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14 chap 10 a system of dosimetric calculations

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14 chap 10 a system of dosimetric calculations

  1. 1. Chapter 10 A System of Dosimetric Calculations• Percent depth dose is suitable for SSD treatment technique.• Tissue-air-ratios (TAR) suitable for SAD treatmenttechnique, but limited to energies no higher than Co-60.• Tissue-maximum-ratios (TMR) and Tissue-phantom-ratios(TPR) were designed to overcome the limitations of the TAR. 1
  2. 2. 10.1 Dose Calculation Parameters• Dose = Primary dose + Scattered dose• Primary dose (includes those scatteredin the head, so-called collimator scatter)= dose due to unscattered (in phantom)photons. Conceptually, primary dose canbe achieved with a beam of zero field Psize, or a phantom column of zero S Sradius. (In both cases, to eliminatescatter in the phantom.)• (phantom) Scattered dose = dose due tophotons scattered at least once in thephantom. 2
  3. 3. 10.1 Dose Calculation Parameters (collimator & phantomscatter factor) SAD ReferenceBuildup Reference depth cap field air phantom Sc,p Sc Sp = Sc,p / Sc Reference field Field size 3
  4. 4. Phantom Scatter Factor (Sp) keeping same collimator opening block down Reference to reference depth field size r r0 D D0 phantom phantomD (d max ) = D fs (r ) • BSF (r ) D0 (d max ) = D fs (r ) • BSF (r0 ) Dfs unchanged, due to same collimator opening D(d max ) BSF (r ) S p (r ) = = D0 (d max ) BSF (r0 ) 4
  5. 5. 10.1 Dose Calculation Parameters (tissue-phantom &tissue-maximum ratios) d t0 rd rd Dd Dt0 Dd TPR (d , rd ) = where t0 is the reference depth Dt0 Dd TMR (d , rd ) = if t0 = dmax Dmax TAR, TMR, and TPR depends on depth and field size, but is SSD independent. 5
  6. 6. 10.1 Dose Calculation Parameters (properties of TMR)Like TAR, TMR is independent of SSD, increases with energy and field size. TMR 30×30 10×10 TMR (d ,0) = e − µ ( d −t max ) 0×0 Depth in water 6
  7. 7. Relationship between TMR and TAR d dmax rd rd rDfs Dd Dmax d Dd Dd TAR (d , rd ) = TMR (d , rd ) = D fs Dmax Dmax BSF (rd ) = TAR (d , rd ) D fs TMR (d , rd ) = BSF (rd ) 7
  8. 8. Relationship between TMR and PDD f r r0 dmax PDD(d , r , f ) Dd d = D0 dmax 100 D0 rd r Dd Dmax d DdTMR (d , rd ) = TMR (d , rd ) = 2 DmaxPDD(d , r , f )  f + d   S p (r0 )      100  f +d   S p (rd )   f +d  2  S p (r0 )   max    D0   =  Dmax  f + d max     S p (rd )    8
  9. 9. Scatter-Maximum Ratio (SMR) D (d , rd ) − D1 (d ,0) d dmaxSMR(d , rd ) ≡ 1 D2 (d max ,0) rd rd D1 D2 D1 (d , rd ) D (d , r ) D (d , r ) D1 (d ,0) SMR(d , rd ) ≡ × 2 max d × 2 max 0 − D2 (d max , rd ) D2 (d max , r0 ) D2 (d max ,0) D2 (d max ,0) r0 = reference field size S p (rd ) SMR(d , rd ) ≡ TMR(d , rd ) × − TMR (d ,0) S p ( 0) 9
  10. 10. 10.2 Practical Applications (accelerator calculations) SSD technique:calibration field size distanceconditions changed changed SSD SCD SAD  r r  rc rc  = c  x x  SSD SAD  r0 r’ ×Sc(rc)×Sp(r) t0 t0 r 2 t0  SCD  ×  SSD + t    0  d K ×Sc(rc)×Sp(r’) PDD(d , r , SSD) ×  r r  depth 100  = c   SCD − t SAD  changed  0  TD ( target dose) × 100 MU = K × PDD(d , r , SSD ) × S c (rc ) × S p ( r ) × ( SCD SSD + t 0 ) 2 10
  11. 11. 10.2 Practical Applications (accelerator calculations,example 1)SSD technique:Machine: 4 MV photonsCalibration conditions: SSD = 100 cm, dmax = 1 cm, field size = 10 × 10 cm.Calibration dose rate = 1 cGy / MUTreatment conditions: SSD = 100 cm, d = 10 cm, field size = 15 × 15 cm,Sc(15×15)=1.020, Sp(15×15)=1.010, %DD=65.1, TD = 200 cGy.Dose/MU at prescription point= 1 × 1.02 × 1.01 × 65.1/100 = 0.6707MU = 200 / 0.6707 = 298 11
  12. 12. 10.2 Practical Applications (accelerator calculations,example 2)SSD technique:Machine: 4 MV photonsCalibration conditions: SSD = 100 cm, dmax = 1 cm, field size = 10 × 10 cm.Calibration dose rate = 1 cGy / MUTreatment conditions: SSD = 120 cm, d = 10 cm, field size = 15 × 15 cm,Sc(12.5×12.5)=1.010, Sp(15×15)=1.010, %DD=66.7, TD = 200 cGy.Dose/MU at prescription point= 1 × 1.01 × 1.01 × [(100+1)/(120+1)]2 × 0.667 = 0.474MU = 200 / 0.474 = 422 12
  13. 13. 10.2 Practical Applications (accelerator calculations) SAD technique:calibrationconditions field size distance depth SCD changed changed changed SAD t0 t0 r0 t0 r’ d t0 x xr x r rd c d K ×Sc(rc)×Sp(r’) ×Sc(rc)×Sp(rd) ×TMR(d , rd )  r r  2  = c   SCD   SCD SAD  ×   SAD  ID (isocenter dose) MU = K × TMR (d , rd ) × S c (rc ) × S p (rd ) × ( SCD SAD ) 2 13
  14. 14. 10.2 Practical Applications (accelerator calculations,example 3)SAD technique:Machine: 4 MV photonsCalibration conditions: SCD = 100 cm, dmax = 1 cm, field size = 10 × 10 cm.Calibration dose rate = 1 cGy / MUTreatment conditions: SAD = 100 cm, d = 8 cm, field size = 6 × 6 cm,Sc(6×6)=0.970, Sp(6×6)=0.990, TMR(8, 6×6)=0.787, TD = 200 cGy. Dose/MU at prescription point = 1 × 0.970 × 0.990 × 0.787 = 0.756 MU = 200 / 0.756 = 265 14
  15. 15. 10.2 Practical Applications (accelerator calculations, example 4)SAD technique:Machine: 4 MV photonsCalibration conditions: SCD = 101 cm, dmax = 1 cm, field size = 10 × 10 cm.Calibration dose rate = 1 cGy / MUTreatment conditions: SAD = 100 cm, d = 8 cm, field size = 6 × 6 cm,Sc(6×6)=0.970, Sp(6×6)=0.990, TMR(8, 6×6)=0.787, TD = 200 cGy.Dose/MU at prescription point= 1 × 0.970 × 0.990 × [(100+1)/(100)]2 × 0.787 = 0.771MU = 200 / 0.771 = 259 15
  16. 16. 10.2 Practical Applications (Co-60 unit, example 5) SSD technique:Machine: Co-60 photonsCalibration conditions: SSD = 80 cm, dmax = 0.5 cm, field size = 10 × 10 cm.Calibration dose rate = 130 cGy / minTreatment conditions: SSD = 100 cm, d = 8 cm, field size = 15 × 15 cm,Sc(12×12)=1.012, Sp(15×15)=1.014, %DD(8,15 × 15,100)=68.7, TD = 200 cGy. Dose/MU at prescription point = 130 × 1.012 × 1.014 × [(80+0.5)/(100+0.5)]2 × 68.7/100 = 58.80 MU = 200 / 58.80 = 3.40 min 16
  17. 17. 10.2 Practical Applications (irregular fields) S p (rd )SMR(d , rd ) ≡ TMR(d , rd ) × − TMR (d ,0) (from slide #9) S p ( 0) S p ( 0)TMR (d , rd ) = [TMR (d ,0) + SMR(d , rd )] × S p (rd ) BEV [ TMR (d ) = TMR(d ,0) + SMR(d ) × ] S p (0) S p (rd ) 1 n ri SMR(d ) = n ∑ SMR(d , r ) i =1 i n 1 S p (rd ) = n ∑S i =1 p ( ri ) For off-axis [ TMR(d ) = K p × TMR (d ,0) + SMR(d ) × ] S p ( 0) S p (rd ) point: off-axis ratio 17
  18. 18. From slide #8: 2 PDD(d , r , f )  f + d   S p (rt0 )  f TMR (d , rd ) =   f +t     100  0   S p (rd )    r (from slide #9) S p (rd ) t0 SMR(d , rd ) ≡ TMR(d , rd ) × − TMR (d ,0) d S p ( 0) (when d = t0) S p (rd ) rd SMR(t 0 , rd ) ≡ 1× −1 S p (0) 2P (d , r , f ) 100 [ ] = K p × TMR(d ,0) + SMR(d ) × × S p (rd )  f + t 0 S p ( 0) ×  S p (rd ) S p (rt0 )  f + d     2P (d , r , f ) 100 [ = K p × TMR (d ,0) + SMR(d ) × ] 1  f + t0 ×  1 + SMR(t 0 , rt0 )  f + d     18
  19. 19. 10.2 Practical Applications (asymmetric fields) SSD technique: SAD technique:Dose / MU = K Dose / MU = K × Sc (rc) × Sp(r) × Sc (rc) × Sp(rd) × (SSD factor) × (SAD factor) × PDD(d,r)/100 × TMR(d,rd) × OARd(x) × OARd(x)MU = TD / (Dose / MU) MU = ID / (Dose / MU) 19
  20. 20. 10.3 Other Practical Methods of Calculating Depth DoseDistribution (Irregular fields) Collimator field vs. effective fields 20
  21. 21. 10.3 Other Practical Methods of Calculating Depth DoseDistribution (off-axis points) 2a 2d d b b Q Q 2b 2b a a d 1 1 D1 ( 2b × 2d ) D2 ( 2a × 2b ) 4 4 b Q x c P 2a 2d Q Q 2c 2c d a c 1 4 c D3 ( 2a × 2c ) 1 1 D4 ( 2d × 2c ) D3 ( 2a × 2c ) 4 4 21
  22. 22. 10.3 Other Practical Methods of Calculating Depth Dose Distribution (off-axis points) a dLet the dose in free space at P = Dfs(P), then thedose at P(dmax) =Dfs(P)×BSF[(a+d)×(b+c)]. b Q xMoreover, let KQ be the off-axis ratio at Q, then c Pthe dose at Q is: D fs ( P) × K QDQ [ ( a + d ) × ( b + c ) ] = 4 × [ BSF ( 2b × 2d ) • % DD( 2b × 2d ) / 100 + BSF ( 2a × 2b ) • % DD( 2a × 2b ) / 100 + BSF ( 2a × 2c ) • % DD( 2a × 2c ) / 100 + BSF ( 2d × 2c ) • % DD( 2d × 2c ) / 100] % DQ KQ =DP ( d max ) 4 × BSF [(a + d ) × (b + c)] × [ BSF ( 2b × 2d ) • % DD( 2b × 2d ) + BSF ( 2a × 2b ) • % DD( 2a × 2b ) + BSF ( 2a × 2c ) • % DD( 2a × 2c ) + BSF ( 2d × 2c ) • % DD( 2d × 2c ) ] 22
  23. 23. a=10 d=5 Example 6: b=5 For a = 10cm, b = 5cm, c = 10cm, d = 5cm and Q a Co-60 beam with KQ = 0.98 and SSD = 80cm, x c=10 P what is %DQ(d=10cm) relative to DP(dmax)? Given: Field size 10x10 20x10 20x20 15x15 BSF 1.036 1.043 1.061 1.052 %DD(SSD=80cm,d=10cm) 55.6 56.3 60.2 58.4 % DQ 0.98 = × [1.036 • 55.6 + 1.043 • 56.3 + 1.061 • 60.2 + 1.043 • 56.3]DP ( d max ) 4 × 1.052 = 55.8 23
  24. 24. 10.3 Other Practical Methods of Calculating Depth Dose Distribution (points outside the field) a a a c c c b b b a 2c a b 1 − { } 2c b 2 1D[ ( a × b; c ) ] = { D[ ( 2a + 2c ) × b] − D ( 2c × b ) } 2 24
  25. 25. Example 7: a=15 For a = 15cm, b = 10cm, c = 5cm, and a c=5 b=10 Co-60 beam with SSD = 80cm, what is x P Q %DQ(d=10cm) relative to DP(dmax)? Given: Field size 10x10 40x10 15x10 BSF 1.036 1.054 1.043 %DD(SSD=80cm,d=10cm) 55.6 58.8 56.9 % DQ 1 [ BSF (40 × 10) • % DD(40 × 10) − BSF (10 × 10) • % DD(10 × 10)] = ×DP (d max ) BSF (10 × 15) 2 1 [1.054 • 58.8 − 1.036 • 55.6] = × = 2.1 1.043 2 25
  26. 26. 10.3 Other Practical Methods of Calculating Depth DoseDistribution (points under a block) b a c D[ under a block ] = D ( a × b ) − D ( a × c )(1 − t ) t = block transmission 26
  27. 27. b=15 Example 8: For a = 15cm, b = 15cm, c = 4cm, and Co-60 beam with SSD = 80cm. Given: a=15 Dfs(15x15,f=80.5) = 120 cGy/min x P Q Block transmission = 5% Tray transmission = 97%. c=4• What is the treatment time to deliver 200 cGy to P(d=10cm)? (i.e. DP(d=10) = 200 cGy)• What is the %[DQ(d=10)/DP(d=10)]? A/ P15 × 5.5 ⇔ 8 × 8  (a) DP = time × D × BSF × % DD / 100 × trans% DD(10,8 × 8,80) = 54.0 time = 200 / (120 × 1.029 × 0.54 × 0.97 ) = 3.09 minBSF (8 × 8) = 1.029 27
  28. 28. Method 1: Negative weight PDD b=15DQ = DQ (15 × 15) − DQ ( 4 × 15) • (1 − 0.05)DQ (15 × 15) = 3.09 × 120 × BSF (15 × 15) × % DD(15 × 15)DQ ( 4 × 15) = 3.09 × 120 × BSF ( 4 × 15) × % DD( 4 × 15) x a=15 P QDQ = 39.3 cGy 6.3×6.3DQ DP = 39.3 200 ≈ 20% c=4 28
  29. 29. Method 2: Negative weight TAR Field size magnification at d=10cm: (80+10)/10=1.125(15 ×15) cm ×1.125 = (17 ×17 ) cm b=15 A/ P( 4 ×15) cm ×1.125 = ( 4.5 ×17 ) cm ⇔ ( 7 × 7 ) cm A/ P( 5.5 ×15) cm ×1.125 = ( 6.2 ×17 ) cm ⇔ ( 9 × 9) cm x a=15 Q DQ TAR (10,17 × 17 ) − TAR (10,7 × 7 ) • (1 − trans ) P = DP TAR (10,9 × 9 ) c=4 0.771 − 0.667 • (1 − 0.05) = ≈ 20% 0.694 29
  30. 30. Method 3: Negative weight TMR From slide #7: TAR(d,rd) = TMR(d,rd) × Sp(rd) b=15 x a=15 P Q c=4DQ TMR(10,17 × 17 ) • S p (17 × 17 ) − TMR(10,7 × 7 ) • S p ( 7 × 7 ) • (1 − trans ) =DP TAR (10,9 × 9 ) • S p ( 9 × 9 ) 0.733 × 1.02 − 0.651× 0.989 • (1 − 0.05) = ≈ 20% 0.672 × 0.997 30

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