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Question About Heave Calculations

Discussion in 'DIY Motion Simulator Building Q&A / FAQ' started by Marc Turcotte, Feb 13, 2015.

  1. Marc Turcotte

    Marc Turcotte New Member

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    Hello Everyone,

    I'm trying to calculate the acceleration duration for a certain heave displacement in order to generation 0.5G (4.905 m/s²) of acceleration.

    I'm using CKAS U5 3DOF Motion Syteme http://www.ckas.com.au/3dof_systems_64.html to double check my calculations and cannot seem to be able to reproduce their results which are:
    • displacement =0.140 m
    • velocity=0.280 m/s
    • acceleration=4.905 m/s²
    I know that:
    • Time t taken for an object to fall distance d is t=SQRT(2d/g)
    • Instantaneous velocity vi of a falling object after elapsed time t is vi = gt
    • Instantaneous velocity vi of a falling object that has traveled distance d is vi = SQRT(2gd)
    My guest is that we need one half of the displacement to accelerate to speed and the other half to stop moving,

    This would mean that (0.14/2)=0.07 would be the actual length available for the acceleration meaning that
    t = SQRT(2x0.070/4.905) = 0.169 s
    Trying to get back to the CKAS velocity of 0.280 m/s I'm getting
    • vi = gt = 4.905x0.169 = 0.829 m/s
    • vi = SQRT(2gd) = SQRT(2x4.905x0.07) = 0.829 m/s
    0.829 m/s is far for been equal to 0.280 m/s

    I must be missing something so basic. [​IMG]

    Could someone tell me what I'm doing wrong ?
  2. noorbeast

    noorbeast VR - The Next Generation Staff Member Moderator

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    My Motion Simulator:
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    Unless I have made a mistake I make it to be 0.24, not 0.28

    Given:

    vi = 0.280 m/s

    d = 0.140 m

    a = 4.905 m/s2

    Find:

    t = ??
    d = vi*t + 0.5*a*t2
    0.140 m = (0 m/s)*(t)+ 0.5*(4.905 m/s2)*(t)2

    0.140 m = 0+ (2.4925 m/s2)*(t)2

    (0.140 m)/(2.4925 m/s2) = t2

    0.0561685055165496 s2 = t2

    t = 0.2369989567836736 s
    Last edited: Feb 13, 2015
  3. bsft

    bsft

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  4. value1

    value1 Nerd Staff Member Moderator SimAxe Beta Tester SimTools Developer Gold Contributor

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    My Motion Simulator:
    2DOF, DC motor, JRK, Joyrider
    This is the speed profile with the given data:
    Dynamic Profile CKAS Time.png
    With 0.5 g acceleration you reach the target speed of 0.28 m/s after approximately 50 ms. For the next half second you travel at 0.28 m/s speed and then decelerate during another ~50 ms to v = 0 m/s
    The whole motion lasts about 0.6 seconds. Its more like a hit than a continous movement.
    • Agree Agree x 1
    • Winner Winner x 1
  5. Marc Turcotte

    Marc Turcotte New Member

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    Hello Value1,

    A graph is worth a thousand word; I now know where my confusion came from. :)

    I presumed they needed the full half to get to 0.28 m/s which is not the case since 0.5 G will get you at that speed much sooner.

    What software did you use the generate that graph ?

    Lets assume we are simulating a drop cue :
    • The drop cue will last ~57 ms/8 mm and will be felt by the operator
    • The remaining ~500 ms will be done a constants speed so the operator wont' fill a thing
    • Obviously we have to stop within that last ~57 ms/8 mm and will be felt by the operator
    Why wait up to ~ 500 ms/124 mm to stop ?

    Can I presume it will be for making the drop cue feel/last longer ?

    I know from experience that you can fall for what seems an eternity while flying is bad weather condition.

    Is there some king of relation between the heave displacement, speed and acceleration ?

    Home much heave is not enough or to much ?
    Last edited: Feb 13, 2015
  6. value1

    value1 Nerd Staff Member Moderator SimAxe Beta Tester SimTools Developer Gold Contributor

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    My Motion Simulator:
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    Excel is your friend emoticon-00139-bow.gif
    Not sure about that. I think you can easily tell, if your driving constant 120 km/h on a motor bike or if you're standing still emoticon-00105-wink.gif
    I would say the acceleration is not 0.5 g from the very beginning but increasing from 0 g up to 0.5 g
    And therefore you have a longer sensation, if the displacement is 140 mm instead of 50 mm.
    • Informative Informative x 2