Or anyone.
In the Paper titled Accuracy of Elevation Reduction Factor it states "The earth is not quite spherical, but flattened at the poles. Students of geodesy soon learn that the earth’s radius in the north-south direction changes from a smaller value at the equator to a larger value at the poles."
Later in the paper there is a table to demonstrate this.
Latitude Mean Radius Mean Radius
(degrees) (meters) (U.S. Survey Feet)
0 6,356,752.314 20,855,444.88
90 6,399,593.626 20,996,000.09
Is this correct? If so please provide references and/or an explaination. We understood it to be the opposite.
Thanks
Good question. At first it seems like a mistake, but I could see a logical explanation. As the curve gets flatter at the pole, the true radius gets longer (but the radius point would not be the center of earth). You are comparing axises to radius.
> Or anyone.
>
> In the Paper titled Accuracy of Elevation Reduction Factor it states "The earth is not quite spherical, but flattened at the poles. Students of geodesy soon learn that the earth’s radius in the north-south direction changes from a smaller value at the equator to a larger value at the poles."
>
> Later in the paper there is a table to demonstrate this.
> Latitude Mean Radius Mean Radius
> (degrees) (meters) (U.S. Survey Feet)
> 0 6,356,752.314 20,855,444.88
> 90 6,399,593.626 20,996,000.09
>
>
> Is this correct? If so please provide references and/or an explaination. We understood it to be the opposite.
>
>
>
> Thanks
Guessing it's a typo.
larger value at the equator to a smaller value at the poles
(If we are truly talking about radius.)
Additionally Later in the article the same typo seems to creep into the conclusion:
Conclusions and final comments are:
1. The average earth radius commonly used is acceptable for all but the most
demanding applications. Values of earth radius in Table 1 go from 6,356,752 m
at the equator to 6,399,594 m at the pole. Using an average earth radius of
6,372,000 m and an earth radius uncertainty of 20,000 meters, the right-most
column of Table 2 includes all latitudes from S 65° to N 65°.
Should be from 6,356,752 m at the pole to 6,399,594 m at the equator.
Table 1, Geometrical Mean Radius of Curvature at Various Latitudes
Latitude Mean Radius
(deg) (meters)
0 6,356,752.314
5 6,357,075.580
30 6,367,408.778
35 6,370,783.223
70 6,394,552.344
75 6,396,705.765
90 6,399,593.626
I think it's trying to say that your semi-minor axis becomes smaller in a north south direction as you leave the poles and go east or west. As you run perpendicular to the equator that is.
> Guessing it's a typo.
Burkholder's statement is actually correct. Radius of curvature and semi-major axis of an ellipse aren't comparable quantities.
A simple way to visualize radius of curvature is to imagine fitting a circle to successively smaller arcs of an ellipse. As the arc length approaches zero, the value of the circle that fits the arc converges to the radius of curvature at the point the arc is centered upon.
I'm seeing the phrase "earth’s radius in the north-south direction"
I see what you're saying, and I am seeing that phrase. Not sure exactly what is meant.
If I looked at the shape of the earth in a cross-section looking at it from the north (where you don't see the north-south "squash", but see a more consistent curve around the equater) I am thinking I would be looking a close to a circle. we are looking at it from a "side-view" cross-section where the north-south pole is a shorter distance. But if you look at the tight curve on the right and left (and ignoring the distance to the "center" of the ellipsoid) you would see where at any given point the rate of curve would have a shorter radius. from the north pole down, you would have a huge distance to a radius point that is below the south pole (with the exaggerated squash as drawn). Maybe that is what you are talking about and also being the "north-south" radius as opposed to looking at the radius from a overhead cross-section.
??
:good:
I like your explanation. I tried some convoluted explanation above but your's sounds much better.
Here's a discussion from a math forum. There is a table posted 3/4 down. Note, the radius does get longer as the latitude increases.
http://mathforum.org/library/drmath/view/61089.html
>The earth is flattened at the poles. The flatter a curve (or a surface containing a curve), the greater its radius of curvature - the radius of the circle that most closely approximates the curve. So the radius of curvature at the poles will be greater than the equatorial radius...
>You will see there that the radius of curvature is GREATEST at the poles and LEAST at the equator, contrary to your calculation based on distance from the center of the earth.
It's a table of the mean radius of curvature at various latitude values.
meridian radius of curvature
M = a*(1.0 - e*e) / (1.0 - e*e*sin(phi)*sin(phi))^3/2
radius of curvature in the prime vertical:
N = a / (1.0 - e*e*sin(phi)*sin(phi))^1/2
They both extend past the center of the ellipsoid except at phi = 0.
a = semimajor axis
e = eccentricity
phi = geodetic latitude
Thanks
Learn something new every day. We were trying to figure out why some softwares use a mean radius to compute elevation factor and others use the latitude formula. It makes about .03 in a mile difference at 41 30.
Computing elevation factor is using the Gaussian Mean Radius which is the square root of the product of the two radii of curvature that Melita just listed above.
Agreed
NGS data sheets and Topcon to name two use the mean value however I know of some other softwares that use a formula based on actual latitude to compute elevation scale factor. I was questioned by a user of the "other" software re my published elevation factor.
Same thing. Note that Melita's equations have a single variable: geodetic latitude. To compute the Gaussian Sphere, you need to plug in a specific latitude.
All I know is you can get different answers depending on your software.
NGS datasheets and Topcon use a constant 6372000m for r in EF=r/(r+h). Some others use a variable for r based on latitude.
Of course all this debate errr discussion with an engineer started over an EF the same out to the 6th decimal on the engineer's software. The debate is that the answer should always be the same for all softwares out to however many decimals.
The radius of the surface is entirely different than the distance from the surface to the center of the ellipsoid.
The thread is about the different ways r is determined in the elev scale factor
You all are missing the most important question
What time is it at the North Pole?
The flatter the arc, the longer the radius. Think about spiral curves. If the formula uses a mean value it will have to depart from a specific value solution eventually. Used to be that approximation of parabolic arc was used for ease of computation where the difference was negligible for the purpose and could be ignored. Not sure why one would program that way now though.
i just wrote a program for my hp33s, and checked in on some of dr. burkholder's values. they close.
here it is for those that want to try it:
lbl r
input a
input e
input l
1
rcl e
-
sqrt x
rcl a
x
sto n
rcl l
sin
x squared
rcl e
x
1
x >< y
-
sto d
divide
sto r
view r
rtn
a is based on 6378137 m
e squared is 0.00669438002290 per dr burkholder's MEAN RADIUS OF CURVATURE
pages three and four.
so, for me, that begs the question: is it the definition of mean radius of earth that i/we don't comprehend? seems to be not a simple center of the ellipsoid to a point on surface at given latitude. typo? geometric nomenclature?
So the r of the curvature as used in EF is not the same as the point of origin in ECEF as I always assumed for some stupid reason.
