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# Fantasy Engineering

An engineer can do for a dime
what any fool can do for a dollar
.

### Related Links (Outside this Site)

John Archibald Wheeler (1911-2008) on Wikipedia.
Mysteria Onera, Technology of Dreams: Ultrahard fullerite "light saber".
Temperature Gradient above the Deep-Sea Floor  by Mark Wimbush  (1970).

### Videos :

How High Can We Build?  by  Michael Stevens   (Vsauce, 2012-06-03).
Real  Possibilities for Interstellar Travel  by  Matt O'Dowd   (PBS Space Time).
Is the Alcubierre Warp Drive Possible?  by  Matt O'Dowd   (PBS Space Time).
Classified Project Orion Interplanetary Space Flight   (BBC 4).

## Impractical Engineering

(D. M. of Titusville, FL. 2001-02-11)
How much hydrogen would you need to raise the Titanic?

How do you want to do it?  I suppose you don't want the wreck airborne  (that would be both silly and unpractical).  I'm guessing that you are not considering inflating some device at the surface of the ocean with long (3800 meters!) cables to pull the wreck.  (Besides, hydrogen is not much better than air for this application.)  So, I assume that your idea is probably along the following lines:

Tie some kind of deflated balloons to the wreck  (or put them inside the hull)  and let some kind of liquefied gas evaporate into these until the whole thing lifts up.  Lowering the gas in liquid form  (in containers)  avoids difficult pumping problems when it comes to overcoming the tremendous pressure difference (about 380 times the atmospheric pressure!).  Once the balloon starts going up, its volume will expand greatly because the surrounding pressure decreases.  In practice (?), you'd have to let some gas out during the ascent so the balloons won't explode.  The maximum amount of gas would be needed at the very bottom, at a depth of about 3800 meter (12500 ft).

Let's assume the necessary work can be carried out at such depth.  The problem is that the gas is under a pressure equal to about 3800*1025*9.8 (depth x density x gravitation) or about 38.2 MPa (377 atm).  What's the density of hydrogen at such a pressure?  Well, hydrogen  cannot  be considered an ideal gas  (pV=RT)  at that pressure, but we may still use the Van der Waals approximation, which reduces to:

(p + 3/V2 ) (3V-1)   =   8 T

Where  p,V,T  are expressed using as units the  p,V,T  values at the critical point of hydrogen  (p=1.293 MPa and T=32.98 K).  The bottom of the ocean is probably at a temperature near the densest point for water (3.98°C) under normal pressure, so let's us  T=277.13 K.  The volume  V  of a mole of hydrogen at  P=38.2 MPa  and  T=277.13 K  is thus about  x  times the critical molar volume of hydrogen, if  x  is solution of:

(38.2/1.293+3/x2)(3x-1) = 8(277.13/32.98)

Which gives  x = 1.024948...  The density of hydrogen near the Titanic wreck would thus be 1/x times the density at the critical point of hydrogen, which I found listed as 0.031 g/cc.  It's therefore roughly equal to 0.03 g/cc  (which is more than 300 times the density of hydrogen at 1 atm and 0°C).

The density of seawater at the surface is about 1.025 g/cc.  Under 377 atm of pressure, this would increase only by about 2% to about 1.043 g/cc.  All told, 0.03 grams of hydrogen generate 1.013 grams of buoyancy at the bottom of the ocean and the mass of hydrogen required is 30/1013 or about 3% of the mass of the whole wreck, if we neglect the self-buoyancy of the wreck.  Taking into account the self-buoyancy of the wreck would probably bring the above number down to about 2.5%.  I am guessing the mass of the Titanic to be about 10,000 metric tons  (if anybody knows better, please let me know); 2.5% of that is 250 metric tons.  That's a lot of hydrogen!  It would be a lot less if the Titanic was in shallower waters, but the huge pressure is the killer here!

 On 2001-03-08, Bob Cat (USA) made the following suggestion: Rather than using a balloon to lift the wreck, run an electrical cable into the ship and use [electrolysis] to fill it with your choice of hydrogen or oxygen.

You do need the same amount of hydrogen whether you put it in a balloon or in the hull of the Titanic (assuming a broken hull at such a depth could somehow be fixed into something airtight). You would also need to let gas out during the ascent, or the hull would almost immediately burst open (just like a balloon would). The solubility of hydrogen in water is rather low under normal pressure  (saturating at 2.14 volumes of hydrogen in 100 volumes of water),  but I would venture a guess that it's much higher at 380 atm of pressure.  If hydrogen is significantly soluble, it can't be used to expel water from the hull, since whatever hydrogen is produced gets dissolved immediately (and oozes out of the wreck through whatever opening is left for water itself to be expelled). Let's not dwell on that particular objection, which may or may not be valid, and focus only on the main aspect of Bob's comment:

Bob's interesting proposition is to use electricity to make 250 metric tons of hydrogen from seawater, instead of bringing it to the site in liquid form. Let's see how much electricity would be needed:  To make a mole of  H 2 gas, it takes 2 faradays of electric charge, or 192970 C [a coulomb (C) is the charge delivered by a current of one ampere in a time of one second ] which is roughly 53.6 A-h (ampere-hour).  What this gives you is just 2 grams of hydrogen.  What you need is about 250 000 000 grams, corresponding to 6.7 GA-h (6700 000 000 A-h).

To complete the job in a century  (876600 hours)  you'd have to run a current of about 7643 A, 24 hour a day 7 days a week.

The  delivery voltage  at the sea floor should be at least a few volts  (I'll be more precise if/when I find the time to derive the proper dependence on pressure for the electrolysis of seawater; I've not done this in 20 years and I am rusty on this particular issue).  Using transformers and AC to DC converters at the sea floor is something you could consider  (it would not be technologically easy)  but let's say we stick with low-tech delivery of straight DC current from the surface...

A length of 3800 m of hefty copper cable ½" in diameter  (12.7 mm)  would have a resistance of about 0.5 W.  Even if we use the ocean as a low-resistance return path  (so that we only need a single wire)  this means a voltage drop of about 3800 volts!  So, you're gonna use about 30 megawatts  (30 MW)  of power, most of it spent warming up seawater...  This is 150% of the power delivered by each of the 4 engines of a 747 jetliner, 24 hours a day for a hundred years!

### A Jumbo Jet for 10 Years

If you have the power of the 4 engines of a jumbo jet  (80 MW)  you'll run a current 63.3% larger  (1.633 is the square root of 80/30)  and do the job in "only" 61 years.  You may want to save time and energy by running N identical cables fed with 1/N of your power:  Each cable would be running a current ÖN smaller than before under a voltage also ÖN times smaller.  The total current in the N cables is thus ÖN bigger for the same power than it would be with a single cable.  Therefore, the whole job may be completed ÖN times as fast.  With N=100 cables, the job would thus apparently be completed in "only" 6 years  (and 45 days).  Actually, the power lost to the return path becomes very much relevant at this point  (see next comment by Edgar Bonet)  so we would need at least 10 solid years, even with a bigger boat to lessen the return resistance...

After viewing this page  (on 2004-06-14)  French physicist Edgar Bonet remarked that the resistance of the ocean on the return path could be estimated by considering the hull of the surface ship through which it goes to be roughly a half-sphere of radius  R = 5 m  (we are gold-plating the thing to avoid corrosion   ).  This gives a resistance of about 6 mW (namely, r/2pR,  where r » 0.2 m is the resistivity of seawater).

This is negligible when we have but a single cable, but roughly doubles the total resistance when we have 100 of them in parallel.  This also means that there's little to gain by adding many more cables beyond this point, unless we're willing to use about half of them on the return path...  Lest they become a critical bottleneck, the bottom electrodes should be quite large too...

Note that each cable has a mass of about 4.3 metric tons, so that the total mass of copper you need to bring on site to produce the hydrogen (in ten years) is about twice the mass of the liquid hydrogen you could have brought.

### A Jumbo Jet for One Hour

Notice also how terribly inefficient the whole thing is.  Lifting a wreck of 10000 metric tons from a depth of 3800 meters requires "only" about 300 gigajoules  (allowing for helpful self-buoyancy).  Using the power of a jumbo jet  (80 MW)  for about an hour would do the trick.  (Actually you don't want to do it that fast because you'd have to fight quite a drag if the wreck is being lifted at 3.8 km/h!)  The hydrogen idea is very wasteful to begin with  (because you end up releasing on the way up almost all the hydrogen you needed to get some buoyancy at the bottom).  If, on top of that, you insist on running long copper cables (presumably from a nuclear-powered ship) to produce the hydrogen where it's needed, you end up with a combined efficiency of roughly 0.001%.

Well, that was a nice excuse for some "fantasy engineering"...  Thanks, Bob.

(2006-09-16)   A "free" gravitational subway system :
Commuting between any pair of destinations on Earth in 42 minutes.

"Boomeranging through the Earth" is the title of the fourth chapter of a 1990 book by  John Archibald Wheeler  (1911-2008)  entitled  A Journey into Gravity and Spacetime  (Scientific American Library, ISBN 0-7167-6034-7).  In it, Wheeler discusses the motion of two  earthships  falling through a shaft drilled through the center of the Earth.  He explains at length that the passengers of each  earthship  would experience the weightlessness associated to a local flatness of their spacetime.  However, they would find out about the  curvature  of their surroundings by tracking the relative motion of the neighboring  earthship  which shares their fate, 2 seconds ahead (or behind).

On a less philosophical level, the gravitational "magic" of a shaft drilled through the Earth is also explored in Chapter 11 of a 1999 book by Robert B. Banks  Slicing Pizzas, Racing Turtles, and Further Adventures in Applied Mathematics  (Princeton University Press  ISBN 0-691-05947-0).  Unlike Wheeler (who has the loftier motivations described above) Banks investigates a shaft which need not go through the center of the Earth.  He quotes the example of a  straight  tunnel from Washington to Boston, 631.23 km in length, reaching its maximum depth of 7.82 km at its midway point, directly under New York city.  Banks needlessly skips some of the elementary theoretical part, which we outline below.

The basic idea is a part of scientific folklore whose exact origins remain unknown.

Let's ignore completely the rotation of the Earth and assume it's simply a perfect homogeneous sphere of radius R whose surface gravity is g  (in the absence of any rotation, g would be close to 9.82 m/s2; slightly  higher  than the standard value).

Before you start raising funds for the actual project, consider just one of the many issues which make it absolutely impractical:  The center of the Earth is actually  hotter  than the surface of the Sun.

By Gauss Theorem applied to gravity (instead of the mathematically similar electrostatic field) we see that the centripetal gravitational field inside the Earth is proportional to the distance  r  from the center.  It is equal to  g r / R.

Assume a  straight  tunnel exists (shown in red in the picture below) between two points on the surface of the Planet and that we can drop an object (a marble or a passenger vehicle) which can slide without any friction along this tunnel.  This object starts with zero speed and can be later retrieved as it emerges on the other side of the tunnel, also with zero speed.  Gravity accelerates the thing until the tunnel's maximum depth is reached (at a point H located at a distance  a  from the center O of the Earth) and decelerates it beyond that point.  This looks like a nice fantasy ride...  Let's see how long it would take:

Let M be the position of the vehicle.  If q is the angle from OH to OM, the vehicle's distance to the center of the Earth is  r = a / cos(q).  The component of the gravitational acceleration along the direction of the the tunnel is:

- g r sin(q) / R     =     - g a tan(q) / R

The above is the second derivative of   x = HM = a tan(q).

Therefore,   x''  +  (g/R) x   =   0

Surprisingly enough, the motion inside the tunnel is thus found to be an harmonic oscillation which does not depend on the parameter  a  (which characterizes the position of the tunnel).  The motion's "period" (twice the travel time from one end of the tunnel to the other)  does  not  depend on the "amplitude" (which is half the length of the tunnel).  In other words, this type of transportation allows "free" travel between two points on the surface of the Earth in a  constant time  equal to:

p Ö(R/g)   =   2530.4187 s

That's about 42 minutes and 10 seconds from  here  to any point on the Globe.

John R. (Yahoo! 2007-07-18)   Reaching the center of the Earth
How long would it take for a ball dropped in a vacuum tube
to reach the center of the Earth?  [Assuming homogeneous density.]

Answer :   Half  of the above time, namely  21 minutes and 5 seconds.

At the center of the Earth, that ball would have a speed  Ö(R g) = 7900 m/s.

Coincidentally,  Ö(R g)  is also the speed of satellites in low Earth orbit.  In either case,  g  denotes the  pure  gravitational field of about  9.82 m/s2  (slightly more than the "standard" acceleration, which includes an extra centrifugal component).

More precisely, a sattelite in a circular orbit of radius  R+z  travels at a slightly lower speed, whose  square  is  R g / (1+z/R)

(2014-10-18)   Detecting a Graviton

Is a Graviton Detectable?   by  Freeman Dyson  at Nanyang (August 2013).
Have Gravitational Waves Been Discovered?  by  Matt O'Dowd   (PBS Space Time, 2015-10-22).

(2014-10-18)   Controled Nuclear Fusion
The reason why the Tokamak approach will never produce energy.

For the record, the author would like to apologize for apparently killing some of the most attractive types of fusion reactors which have been proposed.  He advises future graduate students working on their theses to avoid accidentally demolishing the area of research in which they plan to work after graduation.
(from the doctoral dissertation of  Todd H. Rider  at MIT, 1995)