Finding the coordinates of a points of 2D projection in 3D volume

Hi,

I already got the projection of a 3D image on the axial plane, so now, I have the 2D projection of that image. I selected some points on the 2D projection and now I want to find the location of those points on the 3D image. Does any one know how I can find the coordinates of those points?

[darova]

Can you attach the images and sketch? What points do you want to select?

[Bahareh Salafian Esfahani]

The image with star points are those I want to find in 3D image and the image that I showed it in 3D slicer is my 3D image. I had coordinates of star points.

[darova]

What about interp3?

Answer 1

[Charles Mistretta]

I am not good matlab programmer but maybe this will get you started. I hav a 3d volume called temp

I make mips in all three directions and project the intensity values (1) throughout the volume

When I add up all these backprojected values I get three where the bacprojections overlap

Then I apply a function that gives the coordinates at the minimum

%find 3d coordinates of point identfied in mipt images through volume

% multiply through the volume at point identified in three mip views

%start wth volume called temp

load temp

tempmip2=squeeze(max(temp,[],2));

figure, imshow(tempmip2(:,:),[])

title 'temp2'

tempmip1=squeeze(max(temp,[],1));

figure, imshow(tempmip1(:,:),[])

title 'temp1'

tempmip3=squeeze(max(temp,[],3));

figure, imshow(tempmip3(:,:),[])

title 'temp3'

drawnow

%make all necessary volumes the same

tempmult1=zeros(512,512,394);

tempmult2=zeros(512,512,394);

tempmult3=zeros(512,512,394);

tempmult12=zeros(512,512,394);

tempmult123=zeros(512,512,394);

%multiply throughout volume by value in mip

for x=1:512

x

tempmult1(x,289,15)=tempmip1(289,15);

end

max(max(max(tempmult)))

for y=1:512

y

tempmult2(196,y,15)=tempmip2(196,15);

end

max(max(max(tempmult)))

tempmult12(:,:,:)=tempmult1(:,:,:)+tempmult2(:,:,:);

max(max(max(tempmult12)))

for z=1:394

z

tempmult3(:,:,z)=tempmip3(197,296);

end

max(max(max(tempmult3)))

% add all the volume They should add at the point related to the points

% chosen in the mips

tempmult123(:,:,:)=tempmult1(:,:,:)+tempmult2(:,:,:)+tempmult3(:,:,:);

max(max(max(tempmult123(:,:,:))))

%this was a function previously given to me someone named Walter-sorry -lost

%last name

[~, linearidx] = max(tempmult123(:));

[r,c,p] = ind2sub(size(tempmult123), linearidx);

%this give 3d coords corresponding to points chosen in MIPs

[r,c,p]

% the points identified in the three mip views might not be consistent so it might be good to do some blurring along the way

Answer 2

[Charles Mistretta]

send e-mail to [e-mail] and I will send you the volume called temp which is an angiogram

Answer 3

[Charles Mistretta]

The script I sent was probably too complicated.You can just get the coordinates by placing your points in the three views. The key thing is that you need three views to tease ut the coordinates. One 2d image wont provide that informationn

Answer 4

[Tim]

What kind of projection are you using? Only some projections allow for direct correspondence between values in the projection and locations in the volume (e.g. MIP values can be associated with volume indices, assuming there are no repeat values, but averages cannot). If you are using MIP's, the easiest way is simply to do something like:

[Projection, Indices_of_max_along_MIP_axis] = max(volume, [], MIP_axis);

Use the indices of the feature selected in the projection to look up the indices along the MIP axis. The other indices are just the indices along the first and second dimensions of the projection for the value you are selecting.

But this only works for MIP's, and not even then sometimes (imagine a uniform value cube). The best way to locate a volume is to use an actual volume visualizer, vs. working with a projection.