INGENIA

RTH-10

Tissue–phantom ratio

TPR(d,dref) = D(d)/D(dref) at fixed SAD. TPR20,10 indexes beam quality.

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Beam dataIAEA TRS-398TPR20,10

Governing equation

TPR(d,dref)=e−μ(d−dref)\mathrm{TPR}(d,d_{\mathrm{ref}})=e^{-\mu(d-d_{\mathrm{ref}})}

d∈[1,30]d \in \left[1, 30\right]

where

\mu
Attenuation μ (1/cm)
d
Depth (cm)
d_{ref}
Reference depth (cm)
TPR
Tissue–phantom ratio (—)

The figure, part by part

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RTH-10 · dose
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①

Governing law — red

TPR(d,dref)=e−μ(d−dref)\mathrm{TPR}(d,d_{\mathrm{ref}})=e^{-\mu(d-d_{\mathrm{ref}})}

d∈[1,30]d \in \left[1, 30\right]

d=1d = 1TPR=1.568TPR = 1.568

②

Domain of the variable — blue

dmin⁡=1,dmax⁡=30d_{\min} = 1,\quad d_{\max} = 30

d∈[1,30]d \in \left[1, 30\right]

d=1d = 1d=1d = 1

③

Response curve — green

TPR(d,dref)=e−μ(d−dref)\mathrm{TPR}(d,d_{\mathrm{ref}})=e^{-\mu(d-d_{\mathrm{ref}})}

d∈[1,30]d \in \left[1, 30\right]

d=1d = 1TPR=1.568TPR = 1.568

④

Rate of change — amber

dTPRdd\frac{\mathrm{d}TPR}{\mathrm{d}d}

d∈[1,30]d \in \left[1, 30\right]

d=1.182d = 1.182TPR′=−0.078TPR^{\prime} = -0.078

Lecture brief

Historical brief

The linear-quadratic model, BED, tissue-maximum ratio and Clarkson scatter turned radiotherapy into fraction arithmetic. MU calculation on these sheets is that clinical closed form. This sheet (RTH-10 — Tissue–phantom ratio) is the form associated with IAEA TRS-398 · TPR20,10. Working symbols: μ\mu, dd, drefd_{ref} →\rightarrow TPRTPR. Unlike PDD there is no inverse-square change of SAD. TPR20,10 ≈ 0.67–0.80 for clinical x-rays.

Purpose

Purpose: compute TPRTPR from μ\mu, dd, drefd_{ref} in Radiotherapy via TPR(d,dref)=e−μ(d−dref)\mathrm{TPR}(d,d_{\mathrm{ref}})=e^{-\mu(d-d_{\mathrm{ref}})} TPR(d,dref) = D(d)/D(dref) at fixed SAD. TPR20,10 indexes beam quality. Use it when a real radiotherapy question must be answered in SI before a code check.

Live realistic example

In symbols

Live case. Given μ=0.050 1/cm\mu = 0.050\,\mathrm{1/cm}, d=20.000 cmd = 20.000\,\mathrm{cm}, dref=10.000 cmd_{ref} = 10.000\,\mathrm{cm}, the governing relation TPR(d,dref)=e−μ(d−dref)\mathrm{TPR}(d,d_{\mathrm{ref}})=e^{-\mu(d-d_{\mathrm{ref}})} yields TPR=0.6065 —TPR = 0.6065\,\mathrm{—}. Two phantom depths at isocentre. Move a slider: the numbers are this situation, not a canned story.

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Inputs

Outputs

  • Tissue–phantom ratio TPR0.6065 —
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Narration of this film

Two phantom depths at isocentre.

Unlike PDD there is no inverse-square change of SAD. TPR20,10 ≈ 0.67–0.80 for clinical x-rays.

Reading speed

Watch on YouTube