### Ping Pong Ball Hit

 Ping Pong Ball Hit: 250 Frames/second Ping Pong Ball Hit 2: 250 Frames/secondPing Pong Ball Hit 2 Analysis:Ball Before Hit Position Graph (x-axis)The above graph is a position graph of the ping pong ball (with respect to the x-axis) before it hits the paddle. Let’s take a look at this graph before we move on to the velocity graph. So the first thing we can notice about this graph is the slope. The autofit function shows us that the slope = -1.548. Now, the slope of the position graph gives us the change in velocity. So that means that the change in velocity = -1.548m/s. (If we were to calculate this out manually it would be   Change in velocity = [(Final velocity – Initial velocity) / (Final time – Initial time)]. In this case, Change in velocity = [((0.29m) – (-0.699m)) / ((0.012s) – (0.0636s))] = -1.58 m/s, which is close enough to the autofit to be considered accurate. From this graph we can also learn displacement. To find displacement (how far from the origin the ball moved) we simply take the initial distance and subtract it from the final distance (Displacement = final distance – initial distance). In this case we get (-0.699m) – (0.29m) = -0.989m. Ball Before Hit Velocity Graph (x-axis) Now we can move on to the velocity graph. The above graph is the velocity graph of the ping pong ball (with respect to the x-axis) before it hits the paddle. The first thing you may notice about this graph is that we used the area feature to figure out the area above this graph. The area under a velocity graph is equal to the total displacement of an object. Knowing this, we can compare our displacement we calculated earlier against the area of this graph as a check to see how accurate we are. Since -0.989 and -0.941 are reasonably close, we are good in our calculations. Now in a position graph, the slope was equal to the change in velocity. The slope of a velocity graph, however, is equal to the change in acceleration. So using the autofit feature we get 0.905 m/s^2. If we were to calculate this out manually it would be  change in acceleration = [(Final velocity – Initial velocity) / (Final time – Initial time)]. In this case it would be [((-1.154m/s) – (-2.693m/s)) / ((.0624s) – (0.24s))] = 2.557 m/s^2. Wait, that doesn’t seem right does it? That’s because if you look at the graph more of the points are above the initial velocity. So to get a more accurate slope manually we should use the fourth point up, which is more in line with the rest. Using this point we get [((-1.714m/s) – (-1.159m/s)) / ((0.06s) – (0.624s))] = 0.984 m/s^2. Granted, we didn’t use three points (which would factor into the slope of the line) when we solve manually but to keep things simple 0.984 is close enough to 0.905 to be considered accurate (the three points we excluded would lower the value of our slope we found out manually bringing the 0.984 closer to the 0.905, this is why we can say it's close enough to be accurate). Ball Before Hit Acceleration Graph (x-axis) Finally! We’re on the acceleration graph. The above graph is the acceleration graph of the ping pong ball (with respect to the x-axis) before it hits the paddle. If you notice, we again have the area of the graph calculated. While the area of the velocity graph equaled the total displacement, the area of the acceleration graph equals the change in velocity. So according to our graph the area = 1.32 m/s. Now this doesn’t match our previously calculated velocity. However, if you look carefully in the original video, the camera shakes. This causes the axis to shift suddenly in relation to the ping pong ball. As a result, the tracker program used to analyze these videos thinks the ball has an acceleration faster (and in some cases slower) than what it really has. This error in acceleration causes there to be an error in finding the area which is why 1.32m/s doesn't match the value we solved for previously. We can still make some observations from this graph, however. The most important things we can see is that the ping pong ball does have acceleration and that this acceleration is not constant. So now that we have the background on what all this means, I’ll use numbers for the following graphs. Ball After Hit Position Graph (x-axis) The above graph is a position graph (with respect to the x-axis) of the ping pong ball after it hits the ball. Autofit Slope = 6.011 m/s Manual Slope = [((0.556m) – (-0.666m)) / ((0.852s) - (0.648s))] = 5.99 m/s 6.011 is close enough to 5.99 to be considered accurate. Displacement = (0.556m) - (-0.666m) = 1.22m Ball After Hit Velocity Graph (x-axis) The above graph is a velocity graph (with respect to the x-axis) of the ping pong ball after it hits the ball. Area = 1.08m (1.22m is close enough to 1.08m to be considered accurate) Autofit Slope = -7.941 m/s^2 Manual Slope = [((5.21m/s) – (6.533m/s)) / ((0.84s) – (0.66s))] = -7.35m/s^2 -7.941m/s^2 is close enough to -7.35m/s^2 to be considered accurate. Ball After Hit Acceleration Graph (x-axis) The above graph is an acceleration graph (with respect to the x-axis) of the ping pong ball after it hits the ball. Area = -1.23m/s (due to camera shake, not accurate) Key points: it has acceleration that is not constant Paddle Before Hit Position Graph (x-axis) The above graph is a position graph (with respect to the x-axis) of the paddle before it hits the ball. Autofit Slope = 2.872 m/s Manual Slope = [((-0.764m) – (-1.006m)) / ((0.636s) – (0.552s))] = 2.88 m/s 2.872m/s is close enough to 2.88m/s to be considered accurate Displacement = (-0.764m) – (-0.885m) = .121m Paddle Before Hit Velocity Graph (x-axis) The above graph is a velocity graph (with respect to the x-axis) of the paddle before it hits the ball. Area = 0.0739 m .121m is close enough to .0739m to be considered accurate. Autofit Slope = 46.13 m/s^2 Manual Slope = [((4.006m/s) – (0.905m/s)) / ((0.624s) – (0.552s))] = 43.1 m/s^2 46.13m/s^2 is close enough to 43.1m/s^2 to be considered accurate. Paddle Before Hit Acceleration Graph (x-axis) The above graph is an acceleration graph (with respect to the x-axis) of the paddle before it hits the ball. Area = 3.84m/s (due to shaking, not accurate)
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Steve Dickie,
Oct 27, 2009, 8:13 PM
ċ
Steve Dickie,
Oct 27, 2009, 8:12 PM
ċ
Steve Dickie,
Oct 27, 2009, 8:13 PM
ċ
Steve Dickie,
Oct 27, 2009, 8:12 PM
ċ
Steve Dickie,
Oct 27, 2009, 8:12 PM
ċ
Steve Dickie,
Oct 27, 2009, 8:12 PM
ċ
Steve Dickie,
Oct 27, 2009, 8:13 PM
ċ
Steve Dickie,
Oct 27, 2009, 8:13 PM
ċ
Steve Dickie,
Oct 27, 2009, 8:13 PM
ċ
Steve Dickie,
Oct 27, 2009, 8:11 PM