Reef Aquarium Tipping Force Help: Tank, Sump, and Stand Dimensions

  • Thread starter Bilk
  • Start date
  • Tags
    Force
In summary, the conversation discusses the process of building a reef aquarium with a movable sump. The tank weighs 100lbs and is 24"x24"x20", while the saltwater adds 385lbs. The stand weighs approximately 45lbs and has the same dimensions as the tank, with the sump measuring 20"x15"x16". The sump, when filled to capacity, holds 20 gallons of water at 172lbs and sits 7" above the floor inside the cabinet. The sump rests on a sliding TV base rated for 225lbs. The conversation also involves a discussion on calculating the tipping force of the tank and sump assembly, taking into account the center of gravity and torque. The dimensions and
  • #1
Bilk
9
0
I'm in the process of building a reef aquarium. It will have a movable sump which will slide out from under the tank. I would like to know if this will exceed the tipping force and cause the tank to fall over.

Components:

  • tank=100lbs 24"x24"x20"
  • 45g salt water=385lbs
  • stand=approx. 45lbs 24"x24"x37"
  • tank and stand dimensions=24"x24"x57"
  • sump dimensions=20"x15"x16"

The sump when filled to the brim holds 20g @ 172lbs and sits 7" above the floor inside the cabinet. It extends two thirds of it's length from the cabinet. Below is a picture of the assembly. I'd like to know if the sump when filled will cause the tank to tip when extracted from the cabinet below. The sump will never be filled to capacity, but this would be the most extreme scenario and I should account for it. Any and all help is greatly appreciated.

DSC00075.jpg


Note: the sump sits atop a sliding TV base rated for 225#. The tank will also contain sand and rock structure, but were not included as the bare tank with water would be the minimum load for the calculation.
 
Physics news on Phys.org
  • #2
Im wondering, did I post this to the wrong forum?
 
  • #3
I'm not sure there is quite enough information regarding the dimensions (eg which is the width vs length).

The way to approach the problem is to think of it as a teeter totter/seesaw with the pivot where the front wheels are (or front edge if no wheels). Then you work out the torques acting about that point. For example see this diagram which shows the set up from the side..

Mm = Mass of Main tank
Ms = Mass of sump
dm = distance from center of gravity of main tank to pivot
ds = distance from center of gravity of sump to pivot.

For most shape tanks the center of gravity will be in the middle of the tank (eg mid way front to back).

Then if we define clockwise as positive the torque equation looks like this:

Net torque = (Mm * g * dm) - (Ms * g * ds)

If the result is +ve then the unit will try to rotate clockwise about the pivot (eg it's stable). If the result is negative the unit will try to rotate anticlockwise and will fall to the left (eg forwards).

I've only shown the maths for the main tank and sump tanks but you can factor in the other items the same way. For example if the mass of the salt is Msalt and it is distance dsalt to the right of the pivot then the equation becomes..

Net torque = (Mm * g * dm) + (Msalt * g * dsalt) - (Ms * g * ds)

Note that the height of items isn't a factor, or at least it isn't until the unit starts to tip over. If tipped over a few degrees that would move the center of gravity of the tanks and affect the calculation. I'm assuming that no amount of tipping is allowed!
 

Attachments

  • Tank.jpg
    Tank.jpg
    8.3 KB · Views: 485
  • #4
oops sorry I miss read your post. I assumed there was also bag of salt that had to be stored in the unit.
 
  • #5
CWatters said:
I'm not sure there is quite enough information regarding the dimensions (eg which is the width vs length).

The way to approach the problem is to think of it as a teeter totter/seesaw with the pivot where the front wheels are (or front edge if no wheels). Then you work out the torques acting about that point. For example see this diagram which shows the set up from the side..

Mm = Mass of Main tank
Ms = Mass of sump
dm = distance from center of gravity of main tank to pivot
ds = distance from center of gravity of sump to pivot.

For most shape tanks the center of gravity will be in the middle of the tank (eg mid way front to back).

Then if we define clockwise as positive the torque equation looks like this:

Net torque = (Mm * g * dm) - (Ms * g * ds)

If the result is +ve then the unit will try to rotate clockwise about the pivot (eg it's stable). If the result is negative the unit will try to rotate anticlockwise and will fall to the left (eg forwards).

I've only shown the maths for the main tank and sump tanks but you can factor in the other items the same way. For example if the mass of the salt is Msalt and it is distance dsalt to the right of the pivot then the equation becomes..

Net torque = (Mm * g * dm) + (Msalt * g * dsalt) - (Ms * g * ds)

Note that the height of items isn't a factor, or at least it isn't until the unit starts to tip over. If tipped over a few degrees that would move the center of gravity of the tanks and affect the calculation. I'm assuming that no amount of tipping is allowed!

Yes I wasn't clear with the dimensions. The tank footprint is 24" square and 20" high. The supporting stand has the same footprint and stands 37" high. The sump is 20" from front to back, 15" wide and 16" tall.

Am I correct in thinking the center of mass for the tank sitting on the stand is different than the center of mass for just the tank? Hope that makes sense. I'm also confused regarding the height of the sump in relation to the full assembly. Does the force required to tip the assembly change as the external mass moves upward on the assembly? In other words, if the sump were at the top, aligned with the top of the tank, would that change the equation? (assume the tank was affixed to the supporting base).

I appreciate you help. Thanks.
 
  • #6
CWatters said:
oops sorry I miss read your post. I assumed there was also bag of salt that had to be stored in the unit.
No the water is seawater. It has an approximate weight of 8.55lbs/gal. The volume of the tank to the overflow is 45gal, yielding a weight of 385lbs. The glass tank weight approximately 100lbs. The stand approximately 45lbs.

Yes no movement or tipping is desired :)
 
  • #7
The height of items only comes into play if the unit actually starts to tip. So raising the sump shouldn't make any difference. If the floor was soft carpet or something that might allow it to tip a little and then you might need to do a more detailed calculation involving the heights.

As I see it the numbers work out like this..

The center of gravity for stand, main tank and water is 12" behind the pivot point.
The center gravity of the sump tank and it's water is 5" in front of the pivot point (when extended).

Net Torque = {(100+385+45) * g * 12"} - {172 * g* 5"}
= (6360 - 860) * g

which is clearly +ve so it appears the sump can't tip it over.

Looks like you would have to pull the sump out until the center of the sump is about three feet in front to the pivot before it might tip over (6360/172 = 37").
 
  • #8
This problem was originally solved in the sixteenth century by Sir Isaac the Newt. :wink:
 
  • #9
I just noticed that the weight of the empty sump tank hasn't been factored in. I don't think it will make a difference but do you know the empty weight?
 
  • #10
CWatters said:
The height of items only comes into play if the unit actually starts to tip. So raising the sump shouldn't make any difference. If the floor was soft carpet or something that might allow it to tip a little and then you might need to do a more detailed calculation involving the heights.

As I see it the numbers work out like this..

The center of gravity for stand, main tank and water is 12" behind the pivot point.
The center gravity of the sump tank is 5" in front of the pivot point (when extended).

Net Torque = {(100+385+45) * g * 12"} - {172 * g* 5"}
= (6360 - 860) * g

which is clearly +ve so it appears the sump can't tip it over.

Looks like you would have to pull the sump out until the center of the sump is about three feet in front to the pivot before it might tip over (6360/172 = 37").
Ok thank you very much.

My reasoning behind thinking the height of the sump would matter was in terms of the masses vertical distance above the pivot point acting as a moment arm or lever. The closer to the floor the shorter the lever, the higher, the longer the lever. I guess that doesn't come into play here?

Again, than you for your help. I didn't want to move ahead unless I had a better understanding of the physics. I basically want to avoid disaster :)
 
  • #11
I have done some experiments with this sort of thing and found that in practice, theory and practice can differ. I would recommend using a safety factor, partly because the theory contains assumptions that may not be true, but also because the operative might stand on the open tv shelf without thinking.
 
  • #12
CWatters said:
I just noticed that the weight of the empty sump tank hasn't been factored in. I don't think it will make a difference but do you know the empty weight?
The sump weight is approx. 32lbs.
 
  • #13
pongo38 said:
I have done some experiments with this sort of thing and found that in practice, theory and practice can differ. I would recommend using a safety factor, partly because the theory contains assumptions that may not be true, but also because the operative might stand on the open tv shelf without thinking.

Yes I see what you mean.
 
  • #14
I see a similar question was posed before. Link That scenario took into account the height of all the components above the fulcrum (where the front legs meet the floor). However they were looking at lateral force exerted on the tank and structure. Does that change anything?
 
  • #15
Bilk said:
The sump weight is approx. 32lbs.

That makes my equation..

Net Torque = {(100+385+45) * g * 12"} - {(172 + 32) * g* 5"}
= (6360 - 1020) * g

Still well on the safe side. That main tank is heavy!

Lets see what happens if someone (190lbs) pulls out the sump and stands on the TV shelf say 10" from the front pivot..

Main tank - Sump - Person

= {(100+385+45) * g * 12"} - {(172 + 32) * g* 5"} - {190 * g * 10")
= (6360 - 1020 - 1900) * g

Still seems ok.

What happens when the main tank is empty and the sump tank is full...

{(100+45) * g * 12"} - {(172 + 32) * g* 5"}
= (1740 - 1020) * g

Even that seems ok.
 
  • #16
CWatters said:
That makes my equation..

Net Torque = {(100+385+45) * g * 12"} - {(172 + 32) * g* 5"}
= (6360 - 1020) * g

Still well on the safe side. That main tank is heavy!

Lets see what happens if someone (190lbs) pulls out the sump and stands on the TV shelf say 10" from the front pivot..

Main tank - Sump - Person

= {(100+385+45) * g * 12"} - {(172 + 32) * g* 5"} - {190 * g * 10")
= (6360 - 1020 - 1900) * g

Still seems ok.

What happens when the main tank is empty and the sump tank is full...

{(100+45) * g * 12"} - {(172 + 32) * g* 5"}
= (1740 - 1020) * g

Even that seems ok.
Well sir I thank you for your help. Even though my rough math and common sense told me this would work, your confirmation makes me feel much better about this design.

Again thank you for taking the time to respond and best of everything to you!
 
  • #17
Bilk said:
I see a similar question was posed before. Link That scenario took into account the height of all the components above the fulcrum (where the front legs meet the floor). However they were looking at lateral force exerted on the tank and structure. Does that change anything?

As I said the height matters if the tank starts to tilt. Take a look at this diagram. On the left the tank is level and raising or lowering the tank doesn't change the distance between the center of gravity of the tank and the pivot (dm).

On the right you see the tank tipped over a bit. Now if you raise and lower the tank in the stand the vertical arrow acting through the center of gravity of the tank moves from one side of the pivot to the other. The diagram shows it on the left of the pivot contributing to tipping. If the tank was lowered the vertical arrow would move to the right.

What this means is that the higher the tank the less you have to tilt it before it falls...but I think it will be fine.
 

Attachments

  • Tank1.jpg
    Tank1.jpg
    16.4 KB · Views: 421

1. What is the purpose of calculating the tipping force in a reef aquarium?

The tipping force in a reef aquarium helps determine the stability and safety of the tank, sump, and stand dimensions. It takes into account the weight of the water, rocks, and equipment, and ensures that the structure can support the weight without tipping over.

2. How do I calculate the tipping force in my reef aquarium?

To calculate the tipping force, you will need to determine the total weight of your tank, sump, and stand, which includes the weight of the water, rocks, and equipment. Then, you will need to multiply the weight by the distance from the center of gravity to the edge of the stand. This will give you the tipping force.

3. What are some factors that can affect the tipping force in a reef aquarium?

The weight of the water, rocks, and equipment are the main factors that affect the tipping force. The dimensions and construction of the tank, sump, and stand also play a role. Additionally, the placement of the equipment and the water level in the tank can impact the tipping force.

4. How can I ensure that my reef aquarium has enough tipping force?

To ensure that your reef aquarium has enough tipping force, you can use a tipping force calculator or consult with a professional. You can also choose to overestimate the weight and dimensions of your tank, sump, and stand to add an extra margin of safety.

5. What should I do if my reef aquarium does not have enough tipping force?

If your reef aquarium does not have enough tipping force, you may need to reinforce the structure or consider downsizing your tank or equipment. It is important to address this issue to prevent any potential accidents or damage to your aquarium and home.

Similar threads

  • DIY Projects
Replies
2
Views
5K
Replies
4
Views
1K
Back
Top