Photon Beam MonitorUnit Calculations
Introduction to Medical Physics III:
Therapy
Steve Kirsner, MS
Overview
 Introduction
 General Formalism for MU Calculations
 Linear Accelerator MU Calculations
 SSD Formalism- Equations
 SAD Formalism - Equations
 Important Facts
 Summary: Equations SSD and SAD
 Examples
Introduction
 Standard calibration geometry
 Linear Accelerators are calibrated under standard
conditions. These standard conditions enable us to
know the absolute dose under these conditions
 At MD Anderson, and commonly elsewhere, this point
is located at dmax in a water phantom, 100 cm SSD,
along the central axis of an open 10x10 field. Other
option is defined at 100 cm SAD.
 At this point, the unit is calibrated such that 1 monitor
unit (MU) is equal to 1.0 cGy muscle
Introduction
 Corrections Needed if not at Standard Geometry
 Depths other than dmax, and SSDs other than 100
cm, and for field sizes other 10x10, and points off the
central axis, corrections become necessary
 These corrections are found in an institutions clinical
data tables. Where these relationships to the standard
geometry have been established
 Corrections are also necessary to account for anything
that is placed in the beam that will attenuate the
radiation. Wedges, compensators, blocks.
Corrections to standard geometry

 Depth corrections
 Field-size corrections 
PDD, TAR, TMR,TPR
Output (scatter factors)
 ST, SP, SC
 Distance corrections
 Off-axis corrections
 Attenuation
corrections
 Inv. Sq.
 “SAD Factor”
 OAFs-wedge and open
 WFs, TFs, compensator
factors
Formalism
 In general, the dose (D) at any point in a water
phantom can be calculated using the following
formalism:
D  MU  O  OF  ISq DDF  OAF  TF
 Where:




MU = monitor-unit setting for given conditions
O = calibrated output (cGy/MU) for standard conditions
OF = output (scatter) factor(s): SC, SP, ST
ISq = inverse-square correction (as needed) depending on
calibration conditions and treatment conditions.
 DDF = depth-dose factors (PDD, TMR, TPR, TAR)
 OAF = off-axis factors, open and wedge
 TF = transmission factors-attenuation
SSD Treatments and Calibration
 When the treatment unit is calibrated in a “SSD”
geometry, then for “SSD” beams, the formalism
becomes:
D  MU  SC  SP  PDD  OAF  TF
 where it is assumed that output (scatter) factors are
given by SC and SP, and where it is also assumed that
the calibrated output = 1.0 cGy/MU for a 10 x 10 field at
dmax
 Note that no inverse-square term is needed since the
distance to the point of dose normalization (SSD + dmax)
is equal to the distance to the point of dose calibration.
This is true unless treating at extended ssd. Then inverse
square is given by: (scd/(ssd +dmax))2
SAD Setup-SSD Calibration
 When the treatment unit is calibrated in a “SSD”
geometry, then for “SAD” (isocentric) beams, the
formalism becomes:
D  MU  SC  SP  ISq TMR  OAF  TF
 where the inverse-square factor accounts for the change
in output produced by the differences in the distances
between the source and the point of calibration (SCD)
and between the source and the point of normalization
(SPD):

ISq  SCD

SPD
2
Important Facts to Remember
 The inverse-square term of the SAD equation
accounts for the increased output that exists at the
isocenter distance relative to the output that exists at
“isocenter + dmax” (where the machine output is 1
cGy/MU) due to calibration conditions chosen.
 This inverse square factors is sometimes called the
“SAD Factor”, not to be confused with other inverse
square factor. This is solely adjust for calibration
conditions.
 For 6 MV, the SAD Factor is:

ISq  FSAD  SCD
 

 100 1.5
 1.030
SPD
100
2
2
More Important Facts
 Field sizes, unless otherwise stated, represent
collimator settings
 For most accelerators, field sizes are defined at 100 cm
(the distance from the source to isocenter)
 For SSD beams, field sizes are defined at the surface
(normally 100 cm SSD)
 For SAD beams, field sizes are defined at the depth of
dose calculation (normally 100 cm SAD)
 For field sizes at distances other than 100 cm, field sizes
must be computed using triangulation:

FSSSD , d  FS100  SSD  d

100
Points to Remember
 Depth Dose and Scatter Factors
 SC is a function of the collimator setting
 SP is a function of the size of the field:
 at the phantom surface for SSD beams
 at the depth of calculation for SAD beams
 Depth-dose factors are a function of:
 field size at the phantom surface for SSD beams
 field size at depth for SAD beams
Prescription Dose
 Calculate Monitor units per field for a given
Prescription dose.
 This dose is “prescribed” by the Physician.
 Value must be known at the point of calculation.
 With multiple fields, the dose per field is calculated
using the beam weights.
 If a dose DRx is prescribed through multiple fields
i each having a relative weight wti, then the dose
Di from each field is:
 wti


Di  DRx  
 i wt 
Prescription Dose
 If the physician then prescribes the dose
to a specific isodose line. The dose per
field then becomes:
 Di/IDL where IDL is the isodose line that is
prescribed to.
Calculation Equations
 For SSD beams: Inverse square if extended
SSD not shown. If extended SSD add inverse
square. (scd/(ssd +dmax))2
Di
MU i 
SC  SP  PDD  OAF  TF
 For SAD beams:
Di
MUi 
SC  SP  ISq TMR  OAF  TF
Examples
 A patient is planned to deliver a four field
box. The weightings of the beams are as
follows:
 AP=25%, PA = 20%, Rt lat=25%, Lt. Lat=
30%
 What is the dose per field if the Physician
prescribes 180 cGy to the 95 % isodose
line.
Examples
 AP and Rt Lateral – ((180 x .25)/(0.95))=47.4 cGy
 PA = ((180 x .30)/(.95)) = 56.8 cGy
 Lt. Lateral = ((180 x .20)/(0.95)) = 37.9 cGy
Examples
 A patient is to be treated with parallel
opposed fields that are weighted 3:2
Anterior to Posterior. The prescription
dose is 200 cGy to isocenter. What is the
dose per field?
Examples
 Total weight is 5.
 Dose from anterior is 200 x 3/5 = 120 cGy
 Dose from posterior is 200 x 2/5 = 80 cGy
Examples
 What monitor-unit setting is necessary to
deliver 200 cGy to a point at 5 cm depth in
a water phantom. The field size is 12 x 20.
Energy used is 6 MV. 100 cm SSD is set
to the surface of the phantom.
Examples
 First calculate the equivalent square it is
15.0 .
 Next determine which formula to use
based on SSD or SAD set-up.
 This is an ssd set-up so pdd will be used
and the ssd equation.
Examples





Sc for 15 = 1.021
Sp for 15 = 1.013
PDD for 15 at 5cm = 0.87
Dose = 200 cGy
MU = (200)/(1.021 x 1.013 x 0.87)) = 222
Di
MU i 
SC  SP  PDD  OAF  TF
Examples
 What monitor-unit setting is necessary to
deliver 200 cGy to a point at midseperation in a phantom 10 cm thick. The
phantom is irradiated with parallel
opposed fields with a collimator setting of
12 x 20 cm. Fields are blocked to a 10 x
16 cm field using MLC. A 6 MV beam is
used. Fields are weighted 3 to 1 and the
dose is prescribed to the 98% idl.
Examples
 First calculate the equivalent squares:
 12 x 20 = 15 ; 10 x 16 = 12.3
 Next determine from type of set-up which
equation will apply.
 Next determine which field size is used to
look up each factor.
 Calculate dose per field.
Examples
 First Field: Dose = ((200 x ¾)/(0.98) = 153.1
 Second Field : Dose = ((200 x ¼)/(0.98) = 51.0
 Sc for 15 = 1.021
 Sp for 12.3 = 1.007
 DD at 5cm for 12.3 = .8664
Examples
 MU field 1 = ((153.1)/ (1.021 x 1.007 x.8664)) = 172
 MU field 2 = ((51) / (1.021 x 1.007 x .8664)) = 57
Di
MU i 
SC  SP  PDD  OAF  TF
Examples
 Recalculate the monitor units necessary
from the previous problem. If now the
blocking is done with cerrobend. The
prescription point at mid seperation is now
5 cm off-axis.
 Only difference is need to look up off axis
factor for 5 cm off axis at 5 cm depth and
include tray factor for blocks.
Examples
 MU field 1 = ((153.1)/ (1.021 x 1.007 x.8664 x 1.019 x .97))
= 174
 MU field 2 = ((51) / (1.021 x 1.007 x .8664 x 1.019 x .97)) =
58
Di
MU i 
SC  SP  PDD  OAF  TF
Examples
 A patient is to be treated with an isocentric
wedged pair on a Varian Linac with 6 MV.
Field 1 is 8 x 14 tht is blocked to a 6.5 x
14. The SSD for this field is 94 cm. Field 2
is 12 x 14 that is blocked to a 7 x 14, its
SSD is 88cm. Both fields use 30 degree
dynamic wedges. The prescribed dose is
to isocenter and is 180 cGy, beams are
weighted 2 to 1.
Examples
 Calculate Dose per field:
 Field 1: Dose = ((180 x 2/3)/(0.95)) = 126.3
 Field 2 : Dose = ((180 x 1/3)/(0.95) = 63.2
 Determine depths of treatment per field.
 Field 1: SSD= 94 therefore depth = 6 cm
 Field 2: SSD= 88 therefore depth = 12 cm
Examples
 Calculate Equivalent Squares for each
field.
 Field 1: 8 x 14 = 10.1; 6.5 x 14 = 8.9
 Field 2: 12 x 14 = 12.9 ; 7 x 14 = 9.3
 Determine which depth factor to use:
isocentric set-up indicates TMR.
Examples
 Field 1: TMR at 6 cm for 8.9 field = ..8912





Wedge factor for 8.9 field = 0.872
Sc for 10.1 field is 1.0
Sp for 8.9 field is 0.996
Inverse square = (101.5/100)2 = 1.030
Tray factor for 8.9 field = 0.97
Examples
 Monitor Units from field Number 1
 (126.3)/ (1.0 x .996 x 1.03 x .8912 x (0.97 x .872)) = 163
Di
MUi 
SC  SP  ISq TMR  OAF  TF
Examples
 Field 2






TMR at 12 cm for 9.3 field = 0.7161
Wedge factor for 9.3 field = 0.865
Sc for 12.9 field = 1.025
Sp for 9.3 field = 0.998
Inverse square = 1.03
Tray factor for 9.3 field= 0.97
Examples
 Monitor Units for field number 2
 (63.2)/ (1.025 x 0.998 x 1.03 x 0.7161 x (0.97 x 0.865))= 100
Di
MUi 
SC  SP  ISq TMR  OAF  TF
Examples
 A 30 x 30 x 30 cm3 water phantom is centered at isocenter in a
pair of Varian 6 MV x-ray beams, a “right lateral” and a “left lateral”.
Each field has a collimator setting of 12x18 and is further
collimated to a 10x14 using the MLC.
 (a) What are the MU settings of each field if a total dose of 200 cGy is
to be delivered using a relative weighting of 2:1 with the right lateral
having the higher weight?
Make a
picture!
Examples
 First compute the relative doses of the right- and
left-lateral fields:
 Rt Lat (wt = 2):
 Lt Lat (wt = 1):
 


  200 2  133
Di  DRx   wti
3
 i wt 
 


  200 1  67
Di  DRx   wti
3
 i wt 
Examples
 Then compute the equivalent squares of the open
and blocked fields:


  14.4
 2 1218
L W
12  18


  11.7
 2 1014
L W
10  14
 12x18:
EqSq  2 LW
 10x14:
EqSq  2 LW
Examples
 Determine equation (for “SAD” beams):
Di
MUi 
SC  SP  ISq TMR  OAF  TF





SC (for 14.4) = 1.019
SP (for 11.7) = 1.005
ISq = 1.030
TMR (depth 15, for 11.7) = 0.651
OAF and TF = 1.0
Examples
 Rt Lat:
133
MU i 
 194
1.019 1.005 1.030  0.651 1.0 1.0
 Lt Lat:
MU i 
67
 98
1.019 1.005 1.030  0.651 1.0 1.0
Examples
 Calculate the Monitor Units Necessary to
deliver a dose of 200 cGy to a depth of 8
cm from parallel opposed fields equally
weighted. The field size is 15 x 15 blocked
to an 8 x 8 field. The patient has to be
treated with an extended distance of 120
cm SSD. Assume that the field size given
is defined at 120 cm. The energy that is
used is 6 MV.
Examples




Field size at 100 cm is 15 x (100/120)= 12.5
Sc for 12.5 = 1.023
Sp for 8 cm field = 0.993
PDD for 8 x 8 field = .732 at 100 cm which
equals .732 x 1.02 = .747 at 120 cm.
 Inverse square factor for extended SSD is
(101.5/121.5)2 = 0.698
 Dose per field is 200/2 = 100 cGy
Examples
 Monitor Units per field
 (100/(1.023 x 0.993 x .747 x 0.698)) = 189
Di
MU i 
SC  SP  PDD  OAF  TF