RE: MI Area calculations
The spherical algorithm is NOT the proper way. The UK National Grid is based on the Gauss-Kruger Transverse Mercator fully-conformal projection. Therefore, you: (1.) MUST use the cartesian algorithm, and (2.) MUST correct for the scale factor squared. (Length x scale factor x width x scale factor = area x scale factor squared.) If the referenced ArcView data gives the identical answer to what you get with step (1.) above, then the ArcView data is incorrect. You must correct for this non-trivial source of systematic error each time you measure an area, whether it by a fancy software package or with your own two hands on a paper map. (Computers just make it easier to generate garbage data in a more efficient manner.) Yes this should (and will) compare favorably with the methods used with paper maps. (Such as with a compensating polar planimeter.) Consider: (1.) Purchasing a copy of "Measurements from Maps," by D.H.Mailing, Pergamon Press, ISBN 0-08-030289-0, and (2.) Reading it. The folks at Ordnance Survey will be happy to help you also, they just won't be familiar with the esoterics of MI. (Nor am I.) Good luck! Clifford J. Mugnier ([EMAIL PROTECTED]) The Topographic Engineering Laboratory Department of Civil and Environmental Engineering UNIVERSITY OF NEW ORLEANS New Orleans, Louisiana 70148 Voice and Facsimilie: (504) 280-7095 On Thursday, 23 September, 1999 4:46 AM, Ian Macey [SMTP:[EMAIL PROTECTED]] wrote: > I have been using the spherical algorithm to calculate the > area of field > polygons since upgrading to MI Pro 5.5. Is this the > correct way for use with > Ordnance Survey data which is tied in to the British > National Grid? The > reason that I ask is because I have been sent an Arc View > shape file where > the person quoted the area as 546.01 ha. This is the area > that I get when > using the cartesian algorithm. When using the spherical > option I get an area > of 544.48 ha which is causing some contention. > > Also how should these compare with the traditional ways > that the OS use > calculate their areas on old paper plans. > > Any help is much appreciated > > IAN > --- > Ian Macey > Dreweatt Neate > > mailto:[EMAIL PROTECTED] > http://www.dreweatt-neate.co.uk > Telephone: 01635 263050 > Facsimile: 01635 263090 > > > ** > > EMAIL DISCLAIMER: > The information in this e-mail is confidential and may be > legally > privileged. It is intended solely for the addressee and > access to the e-mail > by anyone else is unauthorised. If you are not the > intended recipient, any > disclosure, copying, distribution or any action taken or > omitted to be taken > in reliance on it, is prohibited and may be unlawful. > > If you have received this email in error please notify: > [EMAIL PROTECTED] > ** > > > -- > > To unsubscribe from this list, send e-mail to > [EMAIL PROTECTED] and put > "unsubscribe MAPINFO-L" in the message body, or contact > [EMAIL PROTECTED] -- To unsubscribe from this list, send e-mail to [EMAIL PROTECTED] and put "unsubscribe MAPINFO-L" in the message body, or contact [EMAIL PROTECTED]
RE: MI Area calculations
Ian, I contacted our expert on geodesy, Dr Phil Davies, and he replies: "Yes, a 2D Cartesian method would usually be used when working on the National Grid. However, this will cause an error due to the projection scale factor distortion which can be corrected by using the local scale factor for the location (it must be squared to correct a measured area). An Excel spreadsheet projcal.xls which gives the local scale factor for any national grid easting and northing is available from our Web Site." If you need to you can email Phil directly on [EMAIL PROTECTED] Our web site address is http://www.ordsvy.gov.uk I hope this is of some help, regards, Pete, Ordnance Survey, UK > -Original Message- > From: Thake, Peter > Sent: 23 September 1999 13:52 > To: Davies, Phil > Subject: FW: MI Area calculations > > > Phil, > I'd say cartesian is best for use with Nat Grid? It is a > 'flat' projection, isn't it? -Original Message- From: Ian Macey [mailto:[EMAIL PROTECTED]] Sent: 23 September 1999 10:46 To: 'MI List' Subject: MI Area calculations I have been using the spherical algorithm to calculate the area of field polygons since upgrading to MI Pro 5.5. Is this the correct way for use with Ordnance Survey data which is tied in to the British National Grid? The reason that I ask is because I have been sent and Arc View shape file where the person quoted the area as 546.01 ha. This is the area that I get when using the cartesian algorithm. When using the spherical option I get an area of 544.48 ha which is causing some contention. Also how should these compare with the traditional ways that the OS use calculate their areas on old paper plans. Any help is much appreciated IAN --- Ian Macey Dreweatt Neate mailto:[EMAIL PROTECTED] http://www.dreweatt-neate.co.uk Telephone: 01635 263050 Facsimile: 01635 263090 ** EMAIL DISCLAIMER: The information in this e-mail is confidential and may be legally privileged. It is intended solely for the addressee and access to the e-mail by anyone else is unauthorised. If you are not the intended recipient, any disclosure, copying, distribution or any action taken or omitted to be taken in reliance on it, is prohibited and may be unlawful. If you have received this email in error please notify: [EMAIL PROTECTED] ** -- To unsubscribe from this list, send e-mail to [EMAIL PROTECTED] and put "unsubscribe MAPINFO-L" in the message body, or contact [EMAIL PROTECTED] *** * This email/attachment(s) has been virus checked at OS * *and is free of all known viruses. * * All non work related emails/attachments should be deleted *to comply with 'Internet code of practice' * as per office notice 524/98. * *** -- To unsubscribe from this list, send e-mail to [EMAIL PROTECTED] and put "unsubscribe MAPINFO-L" in the message body, or contact [EMAIL PROTECTED]
Re: MI Area calculations
Hi, I am using OS data to digitise agricultural field polygons and then obtain their areas in Mapinfo. I believe you should use Options/Preferences/Map Window and select the Cartesian Distance/Area option. Good luck Rod Ellis GIS Unit ADAS Aberystwyth Tel 01970 621438 Fax 01970 617798 Email [EMAIL PROTECTED] >>> Ian Macey <[EMAIL PROTECTED]> 09/23/99 10:46am >>> I have been using the spherical algorithm to calculate the area of field polygons since upgrading to MI Pro 5.5. Is this the correct way for use with Ordnance Survey data which is tied in to the British National Grid? The reason that I ask is because I have been sent and Arc View shape file where the person quoted the area as 546.01 ha. This is the area that I get when using the cartesian algorithm. When using the spherical option I get an area of 544.48 ha which is causing some contention. Also how should these compare with the traditional ways that the OS use calculate their areas on old paper plans. Any help is much appreciated IAN --- Ian Macey Dreweatt Neate mailto:[EMAIL PROTECTED] http://www.dreweatt-neate.co.uk Telephone: 01635 263050 Facsimile: 01635 263090 ** EMAIL DISCLAIMER: The information in this e-mail is confidential and may be legally privileged. It is intended solely for the addressee and access to the e-mail by anyone else is unauthorised. If you are not the intended recipient, any disclosure, copying, distribution or any action taken or omitted to be taken in reliance on it, is prohibited and may be unlawful. If you have received this email in error please notify: [EMAIL PROTECTED] ** -- To unsubscribe from this list, send e-mail to [EMAIL PROTECTED] and put "unsubscribe MAPINFO-L" in the message body, or contact [EMAIL PROTECTED] The information in this electronic mail message is confidential and intended only for use by the addressee. Any view expressed are those of the individual sender and not of ADAS unless otherwise stated. This message has been checked for known virus. ADAS Postmaster [EMAIL PROTECTED] -- To unsubscribe from this list, send e-mail to [EMAIL PROTECTED] and put "unsubscribe MAPINFO-L" in the message body, or contact [EMAIL PROTECTED]
MI Area calculations
I have been using the spherical algorithm to calculate the area of field polygons since upgrading to MI Pro 5.5. Is this the correct way for use with Ordnance Survey data which is tied in to the British National Grid? The reason that I ask is because I have been sent and Arc View shape file where the person quoted the area as 546.01 ha. This is the area that I get when using the cartesian algorithm. When using the spherical option I get an area of 544.48 ha which is causing some contention. Also how should these compare with the traditional ways that the OS use calculate their areas on old paper plans. Any help is much appreciated IAN --- Ian Macey Dreweatt Neate mailto:[EMAIL PROTECTED] http://www.dreweatt-neate.co.uk Telephone: 01635 263050 Facsimile: 01635 263090 ** EMAIL DISCLAIMER: The information in this e-mail is confidential and may be legally privileged. It is intended solely for the addressee and access to the e-mail by anyone else is unauthorised. If you are not the intended recipient, any disclosure, copying, distribution or any action taken or omitted to be taken in reliance on it, is prohibited and may be unlawful. If you have received this email in error please notify: [EMAIL PROTECTED] ** -- To unsubscribe from this list, send e-mail to [EMAIL PROTECTED] and put "unsubscribe MAPINFO-L" in the message body, or contact [EMAIL PROTECTED]
RE: MI area calculations broken
Thanks to all for the suggestions so far. Here's some more information from further experimenting. The problem seems to be restricted to objects that cross or touch the 180 degree line in any way. If you click on the triangles and drag them one pixel away from the 180degree line and the pole, the calculated areas change to be (presumably) correct. In the objects in the sample I sent out, the correct size is approx 2.3 million sq km. The size of 5600 sq km is incorrect. I have tried creating the objects using different "renditions" of those geographic points, i.e. using all positive longitudes, all negative, mixed positive and negative, and using lambert azimuthal coordinates (metres). None of these changes make any difference. For the work I'm doing, mapping fishing activity in the seas around Antarctica, this is quite a problem. Obviously the Antarctic continent unavoidably crosses the 180 meridian, and so do various other areas we are monitoring (fisheries, treaty areas, etc). I suspect this problem is related to (it may even be the same problem?) one I came across last year, where the Intersects operator fails in certain situations, I think objects crossing the 180 degree line was one condition. Hypothesis: even when working entirely in rectangular coordinates (metres) in polar projections, all of MI's geographic functionality (geographic operators e.g. "intersects", geographic functions e.g. Area(), geographic procedures e.g. "Overlay Nodes") is liable (likely?) to fail when working with objects that cross the 180 degree meridian. Any takers? Maybe this should be called "Mark's theory of circular failure", that sounds fun! ** Note to MapInfo corp ** : If you could document and publish any limitations (such as these) that we need to be aware of, it would be appreciated, saving us having to go through the painful trial-and-error process of testing pieces of functionality to see what works and what doesn't. Cheers Mark O -- To unsubscribe from this list, send e-mail to [EMAIL PROTECTED] and put "unsubscribe MAPINFO-L" in the message body, or contact [EMAIL PROTECTED]
RE: MI area calculations broken
Mark, I note that you have used the Lambert Azimuthal Equal-Area (South Pole) projection. If you calculate the area using the cartesian option you get an exact match on area (2,725,144.9 sq km) Given that your triangle is 3,300 km by 1,700 km and originating at the south pole and heading due north my guess is that the coordinate system is not appropriate for the work you are doing, rather than 'broken'. Perhaps someone with an understanding of the limitations of the different coordinate systems would be able to enlighten us. Cheers Martin === Martin Roundill GIS Manager Waitakere City Council Private Bag 93109 Henderson Waitakere City New Zealand Ph +64 9 836 8000 ext 8344 Fax +64 9 836 8001 email [EMAIL PROTECTED] === > -Original Message- > From: Ogilvie, Mark [SMTP:[EMAIL PROTECTED]] > Sent: Wednesday, 4 August 1999 12:25 > To: '[EMAIL PROTECTED]'; '[EMAIL PROTECTED]' > Subject: MI area calculations broken > > I have been making comparisons of the area of various objects, and have > found MapInfo's area calculating function to be seriously broken, at least > in some projections (Polar Azimuthal projections). > > Attached is the MB code and an MBX (both are tiny) which demonstrates the > problem. To summarise what the programme does: > 1/ it draws two triangles of equal size, > 2/ labels the triangles with their area, and > 3/ displays the area in a browser. > > First problem: the labels don't show on the map because of an apparent bug > in the label positioning function. You have to zoom out or pan up to see > where the labels have gone. > > Second problem is the areas: one area calculated as 2,375,776 sq km, the > other area as 5,688 sq km. Three orders of magnitude difference. > > I suspect the problem is related to projection. If anyone knows what is > happening here, could you please let me know? What projections are "safe" > and which are "broken"? > > Cheers > Mark O. > > > > <> <> > > > -- > To unsubscribe from this list, send e-mail to [EMAIL PROTECTED] and put > "unsubscribe MAPINFO-L" in the message body, or contact [EMAIL PROTECTED] << > File: area_bug.MB >> << File: area_bug.MBX >> -- To unsubscribe from this list, send e-mail to [EMAIL PROTECTED] and put "unsubscribe MAPINFO-L" in the message body, or contact [EMAIL PROTECTED]
Re: MI area calculations broken
Hello Mark, > I have been making comparisons of the area of various objects, and have > found MapInfo's area calculating function to be seriously broken, at least > in some projections (Polar Azimuthal projections). MapInfo Pro's area calculation is independent of the current map projection. It just assumes a spherical Earth, with the option of Cartesian in v5.5. You can see this by the fact that the MB area() function does not have any coordsys or map window ID clause. The problem you are having can be seen by simply reverting to a rectangular projection. One of your triangles crosses the 180 degree meridian. While this looks perfectly fine on the polar projection, it is in fact implemented by a massive triangle crossing the entire world and this is the reason for your outsized area. Try breaking your trans-meridian region into a two polygon region, broken at 180deg. You should find that the area are then equal, but they may still be incorrect. I rarely trust MI's distance and area calculations these days. One additional point. Having polygons with such long edge lengths doesn't really make sense since they are straight lines no matter what the projection is. It's better to decide on an acceptable resolution for your "lines" and insert nodes accordingly. You can then decide whether these lines are great circles, rhumb lines, etc. and be in better control of the areas definition. Regards, Warren Vick Europa Technologies Ltd, U.K. http://www.europa-tech.com -- To unsubscribe from this list, send e-mail to [EMAIL PROTECTED] and put "unsubscribe MAPINFO-L" in the message body, or contact [EMAIL PROTECTED]
MI area calculations broken
I have been making comparisons of the area of various objects, and have found MapInfo's area calculating function to be seriously broken, at least in some projections (Polar Azimuthal projections). Attached is the MB code and an MBX (both are tiny) which demonstrates the problem. To summarise what the programme does: 1/ it draws two triangles of equal size, 2/ labels the triangles with their area, and 3/ displays the area in a browser. First problem: the labels don't show on the map because of an apparent bug in the label positioning function. You have to zoom out or pan up to see where the labels have gone. Second problem is the areas: one area calculated as 2,375,776 sq km, the other area as 5,688 sq km. Three orders of magnitude difference. I suspect the problem is related to projection. If anyone knows what is happening here, could you please let me know? What projections are "safe" and which are "broken"? Cheers Mark O. <> <> begin 600 area_bug.MB M9&5C;&%R92!S=6(@;6%I;@T*)W!R;V=R86UM92!T;R!D96UO;G-TFEM=71H86P@97%U M86P@87)E82!3;W5T:"!0;VQA2X@3F]T92!T:&%T(&QA8F5L65R(#$-"@T*@#Q"H$`]3$``O\9!C<$_OH7`/P_!!\%9P3]]14``/]-!`4%3@3]\0)"`/_^ M^AD`_#X$_?$8"@#_`P3]\1AJ`/$*@0#U,0`"_RH$_?$8``#_,@3]\1B'`/_^ M^AH`_!T$#@`J!/WQ&(<`__[X87)E85]T97-T`&YA;64`)$`` M$$!/0&T``(!60+'-&%! M\#\(0```@&9```#`8D!.0&]B:@!A@`!@0`5 MAP`-`/__`$`!```$;6%I;@``N00``,T$```!0`$``,T$ M```!!W1E;7!O8FHV!0```0```"]&.EQ34D-<5DU37%!H87-E,5Q$979< M36%P26YF;UQM:7-C7&%R96%?8G5G+DU"6#!&.EQ34D-<5DU37%!H87-E,5Q$ =979<36%P26YF;UQM:7-C7&%R96%?8G5G+DU"```= ` end -- To unsubscribe from this list, send e-mail to [EMAIL PROTECTED] and put "unsubscribe MAPINFO-L" in the message body, or contact [EMAIL PROTECTED]
