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Homework Help: Help Projectile Motion and High Speed Camera

  1. Jun 26, 2008 #1
    We are trying to calculate the acceleration of gravity using a quarter inch ball bearing and a high speed video camera. The ball bearing it shot at 6 m/s across a grid .5 meters long. The high speed camera is 1.8 meters away and records the position of the ball at 250 frames per second. We thought that we could determine how fast the ball was accelerating downwards by analyzing the position of the ball bearing on each frame. However, we are systematically getting lower than usual results (9 m/s/s instead of 9.8 m/s/s). We have made sure that the set up is as precise as possible using laser levels. We have also checked the timing mechanism of the camera's clock. Does anyone have any suggestions?
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  3. Jun 26, 2008 #2
    It could possibly be the camera placement. I guess your looking at an overall drop of about 34cm across the grid. The further the ball is from the center of the lens, the more error will be introduced to the measurement...but I don't see how it would effect your measurement that much. How did you measure the inital velocity of 6 m/s? Are you using NI's Vision Assistant? What do you measure if you just drop the ball bearing straight down?
  4. Jun 26, 2008 #3


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    could you please describe in detail the method you use to calculate the acceleration based on the images recorded by the camera?
  5. Jun 26, 2008 #4
    We are using Logger Pro Software to analyze the position of the ball bearing per time period. The software allows us to calibrate the length that the ball travels by setting a length equal to a certain number of pixels. Once we have done this, we simply mark the center of the ball on each frame. This gives us a parabolic shape of the balls trajectory. The software now has a coordinate for each of the ball positions per time period. We then apply a best fit line to curve of the balls trajectory using the At^2 + Bt + C exponential equation. Logger Pro then gives values for A, B, and C based on the data collected. Since the position function for projectile motion is .5gt^2 +vt, we are able to determine what g should be based on the value of A that the software gives us. We generally get A= 4.5. What we need is A= 4.9. We are aware that the there is distortion caused by the spherical nature of the lens at the tops and bottoms of the lens. We have determined that this is not the cause of the problem.
  6. Jun 26, 2008 #5
    As to how we measure the initial velocity, Logger Pro also can determine the velocity by taking the differences of the first two points and dividing by the time between those two points. We are able to get better results when we simply drop the ball bearing with no horizontal velocity. We thought that having a larger drop in the y direction would be easier to calculate.
  7. Jun 26, 2008 #6


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    Just how close is the bearing to the grid? If you're comparing the ball's position to the grid, from the lens of the camera, apparent position on the grid will be offset relative to distance between bearing and grid.

    Shot across the grid? If the bearing is being shot horizontally, then you only have 1/12th of a second of images in front of the grid, or 20.83 frames, and the ball is not going to drop very much in this short of a period of time.

    If you shot the bearing nearly vertical at 3 m/s, it would stay within the bounds of the grid and give you over 1/2 second or 125 frames to use for your experiment.

    pixels? Do you know for certain that the pixel aspect ratio is 1:1 and not some other ratio (4:3, 16:9, ...)?
    Last edited: Jun 26, 2008
  8. Jun 26, 2008 #7
    The bearing is 4 cm in front of the gridl. Not far enough to cause this much distortion we checked. We even used the actual distance the bearing travels instead of the relative distance but to no avail. We realize that the ball is not going to drop very much but we need to shoot the bearing at this speed in order to generalize our results to that of seeds being released ballistically in nature. We had hoped to use spin on our seeds and compare the resulting y accelerations to that of gravity. I am not familiar with pixel aspect ratios. I will check this, but our resolution is 640 by 480. We have mapped the grid in pixels and found that it is flat and distances are symetrical to within 2 pixels.
  9. Jun 26, 2008 #8


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    Assuming a 4:3 image, then the pixel aspect ratio would be 1:1, in otherwords, the pixels are square, and the inner 480x480 pixels take an image of a square.

    I doubt this is an issue with a heavy ball bearing, but could spin be creating an aerodynamic issue? This doesn't seem likely, because what I would expect is since the ball is resting on the bottom of the barrel when shot, there will be a bit of top spin, causing downward curvature, which would seem to increase downwards acceleration, not decrease it (backspin would be required to decrease downwards acceleration).

    Am I correct in assuming it's a horizontal speed of 6m/s across a .5m grid or time is 1/12 of a second, which would mean the distance dropped is 1/2 9.8 (1/12)2 = 3.4 cm. How many pixels does this translate into?
  10. Jun 26, 2008 #9
    oops..derr.....34mm. Yea, lens distortion wouldn't effect such a small height. When I did video measurements (crash tests), we would sometimes use a 4 or 5 point moving average, which might be useless with 20 frames.
    Last edited: Jun 26, 2008
  11. Jun 27, 2008 #10
    There should not be any spin with the ball bearing. The ball bearing it mounted on a pin and the pin is inserted horizontally onto the gun. When the gun launches, the pin stays put and the ball leaves the pin and travels across the grid. The grid is longer than .5 m, but the usual length the ball travels is around .5-.6m. On my last video, the x distance was .604m, the y drop was .102m, and the time period was .108 seconds. The initial velocity was 5.915 m/s and the acceleration in the y direction was -8.8 m/s/s and -3.4 m/s/s in the x direction. The drag is much too high and gravity is much too low. They should be -.3m/s/s and -9.8 m/s/s respectively.
  12. Jun 27, 2008 #11
    As for pixels, 10 inches or 25.4 cm equals 254 pixels. I have no idea why it works out so nicely.
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