Who Prefers UTM?

[+bigredmed]

OK for a UTM novice, Omaha is on the western border of grid 16 and the eastern border of grid 17 (96th meridian runs right through east central Omaha). So if I were to be at the Missouri River which is east of the border and I wanted to calculate how many meters I would have to go west to get home, I couldn't just measure the meters on the way to the river and then add these to the easting that I had on my GPS?

[+Touchstone]

Hah! I cheated. I used my brainy wife to figure it out. Here's her reply:

 

That's not so hard. Go due north - you will stay on the same longitude -

E737152. Your latitude will change, but since we're headed north, we don't

have to worry about convergence of the meridians. So, take the radius of

the earth, multiply by 2 pi to get circumference, divide by 360 to get the

number of km per degree latitude, then divide by 15. Actually, this is

easier.

 

(360/2*pi*radius of earth in km)*15 km = fraction of degree to go north.

Should get an answer of 0.8477 degrees (I used 6.37 x10exp6 meters as

radius).

 

UTM is nice, because I think you don't have to convert those silly degrees

and seconds to fractions.

 

Answer should be close to 50.3472, but you will want to get a more accurate

estimate of the earth's radius first.

 

Well, that's a start anyway. How's that Fizzy? Who needs Geocalc anyway

Edited December 31, 2003 by Touchstone

[+fizzymagic]

Once again with a quick mental calc, I get 16T 737152 N 4964950. I just added 15000 meters to the Northing.

Hey, MarinerBC!

 

It turns out that just adding 15000 to the Northing gives a very wrong answer. In fact, your answer is off from the correct one by more than 500 meters!

[+fizzymagic]

Hah! I cheated. I used my brainy wife to figure it out.  Here's her reply:

 

[cut to save space]

 

Well, that's a start anyway.  How's that Fizzy?  Who needs Geocalc anyway

You're on the right track, Touchstone.

 

I posed this problem because many people on this thread were talking about how much easier UTM is to use than Lat/Long. In fact, that is generally not true. UTM is great for estimating distances, but it is lousy for doing projections, because the angles are wrong.

 

This problem is very easy to solve using Lat/Long. Like you said, you stay on the same longitude. For latitude, one minute equals one nautical mile. And a nautical mile is 1.852 kilometers. So 15 km / 1.852 = 8.0994 nm. So just add that to the minutes and you get the answer.

 

What I am anxious to see is how to solve the problem using UTM, which is supposed to be so easy.

[+fizzymagic]

Hmm...

 

It's starting to look as if the prize will go unclaimed. Oh, well, I can keep the problem around for the next "UTM is better" thread. :-)

[+Touchstone]

Hey, what’s the big rush! OK, SH (Smarter Half) and I have combined our collective brain power and come up with an outstanding 1.5 brain cells for this one. Just let me get this out of my system, then I’ll slink back to my hole and lick my wounds in peace.

 

First off, we can see the point that Fizzy is trying to make here. What we have, basically, is an ugly flat planar projection (i.e. UTM), over a beautiful curved surface (i.e. the Earth). Well, what do you expect, UTM was invented by the military! The military likes quick and dirty solutions, and UTM gave a very quick method of locating your position within one of those UTM grids. Now here’s the kicker: each grid square is 100 KM on a side, centered about a Meridian. Heading north when you’re near that central Meridian is fine. The solution suggested by a few of us would work out perfectly (i.e. just head north 15 KM on the same longitude). It’s at the edges of the grid square that you start to run into difficulties due to the distortion that the planar projection makes on that beautiful curved surface. You can even see this fact alluded to at the bottom of each USGS topo map where they give you the declination (for lack of a better word) between the “grid north” and “true north”. Which leads me to my second point.

 

We had to make an assumption. I’m assuming that Fizzy posed the problem in order to accentuate this problem of distortion at the edges of the UTM grid. Therefore, I assume that the coordinate presented in the problem is on the edge of T16 where the distortion is the greatest. Alas, I don’t have any maps of MI, so I’m going to have to guess at the difference between grid north and true north. I know that the difference is 0 at the equator, and that at 80 degrees (where UTM basically ends and polar stereographic begins) it is 3 degrees. Fizzy’s coordinate is pretty close to the 45th parallel, so we took 2/3rds of the value of the greatest error, or 2 degrees.

 

Next, we had to visualize the problem. Heading north from Fizzy’s coordinate will not take you in a straight line. Rather, you’ll be making a curve (albeit, a rather gentle curve). So you’ll be heading north some distance, but also slightly west as well. If you can imagine, the distance of the northerly and the westerly directions make up two sides of a triangle, however, the hypotenuse will be a curve, rather than a straight line. For all intents and purposes, the hypotenuse is pretty much a straight line for this small a distance. Now, this is where it gets a bit messy, and where SH and I had to go into the dark recesses of our brains which we would rather have left dark. Using the Law of Sines we come up with the following:

 

Sin A/A = Sin B/B = Sin C/C

 

Or:

 

Sin 90/ 15000 = Sin 2/ X = Sin (90-2)/Y

 

Solving for X we get: 523.49

 

Solving for Y we get: 14990.86

 

As a sanity check: X(2) + Y(2) = 15000(2)

 

Therefore (drumroll please), we came up with:

 

16T E 736628.5 N 4964940.9

 

Now, as someone stated on CJ’s Chatroom, I think it’s time to find another dead horse to beat to death

[+fizzymagic]

Therefore (drumroll please), we came up with:

 

16T E 736628.5 N 4964940.9

 

Pretty close. Off by only 27 m. Last night, Marty Fouts discovered that you can get the Grid North declination off of TopoZone. For this quadrangle, it's 2.105 degrees. That was key. Using that, you can get the coords pretty well:

 

Delta-Easting = 15000 * sin(-2.105) = -551 meters

Delta-Northing = 15000 * cos(-2.105) = 14991 meters

 

So the new coords would be:

 

16T E 736601 N 4964941

 

The correct answer, as I mentioned above, would be to add 8.099 to the minutes of the latitude:

 

N 44 39.824 + 8.099 = N 44 47.923

W 84 00.510 + 0 = W 80 00.510

 

The UTM of that spot is 16T E 736601 N 4964945

 

So the answer I got was about 4 meters off.

 

Thus, my question of how to do this simple projection using UTM has been answered. I still don't understand why it's easier than using Lat/Long, but I suspect that none of the UTM advocates are going to tell me. Touchstone was close enough to win the prize! Email me your snailmail address and I'll send it along...

[+Touchstone]

WaHoo! Thanks Fizzy.

 

There's got to be a cache concept in there somewhere. Oh wait, I'm so late to this game it's probably already been done

 

Sure learned quite a bit from this problem Can you tell I was googling alot!

[+Klemmer]

WOW! I knew I liked Lat/Long but wasn't sure why. Now I'm sure!

I'm just an angle kind of guy, I guess.