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Re: Taking four stars for checking accuracy of fix - and "Cocked Hats"
From: Paul Hirose
Date: 2008 Aug 03, 22:14 -0700
From: Paul Hirose
Date: 2008 Aug 03, 22:14 -0700
George Huxtable wrote: > 34º 16.7', 34º 21.0, 34º 25.4, 34º 32.0, 34º 38.6, 34º 46.3, 34º 49.9, 34º > 58.3, 34º 58.8. I did a least squares solution for altitude at T0 and rate of change with my HP 49G. Like many calculators nowadays, it can solve systems of equations. Here, we have are nine equations of y = xm + b, where y is the observed altitude, x the time offset of the observation from the nominal time T0, m the unknown rate of altitude change, and b the unknown "best fit" altitude at T0. For example, for the first two observations, the equations are: 34º16.7' = -4 * m + b 34º21.0' = -3 * m + b After setting up all nine equations and solving by the method of least squares, the first data set gives: 5.727'/minute, 34°38.56' at T0 (38.56) > 76º 29.7, 76º 39.2, 76º 42.8, 76º 48.2, 76º 57.8, 77º 03.1, 77º 05.9, 77º > 12.8, 77º 19.5. 6.018'/minute, 76°55.44' at T0 (55.44) > 42º 47.3, 42º 53.3, 42º 54.9, 43º 05.8, 43º 12.4, 43º 16.9, 43º 24.4, 43º > 29.2, 43º 37.9. 6.337'/minute, 43°11.34' at T0 (11.34) > 11º 18.0 , 11º 25.9, 11º 30.8, 11º 39.8, 11º 42.7, 11º 44.1, 11º 51.3, 11º > 58.6, 12º 05.3. 5.543'/minute, 11°41.83' at T0 (41.83) > Yes, it's better to calculate the slope thanc trying to determine the slope > of the fitted straight line from the data points themselves; no question > about that. Are you sure? My results fail to confirm that. The procedure above is equivalent to plotting the points, then drawing the straight line that best fits the points. Both the slope and the vertical position of the line are adjusted to achieve that fit. What you seem to be saying is that if the slope is constrained to the known correct value of 6'/minute, and I'm only free to move the best fit line up or down, the result is more accurate. So I tried it. Substituting 6'/minute for m, the first two equations in the first data set become: 34º16.7' = -4 * 6.0 + b 34º21.0' = -3 * 6.0 + b Continuing in that manner, we get nine such equations for each data set. Each one gives a value for b. Finding the "best fit" simply requires taking a mean. What I get, for all four data sets, are altitudes identical to the previous ones! (By "identical" I mean they agree to the .01'; I didn't write down the values any more precisely.) At least in this case, it seems that knowing the correct slope of the line is no benefit. > And it's better than simply averaging those changing altitudes > over the time period, if the observations were unequally-spaced in time. (If > they were equally spaced, you would arrive at exactly the same answer). These observations are equally spaced, and I verified that a simple mean gives the same result as a least squares fit. -- I block messages that contain attachments or HTML. --~--~---------~--~----~------------~-------~--~----~ Navigation List archive: www.fer3.com/arc To post, email NavList@fer3.com To , email NavList-@fer3.com -~----------~----~----~----~------~----~------~--~---