+walkergeoff and wife Posted February 22, 2005 Share Posted February 22, 2005 Hi All When I try to log a trig point on the Trigpointing web site http://www.trigpointinguk.com/ It always seems to give a huge error (10s of metres) compared with the stated location of the trig point. But my GPS will get me within 1 metre of caches, so it can't be my GPS that is out. Could it be that the trig points are not where the maps and/or the OS think they are? Regards Geoff Quote Link to comment
+The Forester Posted February 22, 2005 Share Posted February 22, 2005 The problem you describe looks exactly like what would happen if you are using the wrong geodetic datum and/or spheroid. The trig pillar grid co-ordinates are on the Ordnance Survey Great Britain 1936 datum on the Airey spheroid. Geocaching's standard datum/spheroid is WGS84. The difference is about a hundred metres. You can get an extremely accurate conversion between the two systems at the Ordnance Survey webpage. You will need to register, but that's easy and free. Cheers, The Forester Quote Link to comment
interpleb Posted February 22, 2005 Share Posted February 22, 2005 (edited) Newbie question here - I just checked out my local trig point and took a fix using GPS Tuner on my Pocket PC. How do I find how far that is from the specified co-ordinates? The trig point is quoted as being at: 51.459751 N, 0.023914 W My GPS reported 51.45979641 N, 0.02386053 W Thanks. Edited February 22, 2005 by interpleb Quote Link to comment
+The Forester Posted February 22, 2005 Share Posted February 22, 2005 The distance between the co-ords which you mention is 6.23 metres. The distance from the listed co-ords of that trig and your as-found co-ords is 5.38 metres. To perform the distance calc for yourself, probably the easiest way would be for you to have your machine convert your co-ords to grid format (any grid, doesn't matter which one) and then for you to hand calculate using Pythagoras's Theorem. Subtract the Eastings of one point from the Eastings of the other, then square the difference. Subtract the Northings of one point from the Northings of the other, then square the difference. Add those two numbers together and then take the square root of the result. That gives you the distance in metres, to a very high level of precision. Cheers, The Forester Quote Link to comment
interpleb Posted February 22, 2005 Share Posted February 22, 2005 The distance between the co-ords which you mention is 6.23 metres. The distance from the listed co-ords of that trig and your as-found co-ords is 5.38 metres. To perform the distance calc for yourself, probably the easiest way would be for you to have your machine convert your co-ords to grid format (any grid, doesn't matter which one) and then for you to hand calculate using Pythagoras's Theorem. Subtract the Eastings of one point from the Eastings of the other, then square the difference. Subtract the Northings of one point from the Northings of the other, then square the difference. Add those two numbers together and then take the square root of the result. That gives you the distance in metres, to a very high level of precision. Thanks for that reply, although I have to admit that everything except Pythagoras went completely over my head - I think I will have to go and do some research (I was kind of hoping there was a handy website with such a calculator on it!). I do have one bit that really puzzled me though - why do you say the difference is 6.23M and then 5.38M, I don't see the distinction between the two sentences. Either way ~6M out isn't very good as it was an average over 125 samples. Quote Link to comment
+wildlifewriter Posted February 22, 2005 Share Posted February 22, 2005 I do have one bit that really puzzled me though - why do you say the difference is 6.23M and then 5.38M, I don't see the distinction between the two sentences. The Forester may have been pointing out - subtly - that the co-ordinates which you quoted as "quoted", are not the ones actually listed. Trig Points are listed by OS grid reference. The one you attended is listed at TQ 37381 75271... ... which converts to N 51.459751, W 00.023887 (WGS84) So: there was a difference between your reading and where you were. There was also a difference between your reading and where you thought you were. Hope that makes sense. Quote Link to comment
interpleb Posted February 22, 2005 Share Posted February 22, 2005 The Forester may have been pointing out - subtly - that the co-ordinates which you quoted as "quoted", are not the ones actually listed. Aha, that was a bit subtle for this newbie, thanks for the explanation :-) Quote Link to comment
+The Forester Posted February 22, 2005 Share Posted February 22, 2005 Either way ~6M out isn't very good as it was an average over 125 samples. That depends whether your system was using WAAS in the fixes. Without WAAS, ±5m is about right for only 125 fixes. Modern GPSrs have a fix update rate of about 1Hz, so 125 fixes is only a couple of minutes, I was kind of hoping there was a handy website with such a calculator on it! A flying chum of mine, Ed Williams, has put together a lovely compendium of lots of calcs and has also put most of them onto a simple to use spreadsheet. Cheers, The Forester Quote Link to comment
interpleb Posted February 22, 2005 Share Posted February 22, 2005 A flying chum of mine, Ed Williams, has put together a lovely compendium of lots of calcs and has also put most of them onto a simple to use spreadsheet. Excellent web page, thank you very much The Forester :-) Quote Link to comment
barryhunter Posted February 22, 2005 Share Posted February 22, 2005 (I was kind of hoping there was a handy website with such a calculator on it!). The grid conversion page on streetmap can tell you the direct line distance between two coordinates: http://www.streetmap.co.uk/streetmap.dll?GridConvert Just enter the first coordinate* then the second and it will show the distance between them. *Works with Lat/Long, Grid References or even postcodes/placenames Hope that helps, Barry Hunter Quote Link to comment
interpleb Posted February 22, 2005 Share Posted February 22, 2005 http://www.streetmap.co.uk/streetmap.dll?GridConvert Just enter the first coordinate* then the second and it will show the distance between them. Cool, thanks Barry :-) Quote Link to comment
+John Stead Posted February 23, 2005 Share Posted February 23, 2005 That is really useful - but can anyone tell me what is the "M Grid" referred to on the StreetMap page? Quote Link to comment
+wildlifewriter Posted February 23, 2005 Share Posted February 23, 2005 That is really useful - but can anyone tell me what is the "M Grid" referred to on the StreetMap page? It's a single proprietary grid system which references all Bartholomew map products. I don't think you can search by this code - even though it says you can. Quote Link to comment
+John Stead Posted February 23, 2005 Share Posted February 23, 2005 That is really useful - but can anyone tell me what is the "M Grid" referred to on the StreetMap page? It's a single proprietary grid system which references all Bartholomew map products. I don't think you can search by this code - even though it says you can. Many thanks - had me puzzled! Quote Link to comment
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