That's incorrect. From another post about scale weight...
Using the formula given by WindUpGuy:
To find the scale weight of an O scale model - assuming it was built using exactly the same materials in perfect scale dimensions - you would divide the prototype weight by the scale factor cubed, i.e. 48^3 or 110592.
So, 432,000 lb (max weight of an AC6000) / 110592 = 3.9 lbs not 8000 lbs.
Jerry
I've seen that posted before by others more than once over the years, and just like then as well as now, I still don't buy it (and physics doesn't either). It's kind of like that scene in the James Bond movie Thunderball where a full-size sedan is put in a crusher then you are supposed to believe that since it's now smaller, it's also lighter and therefore it can be put in the back of Oddjob's Ford Ranchero without collapsing the suspension & blowing out the tires. 
If you were able to take a prototype engine that is x- number of pounds in the real world and magically shrink it down to 1/48 scale, it would still be the same weight as it was in the 12" = 1' world, although significantly more dense (obviously) and weigh ever-so-slightly less since there would also less air occupying empty spaces in the loco (air has weight too). So a locomotive that's 432,000 pounds would still be 432,000 lbs, no matter if it was half it's original size or if it were as small as a sugar cube.
Truth is that performance (adhesion, rolling resistance, stability) is what is factored when designing scale model and toy trains, not by "scaling down" what the weight would be if it were 12 inches = 1 foot.
You're wrong John, we're not crushing a full size and weight locomotive down to a block the size of an O scale model. A 48 (48x48x48) square foot cube of steel (real world size) would not weigh the same as a 1 (1x1x1) square foot cube of steel(O gauge size)or vice versa.
Jerry
