INVERSE - Ellipsoidal distance

[fivethings]

I have a question about the INVERSE calculator of ellipsoidal distance (at INVERSE ) .

 

Is my understanding correct - that "ellipsoidal distance" is, essentially, the same as or very similar to the distance between two points as measured along a segment of the arc of a great circle?

 

And, if that's correct, then "ellipsoidal distance" and "great circle distance" would be the same if the two points were at the same longitude, but the calculated "ellipsoidal distance" would be a lot shorter for two points at the same latitude as compared to the distance measured along that parallel of latitude?

 

And, if that's correct, is there a calculator for measuring the distance between two points of equal latitude along that parallel of latitude? or, alternatively, is there a simple correction for converting between ellipsoidal distance and (whatever the technical term is for) distance along a parallel of latitude?

 

Thanks,

5

 

p.s. Are the distances for points along a calibration baseline measured as "ellipsoidal distances", or are they determined as if the world were flat?

[Papa-Bear-NYC]

I have a question about the INVERSE calculator of ellipsoidal distance (at INVERSE ) .

...

p.s. Are the distances for points along a calibration baseline measured as "ellipsoidal distances", or are they determined as if the world were flat?

I have read a bit about methods used in early triangulations. The base lines are measured with calibrated steel tapes and compensated for temperature. This distance is then reduced to "sea level" which in effect means an elipsoid. For a 3 mile base line, it hardly matters when the tape accuracy is say 1:500,000

 

But ... The triangulation is usually done on the basis of tangent planes, especially for small triangulations such as for a city. The difference is too small to matter. It would be very hard to solve say 600 triangles (for a big city like NYC) using ellipsoidal trignometry rather than plane trigonometry. These calculations took years and were done by hand. Remember a "computer" in 1910 was a very adept person.

 

For large traingulations (the East Coast) the more complicated eliptical calculations were used (or else the local portions would not match up).

 

Once everything got adjusted in the early 20th century, everything was reduced to the ellipsoid.

 

For modern day triangulations (ddoes anyone still do these? Or do they just use GPS>) calculations are easy, hence INVERSE and FORWARD programs abound.

[Bill93]

>For a 3 mile base line, it hardly matters when the tape accuracy is say 1:500,000

 

Maybe not in NYC, but probably so in Denver.

 

My old surveying book gives the correction to be subtracted rom the measured length to get equivalent length at sea level, as

C = LA/R

 

where L is the measured length, A is the altitude, and R is the radius of the earth, all in the same units. So if we take

A=1 mile

R=4000 mi

then the correction is C/L = A/R = 1:4000.

 

If you allow 1:500,000 the elevation can be up to 1/125 mile or 42 ft.

-----------

The difference between arc length and chord length is much smaller than this unless you are measuring very long distances.

The difference between arc length S and chord C is

(S-C)/S = 1- [sin(a/2) / (a/2)]

 

where a = radian angle as measured at the center of the curve. So if you have an arc that is 1 degree of a great circle (about 60 miles) the difference is about 1:78,786

 

and if it is only 5 minutes of arc (about 5 miles) then 1:11 million.

 

Thus for most optical triangulations the arc-vs-chord question is not significant.

-----------

Of course, you probably should check another source before you use my computations to base your bridge layout on.

[Papa-Bear-NYC]

>For a 3 mile base line, it hardly matters when the tape accuracy is say 1:500,000

 

Maybe not in NYC, but probably so in Denver.

 

My old surveying book gives the correction to be subtracted rom the measured length to get equivalent length at sea level, as

C = LA/R

 

where L is the measured length, A is the altitude, and R is the radius of the earth, all in the same units. So if we take

A=1 mile

R=4000 mi

then the correction is C/L = A/R = 1:4000.

 

If you allow 1:500,000 the elevation can be up to 1/125 mile or 42 ft.

The primary base line used in NYC was along ocean Parkway in Brooklyn (Map). The map shows an elevation between around 15 and 35 feet. BTW, Isn't the radius of the earth more like 6000 miles?

The difference between arc length and chord length is much smaller than this unless you are measuring very long distances.

The difference between arc length S and chord C is

(S-C)/S = 1- [sin(a/2) / (a/2)]

 

where a = radian angle as measured at the center of the curve. So if you have an arc that is 1 degree of a great circle (about 60 miles) the difference is about 1:78,786

 

and if it is only 5 minutes of arc (about 5 miles) then 1:11 million.

 

Thus for most optical triangulations the arc-vs-chord question is not significant.

-----------

Of course, you probably should check another source before you use my computations to base your bridge layout on.

That probably means the use of plane versus ellipsoidal trigonometry was the correct choice. The book said everything was within a 20 mile radius of the center which justified using that approximation. Edited February 7, 2007 by Papa-Bear-NYC

[+Ernmark]

Earth' equatorial radius is 3963 mi....just to make things worse, Earth isn't a true sphere..

[CallawayMT]

I have a question about the INVERSE calculator of ellipsoidal distance (at INVERSE ) .

 

Is my understanding correct - that "ellipsoidal distance" is, essentially, the same as or very similar to the distance between two points as measured along a segment of the arc of a great circle?

 

And, if that's correct, then "ellipsoidal distance" and "great circle distance" would be the same if the two points were at the same longitude, but the calculated "ellipsoidal distance" would be a lot shorter for two points at the same latitude as compared to the distance measured along that parallel of latitude?

 

And, if that's correct, is there a calculator for measuring the distance between two points of equal latitude along that parallel of latitude? or, alternatively, is there a simple correction for converting between ellipsoidal distance and (whatever the technical term is for) distance along a parallel of latitude?

 

Thanks,

5

 

p.s. Are the distances for points along a calibration baseline measured as "ellipsoidal distances", or are they determined as if the world were flat?

 

I assume that this is for Will, but either way you guys sure ask some interesting questions. I just ran a report using two very different elevations to show the difference that the elevation creates. I just went 10 minutes North for each comparison and a 0' elevations and around 9000' in elevation. As can be expected at 0' the ellipsoid and ground distances are one in the same. At around 9000' the ground distance is 26.7' longer than the ellipsoid distance.

 

Inverse Report

 

Project : Benchmarks

 

User name kurt Date & Time 1:15:13 PM 2/7/2007

Coordinate System US State Plane 1983 Zone Montana 2500

Project Datum NAD 1983 (Conus)

Vertical Datum Geoid Model GEOID99 (Conus)

Coordinate Units International feet

Distance Units International feet

Height Units US survey feet

 

 

--------------------------------------------------------------------------------

 

Grid Local Cartesian

(WGS-84)

From: AUTO0001

 

Northing: 1133061.4546ift Latitude: 47°18'35.72067"N X: -5517178.0444ift

Easting: 1143391.3660ift Longitude: 112°49'44.25361"W Y: -13106300.5338ift

Elevation: 9208.3149sft Height: 9167.8196sft Z: 15312496.3958ift

Convergence: -2°26'06.531895"

 

Inverse:

Grid Azimuth: 2°26'07" NS Fwd Azimuth: 0°00'00" Delta X: 17367.769ift

Grid Distance: 60757.585ift NS Back Azimuth: 180°00'00" Delta Y: 41257.903ift

Delta Elevation: 0.9810sft Ellipsoid Dist: 60793.148ift Delta Z: 41172.727ift

t-T Correction: 0°00'00.000000" Ground Dist: 60819.816ift Slope Dist: 60819.795ift

Elevation Scale Factor: 0.99956152 Delta Height: 0.0000sft

Grid Scale Factor: 0.99941503

Combined Factor: 0.99897680

 

To: AUTO0002

Northing: 1193764.1732ift Latitude: 47°28'35.72067"N X: -5499810.2755ift

Easting: 1145972.8678ift Longitude: 112°49'44.25361"W Y: -13065042.6304ift

Elevation: 9209.2959sft Height: 9167.8196sft Z: 15353669.1224ift

Convergence: -2°26'06.531895"

 

Grid Local Cartesian

(WGS-84)

From: AUTO0003

 

Northing: 1133061.4546ift Latitude: 47°18'35.72067"N X: -5514766.3163ift

Easting: 1143391.3660ift Longitude: 112°49'44.25361"W Y: -13100571.3670ift

Elevation: 40.4956sft Height: 0.0003sft Z: 15305757.7416ift

Convergence: -2°26'06.531895"

 

Inverse:

Grid Azimuth: 2°26'07" NS Fwd Azimuth: 0°00'00" Delta X: 17360.154ift

Grid Distance: 60757.585ift NS Back Azimuth: 180°00'00" Delta Y: 41239.813ift

Delta Elevation: 0.9810sft Ellipsoid Dist: 60793.148ift Delta Z: 41154.673ift

t-T Correction: 0°00'00.000000" Ground Dist: 60793.148ift Slope Dist: 60793.126ift

Elevation Scale Factor: 1.00000000 Delta Height: 0.0000sft

Grid Scale Factor: 0.99941503

Combined Factor: 0.99941503

 

To: AUTO0004

Northing: 1193764.1732ift Latitude: 47°28'35.72067"N X: -5497406.1627ift

Easting: 1145972.8678ift Longitude: 112°49'44.25361"W Y: -13059331.5542ift

Elevation: 41.4766sft Height: 0.0003sft Z: 15346912.4148ift

Convergence: -2°26'06.531895"

 

Back to top

 

CallawayMT

[holograph]

Is my understanding correct - that "ellipsoidal distance" is, essentially, the same as or very similar to the distance between two points as measured along a segment of the arc of a great circle?

 

And, if that's correct, then "ellipsoidal distance" and "great circle distance" would be the same if the two points were at the same longitude, but the calculated "ellipsoidal distance" would be a lot shorter for two points at the same latitude as compared to the distance measured along that parallel of latitude?

 

And, if that's correct, is there a calculator for measuring the distance between two points of equal latitude along that parallel of latitude? or, alternatively, is there a simple correction for converting between ellipsoidal distance and (whatever the technical term is for) distance along a parallel of latitude?

 

The distance along a constant latitude would be a rhumb line at bearing 90 or 270 degrees. The great circle would be shorter, as you say. The rhumb distance at a constant latitude is fairly easy to calculate, since it is simply the length of a circular arc. What you need is to use the ellipse to find the radius of the arc. The formula is below. There might be some online calculator somewhere to do the calculation.

 

You can find the parameters of some ellipsoids here.

 

L = a * cos(phi) / sqrt(1 - e^2 * sin(phi)^2 )

L = length of a radian of longitude
phi = latitude
a = semi-major axis of ellipsoid
e = eccentricity of ellipsoid

 

For base line measurements where the desired accuracy may be 1:1,000,000, or a few MM over a few KM, almost every factor that can be considered is considered. All geodetic distances are usually reduced (projected) to the surface of an ellipsoid.

 

edit: FYI, at latitude 41 degrees, you have to go about 10 km before there is a difference of more than 1 mm between the ellipsoid great circle distance and the rhumb distance along the constant latitude. The simple formula above works quite well over modest distances.

Edited February 8, 2007 by holograph

[+TheBeanTeam]

I think I just slipped into the Greek Forum by accident.

[+Holtie22]

Holograph is absolutely correct and exactly on point with fivethings' question. To take it further, any line of constant bearing, (with the exception of a true meridian and a line along the equator), is by definition a curved line. The best way to visualize this is to examine a map utilizing a polar projection whereby the north (or south) pole is in the center of the map, and lines of latitude are concentric circles around the pole. Any line that you tried to plot at a constant bearing (say N45E) would be curved since you are maintaining an angle of 45 degrees between your position and the pole.

 

This is one of the reasons why when you examine the datasheets for two stations that reference each other in the box score, the geodetic azimuths between the stations do not differ by exactly 180 degrees as you might expect. For example, look at the geodetic azimuth from 8 Mile to Pinon 2, JM0516 then from Pinon 2 to 8 Mile, JM0517 and note that, while close, they do not differ by exactly 180 degrees. (They are only about 4 miles apart, and oriented more N-S than E-W, otherwise the difference would be greater.)

[fivethings]

Thanks, all, for the replies; my understanding is confirmed, and the (physical) world makes more sense to me now.

 

Thanks, especially, to Holograph - your "FYI" about the difference between the ellipsoidal distance between two points along the 41st Parallel of latitude and the distance between those two same points as measured along the Parallel was particularly on point. The question arose in my mind when I compared the INVERSE distance along the southern boundary line of Pennsylvania (at about N39-43.330) between Mile 0 and Mile 10 to the cumulative distance between Mile 0 and Mile 1, plus Mile 1 and Mile 2, etc. (It's a really small distance.)

 

And Holti22's observation about the reciprocal geodetic azimuths between two points never being exactly = 180 (except for true meridians and the Equator) confirms my understanding of that property of a straight line.

 

So, if I started at Point A (such Point NOT lying on the Equator, but somewhere in the Northern Hemisphere), determined a bearing of true West and ran a perfectly straight line to the horizon, the point at which my line intersected a true meridian at the horizon would lie at a more southerly latitude than Point A. Furthermore, if I used INVERSE to calculate the starting bearing between my Point A and a point at the exact same latitude on the horizon, that bearing would be greater than 270.00000.

 

Thanks, again.

5

[holograph]

So, if I started at Point A (such Point NOT lying on the Equator, but somewhere in the Northern Hemisphere), determined a bearing of true West and ran a perfectly straight line to the horizon, the point at which my line intersected a true meridian at the horizon would lie at a more southerly latitude than Point A. Furthermore, if I used INVERSE to calculate the starting bearing between my Point A and a point at the exact same latitude on the horizon, that bearing would be greater than 270.00000.

 

Yes, if your horizon was 8 miles, the point you sighted from A at a bearing of 270 degrees would lie about 35.5 feet south of the latitude of A. That's exactly the problem that Mason and Dixon had to solve in order to survey the line, and if I recall that's very nearly the way they solved it -- they ran the line by sightings and then computed the corrections that were necessary to bring it back to the parallel. The book "Drawing the Line" by Edwin Danson has some good information about the techniques used.

[+fizzymagic]

You might want to try my little GeoCalc program that will calculate ellipsoidal, great circle, and rhumb line distances between two points. I made some nice pictures that illustrate the differences; I put them here.

[68-eldo]

You might want to try my little GeoCalc program that will calculate ellipsoidal, great circle, and rhumb line distances between two points. I made some nice pictures that illustrate the differences; I put them here.

 

Great pictures. The line extensions (Great circle around the word, and Rhumb line spirals in to the pole) make it even clearer.