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Thread: This could work

  1. #11
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    Quote Originally Posted by mike_biddell View Post
    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

  2. #12
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    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.

  3. #13
    Quote Originally Posted by mike_biddell View Post
    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.
    Quote Originally Posted by Slatye View Post
    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).

  4. #14
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    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.

  5. #15
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    Quote Originally Posted by Feign View Post
    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.

  6. #16
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    Quote Originally Posted by mike_biddell View Post
    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.)

  7. #17
    Quote Originally Posted by Feign View Post
    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

  8. #18
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    Quote Originally Posted by masterada View Post
    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
    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.

  9. #19
    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?

  10. #20
    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

    Edit: forget it, this is way more simple (metal rails are fixed on the outer container):
    printersketchabove2.jpg
    Last edited by masterada; 01-21-2014 at 04:18 PM.

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