Hi, Micha.
Is the picture correct?

Hi lyoyha,

it is a Cartesian system defined by the principal point P0, cf. Wikipedia: From ECEF to ENU.

P0 is the only point that is orthogonal to both, the local system as well as the Earth surface.

regards
Micha

It turns out, if we consider this local coordinate system strictly, then for any point of standing (except for P0) - there must be vertical deflections?

Hello,

It turns out, if we consider this local coordinate system strictly, then for any point of standing (except for P0) - there must be vertical deflections?

Yes, the vertical of P0 is orthogonal to the local plane (the vertical axis is the normal axis) and the vertical is orthogonal to the surface of the ellipsoid. Each point is transformed to local plane. The tilt of the vertical of such points must be taken into account during the adjustment.

Kind reagrds
Micha

Then another question arises:

Why, in this case, do you not take into account the vertical deflections from the local plane of points A, B and M when adjusting networks 30m, 60m ... 40000m?

Hello,

Why, in this case, do you not take into account the vertical deflections from the local plane of points A, B and M when adjusting networks 30m, 60m ... 40000m?

I'm not sure, what you are meaning. You are talking about the Evaluation of Compatibility among Network Adjustment Software, right? The used JAG3D projects are available. In all projects, an ellipsoidal Earth model with local Cartesian coordinates is used, see the screenshot below. If the option "local ellipsoidal system" is selected, the tilt is known by definition.

Kind regards
Micha

Thanks for answers.
It turns out that Jag3d "knows" that the angle between the plumb lines at points A and B at a distance between them, say, 1000m, is about 33 ". And it uses it in the calculations.

Hello,

It turns out that Jag3d "knows" that the angle between the plumb lines at points A and B at a distance between them, say, 1000m, is about 33 ". And it uses it in the calculations.

Yes. The tilt can be derived by some geometric calculations. Just try it out and chose the option "local Cartesian system" to see the difference.

Kind regards
Micha

Hi.
Should there be such differences between the elevation marks ???
https://ru.files.fm/u/rc2mcxumx

Hello,

Should there be such differences between the elevation marks ???

The approaches differ and, thus, the result depends on your selected option. Or the other way around: If the results would not differ, why we need these options?

Kind regards
Micha

Hi,

you can reproduce the results of the elliptical approach using the local Cartesian option. In this case, you have to specify the deflection parameters, cf.

Kind regards
Micha

Why then in any of the projects (JAG3D_Results_of_Compatibility_among_Network_Adjustment_Software):
Local cartesian system + Earth Curvature = local ellipsoidal system (with a=b)?

In the above example:
Local cartesian system + Earth Curvature don`t= local ellipsoidal system (with a=b)

Hello,

Why then in any of the projects (JAG3D_Results_of_Compatibility_among_Network_Adjustment_Software):
Local cartesian system + Earth Curvature = local ellipsoidal system (with a=b)?

That is simply not true! If one adjust the 40 km network using the local Cartesian system and select the option Earth Curvature, the coordinates of the point M reads

M 3.8567191065322965E-5 23094.010829119456 0.007236303280461759

If one apply the ellipsoidal approach the result is

M -2.3686794627367602E-5 23094.01077588917 -0.001651724552338522

Both results differ especially in the up component.

Kind regards
Micha

FYI: There is a formatting error in sexagesima degree representation. The sign is missing, if the value is between -1° and 0°. I fixed the sources. Please note: The value is correctly used because it is only a formatting issue.

Hello,

Why then in any of the projects (JAG3D_Results_of_Compatibility_among_Network_Adjustment_Software):
Local cartesian system + Earth Curvature = local ellipsoidal system (with a=b)?

That is simply not true! If one adjust the 40 km network using the local Cartesian system and select the option Earth Curvature, the coordinates of the point M reads

M 3.8567191065322965E-5 23094.010829119456 0.007236303280461759

If one apply the ellipsoidal approach the result is

M -2.3686794627367602E-5 23094.01077588917 -0.001651724552338522

Both results differ especially in the up component.

Kind regards
Micha

Yes you are right. I should have written "almost equal".
But in my example the difference is huge - 1250mm for a 4000m line.
(and this is taking into account the curvature of the Earth).
see 1.jpg; 2.jpg

Hello,

Yes you are right. I should have written "almost equal".

Okay.

But in my example the difference is huge - 1250mm for a 4000m line.

Yes, that is right. If you estimate the influence of the curvature of the Earth, the result is

$\frac{4000^2}{2R} = 1.256~\mathrm{m}$

This is the different in the up component you already mention. The equation (approximately) corrects the zenith angle to the Earth surface.

However, if the ellipsoidal approach is applied, the coordinates are related to a local planar system defined by the principal point, cf. the figures in the documentation. That is the different. In your case, the principal point is identical to the station position.

Kind regards
Micha

Thanks for answers.
If I finally came to understand the difference between these coordinate systems, then please check me:

if several (or many) points lie on the same level surface (for example, at sea level), then

1) then they will necessarily have a different "Z" coordinate in jag3d, provided that "local elipsoidal system" is selected;

2) then they will necessarily have almost the same "Z" coordinate in jag3d, provided that "local Cartesian system + Earth curvature" is selected.

Hello,

1) then they will necessarily have a different "Z" coordinate in jag3d, provided that "local elipsoidal system" is selected;

Yes, that is right. The local system has a direct link to a global (Earth fixed) frame and it is possible to project the points (using UTM, for instance). In your case, the defined local system is identical to the instrument/station frame. So, there is no need to reduce observations w.r.t. the frame.

2) then they will necessarily have almost the same "Z" coordinate in jag3d, provided that "local Cartesian system + Earth curvature" is selected.

If the network is quite small, the solution may be sufficient. The error depends on the network extent.

Kind regards
Micha

Hello. Is there a mistake in the formula?

Hello lyoyha,

Is there a mistake in the formula?

Thank you for critical read the equation. This equation seems to be valid and can be found at Wikipedia, cf. From ECEF to ENU. The order is, however, east, north, up (which should be y, x, z instead of x, y, z). I changed the order to:

$\begin{pmatrix}y_i \\ x_i \\ z_i\end{pmatrix} = \begin{pmatrix} -\sin\lambda_0 & \cos\lambda_0 & 0 \\ -\sin\phi_0\cos\lambda_0 & -\sin\phi_0\sin\lambda_0 & \cos\phi_0 \\ \cos\phi_0\cos\lambda_0 & \cos\phi_0\sin\lambda_0 & \sin\phi_0 \end{pmatrix} \begin{pmatrix} X_i - X_0 \\ Y_i - Y_0 \\ Z_i - Z_0 \end{pmatrix}$

(EDIT: You are referred to the subindex, right? I corrected the equation and replaced the r by 0. Thanks!)

Can you confirm this equation?

Kind regards
Micha

What is ϕr in this formula?

Probably should be ϕ0...

Hello,

What is ϕr in this formula?

Probably should be ϕ0...

Ah, that is the point. Yes, you are right. I corrected my posting and the equation in the documentation. Thank you!

/Micha

These are the results I got.
https://ru.files.fm/u/xfrnqtm6n

Hello,

These are the results I got.

Okay?! I do not know what you expect from me. Is there a problem? It is unclear to me.

I'm sorry. It is hard to support you, if you do not describe the problem in detail.

Kind regards
Micha

I tried to answer this question of yours:

Can you confirm this equation?

аnd made calculations for the 5000m network using this formula. The results match yours with an accuracy of about 0.01mm.

Hello,

аnd made calculations for the 5000m network using this formula. The results match yours with an accuracy of about 0.01mm.

Perfect. Good work and thank you for re-calculating the local coordinates.

/Micha