OK, I'll show my work...

40 ft * 60 ft == 2,400 ft**2 (area of the tunnel face)

1,400 mi X 5,280 ft / mi == 7,392,000 ft (length of the tunnel in feet)

7,392,000 ft * 2400 ft2 == 17,740,800,000 ft3 (area of the base X
height == volume of a
rectangular prism)

147,197,952,000 ft3/mi3

17,740,800,000 ft3 / 147,197,952,000 ft3/mi3 == 0.12 mi3

that's right around 197,120,000 truckloads of spoil.

I'm pretty sure that an eighth of a cubic mile of rock and dirt and almost 2 million truckloads would be easily noticed.

We are, of course, disregarding the true length of the tunnel, the volume of any vertical drifts, the difficulties of tunneling under the sea, the amount of cribbing needed and whatever means necessary to control water infiltration into the bore.

As you see, I did do the math and a little tiny bit of engineering.

It's still bullshit.

OK, I'll show my work...

40 ft * 60 ft == 2,400 ft**2 (area of the tunnel face)

1,400 mi X 5,280 ft / mi == 7,392,000 ft (length of the tunnel in feet)

7,392,000 ft * 2400 ft2 == 17,740,800,000 ft3 (area of the base X
height == volume of a
rectangular prism)

147,197,952,000 ft3/mi3

17,740,800,000 ft3 / 147,197,952,000 ft3/mi3 == 0.12 mi3

that's right around 197,120,000 truckloads of spoil.

I'm pretty sure that an eighth of a cubic mile of rock and dirt and almost 2 million truckloads would be easily noticed.

We are, of course, disregarding the true length of the tunnel, the volume of any veritcal drifts, the difficulties of tunneling under the sea, the amount of cribbing needed and whatever means necessary to control water infiltration into the bore.

As you see, I did do the math and a little tiny bit of engineering.

It's still bullshit.