Hi Bing Wang, As I understand your question, you should actually be measuring the distance between a plane and a centroid rather than another plane. The planes in chimera, while shown as finite disks, are actually infinite internally and thus only perfectly parallel planes will have a non-zero distance. On the other hand, if you are measuring the distance between plane p1 and a centroid c2 lying on plane p2, you specify that chimera should measure the shortest distance between a plane p1 and a defined point on plane p2. This distance is measured along the perfectly "vertical" (e.g. perpendicular to the plane p1) line from the plane p1 to the centroid c2. Best, Matej ------------------------------------------------------ Dr. Matej Repic Ecole Polytechnique Fédérale de Lausanne Laboratory of Computational Chemistry and Biochemistry SB - ISIC LCBC BCH 4108 CH - 1015 Lausanne ------------------------------------------------------ On 8/17/15, 20:12, "chimera-users-bounces@cgl.ucsf.edu on behalf of Elaine Meng" <chimera-users-bounces@cgl.ucsf.edu on behalf of meng@cgl.ucsf.edu> wrote:
Hi Bing Wang, It depends what you mean by ³vertical². You can get the closest distance of two planes (considering them as infinite) by selecting both their rows in the axes/planes/centroid tables, or with the distance command, for example:
distance p1 p2
However, there is no simple way to get the distance only along a specific direction. I hope this helps, Elaine ---------- Elaine C. Meng, Ph.D. UCSF Computer Graphics Lab (Chimera team) and Babbitt Lab Department of Pharmaceutical Chemistry University of California, San Francisco
On Aug 13, 2015, at 7:54 PM, Wang, Bing <bingwang@ou.edu> wrote:
Hi Chimera,
Is there a way to measure the vertical distance of two planes in the structure? I do know how to measure the angle between two planes or one line and one plane by the Axes/Planes/Centroids in Chimera. I don't need the distance of two centroid of two planes. I do need the vertical distance of two planes. Could you please show me in detail?
Thanks a lot!
Bing Wang
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