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phy123summer:lab_1 [2010/06/04 08:29]
mdawber
phy123summer:lab_1 [2010/06/06 17:15] (current)
mdawber
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 {{lab1plotfig.png}} {{lab1plotfig.png}}
  
-The data is clearly quite linear when plotted like this, so it gives us an indication that our formula at the least has the right form. (Maybe you would like to try plotting $L$ directly against $T$ and see what looks like). Notice that you can't see the y error bars because they are very small. The program has fitted the data using a least-squares fitting approach. This means that it has calculated for each data point the square of the difference between the data point and the line. It then adds up all these "​squares"​ and uses this number to determine how good the fit is. The computer tries to find the line that gives the smallest sum of squares and calls this the line of best fit. It's drawn this on the graph and called it "​y=a*x"​. It's also given you the value of a and its estimate for the uncertainty in a. The value the program gives for the error in a is often fairly small, it relies mostly on the scatter in a and only uses the errors you enter to weight the points differently in its fit.+The data is clearly quite linear when plotted like this, so it gives us an indication that our formula at the least has the right form. Notice that you can't see the y error bars because they are very small. The program has fitted the data using a least-squares fitting approach. This means that it has calculated for each data point the square of the difference between the data point and the line. It then adds up all these "​squares"​ and uses this number to determine how good the fit is. The computer tries to find the line that gives the smallest sum of squares and calls this the line of best fit. It's drawn this on the graph and called it "​y=a*x"​. It's also given you the value of a and its estimate for the uncertainty in a. The value the program gives for the error in a is often fairly small, it relies mostly on the scatter in a and only uses the errors you enter to weight the points differently in its fit.
  
 Another technique you can use to estimate the error in the slope is to draw "​max-min lines"​. Here we draw in two lines, one that has the maximum slope that seems reasonable, the "​max"​ line, and another with the smallest slope that seems reasonable, the "​min"​ line. Normally we do these exercises on paper, but you can probably do it simply by holding a clear plastic ruler up to the screen to decide where you think the max-min lines should be (please DON'T draw on the screen!!). A line is reasonable if it just passes within //most// of the error bars. You then just take two convenient points on the line, and find the change in y over the change in x to calculate the slope. ​ You can then work out the slope of both lines to give yourself an estimate of the error in the slope. In the example below you can calculate that the "​max"​ line has a slope of about 90/3.6=25 cm/​s<​sup>​2</​sup>,​ and the "​min"​ line has slope of about 90/3.8=23.7 cm/​s<​sup>​2</​sup>,​ therefore if you used this method you would conclude that the value of the slope is 24.4+/-0.7 cm/​s<​sup>​2</​sup>,​ as compared to the computers estimate of 24.41+/​-0.16 cm/​s<​sup>​2</​sup>​. ​ Note that I drew the lines through (0,0) which we can consider as an error free point, i.e. the fits **must** go through this point. Another technique you can use to estimate the error in the slope is to draw "​max-min lines"​. Here we draw in two lines, one that has the maximum slope that seems reasonable, the "​max"​ line, and another with the smallest slope that seems reasonable, the "​min"​ line. Normally we do these exercises on paper, but you can probably do it simply by holding a clear plastic ruler up to the screen to decide where you think the max-min lines should be (please DON'T draw on the screen!!). A line is reasonable if it just passes within //most// of the error bars. You then just take two convenient points on the line, and find the change in y over the change in x to calculate the slope. ​ You can then work out the slope of both lines to give yourself an estimate of the error in the slope. In the example below you can calculate that the "​max"​ line has a slope of about 90/3.6=25 cm/​s<​sup>​2</​sup>,​ and the "​min"​ line has slope of about 90/3.8=23.7 cm/​s<​sup>​2</​sup>,​ therefore if you used this method you would conclude that the value of the slope is 24.4+/-0.7 cm/​s<​sup>​2</​sup>,​ as compared to the computers estimate of 24.41+/​-0.16 cm/​s<​sup>​2</​sup>​. ​ Note that I drew the lines through (0,0) which we can consider as an error free point, i.e. the fits **must** go through this point.
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