What two variables need to be set on the block cutter, prior to cutting a block?

On the block cutter you must set the target to tray distance (usually stays the same, unless the clinic has two different linacs), and you must set the target to image distance for the film you are cutting the block from (may vary from film to film).

To make things easier most clinics use a "standard" magnification factor (your target to image distance). If you are using a traditional simulator, sometimes it is more advantageous to use a different target to image distance. If you ever have to cut a block make sure you check the target to image distance for each film you are cutting from.

You need to cut a block. The target to tray distance is already set on the block cutter. The Simulation Technologist forgot to write down the target to image distance. The film has hash marks that represent a 1 cm scale at a distance of 100 cm.

Yes, there is away to find the target to image distance.

There are a number of ways to calculate this. One way is to take 8.0 cm represented on the film and with a ruler measure a distance of 12.0 cm. What is the target to image distance?

Your target to image disance would be 150 cm.

To do this problem all you have to do is set up a simple proportion. One was to do this is:

--------- = ---------

8 cm 12 cm

The 100 cm is your target to isocenter distance and, in this case the hash marks represent 1 cm at a distance of 100 cm. If you count out 9 hash marks, this correlates to 8 cm at 100 cm distance. If you measure the 9 hash marks and get an actual measurement of 12 cm on the film, then you can use the above proportion to solve for x and come up with the correct target to image distance.

On the film you have 11 hash marks, that represent 1.0 cm apart at a distance of 100 cm. With your ruler you measure 14.5 cm on the film for the 11 hash marks. What is the target to image distance?

For this one the target to image distance is 145 cm.

The reason I had you do a problem like this was to show you a trick. If you measure out 10 cm for a distance of 100 cm (pretty much all simulators), then your measurement is simply moving over a decimal place to get your target to image distance.

To do this problem all you have to do is set up a simple proportion. One was to do this is:

--------- = ---------

10 cm 14.5 cm

In the future when you measure 10 cm at 100 cm, you can just take your actual measurement (in this case 14.5 cm) and multiply it by 10, to give you 145 cm for your target to image distance.

Is the block you insert into the linear accelerator always going to be bigger or smaller then the block drawn on the film?

The block that goes into the machine is ALWAYS going to be smaller then the block drawn on the film.

Just think about divergence.

The fudical scale on the film indicates that the block projects to a size of 3.0 X 6.3 cm at 100 cm. What size will the physical size of the block be that goes into the linear accelerator, if that machine has a target to tray distance of 56 cm?

The physical block would measure 1.7 x 3.5 cm.

You can do this problem two ways, you can first set up a proportion and solve for each part of the block.

--------- = ---------

X 56 cm

--------- = ---------

X 56 cm

This gives you your 1.7 x 3.5 block. Another way to do this is to take 56 cm and divide that by 100 cm. This gives you a factor of .56, take this factor and multiply .56 by 3.0 cm and .56 by 6.3 cm. You will come up with the same answer.

(3.0 cm)(.56)= 1.7 cm

(6.3 cm)(.56)= 3.5 cm

Everything being equal, will a block in the tray holder be bigger, smaller, or the same size for a linear accelerator with a target to tray distance of 56 cm, when compared to a linear accelerator with a target to tray distance of 61 cm?

Everything being equal the block in the tray holder would be SMALLER for the target to tray distance of 56 cm.

This is because the 56 cm target to tray distance is farther away from the patient, and thus needs to be smaller if everything is equal compared to the 61 cm target to tray distance, which would be closer to the patient.

For a block to give very little penumbra, what characteristic must the block have?

The block must diverge to give very little penumbra.

How many half value layers must a block be?

The block must be at least 5 half value layers.

About what percent of dose gets though the block if the block is exactly 5 half value layers (ignore scatter)?

About 3%.

1 half value layer is 100%/2= 50%

2 half value layer is 50%/2= 25%

3 half value layer is 25%/2= 12.5%

4 half value layer is 12.5%/2= 6.25%

5 half value layer is 6.25%/2= **3.125%**

If 180 cGy is given to d-max, about what dose is the blocked tissue getting at d-max if the block is exactly 5 half value layers (ignore scatter)?

The blocked tissue would receive about 5.6 cGy. This is the dose from transmission only and is ignoring any scatter that might also contribute to this area.

Simply take 180 cGy and multiply it by the 3.125%.

(180 cGy)(.03125)= 5.6 cGy

THE END

*Author:* Adam Buell
*Editors:* Mary Hare
*Date Created:* March 2005