EPSG

EPSG guidance note #7-2, http://www.epsg.org

2017-06-13

Reversibility is subject to constraints. See Guidance Note 7 for conditions and clarification.

Note: These formulas have been transcribed from EPSG Guidance Note #7-2. Users are encouraged to use that document rather than the text which follows as reference because limitations in the transcription will be avoided. See method code 9647 for description of general polynomial formula. A general polynomial transformation is reversible only when the following conditions are met. 1. The co-ordinates of source and target evaluation point are (numerically) the same. 2. The unit of measure of the coordinate differences in source and target coordinate reference system are the same. 3. The scaling factors applied to source and target coordinate differences are the same. 4. The spatial variation of the differences between the coordinate reference systems around any given location is sufficiently small. Clarification on conditions for polynomial reversibility: Regarding 1 and 2 - In the reverse transformation the roles of the source and target coordinate reference systems are reversed. Consequently, the co-ordinates of the evaluation point in the source coordinate reference system become those in the target coordinate reference system in the reverse transformation. Usage of the same transformation parameters for the reverse transformation will therefore only be valid if the evaluation point coordinates are numerically the same in source and target coordinate reference system and in the same units. That is, XS0 = XT0 = X0 and YS0 = YT0 = Y0. Re 3 - The same holds for the scaling factors of the source and target coordinate differences and for the units of measure of the coordinate differences. That is, mS = mT = m. Re 4 - If conditions 1, 2 and 3 are all satisfied it then may be possible to use the forward polynomial algorithm with the forward parameters for the reverse transformation. This is the case if the spatial variations in dX and dY around any given location are sufficiently constant. The signs of the polynomial coefficients are then reversed but the scaling factor and the evaluation point coordinates retain their signs. If these spatial variations in dX and dY are too large, for the reverse transformation iteration would be necessary. It is therefore not the algorithm that determines whether a single step solution is sufficient or whether iteration is required, but the desired accuracy combined with the degree of spatial variability of dX and dY. An example of a reversible polynomial is transformation is ED50 to ED87 (1) for the North Sea. The suitability of this transformation to be described by a reversible polynomial can easily be explained. In the first place both source and target coordinate reference systems are of type geographic 2D. The typical difference in coordinate values between ED50 and ED87 is in the order of 2 metres (approximately 10E-6 degrees) in the area of application. The polynomial functions are evaluated about central points with coordinates of 55 deg N, 0 deg E in both coordinate reference systems. The reduced coordinate differences (in degrees) are used as input parameters to the polynomial functions. The output coordinate differences are corrections to the input coordinate offsets of about 10E-6 degrees. This difference of several orders of magnitude between input and output values is the property that makes a polynomial function reversible in practice (although not in a formal mathematical sense). The error made by the polynomial approximation formulas in calculating the reverse correction is of the same order of magnitude as the ratio of output versus input: (output error / input error) = ( output valu/ input value) which is approximately 10E-6 As long as the input values (the coordinate offsets from the evaluation point) are orders of magnitude larger than the output (the corrections), and provided the coefficients are used with changed signs, the polynomial transformation may be considered to be reversible.

For geodetic transformation ED50 to ED87 (1) Offset unit: degree Ordinate 1 of evaluation point X0 = 55° 00' 00.000"N = +55 degrees Ordinate 2 of evaluation point Y0 = 0° 00' 00.000"E = +0 degrees Parameters: A0 = -5.56098E-06 A1 = -1.55391E-06 ... A14 = -4.01383E-09 B0 = +1.48944E-05 B2 = +2.68191E-05 ... B14 = +7.62236E-09 Forward calculation for: ED50 Latitude = Xs =52* 30’30"N = +52.508333333 degrees ED50 Longitude = Ys = 2*E= +2.0 degrees U = XS - X0 = * ED50 - X0 = 52.508333333 - 55.0 = -2.491666667 degrees V = YS - Y0 = * ED50 - Y0 = 2.0 - 0.0 = 2.0 degrees dX = A0 + A1.U + ... + A14.V4 = -5.56098E-06 + (-1.55391E-06 * -2.491666667) + ... + (-4.01383E-09 * 2.0^4) = -3.12958E-06 degrees dY = B0 + B1.U + ... + B14.V4 = +1.48944E-05 + (2.68191E-05 * -2.491666667) + ... + (7.62236E-09 * 2.0^4) = +9.80126E-06 degrees Then ED87 Latitude = XT = XS + dX = 52.508333333 - 3.12958E-06 degrees = 52* 30’ 29.9887" N ED87 Longitude = YT = YS + dY = 2* 00’ 00.0353" E Reverse calculation for transformation ED50 to ED87 (1). The transformation method for the ED50 to ED87 (1) transformation, 4th-order reversible polynomial, is reversible. The same formulas may be applied for the reverse calculation, but coefficients A0 through A14 and B0 through B14 are applied with reversal of their signs. Sign reversal is not applied to the coordinates of the evaluation point. Thus: Ordinate 1 of evaluation point X0 = 55° 00' 00.000"N = +55 degrees Ordinate 2 of evaluation point Y0 = 0° 00' 00.000"E = +0 degrees A0 = +5.56098E-06 A1 = +1.55391E-06 ... A14 = +4.01383E-09 B0 = -1.48944E-05 B1 = -2.68191E-05 ... B14 = -7.62236E-09 Reverse calculation for: ED87 Latitude = XS = 52° 30’29.9887"N = +52.5083301944 degrees ED87 Longitude = YS = 2° 00’ 00.0353" E = +2.0000098055 degrees U = 52.5083301944 - 55.0 = -2.4916698056 degrees V = 2.0000098055 - 0.0 = 2.0000098055 degrees dX = A0 + A1.U + ... + A14.V4 = +5.56098E-06 + (1.55391E-06 * -2.491666667) + ... + (4.01383E-09 * 2.0000098055^4) = +3.12957E-06 degrees dY = B0 + B1.U + ... + B14.V4 = -1.48944E-05 + (-2.68191E-05 * -2.491666667) + ... + (-7.62236E-09 * 2.0000098055^4) = -9.80124E-06 degrees Then ED50 Latitude = XT = XS + dX = 52.5083301944 + 3.12957E-06 degrees = 52° 30’ 30.000" N ED50 Longitude = YT = YS + dY = 2° 00’ 00.000" E