This could work

[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.

[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.

[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.

[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.