Why does it matter what the exact translation is -- the exact origin?
Why not just arbitrarily assign (0,0,0) to A0'?
Bob
Timothy Driscoll wrote:
>On May 11, 2007, at 2:09 PM, Bob Hanson wrote:
>
>
>
>>>original
>>> ATOM 1ATOM 2d(2-1)M*d
>>>x 3.95 3.86 -0.08 -1.25
On May 11, 2007, at 2:09 PM, Bob Hanson wrote:
>> original
>> ATOM 1ATOM 2d(2-1)M*d
>> x 3.95 3.86 -0.08 -1.25
>> y 38.05 36.58 -1.48 0.59
>> z 17.44 17.32 -0.11 0.53
>>
>>
>> transformed
>> ATOM 1ATOM 2d(2-1)
>> x 52.94 51.69
I think you've got it now.
Timothy Driscoll wrote:
>
>original
> ATOM 1ATOM 2d(2-1)M*d
>x 3.95 3.86 -0.08 -1.25
>y 38.05 36.58 -1.48 0.59
>z 17.44 17.32 -0.11 0.53
>
>
>transformed
> ATOM 1ATOM 2d(2-1)
>x 52.94 51.69 -1.25
>y
On May 11, 2007, at 7:46 AM, Bob Hanson wrote:
> what you will need to know are two points in each case, not one. What
> you don't know or aren't figuring in is the possibility that the
> origin
> has been shifted. Rotation requires a center. Maybe it's (0,0,0),
> maybe
> not. In Jmol what we
what you will need to know are two points in each case, not one. What
you don't know or aren't figuring in is the possibility that the origin
has been shifted. Rotation requires a center. Maybe it's (0,0,0), maybe
not. In Jmol what we have to do is:
1) apply a translation to set the "center of
On May 11, 2007, at 1:37 AM, Bob Hanson wrote:
> See if by any chance going the other way works -- from the verified
> coordinates to the original. If that's the case, then you need the
> inverse of this matrix.
>
>
yeah, I've been trying stuff like this, kinda haphazardly, without
luck. for ex
See if by any chance going the other way works -- from the verified
coordinates to the original. If that's the case, then you need the
inverse of this matrix.
Timothy Driscoll wrote:
>On May 10, 2007, at 3:54 PM, Bob Hanson wrote:
>
>
>>This is always a royal pain to get right. Certainly lo
give us a hint -- a couple of initial coordinates, a couple of final
coordinates, the matrix
Bob
Timothy Driscoll wrote:
>On May 10, 2007, at 3:54 PM, Bob Hanson wrote:
>
>
>>This is always a royal pain to get right. Certainly looks like a
>>rotation matrix to me.
>>It's got the right for
On May 10, 2007, at 3:54 PM, Bob Hanson wrote:
> This is always a royal pain to get right. Certainly looks like a
> rotation matrix to me.
> It's got the right form. My guess is you put the vector on the
> right and
> multiply as:
>
> x' = m11*x + m12*y + m13*z
>
> across the top row of that matr
This is always a royal pain to get right. Certainly looks like a
rotation matrix to me.
It's got the right form. My guess is you put the vector on the right and
multiply as:
x' = m11*x + m12*y + m13*z
across the top row of that matrix. Likewise for y' across the second.
Does that not work?
Bo
hi,
I have two structures that I have aligned using a program called
MUSTANG. the program gives me a 3x3 transformation matrix like this
one:
0.58 0.805 0.125
0.521-0.484 0.703
0.626-0.342 -0.7
in order to superimpose my second structure onto the
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