"dixon" wrote
I wanted to build a 1000 plus gallon aquarium in the new house I was
having
built. I wasn't sure how thick to make the front glass.
.... To figure the total
"push" on the front glass, you just need to find the pressure at the
halfway
point(19 inches in this case) and multiply by the total sq. in. of the
glass
(3,648).

There's part of your problem. You need to size the glass for the total
force across the span of the glass, at max pressure ie along the bottom
edge.


... I believe
at 19" the pressure was about .686 or so psi. It figured out to be around
2,500 lbs. of force on the glass. It would be about 9,480 lbs of water.
As
I
was filling the tank, a straight edge laid against the front glass showed
the glass bowing outward very noticeably even at 1/4 full. I nervously
filled it full. Now it was very bowed even to the naked eye. The 2,500 lbs
was definitely there on the glass.

Yep. Just like magic! I don't see where there is any contradiction here
bewteen force on the glass and measurement of the water column.

P.Fritz wrote:
You are mixing apples and oranges. The force of water at a certain depth
remains the same......Xpsi.

When you take that force and apply it over an area, the cumulative amount of
force reacts against the plane of glass or plexi...

In your thimble example....you only have 1 sq. in of force at 19" so the
force remains .686 psi.....

The plane of glass has to resist two basic forces.......shear and bending
moment.......typically the thickness required to resist the bending is
greater than that required for shear. Bending moment is a relationship of
the total force and the span of the material.

Correct. The glass should be sized so that the spanwise loading results
in little or no deformation (or elongation as some term it) of the
material. The easiest way to figure it, and a way that gives a safe
result, is to figure the total force on the glass as acting on a point
at the center, and size/spec the glass to withstand this force. It might
be a little overkill though and a bit more expensive.

But it sounds to me like you've already got a piece of glass that will
withstand the force, it's just a question of how long it will continue
to do so, does it have enough safety margin, and do you mind it looking
all bulged out like that

Regards
Doug King

"DSK" wrote in message
.. .
"dixon" wrote
I wanted to build a 1000 plus gallon aquarium in the new house I was
having
built. I wasn't sure how thick to make the front glass.
.... To figure the total
"push" on the front glass, you just need to find the pressure at the
halfway
point(19 inches in this case) and multiply by the total sq. in. of the
glass
(3,648).

There's part of your problem. You need to size the glass for the total
force across the span of the glass, at max pressure ie along the bottom
edge.


... I believe
at 19" the pressure was about .686 or so psi. It figured out to be
around
2,500 lbs. of force on the glass. It would be about 9,480 lbs of water.
As
I
was filling the tank, a straight edge laid against the front glass
showed
the glass bowing outward very noticeably even at 1/4 full. I nervously
filled it full. Now it was very bowed even to the naked eye. The 2,500
lbs
was definitely there on the glass.

Yep. Just like magic! I don't see where there is any contradiction here
bewteen force on the glass and measurement of the water column.

P.Fritz wrote:
You are mixing apples and oranges. The force of water at a certain
depth
remains the same......Xpsi.

When you take that force and apply it over an area, the cumulative
amount of
force reacts against the plane of glass or plexi...

In your thimble example....you only have 1 sq. in of force at 19" so the
force remains .686 psi.....

The plane of glass has to resist two basic forces.......shear and
bending
moment.......typically the thickness required to resist the bending is
greater than that required for shear. Bending moment is a relationship
of
the total force and the span of the material.

Correct. The glass should be sized so that the spanwise loading results
in little or no deformation (or elongation as some term it) of the
material. The easiest way to figure it, and a way that gives a safe
result, is to figure the total force on the glass as acting on a point
at the center, and size/spec the glass to withstand this force. It might
be a little overkill though and a bit more expensive.

But it sounds to me like you've already got a piece of glass that will
withstand the force, it's just a question of how long it will continue
to do so, does it have enough safety margin, and do you mind it looking
all bulged out like that

Regards
Doug King

The engineering data I read on glass was interesting. A piece of steel is
very consistant in strength from piece to piece. Not so with glass. Because
of microscopic imperfections one piece of glass might be only 10% the
strength of another. A small scratch goes a long way to weaken glass. It's
possible my aquarium is at 99% of its breaking point, and has been for
years. The bowing is not really noticeable unless sighting down the glass
from an end.
Dixon

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

I think you should enlist the aid of an expert. If the glass was
bowed to begin with, I suspect it had bowed more over time. It will
continue to bow until it bursts.
But, good luck.

On Tue, 30 Nov 2004 14:47:25 GMT, "dixon"
wrote:

(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.

dixon wrote:
I wanted to build a 1000 plus gallon aquarium in the new house I was having
built. I wasn't sure how thick to make the front glass. The rest of the tank
would be fiberglass covered steel. The dimensions I arrived at were 8 ft.
across, 6 ft. front to back and 38 in. tall. This would be 1137 gallons. A
friend had a tank that was 38 in. tall but only 18 in. front to back. It was
also 8 ft. long. (284 gallons). The glass in his tank was 3/4 in thick. My
tank would be identical except the front to back dim. would be 72 in.
At first I was concerned that there would more pressure on the front glass.
After much research I became confident there would be no difference
regardless of front to back depth. For simplicity lets call depth the front
to back dim. and height the top to bottom dim. I found charts that told the
pressure at every height of water in inch increments. To figure the total
"push" on the front glass, you just need to find the pressure at the halfway
point(19 inches in this case) and multiply by the total sq. in. of the glass
(3,648). I believe
at 19" the pressure was about .686 or so psi. It figured out to be around
2,500 lbs. of force on the glass.

In fresh water, I guess because that seems a low number. Anyway the
force isn't "all" over the glass it's a water column 38" deep so along
the bottom of the front glass the pressure is what?? say 1.37 psi on
your figure & probably more like 1.55 psi on my figures.



It would be about 9,480 lbs of water.

The total weight of the water in the tank isn't relevant all the glass
sees is a PSI pressure starting at nothing at the surface, graduating to
say 1.55 psi along the bottom edge.

As
I
was filling the tank, a straight edge laid against the front glass showed
the glass bowing outward very noticeably even at 1/4 full.

How so??, did it start leaking?? what is the joint along the bottom
edge doing?? for the glass to "bow" it must be moving relative the the
tank bottom?? The bottom edge of the glass is about the only part that
sees full pressure & you should have that well tied to the bottom of the
tank (angle iron frame??? or similar) The top of the glass should have a
frame also but it's not as important as the bottom, because at the top
the glass sees very little psi of pressure.

I nervously
filled it full. Now it was very bowed even to the naked eye. The 2,500 lbs
was definitely there on the glass. Now, try to visualize the front glass as
a giant rectangle piston with 2,500 lbs of force on the inside. If you put a
4"x4" in the center of the glass that ran across the room to an opposite
wall and put a bathroom scale (quite a scale!) against the wall or better
yet, an "I" beam, the scale after filling, would read a total of 2500 lbs.
Or 25 men each pushing on the outside at 100 lbs would cancel the pressure
on the glass. 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?

Again the total volume of water in the tank is not relevant, what does
matter is that the glass be strong enough to resist the pressure at the
deepest point, if it's OK there then it's OK at any lesser depth. At
6ft X 3ft+ it's a big sheet of heavy glass.

Most boat windows are not 6ft long in one go & if built to survey
standards are surprisingly thick, the forward facing windows in my boat
are 3/8+" thick & nothing like 6 ft in any one direction.

I'm sure you've noticed most "in survey" paying passenger carrying open
water vessels don't have large areas of unsupported glass, even
laminated etc?? There are specific rules which govern max window size vs
material, thickness & support in the USL code & it seems smaller windows
are a safer, lighter (no pun intended) & probably a cheaper choice than
trying to persist with huge picture windows that pleasure craft can get
away with.

I think 3/4" glass is just not thick enough unsupported over that
length (again I'm still wondering how it bows away from the bottom of
the tank??) & probably thicker wouldn't help much (as glass gets thicker
it doesn't get stronger in proportion, it's still the outer skin that
carries the load & it fails in tension; also it starts to have problems
supporting it's own weight), so maybe consider some better framing??

Again the pressure is mostly along the lower edge.

K

Dixon

"Harry Krause" wrote in message
...

You know, for a few more bucks, you could have gone with...

transparent aluminum



Have you been talking to your mouse again??

Computer....Computer!

--

"WaIIy" wrote in message
...
On Tue, 30 Nov 2004 13:33:46 -0700, James Gemmill
gemmilljim@hotmail,com wrote:

I think you should enlist the aid of an expert. If the glass was
bowed to begin with, I suspect it had bowed more over time. It will
continue to bow until it bursts.
But, good luck.

and don't sit in the chair near aquarium.

The couch is right in front of the tank. When I had to sleep on the couch
( no jokes ) it was a little scary. The living room is 14' x 28'. If the
glass broke the water would be 4" deep.
Dixon

"K. Smith" wrote in message
...
dixon wrote:
I wanted to build a 1000 plus gallon aquarium in the new house I was
having
built. I wasn't sure how thick to make the front glass. The rest of the
tank
would be fiberglass covered steel. The dimensions I arrived at were 8
ft.
across, 6 ft. front to back and 38 in. tall. This would be 1137 gallons.
A
friend had a tank that was 38 in. tall but only 18 in. front to back. It
was
also 8 ft. long. (284 gallons). The glass in his tank was 3/4 in thick.
My
tank would be identical except the front to back dim. would be 72 in.
At first I was concerned that there would more pressure on the front
glass.
After much research I became confident there would be no difference
regardless of front to back depth. For simplicity lets call depth the
front
to back dim. and height the top to bottom dim. I found charts that told
the
pressure at every height of water in inch increments. To figure the
total
"push" on the front glass, you just need to find the pressure at the
halfway
point(19 inches in this case) and multiply by the total sq. in. of the
glass
(3,648). I believe
at 19" the pressure was about .686 or so psi. It figured out to be
around
2,500 lbs. of force on the glass.

In fresh water, I guess because that seems a low number. Anyway the
force isn't "all" over the glass it's a water column 38" deep so along
the bottom of the front glass the pressure is what?? say 1.37 psi on
your figure & probably more like 1.55 psi on my figures.

The info I am using for fresh water says; 2'=.87 psi / 3'=1.30 psi / 4'=1.73
psi. It seems linear. I don't have salt figures, but I wouldn't think there
would the that much diff.
An interesting note, after the tank was full I added the commercial salt mix
and it mixed with the water but didn't noticeably raise the level. I guess
this explains why salt water is heavier. I believe it took 3 or 4 hundred
pounds of salt.

For you folks that take salt water for granted, it cost me almost $500.00 to
turn 1140 gallons of fresh into salt. So much for living in Michigan!



It would be about 9,480 lbs of water.

The total weight of the water in the tank isn't relevant all the glass
sees is a PSI pressure starting at nothing at the surface, graduating to
say 1.55 psi along the bottom edge.

As
I
was filling the tank, a straight edge laid against the front glass
showed
the glass bowing outward very noticeably even at 1/4 full.

How so??, did it start leaking?? what is the joint along the bottom
edge doing?? for the glass to "bow" it must be moving relative the the
tank bottom?? The bottom edge of the glass is about the only part that
sees full pressure & you should have that well tied to the bottom of the
tank (angle iron frame??? or similar) The top of the glass should have a
frame also but it's not as important as the bottom, because at the top
the glass sees very little psi of pressure.

I nervously
filled it full. Now it was very bowed even to the naked eye. The 2,500
lbs
was definitely there on the glass. Now, try to visualize the front glass
as
a giant rectangle piston with 2,500 lbs of force on the inside. If you
put a
4"x4" in the center of the glass that ran across the room to an opposite
wall and put a bathroom scale (quite a scale!) against the wall or
better
yet, an "I" beam, the scale after filling, would read a total of 2500
lbs.
Or 25 men each pushing on the outside at 100 lbs would cancel the
pressure
on the glass. 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?

Again the total volume of water in the tank is not relevant, what does
matter is that the glass be strong enough to resist the pressure at the
deepest point, if it's OK there then it's OK at any lesser depth. At
6ft X 3ft+ it's a big sheet of heavy glass.

Most boat windows are not 6ft long in one go & if built to survey
standards are surprisingly thick, the forward facing windows in my boat
are 3/8+" thick & nothing like 6 ft in any one direction.

I'm sure you've noticed most "in survey" paying passenger carrying open
water vessels don't have large areas of unsupported glass, even
laminated etc?? There are specific rules which govern max window size vs
material, thickness & support in the USL code & it seems smaller windows
are a safer, lighter (no pun intended) & probably a cheaper choice than
trying to persist with huge picture windows that pleasure craft can get
away with.

I think 3/4" glass is just not thick enough unsupported over that
length (again I'm still wondering how it bows away from the bottom of
the tank??) & probably thicker wouldn't help much (as glass gets thicker
it doesn't get stronger in proportion, it's still the outer skin that
carries the load & it fails in tension; also it starts to have problems
supporting it's own weight), so maybe consider some better framing??

Again the pressure is mostly along the lower edge.


The bowing is in the center of the glass, the frame is to rigid to flex.
Dixon

Judging from all the various viewpoints about how water can affect an
aquarium glass, it's a good thing we all aren't hired to build a submarine
or an airplane. This problem is old science.