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CCP4BB  January 2015

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Subject:

Re: Redundancy vs no of frames

From:

James Holton <[log in to unmask]>

Reply-To:

James Holton <[log in to unmask]>

Date:

Mon, 19 Jan 2015 17:15:19 -0800

Content-Type:

text/plain

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Parts/Attachments

text/plain (56 lines)

If you spin the crystal through a full 360 degrees then any given "relp" 
(unique reciprocal lattice point) will pass through the Ewald sphere no 
more than twice.  This is because the Ewald sphere is a surface and the 
relp is taking a circular path.  It is either "inside" the Ewald sphere 
or "outside", and transitions once in-to-out and once out-to-in in a 
closed loop.  Some relps are always outside the Ewald sphere and their 
circular paths are too small to ever enter into it.  These tend to be 
close to the rotation axis, and the reason why it is formally impossible 
to get 100% completeness when your space group is P1 and you only have 
one rotation axis.  You can keep spinning, of course, and then your 
"redundancy" is 2x the number of spins, but since you are using the same 
pixels over and over again, it isn't really "multiplicity", to use the 
flippant definition from my previous post.

Now, of course, multiplicity/redundancy generally includes symmetry 
mates, and even in P1 you have Friedel symmetry.  The circle traced out 
by the Friedel mate (-h,-k,-l) is a mirror image of the (h,k,l) circle 
in the plane normal to the rotation axis. That is, -h,-k,-l is always on 
the opposite side of the origin from h,k,l and also at the same 
d-spacing, so its circle is the same radius and the same distance from 
the origin as the h,k,l circle, just on the opposite side of the beam.  
This circle also crosses the Ewald sphere twice, and always on different 
pixels than h,k,l.

If your space group is not P1, then your multiplicity per revolution 
will therefore be 4*n, where n is the number of real-space symmetry 
operations.  That is, the number of x,y,z-ish lines you see under the 
space group in ${CLIB}/symop.lib is the "n" that you want.

If you don't do a full 360-degree rotation, then the relationship to 
multiplicity gets "noisier", you may see some spots more than once long 
before you have even one of another.  But, if you want a rule of thumb, 
your "coverage" of reciprocal space is roughly 4*n*revolutions, where 
"revolutions" can be a fraction.

As for the "right" definition of redundancy vs multiplicity, people seem 
quite adamant to stick with whatever term is used in the log file of 
their favorite processing program denzo uses "redundancy", but 
scala/aimless use "multiplicity". Funny how terms get "defined" this 
way.  Perhaps the best way to change terminology for good is to write a 
really amazing computer program that everyone will have to use and make 
it print out the term you like?

-James Holton
MAD Scientist

On 1/17/2015 1:05 PM, rohit kumar wrote:
>
> Dear all,
>
> Can anyone tell me how to calculate number of frames from redundancy 
> or vica versa
>
> Thank you
>

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