The same patient above also needs to be treated at the right shoulder with 800 c

Question

The same patient above also needs to be treated at the right shoulder with 800 cGy in one single fraction, with 6MV POP and two fields equally weighted. The prescription point is at mid-plane and the separation is 16.4 cm. The collimator settings are (in cm. see the figure below). AP field: X1 = 5.7, X2 = 8.9; Y1 = 14.6, Y2 = 3.6; PA field: X1 = 10.5, X2 = 5.7;Y1 = 14.6, Y2 = 4.2; Due to multileaf collimator (MLC) shielding, X1 of the effective field for AP is reduced to 0.6 cm while X2 of the effective field for PA is reduced to 0.1 cm. Calculate effective width (X) and length (Y) of each field; Calculate equivalent square of each field; With TMR (12.5 times 12.5. d = 8.2) = 0849 and ROF (12.5 times 12.5) = 1.018, calculate MU for the AP field; With TMR (13.6 times 13.6, d = 8.2) = 0.852 and ROF (13.6 times 13.6) = 1.025, calculate the MU for the PA field; Try to explain why a higher MU is needed for the AP field than for PA, despite the same dose contribution of 400 cGy.

Explanation / Answer

Dose to deliver is,D= 800 cGy

Energy,E= 6 MV

depth is ,d=8.2 cm

X1=5.7 cm X2=8.9 cm, So X=X1+X2=5.7+8.9=14.6 cm

Y1=14.6 cm, Y2=3.6 cm, Y=14.6+3.6=18.2 cm

For PA

X1=10.5 cm X2=5.7 cm, So X=X1+X2=10.5+5.7=16.2 cm

Y1=14.6 cm, Y2=4.2 cm, Y=14.6+4.2=18.8 cm

effective width of X and Y of each field is

1) X1 is reduced to 5.1 cm.

So, X1=5.1 cm X2=8.9 cm, So X=X1+X2=5.1+8.9=14 cm

Y=18.2 cm

For PA

X1=10.5 cm X2=5.6 cm, So X=X1+X2=10.5+5.6=16.1 cm

2) Equivalent Square for AP field is,S=2*14*18.2/(14+18.2)=15.82

Equivalent Square for PA field is,S=2*16.1*18.8/(16.1+18.8)=17.34

3)MU=800 cGy/(0.849*1.018)=925.6 MU

MU=800 cGy/(0.852*1.025)=916.06 MU

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.