AMBER Archive (2009)Subject: Re: [AMBER] Need help with point to plane distance calculation in nab
From: M. L. Dodson (activesitedynamics_at_comcast.net)
Date: Thu Jan 08 2009  13:54:15 CST
IN SUK JOUNG wrote:
> I think I made a mistake here.
>
> The distance should be multiplied by sqrt( A^2 + B^2 + 1 ).
>
> That is, distance = abs( ( z  Ax  By  C ) / ( A^2 + B^2
> + 1 ) ) * sqrt( A^2 + B^2 + 1 );
>
Thanks, between this and your original reply, I should be able to
write a brief paragraph for my materials and methods. Thank you very
much.
Bud Dodson
> On Thu, Jan 8, 2009 at 12:16 PM, M. L. Dodson <
> activesitedynamics_at_comcast.net> wrote:
>
>> IN SUK JOUNG wrote:
>>
>>> My rough calculation gives...
>>>
>>> When the point is (x,y,z), distance = abs( ( z  Ax  By  C ) / ( A^2 +
>>> B^2
>>> + 1 ) )
>>>
>>> You should check if it is valid by yourself.
>>>
>>>
>> Thank you very much. Could I impose on you some more to describe in
>> words where the two terms come from? In other words, how you derived
>> the formula (or give me a reference if this is a well known
>> relationship)? I would be grateful for this as I want to understand
>> how it comes about. This will be a (minor) point in a publication,
>> and I need to justify the algorithm.
>>
>> Also, let me be sure I understand your nomenclature: In the expression
>> ( z  Ax  By  C ), are x, y, and z the coordinates of the single
>> atom (given by aex2) that I want to know the distance from the plane
>> for? In other words, what I am calling X and Y? (Z would be the z
>> coord for aex2.)
>>
>> Hope I am clear.
>> Thanks again.
>>
>> Bud Dodson
>>
>>
>> On Thu, Jan 8, 2009 at 11:39 AM, M. L. Dodson <
>>> activesitedynamics_at_comcast.net> wrote:
>>>
>>> Hello all,
>>>> I'm having a brain cramp coming up with the correct algebra for an
>>>> algorithm giving the distance from a point to a plane along a normal
>>>> to the plane. In nab, I can calculate the least squares best fit of a
>>>> plane of form z = Ax + By + C to a series of atomic positions
>>>> (represented by a nab atom expression, complete_aex1, in the molecule
>>>> mol) by using:
>>>>
>>>> plane(mol, complete_aex1, A, B, C),
>>>>
>>>> where complete_aex1 is an atom expression identifying the atoms whose
>>>> positions are to be fit by the plane.
>>>>
>>>> I want to calculate the distance from that plane to another atom
>>>> (given by aex2) ALONG A NORMAL to the plane.
>>>>
>>>> I need the correct POINT in the plane where the normal intersects. At
>>>> first I set the x and y coords of the POINT to be the x and y coords
>>>> of the atom given by aex2, say X and Y. Then set z coord of the POINT
>>>> to be z = AX + BY + C. Then I calculated the distance between the
>>>> position of atom aex2 and that point in the plane. But this (pretty
>>>> clearly, it seems to me), is NOT along a normal to the plane.
>>>>
>>>> Can any of you geometers out there hit me with a clue bat? An
>>>> algorithm will do. You do not need to know nab to answer.
>>>>
>>>> Thanks,
>>>> Bud Dodson
>>>>
>> 
>> M. L. Dodson
>> Business Email: activesitedynamicsatcomcastdotnet
>> Personal Email: mldodsonatcomcastdotnet
>> Phone: eight_three_two56_three386_one
>>
>
>
>

M. L. Dodson
Business Email: activesitedynamicsatcomcastdotnet
Personal Email: mldodsonatcomcastdotnet
Phone: eight_three_two56_three386_one
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