Re: problem in plotting trajectory using global data

Hi Ufuk,

For the first set of longitudes, what about something like this:

   lon = where(lon.gt.180,360-lon,lon)

See the attached PNG file.

--Mary

On Dec 28, 2009, at 12:30 AM, Ufuk Utku Turuncoglu wrote:

> TRACK_ID 421
> POINT_NUM 28
> 44 0.502097 73.913086 2.791847e+00
> 45 1.147323 73.242691 6.004342e+00
> 46 1.515933 71.030869 8.055365e+00
> 47 1.298914 68.667351 8.317214e+00
> 48 359.402008 64.137444 6.039263e+00
> 49 357.505035 59.299000 5.366868e+00
> 50 357.192780 55.944424 3.733113e+00
> 51 356.532043 53.748260 3.187781e+00
> 52 357.106659 50.272743 4.486194e+00
> 53 358.536896 47.064690 7.892114e+00
> 54 1.560772 44.485828 1.053436e+01
> 55 3.515971 42.982204 1.313965e+01
> 56 4.783206 41.885006 1.440779e+01
> 57 5.995381 41.050323 1.705493e+01
> 58 6.665836 39.986656 1.876119e+01
> 59 8.133615 39.737965 1.857688e+01
> 60 10.046181 39.868969 1.736890e+01
> 61 11.255802 40.183357 1.643047e+01
> 62 13.257846 40.473854 1.531118e+01
> 63 14.234121 40.869377 1.190000e+01
> 64 11.396882 41.550262 9.673411e+00
> 65 8.818160 40.083595 7.746586e+00
> 66 9.923666 39.717323 1.048592e+01
> 67 10.582663 40.126450 1.282651e+01
> 68 13.352924 39.639008 1.332119e+01
> 69 15.406126 39.694725 1.596746e+01
> 70 17.604263 40.084320 1.627496e+01
> 71 20.176548 40.460274 1.413614e+01
>
> TRACK_ID 535
> POINT_NUM 16
> 84 152.471680 40.310097 1.126093e+01
> 85 156.280151 40.776115 1.342023e+01
> 86 160.719894 40.529774 1.639741e+01
> 87 164.660049 40.821220 1.767797e+01
> 88 168.426666 41.405369 1.843181e+01
> 89 172.029724 42.594719 2.017140e+01
> 90 176.103165 43.760841 2.256382e+01
> 91 179.600800 45.465179 2.235561e+01
> 92 182.329681 47.291069 2.268816e+01
> 93 183.897781 49.648006 2.256968e+01
> 94 185.061844 52.251072 2.102557e+01
> 95 185.301468 55.234501 2.045463e+01
> 96 182.694244 58.037880 2.110939e+01
> 97 179.180893 59.926064 2.139341e+01
> 98 179.564819 61.041470 1.708533e+01
> 99 181.582291 61.607300 1.346206e+01

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Received on Tue Dec 29 10:26:41 2009