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Is this design possible to get an axcess energy?

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  • Is this design possible to get an axcess energy?

    Dear All,

    Please see my drawing in the attachment.

    The idea to create an excess energy. Not sure if this theory are work.

    Example ;
    • Big Cylinder Diameter 4 Inch
    • Small Cylinder Diameter 2 Inch
    • Shaft Diameter 1 Inch
    • Pump Pressure 1000 PSI

    When Piston reach top stroke, pressure will raise on Pilot Port A and pilot control (not shown) will joint Cylinder port A and Cyl Port B. This action will increase pressure on big cylinder and same time reduce pressure on small cylinder to equal pressure (if my formula are correct, with 4” stroke length the pressure will be 133 PSI). The load (estimate 1 Tone) will push down by gravity and pressure on big cylinder will also push down in other word the force of down stroke much bigger than the load (about 1500kg).

    When all piston down completely, the pressure will increase on Pilot port B and then pilot control will connect the Cylinder Port B to “Pump” and at same time connect Cylinder Port A to Tank. So the Piston will lift up until next step and so on.

    Please comment.......

    Regards,
    Selamatg
    Attached Files

  • #2
    Originally posted by selamatg View Post
    Dear All,

    Please see my drawing in the attachment.

    The idea to create an excess energy. Not sure if this theory are work.

    Example ;
    • Big Cylinder Diameter 4 Inch
    • Small Cylinder Diameter 2 Inch
    • Shaft Diameter 1 Inch
    • Pump Pressure 1000 PSI

    When Piston reach top stroke, pressure will raise on Pilot Port A and pilot control (not shown) will joint Cylinder port A and Cyl Port B. This action will increase pressure on big cylinder and same time reduce pressure on small cylinder to equal pressure (if my formula are correct, with 4” stroke length the pressure will be 133 PSI). The load (estimate 1 Tone) will push down by gravity and pressure on big cylinder will also push down in other word the force of down stroke much bigger than the load (about 1500kg).

    When all piston down completely, the pressure will increase on Pilot port B and then pilot control will connect the Cylinder Port B to “Pump” and at same time connect Cylinder Port A to Tank. So the Piston will lift up until next step and so on.

    Please comment.......

    Regards,
    Selamatg
    Your diagram is a basic hydraulic/pnuematic pressure intensifier, you will indeed get a gain in pressure in the B chamber but when this pressure is applied to the A chamber which has a larger volume the best you can expect is an equalisition of pressure.
    Force = Pressure * Area
    There is no gain in this configuration.

    Comment


    • #3
      @HiggsBoson

      Thanks for your comment.

      You are right for the on the equalisition of pressure.
      Since, Force = Pressure * Area, so that mean the force down of big cylinder much bigger than force up of small cylinder.

      In my calculation (if my formula are correct) then the lowest pressure of both cylinder (when joinned) is 133 PSI, it will about 500kg push down excess.


      selamatg

      Comment


      • #4


        It is possible ????

        Please advice....


        selamatg
        Attached Files

        Comment


        • #5
          How to raise the 2 tons up . . .
          "Amy Pond, there is something you need to understand, and someday your life may depend on it: I am definitely a madman with a box." ~The Doctor

          Comment


          • #6
            Mouse Power


            Click For Larger Picture


            So the mouse eats and sleeps. And when he does, it shifts the weights
            "Amy Pond, there is something you need to understand, and someday your life may depend on it: I am definitely a madman with a box." ~The Doctor

            Comment


            • #7
              Maybe a Hydraulic ram could help you?

              Hydraulic ram - Wikipedia, the free encyclopedia

              Home-made Hydraulic Ram Pump

              Comment


              • #8
                *Removed to prevent BIG misunderstandings*
                Last edited by Harvey; 04-25-2010, 02:30 AM.
                "Amy Pond, there is something you need to understand, and someday your life may depend on it: I am definitely a madman with a box." ~The Doctor

                Comment


                • #9
                  That's my signature Harvey, it's not a part of my post.

                  Comment


                  • #10
                    *Removed to Prevent BIG misunderstandings*
                    Last edited by Harvey; 04-25-2010, 02:31 AM.
                    "Amy Pond, there is something you need to understand, and someday your life may depend on it: I am definitely a madman with a box." ~The Doctor

                    Comment


                    • #11
                      Is there something I wrote to offend you Harvey? Is my original post about a ram pump not pertinent? I was simply giving possible advice to the person asking. In no way was I replying or even insinuating anything towards your post and now you seem to be attacking me with... Dull wit?

                      So yeah, Harvey, either your confused about my intentions or your looking to pick an argument with me. Which is it?

                      Comment


                      • #12
                        @Harvey

                        Originally posted by Harvey View Post
                        How to raise the 2 tons up . . .
                        Just increase the pump pressure.

                        Force = Pressure * Area
                        The efective area of small cylinder is 2.36 inch square
                        The efective area of big cylinder is 11.79 inch square
                        =2.36 x 2100 (psi) x 0.9 (power factor) = 2,025 kgs

                        On stroke down :

                        If my calculation is right then the pressure of both cylinder when joinned is 420 PSI.
                        The force down will be ;
                        The load weight + the force down of big cylinder - the force up small cylinder
                        = 2 tons + (11.79 x 420 x 0.9) - (2.36 x 420 x 0.9)
                        = 2 tons + 2 tons - 405 kgs
                        = 3.62 tons force down

                        am i right?

                        Regards,
                        Selamat

                        Comment


                        • #13
                          Originally posted by HairBear View Post
                          Is there something I wrote to offend you Harvey? Is my original post about a ram pump not pertinent? I was simply giving possible advice to the person asking. In no way was I replying or even insinuating anything towards your post and now you seem to be attacking me with... Dull wit?

                          So yeah, Harvey, either your confused about my intentions or your looking to pick an argument with me. Which is it?
                          You clearly have a problem - and no, I do not wish any ill will toward any member here.

                          You may wish to choose your signatures more wisely prior to offending people with them - I see you have found it proper to change it.
                          "Amy Pond, there is something you need to understand, and someday your life may depend on it: I am definitely a madman with a box." ~The Doctor

                          Comment


                          • #14
                            Originally posted by selamatg View Post
                            Just increase the pump pressure.

                            Force = Pressure * Area
                            The efective area of small cylinder is 2.36 inch square
                            The efective area of big cylinder is 11.79 inch square
                            =2.36 x 2100 (psi) x 0.9 (power factor) = 2,025 kgs

                            On stroke down :

                            If my calculation is right then the pressure of both cylinder when joinned is 420 PSI.
                            The force down will be ;
                            The load weight + the force down of big cylinder - the force up small cylinder
                            = 2 tons + (11.79 x 420 x 0.9) - (2.36 x 420 x 0.9)
                            = 2 tons + 2 tons - 405 kgs
                            = 3.62 tons force down

                            am i right?

                            Regards,
                            Selamat
                            The area of your pistons is correct.

                            I assume the power factor is to offset frictional losses ?

                            To convert from PSI to kg / cm² we divide PSI / 14.22. So, (2.36 * 2100) / 14.22 = 348.52 kg / cm²

                            Another thing to consider is the volume differential when Port A and Port B are connected. The pressure in the smaller cylinder will drop as the overall volume increases which it will as the top piston drops opening up that space. So the downward force needs to be integrated for that volume differential.

                            Last edited by Harvey; 04-25-2010, 04:37 AM.
                            "Amy Pond, there is something you need to understand, and someday your life may depend on it: I am definitely a madman with a box." ~The Doctor

                            Comment


                            • #15
                              Hi Harvey,

                              To convert from PSI to kg / cm² we divide PSI / 14.22. So, (2.36 * 2100) / 14.22 = 348.52 kg / cm²
                              sorry i forgot to write divide by 2.2 to convert lb to kg. but the result are include with divide 2.2.

                              My understanding the basic formula as below :

                              Force (lb) = Area (in2) x Pressure (PSI) x Mechanical Efficiency (0.9)

                              Please correct me if i'm wrong.

                              Another thing to consider is the volume differential when Port A and Port B are connected. The pressure in the smaller cylinder will drop as the overall volume increases which it will as the top piston drops opening up that space. So the downward force needs to be integrated for that volume differential.
                              You right, both cylinder will keep same pressure when connected. That point in my understanding, cause the top cylinder much bigger the effective area than small cylinder will add the downward force.

                              The pressure will come down gradually.

                              Again...please correct me if i'm wrong.


                              selamatg

                              Comment

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