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  • Continuous Change of Direction of Rotation

    Discussion in 'The main mechanical design forum' started by SachinHB, Feb 6, 2016.

    1. robertjeffery

      robertjeffery Active Member

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      Right.... Hopefully got my head round it for another attempt. A jerk would be created by a change in acceleration, and the changes of acceleration are in the centre of the stroke of the rack. It goes from a positive acceleration till the centre, then switches to a negative acceleration untill it reaches the center point on the return journey, in which it then turns back into a positive acceleration again.....
       
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    3. Bill @ ERG

      Bill @ ERG Member

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      Exactly. There should be no "jerk". Acceleration is as smooth as the displacement curve, but opposing (off by 180°). Piston velocity is also sinusoidal, but offset by 90°. There is a bit of sport in getting the structure and bearing loads sorted out. Essentially this is very similar to a recip engine, except the effective mass of the "piston" is increased with the geared flywheel assembly.
       
    4. SCIYER

      SCIYER Well-Known Member

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      Motion being a SHM, when we differentiate the motion thrice (First for Velocity, second for Acceleration, third one for Jerk or Rate of change of Acceleration) we get Jerk. When we consider the instantaneous values, for the figure, acceleration at theta = 0 (at x-axis) and theta = pi, turns from a Finite +ve to Finite -ve. It is at this instant, the rate of change of acceleration turns into infinite Jerk, though for a moment of time.

      I dont know how to paste a Mathematical notation here. Tried but failed.

      I hope I could explain my thoughts.
       
      Last edited: Feb 28, 2016
    5. s.weinberg

      s.weinberg Well-Known Member EngineeringClicks Expert

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      I think Bill is correct. Maximum acceleration is at the end of stroke, but motion should still be pretty smooth. Easy to see this from simple analysis.

      For circular motion, there is a constant acceleration towards the center, and a constant velocity tangent. At the end of stroke, the horizontal component of velocity is 0, because the tangent f circular motion is perpendicular to the linear motion. At the same time, acceleration is parallel to the linear motion and linear acceleration equals the acceleration of the rotational link.

      From a mathematical perspective, the derivative of sin(x) is cos(x), and at 0 and 180 degrees, it equals 1.

      So you get a nice, smooth, ramp up of acceleration and vice versa, with a maximum equaling that experienced by your rotational member.
      It functions, as others have pointed out, just like the cylinder in an engine.
       
    6. SCIYER

      SCIYER Well-Known Member

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    7. Bill @ ERG

      Bill @ ERG Member

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      The loading on this rack will be a function of the peak acceleration rather than the rate of change of acceleration. Yes at the ends of the stroke the peak acceleration change reversal occurs, but the direction of acceleration does not, reversing at the nominal mid-stroke locations passing through zero and smoothly reversing until peak accelerations at the other end of the stroke. Because of the connecting rod geometry, the motion of the rack will not be a true sinusoidal motion, but a close approximation, with smoothing of the rate of change due to the angular displacement of the rod. Depending on the rod length / stroke ratio, considerable change to the low velocity period can take place, and increases in the velocity near mid stroke. The fact that piston engines can operate at very high rpm without self-destructing (in fact it is much easier to get the crank / rod / piston to live than the cam / valve system) and that going to a fully controlled valve mechanism such as a desmodromic valve arrangement can tolerate substantially higher rpm and more aggressive valve profiles without consequent valve train damage is proof of this.
       
    8. robertjeffery

      robertjeffery Active Member

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      I agree with Bill, the comment above.
      the initial post asks "Will there be any problem caused due to continuous change of rotation direction"
      I don't see any mechanical faults with this concept, the only slight thing that concerns me....
      Is at Mid travel of the rack, where the output shaft will go from a positive acceleration to a negative acceleration. How will pinion take up the slack.

      eg, if you just imagine the pinion as a flywheel, then the flywheel will be driven up to Mid stroke, then effectively overrun until the clearances between the gear faces allow the pinion to make contact with the other side of the gear on the rack and therefore start the de-acceleration stroke.

      but now think that its not a flywheel, but it is a propeller, eg constant load.
      So the rack is driving the pinion up to the Mid point, and it will have to continue to drive it even as it de-accelerates, up to where it changes direction at which the clearances in the rack and pinion are taken up as it will then have to be driven back in the opposite rotation.

      My conclusion. No issue with the concept, but use a helical cut rack and pinion to eliminate backlash.

      I really don't like scrap heap challenges for engineering, but to make a really simple prototype.
      you could just obtain a conventional car rack and pinion, rip out the hydraulic gubbins which will add drag. leave one side track rod arm attached, connect the track rod end that usually fixes to the hub, to a lathe faceplate or equivalent speed adjustable machine. and put a load on the output eg where the steering wheel goes.
       
      Last edited: Mar 2, 2016
    9. Strambo

      Strambo Member

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      No, it doesn't. That much should be obvious. Look at the curve.

      In what sense is a cosine "discontinuous"? If v follows a sine curve, dv/dt follows a cosine, so da/dt follows a -sine curve and so forth. There is no step function. Sine waves are point, slope, and curvature continuous last time I checked...
       
    10. SCIYER

      SCIYER Well-Known Member

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      You are looking at ONE cycle. Dv/dt though cosine, is negative max at the end of the cycle and positive max at the beginning of next cycle. To say, acc is tending to -ve max as angle tends to pi ( say 179 deg) and suddenly changes to +ve at an angle little more than pi. Say 181 deg. For a half cycle there is no Jerk. Jerk is only at the beginning and ends of every half cycle.
      A kinematic analysis for the Full cycle would reveal that.
      Thank you all. That was interesting.
       
      Last edited: Mar 5, 2016
    11. Strambo

      Strambo Member

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      I don't think you are looking at a sine function when you claim that a portion of that curve goes vertical (infinite slope), because that simply doesn't happen. If the follower is following the surface and doesn't bounce, The curve is NOT a step function.

      The radial displacement follows a sine function. The velocity is the derivative and follows the cosine function.

      As the cam goes from the +radial vel lobe to the -radial vel lobe, it passes through the vel=0 point, and the acceleration is max negative. Isn't that the basic behavior of SHM? The functions are sinusoidal and 90° out of phase. No flat spots, no steps, no spikes and no infinite slope. There is no point where any of the functions derived from the dr/dt function "suddenly changes" from zero to some finite value.

      With me so far? The acceleration dv/dt follows a -sine function! There are no vertical portions! You can take as many derivatives as you like and the result is still sinusoidal with no jump discontinuities. The same goes for j=da/dt. It, also, is a CONTINUOUS SINUSOIDAL FUNCTION> If you cannot understand that, you might benefit from a college course in vibrations.
       
      Last edited: Mar 5, 2016

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