Appendix F
Additional Calculations
This appendix gives more detailed procedures for working radiological fallout and decay problems discussed in Chapter 6.
Determining Decay Rate
Pocket Calculator Method
Assume that you are required to find the logarithm of 12.85. Reading down Column A in Table 64, you find that a value exists for 12.8 and 12.9, but not 12.85. How do you find the log of 12.85?
Set the problem up like this: 12.85 log = 1.109 enter 12.85 hit log = 1.109.
Graphical Method
When a sequence of dose rates (NBC 4 nuclear series reports) from one location is plotted on loglog graph paper, the decay rate of the contamination causes the line plotted to be a straight line, inclined at a slope (n) to the axes of the graph.
Suppose you have readings from a set of NBC 4 nuclear series reports (Table F1) are received for decayrate determination. Hhour is known or determined to be 0930.
Figure F1 shows this data plotted on loglog graph paper. The time is used as the number of hours past Hhour. Three lines are drawn through the points. The slope of these three parallel lines is n, the decay exponent.
Remember, this is an example to demonstrate a procedure. In actual practice, the points will not likely fall exactly in straight lines. In actual practice, the best straight line is fitted to the points. The value of n may then be determined for each location and an average n determined as follows:
Place a piece of acetate, overlay paper, or other transparent material over Figure F2 and trace it. Next, orient the transparent material over the loglog paper. Position the arrow on the transparent device at the point where the slope intersects the xaxis. Holding this position, align the yaxis indicator so that it is parallel with the yaxis of the loglog paper.
Note which slope on the transparency most closely matches the slope on the loglog paper. The slope which most closely matches has an n value printed along the left side of the transparency. This is the decay rate for these plotted dose rates.
Decay rate may be calculated from the plotted slope by measuring each axis in centimeters and using the formula
(Greek letter Delta = percent of change.) 
Once the decay rate (n) is determined, the radiological reading may be normalized to H + 1 readings. This reading is commonly referred to as the R_{1} reading. This is nothing more than determining, mathematically what the dose rate reading was at any given location, one hour after the burst. Survey teams and monitors enter an area and take readings at various times after the burst (Hhour). These readings may be 15 minutes or 10 hours after the burst.
Any reading that is not recorded 1 hour (H + 1) after a burst is commonly referred to as an Rt reading. To perform radiological calculations and make decisions on the nuclear battlefield, all readings must be represented using the same time reference. If this is not done, the radioactive elements will decay and a true representation of the hazard, past and present, cannot be made.
Determining Dose Rates
Situation 1: A monitor reports a dose rate of 100 cGyph 5 hours after the burst. The decay rate is unknown so the monitor assumes standard decay (n = 1.2). What was the dose rate at the monitor's location at H + ?
This can be determined mathematically, using a handheld pocket calculator that as a power function, which is represented by a button labeled either "Y^{x}" or "X^{y}".
The following formula is used to solve the problem mathematically:
R_{1} = Rt ÷ t (Y^{x}) n +/.
(R_{1} may be calculated by R_{1} = R_{t} x ty^{xn}.
Step 1. Turn on calculator; and punch in the Rt reading be of 100 cGyph; press the ÷ key.
Step 2. Push 5 (for H + 5 when the reading was taken). Push Y^{x} or X^{y} fiction key, then the n value of 1.2.
Step 3. Press the +/ key, and then the equals (=) key. The answer should be 689.86, or 690 cGyph.
This method is the most accurate. The answer may slightly different from that found using the nomogram method discussed in Chapter 6. That is because nomograms are subject to operator error and interpretation.
Situation 2. Further monitoring determines the decay rate to be 0.9. The monitor's reading, using the procedure of Situation 1, is normalized to a new R_{1} (H + 1) of 426 cGyph. The commander wants to know what the reading will be at the monitor's location at H + 8 hours.
The formula used in Situation 1 also can be written or used as  R_{t} = R_{1} x t (Y^{x}) n (+/).
Step 1. Turn on calculator, and punch in the R_{1} reading 426 cGyph; press the multiplication key (x).
Step 2. Push 8 (for H + 8). Press the Y^{x} or X^{y} key, and enter the n value of 0.9.
Step 3. Press the +/key and then equals (=). The answer should be 66 cGyph.
Normalizing Factors
To compute normalizing factors without using the table of values method discussed in Chapter 6, use either the mathematical or graphical method.
Mathematical Method
The normalizing factor is the ratio of the ground dose rate at a reference time to the ground dose rate at any other known time. It can be expressed as 
The normalizing factor is computed using the Kaufman equation, R_{1}T_{1}n = R_{2}T_{2n }. This is the mathematical method. Subscript 1 denotes the reference time, and subscript 2 denotes any other known time a dose rate is determined.
Since:
Graphical Method
The graphical method is used when it is necessary to determine a large number of normalizing factors or to extend the time scope of an existing table. You have to know both n and Hhour.
Figure F3 shows NF plots for H + 1 and H + 48 hours. To use these plots, enter the bottom of the plot with time the dose rate was measured. Read up to the appropriate decay slope. At this intersection, read left to the lefthand scale for the NF.
Multiple Burst Procedures
Calculating Fallout of One Burst
Fallout has been received from two detonations, one at 0800Z and one at 1100Z, resulting in the readings shown in Figure F4.
Predict the dose rates for the 0800Z burst at this location 24 hours after the burst. Sufficient data is available to separate the two bursts.
Use this method if your calculator has a logarithm function, or log button and a power key, or X^{y}, or Y^{x} button.
Step 1. Calculate the decay exponent for the first burst. Divide 100 by 27 and push the log button on your calculator. Store this information in memory. Divide 3 by 1; push the log button. Divide the log of 100 ÷ 27 (stored in memory) by the log of 3 ÷ 1. The answer is 1.1918.
Step 2. Determine the decay rate for the second burst. First determine the contribution of the first burst dose rate to the 1200 hour reading of 219 cGyph. This is determined with the formula R_{1} ÷ ty^{x} n = R_{ t}:
R_{1} = H + 1 reading for the first burst
t = time in hours from the H = 1 value of the first burst to H + 1 for the second burst
n = decay rate for the first burst calculated in Step 1
Y^{x} = power button on calculator 100 ÷ 4Y^{x} 1.2 = 1.9 cGyph.
To determine the reference or peak reading of the second burst 
219 cGyph  19 cGyph = 200 cGyph at 1200 from the second burst
R_{a }= H + 1 reading of second burst
R_{b} = Last reading minus contribution from first burst
T_{b} = Time, in hours, of last reading from detonation time of second burst
T_{a} = Time, in hours, of reference reading for second burst
Rt value for the first burst: Rt 
= (R1 ÷ ty^{x} n) 
= (100 ÷ 5y^{x} 1.2) 

= Rt = 14.5 cGyph. 
The R_{1} value in this case is 100 cGyph. Use the value of 5 hours for t, because the reading of 108 cGyph occurred at 1300, which is 5 hours after the first burst. Push the y^{x} power key on the calculator, then the decay rate of 1.2 for the first burst. The answer, in this case, is 14.5 cGyph contribution from the first burst.
Using the log button on the calculator, find the log of the top and bottom numbers, then divide 0.3302 by 0.3010; and your answer is 1.097, or 1.1.
The decay rate, therefore, after rounding to the nearest tenth, for the second burst is 1.1.
Step 3. Calculate the 0800Z dose rate 24 hours after the first burst. Remember, the T_{b} for 1.1 is only 5 hrs. You have to use 1.2 or get another series report.
R24_{x} = R_{1}÷ 24Y^{x} 1.2
R24_{x }= rate 24 hours after first burst
R_{1} = R_{1} reading for first burst
Y^{x} = power key
R24_{x} = 100 ÷ 24y^{x} 1.2
R24_{x} = 2.2067 cGyph
R21_{y} = R_{1}÷ y^{x}, 1.1
R21_{y} = rate of second burst at 0800Z
R_{1} = R_{1 }or reference reading for the second burst
21 = 0800Z is 21 hours after the detonation time of the second burst
Y^{x} = power key on calculator
1.1 = decay rate for second burst
R21_{y} = 7.02 cGyph
R21_{y} = 7.02 cGyph.
Add the reading for 24 hours after the first burst to the reading 21 hours after the second burst:
2.2067 + 7.02 = 0.23, or 9 cGyph.
By using these procedures you can determine any dose rate, at any particular time with your calculator.
R_{1} = Rt (t) y^{x} n
Rt = R_{l} ÷ t y^{x} n
t = Rt ÷ R_{1} = INV y^{x} n t/.
To solve for t is a little more complicated than the other procedures.
Given: R_{t} = 7
R_{1}= 200
n = 1.1
Find: t.
Divide 7 by 200. Push equals. Push the INV button, then 1.1, then the +/button; then the equals button. In other words, t = R_{t} ÷ R_{l} = INV Y^{x}n + 1 .
Graphical Method
Use the graphical method when sufficient data is not available to separate the multiple burst readings.
Fallout has been received from two detonations. Doserate measurements were made at the intervals shown in Figure F5. The time of the second burst is 0800. Time of the first burst is not known.
After receiving the measurement made at 1100, predict the dose rate at that location at 2000 on the following day (36 hours after the burst). After receiving each succeeding doserate measurement, update this prediction. Sufficient data are not available to separate the two bursts.
Step 1. Plot on loggraph paper the 0900 and 1100 doserate measurements against the time after the second burst.
Step 2. Draw a straight line through these points and extrapolate the line past H + 36 hours (see Figure F6).
Step 3. As a first approximation, determine a dose rate of 28.0 cGyph for 2000 on the day following the burst (R_{36}) directly from the graph.
Step 4. Upon receiving the 1300 measurement, plot this reading on the graph.
Step 5. Draw a new straight line through the 1100 and 1300 points; and extrapolate the line past H + 36 hours.
Step 6. As a second (and better) approximation, determine a dose rate of 20.5 cGyph for H + 36 hours directly from the second extrapolation.
Step 7. Repeat the procedure described in steps 4 through 6. Upon receipt of the 1500, 0200, and 1200 measurements, update the prediction for R36 to 18.0, 14.5, and 12.5 cGyph, respectively.
Step 8. See Figure F6 for an illustration of the doserate calculation. Reading from this figure, the true dose rate encountered at H + 36 hours at that location is 12.5 cGyph.
Calculating Overlapping Fallout
Hhour is known for each burst. At 251500, a 20KT nuclear weapon was detonated on the surface "near" your position. Sometime later, fallout arrived on your position. At 1630, a peak dose rate of 126 cGyph was measured. Subsequent readings indicated that n = 1.4. At 251700, another weapon was detonated close to your area, and fallout arrived soon after. At 251730, a second peak dose rate of 300 cGyph was measured.
Assuming that n = 1.2 for the second weapon, what will the dose rate be at 1900? This may be calculated using Step 3 of the calculator method for determining an R_{1} value.
When Hhour for each detonation is known, calculate the dose or dose rate for each event and add them together to get the total dose or dose rate received.
Step 1. Find R_{1} for the first detonation.
R_{1} = [Rt (t), y^{x}n]
R_{1} = [126 (1.5) y^{x}, 1.4]
R_{1} = 222.3 cGyph.
Step 2. Find R_{t} at 1730 for first fallout only.
R_{t} = R_{1} ÿ t Y^{x}n
R_{t} = 222.3 ÷ 2.5 Y^{x} 1.4
R_{t} = 61.6, or 62 cGyph.
Step 3. Find the dose rate contribution at 1730 from the second burst.
Rt_{a} = R_{t}  R_{t}1
R_{ta} = 300 cGyph  62 cGyph
R_{ta} = 238 cGyph.
Step 4. Find R_{1a} for the second burst only.
R_{1} = R_{t} (t), Y^{x}. n
R_{1} = 238 (0.5), Y^{x}, 1.2
R^{1} = 103.5, or 104 cGyph.
Step 5. Find R_{t} at 1900 for each burst. For the first burst
R_{1} = 222 cGyph
t = H + 4 hours
R_{t} = (R_{1} ÿ t, Y^{x}, n)
R_{t} = 222 ÿ 4, Y^{x}, 1.4
R_{t} = 31.8, or 32 cGyph.
For the second burst
R_{1} = 104 cGyph
t = H +2
R_{t} = R_{l} ÷ t, Y^{x}, n
R_{t} = 104 ÷ 2, Y^{x}, 1.2
R_{t} = 45.3, or 45 cGyph.
Step 6. Find the total dose rate at 1900. Total dose rate is the sum of dose rates at that time.
R total = R_{t} + R_{t2}
R total = 32 + 45
R total = 77 cGyph.
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