# Specific 3D Printers, Scanners, & Hardware > Peachy Printer Forum >  This could work

## mike_biddell

How about this idea to overcome laser intensity and focus problems, due to distance from print to peachy changing as the print progresses. Dump the upper tank completely and in the lower tank have a telescopic platform, say 8 inches in diameter and with nesting cylinders with a weight and an O ring. Place a valve at the base of the cylinder and open it as the print starts. the cylinders are full of saline and fully extended to contact the floating resin. When the valve is open, the weight pushes the top cylinder downwards and expels the saline into the tank, thru the orifice (so the level in the tank never varies). The platform will fall under gravity in an entirely deterministic way for a given orifice. So once calibrated, Z is deterministic. The cost of two nesting cylinders with an O ring is minimal and hence we could have a very cheap lowering platform, with a decent print size, and peachy would remain at a fixed distance from the print. This idea overcomes problems associated with floating solutions with the wobble factor and overcomes focus problems.

----------


## Anuvin

I think I see where you are going with this idea, and I do like it, but I am having a hard time conceptualizing it. Any chance you could doodle up a design?

----------


## mike_biddell

> I think I see where you are going with this idea, and I do like it, but I am having a hard time conceptualizing it. Any chance you could doodle up a design?


two cylinders nest into each other and are sealed with an O ring. The top cylinder orifice is to allow the cylinders to be full of saline and is closed when printing starts (using a small bung). The shaky rectangle on the top cylinder is a weight. You simply pull the cylinders apart until the top cylinder just touches the resin, then close the top orifice. The platform descends, expelling saline out of the bottom orifice, back into the tank. The descent of the top cylinder for a given orifice size and weight should be reliable and reproducible, as long as the O ring is not too worn. Sorry for the rubbish sketch.

----------


## Slatye

Build_platform.jpg

Something like this? So the green cylinder slowly displaces the water in the blue one, at a rate determined by how far open the valve is (and obviously the Peachy's water drop sensor gets attached to the valve). 

There might be some issues getting the friction low enough on the O-rings; you might have to grease them before each job to keep them sliding nicely.

----------


## mike_biddell

> Build_platform.jpg
> 
> Something like this? So the green cylinder slowly displaces the water in the blue one, at a rate determined by how far open the valve is (and obviously the Peachy's water drop sensor gets attached to the valve). 
> 
> There might be some issues getting the friction low enough on the O-rings; you might have to grease them before each job to keep them sliding nicely.


slatye.... your drawing is a million times better than mine....but the bottom cylinder must also be contained in the tank such that the when the saline is expelled it leaves the volume in the tank unchanged. If the weight is large, the O ring friction becomes a negligible factor. But thanks for grasping the concept so quickly as it is difficult to describe.

----------


## Slatye

Hmm, not quite sure that I fully understand that bit. I was assuming that you'd have an O-ring around the bottom of the main tank so that the "printing water" (in that tank) is completely separated from the "support water" (in the lower cylinder). You use up the support water, but the printing water remains fixed so you've got a constant volume of that. This should just result in the green cylinder/piston moving down, lowering the print bed into the printing water and resin.


Regarding the weight, there are two problems with making it really heavy. One is obviously that it'll be a pain to ship it - you might have to ask end users to attach big steel bolts or something to make it heavy enough. The other is that there'll be a lot of pressure in the support cylinder which makes it difficult to find a suitable valve setting. Still, it might be that this is just what has to be done to make it work. It'd be a small price to pay to get a nice printing platform.

----------


## mike_biddell

> Hmm, not quite sure that I fully understand that bit. I was assuming that you'd have an O-ring around the bottom of the main tank so that the "printing water" (in that tank) is completely separated from the "support water" (in the lower cylinder). You use up the support water, but the printing water remains fixed so you've got a constant volume of that. This should just result in the green cylinder/piston moving down, lowering the print bed into the printing water and resin.
> 
> 
> Regarding the weight, there are two problems with making it really heavy. One is obviously that it'll be a pain to ship it - you might have to ask end users to attach big steel bolts or something to make it heavy enough. The other is that there'll be a lot of pressure in the support cylinder which makes it difficult to find a suitable valve setting. Still, it might be that this is just what has to be done to make it work. It'd be a small price to pay to get a nice printing platform.


slatye..... there is no problem with the user adding the weight, so it is not shipped.  You will see with your sketch that the volume in the tank will drop as the platform drops.... the whole concept is that the level in the tank does not drop.... therefore the expelled volume must be into the tank, so that the level in the tank does not change. The heavy weight takes friction out of the equation, and to compensate, the orifice is quite small.

----------


## masterada

Hey guys!
What about this sollution?
printersketch.jpg
If  the size of the 2 containers is right, the resin should stay in place,  while the inner container floats down (didnt do the math). The z-speed  remains measurable the same way it was before.
The sketch misses some  stuff (like some supporter rods or something for the inner container so  it wont float randomly in the x-y plane, and a plug from the inner  container to reset the setup). 
The laser is in a fixed position above the whole setup.

----------


## Feign

Masterada, I actually really like that one.  It allows your outer tank to be vary heavily built up to hold the hydrostatic pressure of a really deep resivoir and the inner tank doesn't need much structure at all.

Looks like a good alternative to the floating laser base that was talked about in earlier threads.

----------


## mike_biddell

Masterada, not quite sure how the fluid gets from the outer container into the inner. If it's syphonic action, how do you start the syphon? I believe your solution would work if you just retain the upper tank and drip from that. Your inner tank should have arms which touch the outer tank, to prevent relative movement of the inner tank, with respect to the  outer tank.

----------


## mike_biddell

> Masterada, not quite sure how the fluid gets from the outer container into the inner. If it's syphonic action, how do you start the syphon? I believe your solution would work if you just retain the upper tank and drip from that. Your inner tank should have arms which touch the outer tank, to prevent relative movement of the inner tank, with respect to the  outer tank.


Finally done a full sketch of my idea. The Z axis is drip monitored in exactly the same way as the original Peachy. The two cylinders nest to form an hydraulic cylinder. The saline is forced back into the same tank and therefore the level never changes as the platform drops.3d.jpg

----------


## Slatye

Mike, now I understand. Good point. Would need to rework that.

Masterada - I like that design, mainly because the moving parts (ie inner container moving inside the outer container) don't need a watertight seal - which means you can use proper bearings. Less likely to have friction problems than a cylinder sliding on an O-ring. As Mike has said, starting the syphon would be a bit of a challenge, but once it's going it should be fine. I think the requirement on the container size is just that the surface area of the liquid in each is the same - otherwise one rises/falls faster than the other falls/rises.

----------


## masterada

> Masterada, not quite sure how the fluid gets from the outer container into the inner. If it's syphonic action, how do you start the syphon? I believe your solution would work if you just retain the upper tank and drip from that. Your inner tank should have arms which touch the outer tank, to prevent relative movement of the inner tank, with respect to the  outer tank.





> Mike, now I understand. Good point. Would need to rework that.
> 
> Masterada - I like that design, mainly because the moving parts (ie  inner container moving inside the outer container) don't need a  watertight seal - which means you can use proper bearings. Less likely  to have friction problems than a cylinder sliding on an O-ring. As Mike  has said, starting the syphon would be a bit of a challenge, but once  it's going it should be fine. I think the requirement on the container  size is just that the surface area of the liquid in each is the same -  otherwise one rises/falls faster than the other falls/rises.


You guys make a point. I updated my sketch to solve the problem:
printersketch.jpg
Open tap1, close tap 2, suck the water in via the rubber tube, close tap1 then remove the tube so its not in the way during the print. An alternative could by using a large syringe instead of the tube, so u wont have your mouth full of salt water. If the syringe is not big enough, you can open tap, siphon, close tap, empty syringe, and repeat the process as many times as it takes (a 100ml syringe costs less than 2$ on ebay). 

Mike:
If i drip from the upper tank, and leave the fluid in the outer tank unchanged, the level of the resin will rise. This can be minimalized if the space wetween the outer and the inner tank is minimal, but then the level of the water in the outer tank would rise way too fast.

Slatye:
The requirement you mentioned is only true if the thickness of the inner container is negligible. Otherwise, it messes up things a bit. (If you want the resin level to stay unchanged, the water in the outer container must rise a little as the inner container floats down).

----------


## Feign

Ah, Mike, now I see how your idea works, and it seems like it actually would work mechanically. There's actually a lot of conceptual similarities between Mike's and  Masterada's designs.  Just that one is pushing the water through a high pressure end, and the other is pulling the water therough a low pressure end.

----------


## mike_biddell

> Ah, Mike, now I see how your idea works, and it seems like it actually would work mechanically. There's actually a lot of conceptual similarities between Mike's and  Masterada's designs.  Just that one is pushing the water through a high pressure end, and the other is pulling the water therough a low pressure end.


Feign, it definitely would work, and negate the need for auto focus laser. I cant quite get my head around the linearity of Masterada's  design.... I have a feeling (need to do the maths), that the resin level would fall...... but both ideas need exploring.

----------


## Feign

> Feign, it definitely would work, and negate the need for auto focus laser. I cant quite get my head around the linearity of Masterada's  design.... I have a feeling (need to do the maths), that the resin level would fall...... but both ideas need exploring.


It's understandable that it looks like that, but you'll find that the water level in the inner container relative to the top of the outer container actually rises a negligible amount, while your design indeed keeps the level completely constant.  (though the need for a gasket introduces some potential for mechanical problems.)

----------


## masterada

> It's understandable that it looks like that, but you'll find that the water level in the inner container relative to the top of the outer container actually rises a negligible amount, while your design indeed keeps the level completely constant.  (though the need for a gasket introduces some potential for mechanical problems.)


That's not necessarily true. The weight of the inner container (including the weight of the water in it) is always equal to the weight of the water in the outer container surrounding the inner container. (http://en.wikipedia.org/wiki/Archimedes%27_principle)

_Ignore the italic part, the next part is clearer
Imagine the inner container is very thin. In this case, dropping a very small amount of water will not effect the weight of the inner container (so the container itself doesnt move at all), but will raise the level of the water, thus the resin will be higher the originally.
Now imagine the inner container is just marginally smaller than the outer container (both containers have very high walls). Now siphon 10mm height of water from the bottom of the outer container. The resin falls by 10mm (there is 10mm less water below the inner container). Hold the inner container, so that it cant sink. Pour the siphoned water into the inner container. The resin level rises almost 10mm (you just poured that much water into the inner container). Let go off the inner container. As it is now heavier by 10mm worth of water, it will displace that amount of water from the outer container (Archimedes' principle). That means 10mm water from below the inner container will move to the gap between the inner and the outer container, effectively shrinking the inner-container and the resin for another 10mm.

So: 
very thin inner container -> resin level rises
very thick inner container -> resin level falls

My head starts hurting._

Lets imagine there are 3 containers. "A" is the inner container. "B" is the water in the outer container, thats near the inner container. "C" is the water in the outer container thats below the level of the bottom of the inner container. Due to Archimedes' principle, "A" and "B" must always weight the same. If you move x amount of water from the "C" container into the "A" container, then you must also move the same amount of water from "C" to "B". That means, that regardless of the size of the inner container, for every mm of water moved from the bottom of the outer container into the inner container, the inner container will shrink 2 mm. 
*So if the inner base territory of the inner container is exactly half of the inner base territory of the outer container, then the resin never moves. The thickness of the wall of the inner container doesnt matter, neither does its weight.
*
I thought it would be more complicated then that  :Smile:

----------


## mike_biddell

> That's not necessarily true. The weight of the inner container (including the weight of the water in it) is always equal to the weight of the water in the outer container surrounding the inner container. (http://en.wikipedia.org/wiki/Archimedes%27_principle)
> 
> _Ignore the italic part, the next part is clearer
> Imagine the inner container is very thin. In this case, dropping a very small amount of water will not effect the weight of the inner container (so the container itself doesnt move at all), but will raise the level of the water, thus the resin will be higher the originally.
> Now imagine the inner container is just marginally smaller than the outer container (both containers have very high walls). Now siphon 10mm height of water from the bottom of the outer container. The resin falls by 10mm (there is 10mm less water below the inner container). Hold the inner container, so that it cant sink. Pour the siphoned water into the inner container. The resin level rises almost 10mm (you just poured that much water into the inner container). Let go off the inner container. As it is now heavier by 10mm worth of water, it will displace that amount of water from the outer container (Archimedes' principle). That means 10mm water from below the inner container will move to the gap between the inner and the outer container, effectively shrinking the inner-container and the resin for another 10mm.
> 
> So: 
> very thin inner container -> resin level rises
> very thick inner container -> resin level falls
> ...


Masterada... you are correct, it is Archimedes principle..... a floating body displaces its own weight of 'water'.  Whether it works or not does depend upon the relative areas/sizes of the inner and outer.

----------


## bovalis2037

Masterda - The idea is awsome but the neglected rods to support it on its way would prove to be a big issue. If we're planning on printing something to the 64th of an inch detail the rods would have to move perfectly right...

Also how would you precisely control the flow of water?

----------


## masterada

The rods are just not in the sketch. There are many ways to keep it in place, i just havent really thought about what could be the best.
The flow of water can be controlled the same way the original does. (Dont ask me how, that and the mirror movement are the two things i havent really grasped yet.) The same sensor and tap are in place, so probably just manually setting the tap to a certain drop rate and measuring it with the mic.

Wait, i just thought about something:

printersketchabove.jpg

Its the two containers from above. Assuming its two plastic container, will probably need to place some metal L plates to the four sides of the outer container, where we cut into the plastic.
Ps: as i stated before, the territory of the outside container must be twice of the territory of the inside container  :Wink: 

Edit: forget it, this is way more simple (metal rails are fixed on the outer container):
printersketchabove2.jpg

----------


## 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)
> 
> ...


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  :Smile:

----------


## 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.

----------


## masterada

> 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.


Oh i get it now. I was trying to calculate the "Volume occupied by the inner container" with a more direct aproach: (A_inner,internal + A_inner,wall)*h_sink - which is the same as A_outer-A_inner,external. But yes, your way is right too, my bad for that. Shouldnt change the result anyway.




> 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.


Read a couple post back (http://3dprintboard.com/showthread.p...ull=1#post5250). Thats the same conclusion i got from my theoretical modelling. I was surprised too first, but if you think about it a while, it starts to make sense (read my other post).

----------


## Slatye

Well, I just tried a different approach and got much the same answer that I had last time (just with a minus sign, due to the mistake you pointed out). The new approach is similar to the one you just linked (more on this below). Unfortunately I can't post a direct attachment here, but if you've got TeX installed then you can turn this into a pretty PDF of the maths:


```
\documentclass{article} \usepackage{amsmath}
 \usepackage[margin=0.5in]{geometry}
 \usepackage{parskip}
\usepackage{amsmath}
 \usepackage{amssymb}
  \begin{document}
  Definitions:
  \begin{tabular}{rl}
 $h_\text{inner}$ & Height of water in the inner container, initially zero\\
 $h_\text{sink}$ & Height that the inner container sinks into the water in the outer container\\
 $h_\text{support}$ & Height of water underneath the inner container\\
 $h_\text{base}$ & Thickness of the base of the inner container, constant of course\\
 $M_\text{inner}$ & Mass of the inner container (total, including transferred water)\\
 $V_\text{transfer}$ & Volume of water transferred\\
 $V_\text{outer,1}$ & Volume in the outer container surrounding the inner container (ie not the water below the container)\\
 $\rho$ & Density of water
 \end{tabular}
   First things first, what's the height of the water in the inner tank?
 \begin{align}
 h_\text{inner} &= \frac{V_\text{transfer}}{A_\text{inner,internal}}
 \end{align}
  How far will the inner tank sink into the outer tank? Well, its mass $M_\text{inner}$ is going to be:
  \begin{align}
 M_\text{inner} &= \rho V_\text{transfer} + M_\text{constant}
 \end{align}
  How far will that sink into the inner tank? Well, that's easy.
  \begin{align}
 h_\text{sink} &= \frac{M_\text{inner}}{\rho A_\text{inner,external}}
 \end{align}
  How high is the water in the outer tank? First we find the volume surrounding the inner container.
 \begin{align}
 V_\text{outer,1} &= h_\text{sink}(A_\text{outer} - A_\text{inner,external}) 
 \end{align}
 The volume remaining (ie $V_\text{initial} - V_\text{transfer} - V_\text{outer,1}$) fills the space underneath the inner container. We can find the height of this
  \begin{align}
 h_\text{support} &= \frac{V_\text{initial} - V_\text{transfer} - V_\text{outer,1}}{A_\text{outer}}
 \end{align}
  Now, the print height is just going to be the water under the outer container ($h_\text{support}$) plus the height of water in the inner container ($h_\text{inner}$) plus (constant) base thickness $h_\text{base}$.
  \begin{align}
 h_\text{print} &= h_\text{support} + h_\text{inner} + h_\text{base}\\
 &= \frac{V_\text{initial} - V_\text{transfer} - V_\text{outer,1}}{A_\text{outer}} + \frac{V_\text{transfer}}{A_\text{inner,internal}} + h_\text{base}\\
 &= \frac{V_\text{initial} - V_\text{transfer} - h_\text{sink}(A_\text{outer} - A_\text{inner,external}) }{A_\text{outer}} + \frac{V_\text{transfer}}{A_\text{inner,internal}} + h_\text{base}\\
 &= \frac{V_\text{initial}}{A_\text{outer}} - \frac{V_\text{transfer}}{A_\text{outer}} - h_\text{sink} - \frac{h_\text{sink} A_\text{inner,external}}{A_\text{outer}} + \frac{V_\text{transfer}}{A_\text{inner,internal}} + h_\text{base} \\
 &= \frac{V_\text{initial}}{A_\text{outer}} - V_\text{transfer}\left(\frac{1}{A_\text{outer}} - \frac{1}{A_\text{inner,internal}}\right) - h_\text{sink}\left(1 + \frac{A_\text{inner,external}}{A_\text{outer}}\right) + h_\text{base}\\
 &= \frac{V_\text{initial}}{A_\text{outer}} - V_\text{transfer}\left(\frac{1}{A_\text{outer}} - \frac{1}{A_\text{inner,internal}}\right) - \frac{\rho V_\text{transfer} + M_\text{constant}}{\rho A_\text{inner,external}}\left(1 + \frac{A_\text{inner,external}}{A_\text{outer}}\right) + h_\text{base} \\
 &= \frac{V_\text{initial}}{A_\text{outer}} - V_\text{transfer}\left(\frac{1}{A_\text{outer}} - \frac{1}{A_\text{inner,internal}}\right) - \frac{V_\text{transfer}}{A_\text{inner,external}} - \frac{V_\text{transfer}}{A_\text{outer}} - \frac{M_\text{constant}}{\rho A_\text{inner,external}} - \frac{M_\text{constant}}{\rho A_\text{outer}} + h_\text{base} \\
 &= \frac{V_\text{initial}}{A_\text{outer}} + V_\text{transfer}\left(\frac{1}{A_\text{inner,internal}} - \frac{1}{A_\text{inner,external}}  \right) - \frac{M_\text{constant}}{\rho A_\text{inner,external}} - \frac{M_\text{constant}}{\rho A_\text{outer}} + h_\text{base} 
 \end{align}
  Let's make a constant called $h_\text{constant}$ with the definition 
 \begin{align}
 h_\text{constant} = \frac{V_\text{initial}}{A_\text{outer}} - \frac{M_\text{constant}}{\rho A_\text{inner,external}} - \frac{M_\text{constant}}{\rho A_\text{outer}} + h_\text{base} 
 \end{align}
  The result looks like what I had before, although with a minus sign due to the mistake pointed out earlier.

 \begin{align}
 h_\text{print} &= V_\text{transfer}\left(\frac{1}{A_\text{inner,internal}} - \frac{1}{A_\text{inner,external}}  \right) + h_\text{constant}
 \end{align}
  \end{document}
```

I can see a few issues with your explanation.



> Lets imagine there are 3 containers. "A" is the inner container. "B" is  the water in the outer container, thats near the inner container. "C" is  the water in the outer container thats below the level of the bottom of  the inner container. Due to Archimedes' principle, "A" and "B" must  always weight the same.


I think you've misunderstood Archimedes' principle.

Archimedes principle does not mean that the mass of the object must  equal the mass of the water around it; otherwise you'd be able to walk  on water. Your mass is maybe 100kg at most. If you step out onto  a 100m square lake and you sink 0.01mm into the water, then the water  around your feet has volume 100*100*0.00001 = 0.1 cubic metres, with a mass of 100kg. Equal weights would suggest, based on your reading of Archimedes' principle, that you'd float around like that, never sinking more than 0.01mm into the water surface. Furthermore, if you went to a bigger lake (say, 1km square) then you'd float even higher. Needless to say, we'd skip cargo ships and just send trains between continents.

In fact, Archimedes principle means that the weight of the water you've displaced must equal the weight of the object floating in the water. The two scenarios are only equal when the two things are exactly the same size. That's manageable here, but it might be difficult. We have to add the restriction that the area of A and B are the same, and that does require taking into account the thickness of the sides of B.




> If you move x amount of water from the "C"  container into the "A" container, then you must also move the same  amount of water from "C" to "B". That means, that regardless of the size  of the inner container, for every mm of water moved from the bottom of  the outer container into the inner container, the inner container will  shrink 2 mm.


Hmm, not sure about that. Talking about millimetres of water doesn't make sense unless you define areas for the containers. Using the above requirement, [area of A = area of B], and if we assume zero wall thickness then [area of C = area of A + area of B] (which is the same as [2 * area of A]). If we take 1mm of water from C, then that's got enough volume to raise A by 1mm and raise B by 1mm. Raising B causes the inner tank to rise 1mm (obviously if the water it's floating in rises, so does the container). Raising A causes it to go back down 1mm, but, crucially, doesn't cause the resin height to go down (because the resin is on top of the water). However, C has now decreased by 1mm too, and that's what everything else is sitting on, so that pulls everything down 1mm. The result is that the top of the water in the outer container is unchanged, the top of the water in the inner tank is unchanged, but the inner tank itself has moved down 1mm.

That's pretty much what we want.


The problem is what happens when the wall thickness is not zero. Now [area of C = area of A + area of B + area of walls]. Now if you take 1mm of water from C, you put slightly more than 1mm of water into each of A and B, but you only move them down by exactly 1mm. The result is that the print level rises slowly.

----------


## masterada

Now we are getting somewhere  :Smile: 

Ye, i totally messed up that ABC model.
I finally understand what A_inner,external means to you. (For some strange reason i thought it was the area in the outer container outside of the inner container).

So the result is A_inner,external = A_inner,internal. Makes much more sense. Could that mean the area of the inner container can be almost as big as the outer? 
Now onto the next problem at hand. Lets make a hole! 
printersketch.jpg
I will pull up some math on it tomorrow or during the weekend, after 36hours of not sleeping im not really up to the task now.

Edit: might actually be able to switch back the above tank, so theres no need for that long pipe

Edit2:
Thats not enough:


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

Meaning that (A_innier,internal/A_inner,external) must be 1 for it to work.
So lets switch back to the tank above:

printersketchv2.jpg
And lets put less salt in the water in the outer container. (d_outer < d_inner)



```
h_sink = M_inner / (d_outer*A_inner,external)
          = (M_inner,initial + V_inner * d_inner) / (d_outer*A_inner,external)
          = M_inner,initial/(d_outer*A_inner,external) + h_inner*(A_innier,internal/A_inner,external)*(d_inner/d_outer)
```

If the inner container is a 700mm*700mm base column with 5mm wall thickness, then A_innier,internal/A_inner,external = 1,014, so its highly likely is possible to get the right densities. (Dont know the density of resin).
PS: im not fully statisfied with the design yet, especially the "some system to avoid waves from drops part". The long tubes are too vulnerable.

----------


## Slatye

Great, sounds like we are indeed getting somewhere. As you've said, A_inner,internal = A_inner,external is the key requirement for it to work using the original syphon system. Unfortunately that means that the walls of the inner container need to have zero thickness - impossible.


However, I did a quick check and actually it's not as bad as it looks. An old margarine container or something similar as the inner container gives you a wall thickness under 0.5mm. Say A_inner,internal = 100x100mm = 10,000mm². Then A_inner,external is 10,100.25mm² (assuming 0.5mm wall thickness). The resulting equation looks like:


```
h_print = V_transfer * 0.000001 + h_constant
```

If you transfer 1L of water between the containers (V_transfer = 100*100*100 = 1000,000mm³) then the print height changes by a whole 1mm. In the mean time, the bottom of the container has fallen 100mm, so you've printed a 100mm high object. The focus change with a 1mm change in height isn't worth worrying about.




The new idea is intruiging. I hadn't thought about using different density water in the containers. I think that'll definitely work, although getting the right densities could be rather painful (measuring density to high accuracy is difficult if you can't measure volume perfectly).

For avoiding waves from the droplets, maybe what you need is a sort of Z-bend near the end of the pipe. That way, if a drop is falling freely down the middle of the pipe, it now gets attached to the wall and slides down there at a reasonably low speed.

----------


## rylangrayston

Wow .... kind of speachless right now... 
LOVE This Thread 

Keep Running with this you guys are abviusly going to make the peachy printer way better that it already is.

----------


## masterada

> Great, sounds like we are indeed getting somewhere. As you've said, A_inner,internal = A_inner,external is the key requirement for it to work using the original syphon system. Unfortunately that means that the walls of the inner container need to have zero thickness - impossible.
> 
> 
> However, I did a quick check and actually it's not as bad as it looks. An old margarine container or something similar as the inner container gives you a wall thickness under 0.5mm. Say A_inner,internal = 100x100mm = 10,000mm². Then A_inner,external is 10,100.25mm² (assuming 0.5mm wall thickness). The resulting equation looks like:
> 
> 
> ```
> h_print = V_transfer * 0.000001 + h_constant
> ```
> ...



So, if we calculate with a 500*500*700mm inner container, with 5mm wall, then


```
V_transfer = 500*500*700 = 1.75*10^8 mm^3
h_print = V_transfer * (1/A_internal - 1/A_external) + h_constant
V_transfer * (1/A_internal - 1/A_external) = 1.75*10^8 * (1/250000 - 1/255025) = 13.79mm
```

If we assume the avarage "focus distance" at the middle of the surface is 500mm, then at the side of the surface its (500^2+250^2)^0.5 = 559.02mm, and at the corner its 612.37mm. (Its even worse if the printer is closer to the surface)
In other words, the distance change from moving in the x-y pane is much more than what comes from the z-axis. So theres no point in making my second type, its too complex anyway.

----------


## Slatye

Sounds about right, and that's a pretty 'harsh' scenario (5mm wall thickness is really, really thick).

I suspect that any Peachy installation is going to be pretty 'tall'. If the bath is 500mm wide, you'll probably want the Peachy at least 1000mm vertically up. That gives you 1000mm focal length in the centre and 1031mm at the sides - so about half the deviation for double the height. It's been a while since I've looked at optics, but intuition says that we're more concerned about the ratio of deviation to total length (ie 31mm/1000mm = 0.031, compared to 59mm/500mm = 0.118) so that would mean that making it even a little bit taller helps a lot.

----------


## rylangrayston

We have made it really easy to change the spring force on the mirror, this means you can play with the traids off between spring force, speed, and deflection. 

more spring force = more speed but less deflection  and possibly a taller printer to make up for the lack in deflection

We have done alot of work to make sure when you get your peachy you wont really be stuck with how we built it, were doing our best to make all the important stuff adjustable, easy to swap hack up and change.
The Idea is to make a printer that can be made to work in many situations. This is a big part of why its hard to give good numbers on the peachy, there is no set in stone deflection or build volume, laser power, aperture size.  Its possible and in many cases easy to change these things.

----------

