Talk:Gravity train

Error

The error has now been corrected (see previous version of talk page).--New Thought 10:16, 6 September 2005 (UTC)[reply]

I think there is some error in calculation about tunnel from London to Paris (Mathematical considerations section). It is written "would need the creation of a 55,704-metre-deep hole" 55 km, but my simple calculations from here http://physics.stackexchange.com/questions/296949/why-we-still-do-not-have-gravity-train/ shows that it will be only 2.4 km and it looks like possible. — Preceding unsigned comment added by Zlelik2000 (talk • contribs) 14:40, 6 December 2016 (UTC)[reply]

Small deep tunnel

I suggest to describe more the small deep tunnel (less than 5 km deep). Because Objections section does not look relevant for small size tunnel. Like I calculated here http://physics.stackexchange.com/questions/296949/why-we-still-do-not-have-gravity-train/ 1-5 km deep tunnel can give us distance travel up to 500 km and average speed 700 km/h. What are the objections in this case?

Hilarious

The gravity train graphic is the most unintentionally hilarious thing I've seen for a long time. On the other hand, it's supremely distracting. Does the concept even need illustrating?

I love that graphic, thanks to the guy or gal who did it. Maikel (talk) 13:19, 23 May 2009 (UTC)[reply]

The graphic is worse than distracting, it apports no information while it is misleading people into thinking that such a tunnel must pass through the center of Earth, I vote for its deletion.

Did you wish to provide a better one? Friendly Person (talk) 02:54, 10 April 2014 (UTC)[reply]

No. But if in the page for the PI number I saw a picture that shows "PI = 3", I would vote for deletion, too (even if I do not provide an alternate image with all the digits of PI). And at least, "PI = 3" is an approximation that may be valid for some calculation (which is more what can be said of the image at the page). — Preceding unsigned comment added by 79.109.218.80 (talk) 18:06, 7 May 2014 (UTC)[reply]

Dumb

Dumb idea, what about the magma?

That's why it's a "theoretical" means of travel.

Dumb issue, you do not need to build a "straight" tunnel to get a gravity train. You may keep the tunnel in the Earth crust.

Earth's rotation

How does the earth's rotation affect the line of travel? Particularly when the train would travel east/west as opposed to directly north/south. 207.245.73.241 13:53, 9 October 2006 (UTC)[reply]

Vacuum

The comment "Evacuating the air to eliminate drag would require additional power." seems to over-state the difficulty of such a task. Once an extremely strong sealed pipe has been 'laid' through the earth, evacuating it and providing airlocks at each end seems a relatively trivial engineering feat. In fact it would probably be required to avoid the immense air pressure at the bottom of the hole (or even to stop it liquefying?).

I assumed, but didn't know the Gravity Train concept had been thought of before. As a kid I used to think about such things. I only considered the hole through the middle of the earth, the entire journey would be weightless. It seems to be on the limits of physical practicality, the main problem being the immense pressure (and currents), not the temperature which is (probably) much more reasonable. Perhaps someone could expand on just how immense this pressure is.

It's similar to my low earth orbit idea, where an evacuated pipe is laid on or under the sea, and vehicles are accelerated to mach 22 (or whatever LEO speed is) inside this pipe, where they will levitate naturally. It would take the same time to get to the other side, and assuming the energy is recovered, also lossless (and weightless of course). Again, not the safest form of travel. --Adx 01:31, 23 October 2006 (UTC)[reply]

I don't think the trip would be weightless, unless the trip was directly through the centre of the world. The gravity vector would be on an angle which could be resolved into two forces only one of which would provide forward motion, the other would still act perpendicular to that motion. The ratio of those forces would make you weigh less, but not weightless. Vespine 02:47, 8 February 2007 (UTC)[reply]

In my last paragraph I was referring to an orbital trajectory (weightless by definition), only at the surface of the earth rather than slightly above it (low earth orbit). Probably "ballistic trajectory" is the better term. One example goes around the earth, the other straight through it. I hadn't thought about it, but it should be possible to have a weightless trip using any mix of these two paths, provided you are shot down the hole with some initial speed. For example, if the path bent so that at the midpoint of your trip you were halfway between the surface and the centre, then you would be orbiting what was effectively a half diamater earth at that point. I wonder what the path would look like. --Adx (talk) 15:02, 22 November 2007 (UTC)[reply]

The other Gravity Train

The non theoretical gravity train also needs an entry. It was used a lot in the 1800's a horse would pull several cars to the top of a hill, it would be loaded with cargo and the train would coast down the hill with a brakeman or two (and the horse - see Dandy waggon). Does anyone have thoughts on if both should be discussed in this article or if they should have separate articles and if separate how to handle the disambiguation? Jeepday 03:59, 19 November 2006 (UTC)[reply]

I think they should have separate articles. Probably call your article on the other practical one "Gravity train" because that seems to be the definite name of it. But I dunno what the best name for this theoretical article is. Took me a while to find that this article actually existed. As for disambiguation, all we'd need is links at the top of each article. There are templates, have a look at Wikipedia:Disambiguation. Vadmium 14:27, 17 June 2007 (UTC)[reply]

We already have an article, see Gravity railroad. Biscuittin 12:44, 4 November 2007 (UTC)[reply]

External links

I just thought I would add an external links section

will add this link

http://www.damninteresting.com/?p=696

--Mrebus 19:46, 17 October 2006 (UTC)[reply]

Important Fact Ignored?

Since gravity is a function of mass and distance from the center of gravity of that mass, an interesting thing would happen at the center of the earth: If you could hollow out the center, and position yourself in the hollow space, you would be pulled simultaneously in all directions, and float weightless.

What I'm wondering, is whether the acceleration assumed by the takes this into account. We think of acceleration due to the earth's gravity to be 32 feet/second/second, but this is not constant. As you proceed toward the center of the earth, the mass above you increases, and begins to offset the gravitational pull of the mass below you. As a result, the rate of acceleration will decrease over time.

Any thoughts on this?

hubby2debbie--Hubby2debbie 03:36, 5 December 2006 (UTC)[reply]

I believe that's why the train accelerates until it's in the center, and then decelerates until it comes to a stop at the other side. Something like a parabolic acceleration but with one side negative. -72.87.188.204 19:20, 11 February 2007 (UTC)[reply]

At school I was taught that the gravitational field inside a solid sphere is equivalent to being on the surface of a smaller sphere at the same centre, ie ignore all the matter at a radius 'above' your position. (That was a decade or 3 ago though so you would want to verify this for yourself.) Also see my comment above. Adx (talk) 00:49, 19 October 2009 (UTC)[reply]

That is correct. Anwhere inside a hollow sphere of uniform thickness, gravity is zero. There's a simple rationale for this, see a physics textbook. If we view the earth as a series of concentric spheres, then supposing this tunnel has been bored thru the earth, as we climb down the ladder, we weigh less and less. We are on the surface of a solid ball, with the rest of the earth outside that forming a hollow sphere (so zero pull.) The effect is linear with depth, at the center we weigh zero. Friendly Person (talk) 03:01, 10 April 2014 (UTC)[reply]

Solid moon

Several articles, such as Moon and Lunar_Laser_Ranging_Experiment both say the moon has a liquid core. -72.87.188.204 19:15, 11 February 2007 (UTC)[reply]

Point taken. Can anybody think of a substitute? Maikel (talk) 13:18, 23 May 2009 (UTC)[reply]

"and Wes"?

Robert Hooke and Wes? Wes who? —The preceding unsigned comment was added by 66.27.69.167 (talk) 04:05, 9 March 2007 (UTC).[reply]

Friction

Quote: All straight-line gravity trains on a given planet take exactly the same amount of time to complete a journey (that is, no matter where on the surface the two endpoints of its trajectory are located). For Earth, this time would equal 42 minutes and 12 seconds if it were a perfect sphere.

Where does Friction come into this? For this example, assume friction is measured in units of Kevin (Kev), the amount of friction required to decelerate one clothed, 100kg human 1m/s. If System A has a friction of 25 kKev over the entire route, and System B has a friction of 30 kKev, doesn't system A run faster? —Preceding unsigned comment added by 24.205.50.170 (talk) 10:41, 10 October 2008 (UTC)[reply]

I'd guess that neither system would run very fast. Not sure how you're relating your Kevin unit to mass, speed, body shape, viscosity, etc. but maybe it's just intended to be a kind of qualitative idea? 25 or 30 kKev would slow a human being down by 25 or 30 thousand meters per second (over some period of time, hopefully not 1 second) Friendly Person (talk) 03:12, 10 April 2014 (UTC)[reply]

Perfect sphere?

"All straight-line gravity trains on a given planet take exactly the same amount of time to complete a journey (that is, no matter where on the surface the two endpoints of its trajectory are located). For Earth, this time would equal 42 minutes and 12 seconds if it were a perfect sphere." Is the first sentence true even if the planet is not a perfect sphere? --NE2 06:42, 27 October 2008 (UTC)[reply]

No. Divide the height variation by the peak speed, and you'll get the time variation, or at least that's my first guess as to how to calculate it. Jim.henderson (talk) 02:50, 1 November 2008 (UTC)[reply]

Total Recall 2012

Several flaws I saw in the movie.

Gravity - Without external accelleration, you should be in free fall for the entire trip.

Time - Since they reduced the time from 42 minutes to 17 minutes, sounds like the added a 1 G acceleration to the core beyond what gravity force you are experiencing.

Reversals - You should have an up / down reversal when you accelerate from surface steady to 2g down when you start out, another reversal when you switch from 1g to destination to 1g away from destination at the core, and a 3rd reversal when you arrive at the other side and reverse from 2g down to surface steady.

Air speed - several times the speed of sound only blows their hair a little bit? 75 mph hurricanes blow people off their feet. Or are they pumping the air along with them.

Pressure - If I change the barometer reading in M$ Flight Simulator X from 29.92 to 30.92, the altitude reading changes by 1,000 ft. 7,000? miles * 5,000(aproximate) ft / mi = 35,000,000 ft so the barometer reading would be 35,029.92, or about 1000 atmospheres. But that is assuming a linear pressure increase, I think it increases more than that. But the reducing gravity may balance that out some.

Lining - That must be some plexiglass lining. 5 million atmospheres and 10,000K. At least they showed lava-like conditions. And how do they keep the air cool? — Preceding unsigned comment added by Maschwab (talk • contribs) 21:33, 10 August 2012 (UTC)[reply]

For a straight line gravity train that passes through the core, during the first half of the trip would be free fall (thus appear to be weightless). The second half would be in deceleration combined with an increase in gravitational acceleration. Only the first half will be seem weightless while the second half will increasingly return to normal feeling of weight.

In that movie, the gravity train is not straight line, nor does it pass through the core (it curves around it). The movie does not depict the actual sense of weight for a straight line through-core gravity train, nor for the curved-around-core gravity train of the movie.

This current article does not discuss the curved-around type used in the movie, nor 2 other types of gravity trains (straight-line chord gravity trains, and "arc-chord" gravity trains — which the GT in the movie is an inverse version of, ie, the movie version curves away from the core which a typical arc-chord GT would curve toward the core). All but the straight-line through the core GT act somewhere in between the description I gave in the first paragraph of this response, and that of a conventional roller coaster. How much between those two behaviours depends on how it curves and how much curve (if any), as well as how close it approaches the center of the planet. A straight line GT from Los Angeles to New York (a chord GT) would behave like a rather overly drawn-out roller coaster and there will be no free fall involved. — al-Shimoni (talk) 10:02, 1 December 2012 (UTC)[reply]

"while the second half will increasingly return to normal feeling of weight." - that isn't true al-Shimoni. You would be weightless during that part as well, you will only feel weight if the train is stopped by a platform of some kind during the exist. Ariel. (talk) 23:33, 14 December 2015 (UTC)[reply]

Removing argument about "pull to the side"

I've removed this argument: <blockqote> Another objection (argument) could be that the portions of the planet which are at the sides of the tunnel (see graphic annimation above) would also exert its gravitational pull on the sides of the train,which would be in opposite directions, thus slowing down the train. In a graphic as shown above we are guided by our concept of UP and DOWN. But as far as laws of gravity are considered there is no up or down. There is only gravitational pull of earth inside the tunnel from all the directions which has to be considered. It is possible that the train may come to a halt much before it reaches the centre. It may go on moving only as long as the mass and gravity of the earth in the direction in which it is moving is strong enough. After that it may come to a halt.

It seems to be based on a totally confused opinion about how one calculates the total force of gravity, as well as misunderstandings about how friction works; I thought about adding "citation needed", but decided to WP:BB and just remove it. --Alvestrand (talk) 06:09, 16 June 2013 (UTC)[reply]

Through the center of Earth? Straw man argument

An important missunderstanding that this page shows is the idea that a gravity train has to be a "straight" line from A to B. A gravity train from London to Los Angeles could be build inside the Earth crust (up to 40 kms depth); assuming the energy lost due to friction is recovered with motors, as described:

The train is left at London to "fall" into the tunnel, getting speed. 30 km of vertical, frictionless fall provide quite a good final speed of 766 m/s (twice the speed of sound)

Once the "fall" has stopped the train would keep moving in a "flat" trajectory (with respect to the gravitatory potential), keeping the speed.

When the train approachs the destination it starts climbing against the gravitatory field, losing speed. If the destination has the same gravitatory potential that the source, it stops exactly when it arrives at the surface; if it has more (is "higher") additional energy has to be supplied just as a train needs more energy to climb a hill.

In essence it could be compared with a skateboarder doing stunts in an "U" shaped wall, he uses the gravity pull from move from one point of the wall to another but does not need to go in an straight line.

Of course, the issues of cost of the infrastructure and the cost of keeping the vacuum in the tunnel are valid ones, but the "you cannot make a tunnel through the mantle!!!" opinions are without merit because that is not needed.

Please mention and cross reference Concrete Gravity Trains used for energy storage

Please mention and cross reference Concrete Gravity Trains used for energy storage. See, for example: http://interestingengineering.com/concrete-gravity-trains-may-solve-energy-storage-problem/ Thanks! --Lbeaumont (talk) 18:20, 20 August 2017 (UTC)[reply]

Although it has a similar name, this article is about a specific type of train that uses gravity to travel through a tunnel deep below the surface of the Earth. ARES is already mentioned at Energy_storage#Mechanical_storage. Dlthewave (talk) 18:44, 20 August 2017 (UTC)[reply]

Atmospheric Pressure in the Tunnel

Based on ideal gas law and constant temperature at 15°C as well as an Earth with uniform density, my estimate of the air pressure at the center of the Earth is about 10¹⁶⁴ (1 followed by 164 zeros) times that of the atmospheric pressure at the Earth's surface. Apparently, this number is too large to be realistic, but it implies that the pressure in the tunnel would be so large that the air would probably be in liquid or even solid or some supercritical state. (Would someone make a more realistic estimate?) This would make the environment for the digging impossibly difficult. --Roland (talk) 21:13, 19 March 2019 (UTC)[reply]

Less ambitious, more practical versions ?

I can envision a variant where something like subway cars are used, in short, non-evacuated tunnels, and the initial angle is just a bit more than needed to overcome rolling friction and air resistance. The tunnel could then level off, near the bottom, allowing the trains to coast to a stop at a station there, where the passengers could switch trains. Power would be needed to lift them back up to the surface from there, for the 2nd half of the trip. The advantage would be only needing to provide power for half the trip. Presumably more energy would be needed on the trip back up, than in a traditional subway, and the total would be comparably to a regular subway, but there is the initial cost savings of not having to lay down a third rail or other power source on half the tunnel, and the need for braking could be greatly reduced for both halves, perhaps saving a bit of energy there. So, is there such a concept ? And is this within the scope of this article ? SinisterLefty (talk) 15:59, 22 March 2019 (UTC)[reply]

Comment

I think this article needs to be rewritten, either to describe "a gravity train as a theoretical means of transportation", which is what the introduction says, or to describe the construction of such a means of transport. At the very least, the title of the first section should be changed to something like "Practical Engineering Difficulties" and it should be moved to the end of the article after the concept has been defined in detail. It is jarring to say the least to find that the first section "objects" to a perfectly valid, and interesting, "theoretical means of transport". Some diagrams would help as well.81.154.119.34 (talk) 11:38, 15 November 2019 (UTC)[reply]