Strictly speaking this is not a GPS tutorial, but rather?ÿ a projection system tutorial.?ÿ
So how about a little quiz.
1.?ÿ Using NGS data sheets and a calculator or a blank computer spreadsheet, compute the grid azimuth from FY2147 to EZ0604 on the North Carolina SPCS.?ÿ
2. Using the same resources as in (1) and your computed grid azimuth, compute the geodetic azimuth.
3. Check your answer for the geodetic azimuth using NGS Inverse.
Sorry, no CEUs, but it was engaging, wasn't it?
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And for more fun, calc the azimuth in the opposite direction.?ÿ
So how about a little quiz.
1.?ÿ Using NGS data sheets and a calculator or a blank computer spreadsheet, compute the grid azimuth from FY2147 to EZ0604 on the North Carolina SPCS.?ÿ
2. Using the same resources as in (1) and your computed grid azimuth, compute the geodetic azimuth.
3. Check your answer for the geodetic azimuth using NGS Inverse.
Sorry, no CEUs, but it was engaging, wasn't it?
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And for more fun, calc the azimuth in the opposite direction.?ÿ
For some real fun, Ask a surveyor what their basis of bearings is....
And, thanks Mr. Math teacher.
I'm no Kent Mc.?ÿ
I can run out of stuff to teach. But most of what I've got is practical.
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And, Loyal, yes tis true.
I think an understanding of projections is fundamental to practical GPS?ÿabuse. Er.. Use.
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Both azimuths come from the same formula, given by James Stem on page 29 of NGS Manual 5. His version is: t = alpha - gamma + delta, where t is the grid azimuth, alpha is the geodetic azimuth, gamma is the convergence angle, and delta is the arc-to-chord correction. Arc-to-chord is diagrammed on page 19 of the same publication. As @Bill93 pointed out, it becomes significant as lines become longer and in the case of Lambert projections, as azimuths approach east-west.
Solving for alpha gives?ÿalpha = t + gamma - delta which is the form that solves the quiz.
The grid azimuth in the opposite direction is 180 degrees + the original grid azimuth and the geodetic azimuth is the original back azimuth from NGS Inverse. Those two numbers, along with a new delta calculated with the latitudes interchanged in the formula on Stem's page 30, plugged into the formula for?ÿalpha would do it.
You don't really have to understand it in order to calculate it, so, in a way, Stem helped to make us all button-pushers. But, his buttons required a bit more effort to push.
@Nate The Surveyor did a great job of pointing out how the differences in azimuths between two-dimensional coordinate systems and three-dimensional coordinate systems arise. Then, as he predicted and my students often pointed out, I went and complicated it.
I hope someone found it valuable.
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That pesky arc-to-chord correction appears to be about 7.3 arc seconds in this case (what I computed by comparing reverse bearings by SPC+theta versus NGS Inverse FAZ and BAZ, not by Stem's formula).?ÿ
It's not much of an issue on a 1-mile survey, but grows fast with large distances, and depends on the direction relative to the projection axis.
Bowring's formula works well for sub 150km measurements as well.
The arc-to-chord correction for the quiz is 3.7 seconds. Note that?ÿarc-to-chord = grid az + convergence - geodetic az by rearranging Stem's formula above. You only need the geodetic azimuth, the grid azimuth, and the convergence angle from the beginning point.
When you take the difference between the geodetic back azimuth and forward azimuth, and then subtract the differences between the grid azimuths adjusted for the convergence angle, you're left with the difference between the arc-to-chord adjustments at each end of the line. Since they are almost numerically equal but of opposite signs, the result is close to double the desired value.
@steven-metelsky what does Bowring's formula produce?
