# How to convert ITRF to ITRF

``````ITRF 96 1998.0 epoch

X          Y           Z         Vx(m/y)   Vy(m/y)  Vz(m/y)
4059623.25 2764405.93 4056720.43    -0.0079   -0.0007  0.0132
``````

``````Transformation parameters from ITRF1996 to ITRF2014.(epoch:2010)

Solut.  Tx      Ty      Tz     D       Rx        Ry        Rz
(mm)    (mm)    (mm)  (ppb)   (,001)    (,001)    (,001)
Rates   Tx      Ty      Tz     D       Rx        Ry        Rz
(mm/y) (mm/y) (mm/y) (ppb/y) (0,001/y) (0,001/y) (0,001/y)
ITRF96 -7.4     0.5    62.8  -3.80      0         0       -0.26
rates  -0.1     0.5    3.3   -0.12      0         0       -0.02
``````

``````GRS 80 ellipsoid a=6378137.00 b=6356752.314m 1/f=298.257222101
``````

Question is 2017,8911704 epoch ITRF2014 XYZ coordinates

For now I know how to convert epoch but I don't know how to convert datum(ITRF to ITRF). This is my quiz question. Is there anyone who can help me?

• What do you mean by "quiz question"? – PolyGeo Dec 1 '19 at 8:12
• It looks like the base epoch of the 14-parameter transformation is 2010.0. You can calculate the parameters using the rates to 2017.8911704 (that precise? that's unusual). You can also move the coordinates of the XYZ position to 2017.8911704 using the point's velocity values. – mkennedy Dec 2 '19 at 22:06
• This is useful, thank you – Halil Yilmaz Dec 4 '19 at 4:21