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orbital valve
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<blockquote data-quote="TBItoy" data-source="post: 167761" data-attributes="member: 1384"><p>pretty simple to figure out your displacement and size your orbital to your ram.</p><p></p><p>On the "big side" (the side that doesn't have the shaft on a single ended ram) you use the area of the piston x the length. So: 3.14 x (diameter/2)^2 x length</p><p></p><p>So a 2"x10" ram would be: 3.14 x (2/2)^2 x10 = 31.4 in^3 (cubic inches) of fluid to go from lock to lock. So a 10 in^3/rev valve (moves 10in^3 of fluid per revolution) will put you right at 3 turns lock-lock one way.</p><p></p><p>Now on the "small side" you have to subtract the volume of the rod. Say the rod is 1.25" and the same 2x10 cylinder.</p><p></p><p>3.14 x (1.25/2)^2 x 10 = 12.265 in^3, Now you gotta subtract that from the cylinder volume so: 31.4 -12.265 = 19.135 in^3, </p><p></p><p>So with the same 10in^3/rev orbital, you'll be a little more than 3 turns lock-lock one way and a bit less than 2 turns lock-lock the other.</p></blockquote><p></p>
[QUOTE="TBItoy, post: 167761, member: 1384"] pretty simple to figure out your displacement and size your orbital to your ram. On the "big side" (the side that doesn't have the shaft on a single ended ram) you use the area of the piston x the length. So: 3.14 x (diameter/2)^2 x length So a 2"x10" ram would be: 3.14 x (2/2)^2 x10 = 31.4 in^3 (cubic inches) of fluid to go from lock to lock. So a 10 in^3/rev valve (moves 10in^3 of fluid per revolution) will put you right at 3 turns lock-lock one way. Now on the "small side" you have to subtract the volume of the rod. Say the rod is 1.25" and the same 2x10 cylinder. 3.14 x (1.25/2)^2 x 10 = 12.265 in^3, Now you gotta subtract that from the cylinder volume so: 31.4 -12.265 = 19.135 in^3, So with the same 10in^3/rev orbital, you'll be a little more than 3 turns lock-lock one way and a bit less than 2 turns lock-lock the other. [/QUOTE]
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