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The Right Radius for the Projection

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davidgstoll
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Page 16 of Michael L. Dennis' excellent White Paper, GPS, Geodesy, and the Ghost in the Machine, gives several formulas for figuring a Radius of Curvature of the Ellipsoid:

I plugged those 4 formulas into Excel and did me some figurin'. The Prime Vertical Radius, at 36.25 degrees latitude, is 17+ miles longer than the Meridian Radius. Michael Dennis, along with every Beerlegger that chimed in on my [msg=279764]Calculating the Radius of Curvature[/msg] post, says to use the Geometric Mean. I thought I'd put it in graph form to help me understand.

My aim was to use the Radius of Curvature to figure Scale Factor on Low Distortion Projections, using Scale Factor = (R + P) / R. I've come to some conclusions. I think the Geometric Mean Radius will best be used for 45°, 135°, 225°, and 315° Oblique Mercator Projections, the Prime Vertical Radius best used for Lambert Conformal Conics, and the Meridian Radius best used for Transverse Mercator. N'est-ce pas?

Dave


 
Posted : November 20, 2014 11:01 am
Cliff Mugnier
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Your "Geometric Mean Radius of Curvature" is known as the Radius of the Gaussian Sphere. That is what is used to determine scale factors versus average ellipsoid height for the "Sea Level Correction." Originally thunk up by ole Carl Frederich Gauss in the early 1800's in Hannover, Germany. Also used as the mean radius of a province when using spherical approximations for cadastral map projections and occasionally for the size of the Earth for the Gauss Hannoverian Transverse Mercator and sometimes for the Gauss Schreiber Transverse Mercator (as implemented by the USC&GS for NAD27 Transverse Mercator projection tables & "blue books.")

Equation 1.5 is also referred to as the "ellipsoid normal terminated by the semi-minor axis." It's the workhorse for geocentric datum shift equations, ECEF transformations, etc.


 
Posted : November 20, 2014 2:20 pm
Martin F
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Thanks for posting those equations and for making and posting the lovely graph!

I'm not sure if you're right or not about which type of radius of curvature is a good average for a particular projection, but it certainly seems intuitively that you could be right.

However, if you have the ability to calculate the actual radius of curvature in any given direction -- clearly you did that inside the spreadsheet -- then why not go with that? On the other hand, maybe the differences between the two extremes (NS radius vs EW radius) is too small to make a significant difference.

I do think it sheds a little light on one issue that arose in the debate [msg=283201]Geodesy article - How many errors can you find[/msg]


 
Posted : November 20, 2014 2:47 pm
davidgstoll
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Martin,

I've never been able to figure out the how or why of the different Radius of Curvature equations. I have absolutely no clue why a particular azimuth produces a unique Radius. It just doesn't make sense to me. Putting those formulas in graph form helped me visualize the differences between the different Radius equations, but I think most people would just accuse me of big-dealin' it half to death. And they're probably right.

Dave


 
Posted : November 20, 2014 4:19 pm
davidgstoll
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Cliff,

"...ellipsoid normal terminated by the semi-minor axis."

I thought that was the point of ALL of those equations. And that is exactly why I don't understand the need for all those formulas. I'm just about ready to throw up my hands and use "any old number" to figure my scale factor.

Dave


 
Posted : November 20, 2014 4:29 pm

Cliff Mugnier
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Try looking up the "Principal Problem of Geodesy." It has a little bit to do with elliptic integrals of the second kind. Clairaut's theorem is used where the geodesic azimuth has a constant rate of change.

🙂

Note that the "Gaussian Sphere" was the standard way of doing things back when it was only possible with tables of logarithms and trig functions rather than just pushing buttons.

If all else fails, go to the NGS site and read Jim Stem's stuff on the State Plane Coordinate Systems of NAD83. That's the bible.


 
Posted : November 20, 2014 4:39 pm
davidgstoll
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Cliff,

Stem is impenetrable. My every attempt to read him ended up with me fast asleep and drooling on my desk.

Math is not the problem. Defining the normal is where I break down. A line can only be perdendicular to another line or a plane. A line cannot be perpendicular to a Point on an ellipse.

Dave


 
Posted : November 20, 2014 5:09 pm
Cliff Mugnier
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Sure it can. Consider an infinitesimal surface at the point. Similar to how many angels that can dance on the head of a pin.

🙂


 
Posted : November 20, 2014 5:19 pm
shawn-billings
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You can calculate normal to a circular arc (it passes through the radius point). First derivative calculus gives the instant slope of a function. I tried that with an ellipse ten years ago and couldn't crack it. May be time to pick it up again.


 
Posted : November 20, 2014 5:25 pm
mathteacher
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It is a matter of definition. To any point on a surface, we can construct a line tangent to that surface. A line perpendicular to that tangent line at that point of tangency is said to be perpendicular to the surface at that point.

What's worse is the azimuth; it's good only for an infinitesimal distance, and then it changes.

The great thing about studying this stuff is that persistence pays off in both knowledge and insight, and those are two fantastic rewards.


 
Posted : November 20, 2014 5:40 pm

davidgstoll
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Cliff,

Angels may be able to dance on the head of a pin, but where I learnt it all they told me that a line requires at least two points and a plane requires three. So technically, a normal off a point on an ellipse is undefined. And that is where I'm utterly failing to grasp a necessary concept in map projections.

Dave


 
Posted : November 20, 2014 6:14 pm
Cliff Mugnier
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Better explanation than mine.


 
Posted : November 20, 2014 6:17 pm
davidgstoll
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MT,

"To any point on a surface, we can construct a line tangent to that surface."

Yes, we can construct any number of arbitrary tangent lines, but they are arbitrary constructs, and most of them fail the "ellipsoid normal terminated by the semi-minor axis" definition. The only two R's that meet that definition are the Meridian and the Prime Vertical Radius. That's why I think they should be used for the Scale Factors of the Transverse Mercator and the Lambert Conformal Conic Projections, respectively.

OK, picture this:
A stiff cardboard plane sitting on a rubber ball. It wobbles, right? But a point on that plane will only form a Normal to the Minor Axis along the meridian line defined by the point. If you wobble it anywhere else, the Normal won't point to the Minor Axis.

Dave


 
Posted : November 20, 2014 6:41 pm
davidgstoll
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Cliff,

There was nothing wrong with your explanation. The problem is that my skull is about a tenth of a US Survey Foot thick.

Dave


 
Posted : November 20, 2014 6:45 pm
davidgstoll
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Shawn,

The problem with Limits is that they can never be reached. Same thing with my understanding of some things.

Dave


 
Posted : November 20, 2014 6:47 pm

mathteacher
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You're absolutely correct. The tangent line we are talking about is not arbitrary. For the Lambert, it is the line tangent at the central parallel of the projection, either SPCS or LDP. And that makes it perpendicular to the normal at that point.

For SPCS, the tangent line is a secant line instead, intersecting the ellipsoid at the northern and southern standard parallels.. For LDP, it is raised above the ellipsoid and is neither tangent nor secant since it doesn't touch the ellipsoid at all. The point is, you can put it wherever you want it, but the perpendicular property is preserved.

Calculus gives us the slope of the tangent line. The slope of the normal is the negative reciprocal of the slope of the tangent line. The whole construct is a beautiful exercise in applied mathematics.

Having been down virtually the same path you're following, I admire your efforts greatly.


 
Posted : November 20, 2014 7:17 pm
EFBURKHOLDER
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Sometimes the answer is confusing because we've not asked the right question. . . .

How accurately do you need the answer? How much can the radius of curvature change without exceeding the tolerance you've set for yourself?

Those questions can be answered using equation (6) of a paper on my web site.

As it turns out, you do not need the radius of curvature very precisely at all. But you should know what the impact of your approximation really is. Or, like others suggested, compute/use the geometrical mean radius at the point in question.


 
Posted : November 20, 2014 7:20 pm
davidgstoll
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EF,

"...the right question. . . ."

Indeed. I often find myself in the position of not knowing enough to even ask an intelligent question. And on any subject, more questions can always be asked. There's just no end of questions.

"How accurately do you need the answer?"

I'd like to "know" the answer to 100 decimal places, even though I'll probably lop off 99 of them. The age-old accuracy/precision question is what I'm after. Would it do any good to know an answer to 100 decimal places if my data or formulas are wrong?

Your paper is a very good read, even though equation (6) is a bit over my pay grade. I cowboy'ed some numbers, using the extremes of Prime Vertical and Meridian Radii at the average ellipsoid elevation of my project:

R/R+h
20950138/20950138+743=0.99996453610
20858713/20858713+743=0.99996438066

0.99996453610 - 0.99996438066 = 0.00000015544

1 mile X 0.00000015544 = 0.00082070736 Feet

Conclusion? Quit Big-Dealin' it, Dave!

Dave


 
Posted : November 21, 2014 3:52 am