This could work

[Anuvin]

I don't know if you can maintain a siphon with a drip. Any thoughts on the issue?

[harpo99999]

it might be possible IF the drip point is below the water level of the outer container

[Slatye]

Harpo is correct. A simple syphon requires that the outlet (ie level of the valve that turns it into a drip) is lower than the inlet (level of the water in the outer tank). With a fixed syphon pipe, this is a challenge: it means that the initial difference in water level has to be more than twice the height of the build. Yhe inner water has to rise exactly the height of the build while still remaining below the outlet of the valve, and the outer water has to fall the same height while still remaining above the level of the valve. I suspect that there'll also be a slight issue as the pressure in the syphon will decrease as the height difference decreases, so you'll tend to have a high flow rate at first and a low flow rate at the end.

Incidentally, if you can live with these issues then you can get rid of the syphon completely.

Peachy2.PNG

However, I suspect that the pressure issue in particular will take some work.

(Edit: note that the water in the inner container is only a different colour to make it clear where the containers are. It's the same water).

[masterada]

1. The water level of the inner container relative to the bottom of the outer container never changes.
2. The water level of the outer container relative to the bottom of the outer container is also amlost constant (its raises very slowly, and only because of the walls of the inner container).
3. Its true, that the outlet must be lower than the water level in the outer container, but they can be connected via a fixed pipe(see my original design). To start the siphon (ie. fill the pipe with water), see my 2nd design (the one with two valves).

I will make a 3d design when i have some time.

I know its hard to believe, but its true.

[Slatye]

Well, I've just done a bunch of maths, and come to the conclusion that a syphon system won't work.

V_outer = volume of water in the outer container
V_outer,0 = initial volume of water in the outer container
V_inner = volume of water in the inner container
V_inner,0 = 0 (no water in there initially)

A_outer,internal = area of the outer container, disregarding space occupied by the inner container
A_inner,external = area of the inner container that it occupied in the outer container 
A_inner,internal = area of the inner container that can hold liquid (ie external area minus wall thickness)

d = density of water

M_inner,initial = mass of the inner container at the start (resin + container + weights)
M_inner = mass of the inner container with whatever water it has in it. 
        = M_inner,initial + V_inner*d

h_outer = height of the water in the outer container, relative to the bottom of the outer container
h_inner = height of the water in the inner container, relative to the bottom of the inner container
        = V_inner/A_inner,internal 
h_diff  = difference in height between the water in the outer container and the inner container
h_print = height of the printing surface (ie top of the water in the inner container) with respect to the bottom of the outer container
h_sink  = height of the bottom of the inner container with respect to the water in the outer container - how far the inner one sinks.

Okay, so now we need some equations to relate these. Obviously what we want is for h_print to stay constant. We can freely adjust the container areas, but not really anything else.

First up, how far down does the inner container sink in the outer one? Well, we know its mass, and we know the density of the water, and we know its external area. Therefore we can say:

h_sink = M_inner / (d*A_inner,external)
       = (M_inner,initial + V_inner * d) / (d*A_inner,external)

We know the height of the water in the inner container, so now we can find h_diff.

h_diff = h_sink - h_inner
       = (M_inner,initial + V_inner * d) / (d*A_inner,external) - V_inner/A_inner,internal

If we know the height difference and we know the height of water in the outer container, then that's all good. We can easily compute h_print. So, what's the height of water in the outer container? It's not just volume divided by area because that doesn't take into account volume displaced by the inner container. However, if we add that volume then it'll all work.

h_outer = (V_outer + (h_sink * A_inner,external)) / A_outer

Now, putting that all together and using [V_inner = V_outer,0 - V_outer] we can solve:

h_print = h_outer - h_diff
        = (V_outer + (h_sink * A_inner,external)) / A_outer - (M_inner,initial + V_inner * d) / (d*A_inner,external) + V_inner/A_inner,internal
        = (V_outer,0 + M_inner,initial/d) / A_outer - (M_inner,initial/d + V_outer,0 - V_outer) / A_inner,external + (V_outer,0 - V_outer)/A_inner,internal
        = V_outer,0/A_outer + M_inner,initial/(d*A_outer) - M_inner,initial/(d*A_inner,external) - V_outer,0 / A_inner,external + V_outer/A_inner,external + V_outer,0/A_inner,internal + V_outer/A_inner,internal

So, what's a constant here and can be ignored? Well, V_outer,0 is clearly constant. All the areas are constant. M_inner,initial is a constant. All that's left over is V_outer! So, rearranging with that we have:

h_print = V_outer * (1/A_inner,internal + 1/A_inner,external) + <a whole bunch of constants>

The problem here is that for constant print height, we need to make that term constant. We can't keep V_outer constant because the whole point of this design is that you funnel water from the outer container to the inner one. Therefore the only option is to multiply it by zero - which basically means that A_inner,internal and A_inner,external are going up to infinity. Clearly this is not a very practical concept, and even if it was it wouldn't actually work (it keeps the print height constant, but the idea is that the inner container falls as it fills with water. If it's infinitely large it's never going to move much at all).

Back to Mike's idea, which I think might work.

[mike_biddell]

Well, I've just done a bunch of maths, and come to the conclusion that a syphon system won't work.

V_outer = volume of water in the outer container
V_outer,0 = initial volume of water in the outer container
V_inner = volume of water in the inner container
V_inner,0 = 0 (no water in there initially)

A_outer,internal = area of the outer container, disregarding space occupied by the inner container
A_inner,external = area of the inner container that it occupied in the outer container 
A_inner,internal = area of the inner container that can hold liquid (ie external area minus wall thickness)

d = density of water

M_inner,initial = mass of the inner container at the start (resin + container + weights)
M_inner = mass of the inner container with whatever water it has in it. 
        = M_inner,initial + V_inner*d

h_outer = height of the water in the outer container, relative to the bottom of the outer container
h_inner = height of the water in the inner container, relative to the bottom of the inner container
        = V_inner/A_inner,internal 
h_diff  = difference in height between the water in the outer container and the inner container
h_print = height of the printing surface (ie top of the water in the inner container) with respect to the bottom of the outer container
h_sink  = height of the bottom of the inner container with respect to the water in the outer container - how far the inner one sinks.

Okay, so now we need some equations to relate these. Obviously what we want is for h_print to stay constant. We can freely adjust the container areas, but not really anything else.

First up, how far down does the inner container sink in the outer one? Well, we know its mass, and we know the density of the water, and we know its external area. Therefore we can say:

h_sink = M_inner / (d*A_inner,external)
       = (M_inner,initial + V_inner * d) / (d*A_inner,external)

We know the height of the water in the inner container, so now we can find h_diff.

h_diff = h_sink - h_inner
       = (M_inner,initial + V_inner * d) / (d*A_inner,external) - V_inner/A_inner,internal

If we know the height difference and we know the height of water in the outer container, then that's all good. We can easily compute h_print. So, what's the height of water in the outer container? It's not just volume divided by area because that doesn't take into account volume displaced by the inner container. However, if we add that volume then it'll all work.

h_outer = (V_outer + (h_sink * A_inner,external)) / A_outer

Now, putting that all together and using [V_inner = V_outer,0 - V_outer] we can solve:

h_print = h_outer - h_diff
        = (V_outer + (h_sink * A_inner,external)) / A_outer - (M_inner,initial + V_inner * d) / (d*A_inner,external) + V_inner/A_inner,internal
        = (V_outer,0 + M_inner,initial/d) / A_outer - (M_inner,initial/d + V_outer,0 - V_outer) / A_inner,external + (V_outer,0 - V_outer)/A_inner,internal
        = V_outer,0/A_outer + M_inner,initial/(d*A_outer) - M_inner,initial/(d*A_inner,external) - V_outer,0 / A_inner,external + V_outer/A_inner,external + V_outer,0/A_inner,internal + V_outer/A_inner,internal

So, what's a constant here and can be ignored? Well, V_outer,0 is clearly constant. All the areas are constant. M_inner,initial is a constant. All that's left over is V_outer! So, rearranging with that we have:

h_print = V_outer * (1/A_inner,internal + 1/A_inner,external) + <a whole bunch of constants>

The problem here is that for constant print height, we need to make that term constant. We can't keep V_outer constant because the whole point of this design is that you funnel water from the outer container to the inner one. Therefore the only option is to multiply it by zero - which basically means that A_inner,internal and A_inner,external are going up to infinity. Clearly this is not a very practical concept, and even if it was it wouldn't actually work (it keeps the print height constant, but the idea is that the inner container falls as it fills with water. If it's infinitely large it's never going to move much at all).

Back to Mike's idea, which I think might work.

OOOhhh you are cheeky LOL...... Mike's idea which will work !!!!!!

[Slatye]

If I've got time tomorrow then I'll sit down and figure out the maths behind your idea. I can see a way that'll definitely make it work, and hopefully that won't be necessary.

[mike_biddell]

More seriously, I have no misgivings about the seal. Hydraulic cylinders can go years without leaking and we are in a relatively much lower pressure situation. The drip feed should work well as the pressure is constant (force/per unit area) as the weight remains constant and the area remains constant.

[masterada]

Your magic does not work on me!

Mistake 1: h_outer = (V_outer + (h_sink * A_inner,external)) / A_outer
Mistake 2:

    = (V_outer,0 + M_inner,initial/d) / A_outer - (M_inner,initial/d + V_outer,0 - V_outer) / A_inner,external + (V_outer,0 - V_outer)/A_inner,internal         
    = V_outer,0/A_outer + M_inner,initial/(d*A_outer) - M_inner,initial/(d*A_inner,external) - V_outer,0 / A_inner,external + V_outer/A_inner,external + V_outer,0/A_inner,internal + V_outer/A_inner,internal

Correctly:

h_sink = M_inner / (d*A_inner,external)
       = (M_inner,initial + V_inner * d) / (d*A_inner,external)

h_diff = h_sink - h_inner
       = (M_inner,initial + V_inner * d) / (d*A_inner,external) - V_inner/A_inner,internal

h_outer = (V_outer + (h_sink * (A_outer-A_inner,external))) / A_outer

V_inner = V_outer,0 - V_outer

h_print = h_outer - h_diff
        = (V_outer + (h_sink * (A_outer-A_inner,external))) / A_outer - (M_inner,initial + V_inner * d) / (d*A_inner,external) + V_inner/A_inner,internal
        = (V_outer + ((M_inner,initial + V_inner * d) / (d*A_inner,external) * (A_outer-A_inner,external))) / A_outer - (M_inner,initial + V_inner * d) / (d*A_inner,external) + V_inner/A_inner,internal
        = (V_outer + ((M_inner,initial + (V_outer,0 - V_outer) * d) / (d*A_inner,external)  * (A_outer-A_inner,external))) / A_outer - (M_inner,initial + (V_outer,0 - V_outer) * d) / (d*A_inner,external) + (V_outer,0 - V_outer)/A_inner,internal
        = (V_outer + ((M_inner,initial/d + V_outer,0 - V_outer) /  (A_inner,external)  * (A_outer-A_inner,external))) / A_outer - (M_inner,initial/d +  (V_outer,0 - V_outer)) / (A_inner,external) + (V_outer,0 -  V_outer)/A_inner,internal
lets call (A_outer-A_inner,external) / A_inner,external = A_correction
        = (V_outer + ((M_inner,initial/d + V_outer,0 - V_outer) * A_correction)) / A_outer - (M_inner,initial/d +  (V_outer,0 - V_outer)) / (A_inner,external) + (V_outer,0 -  V_outer)/A_inner,internal
        = (V_outer * (1-A_correction)/A_outer + V_outer,0/A_outer + M_inner,initial*A_correction/(d*A_outer) - M_inner,initial/(d*A_inner,external) - V_outer,0/A_inner,external + V_outer/A_inner,external + V_outer,0/A_inner,internal - V_outer/A_inner,internal

h_print = V_outer * ((1-A_correction)/A_outer - 1/A_inner,internal + 1/A_inner,external) + <a whole bunch of constants>

So:

(1-A_correction)/A_outer - 1/A_inner,internal + 1/A_inner,external = 0
(1-((A_outer-A_inner,external) / A_inner,external))/A_outer - 1/A_inner,internal + 1/A_inner,external = 0
1/A_outer - 1/A_inner,external + 1/A_outer - 1/A_inner,internal + 1/A_inner,external = 0
2/A_outer = 1/A_inner,internal
2*A_inner,internal = A_outer

Dont get me wrong. I love that you did the math, never would have done it myself otherway

[Slatye]

I'm pretty sure that Mistake 1 is not actually a mistake. I'm trying to find the volume of stuff in the outer container that's at or below the water level. That'll be equal to the volume of the water in there, plus the volume occupied by the inner container. Volume occupied by the inner container is A_inner,external * h_sink. Then divide by the area of the outer tank to get the water level. I've spent a while trying to understand how your replacement expression would work, but I haven't figured it out yet - any chance of an explanation?

The second mistake is indeed a mistake. Now I'm working through the maths again, because while your final answer does seem much more reasonable than mine, I can't see how you got to the h_outer expression.

Edit: actually, now your answer is not making sense to me either! It has no dependence on A_inner,external, which implies that it doesn't care at all about wall thickness of the inner container. That doesn't seem right.