antipodal points

[RobertLG]

Anyone know how to accurately calculate antipodal coordinates? Antipodals are the opposite coordinates, either in the same northern or southern hemisphere, or the exact opposite side of the earth.

 

I have discovered that to determine the antipodal degrees of longitude you simply subtract your degrees of longitude from 180.

 

However, I am having trouble getting my light-weight mind around part of that process, and especially how to determine the minutes and seconds of longitude.

 

For example:

The longitude I'm working on is 122º 56' 44"

 

If I subtract the 122º from 180º I get 58º

 

Now, to get minutes and seconds, am I supposed to

subtract 56' 44" from 60' 60"? If so, then it appears that I'm supposed to end up with 04'16" of longitude.

 

If I have this right, then the antipodal of

 

122º 56' 44"

would be

58º 04' 16"

 

HOWEVER, I would think that since my first figure, 122º is a full 122 degrees, and Then there are very nearly another sixty minutes of the next degree, wouldn't the exact opposite be something closer to 57º 04' 16"? Hmm. I'm losing it now. Help?

 

 

Does anyone know how to make a simple spreadsheet for this?

 

Thanks for any help.

 

RobertLG

[+Jamie Z]

Easiest method, in my opinion? Change to decimal degrees.

 

Your example: 122º 56' 44" converts to 122.9456°.

 

180 - 122.9456 = 57.0544

 

Convert back to DDDMMSS = 57° 03' 16"

 

Of course, there are methods to subtract directly, but this seems to be the least confusing to me.

 

Jamie

[GLSailor]

I think you have to "borrow" a degree from 180 to get 60'60". If you do that, you get your 57*04'16"

179*60'60"

-122*56'44"

 

57*04'16"

 

QED

 

GLSailor

[Eeyore and Shadow]

If you wanted to stay in the same hemisphere (subtracting from 180°) your equasion would be 180°-122º 56' 44" which equals 57°03'16" But a true antipode would be a complete swap of latitude and 360° minus given longitude.

 

Eeyore

 

My other cachemobile is a broom!

[+Centaur]

If you want to subtract using Deg Min Sec, you have to remeber to subtract 1 from the preceeding unit, just as in base 10 math.

 

180D = 179D 60' 00" = 179D 59' 60"

 

179D 59' 60"

122º 56' 44"

------------

057D 03' 16"

 

[+brdad]

I'd just convert the coords to decimal first!

 

With friends like you, who needs enemas?

[+Planet]

And I thought Anti-Podal was for people who were against podiatrists.

 

Cache you later,

Planet

[RobertLG]

Thank you very much. You have all been very helpful. I appreciate the simplicity of making the coordinates a decimal first. And I also needed the direction of borrowing - that was where I was having troubles.

 

How many "antipodes" are there? I think I can count at least three

(using my coordinates of N46º 28' by W122º 56'):

 

Same northern hemisphere - same latitude/opposite longitude.

N46º 28' by E57º 03')

 

Same western hemisphere - same longitude/opposite latitude.

S46º 28' by W122º 56'

 

"True antipode" exact opposites - opposite latitude and longitude.

S46º 28' by E57º 03'

 

I'm sure I don't have the above antipodes quite right, but I think they are close. What do you think?

 

RobertLG

[Cholo]

It appears that you're trying to figure out the opposite spot on the Earth from your location. Years ago it was said one would end up in China. I calculated that from my location it would either be in Mongolia or the Indian Ocean, depending on what angle I dug at. If you're in southwest Washington, I truly believe your opposite spot would be somewhere in south Los Angeles.

[+erik88l-r]

I posted an antipodal cache, with nice easy coordinates to simplify the math:

 

http://www.geocaching.com/seek/cache_details.aspx?ID=14485

 

I found out the hard way that travel bugs should be excluded though!

 

~erik~

[dave and jaime]

in a mathematical sense, assuming the earth represents a sphere each point on earth has exactly one antipodal point by defination. antipodal points are points on a line passing throught the geometrical center of earth and intersecting the surface at right angles on opposite sides of the globe. see http://astronomy.swin.edu.au/~pbourke/geometry/sphere/ for more details.

 

for navigation purposes to calculate these points you must add or subtract 180d from both lat and long. the true north and true south poles would be examples of antipodal points

[Couch_Potato]

quote:

---

for navigation purposes to calculate these points you must add or subtract 180d from both lat and long.

---

 

Ummm, no.

As previously mentioned, subtract the longitude from 180d and switch hemispheres and only switch signs (or hemispheres) on the latitude. Subtracting 180 from the latitude would put you completely off the map.

 

I'm not lost!

I just don't know where I am.

[dave and jaime]

sorry, couch potato your right, sorta. latitude goes only to 90d and longitude to 180. therefore you need to subtract 180d from the longitude and from the latitude and then maintain the phase. for the point 67n,55w the antipodal point is 67s, 125e (ie 67d-180d=-113d, -113d=-67d; -55d-180d=-235d, -235d=+125d).

 

[This message was edited by dave and jaime on October 29, 2002 at 03:02 PM.]

[+Stunod]

Latitude only requires you to switch signs (N to S, S to N).

 

Think of it this way...there would be a line that runs thru the center of the earth. If you are 67 deg North of the equator, the line would end 67 deg south of the equator. North pole is 90 N, South Pole is 90 S.

 

Just swap the N/S. See this picture...

 

 

EDIT....You edited your post just as I corrected you, you sly dog you!!!

 

"Just because I don't care doesn't mean I don't understand."

 

[This message was edited by Stunod on October 29, 2002 at 03:02 PM.]

 

[This message was edited by Stunod on October 29, 2002 at 03:04 PM.]

[dave and jaime]

i havent done spherical geometry in several years and some of it eludes me now, as it did then. as i wrote both posts i didnt feel right as i was typing but did anyway. i like your graphic stunod. while it works, simply switching the signs is not the proper way to solve this problem and only works because latitude consists of180d {90d-(-90d)}.

[+Jamie Z]

A bit off-topic, but since the antipodal question has been answered, I thought I'd throw in this little story.

 

Take this little problem:

 

At noon, ship A is 150 km west of ship B. Ship A is sailing east at 35 km/h and ship B is sailing north at 25 km/h. How fast is the distance between the ships changing at 4:00pm.

 

It was an exercise in my calculus homework that the professor worked in class. After he arrived at the answer, I was very tempted to submit that it wasn't quite right due to the fact that the ships were on a sphere-like object, namely the earth, and great-circle calculations should have been used. I kept my mouth shut, though. I need a good grade in the class.

 

Jamie

[ArchieDH]

quote:

---

Originally posted by Jamie Z:

I kept my mouth shut, though. I need a good grade in the class.

 

Jamie

---

 

I'm soooo sorry Jamie! I just can't help myself...

 

BUT....

 

YOU kept your mouth SHUT????

 

YOU really must have wanted a GOOD grade REALLY, REALLY bad, huh?

 

I must pass this on the all your Mississippi buddies...They will be pretty impressed...

 

As side note: have you check out the cach count standing lately in Mississippi..

 

I'm thinking you better get in that ole truck and get back here and soooooo...

 

ArchieDH

[+Team Screamapillar]

"while it works, simply switching the signs is not the proper way to solve this problem and only works because latitude consists of180d {90d-(-90d)}."

 

If it "works" how is it "improper"? And what difference does it make if we measure a circle (a plane of the sphere) in 2 traditional 180 degree arcs or in any other numeric system that incorporates two symmetrical halves? The plane geometry involved will not change, whether we use 86 degrees of measurement or 1598234 degrees.

 

"Everybody wants to save the world, but nobody wants to help mom with the dishes," -P.J. O'Rourke

[+Jamie Z]

quote:

---

Originally posted by ArchieDH:

As side note: have you check out the cach count standing lately in Mississippi..

---

Yup. Last I checked I was still up by one. There's a new MS cache just down the road from me, too. No worries.

 

Jamie

[King Pellinore]

but how far away is the antipodal point? What's the maximum my GPS could ever read (on earth)? What would your gps read on the moon, or any other point outside the satellite sphere?

 

King Pellinore

[+Peanuthead]

To heck with this...

 

This is the information age..

 

Somebody find me a website that calculates the

 

antipode for you, and send me the link!

 

[+Peanuthead]

FOUND AN ANTIPODES MAP:

 

http://www.wendycarlos.com/maps/nadirs.jpg

 

Enjoy.