GS020113: Introduction to Medical Physics III: Therapy
Answers to home work problem set assigned on 3/22/11
1. Question
A patient is set up at 100 cm SSD on a 6 MVX machine. The dose rate at 10 cm in
phantom is 1 cGy/MU; F.S. = 10 cm x 20 cm @ 10 cm. What is the dose rate at 5
cm? Use each method (%DD, TMR)
Answer
(A) %DD method
(a) Equivalent square for FS (field size) of 10 cm x 20 cm = 2 x (20 x 10) / (20+10) =
13.3 cm at SAD of 110 cm
(b) Field size at 100 cm SSD = 13.3 x 100 / 110 = 12.1 cm
(c) PDD (r=12.1 cm, d= 10 cm, SSD = 100 cm) = 67.9
(d) dose rate at dmax = dose rate at 10 cm x 100 / PDD in A.c = 1 x 100 / 67.9 =
1.473 cGy / MU (Note: This dose rate is taken for the given field size and depth, and
is not the dose rate for beam calibration in reference condition)
(e) PDD (r= 12.1 cm , d= 5 cm, SSD = 100 cm) = 86.8
(f) Dose rate at 5 cm = dose rate at dmax x PDD in A.e / 100= 1.473 x 86.8 / 100 =
1. 279 cGy / MU
(B) TMR method
(a) SAD for a point at 10 cm depth with 100 cm SSD = 100 +10 = 110 cm
(b) Equivalent Square FS at 110 cm SAD = 2 x (20 x 10)/ (20+10) = 13.3 cm
(c) TMR (r = 13.3 cm, d=10 cm) = 0.796
(d) Dose rate at dmax with 110 cm SAD = dose rate at 10 cm for 110 cm SAD / TMR
in B.c = 1 / 0.796 = 1.256 cGy / MU and it is for FS of 13.3 cm
(e) SAD at 5 cm depth for 100 cm SSD = 105 cm
(f) FS at 105 cm SAD for FS of 13.3 at 110 cm SAD = 13.3 x 105 / 110 = 12.7 cm
(g) Dose rate at dmax for FS of 12.7 cm at 110 cm SAD obtained from dose rate for
FS of 13.3 cm at 110 cm SAD = Sp (12.7 cm) x 1. 256 / Sp (13.3) = 1.007 x 1.256 /
1.009 = 1.254 cGy / MU (Note: this is a very small correction for the FS and is often
neglected. )
(h) Dose rate at dmax for 105 cm SAD for FS of 12.7 cm calculated from dose rate at
110 cm SAD for FS 12.7 cm SAD = 1.254 x (110/105)2 = 1.376 cGy / MU
(i) TMR (r=12.7 cm, d=5 cm) = 0.929
(j) Dose rate at 5 cm depth for 105 cm SAD for FS 12.7 cm = dose rate from B.h x
TMR in B.i = 1.376 x 0.929 = 1.278 cGy / MU.
2. Question
If 6 MV X-rays deliver 5000 cGy to midplane of a 20 cm thick patient with 2:1
weighted parallel opposed fields, what is the maximum dose in the patient? F.S. = 20
cm x20 cm at 10 cm; SSD = 100cm.
Answer
(a) Dose from the beam with weight of 2 (named as Field 1) is 5000 x 2 / 3 = 3333.3
cGy
(b) Maximum dose from this field will occur at the dmax
(c) FS at the surface calculated from square field size of 20 cm at 10 cm depth and
100 cm SSD = 20 x 100 / 110 = 18.2 cm (square field)
(d) PDD (r= 18.2 cm, d=10 cm, SSD =100 cm) for 6 MVX beam = 69.6
(e) Dose at dmax for the 6 MVX beam with 18.2 cm x 18.2 cm field delivering 3333.3
cGy at 10 cm = 3333.3 x 100 / 69.6 = 4789.2 cGy
(f) Dose at dmax for the 6 MVX beam with 18.2 cm x 18.2 cm field delivering 1666.7
cGy at 10 cm = 1666.7 x 100 / 69.6 = 2394.7 cGy
(g) Table for dose at different depths from the entrance of Field 1
Depth from PDD
Dose (cGy)
Depth from
PDD Dose (cGy)
Total
Field 1
from Field 1 Field 2 entrance
from Field 2
Dose
entrance
= dose at dmax
(cm)
= dose at dmax (cGy)
(cm)
x 100 / PDD
x 100 / PDD
1.4
100
4789.2
18.6
45.4
1087.2
5876.4
2.0
98.8
4731.7
18.0
46.9
1123.1
5854.8
3.0
95.2
4559.3
17.0
49.3
1180.6
5739.9
5.0
87.4
4185.8
15.0
54.5
1305.1
5490.9
7.0
80.1
3836.1
13.0
60.3
1444.0
5280.1
10.0
69.6
3333.3
10.0
69.6
1666.7
5000.0
13.0
60.3
2887.9
7.0
80.1
1918.2
4806.1
15.0
54.5
2610.1
5.0
87.4
2093.0
4703.1
18.6
45.4
2174.3
1.4
100
2394.7
4569.0
(e) Maximum dose occurs at the dmax of Field 1 as shown in the Table above and is
5876.4 cGy
3. Question
If a dose of 100 cGy is delivered from a 6 MVX beam to a point located on the
central axis of the beam at a depth of 15 cm in water with a SAD of 100 cm and field
size of 15 cm x 15 cm at 100 cm, what is the maximum dose along the central axis,
and the dose at a depth 25 cm along the central axis.
Answer
(a) SSD for this field is 100 – 15 = 85 cm
(b) Field size at SSD 85 cm = 15 x 85 / 100 = 12.8 cm
(c) PDD ( d=15 cm, r = 12.8 cm, SSD = 85 cm) for the 6 MVX beam = PDD (d=15
cm, r = 12.8 cm, SSD = 100 cm) x ((85 + 1.5)/ (100+1.5))2 x ((100 +
15)/(85+15))2 = 52.4 x 0.726 x 1.3225 = 50.3, corrected by Mayneord F factor
only, the scatter correction is neglected
(d) The dose at dmax = 100 x 100 / PDD in (c) = 100 / 0.503 = 198.8 cGy
(e) PDD ( d=25 cm, r = 12.8 cm, SSD = 85 cm) for the 6 MVX beam = PDD (d=25
cm, r = 12.8 cm, SSD = 100 cm) X ((85 + 1.5) / (100+1.5))2 x ((100 +
25)/(85+25))2 = 30.5 x 0.726 x 1.291 = 28.6, corrected by Mayneord F factor
only, the scatter correction is neglected
(f) Dose at 25 cm = dose at dmax in (g) x PDD at 25 cm in (e) / 100 = 198.8 x 28.6 /
100 = 56.9 cGy.
Alternatively, one can use the relationship between PDD and TMR, ( PDD (d,r,f) =
TMR (d, rd) x (Sp(rd)/Sp(rdmax)) x ((f+dmax)/(f+d))2) to get the PDD for the desired
SSD, which includes the differences in the scatter.
(g) Thus, PDD (d=15 cm, r = 12.8 cm, SSD = 85 cm) = 100 x TMR (d=15cm, r =
15cm) x (86.5/100)2 x Sp(15 cm)/Sp(13 cm) = 100 x 0.672 x 0.748 x 1.0125
/1.008 = 50.5 (note field size at dmax of 1.5 cm is 13 cm)
(h) Dose at dmax = 100 / 0.505 = 198.0 cGy
(i) SAD at 25 cm depth = 110 cm
(j) FS at 110 cm = 15 x 110 / 100 = 16.5 cm
(k) PDD (d=25 cm, r = 12.8 cm, SSD = 85 cm) = 100 x TMR (d=25 cm, r =16.5 cm)
x (86.5/110)2 x Sp(16.5 cm)/Sp(13 cm) = 100 x 0.463 x 0.618 x 1.016 /1.008 =
28.8
(l) Dose at 25 cm = dose at dmax in (h ) x PDD at 25 cm in (k) / 100 = 198.0 x 28.8 /
100 = 57.0 cGy
4. Question
The PDD at a depth of 10 cm in a Varian 6MV, 100-cm SSD, 10 cm x 10 cm field is
66.8%. What is the PDD at this depth in a 120-cm SSD? (Assume the 10x10 field size
represents the size of the field at the SSD.)
Answer
(a) The field size is assumed to be same at the surface for both the 100 cm and 120
cm SSD
(b) The PDD at the 10 cm depth for the 10 cm x 10 cm field at 120 cm SSD can be
obtained by multiplying the PDD at the 10 cm depth for the 10 cm x 10 cm field
at 100 cm SSD with the Mayneord F factor with the scatter correction being
neglected.
(c) Thus, PDD (d=10 cm, r=10 cm, SSD = 120 cm) = PDD (d=10 cm, r=10 cm, SSD
= 100 cm) x ((120 + 1.5)/(100+1.5))2 x ((100+10)/(120+10))2 = 66.8 x 1.433 x
0.716 = 68.5
Note: As mentioned in page 188 of the Reference text, the scatter correction using the
TAR ratio is small and is often neglected. The TAR ratio can be replaced by TMR
and Sp ratios, and one can use the available values of TMR and Sp ratios from the 6
MVX beam data to derive the PDD with scatter corrections as shown below.
For same energy,
TAR (d, rd) = TMR (d, rd) x PSF (rd)
TAR(d, rd1) / TAR (d, rd2) = TMR(d, rd1) x PSF (rd1) / (TMR(d, rd2) x PSF (rd2))
= TMR(d, rd1) x (PSF (rd1) / PSF (ref)) / (TMR(d, rd2) x (PSF (rd2) / PSF (ref)))
= TMR(d, rd1) x Sp (rd1) / (TMR(d, rd2) x (Sp (rd2))
In this problem, d= 10 cm,
rd1 = 10 x 110 (source to point distance) / 100 (SSD) = 11 cm, rd2 = 10 x 130 (source
to point distance) / 120 (SSD) = 10.8 cm
Scatter correction = TMR (d =10 cm, r =10.8 cm) x Sp(10.8 cm) / (TMR(d = 10 cm, r
= 11 cm) x Sp(11 cm) = 0.785 x 1.002 / (0.786 x 1.003) = 0.998, which is very close
to 1.000.
The PDD in (c) with scatter correction = 68.5 x 0.998 = 68.4.
5. Question
A Varian 6 MV beam has a collimator setting of 15 cm x15 cm and a water phantom
is set beneath it at 100 cm SSD and the dose at 10 cm depth along the central axis is
100 cGy.
a. What is the dose at this depth if the collimator setting remains same, but the field is
blocked to produce a 12 cm x 12 cm field at the surface?
b. What is the dose at 20 cm depth at a point located 5 cm away from the central axis
measured perpendicularly to central axis at 100 cm from the source from the above
blocked field?
Answer
I. As the SSD remains fixed to that of 100 cm SSD for which the beam data for 6
MVX beam is tabulated, the use of PDD method is appropriate for this problem. In
(a), the scatter factor is changed from that for field size of 15 cm x 15 cm to 12 cm
x 12 cm. Assuming that the Sp data is tabulated for field size at the surface for
fixed SSD calibration of output for the reference field and depth, the new Sp (12
cm) = 1.006 as compared to Sp (15 cm) = 1.013 for the open field
II. The PDD for the blocked field is: PDD (d=10 cm, r=12 cm, 100 cm SSD) = 67.9 as
compared to PDD (d=10 cm, r=15 cm, 100 cm SSD) = 68.9
III. Sp (r = 15 cm ) = 1.013, Sp (r = 12 cm ) = 1.006
IV. For the same MU needed to deliver 100 cGy at the depth of 10 cm with the open
field, the dose at 10 cm depth with the blocked field is given by 100 x PDD (d=10
cm, r=12 cm, 100 cm SSD x Sp (r = 12 cm ) / (PDD (d=10 cm, r=15 cm, 100 cm
SSD) x Sp (r = 15 cm )) = 100 x (67.9 x 1.006 ) /(68.9 x 1.013) = 97.9 cGy.
V. For (b), PDD (d=20 cm, r=12 cm, 100 cm SSD) = 39.7
VI. Dose at dmax for the blocked field = 97.9 x 100 / 67.9 = 144.2 cGy
VII. At the 5 cm off-axis distance, the dose will be affected by the off-axis distance
factor. The given table lists the off-axis factors for a 40 cm x 40 cm field at various
off-axis distances and depths. The off-axis factor for the blocked field can be
estimated by using the ratio, off-axis distance / field size = 5 / (12/2 ) = 0.833,
which corresponds to 0.833 x (40/2) = 16.7 cm off-axis distance in the given table.
The OAF at 20 cm depth = 0.964
VIII. The dose at the depth of 20 cm for the blocked field will be dose at dmax x (PDD
(d=20 cm, r=12 cm, 100 cm SSD) / 100) x OAF = 144.2 x (39.7 /100 ) X 0.964 =
55.2 cGy