7
$\begingroup$

In Steven Gould's Jumper series, the main character has the ability to teleport, and he eventually realizes he automatically does frame matching when doing so - no matter where on earth his origin and destination are, when he arrives, he is perpendicular to, and at rest in relation to, the ground.

I'm writing a story with a teleporter without frame matching. This is obviously going to limit how far he can safely teleport without stumbling / breaking a leg / becoming a grease stain; but what are the actual limits? The bulk of the story will be set in and about Seattle, WA, USA; how can I calculate safe distances and directions, or know how fast he'll be sent flying in which direction?

Edit: He can only teleport to locations he has already been or can see; so location isn't going to be an issue, only velocity and orientation.

It might be that I'll get better answers on the Math site, so I'll be posting there in the morning if nothing comes up here; but I thought I'd give Worldbuilding the first stab at it, since I'm more active here. I'll gladly delete here if this is deemed improper here.

Edit: It has been suggested that this is a duplicate of this question, which only asks what potential downsides of retaining momentum might be; but this is not the same question, as I am asking how to determine how to calculate a magnitude and direction for a specific known issue.

$\endgroup$
  • 8
    $\begingroup$ The Mathematics SE would kick your butt if you ask there - it's for pure mathematics, not wild ideas like teleportation. You're actually in the right place here on WB SE. $\endgroup$ – StephenG Apr 17 at 6:49
  • 1
    $\begingroup$ Teleportation has to be set relative to something.... If you use an absolute frame then your going to end up in the middle of space and die. Remember that the earth is rotating on an axis, while spinning around the sun, that is also moving through the milky wave, which itself is moving through space and probably orbiting some huge black hole. $\endgroup$ – Shadowzee Apr 17 at 7:07
  • $\begingroup$ @Shadowzee presumably you do have the ability to match a location that will be on Earth at the moment of jump. The grease stain clause still applies though. $\endgroup$ – John Dvorak Apr 17 at 7:28
  • 2
    $\begingroup$ @Shadowzee Remember that the "jumper" also shares the Earth's motion through space, and as teleportation is essentially instant (speed of light as a limit ? :-) ) the Earth won't have moved significantly in this time. Also note there is no absolute frame of reference - that's relativity. Rotation of the Earth itself is a different issue. $\endgroup$ – StephenG Apr 17 at 7:30
  • 1
    $\begingroup$ This is sort of a dumb question. Once you peel the layers of the physics onion back on teleportation, you quickly get to things with no good answers. How does a human orient itself on a rotating earth with places he cannot see? What is the error in the teleportation (i.e. how far off might you arrive)? How fast does the person "appear" into the space that he is occupying, so how fast to atoms in that space have to vacate? Just let it be magic; Trevortni; teleportation and science do not mix. $\endgroup$ – kingledion Apr 17 at 11:02
11
$\begingroup$

I think that without frame matching there will be severe limitations.

While I am sitting at my desk to write this answer, I feel I am standing still. However:

  • planet Earth is rotating around its axis, and this gives me a certain velocity vector $v_E$. This is 1668 km/h at the Equator, 0 km/h at the poles.
  • the Earth is orbiting the Sun, this gives me another velocity vector $v_S$. This is about 29.5 km/s
  • the Sun is orbiting the galactic core, resulting in another velocity vector $v_g$
  • the Milky Way is also moving ...

Let's focus on the Earth rotation, as I assume you don't want to teleport outside of the planet.

The tangential velocity changes with the latitude L according to the law $v_e(L)=R_{Earth}\cdot \omega_{Earth} \cdot cos(L)=1668 \cdot cos(L)$. If your character moves just 1 grade north, going from 44 N to 45 N, he will experience a difference in velocity of 20 km/h. For reference, crash tests are conducted at 30 km/h.

Moving East or West by an angle $\alpha$ does little better: the module of the vector will stay the same, only the direction will be different, so he will be moving up (or down) with a velocity $1668 \cdot cos(L) \cdot sin(\alpha)$. And the lateral velocity will be diminished to $1668 \cdot cos(L) \cdot cos(\alpha)$.

Let's say he goes from (45 N 0 E) to (45 N 1 E), he will be moving up with a velocity of 20 km/h.

And we are not taking into account the possibility of ending up on a moving object.

Long story short: he can move, but just few tenths of a degree per jump, at most. The closer he is to the poles, the wider is the safe range for the jump.

$\endgroup$
  • 1
    $\begingroup$ The worst part: if you move 1 degree eastwards on the equator, your new velocity relative to the ground will point straight upwards. If you teleport westwards, it will be straight down. Running shoes won't save you. Technically, it will be 0.5 degrees off vertical, but... $\endgroup$ – John Dvorak Apr 17 at 7:16
  • $\begingroup$ Since I had not frame of reference moving from 44 to 45 is a distance of about 110 km or 69 miles. :) $\endgroup$ – genesis Apr 17 at 7:46
  • $\begingroup$ @JohnDvorak: That would be if you were to move 90 degrees. Moving 1 degree eastward would have you moving 1 (or is it .5?) degrees upward from level with the ground, if I'm not overlooking anything. Moving either the origin or destination away from the equator would skew the results, but it starts to get more complicated to model mentally. $\endgroup$ – Trevortni Apr 17 at 19:13
  • $\begingroup$ @L.Dutch: I'm trying to combine the two equations to get a combined equation for points that don't share either latitude or longitude, but the first step, whichever it may be, introduces a non-zero initial velocity to the second step that neither equation accounts for. How to accommodate this? $\endgroup$ – Trevortni Apr 17 at 20:10
  • $\begingroup$ @Trevortni subtract the velocity of the ground you move to (tangent to the latitude circle) from the velocity of the ground you move from. That is your velocity relative to the ground. Draw a diagram with both velocities as arrows, noting they have the same length.. What direction takes you from the end of one to the end of the other? $\endgroup$ – John Dvorak Apr 17 at 20:12
2
$\begingroup$

Let's think about this in terms of the curvature of the Earth firstly; the equator is a bit over 40k Kms in length, meaning that there is approximately 112km per degree of change in aspect in real terms.

If the average person's centre of gravity is around 1m high, let's just say that by the time you get to 5o difference, or around 560km, it's going to be awkward to maintain your balance with that kind of instant change, but actual failure point to remain standing is going to be different for every person depending on how flexible, agile and prepared you are.

The truth is though that points of failure kick in a lot closer because the Earth is not a perfect circle. the ground is uneven, and without being very careful you can just as easily materialise with your foot half in a concrete path because you didn't notice a subtle incline.

Ultimately, your jumper in this context is going to learn a few tricks very quickly; lean forward for long jumps, jump higher than you need because it's safer taking a 30cm drop than it is having half your foot materialise in the middle of the road; that kind of thing.

Another thing he can do is actually start running while he jumps, as that might help with momentum change as well because he can jump while neither foot is actually on the ground. It will take some practice not to break his stride, but it's a skill that can be learned just like normal walking is learned at an early age.

So the actual failure points depend on the terrain, the balance and preparedness of the jumper, and thorough research into the altitude changes (inclines and declines) in the area in which your jumper operates. In other words, pick a flat area to operate in, practice hard, and do your homework. That way, you maximise the distance you can jump without incident.

$\endgroup$
  • $\begingroup$ I should probably add to the original question that the teleporter can only teleport to locations he has already been or can see; so location isn't going to be an issue, only velocity and orientation. $\endgroup$ – Trevortni Apr 17 at 7:22
  • $\begingroup$ @Trevortni even with that restriction, your teleporter guy will be a useful asset in space transportation. Just shove him onto the next Apollo mission, then have him haul material to near-moon space and back. Even though I doubt you can hit the LEO velocity just by a slingshot around the Moon, the benefit is going to be significant one way or another. $\endgroup$ – John Dvorak Apr 17 at 7:27
2
$\begingroup$

I like this. It's a reasonable teleportation limitation, and one with interesting consequences.

As mentioned in other answers, jumps of less than a hundred miles or so shouldn't pose too much of a problem. I suggest wearing a digital compass like a watch so that you know which way to brace yourself after a jump (you should also always aim a few feet off the ground when moving west, since you'll be driven slightly into the ground, which would be hell on your knees otherwise).

But say you want to quickly go larger distances - where could you go safely? Well, here are some cool, albeit specific, places, and how to get there:

  • Get a friend to drive you at 112 mph west on a straight road. You can now safely teleport to the north or south pole. (Wear padded clothing, since in addition to the cold, you'll appear about to fall on your side if you go north, and about to fall straight down onto your shoulder if you go south. Also, no way to get back but the long way - dozens of smaller jumps.)
  • Get a friend to drive you at 110 mph east on a straight road. Jump safely to San Francisco. (You can jump back to Seattle from the passenger seat of a car there moving west.)

Finally, the most helpful equipment for long-distance jumps: skydiving gear and a parachute. Everywhere on Earth less than about 50 degrees north or south (US and Canada, Europe, China, Southern Argentina, and New Zealand) is only moving at about 100 mph at most relative to the Earth's center. That means you can teleport 2 or 3 miles above your destination (that's approximately the cloud layer, so just look up when you visit and you should be able to save that jump location for later) and just skydive down. You'll be yanked sideways when you arrive, but not significantly faster than terminal velocity, so a skilled skydiver with enough time before the ground should be able to land safely. Of course, this task can be made arbitrarily safe by doing several jumps, say, 45 degrees longitude apart from each other, to mitigate this sideways wind.

Note: I've never gone skydiving, and I'm aware that even a 30 mph wind is considered dangerous, let alone 100-200 mph. But my understanding is that this danger is mainly in moving too fast relative to the ground, whereas in this case, the ground and "wind" are moving at the same rate, so it's moot if the diver has enough time to match the "wind" speed. I welcome any skydiver who reads this and wants to chip in - it would make for an interesting and surreal discussion.

$\endgroup$
  • $\begingroup$ The suggestion to drive before jumping can also be extended downward to walking or running, which I am definitely going to include in his standard pre-teleport checklist, as well as jumping before - er, jumping. Thanks! $\endgroup$ – Trevortni Apr 18 at 4:35
  • $\begingroup$ I feel like this is somehow related to the 88mph Delorean..I know it's not. but it feels amusingly BTTF. $\endgroup$ – Ruadhan Apr 18 at 10:28
  • $\begingroup$ @ruadhan coincidental or not, it still works just fine as a callback. $\endgroup$ – John Dvorak Apr 19 at 20:32
1
$\begingroup$

You are describing the "teleporting" flavour of the Azhiri, in Vernor Vinge's The Witling (1976). They can reng objects faster than light, from places they've been, but momentum is conserved, so renging a rock from the farthest Moon (the powerful Federates are an exception to the "You must have been there" rule) allows for nuclear-level strikes.

They have also devised water pools and can jump from one to another using ships in about 80-km jumps, with the pool absorbing the excess of momentum.

In your case, the velocity differential is about 0.6 m/s for each 10 km jump. Assuming our guy can teleport serially, he would need to visualize each successive waypoint and teleport there, then brace himself to adapt the new local frame vector. Up to 1.5 m/s per second is doable; it would become a sort of "speed walking" or "oblique jogging".

But I think the problem will be that he won't be able to visualize exactly all the necessary waypoints (and, if an error is possible, it might have disastrous consequences). He could do "emergency" long jumps at 80-100 km distance, shedding the excess 4-7 m/s with a paratrooper roll.

Or he could try for a Pierson Puppeteer's style of tele-transport: very short jumps to waypoints in one's line of sight. That would be 500m at most, so the velocity differential is completely negligible. Doing two jumps per second will give a virtual speed of 3600 km/h - around Mach 3.

Another possibility is to fly with a wingsuit. Jump high enough in the air, and start falling. Jump higher and farther (still line of sight, maybe two or three kilometers distant). After a while, downward velocity will stabilize on 30-50 m/s depending on the wingsuit affect, and forward "propulsion" will be supplied by teleporting. To land, simply glide down - or bring a parachute. Two teleports per second equal now a speed of Mach 10.

This is also (sort of) how the Earthmen shipwrecked on the Azhiri planet ultimately reach safety, by having an Azhiri reng large quantities of air into an airfoil to give lift to their lifeboat.

$\endgroup$
  • 1
    $\begingroup$ Your best option for quick landing might be to swoop directly upwards and then teleport to the ground as soon as you hit the top of your curve. $\endgroup$ – John Dvorak Apr 19 at 20:36
  • $\begingroup$ @JohnDvorak that's an amazing idea, and it would certainly work! Can I add it to the answer? $\endgroup$ – LSerni Apr 20 at 18:28
0
$\begingroup$

The Earth has an equatorial circumference of 40000km and turns around in a day. That is about 1660km/h. Seattle is at 47 degree latitude, where the radius to the axis of rotation of the Earth is reduced to 70%

The question is now, how much of a difference in velocity does a jump make? The difference in velocity vector jumping east/west is $2*\sin(\frac{a}{2})$ where $a$ is the angle you are covering. A 100km jump gives you a speed of 4km/h, walking speed. You will need some training to not fall over, but that sounds doable.

If you jump north south, what matters is the change in rotation speed of the earth, so $(\cos(\text{new latitude}) - \cos(\text{old latitude})) * 1660\frac{km}{h}$. A 100km jump around Seattle will leave you there at 3km/h, slightly slower than E/W

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.