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I want to measure distance with GPS. However, the GPS is not very precise.

I will do my measurements for repetitions X number of foot lengths. I wonder if I can make better estimations of my position by using the knowledge of taking same lengths between the measurements.

Suggestion 1: $$(d_n - d_0)/n = d_{est}$$ Suggestion 2: $$((d_1 - d_0) + (d_2 - d_1) + .. + (d_n - d_{n-1}) )/n = d_{est}$$

Is this a good idea at all? If so, which one of these methods would be the best? Is there any other suggestions that could be better?

Best regards

tompak
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  • The way you've written it, the two suggestions are algebraically equal. – kimchi lover Jul 04 '17 at 13:35
  • Hehe yea I can see that now, thanks! Do you have any other suggestions or if these even is a valid idea? – tompak Jul 05 '17 at 20:56
  • In general taking averages of multiple measurements is better than using a single measurement, especially when you have some mechanism for discarding or discounting obviously bad measurements. But I confess, I did not really understand the details of what you propose to do. What, for example, is $d_0$ ? Don't GPS devices report Lat and Lon figures? – kimchi lover Jul 05 '17 at 22:52
  • I will measure signal strength for different lengths. I have test equipment for the signal strength measurements but I need to know at which distance I am taking each measurement. I will use my phones GPS which is not very precise (about +-10 meters deviation). Therefore my idea to use the knowledge of using same foot length when going to next measurement position. Example: First 10 signal strength measurements will be spaced with 5-foot lengths, next 10 signal strength measurements will be spaced with 10-foot lengths and so on. – tompak Jul 07 '17 at 07:47
  • $d_0$ would be the first GPS-distance measurement of same foot length and $d_n$ would be the last. For the example above first set of 10 measurements with 5-foot lengths $d_0$ = $d_0$ and $d_n$ = $d_9$, for the next set of 10 measurments with 10-foot lengths $d_0$ = $d_{10}$ and $d_n$ = $d_{19}$ – tompak Jul 07 '17 at 07:52
  • So are the $d_i$ numbers position estimates from the GPS, or differences between successive position estimates? In the first case I think you'd be better off using a formula like $\sum|d_i-d_{i+1}|/n$; in the second, like $\sum d_i/n$. Or maybe $d_i$ is the difference between the $i$th estimated position and the estimated position of a reference place? – kimchi lover Jul 07 '17 at 11:39
  • $d_i$ is the distance from the transmitting unit (reference position) to the i:th measurement position. – tompak Jul 07 '17 at 14:25
  • 2 comments: 1)I think you should move your question to a different forum, https://gis.stackexchange.com/ perhaps, , as it doesn't have much to do with mathematics as practiced here. And 2), is the location of your reference position known exactly, or is it determined by the same GPS means the transmitting locations are known? That is, is part of the error in $d_i$ coming from the inaccuracies in locating the reference position in addition to the obvious errors in the transmitting locations? – kimchi lover Jul 07 '17 at 17:28

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