob will have to begin hurling himself off of the wall and suffering impacts from falling.

One may have an intuition that this would prevent Bob from escaping, but this is an intuition based on our experience living in a world with friction: since friction forces increase based on how hard you're pushing on something, if you're pushing on something with the extremely large forces of crashing into it after falling from a significant height, the frictional forces are very large. In his frictionless hemisphere, Bob cannot faceplant, he can only faceslide. Hopefully Bob is good at frictionless acrobatics, and can land more comfortably than on his face.

Below: two pushes worth of Bob's later trajectory

enter image description here

Bob escapes, but only if he's tough enough to take a few hard thumps on the way out.

$1$: Alice is on a space ship with a large relative velocity $\vec u$ measuring Bob's escape attempt for next semester's homework assignment.

  • 1
    $\begingroup$ When Bob lands back on the ground after leaping, presumably the collision is not perfectly elastic: he is losing some of his energy to deformation. This and the air resistance seem like non-negligible factors for this plan. $\endgroup$
    – amalloy
    Commented May 21, 2023 at 5:24
  • $\begingroup$ You're right that air resistance is nonnegligible. I suspect that Bob's pushing power (order of 1000N over order of 1m once per period) is a bigger than the air's maximal pushing power (order of 10N over order of 10 meters once per period). However, only one order of magnitude on an order of magnitude estimate is a pretty small safety margin for poor Bob. $\endgroup$
    – g s
    Commented May 21, 2023 at 6:28
  • $\begingroup$ @amalloy The question stipulates a frictionless and rigid surface. Since frictionless is a perfect state, presumably rigid is perfect as well. Therefore we can assume perfectly elastic collisions on the part of the bowl at least. Let's just give our poor prisoner perfectly rigid skates to wear as well. $\endgroup$
    – Corey
    Commented May 22, 2023 at 7:07

I don't think a human can escape.

Let us simplify things. You are a two-dimensional person in a frictionless half-pipe. You can place your legs apart and be stable, with your legs at equal angles either side of the vertical. Put your weight on your right foot. The reaction of the half-pipe surface will be perpendicular to the surface. This will give you a slight force to the left. As you move to the left, you are slowed by the curvature. You can do this repeatedly, and put some energy into your oscillation. You can build up some height this way, a bit like pumping on a swing.

You start off effectively being on a 10m swing, so the oscillations are slow, but if you keep at it, you might get somewhere. As you begin to climb the side, the restoring force deviates from simple harmonic motion, and the period becomes longer. As you approach the lip, your pumping efficiency approaches zero, while your speed at the bottom will roughly be given assuming your kinetic energy at the bottom equals your potential energy at the top. With an earth-like g of 10 m/sec^2, you would be going at about 13 m/s, and air resistance would take away any extra energy you could add in an oscillation.

At the edge, you would be going straight up. This does not help but it is not an absolute barrier to getting out. Your centre of gravity may below the edge, but parts of you may briefly get just beyond it. You might even get just beyond the lip of a cauldron if it was a very small one.

  • $\begingroup$ You can get additional force by jumping just before reaching the bottom of the pit and crouching to land slightly up the side. Once you've reached high enough to get above the center of the pit you can get some residual sideways motion by skipping the jump and crouching during the upwards curve. From there it's just a matter of timing. $\endgroup$
    – Corey
    Commented May 22, 2023 at 7:04
  • $\begingroup$ I don't see how the pumping efficiency is related to the oscillation amplitude. You pump at the bottom of the pit, not the lip. A swing with a rigid bar instead of a chain can be swung all the way around just by the rider pumping their legs, but this suggests they would never be able to go above horizontal as their pumping became useless. $\endgroup$ Commented May 22, 2023 at 14:52
  • $\begingroup$ It the balance between pumping energy in and losing it to air resistance. If you get close to the lip, you are travelling at 30 mph at the bottom. Your ability to pump goes down as the surface gets flatter, until it disappears altogether and you can't move on a flat surface. Maybe someone could get out of a 3m hemisphere, but a 10m one does stay the odds against you a lot IMHO. $\endgroup$ Commented May 22, 2023 at 18:40

Yes, of course the prisoner can escape if left unsupervised.

This is a prison, not a deathtrap; it needs to have the capacity to hold a prisoner over the long term. Particularly, prisoners need to be fed and their waste removed. This requires a mechanism to move things in and out of the prison, and once you have a mechanism to move certain things around, tampering with it to move something else (like the prisoner) is just a matter of time.

  • $\begingroup$ I beg your pardon, but no where in the law codes does it stipulate that penitents must be fed. $\endgroup$
    – elemtilas
    Commented May 21, 2023 at 15:45
  • $\begingroup$ Point taken but, theoretically, the magic might exist to move smaller objects but not larger. Having ventilation shafts doesn't mean prisoners can escape using them when they're only a few inches wide. $\endgroup$
    – lly
    Commented May 22, 2023 at 6:00
  • $\begingroup$ I don't understand the logic that if anything can get into or out of the cell, even only under very specific controlled circumstances, that the prisoner can therefore free themselves with no outside help. Obviously someone can let the prisoner out, but I find the conclusion that a tiny hole in the floor makes escape inevitable to be nonsensical. $\endgroup$ Commented May 22, 2023 at 16:19

Combining "Bambi on ice" and a gazelle and the slide-swinging part in the previous answer. The gazelle would swing a bit, to no longer jump straight in the air, but at an angle and jump out.


Since my imagination is running wild in the wizard world you provide. Birds/Birdman/Pegasi/Dragons can definitely escape this.

They would just fly out.


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