Phantom Images ACR CT Accreditation Pitfalls to Avoid in Phantom Testing

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ACR CT Accreditation
Pitfalls to Avoid in Phantom Testing
Phantom Images
• One set of phantom images (2 films) per CT
• Completed data sheets
Tom Payne PhD
Abbott Northwestern Hospital
Minneapolis, MN
First Step: Table 1
• Ensure that the
data matches what
they do clinically
• If AEC, need to
determine average
mAs
• If “effective” mAs,
need to calc mAs
and mA
• N x T = beamwidth
z-axis collimation (T) = the width of the tomographic section along the z-axis imaged by
one data channel. In multi-detector row (multi-slice) CTscanners, several detector
elements may be grouped together to form one data channel.
number of data channels (N)= the number of tomographic sections imaged in a single
axial scan.
maximum numb
er of data channels (Nmax) = the maximumnumber oftomographic
sections for a single axial scan.
increment (I) = the table increment per axial scan or the table increment per rotation of
the x-ray tube in a helical scan.
mA= the actual tube current used (averaged over one rotation)
exposure time = the time required for one complete rotation of the x-ray source.
Pitch
• Use IEC definition
– Pitch = I/N*T
• I = table increment/speed
• N = number of data channels used (not detector elements)
• T = z-axis collimation (thickness of one data channel)
– Not always stated correctly on the CT system
• Know underlying detector configuration
Phantom Positioning
Use CT alignment lasers
Optional base can make
life easier
Use bubble level to verify
pitch and roll
Ensure teflon rings off
module centers
Scan with slice >5mm to
check “longest” wires
High-resolution chest
technique
Must see all four BBs
(in Modules 1 & 4)
Longer wires should be
visible on both ramps
Alignment adjustment
WW = 1000
WL = 0
MODULE 1
Polyethylene
-97 HU
“Bone”
+910 HU
Water
0 HU
Acrylic
+120 HU
If FAIL, do not pass “go”, do not collect accreditation
Film 1, Boxes 4 - 12
CT Number Accuracy and Slice Thickness
• Align Module 1 to lasers
• Scan with “Axial” version Adult Abdomen protocol
• Record slice thickness and all CT numbers
– Water, air, polyethylene, bone and acrylic
• Scan varying slice thickness (Hi Res 3, 5, & 7 mm)
– Measure water CT number and slice thickness
• Scan varying kVp (scan with all available)
– Measure water CT number and slice thickness
Air
-1000 HU
Axial Conversion
•
If an axial acquisition cannot be made using that selection of
N and T, keep T the same as described in Table 1 and use the
next smallest allowed value of N.
Example: Siemens Sensation 16 system with N = 16 and T = 1.5
mm and reconstructed helical scan width = 5 mm. Axial images
cannot be acquired using N = 16. Use the same value of T (1.5
mm) but the next lowest allowed value of N, which would be 12.
Thus the 12 x 1.5 mm detector configuration would be used for
the axial version of the spiral adult abdomen protocol with N =
16 and T = 1.5 mm. This is similarly true for the 16 x 0.75 mm
detector configuration (use an axial 12 x 0.75 mm detector
configuration).
Adult abdomen
technique
Low contrast =
6 HU ± 0.5 HU
Poly
“Bone”
MODULE 2
2 mm
25 mm
“Use 120 kVp”
Provide a note if
regular technique
is 140 kVp
CT number
measured in 5
materials
3 mm
6 mm
Water
4 mm
Acrylic
Air
5 mm
WW = 400
WL = 0
Image Quality – Low Contrast
Low Contrast Resolution
• Scan Module 2
– Adult Abdomen protocol - Helical
– Adult Head protocol - Axial
• Record the diameter (mm) of the smallest set
of LCR rods seen
Adult abdomen
and head
techniques
Record diameter
of the smallest
set of rods for
which all 4 rods
can be seen
ROI check of
absolute contrast
using large rod
WW = 100
WL = 100
High Contrast Resolution
Bar patterns:
lp/cm
MODULE 4
12
4
10
6
• Scan Module 4
8
– Adult abdomen protocol – Helical
– Hi Resolution Chest protocol
9
5
• Record limiting resolution (lp/cm)
7
Adult abdomen,
adult head and
high-resolution
chest techniques
The End
Record the first
highest frequency
bar pattern for
which the bars
and spaces merge
Remember – in completing the ACR Physics Tests
“Artificial intelligence is no match for natural stupidity”
WW = 100
WL 1100
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