Now, here's where the troubling part comes in. Lets say the
back of the tank, instead of being 72" away from the front is moved forward
until it is just 1/8 of an inch away from the front glass. Now there is less
than two gallons of water in the tank. I have trouble seeing the glass (3/4
in. thick) bowing from 2,500 lbs of "push" from less than two gallons of
water(16 lbs). I suppose we could even shrink the 1/8 in. to a few
thousandths and put a thimble of water in. Would there still be 2,500 lbs
of outward force from a gram or two of water? Would the heavy duty scale
across the room be forced all the way to the two thousand, five hundred
pound mark?

Dixon

Dixon,

Good question. Now, I have no physics background whatsoever, so don't
laugh if I'm way off...but regarding decreasing the depth (front to
back) and maintaining the same psi due to the same height...

Isn't the psi on the vertical surface walls based on overall volume of
the tank's contents? As in, it's calculated from a cubic measurement,
then translated to pressure on a square inch? For instance, the
pressure created by the approx. two gallons of water contained in the
very shallow 1/8th inch depth (front to back) would then
understandably be much less than the original 72"' measurement because
the volume is much less? It's not simply just the depth that makes the
difference, but the cubic volume rather, of which depth is only one
third of the equation.

Of course, the density of the contents, in this case water, makes a
big difference too - but we're dealing with the same contents across
the board here.

Anyway....I agree with the poster who said to check with the
experts...but that's my two (less than qualified) cents!

Capt. Rob Welling
Sarasota, FL