Class Handout

Interpolating Instructions

Please note: if mathematics is not your forte, skip down to 4a!! Otherwise, work through the Linear Interpolation portion of this document first, and then refer back to this explanation to see the relationship between my equation and the “shortcut” method.

1)  Find the PDD for the two numbers involved

  1. These will be PDD1 and PDD2

2)  Find the difference between the two numbers

  1. This satisfies the (PDD2 – PDD1) portion of finding the slope m.
  2. Using my FS equation, the answer is positive because PDD1 is smaller than PDD2.
  3. Using my Depth equation, the answer is negative because PDD1 is higher than PDD2.
  4. m = slope = ((y2 – y1)/ ( x2 – x1)) = ((PDD2 – PDD1)/ ( FS2 – FS1)) using FS, and

c.  m = slope = ((y2 – y1)/ ( x2 – x1)) = ((PDD2 – PDD1)/ ( D2 – D1)) using Depth

  1. Notice that you are ignoring the (x2 – x1) portion of this step. This assumes that the difference in FS or D = 1. Dividing by 1 does not change the numerator. This is NOT the case if the FS or D on your PDD chart jumps by an interval not equal to 1 (For example: the FS in the chart jumps from 20 to 22, a difference of 2. Therefore, using the “shortcut” method will not give you an accurate result.)

3)  Multiply the difference by the fraction of the number needed

  1. The fraction of the number needed is (FSN – FS1) using FS or (DN – D1) using Depth
  2. Completing this step gives you the mx of your y = mx + b, (assuming the difference of part 2d above is 1) and again:
  3. Using my FS equation, the answer is positive
  4. Using my Depth equation, the answer is negative

4)  If the first PDD listed in chart is smaller than second, add #3 to it; if larger, subtract from it

  1. If not using my equation, #4 can be rephrased by saying:
  2. If you are interpolating using FS, add #3 to the smaller PDD
  3. If you are interpolating using Depth, subtract #3 from the higher PDD
  4. If using my equation, the fact that mx is positive using FS and negative using Depth takes the brainwork out- just add the b (b = y intercept = PDD1) to mx to finish the y = mx + b equation.

Linear Interpolation with Field Size:

PDD increases with increased field size.

Low Field Size / Field Size Needed / High Field Size
Depth / Low PDD / PDD needed / High PDD

x1 = Low Field Size = FS1

x2 = High Field Size = FS2

x = (Field Size Needed – Low Field Size) = (FSN – FS1)

y = PDD Needed = y

y1 = Low PDD = PDD1

y2 = High PDD = PDD2

m = slope = ((y2 – y1)/ ( x2 – x1)) = ((PDD2 – PDD1)/ ( FS2 – FS1))

b = y intercept = PDD1

y = mx+b

y = ((PDD2 – PDD1)/ ( FS2 – FS1)) * (FSN – FS1) + PDD1

Linear Interpolation with Depth:

PDD decreases with increased depth

Field Size
Low Depth / High PDD
Depth Needed / PDD Needed
High Depth / Low PDD

x1 = Low Depth = D1

x2 = High Depth = D2

x = (Depth Needed – Low Depth) = (DN – D1)

y = PDD Needed = y

y1 = High PDD = PDD1

y2 = Low PDD = PDD2

m = slope = ((y2 – y1)/ ( x2 – x1)) = ((PDD2 – PDD1)/ ( D2 – D1))

b = y intercept = PDD1

y = mx+b

y = ((PDD2 – PDD1)/ ( D2 – D1)) * (DN – D1)+ PDD1

Linear Interpolation Example:

*I find it helpful to organize our information in the same fashion as the information is given in a PDD chart, with the Field Size along the top and the Depth along the side.

Low Field Size / Field Size Needed / High Field Size
Low Depth
Depth Needed / PDD NEEDED
High Depth

*Example: Find the PDD needed to treat an effective field size of 11.3 cm2 at a depth of 7.8 cm with a 6 MV beam at 100 SSD.

11 / 11.3 / 12
7 cm
7.8 cm / PDD NEEDED
8 cm

*Shaded Areas can be found in the 6 MV PDD Chart.

11 / 11.3 / 12
7 cm / .793 / .796
7.8 cm / PDD NEEDED
8 cm / .755 / .759

*Use Linear Interpolation to find values A, B, C and D.

11 / 11.3 / 12
7 cm / .793 / A / .796
7.8 cm / C / PDD NEEDED / D
8 cm / .755 / B / .759

______

Using the Field Size Equation to find A:

11 / 11.3 / 12
7 cm / .793 / A / .796
7.8 cm / C / PDD NEEDED / D
8 cm / .755 / B / .759

A = ((PDD2 – PDD1)/ (FS2 – FS1)) * (FSN – FS1) + PDD1

A = ((.796 - .793)/(12 – 11)) * (11.3 – 11) + .793

A = .7939

Using the Field Size Equation to find B:

11 / 11.3 / 12
7 cm / .793 / .7939 / .796
7.8 cm / C / PDD NEEDED / D
8 cm / .755 / B / .759

B = ((PDD2 – PDD1)/ (FS2 – FS1)) * (FSN – FS1) + PDD1

B = ((.759 - .755)/(12 – 11)) * (11.3 – 11) +.755

B = .7562

Using the Depth Equation to find C:

11 / 11.3 / 12
7 cm / .793 / .7939 / .796
7.8 cm / C / PDD NEEDED / D
8 cm / .755 / .7562 / .759

C = ((PDD2 – PDD1)/ ( D2 – D1)) * (DN – D1)+ PDD1

C = ((.755 - .793)/(8 – 7)) * (7.8 – 7) + .793

C = .7626

Using the Depth Equation to find D:

11 / 11.3 / 12
7 cm / .793 / .7939 / .796
7.8 cm / .7626 / PDD NEEDED / D
8 cm / .755 / .7562 / .759

D = ((PDD2 – PDD1)/ (D2 – D1)) * (DN – D1) + PDD1

D = ((.759 - .796)/ (8 – 7)) * (7.8 – 7) + .796

D = .7664

Now you are ready to find your PDD NEEDED:

11 / 11.3 / 12
7 cm / .793 / .7939 / .796
7.8 cm / .7626 / PDD NEEDED / .7664
8 cm / .755 / .7562 / .759

To find the PDD NEEDED, you can use the Depth equation using values A and B:

11 / 11.3 / 12
7 cm / .793 / .7939 / .796
7.8 cm / .7626 / PDD NEEDED / .7664
8 cm / .755 / .7562 / .759

Depth:

PDD NEEDED = ((PDD2 – PDD1)/ ( D2 – D1)) * (DN – D1)+ PDD1

PDD NEEDED = ((.7562 - .7939)/(8 – 7)) * (7.8 – 7) + .7939

PDD NEEDED = .7637

Alternatively, you can use the Field Size Equation using values C and D:

11 / 11.3 / 12
7 cm / .793 / .7939 / .796
7.8 cm / .7626 / PDD NEEDED / .7664
8 cm / .755 / .7562 / .759

Field Size:

PDD NEEDED = PDD2 – PDD1)/ ( FS2 – FS1)) * (FSN – FS1) + PDD1

PDD NEEDED = ((.7664 - .7626)/(12 – 11)) * (11.3 – 11) + .7626

PDD NEEDED = .7637

Note: Only one of the above methods is needed because they both give the same answer.

PDD NEEDED:

11 / 11.3 / 12
7 cm / .793 / .7939 / .796
7.8 cm / .7626 / .7637 / .7664
8 cm / .755 / .7562 / .759