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(Rob Welling) wrote in message . com...
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

OK everyone, its basically a "statics" problem. Pressure is the same
at any given depth all over the tank regardless of its horizontal
extent. FOR SIMPLICITY and for over engineering, assume the total
outward force is given by pressure at the bottom X the area of the
front glass. With everything in balance, the vertical ends cannot
move and the middle is bowed out. The glass acts as a spring whose
restoring force wants to return to "straight" but it is stretched by
the force. The amount of stretch is given by the "spring constant" X
the horizontal extent of the glass. In other words, a longer span of
glass will stretch more (ie. bow out more). Theoretically, one could
measure the bowing of the glass and resolve the components of the
force at each infinitesimal point and solve for the "spring constant"
of the glass. Failure of materials occurs when they are stretched
beyond their "elastic limit". Unlike other materials that just
stretch more beyond this limit, glass fails catastrophically. Even a
scratch on the glass could result im local stresses beyond failure.
The bottom line is, ask your glass company what kind of load (in
pounds) a given span of your glass can support and make sure it is
less than the force we calculated above.
One analogy is to think of your glass as a very long stiff horizontal
spring supported at the ends. Put many, many, many little weights on
the spring at equal intervals all across it. The weight of the little
weights is the force due to a given pressure against a given area of
glass (spring) so greater depth would be like heavier weights. Due to
the weights, the spring stretches and sags (bows) and will sag more
for more pressure and more for a longer spring. I believe but am not
sure (have drowned too many brain cells with rum in this old punkin)
that the bowing is a catenary.

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