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[Brian] I am visualizing taking an average of the differencesMichael D. Edmiston, Ph.D.
given by posn 1 - posn 0, posn2 - posn1
and so on ....
[Michael] Averaging the differences can't provide a smaller error because it provides exactly the same result.
Explicitly write down the algebra for the average:
[(P1-P0) + (P2-P1) + (P3-P2) + (P4-P3) + (P5-P4)] / 5
Notice that all the positions between P0 and P5 appear in the sum twice; once as positive and once as negative.....You wasted your time obtaining data points P1 through P4, and you wasted your time doing the calculation using them.
The result rests on the first and last points, and the error is totally dependent on how well you established the first and last points.
Indeed, you can fabricate any outrageous data you want for the positions of node1 through node4 and you still get the correct result if the positions of node0 and node5 are correct...