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# errors in graphs

Started by Dave, Mar 20 2005 06:45 PM

8 replies to this topic

### #1

Posted 20 March 2005 - 06:45 PM

i have a graph of 1/V by 1/U. This should give a straight line where i can read off the intercept and get my focal length.

My problem is doing the errors fromt he graph can someone help me

My problem is doing the errors fromt he graph can someone help me

If i am not here i am somewhere else

### #2

Posted 21 March 2005 - 04:21 PM

Not entirely sure what your graph is about, but if it crosses through the origin the centroid method may be the way to go. There's information on that in the SQA uncertainties sheets.

### #3

Posted 21 March 2005 - 05:28 PM

i know you do a parallogram of errors and something about the gradient but thats it

If i am not here i am somewhere else

### #4

Posted 28 March 2005 - 08:14 PM

Ya the parallelogram thing is the centroid method lol. I can't remember exactly what to do with it either. I thiiiiiink its soemthing like:

1) Take an average of all the x-axis points you have and the y-axis points you have then plot that as a point. This is the centroid. Draw a straight line from the origin through this point (it should resemble something similar to a best fit line).

2) Look for the point which is furthest above your centroid line (ie most erroraneous) and draw a line parallel to the centroid line through that point.

3) Do the same again but for the point furthest below the centroid line.

4) Draw a straight vertical line through the largest and smallest x co-ordinate points so that you now have the parallelogram.

5) Find the gradients of the diagonals of the parallelogram (ie the bottom left of the parallelogram to the top right, and the top left to bottom right)

http://www.hitsomebody.com/centroid.gif

(The diagonals are the green line and the pink line)

6) Let these two diagonal gradients be m1 and m2

7) The uncertainty in the gradient is given by

Where n is the number of points on the graph (excluding the centroid).

The uncertainty in the itnercept is found by noting where the two 'worst lines' (diagonals) cut the y axis. let these be c1 and c2.

The uncertainty in the intercept is given by:

1) Take an average of all the x-axis points you have and the y-axis points you have then plot that as a point. This is the centroid. Draw a straight line from the origin through this point (it should resemble something similar to a best fit line).

2) Look for the point which is furthest above your centroid line (ie most erroraneous) and draw a line parallel to the centroid line through that point.

3) Do the same again but for the point furthest below the centroid line.

4) Draw a straight vertical line through the largest and smallest x co-ordinate points so that you now have the parallelogram.

5) Find the gradients of the diagonals of the parallelogram (ie the bottom left of the parallelogram to the top right, and the top left to bottom right)

http://www.hitsomebody.com/centroid.gif

(The diagonals are the green line and the pink line)

6) Let these two diagonal gradients be m1 and m2

7) The uncertainty in the gradient is given by

Where n is the number of points on the graph (excluding the centroid).

The uncertainty in the itnercept is found by noting where the two 'worst lines' (diagonals) cut the y axis. let these be c1 and c2.

The uncertainty in the intercept is given by:

### #5

Posted 29 March 2005 - 08:48 PM

That is right bluehead, i am just doing these on my graphs just now.

I plotted my graphs using Excel, just want to know do these parallelograms have to be done by hand? or can they be drawn using Excel?

I plotted my graphs using Excel, just want to know do these parallelograms have to be done by hand? or can they be drawn using Excel?

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### #6

Posted 29 March 2005 - 10:18 PM

i think its fine using a computer

the computer isn't doing the errors for you after all

all its doing is making sure you get a perfect presentation (3 marks are on presentation)

the computer isn't doing the errors for you after all

all its doing is making sure you get a perfect presentation (3 marks are on presentation)

If i am not here i am somewhere else

### #7

Posted 30 March 2005 - 07:45 AM

Ok cheers

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### #8

Posted 31 March 2005 - 08:53 PM

Can someone tell me How you work out the y-axes values of the points A, and C?

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### #9

Posted 01 April 2005 - 03:22 PM

Where the vertical lines drawn on the highest and lowest x-values cut the two "worst lines". You just read it off the graph.

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