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17862Re: [multimachine] New MM 11/21 (engineer needed)

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  • Pat Delany
    Nov 21, 2013
      Many thanks lance
      Is it possible to figure out how much pressure is needed to push them apart?

      Pat


      On Thursday, November 21, 2013 3:32 PM, Eggleston Lance <wheezer606@...> wrote:
       
      Pat,,
      Some quick calculations:

      Using 3 x 3 x 1 = 9 cu ft /2 =4.5 cu ft for slide

      Density Portland cement  concrete = 145# / cu ft

      4.5 x 145 = 652 # concrete slide

      slide sits on base at 45* angle

      461 # normal to base and 461 # parallel to base angle

      Coef friction LUBRICATED steel to steel is 0.16

      Ff = coef friction x Fn
      Ff= 0.16 x 461#
      Ff = 73 #

      It would take 73# force to overcome the friction (not mass)
      of the slide sitting on the base, assuming it sits on lubricated steel
      on the slide and base.

      Then you'd have to supply 461# parallel to the base angle to overcome the mass.
      So, figure 461+73= 534 # force to move the slide up the base angle.

      For me, I'd use pneumatics or hydraulics, But ropes and pulleys could be rigged
      to lower the force to around 90# using a triple upper and lower pulley block set.

      lance
      ++++
      On Nov 21, 2013, at 3:41 PM, <rigmatch@...> <rigmatch@...> wrote:

      Let us assume this thing is somewhere between 32" and 36" tall and wide and is 1' deep. What would have to be done to the model in the drawing for the vertical slide (and attached concrete lathe cross slide) to make the cross slide press against the (frame?) with at least 50 pounds pressure..



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