Brain Teaser #12
A) Question:
In radiation therapy, when shifting the isocenter location in a patient, how many planes are there that the shift could take place in?
A) Answer:
There are three planes that a shift could occur in.
B) Question:
According to anatomical orientation (not in a coordinate system), what are the three (3) planes?
B) Answer:
The three (3) planes are superior to inferior, left to right, and anterior to posterior.
Follow Up:
All the planes can also be explained in a coordinate system. The coordinate system can change between treatment planning computers, but they usually use x, y, and z. For example:
x may be the right to left plane
y may be the superior to inferior plane
z may be the anterior to posterior plane
Again, this may be dependent on the planning system and institutionaly facility preference.
C) Question:
When looking at a single transverse slice, what two planes are being observed?
C) Answer:
With a single transverse slice, the planes being observed are anterior/posterior and right/left.
Follow Up:
Hopefully this drawing will help. This drawing shows a transverse slice with a patient being supine:
D) Question:
When looking at a single sagittal slice, what planes are being observed?
D) Answer:
When looking at a single sagittal slice, the planes observed are superior/inferior and anterior/posterior.
Follow Up:
Hopefully this drawing will help. This drawing shows a sagittal slice with a patient being supine:
E) Question:
What planes are being observed in a single coronal slice?
E) Answer:
When looking at a single coronal slice, the planes observed are superior/inferior and left/right.
Follow Up:
Hopefully this drawing will help. This drawing shows a coronal slice with a patient being supine:
F) Question:
What planes are observed on a Rt-Lat film, and what planes are observed on an AP film?
F) Answer:
A Rt-Lat film shows the superior/inferior planes and anterior/posterior planes.
An AP film shows the superior/inferior planes and left/right planes.
G) Question:
You are doing a simulation, and the current field size is 10 x 14 (width x length). The doctor wants you to keep the inferior border where it is, and increase the superior border by 2.5 cm. You don't have independent jaws. What will the shift be from the original iso center, and what direction will the shift be?
G) Answer:
The shift will be 1.25 cm superior.
Follow Up:
Hopefully this drawing will help you. To tackle this problem drawing it out may help, it is important to establish what plane we will be making the shift in. Since we will be keeping the inferior boarder, we will be making the shift in the inferior/superior plane. Next we will determine what we will need to open the field length to. The field length was 14 cm, and if we open it up symmetrically to 16.5, it opens up the superior border 1.25 cm, and it opens up the inferior boarder 1.25 cm. We are suppose to keep the inferior boarder where it is. Thus if we shift 1.25 cm superiorly we properly open the superior border 2.5 cm, and keep the inferior border where it was.
H) Question:
You are doing a simulation and the field size for the Rt-Lat is 10 x 12 (width by length). The doctor wants you to keep the posterior border and increase the anterior board by 3.0 cm. Again you have symmetric jaws. What will the new field size be and what will the shift from the original central axis be (please give the direction of the shift, too)?
H) Answer:
The new field size will be 13 x 12 cm (width x length) and the shift would have been 1.5 cm anterior.
I) Question:
When looking at a transverse slice, a patient has a separation of 40 cm in the lateral plane (right to left) and 30 cm in the anterior to posterior plane. Isocenter is currently located in the midplane of the patient. Both field sizes are 10 x 12 (width x length). If the doctor wants to increase the left border on the AP field by 2.0 cm and keep the right border what will the new SSDs be? To make matters worse the doctor wants to change the lateral field size to 14 keeping the anterior border. Please list all the new SSDs and, again, assume that you have symmetric jaws.
I) Answer:
Rt-Lat SSD will be 79
Lt-Lat SSD will be 81
AP SSD will be 83
PA SSD will be 87
Follow Up:
Since we are midplane, we know that we start out with an SSD of 80 cm for Rt and Lt laterals, and a SSD of 85 cm for the AP and PA fields.
First, we will worry about the anterior/posterior move. The doctor wants to keep the anterior border and make the field 14.0 cm wide, thus effectively making the posterior border increase by 4.0 cm. This means we have to open the jaws 4.0 cm, and this opens the jaws equally to 2.0 cm on the anterior border and posterior borders. Again, we want to keep the anterior border where it is. Thus, we must shift the isocenter 2.0 cm posterior to move the anterior back to where it belongs. So, the shift is 2.0 cm. Thus, we know the SSDs will change by 2.0 cm for both AP and PA fields.
Since we are moving the field posteriorly the SSD for the AP will be:
half of the original separation + 2.0 cm
this will be 15 cm + 2 cm = 17 cm.
Take 100 cm - 17 cm and you get an SSD of 83 cm.
The SSD for the PA will be:
half of the original separation - 2.0 cm
this will be 15 cm - 2 cm = 13 cm.
take 100 cm - 13 cm and you get an SSD of 87 cm.
Now we will worry about the left/right move. The doctor wants to keep the right border and make the left border increase by 2.0 cm. This means we have to open the jaws 2.0 cm, and this opens the jaws equally to 1.0 cm on the left border and the right border. Again, we want to keep the right border where it is. Thus, we must shift 1.0 cm left to move the right border back to where it belongs. So, the shift is 1.0 cm. Thus, we know the SSDs will change by 1.0 cm for both Rt-Lat and Lt-Lat fields. Since we are moving central axis towards the left, the SSD for the Rt-Lat will be:
half of the original separation + 1.0 cm
this will be 20 cm + 1 cm = 21 cm.
Take 100 cm - 21cm and you get an SSD of 79 cm.
The SSD for the Lt-Lat will be:
half of the original separation - 1.0 cm
this will be 20 cm - 1 cm = 19 cm.
take 100 cm - 19 cm and you get an SSD of 81 cm.
This is what it looks like when we combine the shifts properly:
J) Question:
If the doctor asks for a patient's field to be shifted 2.0 cm superiorly, will the table be moved towards the gantry or away from the gantry (assume this patient's head is toward the gantry)?
J) Answer:
To move central axis superiorly, the table will have to be moved away from the gantry. When you bring the table out (away from gantry), central axis will then move superiorly.
K) Question:
If the doctor asks for a field shift of 1.0 cm posteriorly, what direction will the table be moved, up or down (assume patient is supine)?
K) Answer:
To move central axis posteriorly, the table will have to move up (vertically).
THE END
Author: Adam Buell
Editors: Mary Hare
Date Created: November 2005
