How much does a seismosaurus weigh?

Answer 1

If a 45 ft. long T-Rex weighs 7 tons - and he is a "mover", it seems reasonable to assume that an animal more than three times as long - which probably doesn't have to move very fast - would weigh at least three times as much and, considering what it ate, T-Rex was probably a lot leaner than a Seismosaur.

3 x 7 (tons) x 2,000 lbs. = 45,000 lbs (roughly equivalent to nine Chevy Blazers).

Perhaps some scientist out there could work backwards from a calculation based on how much it needed to eat and how long that took to break down, but below is my real point - which has more to do with lifestyle and mobility ...

Growing up, I was taught that Brontosaurus's (as Sauropods were called back then) lived in swamps because they couldn't support their own weight on land.

Doing a "back of the napkin" calculation ...

A 40,000 pound animal standing on four legs puts 10,000 lb (two Chevy Blazers) on each leg.

With a typical 18" foot diameter, each foot would have exerted about 40 lbs. per square inch on the ground.

Sand has a load bearing capacity of roughly 21 pounds per square inch.

I suggest that adult Sauropods would have sunk to their hips in sand or soft soil - and stayed there - so, they probably didn't go near swamps or shores or bogs - at least not more than once !

RobinOfTheWest