Measurement Question...

[+Muggermots]

Yeah, this is probably one of the crazier questions. And if this is the wrong place, feel free to move it. I can't seem to find an answer by searching, but it's quite possible I just don't know what to look for.

 

I'm wondering, just how far is it, say from N 45° 41.395 to N 45° 41.396. Or even from N 45° 41.380 to N 45° 41.390.

 

Is it the same or different then it would be from W 121° 32.572 to W 121° 32.573 or from W 121° 32.560 to W 121° 32.570?

 

Are the distances standard, or do they vary from one area to another? If they vary, why, and how much do they vary?

 

And yes, I'm wanting to draw (small area) maps... mostly for kids to use.

 

Any help, or even any efforts to reduce my confusion, are quite welcome.

 

TIA

[+welch]

I think most of your questions are answered in this other thread.

the short: .001 is about five or six foot (five point two or something). go up a decimal, and you add a zero (.010 is 50-60ft, .100 is 500-600ft.)

yes its different for different locations. Latitudes are fairly stable, but longitudes get closer toward the poles.

Edited February 4, 2006 by welch

[+Cardinal Red]

North and South coordinates represent Latitude.

These distances are very nearly consistant anywhere on the earth.

An international nautical mile (int. naut. mile) is defined to be 1852 metres exactly (IHC, Monaco, 1929).

One Minute of Latitude is very nearly equivilent to 1852 metres or 6076.115486 feet.

That makes the distance of .001 Minutes = 6.076115486 feet.

So the distance from 45° 41.395 to N 45° 41.396 is 6.076115486 feet.

 

East and West coordinates represent Longitude.

These distances are NOT consistant and change as you move around the earth.

A rough estimate is the Cosine of your Latitude times 6.076115486 feet.

At the equator (0 Degrees Latitude: Cosine of Zero = 1) Thus 1 times 6.076115486 = 6.076115486 feet.

At 60 Degrees Latitude (Cosine of 60 = .5) Thus .5 times 6.076115486 = 3.038057743 feet.

At either of the Poles (90 Degrees Latitude: Cosine of 90 = Zero) thus Zero times 6.076115486 = Zero feet (touching)

 

So the distance between W 121° 32.572 and W 121° 32.573 is approx:

6.076115486 feet at the Equator.

3.038057743 feet at 60 Degrees North or South Latitude.

Zero feet (touching) at either the North or South Pole.

 

But for actual WGS84 Distances instead of rough estimates go to: The Earth according to WGS 84 (every degree of latitude)

 

Or follow the link from that URL: For those wanting more: A similar table as below for every ten minute of latitude yields finer resolution:

The Earth according to WGS 84 (every ten minute of latitude)

 

Hope some of this helps.

[+ghs]

Unless you need a really accurate distance, I'll describe a process to get the distance between any two points assuming a spherical Earth. (Unless the two points are extremely far apart, it will likely be close enough). First, convert your angles to decimal degrees or radians, whatever your calculator needs. Let the first point have latitude and longitude LAT1 and LONG1, respectively, and the second point be at LAT2 and LONG2. (North latitudes and east longitudes are positive. South latitudes and west longitudes are negative.) Calculate:

 

DOT = COS(LAT1) * COS(LAT2) * COS(LONG1 - LONG2) + SIN(LAT1) * SIN(LAT2)

 

Take the arccosine of DOT to get the angle subtended by the radius vectors to the two points at the center of the earth:

 

ANGLE = ARCCOS(DOT)

 

If your calculator doesn't give ANGLE in radians, convert it to radians (multiply, for example, degrees by PI/180).

 

Multiply ANGLE by 6378.137. That gives the great circle distance between the two points in kilometers.

Edited February 7, 2006 by ghs

[+ghs]

In my previous post, I neglected to mention that you'll get slightly better accuracy if you use "geocentric latitudes," not the geographic ones (e.g., according to the WGS-84 datum) used by GPS receivers and maps. If LATG is the WGS-84/geographic latitude, the geocentric latitude to be used with the formula I posted is

 

LAT = ARCTAN((6356.75231424/6378.137)^2 * TAN(LATG)),

 

(where "^" means exponentiation).

 

If you want to approximately take into account the ellipticity of the earth, first calculate an average of your two geocentric latitudes:

 

AVELAT = (LAT1 + LAT2) / 2.

 

Then, instead of multiplying ANGLE by 6378.137 km, multiply it by the earth radius at latitude AVELAT:

 

R = 1 / SQRT((COS(AVELAT) / A)^2 + (SIN(AVELAT) / ^2),

 

where A = 6378.137 km and B = 6356.75231424 km.

 

(Okay, I think I might have killed that horse.)

Edited February 8, 2006 by ghs

[+fizzymagic]

If you want to approximately take into account the ellipticity of the earth, first calculate an average of your two geocentric latitudes:

AVELAT = (LAT1 + LAT2) / 2.

Then, instead of multiplying ANGLE by 6378.137 km, multiply it by the earth radius at latitude AVELAT:

R = 1 / SQRT((COS(AVELAT) / A)^2 + (SIN(AVELAT) / ^2)

It's probably worth pointing out that this approximate correction only works when both points are in the same hemisphere (northern or southern).

 

Platform-independent C++ source code for all the calculations done by GeoCalc is available as Coord.zip from my website.

[+ghs]

It's probably worth pointing out that this approximate correction only works when both points are in the same hemisphere (northern or southern).

 

I'm not sure why that's true. It will make the approximate correction I intended even if the equator is between the two points. (If, for example, LAT1 = -LAT2, AVELAT = 0 and that will probably give the best radius to use for a spherical Earth calculation.) The real issue is that the two points not be "too far" apart. (But if there's code available at your site that avoids such approximating issues, well, hey, that's certainly a fine idea. )

Edited February 8, 2006 by ghs

[+ewerts]

Just to muddy the waters a little you can also convert you co-ordinates to UTM grid system. From what I've seen UTM isn't too useful for finding things on a globe, but great for up close work, and finding distances between co-ordinates. In your example N 45° 41.395, W 121° 32.572 would be 10T E 613458 N 5060629 in UTM. There is some good info about UTM at Map Tools.

 

Hope that helps.

[+ghs]

Hey, I like muddy waters...just as long as I'm not looking for a geocache in them.

[+fizzymagic]

It's probably worth pointing out that this approximate correction only works when both points are in the same hemisphere (northern or southern).

I'm not sure why that's true. It will make the approximate correction I intended even if the equator is between the two points. (If, for example, LAT1 = -LAT2, AVELAT = 0 and that will probably give the best radius to use for a spherical Earth calculation.)

No, if the equator is between the points the correction will not be even approximately correct, because the Earth's radius is largest at the equator.

 

Here's an analogy: if you are riding your bike from one town at 5000 feet over a 10,000-foot pass and back down to a town at 6000 feet, what is your average altitude? By the method you have given, it would be 5500 feet, while the real answer is probably close to 7700 feet or so.

 

Hope that helps.

[+ThePropers]

I always just use an online calculator such as http://boulter.com/gps/distance/

[+ghs]

It's probably worth pointing out that this approximate correction only works when both points are in the same hemisphere (northern or southern).

I'm not sure why that's true. It will make the approximate correction I intended even if the equator is between the two points. (If, for example, LAT1 = -LAT2, AVELAT = 0 and that will probably give the best radius to use for a spherical Earth calculation.)

No, if the equator is between the points the correction will not be even approximately correct, because the Earth's radius is largest at the equator.

 

Here's an analogy: if you are riding your bike from one town at 5000 feet over a 10,000-foot pass and back down to a town at 6000 feet, what is your average altitude? By the method you have given, it would be 5500 feet, while the real answer is probably close to 7700 feet or so.

 

Hope that helps.

 

(Sigh. ) I realize that the radius of the earth is largest at the equator. That's the point of my "approximate correction." My original formula involved the equatorial radius, regardless of what latitudes were involved. That in fact will generally be a good enough approximation--unless the two points are extremely far apart. Earth's radius is least on the polar axis, but it is only about 22 km less than the equatorial radius. My "approximate correction" simply replaces the equatorial radius with a radius closer to the one that actually exists in the proximity of the points of actual interest. (If the equatorial radius is what corresponds to a point in proximity to the points of interest, then *that* is likely the best radius to use in a spherical Earth approximation.) No single radius will give an exact answer. Earth just isn't spherical. It's just a question of whether or not *any* spherical Earth approximation is accurate enough for what you're doing and what approximate radius you want to use, given that you aren't making an ellipsoidal Earth calculation.

 

Also, nothing in any of my formulas has much to do with effects of different altitudes/elevations. I was just correcting for different latitudes. Elevation effects add a whole new dimension to the calculations. (That wasn't supposed to be a play on words, but oh well, it works.)

Edited February 8, 2006 by ghs

[Stache Rats]

I always just use an online calculator such as http://boulter.com/gps/distance/

 

Bless you. It's been to long since Daddy Stache Rat was in school. And yes, it can be hard to teach an old rat new tricks.

 

Edit Daddy Stache Rat's poor spelling.

Edited February 8, 2006 by Stache Rats

[+fizzymagic]

Also, nothing in any of my formulas has much to do with effects of different altitudes/elevations. I was just correcting for different latitudes.

Sigh. I give up.

 

Never mind.

 

I'll just finish by saying (in the nicest possible way) that if you are going to post equations to a forum, it helps if you understand them.

Edited February 9, 2006 by fizzymagic

[+ghs]

And I'll say in the nicest way possible that if you're going to argue with someone, you might at least first try to understand what he said. You clearly didn't. If you had, you might have had the opportunity to learn something as opposed to come away with an erroneous conclusion. Nothing I said was wrong. (The whole concept of "wrongness" doesn't apply here. I think you need to consider what the term approximation means. ) In my over 20 years of making calculations like this, and similar ones, I can't for the life of me figure out what altitude has to do with where the equator is in regard to anything I posted. (Your last post is why I debated whether or not it was a good idea to even try to answer the original question. Thanks for not surprising me.)

 

And I forgot to mention how active this forum seems. Wow.

Edited February 9, 2006 by ghs

[+fizzymagic]

And I'll say in the nicest way possible that if you're going to argue with someone, you might at least first try to understand what he said.

If you really believed that, you would have noticed when I wrote the word "analogy" in using my altitude example to illustrate your error. Apparently you have the completely incorrect idea that I was saying that the issue had something to do with elevation.

 

But I am done.

Edited February 9, 2006 by fizzymagic

[+ghs]

(Great. You edited your post so now I had to edit mine. I'm still going to respond to one thing you edited out. Sorry if that causes general confusion.) My original use of the equatorial radius wasn't a mistake. It was simply a first approximation--that works quite well for many applications. (I just tested it against MapSource's distance of 4897.2 km from Bogota to Colorado Springs. I got 4912.9 km. That's pretty good considering I would not actually recommend using "my" formula for such a large span and considering that I didn't even correct for the latitudes being geographic instead of geocentric.) (And yes, I saw the word "analogy." It didn't relate.)

 

Using the mean Earth radius is no more a valid general approximation for general points on the earth than the equatorial radius. (There is no absolute definition of "right" that says otherwise.) It will give better results than use of the equatorial radius when you aren't too close to the equator or poles, but may give worse results closer to the equator or poles. My follow-up post simply gave an expression for a better radius to use, taking into account the actual latitudes involved. There is nothing wrong with my formula for the radius of the WGS-84 ellipse. (It was along the same lines as your wanting to use the average Earth radius.) The average latitude formula was just one way of coming up with a single latitude to use to get a single radius value for a spherical Earth calculation. The equator passing between the two points has no bearing. If the two points are near the equator but on opposite sides of it, using an average radius close to the equatorial radius (or = equatorial radius if LAT1 = -LAT2) will give the generally most accurate distance you can get from a spherical Earth approximation.

 

If your point is that spherical Earth calculations aren't accurate enough, well, that's one thing. (But I think an absolute statement like that really depends on the application.) But saying that the equator being in between the two points invalidates use of that average radius, but it's yet somehow okay to use that average radius in the same hemisphere is just plain absurd.

 

(Now, if it's simply a matter that using the average Earth radius is some sort of "standard," well, fine, but standards don't define "right." They define conventions, and I don't have a problem abrogating those in the name of even sometimes superficial increases in accuracy.)

 

Lo and behold, maybe, I think I figured out what you're objecting to. The equatorial radius is larger than the radius to either of two points on opposite sides of the equator, and you think that that radius is therefore a wrong radius to use because it's larger? So what? (Would it be okay if it was smaller???) The average Earth radius you want to use is also larger than the radius to a great many points on the globe that may be of interest. It will be smaller than at other points. Whether it's too large or too small is irrelevant. (A distance that's slightly smaller than what actually exists isn't in any a priori sense any better of an approximation than a distance that's slightly larger.) As I've basically already said, these factors are just a consequence of approximating the earth as spherical for the sake of a simple calculation. If the difference between the equatorial and polar radii was significantly more than 22 km, there might be a big deal here, but then, we wouldn't be discussing any spherical Earth approximation in the first place. As it is, as long as any spherical Earth approximation is sufficient, that 22 km makes it pretty much irrelevant what radius you use. (Just work out some numbers.) I only posted that follow-up message basically because I didn't have anything else to do. (It sure is fun killing horses just before going to bed.)

Edited February 9, 2006 by ghs