finding latitude an longitude of a point

This calculation is a bit more involved, One has to iterate to get correct result.

I will find and post tomorrow.

What projection are you using?

http://jan.ucc.nau.edu/~cvm/latlongdist.html

http://www.movable-type.co.uk/scripts/latlong.html

http://malaysia.answers.yahoo.com/question/index?qid=20071111043539AAhVrTB

http://bbs.keyhole.com/ubb/showflat.php?Cat=&Board=SupportKML&Number=166379&Searchpage=1&Main=166379&Words=calculate+TJ1&topic=&Search=true

one of these sights might help

joshua

Do you need very accurate measurement such as a forward observer determining exact location of enemy target position for calling in an air strike or cruise missile attack on a target you can't get close to but you can use a laser range finder from a known lat/long position.

The US Army guys can tell you. Maybe get hold of one of their manuals. < grin>

You need more than just the distance. With only the distance you have created a circle and you could be any place on that circle.

Can you provie a bearing?

IT CAN'T BE DONE!!!

Lets take a map with location x on it. Take a compass (circle drawing tool), and set it to distance d. Draw a circle with x at the center. The solution for point y is any point on the circle you just drew. If you specify a longitude (or latitude) of point x, you will still get two possible solutions to the problem. Lets assume that point y is west of point x somewhere at the longitude of y. The two solutions are Northwest and Southwest of point x at the longitude of y.

In your last post, I used Pythagoras to solve it to avoid trigonometry (and keep it simple). If you have a bearing of y from x, you can then solve for lat and long using trig. Again, because we are using arc lengths on the same sphere, simple trig works.

Bill

Sciesis2 wrote: IT CAN'T BE DONE!!!

REPLY

Yes it can. revisit Survey 101 to get the solution. How do you think surveyors establish a base line with triangulation?

If you can move away from point x to get a second reading (and more information) certainly it can be done... but with the information given, I cannot see how it can be solved other than it is on a circle d distance from x. Could you explain?

Also... I did some work on his last posting last night using spherical trig, and I have to concede that using Pythagores is only an approximation. The longer the radius r of the sphere, the more accurate the approximation... and as r approaches infinity, the surface of the sphere becomes a plane and Pythagores becomes exact.

I am afraid that in college I was busy pushing electrons and didn't have time to take Survey 101.

Sincerely

Bill

The kid is right.

In survey, traversing from point 1 to point 2 is accomplished by measuring the distance and the azimuth angle from point 1 to point 2. The instrument is set up on point 1 while point 2 is cited. The result of this traverse is the hypotenuse of a triangle. The northerly and easterly can be solved for using trigonometry. Before measuring, zero azimuth is set to face north.

When the coordinates of two points are known, we "inverse" between them. Subtract the northerlies and easterlies (algebraically) paying attention to sign. Now solve for the hypotenuse and the bearing.

Example

Point 1 N5000, E5000

Point 2 - Traverse Point 1-2, found 60d-0m-0s, 100.00 feet.

northerly = 50.00 feet easterly = 86.667 feet.

Point 2 is at N5050.00 E5086.667

Had the problem been such:

Point 1 N5000, E5000

Point 2 N5050.00 E5086.667

We would "inverse" Point 1- Point 2 to get

N60d-0m-0s, 100.00 feet from Point 1 to Point 2.

As stated, coordinates of Point 1 were known. A distance from Point 1 to 2 was known, but no angle. The problem as stated does not have a unique solution.

In his homework, the Prof tried to confuse the kid with Lat/Lon. That is a side issue. You can convert from Lat/Lon to a grid very easily.

JB

ps: Never let a double e survey your lot or steer your boat.

The example I gave applies for cases where point 2 is within the horizon. For larger distances, please do not consider the earth to be flat and most importantly not round. Use your favorite ellipsoid or borrow one from your government. I prefer NAD83 in US.