Comments on: Coordinate Subtraction
http://ask.metafilter.com/210056/Coordinate-Subtraction/
Comments on Ask MetaFilter post Coordinate SubtractionThu, 08 Mar 2012 10:18:35 -0800Thu, 08 Mar 2012 10:34:41 -0800en-ushttp://blogs.law.harvard.edu/tech/rss60Question: Coordinate Subtraction
http://ask.metafilter.com/210056/Coordinate-Subtraction
Which would be the correct way to subtract coordinates that are on different sides of the equator? <br /><br /> I was going off of <a href="http://www.firefightermath.org/index.php?option=com_content&view=article&id=61&Itemid=75">this</a> explanation for my longitude, but now I'm not so sure with my latitude. The first set is South and the intended final set (after the subtraction) is North. What I would like is to be able to give someone a set of Coordinates tell them to subtract this (with another set without North/South) and they would end up with the final set in North<br>
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Here's an example: The original coordinates can be S 90° 42' 30" and I want to tell them to subtract 120° 21' 70". Do I still need to borrow from the 42' so that 30" can become 90" (and the answer being 20")? Or can I just treat each one individually so that the the inches would end up being 40" by treating the Southern coordinates like a negative? (I'm only focusing on the Inches since I assume the technique would be applied to the rest)<br>
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I've tried looking on Google for the answer, but I haven't been able to find it.post:ask.metafilter.com,2012:site.210056Thu, 08 Mar 2012 10:18:35 -0800DeflagrocoordscoordinateslongitudelatitudemathresolvedBy: madcaptenor
http://ask.metafilter.com/210056/Coordinate-Subtraction#3029705
I'm having trouble figuring out what exactly you want to do here. What is the actual problem you're trying to solve?comment:ask.metafilter.com,2012:site.210056-3029705Thu, 08 Mar 2012 10:34:41 -0800madcaptenorBy: FlyingMonkey
http://ask.metafilter.com/210056/Coordinate-Subtraction#3029719
(a) latitude only goes up to 90°. The subunits of degrees, despite the symbol collision, are not feet and inches but rather minutes and seconds, which, as the names imply, only go up to 60 each.<br>
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(b) yes, you can consider the southern coordinates negative; treat them just like you would the western coordinates of a longitude.comment:ask.metafilter.com,2012:site.210056-3029719Thu, 08 Mar 2012 10:51:05 -0800FlyingMonkeyBy: Deflagro
http://ask.metafilter.com/210056/Coordinate-Subtraction#3029726
<a href="http://ask.metafilter.com/210056/Coordinate-Subtraction#3029719">FlyingMonkey</a>: "<i>treat them just like you would the western coordinates of a longitude.</i>"<br>
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So then in the example problem I gave I would need to borrow from the 42' when subtracting a larger Seconds from smaller Seconds?comment:ask.metafilter.com,2012:site.210056-3029726Thu, 08 Mar 2012 10:55:46 -0800DeflagroBy: vacapinta
http://ask.metafilter.com/210056/Coordinate-Subtraction#3029737
You'll be ok as long as you are consistent with whether South is negative and North is positive for example.<br>
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So, if South is negative you want to tell them to add 120° 21' 70" right? Not subtract. In that case, yes it is -30 + 70= 40 seconds (not inches) in your final answer. <br>
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Lets do the math by adding or subtracting each one individually:<br>
(120-90) (21-42) (70-30) <br>
or (30) (-21) (40) is your answer.<br>
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Now, the result is positive but the minutes are negative, so lets add 1 deg=60min to the -21 and thus:<br>
29:39:40 N is your answer <br>
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if you want south positive: 30 - 70 = -40. Or the whole thing: (90-120) (42-21) (30-70) = (-30) (21) (-40)<br>
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Now, you want the answer to be negative, consistent with the first number. So lets subtract 60 from 21 to make it -39. And we add 1 to -30 to compensate:<br>
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The answer is -(29:39:40) S or 29:39:40 N.comment:ask.metafilter.com,2012:site.210056-3029737Thu, 08 Mar 2012 11:04:17 -0800vacapintaBy: aubilenon
http://ask.metafilter.com/210056/Coordinate-Subtraction#3029917
So if you want to actually communicate clearly and effectively, I suggest you take a different approach. This is a weird thing to do and somewhat confusing, because it doesn't hold up well to dimensional analysis.<br>
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Your units in 37N,122W are not just "degrees", they're "degrees north of the equator" and "degrees west of the prime meridian". But if you subtract two amounts of "degrees north of the equator", you don't get a third "degrees north of the equator" you just get an interval with no reference point ("degrees difference"). <br>
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Time of day is an excellent analogy here. 8:15am is 1:45 earlier than 10:00am. It is <i>not</i>1:45am earlier. But like with lat-lon, you are using a similar notation. You certainly could <i>re</i>-interpret your hour and forty-five minute interval as 1:45am, but that's not correct or natural. <br>
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With time you can make an argument that if you go negative you're just wrapping around midnight. And this does work with longitude, but not with latitude (91°N is definitely not 89°S). If 91°N is <i>really</i> 89°N (going over the top), then you have a weird mess where you could justify saying 89°N - 89°N = 2°N. This problem stems from the fact that while your coordinates go from 90N to 90S, your intervals range from 180N to 180S.<br>
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All that said, if your goal is not clear communication but some kind of clever obfuscation (like for a puzzle or something), then this seems 'fair', provided you're consistent about N, and you limit your coordinates to avoid pathological cases that are more than 90° north or south of each other. You don't even have to define positive or negative as one or the other. You can just speak in terms of north / south.<br>
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50°N - 10°N = 40°N, clearly (if you go 10° less north than 50°N, you'll be at 40°). So 10°N - 50°N = 40°S (if you go 50° less north of 10°N, you'll wind up at 40°S). To math it up, you can treat either one as positive, and just convert afterwards.<br>
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It might be a little less confusing to add instead of subtract, and just switch the directions.<br>
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As for the mechanics of adding or subtracting degrees, minutes, seconds, do it the same as you would for hours, minutes, seconds. (e.g, 4° - 2°45'30" = 1°14'30")comment:ask.metafilter.com,2012:site.210056-3029917Thu, 08 Mar 2012 13:17:35 -0800aubilenon