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Acceleration-Constant speed-Deceleration over fixed distance and time

  1. May 8, 2010 #1
    1. The problem statement, all variables and given/known data

    Determine the position of an object at regular intervals as it accelerates at a constant rate for an unknown length of time, then the acceleration falls to 0 for a given time, then abruptly acceleration goes negative till reaching 0.

    the purpose of the calculation is to plot out camera moves for stopmotion animation. the camera begins at a dead stop then is accelerated smoothly to a given speed then decelerated to a stop again. the time over which this takes place is fixed as is the distance the camera travels. the variable here is the number of frames (time intervals of 1/24 second) the accelerations take place over. the are not always the same and one may be zero.

    2. Relevant equations

    velocity & time:
    v=vo+at

    displacement & time:
    x=xo+vot+(1/2)at2

    velocity & displacement:
    v2=vo2+2a[tex]\Delta[/tex]x

    3. The attempt at a solution

    This is an example problem that the formula I'm trying to figure out would be used to solve.

    Ex: The camera move takes place over 15 seconds, or 360 frames (shot at 24 fps) the distance it travels is 29.5 inches (I apologize for the use of imperial units, the dolly rig has a standard threaded leadscrew) The lead in (initial acceleration) takes place over 12 frames (0.5 seconds) the camera moves at a constant speed for a length of time that is dependent upon the 2 periods of acceleration, then the camera has a lead out (deceleration) that takes place over 24 frames (1 second)



    I think I need to break the move into 3 parts, Moves 1,2 and 3 or M1, M2 and M3

    for M1 all I know is Time, that portion of the move takes 0.5 seconds.
    I don't know distance travelled because it's dependent upon M2 and M3 and I don't know acceleration because I don't know displacement

    for M2 I know Time again, 13.5 seconds, Acceleration is zero but displacement is dependent upon M1 and M3

    for M3 the case is identical to M1 with Time being 1 second

    I know the Total Displacement of the 3 moves but have no idea how to solve for so many unknowns with anything but trial and error. This must be possible as I've seen powerful software programs that operate motion control rigs produce results with only the information I have in my problem. any Help here would be greatly appreciated

    Puppetworks
     
  2. jcsd
  3. May 9, 2010 #2

    tiny-tim

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    Welcome to PF!

    Hi Puppetworks! Welcome to PF!! :smile:

    Let's rephrase your question …

    there are three periods, lasting (in seconds) t1 = 0.5, t2 = 13.5, and t3 = 1.0 …

    they have three accelerations, a1 and a3 (unknown), and a2 = 0 …

    and three distances x1 x2 and x3, with x1 + x2 + x3 = 29.5.​

    ok, you know the final speed of part 1 is the initial speed of part 3, and t3 = 2t1, so that tells you the relation between a3 and a1 is … ? :smile:

    From that, you can get a relation between x3 and x1, which should simplify the whole thing enough for you to solve the problem. :wink:
     
  4. May 9, 2010 #3
    boy do I wish it hadn't been 10 years since I was in Physics 30... or Math30C for that matter.

    Thanks Tiny Tim, your restatement made things a lot clearer for me, I think.

    so if t1 is 0.5 and t3 is 1 then the relationship between a1 and a3 is 2:1 making the relationship between x1 and x3 1:2

    so 29.5=x1+x2+x
    or 29.5=x1+x2+2x1
    or 29.5=3x1+x2

    I know I'm missing something here since I still have 2 variables to solve for.

    I may also have taken your help and run off in a completely wrong direction....
     
  5. May 9, 2010 #4

    tiny-tim

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    Isn't it 2:1 ?

    (and you've confused x2 and x3 in your next equations. :redface:)
     
  6. May 9, 2010 #5
    so i had it backwards, relationship of t1 to t3 is 1:2 and relationship of x1 to x3 is 2:1

    I'm worried that I've misunderstood my problem. doesn't this mean that the distance travelled during t3 is less than during t1? even though the displacement takes place over twice the time interval? the acceleration curve I want to achieve would be something like : 0.25;0.5;1;2;4;8;16;32;64 etc. Where each subsequent change in displacement is a constant percentage of it's predecessor. in this case 200%

    so a complete move like the one I'm trying to devise a calculator for might look like the following with the numbers representing change in position, not absolute position:

    1-2-4-8-16-32-32-32-32-32-32-32-32-32-32-24-18-13.5-10.125-7.6-5.7-4.3-3.2-2.4-1.8-1.3-1-0.75

    the rate of increase in the first part is 200% and the rate of decrease at the end is 75%

    have i misunderstood the nature of my problem or the direction given?
     
  7. May 9, 2010 #6

    tiny-tim

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    I'm getting confused. :confused:

    I don't think that's right either.

    Write out all the equations in full, and we'll check. :smile:
     
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