Build a tunnel to Hawaii!

[SLD]

Build a tunnel from Los Angeles, California to Honolulu, Hawaii! I say, to heck with great circle routes, let’s go in a straight line, right through the earth. How long would the tunnel be, and what would be its deepest point under the earth’s surface?



The starting point is 33.75 degrees north, 118.30 degrees west. Your end point is 21.3 degrees north, 157.9 degrees west. Assume the earth is a perfect sphere of radius 6371 kilometers. (Even though it’s really not - this is a math forum, not an engineering forum!)

 

[ideologyhunter]

Just don't give the engineering contract to OceanGate.

 

[SLD]

Ok so for extra credit, assume it’s an oblate spheroid with polar radius of 6357 and equatorial radius of 6378.

That makes the question far, far more difficult.

 

[lpetrich]

That problem has indeed been solved: Oblate Spheroid Geodesic -- from Wolfram MathWorld and (PDF) Geodesic equations and their numerical solutions in geodetic and Cartesian coordinates on an oblate spheroid

I will take on this problem. For typographical convenience, I will make the distance from the center R, the polar angle or colatitude a and the azimuthal angle or longitude p, gives us rectangular coordinates x, y, z

\( x = R \sin a \cos p ,\ y = R \sin a \sin p ,\ z = R \cos a \)

where R is a function of a. The geodesic distance s is given by

\( ds^2 = (R^2 + R'^2) da^2 + (R \sin a)^2 dp^2 \)

Using the geodesic equations of motion, we find

\( \displaystyle{ \frac{d^2p}{ds^2} + 2 \left( \frac{R'}{R} + \cot a \right) \frac{da}{ds}\frac{dp}{ds} = 0 } \)

with solution

\( \displaystyle{ \frac{dp}{ds} = \frac{h}{(R \sin a)^2} } \)

where h is a constant of integration related to the inclination of the geodesic, and

\( \displaystyle{ \frac{da}{ds} = \frac{1}{\sqrt{R^2 + R'^2}} \sqrt{ 1 - \left( \frac{h}{R \sin a} \right)^2 } } \)

This gives us integrals

\( \displaystyle{ s = \int \frac{\sqrt{R^2 + R'^2}}{\sqrt{ 1 - h^2/(R \sin a)^2 }} da } \)

\( \displaystyle{ p = \int \frac{h}{(R \sin a)^2} \frac{\sqrt{R^2 + R'^2}}{\sqrt{ 1 - h^2/(R \sin a)^2 }} da } \)

I'm not sure what to do next. A fully general solution? A perturbative solution around a sphere?

Ellipsoid Geodesic -- from Wolfram MathWorld gives some equations for the triaxial ellipsoid, with all three axes having different lengths.

 

[lpetrich]

To get an idea of how to find the solution, I will set R = 1. Using \( h = \sin a_0 \) the integrals become

\( \displaystyle{ s = \int \frac{\sin a}{\sqrt{(\sin a)^2 - (\sin a_0)^2}} da = \int \frac{d(\cos a)}{\sqrt{(\cos a_0) - (\cos a)^2}} = \arcsin \frac{\cos a}{\cos a_0} } \)

\( \cos a = \cos a_0 \sin s \)

\( \displaystyle{ p = \int \frac{\sin a_0}{\sin a\sqrt{(\sin a)^2 - (\sin a_0)^2}} da = \int \frac{\sin a_0\, ds}{1 - (\cos a_0)^2 (\sin s)^2} } \)

Multiplying the numerator and denominator of the integrand by \( (\sec s)^2 \) gives us

\( \displaystyle{ p = \int \frac{ \sin a_0 \, d(\tan s)}{1 + (\sin a_0)^2 (\tan s)^2} = \arctan (\sin a_0 \tan s) } \)

 

[SLD]

That problem has indeed been solved: Oblate Spheroid Geodesic -- from Wolfram MathWorld and (PDF) Geodesic equations and their numerical solutions in geodetic and Cartesian coordinates on an oblate spheroid

I will take on this problem. For typographical convenience, I will make the distance from the center R, the polar angle or colatitude a and the azimuthal angle or longitude p, gives us rectangular coordinates x, y, z

\( x = R \sin a \cos p ,\ y = R \sin a \sin p ,\ z = R \cos a \)

where R is a function of a. The geodesic distance s is given by

\( ds^2 = (R^2 + R'^2) da^2 + (R \sin a)^2 dp^2 \)

Using the geodesic equations of motion, we find

\( \displaystyle{ \frac{d^2p}{ds^2} + 2 \left( \frac{R'}{R} + \cot a \right) \frac{da}{ds}\frac{dp}{ds} = 0 } \)

with solution

\( \displaystyle{ \frac{dp}{ds} = \frac{h}{(R \sin a)^2} } \)

where h is a constant of integration related to the inclination of the geodesic, and

\( \displaystyle{ \frac{da}{ds} = \frac{1}{\sqrt{R^2 + R'^2}} \sqrt{ 1 - \left( \frac{h}{R \sin a} \right)^2 } } \)

This gives us integrals

\( \displaystyle{ s = \int \frac{\sqrt{R^2 + R'^2}}{\sqrt{ 1 - h^2/(R \sin a)^2 }} da } \)

\( \displaystyle{ p = \int \frac{h}{(R \sin a)^2} \frac{\sqrt{R^2 + R'^2}}{\sqrt{ 1 - h^2/(R \sin a)^2 }} da } \)

I'm not sure what to do next. A fully general solution? A perturbative solution around a sphere?

Ellipsoid Geodesic -- from Wolfram MathWorld gives some equations for the triaxial ellipsoid, with all three axes having different lengths.

The earth has only two axes of different lengths. Aside of course from mountains and such.but for this purpose we only need calculate tunnels from a few feet above sea level.

 

[bilby]

for this purpose we only need calculate tunnels from a few feet above sea level

As long as the spoil heap(s) are kept well away from the tunnel entrances. So you'll need a fleet of trucks to shift that spoil, and a load of unobtainium tunnel linings, plus a couple of TBMs with cutting faces that can penetrate the Mohorovičić discontinuity and drill through mantle rocks under massive pressure.

It's probably easier just to buy a plane ticket.

 

[Elixir]

So you'll need a fleet of trucks to shift that spoil, and a load of unobtainium tunnel linings, plus a couple of TBMs with cutting faces that can penetrate the Mohorovičić discontinuity and drill through mantle rocks under massive pressure.

Jeez. The OP only asked for a few specs of the hypothetical tunnel.

It's probably easier just to buy a plane ticket.

I agree. For me it would be easier to buy a plane ticket than to calculate the answers to the OP question. For that matter it would be easier for me to fly to Hawaii without an airplane than to answer the OP.
But I could give a fairly good guess if I sacrificed my globe for The Cause.

 

[steve_bank]

In La drill down be;ow the lowest part of the ocean crossing to Hawaii.

Assuming there is nothing in the way the runnel is a circular tube at a radius of the Earth.

The distance of the tunnel is the circumference between La and Hawaii at the depth of the tunnel.. The tunnel volume is the volume integral of the circular cross section of the tunnel as dx -> 0. Where dx is the limit as x->0 of the circumference. I don't know LaTex.

Or go down deep enough to make a straight tunnel. For a straight tunnel the distance is the base of an equilateral triangle with the point at the center of the Earth.

Given an average density of martial kg/m^3 the volume tells you how much mass has to be removed.

To build the tunnel in a year you can figure out how much material to remove per day.

Deep sea subs like Alvin use solid titanium shells. A thick titanium tunnel comprised of welded sections might work. The problem would be having enough flex to deal with shifts and quakes.

 

[lpetrich]

Going in a straight line through our planet's interior? Let us get some numbers. The distance between the closest land parts of Los Angeles and the Big Island is 2440 mi / 3926 km, with central LA - central Hilo adding 20 mi / 32 km, and central LA - central Honolulu adding 121 mi / 195 km. That's about 35.31 degrees angle at the Earth's center. For a straight-line path, that gives us a maximum depth of 300 km / 200 mi.

3.3 Earth’s Interior Heat – Physical Geology, First University of Saskatchewan Edition - at 300 km, the Earth's interior temperature is around 1800 C. That's about the maximum rated temperature for some firebricks. Checking on Periodic Table - Ptable - Properties - most common metals are liquid and some are gaseous at that temperature. Looking at pyrolysis temperatures of plastics and wood, where they break down into carbon and volatiles, these are something like 500 C, well below 1800 C. So one will have to use tungsten and silicon carbide and the like - it's close to the melting point of aluminum oxide (alumina, sapphire).

gravity - What is the pressure at the center of the Earth? (Or a neutron star) - Physics Stack Exchange - at 300 km, the pressure is about 12 gigapascals or 120 kilobar.

 

[steve_bank]

Ipetrich is sounding like an engineer ....

Of course it is impossible, but what if exercises like this are informative. Exploring the impossible can lead to an idea unrelated to the impossible problem. It happens.

 

[steve_bank]

A paper napkin approximation, sans calculus.

1. Sketch a circle and put two points L and H approximately proportional to the La to Hawaii distance.
2. Sketch a radius from the center of the Earth to each point labeled r, label the angle phi.
3. Sketch a straight line from L to H, a chord.
4. Sketch a radius from the circle center through the midpoint of the chord.
5 label one half the chord x
6. Label the radius segment from the midpoint to the circle as d
6. Label he radius segment form the midpoint to the center as y.
7. A circle is 360 degrees, phi = (360 degrees)*(distance from L to H on circle )/(circumference of the circle)
8. Two right tangles are formed. Angles are 90 , 90-phi/2,and phi/2 degrees. Find x,y, and d and d max depth.

Python
import math
maxoceadepth = 6 # km
r = 6378 # Earth radiu km
edensity = 5515 # kg/m3
dla2hon = 4119 #km La to Honolulu
ecirc = math.pi*2*r # Earth circumference
phi = 2. * math.pi * (dla2hon/ecirc) # angle in radians
x = r * math.sin(phi/2.)
dtunnel = x * 2.
y = r * math.cos(phi/2.)
depth = r - y
tunpasc = 1e9*depth/30. # change in prssure
tunpsi = tunpasc*6894.76
T = (dtunnel*.5 ) - 273
print("Marianas Trench Pressure Pascals 1e+8 PSI %.1e" %(1e8/6894.76))
print("Tunnel Pressure Pascals %.1e PSI %.1e" %(tunpasc,tunpsi))
print("Distance La Honolulu %.1f km %.1f miles" %(dla2hon,dla2hon*0.621371))
print("Max Depth %.1f km %.1f miles" %(depth,depth*0.621371))
print("Tunnel Length %.1f km %.1f miles" %(dtunnel,dtunnel*0.621371))
print("Temperature Rise %.1f C\n" %T)

Marianas Trench Pressure Pascals 1e+8 PSI 1.5e+04
Tunnel Pressure Pascals 1.1e+10 PSI 7.6e+13
Distance La Honolulu 4119.0 km 2559.4 miles
Max Depth 329.6 km 204.8 miles
Tunnel Length 4047.8 km 2515.2 miles
Temperature Rise 1750.9 C

Submersibles have been to the bottom of the Marianas Trench.

There are several approximations for temperature rise and pressure rise. It would take some reading to get to the basis for he approximations, taken with a grain of salt. It is hard for me to read the low resolution plots.

What is the geothermal gradient in the mantle?
Geothermal gradient - Wikipedia
Thus, the geothermal gradient within the bulk of Earth's mantle is of the order of 0.5 kelvin per kilometer, and is determined by the adiabatic gradient associated with mantle material (peridotite in the upper mantle).

The gigapascal is the conventional unit of pressure when discussing the deep earth. From the surface to the transition zone - lower mantle boundary (670 km depth) we can estimate the pressure by the simple rule of thumb that each 30 km depth change is close to 1 GPa pressure change.

 

[thebeave]

This reminds of a physics problem I had in upper division physics (Dynamics?) from 40 years ago. It was to determine the equation of motion for a spherical chicken, traveling through a tunnel bored directly through the center of the earth. Assume no friction and that pesky molten outer core doesn't exist. Turns out the object will just oscillate back and forth indefinitely from one end of the earth to the other, like a pendulum. Starts out slow, accelerating towards the core, then slows again at the other end and goes back in the same fashion it arrived. Free travel to China...no fuel required! A new project for Elon Musk. He's already halfway there with his Boring Company and hyperloop technology.

 

[bilby]

Free travel to China...no fuel required!

Why do Americans think that the opposite side of the world is in China?

No part of the USA has any part of China as its antipode, and that should be instantly obvious, given that both nations are located in the Northern Hemisphere.

A tunnel drilled through the centre of the Earth from anywhere in the USA would emerge in the Indian Ocean, with the exception of Hawaii (from which the destination would be Botswana), and the far north coast of Alaska (which lies opposite part of the Antarctic coastline).

A tunnel from the Eastern third of China would emerge in Chile or Argentina; From the Western two-thirds, you would reach the South Atlantic.

 

[thebeave]

Free travel to China...no fuel required!

Why do Americans think that the opposite side of the world is in China?

No part of the USA has any part of China as its antipode, and that should be instantly obvious, given that both nations are located in the Northern Hemisphere.

A tunnel drilled through the centre of the Earth from anywhere in the USA would emerge in the Indian Ocean, with the exception of Hawaii (from which the destination would be Botswana), and the far north coast of Alaska (which lies opposite part of the Antarctic coastline).

A tunnel from the Eastern third of China would emerge in Chile or Argentina; From the Western two-thirds, you would reach the South Atlantic.

Why do Austrailians not understand about old timey expressions and dry humor?:

The Hole Truth About Why We 'Dig to China'

Digging to China is a favorite pastime of American children, not to mention a classic parental joke. Any kid with a shovel in their hand has probably been asked if they were digging to China, told to stop trying to dig to China, or perhaps assigned to the impossible task by an exhausted (or devious) caretaker.

In August of 1937 the New York Times published a cute, two-sentence Associated Press wire story with the headline “Boys Uncover Buried Treasure.” Here is the story in its entirety:

Rochester, N.Y., Six small boys, intent on “digging to China” via Homer Hood’s backyard, struck buried treasure. It was a tiny metal box crammed full of old coins and small denominations, sailor buttons and a gift chain.

How Homer Hood felt about the ditch in his backyard or whether he was entitled to any of the booty has been lost to time. But where exactly did this fanciful notion spring from?

 

[bilby]

Why do Austrailians not understand about old timey expressions and dry humor?

I completely understand old timey expressions and dry humour.

But that doesn't answer (nor even address) my question.

Why China?

Your response doesn't provide any answer beyond "We have been making this mistake for so long that it has become traditional". That does't explain the origins of the mistake; Or make it any less incomprehensible.

It just highlights that Americans don't have a clue about how old something must be, in order to become traditional (1937 isn't even close to being a long time ago. My father was alive in 1937 FFS).

 

[thebeave]

Why do Austrailians not understand about old timey expressions and dry humor?

I completely understand old timey expressions and dry humour.

But that doesn't answer (nor even address) my question.

Why China?

Your response doesn't provide any answer beyond "We have been making this mistake for so long that it has become traditional". That does't explain the origins of the mistake; Or make it any less incomprehensible.

It just highlights that Americans don't have a clue about how old something must be, in order to become traditional (1937 isn't even close to being a long time ago. My father was alive in 1937 FFS).

Sorry, dude. Didn't mean to be so flippant about Why China. Sounds pretty important, though. I'll get right on it.

 

[steve_bank]

I would not touch that topic 'with a 10 foot pole' or 'all the tea in China'.

 

[thebeave]

I would not touch that topic 'with a 10 foot pole' or 'all the tea in China'.

If you want to avoid bilby's wrath against the American way, you'd be wise to revise that to "with a 3.048 meter pole". Also be prepared to answer why all the tea is in China and not, say, Japan.

 

[bilby]

If you want to avoid bilby's wrath against the American way, you'd be wise to revise that to "with a 3.048 meter pole".

Metre.