Shifting position of projector or volume

[JohPatata]

I am trying to do projection for a volume of 1024*1024*1024. However, due to the limitation of the GPU I use, only data size of 512*1024*1024 can be handled. Therefore I plan to do the projection for the upper part of the volume and then do the lower part later and eventually combine them at the end.

When I was trying to do so, I struggled with some parameters. I wanted to find out how the parameter Volume.center_in_world is defined. The way I am initialising carm projector is carm = deepdrr.device.MobileCArm(isocenter=volume.center_in_world, ...) and currently the origin of volume is [0., 0., 0.]. I printed volume.center_in_world but could not seem to find out what the value actually means.

The reason why this matter is that I plan to shift either the projector or the volume in a way that only half of the volume is scanned at each time and the other half of the cone beam geometry would not affect the current projection.

[benjamindkilleen]

The center_in_world is the center of the volume. So for your 512*1024*1024 volume, the center would correspond to the IJK location of (256, 512, 512).

What you are describing can actually be accomplished relatively easily with built-in functions. After loading the full volume from disk, but before initializing any projectors, you can use Volume.crop to get downsize to 2 separate volumes and initialize those on the projector separately (not at the same time).

When you try the full volume, what error message do you see? How large is your GPU? It has been a while since I used a GPU that was too small to handle even decently sized volumes. What resolution are you rendering at? If you have a minimal working example that may be helpful.

[JohPatata]

Hi Benjamin,

thank you so much for your solution! It worked to some extend, but the brightness of the two projections does not really match. An example is shown below. I am wondering if you have any insight into it.

Another problem I encountered is that the projector does not produce the correct projection when I rotate the sample. I was using the following code, but the last three images are the same. I have also tried different number of projections num_proj, but the last few images are always the same (for example, when num_proj = 10, the output image start to be the same at 4-th).

fdk = deepdrr.Volume.from_hu(hu_values = fdk, origin=[0., 0., 0.], spacing = [vox_spacing, vox_spacing, vox_spacing], anatomical_coordinate_system='LPS') 
upper = fdk.crop([[0, 513],[0,1024],[0,1024]])

num_proj= 5

with deepdrr.Projector(
    volume=upper,
    carm=carm,
    step=0.001,
    mode="linear",
    max_block_index=200,
    spectrum="90KV_AL40",
    photon_count=100000,
    scatter_num=0,
    threads=8,#8
    neglog=True,
) as projector:
    angles = np.linspace(0, 360, num_proj, endpoint=True)
    for angle in angles:
        carm.move_to(alpha = angle, beta = 0)
        image = projector.project()

Regarding to the error message I encountered when using the full volume 1024*1024*1024, an short example is provided as followed and all the variables are predefined.

fdk = deepdrr.Volume.from_hu(hu_values = fdk, origin=[0., 0., 0.], spacing = [vox_spacing, vox_spacing, vox_spacing], anatomical_coordinate_system='LPS') 

carm = deepdrr.device.MobileCArm(isocenter=fdk.center_in_world, source_to_detector_distance = source_2_detector_distance, source_to_isocenter_vertical_distance = source_2_center, pixel_size = pixel_size, sensor_height = num_pix, sensor_width=num_pix)

num_proj= 10

with deepdrr.Projector(
    volume=fdk,
    carm=carm,
    step=0.001,
    mode="linear",
    max_block_index=200,#200
    spectrum="90KV_AL40",
    photon_count=100000,
    scatter_num=0,
    threads=8,
    neglog=True,
) as projector:
    angles = np.linspace(0, 360, num_proj, endpoint=True)
    for angle in angles:
        carm.move_to(alpha = angle, beta = 0)
        image = projector.project()

The GPU I am currently using has a memory of 11 GB and for the step in deepdrr.Projector, I had to set to 0.001 to achieve desirable resolution. The error message goes as followed, where line 75 of deepdrr_projector_no_metal.py is with deepdrr.Projector(. If the size of input volume is 513*1024*1024, roughly 9 GB would be occupied.

Traceback (most recent call last):
  File "/users/Zijin_Huang/Work/trying_stuff/deepdrr_projector_no_metal.py", line 75, in <module>
    with deepdrr.Projector(
  File "/buffer/miniconda3_python39/envs/deepdrr/lib/python3.10/site-packages/deepdrr/projector/projector.py", line 1729, in __enter__
    self.initialize()
  File "/buffer/miniconda3_python39/envs/deepdrr/lib/python3.10/site-packages/deepdrr/projector/projector.py", line 1507, in initialize
    texobj, texarr = create_cuda_texture(seg)
  File "/buffer/miniconda3_python39/envs/deepdrr/lib/python3.10/site-packages/deepdrr/projector/projector.py", line 201, in create_cuda_texture
    cuda_array = cupy.cuda.texture.CUDAarray(format_descriptor, *(texture_shape[::-1]))
  File "cupy/cuda/texture.pyx", line 298, in cupy.cuda.texture.CUDAarray.__init__
  File "cupy_backends/cuda/api/runtime.pyx", line 519, in cupy_backends.cuda.api.runtime.malloc3DArray
  File "cupy_backends/cuda/api/runtime.pyx", line 146, in cupy_backends.cuda.api.runtime.check_status
cupy_backends.cuda.api.runtime.CUDARuntimeError: cudaErrorMemoryAllocation: out of memory

[benjamindkilleen]

Right now you are applying negative log likelihood to the whole image, which is then normalized by [0,1] by image min/max. You want raw intensity values, so set neglog=False, and make sure intensity_upper_bound is None. You should apply neglog after the fact for visualization.

Regarding the rotation, the MobileCArm class is meant to emulate a specific physical C-arm (the Siemens CIOS Fusion), which has min/max values for rotation. Most likely, you are running into the limit. You can adjust the limits when you instantiate it.

Is this a micro CT? Curious why you need 0.001 mm step size!

[JohPatata]

Thank you so much for your answer! The problem is now solved :)

Yes, my data are micro CT data, that's why I need a smaller step size. Just for curiosity, could you explain a bit more about the meaning of the step size? I did not really understand it from the description.

[benjamindkilleen]

It's the size of dx in the line integral.