This comment was posted to reddit on Jul 26, 2015 at 3:30 pm and was deleted within 1 hour(s) and 15 minutes.

Significant is relative I suppose. I'll try to explain my reasoning.

Firstly, I'm using an ellipsoid as a representation of the sub. It's the simplest shape I can think of to model it in a simple way.

Secondly, to calculate the amount of steel that makes up the hull of the sub I'm using the equations for ellipsoid volume. The volume of the outer hull dimensions minus the volume of the inner bulk should give the volume of the hull, in a rough way.

The equation for the volume of an ellipsoid is:

V=(4/3)*pi*a*b*c

Where A, B, and C are the width, height, and length of your sub.

Let's use small numbers to simplify things, so A= 1 sub length. B =1/10th sub length, and C=1/10th sub length. Volume then equals 0.04186666667

What percentage of the length, width, and height of the sub should be taken up by hull? Now idea, so I'll throw 2% in.

So inner A becomes .98, inner B becomes .098, and inner C become .098. This volume equals 0.0394045717.

Volume 1 - volume 2 = .004. The volume of our example submarine equals .002462095 sub lengths^{3}

Now let's reduce the outer hull by 1/6720th.

Outer A becomes .99985, outer B becomes .099985, and outer C becomes .099985. The volume equals 0.0418478295.

The new volume of your hull would be .0024432578.

So a decrease .001488% of you hull would result in a hull volume loss of .0076508827 sub length's^{3.}

Or, a 1 inch decrease in hull dimensions would result in a loss of 4.2844 cubic feet of steel. Your weight for steel was .28 lbs/inch^{3.} That would be about 483 pounds per cubic foot.

So the actual loss of weight you would get would be closer to 2099 pounds rather than the 5,500 tons you calculated. That's roughly 1/5500th of what you calculated.