X-Ray Machine

Regarding the machine the request specifies:

  • New machine
  • Acteon make
  • Prime model

If it were a replacement, some shielding must be already in place which changes somewhat the nature of the task.

Looking up the web, turns out a website for Acteon. The site does not provide a search feature, which is kind of unusual nowadays. Using the menu, under “PRODUCTS” one finds X-MIND PRIME which seems to be the only option matching the request.

The product page provides information as PDF brochures which can be downloaded. The X-Mind Prime 3D brochure in particular, contains a set of technical specs which are useful for the shielding calculation:

Using “MY DOCUMENTATION” option in the site menu and a selection of options (“Imaging”, “3D & Panoramic Images”, “X-Mind Prime 3D”) then clicking “SEARCH” one can reach more documentation, such as the user manual.

At first sight it is not clear whether the 2D & 3D sets documented in the technical specs belong to the same machine or not. But the weight listed being different, means these are separate machines.

The order does not specify which one will be installed, therefore one must take a conservative approach and use the 3D option for the shielding calculations. The relevant data for shielding is listed below:

  • Tube voltage: 60 - 86 kV
  • Anodic current: 2 - 12.5 mA
  • Exposure time: 7 sec for 3D, however the manual indicates 21.2 sec for CBCT
  • 3D Fields of view: Full dentition (12 x 10cm), Full dentition (8.5 x 9.3cm), Single jaw (8.5 x 5cm), Mandibular teeth (5 x 5cm), Maxillary teeth (5 x 5cm), TMJ (8.5 x 9.3cm)* - Sinus (8.5 x 9.3cm)
  • Total Filtration: \(\geq 2.5\) mm Al @ 86 kVp

The last one above is useful for checking the filtration against “Radiation Protection in Dentistry - Recommended Safety Procedures for the Use of Dental X-Ray Equipment - Safety Code 30”.

Giving that 2.5 mm is listed for 90 kV, and the equipment specs give a value \(\geq 2.5\) mm for 86 kVp, this check passes.

Workload (W)

The request provides the following estimates:

  • 5 exams per week to start with
  • 10 exams per week in the future

Current practice shows that the estimated amount needs to be scaled up two or three times so the number of exams per week becomes \(10 \times 3 = 30\).

According to the manual (page 41, screenshot below) one exam lasts 21.2 seconds and the maximum operating current is 12.5 mA. Therefore the weekly workload becomes: \[W = 30 \times 21.2 \times 12.5 = 7950 \text{ mAs / week}\].

Shielding Considerations

Primary Barrier

The primary barrier provides shielding against the primary beam. However in this instance the primary beam is prevented to reach any of the room limits. Following considerations apply to the floor and ceiling:

  • The patient is standing so the scattered radiation has to pass through a lot of tissue before it meets the floor.
  • The scattered radiation going to the ceiling, must pass through a steel gantry above the patient’s head.

And the next considerations apply to the other room walls:

  • The beams are well collimated.
  • The radiation source is always opposite to the imaging plate during an exam.
  • The imaging plate is generally opaque to the levels of radiation produced by these machines.
  • The primary beam is strongly attenuated by the patient, the imaging plate and the steel in the internal support mechanisms.

Overall, the above effectively rule out the primary beam contribution, therefore it is not necessary to perform a primary barrier calculation.

Secondary barrier

The secondary barrier provides shielding against leakage and scattered radiation. The leakage is whatever radiation comes out of the source enclosure, apart from the primary beam. By law, the manufacturer has to limit the leakage amount to some specified value, therefore the shielding calculations are using it (conservative approach). Directly measured values are even better when possible, but that is typically not the case.

The scattered radiation is whatever originates in the target (patient), imaging plate & even walls. The amount is a function of of energy (kV) and the scattering angle. It also scales linearly with the field size.

Leakage Barrier

The leakage amount is limited (according to NCRP Report 49, page 62) to 1 roentgen per hour at 1 meter distance from the source.

However in this case, the manual indicates the amount of leakage for this particular scanner, on page 29 as being less than 0.5 mGy/hr @ 1 m.

One complication arises from the fact that with the CBCT scanner, the source rotates - this is in contrast to NCRP 49 formalism, where the leakage source is fixed. On top of that the rotation is not even complete, but only \(270^0\) as shown in the image below, reproduced from “Acteon X-Mind Prime - User Manual” page 32.

The derivation of the formula which takes into consideration the leakage source rotation, is detailed in Appendix A. The result is surprisingly simple and elegant for a full rotation, but looks quite acceptable for incomplete rotations too.

According to the same manual, page 31, the distance between the source and the imaging panel is 52 cm (20”). From the image above it would seem that the scatterer (patient’s head) is centered in the middle during exposure. However, the manual (page 4) specifies that the average distance between the focal spot and the patient skin is 264 mm (screenshot included below).

An internet search reveals that according to some study in the US, the average head circumference is about 56 cm. This results in an average head radius of 89 mm - which is roughly confirmed my measuring myself.

Therefore, the distance between the center of rotation and the source must be about \(r = 264 + 89 = 353 \text{ mm} \approx 35 \text{ cm}\). Another fact which corroborates this value is that the maximum field size listed for the machine scales almost perfectly with the listed imager side if the rotation radius is 35 cm.

Based on the image above, the center of rotation is distanced from the mounting wall by \(1107 - 953/2 = 630.5 \text{ mm} \approx 63 \text{ cm}\).

The occupancy factor \((T = 1)\) because it is not known at this time the exact layout of the facility and where the scanner room will be, so we consider the worst case scenario. The usage factor is always unity for leakage radiation \((U = 1)\).

Scatter Barrier

NCRP Report 49, page 66 provides a table of ratios between the scattered and incident exposures for various kV values and scatter angles (reproduced below). In deriving this table, the scatter values were evaluated at 1 meter from the scatterer and \(400 \text{ cm}^2\) field opening, while the scatterer is itself at 1 meter from the source.

A few considerations apply:

  • The scattered radiation is continually blocked (or attenuated) by the opposing source and image enclosures during rotation. Following a conservative line of reasoning we discount this limitation and carry out the calculation as if these attenuators are not present.
  • Following from the above, the scatter from the imaging plate and walls can be neglected (it is compensated by not counting the attenuation from the source and image panel enclosures).
  • The scatter angle variation is averaged by the rotation.
  • The closest upward voltage value listed in the table is 100 kV.

From the above, the patient target is the only source of scattered radiation and the estimate value of the ratio for our case is: \[a = \frac{0.0015 + 0.0012 + 0.0012 + 0.0013 + 0.0020 + 0.0022}{6} = 0.001566667 \approx 0.0016\] The maximum field size in this case is \(F = 12 \text{ cm} \times 10 \text{ cm} = 120 \text{ cm}^2\). Considering a field enlargement of \(4 \text{ cm}\) on both directions, the field size value scales to \(F = 16 \text{ cm} \times 14 \text{ cm} = 224 \text{ cm}^2\) and this value will be used for the purpose of this calculation.

The distance between the source and scatterer is \(d_{sca} = 35 \text{ cm}\) as discussed in the leakage barrier section above.

The occupancy factor \((T = 1)\) because it is not known at this time the exact layout of the facility and where the scanner room will be, so we consider the worst case scenario. The usage factor is always unity for scattered radiation \((U = 1)\).

With these factors equal to unity, the weekly scattered radiation in any direction and at distance \(D\) is \(X_S = \frac{aW}{D^2}\).

Numerical calculations

Before diving into the actual numerical calcs, we must consider the geometry of the install (see next image extracted from the request). The drawing is not scaled well (e.g. the distance marked as 71” is about the same size as the one marked as 58”) and the distance between left and far right is not given. Still, it can be used to some extent.

The machine is going to be installed on the 71” wall, presumably in the middle. Likely the metal mounting provides adequate shielding, but then it does not cover the whole wall - therefore we will do a calculation for this wall too, discounting this shielding. With the floor and ceiling ruled out, we are left with four walls (mounting wall, its opposite and the laterals).

Since the machine is in the middle, the calculation for the laterals would ideally coincide leaving only three calculations. However, because the scanner does not rotate a full circle, this introduces an asymmetry between the lateral walls so we get to do four calculations.

For reference purposes, we will imagine a person standing with its back at the mounting wall & refer to the four walls as the left, right, front & back wall respectively.

One more thing, according to NCRP Report 49, page 4 any person outside the scanning room is assumed to be at least 30 cm (12 inches) away from the walls (screenshot below) and that is where the calculation point will be.

The maximum permissible exposure according to the same report is \(P = 0.1 \text{ R}\). However, nowadays this limit is lower and has a different measurement unit \(P = 1 \text{ [mSv/yr]}\).

For calculation purposes, this must be converted back to \(\text{R/week}\). In the “Safety Code 30” document, in Appendix V, page 69 it is stated that \(\text{1 mGy = 114.55 mR}\) (screenshot included below). Because only photons are involved, the tissue weighting factor is \((T_W = 1)\) and therefore the limit translates to: \[P = 1 \text{ [mGy/yr]} = 114.55 \text{ [mR/yr]}\] One year has 52 weeks so: \[P = 114.55/52 = 2.202885\text{ [mR/week]} \approx 0.0022\text{ [R/week]}\]

Leakage Barrier

The formula used by NCRP 49 for the transmission ratio is:

\[B_{LX} = \frac{Pd_{sec}^2 60I}{LWUT}\] This value is then used to find out the number of half value layers (HVL) or tenth value layers (TVL) - using figure B3 @ page 57 (reproduced below). Finally, with this number one needs to look up table 27 on page 88 for a listing of HVLs & TVLs for various materials (table reproduced below). The barrier thickness is then expressed as: \[S_L = N(HVL) = n(TVL)\]

Left wall

The parameters used by the transmission formula are the following: \[ \begin{aligned} P & = 0.0022\text{ [R/week]} \\ L & = 0.5 \text{ mGy/hr} = 0.5 * 114.55 \text{ mR/hr} = 57.275 \text{ mR/hr} \approx 0.0573 \text{ R/hr} \\ W & = 7950 \text{ mAs / week} = 7950/60 \text{ mA}\cdot\text{min} \text{/week} = 132.5 \text{ mA}\cdot\text{min} \text{/week} \\ T & = 1 \\ U & = 1 \\ I & = 12.5 \text{ mA} \\ d_{sec} & = (71 * 2.54) / 2 + 30 \approx 120 \text{ cm} = 1.2 \text{ m} \end{aligned} \] However, the term \(d_{sec}^2\) is purely geometric and not accurate because of the fact that the leakage source rotates. According to Appendix A, the factor representing \(d_{sec}^2\) is \[\frac{\pi(D^2 - r^2)}{2F(\theta_1, \theta_2)}\] with \(D = d_{sec}\) & \(r = 35 \text{ cm}\).

To evaluate the factor \(F(\theta_1, \theta_2)\) we must take into account that the missing quarter of the circle is towards this wall. Therefore, \[F(\theta_1, \theta_2) = F(\pi/4, 7\pi/4) = 2F(\pi/4, \pi) = \dots\] \[\dots = 2\left\{\pi/2 - \text{arctan}\left[\frac{D+r}{D-r}(\sqrt{2} - 1)\right]\right\}\] Using the known values for \(D\) & \(r\) above: \[F(\theta_1, \theta_2) = \pi - 2\text{arctan}\left[\frac{D+r}{D-r}(\sqrt{2} - 1)\right] = 1.847785\] So \(d_{sec}\) becomes: \[d_{sec} = \sqrt{\frac{\pi(D^2 - r^2)}{2F(\theta_1, \theta_2)}} = 105.83 \text{ cm} \approx 1.06 \text{ m}\] which is less than the original value of 1.2 m.

The transmission value is then: \[B_{LX} = \frac{Pd_{sec}^2 60I}{LWUT} = \frac{(0.0022)(1.06)^2(60)(12.5)}{(0.0573)(132.5)(1)(1)} \approx 0.244\] From table B3, turns out we need \(N = 2\text{ HVL}\) or \(S_L = 2\times 0.27 = 0.54\text{ mm [Pb]}\). Alternatively we’d need \(S_L = 2\times 1.6 = 3.2\text{ cm [Concrete]}\).

Right wall

The parameters used by the transmission formula are exactly the same with those above, except this time the missing quarter of the circle is opposite the wall. Therefore, the factor \(F(\theta_1, \theta_2)\) evaluates this way:

\[F(\theta_1, \theta_2) = F(0, 3\pi/4) + F(5\pi/4, 2\pi) = 2F(0, 3\pi/4) = 2\text{arctan}\left[\frac{D+r}{D-r}(\sqrt{2} + 1)\right]\] Using the known values for \(D\) & \(r\) above: \[F(\theta_1, \theta_2) = 2\text{arctan}\left[\frac{D+r}{D-r}(\sqrt{2} + 1)\right] = 2.694874\] So \(d_{sec}\) becomes: \[d_{sec} = \sqrt{\frac{\pi(D^2 - r^2)}{2F(\theta_1, \theta_2)}} = 87.63266 \text{ cm} \approx 0.876 \text{ m}\] which is even lesser than the original value of 1.2 m.

The transmission value is then: \[B_{LX} = \frac{Pd_{sec}^2 60I}{LWUT} = \frac{(0.0022)(0.876)^2(60)(12.5)}{(0.0573)(132.5)(1)(1)} \approx 0.167\] From table B3, turns out we need \(N = 2.8\text{ HVL}\) or \(S_L = 2.8\times 0.27 = 0.756\text{ mm [Pb]}\). Alternatively we’d need \(S_L = 2.8\times 1.6 \approx 4.5\text{ cm [Concrete]}\).

Back wall

The parameters used by the transmission formula are again the same as above, except this time the geometric term \(d_{sec} = 63 + 30 = 93 \text{ cm}\). Of course, this value needs to be corrected for the rotating leakage source, and to do that we need to observe that the missing quarter of the circle is sideways. Therefore, the factor \(F(\theta_1, \theta_2)\) will evaluate this way:

\[F(\theta_1, \theta_2) = 2F(0, \pi) - F(0, 3\pi/4) + F(0, \pi/4) = \dots\] \[\dots = 2\left(\frac{\pi}{2}\right) - \text{arctan}\left[\frac{D+r}{D-r}(\sqrt{2} + 1)\right] + \text{arctan}\left[\frac{D+r}{D-r}(\sqrt{2} - 1)\right] = 2.496894\] So \(d_{sec}\) becomes: \[d_{sec} = \sqrt{\frac{\pi(D^2 - r^2)}{2F(\theta_1, \theta_2)}} = 68.34 \text{ cm} \approx 0.68 \text{ m}\] which is less that the original value of 0.93 m, as expected.

The transmission value is then: \[B_{LX} = \frac{Pd_{sec}^2 60I}{LWUT} = \frac{(0.0022)(0.68)^2(60)(12.5)}{(0.0573)(132.5)(1)(1)} \approx 0.1\] From table B3, turns out we need \(N = 3.4\text{ HVL}\) or \(S_L = 3.4\times 0.27 = 0.92\text{ mm [Pb]}\). Alternatively we’d need \(S_L = 3.4\times 1.6 \approx 5.5\text{ cm [Concrete]}\).

Front wall

The front side is made of glass and unfortunately the schematic provided does not indicate the distance to the far side. However, it does indicate the distance between the mounting wall and one of the glass wall edges \(74^{"} = 188 \text{ cm}\). We will use this distance.

The other parameters are the same as above, but the geometric term \(d_{sec} = 188 - 63 + 30 = 155 \text{ cm}\). The correction takes into account that the missing quarter of the circle is sideways just as above. Therefore, the factor evaluates to the same value above \(F(\theta_1, \theta_2) = 2.496894\).

The corrected \(d_{sec}\) will be: \[d_{sec} = \sqrt{\frac{\pi(D^2 - r^2)}{2F(\theta_1, \theta_2)}} = 119.76 \text{ cm} \approx 1.2 \text{ m}\] which is of course less that the original value of 1.55 m.

The transmission value is then: \[B_{LX} = \frac{Pd_{sec}^2 60I}{LWUT} = \frac{(0.0022)(1.2)^2(60)(12.5)}{(0.0573)(132.5)(1)(1)} \approx 0.313\] From table B3, turns out we need \(N = 1.4\text{ HVL}\) or \(S_L = 1.4\times 0.27 = 0.38\text{ mm [Pb]}\). Alternatively we’d need \(S_L = 1.4\times 1.6 = 2.24\text{ cm [Concrete]}\).

Scatter Barrier

The formula used by NCRP 49 for the transmission ratio is:

\[K_{UX} = \dot X_NB_{SX} = \frac{P}{aWUT}d_{sec}^2d_{sca}^2\frac{400}{F}\] where:

  • \(d_{sca}\) accounts for the scatterer being at a distance from the source different than 1 m. For the case at hand, this value is 0.35 m.
  • F is the maximum field size of the beam (scaled to account for field enlargement).

This value is then used together with the figures in NCRP 49, Appendix D (pages 91 & 93 in this case) to determine the minimum shielding required (lead and concrete). The figures are reproduced in the Appendix C of this document.

In contrast to the leakage case, the distances \(d_{sec}\) to the four walls do not need to be corrected anymore because the scatterer does not move.

Left & right walls

Due to the machine being mounted in the middle of the back wall, scatter calculations on teh sides coincide. The parameters used by the transmission formula are the following: \[ \begin{aligned} P & = 0.0022\text{ [R/week]} \\ a & = 0.0016 \\ F & = 224 \text{ cm}^2 \\ W & = 7950 \text{ mAs / week} = 7950/60 \text{ mA}\cdot\text{min} \text{/week} = 132.5 \text{ mA}\cdot\text{min} \text{/week} \\ T & = 1 \\ U & = 1 \\ I & = 12.5 \text{ mA} \\ d_{sec} & = (71 * 2.54) / 2 + 30 \approx 120 \text{ cm} = 1.2 \text{ m} \\ d_{sca} & = 0.35 \text{ m} \\ \end{aligned} \] With these values, the transmission value becomes: \[K_{UX} = \frac{(0.0022)}{(0.0016)(132.5)(1)(1)}(1.2^2)(0.35^2)\frac{(400)}{(224)} = 0.003268868 \approx 0.00327\] Using Fig 1 in Appendix D (NCRP 49), also reproduced in this document’s Appendix C, we get 0.8 mm lead. Using Fig 3 we get 8.5 cm concrete.

Back wall

In this case, the only difference is \(d_{sca}\) which is 93 cm. \[K_{UX} = \frac{(0.0022)}{(0.0016)(132.5)(1)(1)}(0.93^2)(0.35^2)\frac{(400)}{(224)} = 0.001963364 \approx 0.002\] Which results in 1.2 mm lead or 9.5 cm concrete.

Front wall

In this case, \(d_{sca}\) is 188 cm. \[K_{UX} = \frac{(0.0022)}{(0.0016)(132.5)(1)(1)}(1.88^2)(0.35^2)\frac{(400)}{(224)} = 0.008023255 \approx 0.008\] Which results in 0.62 mm lead or 6.5 cm concrete.

Appendix A

The following drawing depicts the geometry of the rotating leakage source. To make the image clearer the sizes of the circles were reversed (the largest circle is drawn with a lower radius and the other way around), However, the principle remains the same regardless of this little trick.

Assuming a uniform angular rotation speed \((\omega = \frac{\text{d}\theta}{\text{d}t} = \frac{2\pi}{T} = \text{constant})\) the relationship between the infinitezimal interval of time the source is in \(S\) and the infinitezimal angle variation is: \[\text{d}t = \frac{T}{2\pi}\text{d}\theta\] The infinitezimal contribution of the source to the exposure in point \(P\) during this time is given by: \[\text{d}X_P = \dot X_L \frac{1}{\lvert SP \rvert^2} \text{d}t = \dot X_L \times \frac{1}{\lvert SP \rvert^2} \times \frac{T}{2\pi}\text{d}\theta = \frac{T \dot X_L}{2\pi \lvert SP \rvert ^2} \text{d}\theta\] To find out the whole contribution from leakage at point \(P\) during a full rotation of the scanner, the expression above must be integrated between applicable angles \(\theta_1\) and \(\theta_2\):

\[X_P = \frac{T}{2\pi}\dot X_L \int_{\theta_1}^{\theta_2} \frac{\text{d}\theta}{\lvert SP \rvert^2}\] The cosine theorem applied to the triangle \(\overline {\text{SOP}}\) gives the following relationship: \[\lvert SP \rvert^2 = \lvert OS \rvert^2 + \lvert OP \rvert^2 - 2\lvert OS \rvert \lvert OP \rvert cos\theta\] or using the notations from the drawing \((\lvert OS \rvert = r \text{ & } \lvert OP \rvert = D)\) \[\lvert SP \rvert^2 = r^2 + D^2 - 2rDcos\theta\] Therefore the expression for the exposure in point \(P\) becomes: \[X_P = \frac{T}{2\pi}\dot X_L \int_{\theta_1}^{\theta_2} \frac{\text{d}\theta}{r^2 + D^2 - 2rDcos\theta} = \frac{T}{\pi}\dot X_L \int_{\theta_1}^{\theta_2} \frac{\text{d}\theta}{r^2 + D^2 - 2rDcos\theta}\] To integrate the above we start by changing the variable to \(t = tan\left(\frac{\theta}{2}\right)\) which results in \(cos\theta = \frac{1 - t^2}{1 + t^2}\) and \(\text{d}\theta = \frac{2}{1+t^2}\text{d}t\). After substitutions and some basic algebra we get: \[X_P = \frac{T}{\pi}\dot X_L \frac{2}{(D-r)^2} \int_{\text{tan}(\theta_1/2)}^{\text{tan}(\theta_2/2)} \frac{\text{d}t}{1 + \left( \frac{D+r}{D-r} \right)^2 t^2}\] Changing the variable once again to \(x = \left( \frac{D+r}{D-r} \right)t\) results in \(\text{d}t = \left( \frac{D-r}{D+r} \right)\text{d}x\) which together turn the last expression into: \[X_P = \frac{T}{\pi}\dot X_L \frac{2}{(D-r)^2} \left( \frac{D-r}{D+r} \right) \int_{\frac{D+r}{D-r}\text{tan}(\theta_1/2)}^{\frac{D+r}{D-r}\text{tan}(\theta_2/2)} \frac{\text{d}x}{1 + x^2}\] And after simplifications: \[X_P = \dot X_L \frac{2T}{\pi(D^2 - r^2)}\text{arctan}(x) \bigg|_{\frac{D+r}{D-r}\text{tan}(\theta_1/2)}^{\frac{D+r}{D-r}\text{tan}(\theta_2/2)}\]

Finally, \[\dot X_P = \frac{X_P}{T} = \dot X_L \frac{2}{\pi(D^2 - r^2)}\left\{\text{arctan}\left[\frac{D+r}{D-r}\text{tan}(\theta_2/2)\right] - \text{arctan}\left[\frac{D+r}{D-r}\text{tan}(\theta_1/2)\right] \right\}\] The numerical factor in the curly braces depends on the start and end angles. Let us denote it by: \[F(\theta_1, \theta_2) = \text{arctan}\left[\frac{D+r}{D-r}\text{tan}(\theta_2/2)\right] - \text{arctan}\left[\frac{D+r}{D-r}\text{tan}(\theta_1/2)\right]\] therefore, in general we have: \[\dot X_P = \dot X_L \frac{2}{\pi(D^2 - r^2)}F(\theta_1, \theta_2)\] Next, we will derive a few useful factors.

Half rotation

In this case, \(\theta_1 = 0\) and \(\theta_2 = \pi\):

\[F(0, \pi) = \text{arctan}\left[\frac{D+r}{D-r}\text{tan}(\pi/2)\right] - \text{arctan}\left[\frac{D+r}{D-r}\text{tan}(0/2)\right] = \frac{\pi}{2} - 0 = \frac{\pi}{2}\] For a full rotation, the above result just needs doubled because of the symmetry.

One 8th of a rotation

In this case, \(\theta_1 = 0\) and \(\theta_2 = \pi/4\):

\[F(0, \pi/4) = \text{arctan}\left[\frac{D+r}{D-r}\text{tan}(\pi/8)\right] - \text{arctan}\left[\frac{D+r}{D-r}\text{tan}(0/2)\right]\] \[F(0, \pi/4) = \text{arctan}\left[\frac{D+r}{D-r}\text{tan}(\pi/8)\right] = \text{arctan}\left[\frac{D+r}{D-r}(\sqrt{2} - 1)\right]\] The value of \(\text{tan}(\pi/8)\) above is derived in Appendix B.

Three 8th of a rotation

In this case, \(\theta_1 = 0\) and \(\theta_2 = 3\pi/4\):

\[F(0, 3\pi/4) = \text{arctan}\left[\frac{D+r}{D-r}\text{tan}(3\pi/8)\right] - \text{arctan}\left[\frac{D+r}{D-r}\text{tan}(0/2)\right]\] \[F(0, 3\pi/4) = \text{arctan}\left[\frac{D+r}{D-r}\text{tan}(3\pi/8)\right] = \text{arctan}\left[\frac{D+r}{D-r}(\sqrt{2} + 1)\right]\] Again, the value of \(\text{tan}(3\pi/8)\) above is derived in Appendix B.

Appendix B

Here, for a given angle “\(\theta\)” we derive the relationship between its tangent value “\(\text{tan}(\theta)\)” and the value of its half “\(\text{tan}(\theta/2)\)”.

Let us start by making the following notations - \(m = \text{tan}(\theta)\) & \(z = \text{tan}(\theta/2)\). Then we split the original angle in two halves: \[m = \frac{\text{sin}(\theta)}{\text{cos}(\theta)} = \frac{\text{sin}(\theta/2 + \theta/2)}{\text{cos}(\theta/2 + \theta/2)} = \frac{2\text{sin}(\theta/2)\text{cos}(\theta/2)}{\text{cos}^2(\theta/2)-\text{sin}^2(\theta/2)}\] At this point, let us divide both nominator and denominator by \(\text{cos}^2(\theta/2)\). \[m = \frac{2\text{tan}(\theta/2)}{1-\text{tan}^2(\theta/2)} = \frac{2z}{1-z^2}\] This identity can be re-arranged into a quadratic equation in \(z\): \[mz^2 + 2z-m = 0\] The solutions of these equaion are well known from basic algebra: \[m_{1,2}=\frac{-2\pm\sqrt{4+4m^2} }{2m} = \frac{-1\pm\sqrt{1+m^2} }{m}\] Or reversing back to the original notations: \[{\text{tan}(\theta/2)}_{1,2}=\frac{-1\pm\sqrt{1+{\text{tan}^2(\theta)}} }{\text{tan}(\theta)}\] Now, for the values needed in this document \((\pi/4)\) & \((3\pi/4)\):

\[{\text{tan}(\pi/8)}_{1,2}=\frac{-1\pm\sqrt{1+{\text{tan}^2(\pi/4)}} }{\text{tan}(\pi/4)} = \frac{-1+\sqrt{1+(1)^2} }{1} = \sqrt{2}-1\] Since \(\pi/8 < \pi/2\), its tangent value is positive, therefore in this case we had to choose the solution with “+”.

\[{\text{tan}(3\pi/8)}_{1,2}=\frac{-1\pm\sqrt{1+{\text{tan}^2(3\pi/4)}} }{\text{tan}(3\pi/4)} = \frac{-1-\sqrt{1+(-1)^2} }{-1} = \sqrt{2}+1\] Since \(3\pi/8 < \pi/2\), its tangent value is positive, therefore we had to choose the solution with “-” in this case.

Appendix C

Here we reproduce a few plots and drawings from NCPR 49 so that everything needed is in one place.

>> The geometry used to derive NCRP 49’s basic formulas

>> Table of quantities used by the equations

>> Attenuation in lead

Attenuation in concrete