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Calculating expected accuracy in GPS

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(@alinm)
Posts: 14
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Topic starter
 

Happy Sunday everyone! It is with shame in my heart that my first post on this forum is a request for help, but I have no idea how to approach the problem. I'm currently enrolled in Astronomic Surveying at Purdue Calumet and we are discussing the use GPS as a surveying tool. The problem is I have never used GPS in my life and the online lecture did not cover ANY problems in calculating accuracy. I don't expect anyone to solve this problem for me, but a few explanations and hints that would help me solve it, would be more than appreciated.
this is the problem:
A baseline between two points that are being located by GPS methods is 3.5 miles long.
Determine the difference (in both mm and ft) of the expected accuracies of the baseline using
(a) Static GPS with an accuracy of +/‰Û 5mm + 1 ppm

Thanks in advance, Alin.

 
Posted : April 10, 2016 9:19 am
(@scott-zelenak)
Posts: 601
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the +/- 5mm is a constant independent of the length, plus 1ppm times the length converted to kilometers.

 
Posted : April 10, 2016 9:32 am
(@alinm)
Posts: 14
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Topic starter
 

Scott Zelenak, post: 366432, member: 327 wrote: the +/- 5mm is a constant independent of the length, plus 1ppm times the length converted to kilometers.

Thank you!

 
Posted : April 10, 2016 9:43 am
(@mark-mayer)
Posts: 3370
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AlinM, post: 366428, member: 11332 wrote: A baseline between two points that are being located by GPS methods is 3.5 miles long.
Determine the difference (in both mm and ft) of the expected accuracies of the baseline using
(a) Static GPS with an accuracy of +/‰Û 5mm + 1 ppm

3.5 miles is 18480 feet, is 5632.7 meters, is 5,632,704 mm.

1ppm error in 5.6322704 million millimeters is 5.63 mm of error.

Add the 5mm base error to that to get a total error of +/-10.63mm in the measurement.

To be totally correct in practice you would want to also add in the centering errors. That might add a mm or three, depending on the nature of the instrument setups . But I don't think you are being asked for that today.

 
Posted : April 10, 2016 9:43 am
(@alinm)
Posts: 14
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Topic starter
 

Mark Mayer, post: 366439, member: 424 wrote: 3.5 miles is 18480 feet, is 5632.7 meters, is 5,632,704 mm.

1ppm error in 5.6322704 million millimeters is 5.63 mm of error.

Add the 5mm base error to that to get a total error of +/-10.63mm in the measurement.

To be totally correct in practice you would want to also add in the centering errors. That might add a mm or two. But I don't think you are being asked for that today.

Thank you for your reply. I was doing it right up to 5.63 mm in error. 🙂

 
Posted : April 10, 2016 9:45 am

 ddsm
(@ddsm)
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Do you apply the 5mm base error to BOTH GPS stations or to only the one vector?

DDSMB-)

 
Posted : April 10, 2016 9:51 am
bill93
(@bill93)
Posts: 9867
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And do you add the errors directly or do you do sqrt sum of squares? That answer will depend on whether the errors are directly related or statistally independent. For gps there will probably be a different answer for simuultaneous vs sequentisl observations.

 
Posted : April 10, 2016 10:00 am
(@mark-mayer)
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Dan B. Robison, post: 366443, member: 34 wrote: Do you apply the 5mm base error to BOTH GPS stations or to only the one vector?

The one vector.

 
Posted : April 10, 2016 10:00 am