acceleration question?

Discussion in 'Physics' started by wes, Aug 26, 2010.

  1. wes

    Thread Starter Active Member

    Aug 24, 2007
    Ok , so the question is pretty simple, but I am not sure whether what I think is correct or not.

    so if a say a pulse jet produces a acceleration of 1 m/s2 each pulse and it pulse's 10 times a second then would the total acceleration over a 1 second period be 10 m/s2?

    if not then how does that work exactly?

    I think it is correct because the engine produce's a certain of thrust each pulse and after taking into account the the mass of the engine, it produce's 1 m/s2 of acceleration because each pulse produced the same amount of thrust.

    just for simplicity, I didn't use number's. If you know some calculations that could help then please post.

    Mods: Sorry if this has been posted before, I didn't see one when I searched.
  2. beenthere

    Retired Moderator

    Apr 20, 2004
    This is like PWM. The current pulses deliver maximum power to the motor, but the off time between pulses prevent the motor from running at full speed as long as the duty cycle is less than 100%.

    You have 10 pulses/sec. If you can determine the duration of each pulse, you can get the duty cycle (percentage of on to off time). From that, you can determine the average acceleration.
  3. someonesdad

    Senior Member

    Jul 7, 2009
    The short answer is it depends on the length of each pulse (i.e., the duty cycle). For example, if each pulsed acceleration is 1 ns wide, your physical intuition should tell you there will be almost no change in velocity. Contrast that to each pulse being 0.099 s wide; you can see that this is nearly equal to a constant acceleration of 1 m/s2.

    With python, numpy, and matplotlib, one has free computer tools to integrate the equations of motion and plot the results. Here's a python program that produced the attached graph; it's easy to experiment with different duty cycles and accelerations.

    Code ( (Unknown Language)):
    2. '''
    3. Integrate an acceleration that is pulsed with various duty cycles and
    4. calculate the resulting average acceleration.  For simplicity, assume
    5. the pulse height is 1 and the total time is 1.
    6. '''
    8. from __future__ import division
    9. from pylab import *
    11. def GetWaveform(num_pulses, num_points_per_pulse, duty_cycle, amplitude=1):
    12.     assert num_pulses > 0 and 0 <= duty_cycle <= 1 and num_points_per_pulse > 1
    13.     n, m, s = (int(num_points_per_pulse*duty_cycle),
    14.         int(num_points_per_pulse*(1 - duty_cycle)), [])
    15.     for i in xrange(num_pulses):
    16.         s.append(concatenate((zeros(n) + amplitude, zeros(m))))
    17.     return concatenate(s)
    19. def Integrate(vector, time_step):
    20.     return cumsum(vector*time_step)
    22. def main():
    23.     num_pulses, num_points_per_pulse, duty_cycle = 10, 100, 0.05
    24.     time_step = 1/(num_pulses*num_points_per_pulse)
    25.     a = GetWaveform(num_pulses, num_points_per_pulse, duty_cycle)
    26.     # Plot acceleration
    27.     subplot(311)
    28.     plot(a)
    29.     title("Acceleration (duty cycle = %g)" % duty_cycle)
    30.     xlim(-1, num_pulses*num_points_per_pulse)
    31.     ylim(-0.05, 1.05)
    32.     grid()
    33.     # Plot velocity
    34.     v = Integrate(a, time_step)
    35.     subplot(312)
    36.     plot(v)
    37.     title("Velocity")
    38.     xlim(-1, num_pulses*num_points_per_pulse)
    39.     grid()
    40.     # Show average acceleration; it's delta(v)/delta(t), but delta(t) is 1.
    41.     avg_v = max(v) - min(v)  # Final v minus starting v
    42.     n = num_pulses*num_points_per_pulse
    43.     text(n/20, max(v)/2, r"$a_{avg} = \frac{\Delta v}{\Delta t} = %g$" % avg_v)
    44.     # Plot position
    45.     x = Integrate(v, time_step)
    46.     subplot(313)
    47.     plot(x)
    48.     title("Position")
    49.     xlim(-1, num_pulses*num_points_per_pulse)
    50.     grid()
    51.     if 0:
    52.         show()
    53.     else:
    54.         savefig("pulsed_acceleration.png", dpi=150)
    56. main()
    The abscissa of the graph is the number of points in the simulation; however, note that I set the problem up to consume one second for ease of calculation (and I also used an acceleration of 1).

    Python and numpy are powerful tools for these types of calculations and are worth the effort to learn.
  4. wes

    Thread Starter Active Member

    Aug 24, 2007
    I like how you said what if the pulse was 1 n/s wide and that got me thinking.

    thought experiment!!!!!!!!!

    So if each pulse was 1 n/s wide and produced a certain amount of force that created a acceleration of a tiny amount but it did this 500 million times a second then wouldn't that generate a huge force if say each individual pulse created a acceleration of only small amount but it was done so many times a second it created a large overall force


    engine wight: 1 pound

    each pulse generates a 0.0001 grams of force

    500 million pulses a second

    = eithier overall 0.0001 grams of force or 50,000 grams of force

    50,000 grams = 110.23 pounds

    so 110 pounds with a mass of 1 pound = 1078.73150115 meter/second^2
    or another way is
    0.0001 grams with a mass of 1 pound = 2.1619962434565E-6
    but over 500 million pulse's = 1080.5 meter/second^2

    (roughly equal, not exact cause I did not go to the last digit, only used like the first 5 digits)

    so again yes each pulse does generate a tiny force but since it is done so many times a second then it generates a substantial force. correct or not correct?

    In other words the duty cycle i guess is really high

    So is my thinking correct or am I missing something?
    If anyone know's of some usefull equations (I am not a math genius, got a great imagination though, lol ) then please post and if you could, please explain them, lol
  5. jpanhalt


    Jan 18, 2008
    Nope, I think you are still multiplying instead of looking at the fraction of "on" time (i.e., duty cycle).

    Look at it like this: If the pulse is continuous and gives a certain acceleration, then breaking that pulse up into little pulses will never exceed the acceleration of a continuous pulse. Remember that with each pop of a pulse jet, the acceleration is not instantaneous. Also, air flows back into the tail pipe to fuel the next explosion. That air has mass and has to be stopped and reversed for the nest pulse. Thus, that decreases your net acceleration even more.

    In real life, I have never measured the on time of a pulse jet. I suspect it is relatively short, say <20%, just based on the characteristic sound they make. Mine gives about 4 to 6 pounds of static thrust and goes though alcohol like you are pouring it from a bottle.

  6. DonQ

    Active Member

    May 6, 2009
    Velocity = integral of acceleration over time.

    For a constant acceleration, the integral is simply a multiplication.

    Simply put, this means that (for a constant acceleration) multiply the acceleration by the total time that the acceleration is applied. In your case, the duration of each acceleration pulse, times the number of pulses equals the total duration, and you do have a fixed acceleration. That's the answer.

    You will also see that the units work out... meters/second-squared times seconds equals meters/second.
  7. wes

    Thread Starter Active Member

    Aug 24, 2007
    so then to put it simply, if it has a pulse that times the number of pulses equal half a second then the acceleration is that number times the acceleration created by a single pulse?

    so if each pulse generated .0001 grams of thrust , then the overall thrust a second could never be higher then that?