SPM99 Gem 15: Computing Cerebral Volume (VBM)

Subject:      Re: smoothed modulated image
From: John Ashburner <john@FIL.ION.UCL.AC.UK>
Date:         Fri, 12 Apr 2002 13:45:01 +0000
To: SPM@JISCMAIL.AC.UK

> considering a smoothed modulated image, which is the right
> interpretation of the matrix:  "each value of the matrix denotes the
> volume, measured in mm3, of gray matter within each voxel" or "each
> value of the matrix is proportional to the volume, measured in mm3,
> of gray matter within each voxel" or something else? 

The contents of a modulated image are a voxel compression map
multiplied by tissue belonging probabilities (which range between zero
and one).

The units in the images are a bit tricky to explain easily (so I would
suggest you say that intensities are proportional).  To find the
volume of tissue in a structure in one of the modulated images, you
sum the voxels for that structure and multiply by the product of the
voxel sizes of the modulated image.

The total volume of grey matter in the original image can be
determined by summing the voxels in the modulated, spatially
normalised image and multiplying by the voxel volume (product of voxel
size).

For example, try the following code for an original image and the same
image after spatial normalisation and modulation. Providing the
bounding box of the normalised image is big enough, then both should
give approximately the same answer.

V = spm_vol(spm_get(1,'*.img'))
tot = 0;
for i=1:V(1).dim(3),
        img = spm_slice_vol(V(1),spm_matrix([0 0 i]),V(1).dim(1:2),0);
        tot = tot + sum(img(:));
end;
voxvol = det(V(1).mat)/100^3; % volume of a voxel, in litres
tot    = tot; % integral of voxel intensities

tot*voxvol


I hope the above makes sense.

All the best,
-John

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s