Math Review (Treatment Planning)
RULE OF THUMB: on linacs (around 90-100 cm ssd) a 1cm change in distance= about 2% dose difference (for Co60, with SSD's around 80, a 1cm change in distance = about 2.5% change in dose)
-remember, due to divergence, a change in SSD means the field size is changed so that there might be a geometric miss.
What number represents 83% of the distance from .954 to .987?
.987-.954= 0.033 x 83%, divided by 100 =0.02739+.954=0.98139
At a distance of 95cm, a linac delivers an output of 310 R in 1 minute. What is output of the same machine at a distance of 85cm for 49 seconds? First, do Inv Sq to account for distance. Then do a direct prop to account for the time variation (THIS IS IMPORTANT!!!)
1. Direct proportion: 310/x= 60sec/49sec x= 253 sec 2. Inverse Square: 253/x= (85/95)² x= 316 R
What is the treatment equivalent square if the field size on film measures 16.8x19.6 and a square block on the SIM FILM measures 4.9x4.9, and the TFD=140?
Now we measure the size of our block from the SIM FILM! Aren't sim films magnified? The first step is to de-magnify all the numbers!!! Image/Mag-Factor 1.) Demagnify: 16.8/1.4= 12 19.6/1.4= 14 4.9/1.4=3.5 4.9/1.4=3.5 field size= 12x14 block size= 13.5x13.5 2.) Treatment equivalent square: field size area minus area of object, then square root. field size area: 12x14= 168 area of object: 3.5x3.5= 12.25 subtract: 168-12.25= 155.75 square root: √155.75= 12.5 Answer= 12.5²
Calculate the equivalent square for a 16 x 23 cm rectangular field.
19 x 19
Calculate the mu's for a patient receiving 200 rads/day for a single ap port to a depth of 6cm. The TAR is .783 and the output factor is 1.02
200 / .783 x 1.02 = 250 rads
Calculate the treatment time to deliver 2.0 Gray to midplane on a Co60 80cm SSD unit. The output is 150 R/min, fields size is 10 x 12 open port appa, f-factor is .957, patient separation=18, %DD=86
200 divided by 2 (100 cGy per field) / 150 x .957 x 86 .81 min
ex. If a patient was supposed to be treated at 80 cm SSD and receive 200 rads/day but the SSD indicator was off so that the patients entire treatment was delivered at 77 cm, what was the error in dose and the % error in dose? (this example is from the QA manual right at the beginning of chptr 7)
200/x= (77/80)² x= 216 rads actual dose/prescribed dose x 100= % of dose given 216/200= 1.08 x 100= 108% or 8% overdose
ex. The dmax dose for a single PA spine field is 322 cGy. What is the PDD of the spinal cord which lies at a depth of 6cm and receives 300 cGy/day?
PDD= 300cGy/322cGy x 100 PDD= 93%
Percent Depth Dose (PDD, %dd)
PDD= Dd/Given dose x 100
Penumbra formula
Penumbra at depth=s(SSD +d-SDD)/SDD s= source size d= depth of concern SDD= source to diaphragm distance so, Penumbra at depth= source size × (source to skin distance + depth of concern - source to diaphragm distance) ÷ source to diaphragm distance. cobalt source size= 2cm
A Co60 unit has a source size of 1.5 cm in diameter. The SAD is 80 , the TFD is 112 cm and the source collimator distance is 48cm. What is the penumbra size on the film?
Penumbra formula: 1.5 (80 + {112-80}-48 / 48 2.0cm
F-factor
Roentgen to Rad conversion factor. typical f-factor: 0.957 don't figure this in unless directed.
SSD Technique
SSD=SAD, primarily Cobalt and KV machines. almost exclusively use PDD tables Time=dose in cGy per field/all factors effecting dose (decimals) including PDD, and R/minutes
Target-Film Distance (TFD)
TFD= image/object x TOD (target to object distance) TOD for linac= 100cm TOD for cobalt= 80cm Always expressed in centimeters. TFD can never be less than 100cm if a linac is used
If a patient is to receive 300 rads/day from a single ap SSD port to a depth of 6 cm using 4 mv photons, 6 x 15 field size. What is the d max dose? TAR=.763
393 rads
A dose of 4000 cGy is to be given to a 10cm depth having a PDD of 64%. What is the dose to an underlying critical normal tissue that lies at a depth of 15cm and has a PDD of 45%?
4000cGy/x= 64/45 x= 2,813 cGy
A treatment designed to be given at 120 cm SSD is mistakenly given at 100 cm SSD. What is the error in dose delivered?
44% overdose
If a tumor on film measures 4cm in diameter and the actual diameter of the tumor is 3.2cm, what is the Magnification factor?
4cm/3.2cm= 1.25 1.25 x 100= 125% or 25% larger
If a tumor on film measures 4cm in diameter and the actual diameter of the tumor is 3.2cm, what is the Target-Film Distance?
4cm/3.2cm= 1.25 x 100 (linac)= 125cm
If a tumor on film measures 4cm in diameter and the actual diameter of the tumor is 3.2cm, what is the Target-Film Distance if the patient is being treated on cobalt?
4cm/3.2cm= 1.25 x 80 (cobalt)= 100cm
It is discovered that a patient receiving 200 rads daily has been given 25 fractions at 77 cm SSD instead of 80 cm SSD. Instead of the prescribed 5000 rads, the patient actually received __________ rads.
5395
ex. What number represents 30% of the distance from 447 to 543?
543-447= 96 x 30%, divided by 100 = 28.8+447=475.8 Answer=475.8
An error in looking up the %DD was made. The incorrect value was 86%. The correct value should have been 80%. What is the magnitude of this error (%)?
7.5%
Using the information in the previous 2 problems, what is the true total daily dose:
7.5% overdose: thus 200×1.075 = 215
What is the wedge angle with a 30 degree hinge angle?
75 degrees
ex. if the field size at 80 cm is 8 x 8, what is the field size at 100 cm?
8/x= 80/100 x= 10, field size= 10 x 10 at 100cm
If the dose rate in air at 80.5 SSD is 125 rads/min, what is the dose rate in air at 100.5 cm SSD?
80.2 rad/min
If the dose at the 70% line=88 rads, what is the dose at Dmax?
88rads/70= 1.257 x 100= 126 rads
Wedge Angle
90°- (hinge angle/2)
A patient was to have received 200 rads daily in 20 fractions at 80cm SSD. What is the % error if all treatments were received at a 70 cm SSD?
Actual divided by prescribed: 200/X = 261/200 = 31% overdose 30% overdose
135 mu, mu per degree=1.35, what is the arc angle?
Arc angle= MU/MU per degree x=135/1.35 x=101.35 Answer= 101.35° arc angle
Magnification Factor (M-Factor)
Formula: MF= image (film)/object (actual size) Always expressed as a decimal (a factor used to multiply) MF can never be less than 1.0
A patient is treated with two adjacent Co60 fields. One field is 8cm in length and one is 12 cm in length. Both fields are applied with an SSD os 80 cm. To give a uniform junction dose at a depth of 8cm, the gap between the fields on the skin should be:
Gap 1.0cm
A patient is treated with two adjacent cobalt fields 12cm and 18 cm in length respectively at 80 SSD. Calculate the gap on the skin necessary to give a uniform junction at a depth of 10cm.
Gap Formula 1.87cm
Gap Formula
Gap calculation: calculate a gap on the skin surface that corresponds to fields abutting at depth. Formula: Gap=D/2 (L1/SSD1 + L2/SSD2) D= depth of concern L=field length ("Y" axis measurement)
A spine is to be treated at 100 cm SSD with two adjacent 26 cm field lengths. If the fields match at 7cm depth, the skin gap should be:
Gap formula 1.82cm
A patient is treated with parallel opposed mantle and para-aortic fields with lengths of 15cm and 30cm respectively. Calculate the gap required on the surface for the beams to intersect at a midline depth of 10cm.
Gap=D/2 (L1/SSD1 + L2/SSD2) 10/2 (30/100 + 15/100) = 2.25cm gap
If there's a change in dose with a change in distance (SSD)
Inverse Square Law
Area of a Rectangle=
Length × Width
Area of a Square=
Length² OR Width²
If a field size is 9x7 at isocenter and the field measures 12.6x9.8 on film, what is the mag factor? what is the TFD?
MF= 9.8/7 or 12.6/9= 1.4 TFD= 1.4 x 100= 140cm
Calculating Rotational Therapy Stop Angle
MU per degree= MU/Arc angle Arc angle= MU/MU per degree MU= MU per degree × Arc angle
If 135 mu are set, the arc travels 100', what is the mu/degree?
MU per degree= MU/Arc angle x=135/100 x=1.35 Answer= 1.35 MU per degree
The arc rotation is 100'. mu per degree=1.35, what is the mu setting?
MU= MU per degree × Arc angle x=1.35 × 100° x=135 Answer= 135 MU
Calculate the monitor units required to deliver 100 rads to a depth of 6 cm for a 6 MV accelerator. The TAR is .888 and the AOF is 1.02.
MU= dose/factors 110
The Registry may just give you the distances used with no doses. Ex. A patient is supposed to be treated with an SSD of 100cm but the SSD indicator is off and the patient is actually treated with an SSD of 96cm. What % error in dose does this represent (and state if it is an overdose or underdose)
Make up a dose to find the % error in dose! (typically "100") "100"/x=(96/100)² x= 109 or 9% overdose
The %dd for a 12 x 12 field size Cobalt beam, 8 cm depth and 80 cm SSD is 61.5. Calculate the percent depth dose for the same field size and depth for 100 cm SSD.
Maynard F factor 63.52
The percent depth dose for a 12 x 12 field, 4 MV beam, 5 cm depth and 80 cm SSD is 82.8. Calculate the percent depth dose for the same field size and depth for 100 cm SSD.
Maynard F-factor 84.8%
If there's a change in PDD with a change in distance (SSD)
Mayneord F factor
Inverse Square Law
The intensity of a beam is inversely proportional to the square of the distance Use: to calculate a change in dose or dose rate with a change in distance. If the distance is doubled, the intensity id reduced by a factor of 4 (÷4) If the distance is cut in half, the intensity will be increased by a factor of 4 (x4) I₂ ALWAYS = x
Mayneord F Factor
The inverse square term within PDD. Mayneord F Factor is required in calculating a new PDD when there is a change in SSD with the same depth and dmax (energy). MF = [(SSD2 + dmax)/(SSD1 + dmax)]² x [(SSD1 + d)/(SSD2+d)]² then PDD2 = MF x PDD1
In Air Output Factor
The number of photons (output) varies with field size. Increase field size=Increase number of photons emitted=Increased output Output factors are listed in tables all you have to do is look them up. (some interpolation may be necessary). Output is dependent on collimator field size, and has nothing to do with the blocking, wedges, etc.
The collimator setting to treat a 40 cm field at a 130 cm SSD using an 80 cm SAD Co60 unit is:
This is essentially "a change in field size with a change in distance", which is calculated using a direct proportion 24.6 cm
The output of a Co60 teletherapy unit is 175 rad/min. Calculate the time necessary to deliver a dose of 200 rads at a PDD of 63 and a 1.02 BSF.
Time= dose/factors 200/.63×1.02×175 = 1.78 minutes. Then must convert .78 minutes into a decimal fraction of a minute. .78 x 60 (seconds in a minute) = 1'46.8 seconds = best answer = 1' 47"
TAR
Tissue Air Ratio= dose in air/dose in phantom
Treatment Equivalent Square (Effective Field Size; EFS)
Treatment equivalent square not only takes into account the collimator area but also any blocking or flash in the field. In other words, it gives an equivalent square area for the shape of the beam that hits the patient (while collimator eq. sq. does not take blocking into consideration). Tx eq. sq. is used to look up either TMR, TPR, or PDD, depending on what chart your institution is using. to calculate: field size area minus area of object, then square root
Example: TwoparallelopposedfieldsareusedforCobalttreatment. Patient separation is 12cm, mp dose is 100 cGy/field, the PDD @ 6cm is 70.7%. The PDD at .5cm is 100% while the PDD at 11.5cm depth is 46.3% Calculate the entrance+exit dose.
Use 2 direct proportions Calculate Entrance dose: 100/70.7=x/100 x= 141 cGy Calculate Exit dose: 70.7/46.3=100/x x= 65 cGy entrance + exit dose: 141+65= 206 cGy
Direct Proportion (Law of Similar Triangles)
Use: 1. Calculating a change in field size with a change in distance 2. Calculating a change in dose with a change in PDD%
INTERPOLATION
higher #-lower#= ___ x % desired, divided by100 = _____ + lower #
SAD (isocentric) Technique
primarily linacs, and usually use TAR, TPR, or TMR MU=dose in cGy per field/all factors effecting dose (decimals)
ex. A co60 source is 2.5 cm in diameter, the treatment distance is 80 cm, and the distance from the source to the final diaphragm is 30 cm. The patient separation is 20cm. What is the penumbra size at the skin surface?
s(SSD+d-SDD)/SDD s= source size, 2.5cm SSD= 80cm d= depth of concern, 0cm SDD= source to diaphragm distance, 30cm 2.5(80+0-30)/30 Answer= 4.17cm
This problem will almost certainly be in relation to cobalt: ex. A co60 source is 2.5 cm in diameter, the treatment distance is 80 cm, and the distance from the source to the final diaphragm is 30 cm. The patient separation is 20cm. What is the penumbra size at midplane?
s(SSD+d-SDD)/SDD s= source size, 2.5cm SSD= 80cm d= depth of concern, 10cm SDD= source to diaphragm distance, 30cm 2.5(80+10-30)/30 Answer= 5cm
ex. A co60 source is 2.5 cm in diameter, the treatment distance is 80 cm, and the distance from the source to the final diaphragm is 30 cm. The patient separation is 20cm. For the same problem, the cord lies 15cm deep from the anterior field, what is the penumbra at the cord? (keep in mind that "Depth" means distance past the 80cm SSD)
s(SSD+d-SDD)/SDD s= source size, 2.5cm SSD= 80cm d= depth of concern, 15cm SDD= source to diaphragm distance, 30cm 2.5(80+15-30)/30 Answer= 5.42cm
5100 rads dose is to be delivered 5 days/week in 6 weeks through a single open port. If the TAR is .580 at tumor depth and the in-air factor is 1.10 for the treatment field, then:
the daily mu setting is 266 mu's
Separation
the measurement of the patient thickness from the the beam entry point to exit point. Calipers are used for taking separations. The separation can also be measured indirectly by using SSD readings. Formula: Separation÷2= Answer - 100cm(linac)
Suppose there's a 4x3 rectangular block in a field that is 10x10, what is the Treatment Equivalent Square (Effective Field Size; EFS)?
to calculate: field size area minus area of object, then square root 1.) 10x10= 100 2.) 4x3= 12 3.) 100-12=88 4.) √88= 9.4 Answer= 9.4x9.4 OR 9.4²
What is the Treatment Equivalent Square of an 8x11 field that has a triangular block measuring 1.5x3x4?
to calculate: field size area minus area of object, then square root 1.) 8x11= 88 2.) ½×1.5×3 (½ Base × Height)= 2.25 3.) 88-2.25= 85.75 4.) √85.75= 9.26 Answer= 9.3²
Actual Size of Object Formula
use: to find the actual size of objects projected on a film. Actual Size= Image size/Magnification Factor
A (Cobalt) teletherapy machine has an output of 183 rads/min to a small mass of tissue 80 cm from the source for a 10 x 10 cm square field. Calculate the treatment time needed to deliver a dose of 300 rads to a point 6 cm deep on the central axis through a single field. A compensating filter is used which absorbs 10% of the incident radiation and the PDD at 6 cm is 0.756, BSF=1.113
when it states that a filter is used that absorbs 10% of the radiation, that means that filter would have a factor (for the sake of calculation) of .900, meaning 90% of the beam is transmitted THROUGH the filter, and 10% is attenuated/absorbed. Therefore: 300 / 183×.90×0.756×1.113 = about 2.16 minutes 2.16 min
Area of a Triangle=
½ Base × Height OR L X W/2
Area of a Circle=
π × Radius² (3.14 × Radius²)
What will be the finishing angle of an arc set-up if the starting angle is 270 degrees, the rotation direction is clockwise, mu/degree=2.16 and mu setting=302?
130' Remember: if the question states cobalt or "International Convention" then 0/360 is at the top. Otherwise assume that 180 is at the top.
ex. What is the equivalent square field size for a rectangle that is 14x22.5 cm?
14 × 22.5 × 2/14 + 22.5= 630/36.5 Answer=17.3x17.3 or 17.3²
A field size measures 14.3x12 on a film with a mag factor of 1.4, what is the actual field size?
14.3/1.4 AND 12/1.4= 10.2X8.6
If a field size is actually 10x20 at isocenter and the field size measures 14.5x29 on film, what is the Magnification factor?
14.5/10 or 29/20= 1.45 1.45 x 100= 145% or 45% larger
If a field size is actually 10x20 at isocenter and the field size measures 14.5x29 on film, what is the Target-Film Distance?
14.5/10 or 29/20= 1.45 x 100 (linac)= 145cm
Hinge Angle
180°- (wedge angle x 2)
Here's an important twist on some of the stuff we've done up to this point: Dose is directly proportionate to time! At a distance of 95cm, a linac delivers an output of 310 R in 1 minute. What is output of the same machine at a distance of 85cm for 30 seconds? First, do Inv Sq to account for distance. Then do a direct prop to account for the time variation (THIS IS IMPORTANT!!!)
First, do Inv Sq to account for distance. Then do a direct prop to account for the time variation (THIS IS IMPORTANT!!!) 1. Direct proportion: 310 R/x= 60sec(1min)/30sec x= 155 R/sec 2. Inverse Square: 155/x= (85/95)² x= 194 R
A field size is 12 x 18, a triangular block is in the field that measures 6 x 5 (base and height) what is the collimator and treatment equivalent squares?
For the collimator equivalent square use the Equivalent Square Formula: Length × Width × 2/Length + Width 1.) 12×18×2/12+18 Answer= 14.4² To calculate for the treatment equivalent square: field size area minus area of object, then square root. 1.) 12×18=216 2.) ½×6×5=15 3.) 216-15=201 4.) √201 Answer= 14.2² (answer is ALWAYS equal too or smaller than the collimator equivalent square)
Area/Perimeter Method (AP METHOD)
Formula: 4(Area)/Perimeter OR A/P=Length × Width/2(Length+Width)
Equivalent Square Formula (ESRF, Sterling Formula)
Formula: Length × Width × 2/Length + Width
Dmax Dose (Given Dose, dose at electronic equilibrium, 100% dose line)
Formula: Dmax=Dose at depth/PDD x 100 dmax=Dd/PDD x 100
The start angle for a rotational treatment is 35' and mu/ degree is 1.17 and 100 rads(mu) are given, what is the stop angle if the gantry is rotating CCW (counter clockwise)? (VARIAN = 180 AT THE TOP)
1.) calculate how many degrees the gantry will rotate: Arc angle= MU/MU per degree 100/1.17= 85° 2.) when rotating CCW, you are ADDING numbers! So add, 85° to the stat angle, 35° 85+35= 120 Answer= stop angle is 120°
The start angle for a rotational treatment is 35' and mu/ degree is 1.17 and 100 rads(mu) are given, what is the stop angle if the gantry is rotating CCW for a Cobalt machine (counter clockwise)? (CO60 = 0/360 AT THE TOP)
1.) calculate how many degrees the gantry will rotate: Arc angle= MU/MU per degree 100/1.17= 85° 2.) when rotating CCW, you are ADDING numbers! So add, 85° to the stat angle, 35° start at 35° BUT when you reach 0/360 on the gantry you must then subtract. 85-35= 50 360-50= 310 Answer= stop angle is 310° on a cobalt unit
The start angle for a rotational treatment is 20' and mu/ degree is 1.22 and 150 rads(mu) are given, what is the stop angle if the gantry is rotating CW (clockwise)?
1.) calculate how many degrees the gantry will rotate: Arc angle= MU/MU per degree 150/1.22=123° 2.) start angle is 20° 123-20= 103° 360-103= 257°
ex. A patient is treated isocentrically on a 100cm SAD unit. The field size at isocenter is 10x14. What is the field size at the patients skin surface if the patient's separation is 20cm?
1.) find the SSD using separation; 20÷2=10, 100cm-10= 90cm 2.) x/10= 90/100 x= 9 3.) x/14= 90/100 x= 12.6 Answer= 9x12.6
ex. if the intensity of a beam is 100 R/min and the distance is 20cm, what will be the intensity of the beam at 40 cm?
100/x= (40/20)² x= 25 R/min
The dose to the 80% isodose line is 120 cGy, what is the dose at the 60% isodose line?
120cGy/x= 80/60 x= 90 cGy
If a coin on film measures 12mm in diameter on the film and the mag factor is 1.35, what is the actual size of the coin?
12mm/1.35= 8.89mm
BSF
Back Scatter Factor= Dmax dose/Dose in free space Typical BSF= 1.036
Collimator Equivalent Square (Ceqsq)
Depth dose charts are figured only for exact square fields (2x2,10x10,etc.). Therefore, irregular fields (9x14) must be "converted" into the equivalent area of a square. Collimator equivalent square (Ceqsq) is used to determine air-output factor when performing a dose calculation. The more irregular a field is, the less accurate any correction formula is going to be. There are 2 methods for converting an irregular shape into a square area equivalent: 1. Equivalent Square Formula (STERLING FORMULA or Equivalent Squares of Rectangular Fields (ESRF)) 2. Area/Perimeter Method (AP METHOD)
If there's a change in dose with a change in PDD
Direct Proportion
If there's a change in field size with a change in distance (SSD)
Direct Proportion
A lesion is treated using a 90 degree arc and the midpoint of the tumor volume is 5 cm below the skin surface. The isocenter should be placed at least __________ cm past isocenter
Dosimetry lecture notes don't give us a formula for calculating the exact amount of past pointing, but do describe responses are nearly always somewhere between 3 - 4 cm past isocenter
ENTRANCE + EXIT DOSES (aka total Dmax dose)
Entrance dose: the dose at Dmax depth (Given Dose, 100% line, dose at electronic equilibrium, etc) Exit dose: The dose at the depth of Dmax from the other side. BIG HINT: your answer should be VERY close to the tumor total dose at mid plane. Ex: if treating on cobalt with a Dmax of .5cm, on a patient with a separation of 20cm, the exit dose would be the dose to the point at a 19.5cm depth from the AP (.5cm from the PA field)
How to find the % error in dose?
actual dose/prescribed dose x 100= % of dose given
Timer Error
aka travel time, the time necessary to deliver the source from the off, on position. typical timer(shutter) error: .02 minutes don't figure this in unless directed
The formula commonly employed for calculation dmax dose is:
dmax= Dd/PDD x 100
