New Geometric Theorems Derived from Integral Equations Applied to Radiative Transfer in Spherical Sectors and Circular Segments

1. Introduction

Form factors are a paramount physical manifold which facilitates to intervene in all the processes that involve radiative heat transfer. As this ever-present form of heat transfer takes place without any kind of contact between the intervening bodies, the forms and spatial relationships of the said elements are pivotal to define to which extent energy is emitted and received [1]. In other words, the distribution and appearance of radiation fields is a functional of the volumetric or three-dimensional disposition of the bodies that exchange radiation [2]. It is easy to outline the process for differential elements but if the need arouses to the extend the issue to well defined geometries, like the circle, the spheroid or the triangle [3,4] a great number of drawbacks and shortcomings appears often provoking that we fell short of reliable or robust solutions to the problem. From the mathematical point of view, it is necessary to introduce two rounds of double integrals, very different in nature, that become more complex with every primitive found [4]. Conventional calculus is, so to speak, put at a stake here, but approximate solutions fall of the mark and do not provide a satisfactory output [5]. Therefore, we need to overcome this problem to avoid increasing errors that menace the stability of the whole process.

Innovative tendencies in the matter, like for instance, finite elements or discretization are insufficient because generally speaking there is no adequate discretization of curved bodies like circles or spheres, even in some cases, we can find problems with triangular tiles. Stochastic methods are often as cumbersome as inaccurate. [6,7].

For this manuscript the author proposes a novel methodology based on employing some specific constants subsequently turned into variables, to perform the early parts of the fourfold integral, and completely exact. Previously only the rectangle and the complete circle had been resolved by total integration and the author has collaborated to clarify this especially for the case of cuboids. [8]. It remains surprising the comparatively simple shapes like the semicircle or the derived sector could not be added to the list of exact solutions although mathematical constraints made it difficult to surmount the [9].

In the fashion that we have proposed above, the last phase of integration is prepared to be completed by numerical methods which we will explain in more detail in the corresponding chapter. [10]. We need to consider that in the first stage only the more feasible configuration factor (a double integral) is considered instead of addressing the form factor (fourfold integral) directly. [8]. A significant part of the hindrances found resides in the complexity of spatial bodies that often appear in natural or man-made objects in which the occurrence of radiation is important [11,12].

Until now, diverse approaches to these matters have been attempted without success. The circle and circular sectors and segments are generally speaking indivisible in simpler shapes. [13]. The exact solving of the forms that appear in the problem might be deemed an important finding and an advance by its own merits; it accrues to a number of postulates on thermal radiation proposed thus far [14]. The most straightforward application is radiation on the arched entrances of tunnels. Finally, the formulations in algebraic terms are incorporated to the software which we have developed. In the following chapters we expect to develop in more detail how the whole process has been completed.

2. Materials and Methods. General Expression of Reciprocity in Radiative Exchanges

As it has been enunciated several times since the last quarter of the 18^th century [15] We will begin with the canonical expression that regulates the emission and reception of radiative energy between two non-black bodies of any spatial configuration and regardless of their respective positions as shown in Figure 1, the equation is composed mainly of directional cosines, distances and areas under the following form (Eq.1):

$$d{\varphi }_{12}=\left(E_1-E_2\right)\cos {\theta }_1\cos {\theta }_2{\displaystyle \frac{d{A}_{1}d{A}_2}{\pi r_{12}^2}}$$

(1)

Equation 1 is often termed the reciprocity principle, the first to propose it was Johann H. Lambert [16,17,18] around 1760. It describes the respective quantity of radiative energy by unit area that will strike any of the two surfaces involved, that is, E₁ and E₂. The corresponding incident angles θ₁ and θ₂ which are shown in the figure stand for the inclination to the normal of the distance line that joins two arbitrary points pertaining to each surface called r₁₂, both in the equation and in the figure [19].

In order to solve Eq. 1, which implies finding the amount of a form factor that encompasses the two sources, we need to work out four ascending steps of integration, as explicated in Eq. (2) [20].

$$F_{ij}={\displaystyle \frac{1}{A_i}}{\int }_{A_i}^{\hspace{0.17em}}{\int }_{A_j}^{\hspace{0.17em}}\cos {\theta }_i\cos {\theta }_j{\displaystyle \frac{d{A}_{i}d{A}_j}{\pi r_{ij}^2}}.$$

(2)

Nonetheless, in this manuscript we would focus on the first and second steps (Eq. 3), which are more accessible, leaving a sort of envoy for the next step which adopts the form of a constant (x₀) which turns into a variable in the subsequent phases [21], that is, it becomes active for the last two stages of integration.

$$f_{d{A}_i-A_j}={\int }_{A_j}^{\hspace{0.17em}}{\displaystyle \frac{\cos {\theta }_i\cos {\theta }_j}{\pi r_{ij}^2}}d{A}_j$$

(3)

We propose that the last two operations can be executed using a numerical method that extends the former findings to the surface of emission as a sort of mean or average. In this way we attain the desired figure of the form factor attributed to the two surface sources considered with considerable accuracy. [22]. Afterwards, we would show how to apply the so-found form factors for several three-dimensional configurations that come up in parts of radiation emitting or receiving devices and in openings of buildings. [23,24].

For other different positions of the surface sources, an exact solution has not been achieved yet [5]. Neither for irregular fragments of circular sectors, rectangles and circles or even for the familiar form of the sphere [25,26].

Partially conceived methods to address the preliminary integrals have been presented occasionally [27]. They tend to show the exchanges between the circle and a single point of the receiving surface, sometimes they adopt the form of blurry tables, with confusing entries [28,29]. Instead we will elaborate the complete formulation in standard mathematical language.

In order to develop our method, we start for simplicity with a semicircle of radius a, placed at the intersection of the main axes X, Y and Z (Figure 2).

Under this disposition, the coordinate Z axis is vertical and X is horizontal and perpendicular, Y axis is perpendicular to the semicircle, the radius of the circular sector turns, in Z, equal to rsinα, and in X, to rcosα as in Figure 2

Although it will be demonstrated in the appendix that the integration could be easily performed in cartesian coordinates by bringing in the expression of the semicircle, we would work in this case in polar coordinates to better explain the relevance of changing the angle α [30].

The axis Y in the graph represents the free coordinate perpendicular to X.

Introducing the variable x , limited at first to the semicircle but extended freely in the second part of the proccess, the distance vector (Eq.4) from one receiving point pertaining to the semicircle to another emitting point moving freely over the plane XY, presented as r₁₂ in Eq. 2 above, would be in caertesian coordinates:

$$r_{12}^2=\left(r^2+x^2+y^2-2*x*r*cos\alpha \right)$$

(4)

The respective cosines β and γ, to the normal from Eqs. 1 to 3, would be accordingly y/r₁₂ and z/r₁₂, z in this case is equal to rsinα, as mentioned.

The integral to solve derived from Eq.3, would then result in (Eq. 5):

$$f_{d{A}_i-A_j}=I = {\displaystyle \frac{1}{\pi }}{\int }_0^r{\int }_0^{\pi }{\displaystyle \frac{y*r^2*sin\alpha *d\alpha *dr}{{\left(r^2+x^2+y^2-2*x*r*cos\alpha \right)}^2}}$$

(5)

We need to solve the root equation for a generality of types of circular sectors and segments in diverse positions [31]. To avoid controversial points, we will give the expression for the second step of configuration factor, that is we will solve the double integral in the first place, which give the form factor on a point grid on the surface of a rectangular emitter. When this is achieved, we would use a different property demonstrated by the author [32], by virtue of which the full form factor equates to the mean of the point grid over the emitting surface. Utilizing the tools of numerical methods such planar grid can have any form, rectangular, elliptic or circular, since the average is limited to the points inside the boundary of the shape selected. In a not dissimilar procedure, we can find the total form factors for volumes composed of circular sements, for instance the space determined by an octave of sphere. Internal reflections inside the volumes created (three sides are quarter of circles forming a trihedron and the covering surface is an eight of a sphere) can be determined by special sets of four equations taking into account that one of the surface sources is not planar, see appendix 2 [33].

By combining the circular sectors, with triangles, rectangles or ellipses we can attaing a wider range of forms that were previously unimaginable and not suitable for calculation, which is why the novel procedure that we propose will provide a significant contribution [34].

3. Results. Solutions for various forms of circular sectors

We will summarize and outline below the more important aspects of the integration process. (Figure 3)

In this case the canonical integral in polar coordinates, as mentioned in Eq. 5 turns out to be (Eq.6):

$${\int }_0^r{\int }_0^{\pi }{\displaystyle \frac{y*r^2*sin\alpha *d\alpha *dr}{{\left(r^2+x^2+y^2-2*x*r*cos\alpha \right)}^2}}$$

(6)

As stated, first we would extend the integral to π which would represent the whole semicircle. If we integrate with respect to α , in the numerator we would find the derivative of cosα, -sinα. Therefore by making cosα=t, dt=-sinαdα this integral reduces to,

$${\int }_0^r{\int }_1^{-1}{\displaystyle \frac{-y*r^2*dt*dr}{{\left(r^2+x^2+y^2-2*x*r*t\right)}^2}}$$

(7)

The primitive is, accordingly, just the quotient of the numerator

$${{\displaystyle \frac{-y*r^2}{2*x*r}}\left[{\displaystyle \frac{1}{r^2+x^2+y^2-2*x*r*t}}\right]}_1^{-1}={\displaystyle \frac{y*r}{2*x}}\left({\displaystyle \frac{1}{r^2+x^2+y^2-2*x*r}}-{\displaystyle \frac{1}{r^2+x^2+y^2+2*x*r}}\right)$$

(8)

Then, the last integral to solve is,

$${\int }_0^r{\displaystyle \frac{r*dr}{r^2+x^2+y^2\pm 2*x*r}}$$

(9)

Whose primitives are a combination of a logarithm and an arc of tangent, for the first term,

$${\displaystyle \frac{1}{2}}{\left[{ln(r}^2+x^2+y^2-2*x*r)\right]}_0^r+{\displaystyle \frac{x}{y}}{\left[\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\left({\displaystyle \frac{r-x}{y}}\right)\right]}_0^r$$

(10)

And substituting, we find,

$${\displaystyle \frac{1}{2}}\left({\mathrm{l}\mathrm{n}(r}^2+x^2+y^2-2*x*r\right)-\mathrm{l}\mathrm{n}(x^2+y^2)+{\displaystyle \frac{x}{y}}(arctan{\displaystyle \frac{r-x}{y}}+arctan{\displaystyle \frac{x}{y}})$$

(11)

For the second term,

$$-{\displaystyle \frac{1}{2}}\left({\mathrm{l}\mathrm{n}(r}^2+x^2+y^2+2*x*r\right)+\ln \left(x^2+y^2\right)+{\displaystyle \frac{x}{y}}(arctan{\displaystyle \frac{r+x}{y}}-arctan{\displaystyle \frac{x}{y}})$$

(12)

Multiplying by ${\displaystyle \raisebox{1ex}{$y$}\!\left/ \!\raisebox{-1ex}{$2\pi x$}\right.}$ and grouping, the final result is expressed as,

$${\displaystyle \frac{1}{2\pi }}(arctan{\displaystyle \frac{r+x}{b}}+arctan{\displaystyle \frac{r-x}{b}})+{\displaystyle \frac{y}{4\pi x}}ln{\displaystyle \frac{r^2+y^2+x^2-2rx}{r+y^2+x^2+2rx}}$$

(13)

Or in a more developed form,

$$F_{21}={\displaystyle \frac{1}{2\pi }}\left(arctan{\displaystyle \frac{r+x}{y}}+arctan{\displaystyle \frac{r-x}{y}}\right)+{\displaystyle \frac{y}{4\pi x}}[\ln \left(r^2+y^2+x^2-2rx\right)-ln{(r}^2+y^2+x^2+2rx\left)\right]$$

(14)

The same result has been obtained by the author in Cartesian coordinates, see Appendix A.

From this result onwards, by using different values of the angle α, (see Figure 4), we can obtain a number of form factors as follows.

In time, we have been able to solve the same integral not only for the whole angle π, but in terms of α, that gives

$${\int }_0^r{\int }_0^{\alpha }{\displaystyle \frac{y{r}^{2}sin\alpha d\alpha dr}{{\left(r^2+x^2+y^2-2xrcos\alpha \right)}^2}}$$

(15)

Following, we gather a small but representative collection of new form factors depending on the angle subtended by the circular sector [35].

1.- For the semicircle (angle α = π), as already obtained,

$$F_{21}={\displaystyle \frac{1}{2\pi }}\left(arctan{\displaystyle \frac{r+x}{y}}+arctan{\displaystyle \frac{r-x}{y}}\right)+{\displaystyle \frac{y}{4\pi x}}[\ln \left(r^2+y^2+x^2-2rx\right)-ln{(r}^2+y^2+x^2+2rx\left)\right]$$

(16)

A further simplification is,

$$F_{21}={\displaystyle \frac{1}{2\pi }}\left(arctan{\displaystyle \frac{r+x}{y}}+arctan{\displaystyle \frac{r-x}{y}}\right)+{\displaystyle \frac{y}{4\pi x}}[\ln \left(({r-x)}^2+y^2\right)-\mathrm{l}\mathrm{n}\left(\right({r+x)}^2+y^2\left)\right]$$

(17)

An if we call S = r+x and T=r-x, Eq. 17 is reduced to,

$$F_{21}={\displaystyle \frac{1}{2\pi }}\left(arctan{\displaystyle \frac{S}{y}}+arctan{\displaystyle \frac{T}{y}}\right)+{\displaystyle \frac{y}{4\pi x}}[\ln \left(T^2+y^2\right)-\mathrm{l}\mathrm{n}(S^2+y^2)]$$

(18)

2.- For another typical value, the quarter of circle (angle α = π /2),

$$F_{21}={\displaystyle \frac{1}{2\pi }}\left(arctan{\displaystyle \frac{r-x}{y}}+arctan{\displaystyle \frac{x}{y}}\right)+{\displaystyle \frac{y}{4\pi x}}[\ln \left(r^2+y^2+x^2-2rx\right)-ln{(r}^2+y^2+x^2\left)\right]$$

(19)

3.- For the sixth of circle (angle α = π /3),

$$F_{21}={\displaystyle \frac{1}{2\pi }}\left(arctan{\displaystyle \frac{r-x}{y}}+arctan{\displaystyle \frac{x}{y}}-{\displaystyle \frac{y}{\sqrt{{3x}^2+{4y}^2}}}\left(arctan{\displaystyle \frac{2r-x}{\sqrt{{3x}^2+{4y}^2}}}+arctan{\displaystyle \frac{x}{\sqrt{{3x}^2+{4y}^2}}}\right)\right)+{\displaystyle \frac{y}{4\pi x}}[\ln \left(r^2+y^2+x^2-2rx\right)-ln{(r}^2+y^2+x^2-rx\left)\right]$$

(20)

4.- For the octave of circle (angle α = π /4),

$$F_{21}={\displaystyle \frac{1}{2\pi }}\left(arctan{\displaystyle \frac{r-x}{y}}+arctan{\displaystyle \frac{x}{y}}-{\displaystyle \frac{y}{\sqrt{x^2+{2y}^2}}}\left(arctan{\displaystyle \frac{\sqrt{2}r-x}{\sqrt{x^2+{2y}^2}}}+arctan{\displaystyle \frac{x}{\sqrt{x^2+{2y}^2}}}\right)\right)+{\displaystyle \frac{y}{4\pi x}}[\ln \left(r^2+y^2+x^2-2rx\right)-ln{(r}^2+y^2+x^2-\sqrt{2}rx\left)\right]$$

(21)

5.- For the twelfth of circle (angle α = π /6),

$$F_{21}={\displaystyle \frac{1}{2\pi }}\left(arctan{\displaystyle \frac{r-x}{y}}+arctan{\displaystyle \frac{x}{y}}-{\displaystyle \frac{y\sqrt{3}}{\sqrt{x^2+{4y}^2}}}\left(arctan{\displaystyle \frac{2r-\sqrt{3}x}{\sqrt{x^2+{4y}^2}}}+arctan{\displaystyle \frac{\sqrt{3}x}{\sqrt{x^2+{4y}^2}}}\right)\right)+{\displaystyle \frac{y}{4\pi x}}[\ln \left(r^2+y^2+x^2-2rx\right)-ln{(r}^2+y^2+x^2-\sqrt{3}rx\left)\right]$$

(22)

Thus, for any angular fraction α within the semicircle whose cosine, C goes from 1 (0) to -1 (π), we receive a new Cabeza-Lainez’ postulate number 18 [36]:

$$F_{21}={\displaystyle \frac{1}{2\pi }}\left(arctan{\displaystyle \frac{r-x}{y}}+arctan{\displaystyle \frac{x}{y}}-{\displaystyle \frac{y2C}{\sqrt{4x^2(1-C^2)+{4y}^2}}}\left(arctan{\displaystyle \frac{2(r-Cx)}{\sqrt{4x^2(1-C^2)+{4y}^2}}}+arctan{\displaystyle \frac{2Cx}{\sqrt{4x^2(1-C^2)+{4y}^2}}}\right)\right)+{\displaystyle \frac{y}{4\pi x}}[\ln \left(r^2+y^2+x^2-2rx\right)-ln{(r}^2+y^2+x^2-2Crx\left)\right]$$

(23)

Which in formal notation is,

$$F_{21}={\displaystyle \frac{1}{2\pi }}\left(arctan{\displaystyle \frac{r-x}{y}}+arctan{\displaystyle \frac{x}{y}}-{\displaystyle \frac{y2cos\alpha }{\sqrt{4x^2\left(1-{cos}^2\alpha \right)+{4y}^2}}}\left(arctan{\displaystyle \frac{2(r-xcos\alpha )}{\sqrt{4x^2(1-{cos}^2\alpha )+{4y}^2}}}+arctan{\displaystyle \frac{2xcos\alpha }{\sqrt{4x^2(1-{cos}^2\alpha )+{4y}^2}}}\right)\right)+{\displaystyle \frac{y}{4\pi *x}}[\ln \left(r^2+y^2+x^2-2rx\right)-ln{(r}^2+y^2+x^2-2rxcos\alpha \left)\right]$$

(24)

By subtraction of two different α angles, α₁ α₂ (being α₂ the larger, we would find the form factor for any circular segment at any position within the frame of the semicircle. Substituing (Eq. 25):

$${t=r}^2+y^2+x^2; m=\sqrt{x^2\left(1-C_1^2\right)+y^2;} n=\sqrt{x^2(1-C_2^2)+y^2}$$

$$F_{21}={\displaystyle \frac{y}{2\pi }}\left({\displaystyle \frac{C_1}{m}}arctan{\displaystyle \frac{r}{m+\frac{C_{1}x}{m}\left(C_{1}x-r\right)}}-{\displaystyle \frac{C_2}{n}}arctan{\displaystyle \frac{r}{n+\frac{C_{2}x}{n}\left(C_{2}x-r\right)}}\right)$$

$$+{\displaystyle \frac{y}{4\pi x}}ln\left[{\displaystyle \frac{\left(t-2 C_{1}rx\right)}{\left(t-2 C_{2}rx\right)}}\right]$$

(25)

As the sum of arctan(A) and arctan(B) can be grouped in a single arc of tangent, if AB<1.

But C₁=cos(α₁) and (1- C₁²)=sin²(α₂), therefore we can write, Eq. 26.

$${t=r}^2+y^2+x^2;m=\sqrt{x^2{{sin}^2\alpha }_1+y^2;}n=\sqrt{x^2{{sin}^2\alpha }_2+y^2}$$

$$F_{21}={\displaystyle \frac{y}{2\pi }}\left({\displaystyle \frac{cos{\alpha }_1}{m}}arctan{\displaystyle \frac{r}{m+\frac{cos{\alpha }_{1}x}{m}\left(cos{\alpha }_{1}x-r\right)}}-{\displaystyle \frac{cos{\alpha }_2}{n}}arctan{\displaystyle \frac{r}{n+\frac{cos{\alpha }_{2}x}{n}\left(cos{\alpha }_{2}x-r\right)}}\right)$$

$$+{\displaystyle \frac{y}{4\pi x}}ln\left[{\displaystyle \frac{\left(t-2 cos{\alpha }_{1}rx\right)}{\left(t-2 cos{\alpha }_{2}rx\right)}}\right]$$

(26)

Where α₂ represents the bigger angle (and the lower cosine), while r is the radius of the circular sector.

If we make, $P={\displaystyle \frac{cos{\alpha }_1}{m}}$ and $Q=$ ${\displaystyle \frac{cos{\alpha }_2}{n}}$ ;

$$F_{21}={\displaystyle \frac{y}{2\pi }}\left(P\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\left({\displaystyle \frac{r}{m+m{P}^2{x}^2-Pxr}}\right)-Q\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\left({\displaystyle \frac{r}{n+n{Q}^2{x}^2-Qxr}}\right)\right)$$

$$+{\displaystyle \frac{y}{4\pi x}}ln\left[{\displaystyle \frac{\left(t-2Pnrx\right)}{\left(t-2Qmrx\right)}}\right]$$

(27)

4. Possible Combinations of the Former Results

We have already obtained important and even revolutionary result for some fields of research like for instance the lighting and glare avoidance of tunnels and underground facilities [37]in which the radiation due to the circular opening (or otherwise like elliptic or parabolic was completely unknown). However, by combining these results with other, previously obtained by the author[38] we can extend the findings much further or apparently whithout limits.

In the first place we will incorporate the triangle with the circular sector.

The result for the form factor of the triangle with the point moving freely over x and y is,[12]

$$F_{12}={\displaystyle \frac{1}{2\pi }}\left(arctan{\displaystyle \frac{b-x}{y}}+arctan{\displaystyle \frac{x}{y}}-{\displaystyle \frac{b*y}{\sqrt{x^2{a}^2+y^2(a^2+b^2)}}}\left(arctan{\displaystyle \frac{a^2+b^2-bx}{\sqrt{x^2{a}^2+y^2(a^2+b^2)}}}+arctan{\displaystyle \frac{bx}{\sqrt{x^2{a}^2+y^2(a^2+b^2)}}}\right)\right)$$

(28)

If, for instance, we make a triangle with angle π/4 and we would like the hypotenuse to be r as it happens for instance in the octave of a circle, the short sides of the triangle a=b would logically be equal to r*(√2/2). Substituting a, and b in the above formula it is easy to obtain the following:

$$F_{21}={\displaystyle \frac{1}{2\pi }}\left(arctan{\displaystyle \frac{r\sqrt{2}/2-x}{y}}+arctan{\displaystyle \frac{x}{y}}-{\displaystyle \frac{y}{\sqrt{x^2+{2y}^2}}}\left(arctan{\displaystyle \frac{\sqrt{2}r-x}{\sqrt{x^2+{2y}^2}}}+arctan{\displaystyle \frac{x}{\sqrt{x^2+{2y}^2}}}\right)\right)$$

(29)

However, this is similar to the first term of the result for an octave of circle (Eq. 19) and we have to assume accordingly that the second positive term, plus the new term found gives, when subtracted,

$$F_{21}={\displaystyle \frac{y}{4\pi *x}}[\ln \left(r^2+y^2+x^2-2rx\right)-ln{(r}^2+y^2+x^2-\sqrt{2}rx\left)\right]+{\displaystyle \frac{1}{2\pi }}\left(+arctan{\displaystyle \frac{r-x}{y}}-arctan{\displaystyle \frac{r\sqrt{2}/2-x}{y}}\right)$$

(30)

Or for a general angle α,

$$F_{21}={\displaystyle \frac{1}{2\pi }}\left(arctan{\displaystyle \frac{r-x}{y}}-arctan{\displaystyle \frac{rcos\alpha -x}{y}}\right)+{\displaystyle \frac{y}{4\pi x}}[\ln \left(r^2+y^2+x^2-2rx\right)-ln{(r}^2+y^2+x^2-2rxcos\alpha )$$

(31)

We need to remember that in this particular case, cosα=√2/2 .

Eqs. 28 and 29, represent the factor introduced by the half circular segment (region 1, in Fig.X) that added to the triangle completes the octave. By virtue of which, we have found a new and unbeknownst configuration factor.

If we should add now the form factor of rectangles or squares, as the case may be, we could find the corresponding expression for the upper segment of the circle (region 2, in Figure 5) which, to our knowledge, has not yet been defined in the literature.

Let us continue the example by introducing a radiating rectangle (square, a=b), over a free dimension horizontal plane, the defining equation is, as we known,

$$F_{21}={\displaystyle \frac{1}{2\pi }}\left(arctan{\displaystyle \frac{a-x}{y}}+arctan{\displaystyle \frac{x}{y}}-{\displaystyle \frac{y}{\sqrt{a^2+y^2}}}\left(arctan{\displaystyle \frac{a-x}{\sqrt{a^2+y^2}}}+arctan{\displaystyle \frac{x}{\sqrt{a^2+y^2}}}\right)\right)$$

(32)

The factor for the quarter of the circle was, (Eq. 21)

$$F_{21}={\displaystyle \frac{1}{2\pi }}\left(arctan{\displaystyle \frac{r-x}{y}}+arctan{\displaystyle \frac{x}{y}}\right)+{\displaystyle \frac{y}{4\pi x}}[\ln \left(r^2+y^2+x^2-2rx\right)-ln{(r}^2+y^2+x^2\left)\right]$$

(33)

By subtracting both factors, we receive,

$$F_{21}={\displaystyle \frac{1}{2\pi }}\left(arctan{\displaystyle \frac{a-x}{y}}+arctan{\displaystyle \frac{x}{y}}-{\displaystyle \frac{y}{\sqrt{a^2+y^2}}}\left(arctan{\displaystyle \frac{a-x}{\sqrt{a^2+y^2}}}+arctan{\displaystyle \frac{x}{\sqrt{a^2+y^2}}}\right)\right)$$

(34)

The factor for the horizontal and vertical segments of circle outside of the square is ,

$$F_{21}={\displaystyle \frac{1}{2\pi }}\left(arctan{\displaystyle \frac{r-x}{y}}-arctan{\displaystyle \frac{a-x}{y}}+{\displaystyle \frac{y}{\sqrt{a^2+y^2}}}\left(arctan{\displaystyle \frac{a-x}{\sqrt{a^2+y^2}}}+arctan{\displaystyle \frac{x}{\sqrt{a^2+y^2}}}\right)\right)+{\displaystyle \frac{y}{4\pi x}}[\ln \left(r^2+y^2+x^2-2rx\right)-ln{(r}^2+y^2+x^2\left)\right]$$

(35)

In this particular case r equals a√2 ,

$$F_{21}={\displaystyle \frac{1}{2\pi }}\left(arctan{\displaystyle \frac{\surd 2a-x}{y}}-arctan{\displaystyle \frac{a-x}{y}}+{\displaystyle \frac{y}{\sqrt{a^2+y^2}}}\left(arctan{\displaystyle \frac{a-x}{\sqrt{a^2+y^2}}}+arctan{\displaystyle \frac{x}{\sqrt{a^2+y^2}}}\right)\right)+{\displaystyle \frac{y}{4\pi x}}[\ln \left(2a^2+y^2+x^2-2\surd 2ax\right)-ln{(2a}^2+y^2+x^2\left)\right]$$

(36)

If we want to obtain the value of the upper segment alone we need to subtract the factor previously found before,

$$F_{21}={\displaystyle \frac{y}{4\pi *x}}[\ln \left({2a}^2+y^2+x^2-2\surd 2ax\right)-ln{(2a}^2+y^2+x^2-2ax\left)\right]+{\displaystyle \frac{1}{2\pi }}\left(+arctan{\displaystyle \frac{\surd 2a-x}{y}}-arctan{\displaystyle \frac{a-x}{y}}\right)$$

(37)

The operation follows,

$$F_{21}={\displaystyle \frac{1}{2\pi }}\left({\displaystyle \frac{y}{\sqrt{a^2+y^2}}}\left(arctan{\displaystyle \frac{a-x}{\sqrt{a^2+y^2}}}+arctan{\displaystyle \frac{x}{\sqrt{a^2+y^2}}}\right)\right)+{\displaystyle \frac{y}{4\pi x}}[\ln \left(2a^2+y^2+x^2-2\surd 2ax\right)-ln{(2a}^2+y^2+x^2)-[\ln \left({2a}^2+y^2+x^2-2\surd 2ax\right)+ln{(2a}^2+y^2+x^2-2ax)]$$

(38)

And the final result is,

$$F_{21}={\displaystyle \frac{1}{2\pi }}\left({\displaystyle \frac{y}{\sqrt{a^2+y^2}}}\left(arctan{\displaystyle \frac{a-x}{\sqrt{a^2+y^2}}}+arctan{\displaystyle \frac{x}{\sqrt{a^2+y^2}}}\right)\right)+{\displaystyle \frac{y}{4\pi x}}[\ln {(2a}^2+y^2+x^2-2ax)-ln{(2a}^2+y^2+x^2\left)\right]$$

(39)

With which, we have added a considerable amount of new form factors. Substituting a by rcosα,

$$F_{21}={\displaystyle \frac{1}{2\pi }}\left({\displaystyle \frac{y}{\sqrt{({rsin\alpha )}^2+y^2}}}\left(arctan{\displaystyle \frac{rcos\alpha -x}{\sqrt{{\left(rsin\alpha \right)}^2+y^2}}}+arctan{\displaystyle \frac{x}{\sqrt{{\left(rsin\alpha \right)}^2+y^2}}}\right)\right)+{\displaystyle \frac{y}{4\pi x}}[\ln \left(r^2+y^2+x^2-2rxcos\alpha \right)-ln{(r}^2+y^2+x^2\left)\right]$$

(40)

Eq. 40 coincides with the generic factor, if we remember that a= rsinα = rcosα as the angle is π/4, and also because of that, r=a√2

In a more general case, we have that for any triangle, the form factor is

$$f_{21}={\displaystyle \frac{1}{2\pi }}\left(\left(arctan{\displaystyle \frac{b-x}{y}}+arctan{\displaystyle \frac{x}{y}}\right)-{\displaystyle \frac{by}{\sqrt{a^2{x}^2+y^2\left(a^2+b^2\right)}}}\left(arctan{\displaystyle \frac{a^2+b^2-xb}{\sqrt{a^2{x}^2+y^2\left(a^2+b^2\right)}}}+arctan{\displaystyle \frac{xb}{\sqrt{a^2{x}^2+y^2\left(a^2+b^2\right)}}}\right)\right)$$

(41)

But in relation to the radius r of a circular segment, we known that a=rsinα and b=rcosα, and substituting,

$$f_{21}={\displaystyle \frac{1}{2\pi }}\left(\left(arctan{\displaystyle \frac{rcos\alpha -x}{y}}+arctan{\displaystyle \frac{x}{y}}\right)-{\displaystyle \frac{yrcos\alpha }{\sqrt{({rsin\alpha )}^2{x}^2+y^2\left({\left(rsin\alpha \right)}^2+{\left(rcos\alpha \right)}^2\right)}}}\left(arctan{\displaystyle \frac{({rsin\alpha )}^2+{\left(rcos\alpha \right)}^2-xrcos\alpha }{\sqrt{{\left(rsin\alpha \right)}^2{x}^2+y^2\left(({rsin\alpha )}^2+{\left(rcos\alpha \right)}^2\right)}}}+arctan{\displaystyle \frac{xrcos\alpha }{\sqrt{({rsin\alpha )}^2{x}^2+y^2\left(({rsin\alpha )}^2+{\left(rcos\alpha \right)}^2\right)}}}\right)\right)$$

(42)

And simplifying, it leaves,

$$f_{21}={\displaystyle \frac{1}{2\pi }}\left(\left(arctan{\displaystyle \frac{rcos\alpha -x}{y}}+arctan{\displaystyle \frac{x}{y}}\right)-{\displaystyle \frac{yrcos\alpha }{\sqrt{({rsin\alpha )}^2{x}^2+y^2{r}^2}}}\left(arctan{\displaystyle \frac{r^2-xrcos\alpha }{\sqrt{{\left(rsin\alpha \right)}^2{x}^2+y^2{r}^2}}}+arctan{\displaystyle \frac{xrcos\alpha }{\sqrt{({rsin\alpha )}^2{x}^2+y^2{r}^2}}}\right)\right)$$

(43)

The final result is,

$$f_{21}={\displaystyle \frac{1}{2\pi }}\left(\left(arctan{\displaystyle \frac{rcos\alpha -x}{y}}+arctan{\displaystyle \frac{x}{y}}\right)-{\displaystyle \frac{yrcos\alpha }{\sqrt{r^2(x^2{sin}^2\alpha +y^2)}}}\left(arctan{\displaystyle \frac{r^2-xrcos\alpha }{\sqrt{r^2(x^2{sin}^2\alpha +y^2)}}}+arctan{\displaystyle \frac{xrcos\alpha }{\sqrt{r^2(x^2{sin}^2\alpha +y^2)}}}\right)\right)$$

(44)

For a generic rectangle with the same substitution of a and b, we would have

$$F_{21}={\displaystyle \frac{1}{2\pi }}\left(arctan{\displaystyle \frac{b-x}{y}}+arctan{\displaystyle \frac{x}{y}}-{\displaystyle \frac{y}{\sqrt{a^2+y^2}}}\left(arctan{\displaystyle \frac{b-x}{\sqrt{a^2+y^2}}}+arctan{\displaystyle \frac{x}{\sqrt{a^2+y^2}}}\right)\right)$$

(45)

Applying the said substitution,

$$F_{21}={\displaystyle \frac{1}{2\pi }}\left(arctan{\displaystyle \frac{rcos\alpha -x}{y}}+arctan{\displaystyle \frac{x}{y}}-{\displaystyle \frac{y}{\sqrt{{\left(rsin\alpha \right)}^2+y^2}}}\left(arctan{\displaystyle \frac{rcos\alpha -x}{\sqrt{{\left(rsin\alpha \right)}^2+y^2}}}+arctan{\displaystyle \frac{x}{\sqrt{{\left(rsin\alpha \right)}^2+y^2}}}\right)\right)$$

(46)

The form factor for the generic circular sector was,

$$F_{21}={\displaystyle \frac{1}{2\pi }}\left(arctan{\displaystyle \frac{r-x}{y}}+arctan{\displaystyle \frac{x}{y}}-{\displaystyle \frac{y2cos\alpha }{\sqrt{4x^2\left(1-{cos}^2\alpha \right)+{4y}^2}}}\left(arctan{\displaystyle \frac{2(r-xcos\alpha )}{\sqrt{4x^2(1-{cos}^2\alpha )+{4y}^2}}}+arctan{\displaystyle \frac{2xcos\alpha }{\sqrt{4x^2(1-{cos}^2\alpha )+{4y}^2}}}\right)\right)+{\displaystyle \frac{y}{4\pi *x}}[\ln \left(r^2+y^2+x^2-2rx\right)-ln{(r}^2+y^2+x^2-2rxcos\alpha \left)\right]$$

(47)

Then the vertical segment would turn into,

$$F_{21}={\displaystyle \frac{1}{2\pi }}\left(arctan{\displaystyle \frac{r-x}{y}}-{\displaystyle \frac{y2cos\alpha }{\sqrt{4x^2\left(1-{cos}^2\alpha \right)+{4y}^2}}}\left(arctan{\displaystyle \frac{2(r-xcos\alpha )}{\sqrt{4x^2(1-{cos}^2\alpha )+{4y}^2}}}+arctan{\displaystyle \frac{2xcos\alpha }{\sqrt{4x^2(1-{cos}^2\alpha )+{4y}^2}}}\right)\right)+{\displaystyle \frac{y}{4\pi x}}[\ln \left(r^2+y^2+x^2-2rx\right)-ln{(r}^2+y^2+x^2-2rxcos\alpha \left)\right]-{\displaystyle \frac{1}{2\pi }}\left(\left(arctan{\displaystyle \frac{rcos\alpha -x}{y}}\right)-{\displaystyle \frac{yrcos\alpha }{\sqrt{r^2(x^2{sin}^2\alpha +y^2)}}}\left(arctan{\displaystyle \frac{r^2-xrcos\alpha }{\sqrt{r^2(x^2{sin}^2\alpha +y^2)}}}+arctan{\displaystyle \frac{xrcos\alpha }{\sqrt{r^2(x^2{sin}^2\alpha +y^2)}}}\right)\right)$$

(48)

$$F_{21}={\displaystyle \frac{1}{2\pi }}\left(arctan{\displaystyle \frac{r-x}{y}}-arctan{\displaystyle \frac{rcos\alpha -x}{y}}-{\displaystyle \frac{ycos\alpha }{\sqrt{x^2\left(1-{cos}^2\alpha \right)+y^2}}}\left(arctan{\displaystyle \frac{\left(r-xcos\alpha \right)}{\sqrt{x^2\left(1-{cos}^2\alpha \right)+y^2}}}+arctan{\displaystyle \frac{xcos\alpha }{\sqrt{x^2\left(1-{cos}^2\alpha \right)+y^2}}}\right)+{\displaystyle \frac{ycos\alpha }{\sqrt{(x^2{sin}^2\alpha +y^2)}}}\left(arctan{\displaystyle \frac{r^2-xrcos\alpha }{\sqrt{r^2(x^2{sin}^2\alpha +y^2)}}}+arctan{\displaystyle \frac{xrcos\alpha }{\sqrt{r^2(x^2{sin}^2\alpha +y^2)}}}\right)\right)+{\displaystyle \frac{y}{4\pi x}}[\ln \left(r^2+y^2+x^2-2rx\right)-ln{(r}^2+y^2+x^2-2rxcos\alpha \left)\right]$$

(49)

And simplifying a bit more,

$$F_{21}={\displaystyle \frac{1}{2\pi }}\left(arctan{\displaystyle \frac{r-x}{y}}-arctan{\displaystyle \frac{rcos\alpha -x}{y}}-{\displaystyle \frac{ycos\alpha }{\sqrt{x^2{sin}^2\alpha +y^2}}}\left(arctan{\displaystyle \frac{\left(r-xcos\alpha \right)}{\sqrt{x^2{sin}^2\alpha +y^2}}}+arctan{\displaystyle \frac{xcos\alpha }{\sqrt{x^2{sin}^2\alpha +y^2}}}\right)+{\displaystyle \frac{ycos\alpha }{\sqrt{(x^2{sin}^2\alpha +y^2)}}}\left(arctan{\displaystyle \frac{r-xcos\alpha }{\sqrt{(x^2{sin}^2\alpha +y^2)}}}+arctan{\displaystyle \frac{xcos\alpha }{\sqrt{(x^2{sin}^2\alpha +y^2)}}}\right)\right)+{\displaystyle \frac{y}{4\pi x}}[\ln \left(r^2+y^2+x^2-2rx\right)-ln{(r}^2+y^2+x^2-2rxcos\alpha \left)\right]$$

(50)

$$F_{21}={\displaystyle \frac{1}{2\pi }}\left(arctan{\displaystyle \frac{r-x}{y}}-arctan{\displaystyle \frac{rcos\alpha -x}{y}}\right)+{\displaystyle \frac{y}{4\pi x}}[\ln \left(r^2+y^2+x^2-2rx\right)-ln{(r}^2+y^2+x^2-2rxcos\alpha \left)\right]$$

(51)

Accordingly we can find the upper generic circular segment, to the general expression for the quarter of circle,

$$F_{21}={\displaystyle \frac{1}{2\pi }}\left(arctan{\displaystyle \frac{r-x}{y}}+arctan{\displaystyle \frac{x}{y}}\right)+{\displaystyle \frac{y}{4\pi x}}[\ln \left(r^2+y^2+x^2-2rx\right)-ln{(r}^2+y^2+x^2\left)\right]$$

(52)

We need to subtract the form factor of the rectangle,

$$F_{21}={\displaystyle \frac{1}{2\pi }}\left(arctan{\displaystyle \frac{b-x}{y}}+arctan{\displaystyle \frac{x}{y}}-{\displaystyle \frac{y}{\sqrt{a^2+y^2}}}\left(arctan{\displaystyle \frac{b-x}{\sqrt{a^2+y^2}}}+arctan{\displaystyle \frac{x}{\sqrt{a^2+y^2}}}\right)\right)$$

(53)

But first, we proceed to substitute with the values of a and b as a function of the radius.

$$F_{21}={\displaystyle \frac{1}{2\pi }}\left(arctan{\displaystyle \frac{rcos\alpha -x}{y}}+arctan{\displaystyle \frac{x}{y}}-{\displaystyle \frac{y}{\sqrt{({rsin\alpha )}^2+y^2}}}\left(arctan{\displaystyle \frac{rcos\alpha -x}{\sqrt{{\left(rsin\alpha \right)}^2+y^2}}}+arctan{\displaystyle \frac{x}{\sqrt{{\left(rsin\alpha \right)}^2+y^2}}}\right)\right)$$

(54)

Then we perform the first operation to give,

$$F_{21}={\displaystyle \frac{1}{2\pi }}\left(arctan{\displaystyle \frac{r-x}{y}}-arctan{\displaystyle \frac{rcos\alpha -x}{y}}+{\displaystyle \frac{y}{\sqrt{({rsin\alpha )}^2+y^2}}}\left(arctan{\displaystyle \frac{rcos\alpha -x}{\sqrt{{\left(rsin\alpha \right)}^2+y^2}}}+arctan{\displaystyle \frac{x}{\sqrt{{\left(rsin\alpha \right)}^2+y^2}}}\right)\right)+{\displaystyle \frac{y}{4\pi x}}[\ln \left(r^2+y^2+x^2-2rx\right)-ln{(r}^2+y^2+x^2\left)\right]$$

(55)

And finally we need to subtract the value found for the vertical segment of the circle,

$$F_{21}={\displaystyle \frac{1}{2\pi }}\left(arctan{\displaystyle \frac{r-x}{y}}-arctan{\displaystyle \frac{rcos\alpha -x}{y}}\right)+{\displaystyle \frac{y}{4\pi x}}[\ln \left(r^2+y^2+x^2-2rx\right)-ln{(r}^2+y^2+x^2-2rxcos\alpha )$$

(56)

The new generic form factor thus obtained is,

$$F_{21}={\displaystyle \frac{1}{2\pi }}\left({\displaystyle \frac{y}{\sqrt{({rsin\alpha )}^2+y^2}}}\left(arctan{\displaystyle \frac{rcos\alpha -x}{\sqrt{{\left(rsin\alpha \right)}^2+y^2}}}+arctan{\displaystyle \frac{x}{\sqrt{{\left(rsin\alpha \right)}^2+y^2}}}\right)\right)+{\displaystyle \frac{y}{4\pi x}}[\ln \left(r^2+y^2+x^2-2rxcos\alpha \right)-ln{(r}^2+y^2+x^2\left)\right]$$

(57)

Achieving the final results hereby exposed,

Lower vertical circular segment,

$$F_{21}={\displaystyle \frac{1}{2\pi }}\left(arctan{\displaystyle \frac{r-x}{y}}-arctan{\displaystyle \frac{rcos\alpha -x}{y}}\right)+{\displaystyle \frac{y}{4\pi x}}[\ln \left(r^2+y^2+x^2-2rx\right)-ln{(r}^2+y^2+x^2-2rxcos\alpha )$$

(58)

Upper horizontal circular segment,

$$F_{21}={\displaystyle \frac{1}{2\pi }}\left({\displaystyle \frac{y}{\sqrt{y^2+r^2{sin}^2\alpha }}}\left(arctan{\displaystyle \frac{rcos\alpha -x}{\sqrt{y^2+r^2{sin}^2\alpha }}}+arctan{\displaystyle \frac{x}{\sqrt{y^2+r^2{sin}^2\alpha +}}}\right)\right)+{\displaystyle \frac{y}{4\pi x}}[\ln \left(r^2+y^2+x^2-2rxcos\alpha \right)-ln{(r}^2+y^2+x^2\left)\right]$$

(59)

With adequate adroitness we can find by addition and subtraction an unlimited number of new form factors [39], but the emitting surface must be the same for all the bodies involved, usually the larger one [40]. Consider for instance the annulus in the Figure 7.

We can find the form factor of a circular ring by subtracting the smaller semicircle from the larger one. The two radiuses are r and r₁, with r > r₁ . The horizontal surface on which we proceed with the calculations has to be the same for both semicircles, that is a rectangle or surface larger than the outer half-disk [41].

$$F_{21}={\displaystyle \frac{1}{2\pi }}\left(arctan{\displaystyle \frac{r+x}{y}}+arctan{\displaystyle \frac{r-x}{y}}\right)+{\displaystyle \frac{y}{4\pi x}}[\ln \left(r^2+y^2+x^2-2rx\right)-ln{(r}^2+y^2+x^2+2rx\left)\right]-{\displaystyle \frac{1}{2\pi }}\left(arctan{\displaystyle \frac{r_1+x}{y}}+arctan{\displaystyle \frac{r_1-x}{y}}\right)-{\displaystyle \frac{y}{4\pi x}}[\ln \left({r_1}^2+y^2+x^2-2r_{1}x\right)-ln{(r_1}^2+y^2+x^2+2r_{1}x\left)\right]$$

(60)

5. Advance to Three-Dimensional Results and Calculation of the Total Form Factor

As we have stated in the introduction, to proceed with the toal form factor between two finite surfaces, we need to obtain a sort of mean of the previous formulas over the whole emitting surface (a rectangle in the most simple case). It suffices to compute the values of the configuration factor at each point of the rectangle and calculate the average by numerical methods [42]. This has been mathematicall demonstrated by the author [43].

If the emitting surface is not rectangular, for instance a quarter of circle, there are other numerical procedures to which we can resort, for instance we can determine the transfer form factor of a circular sector over a quarter of circle as in Figure 8 [44].

Ensuing we present the nephographs of some of the main types of circular sectors, calculated above (Figure 9, Figure 10, Figure 11, Figure 12 and Figure 13).

6. Complete three-dimensional curved figures

Having clarified the previous section, we would advance towards more complex three-dimensional figures, like for instance sections of the sphere comprised between two quarters of a circle with a common edge. Previously the author has identified the form factors between semicircles (Figure 14) at different angles sharing a common edge [45] and circles with a tangent point (Figure 15) when their angle of connection is variable [46], but it was not possible to attempt to find the form factors among regions 1 and 3 or 2 and 4 in Figure 16, even less between alternate quarter of circles like 1 to 4.

In order to perform such difficult calculation we should shoud start with a trihedron made of two symmetrical quarters of circles, the base is a circular sector of variable angle α and the cover is a fragment of sphere of corresponding area α [47].

Figure 17. A trihedron made by symmetrical quarter of circles 2 and 3.

Figure 17. A trihedron made by symmetrical quarter of circles 2 and 3.

The proposed new postulate for radiative exchange of the two quarters of circles, that we will elucidate below is, (Eq. 61),

$$F_{23}=1-2F_{21}-{\displaystyle \frac{\alpha }{\pi }}+{\displaystyle \frac{1}{2}}{\left({\displaystyle \frac{\alpha }{\pi }}\right)}^2$$

(61)

As we can see in the figure, surface 4 is the fragment of spherical surface dependent on the angle α, its lateral area is precisely α, the areas of the two symmetrical quarters of circles (named 2 and 3) give an area of π/4 respectively, the area of the circular sector (1) that serves as the basis for the figure is α/2. If α= π/2, we would receive the equilateral trihedron made of three quarters of circle.

Then, as usual, by conservation of radiant energy in an enclosed volume,

$$F_{41}+F_{42}+F_{43}+F_{44}=1$$

(62)

We know, by virtue of Cabeza-Lainez first law, that

$$F_{44}={\displaystyle \frac{\alpha }{4\pi }}$$

(63)

And F₄₂ = F₄₃, as they are symmetrical

$$F_{41}+{2F}_{42}+{\displaystyle \frac{\alpha }{4\pi }}=1$$

(64)

$${2F}_{42}=1-F_{41}-{\displaystyle \frac{\alpha }{4\pi }}$$

(65)

From the downwards side, we have (three planar circular sectors),

$$F_{12}+F_{13}+F_{14}=1$$

(64)

And F₁₂ = F₁₃ , for symmetry reasons,

$$F_{14}=1-{2F}_{12}$$

(65)

$$F_{41}={\displaystyle \frac{A_1}{A_4}}\left(1-{2F}_{12}\right)$$

(66)

$$F_{41}={\displaystyle \frac{1}{2}}-F_{12}$$

(67)

F₁₂ , can be found with a very precise approximation by means of the formula described in section 3 for every angle of the circular sector over a quarter of circle (Figure 18), once we have obtained F₁₂ , with the help of such numerical procedure based on the integral, we could proceed with the deductions.

As anew,

$$F_{21}+F_{23}+F_{24}=1$$

(68)

$$F_{23}=1-F_{21}-F_{24}$$

(69)

But F₄₂ was from Eq. 65,

$${2F}_{42}=1-F_{41}-{\displaystyle \frac{\alpha }{4\pi }}$$

(70)

$$F_{24}={\displaystyle \frac{A_4}{{2A}_2}}\left(1-F_{41}-{\displaystyle \frac{\alpha }{4\pi }}\right)$$

(71)

Substituting,

$$F_{41}={\displaystyle \frac{1}{2}}-F_{12}$$

(72)

$$F_{24}={\displaystyle \frac{A_4}{{2A}_2}}\left(1-{\displaystyle \frac{A_1}{A_4}}\left(1-{2F}_{12}\right)-{\displaystyle \frac{\alpha }{4\pi }}\right)$$

(73)

Thus,

$$F_{24}={\displaystyle \frac{A_4}{{2A}_2}}\left(1-{\displaystyle \frac{1}{2}}+F_{12}-{\displaystyle \frac{\alpha }{4\pi }}\right)$$

(74)

$$F_{24}={\displaystyle \frac{A_4}{{2A}_2}}\left({\displaystyle \frac{1}{2}}+F_{12}-{\displaystyle \frac{\alpha }{4\pi }}\right)$$

(75)

And, including the respective values of the areas in 73

$$F_{24}={\displaystyle \frac{A_4}{{2A}_2}}\left(1-{\displaystyle \frac{A_1}{A_4}}\left(1-{2F}_{12}\right)-{\displaystyle \frac{\alpha }{4\pi }}\right)$$

(76)

$$F_{24}={\displaystyle \frac{2\alpha }{\pi }}-{\displaystyle \frac{\alpha }{\pi }}\left(1-{2F}_{12}\right)-{\displaystyle \frac{{\alpha }^2}{2{\pi }^2}}$$

(77)

$$F_{24}={\displaystyle \frac{\alpha }{\pi }}+{\displaystyle \frac{2\alpha F_{12}}{\pi }}-{\displaystyle \frac{{\alpha }^2}{2{\pi }^2}}$$

(78)

And finally we arrive to,

$$F_{23}=1-F_{21}-{\displaystyle \frac{A_4}{{2A}_2}}\left(1-{\displaystyle \frac{A_1}{A_4}}\left(1-{2F}_{12}\right)-{\displaystyle \frac{\alpha }{4\pi }}\right)$$

(79)

Hower, the respective areas are as follows,

$${\displaystyle \frac{A_4}{{2A}_2}}={\displaystyle \frac{\alpha }{\pi /2}}={\displaystyle \frac{2\alpha }{\pi }}$$

(80)

$${\displaystyle \frac{A_1}{A_4}}={\displaystyle \frac{\alpha /2}{\alpha }}={\displaystyle \frac{1}{2}}$$

(81)

$$F_{23}=1-F_{21}-{\displaystyle \frac{2\alpha }{\pi }}\left(1-{\displaystyle \frac{1}{2}}\left(1-{2F}_{12}\right)-{\displaystyle \frac{\alpha }{4\pi }}\right)$$

(82)

Simplifying with F₁₂ ,

$$F_{23}=1-F_{21}-{\displaystyle \frac{2\alpha }{\pi }}\left({\displaystyle \frac{1}{2}}+F_{12}-{\displaystyle \frac{\alpha }{4\pi }}\right)$$

(83)

Or changing to F₂₁ ,

$$F_{23}=1-F_{21}-{\displaystyle \frac{2\alpha }{\pi }}\left({\displaystyle \frac{1}{2}}+{\displaystyle \frac{{\pi F}_{21}}{2\alpha }}-{\displaystyle \frac{\alpha }{4\pi }}\right)$$

(84)

$$F_{23}=1-F_{21}-{\displaystyle \frac{\alpha }{\pi }}-F_{21}+{\displaystyle \frac{{\alpha }^2}{2{\pi }^2}}$$

(85)

Which yields,

$$F_{23}=1-2F_{21}-{\displaystyle \frac{\alpha }{\pi }}+{\displaystyle \frac{1}{2}}{\left({\displaystyle \frac{\alpha }{\pi }}\right)}^2$$

(86)

If p=α/π ,

$$F_{23}=1-2*F_{21}-p+{\displaystyle \frac{p^2}{2}}$$

(87)

Or

$$F_{23}=1-{\displaystyle \frac{4\alpha }{\pi }}{F}_{12}-{\displaystyle \frac{\alpha }{\pi }}+{\displaystyle \frac{1}{2}}{\left({\displaystyle \frac{\alpha }{\pi }}\right)}^2$$

(88)

Remembering that

$$F_{21}={\displaystyle \frac{2\alpha }{\pi }}{F}_{12}$$

And ntroducing ,$p={\displaystyle \frac{\alpha }{\pi }}$

$$F_{23}=1-4p{F}_{12}-p+{\displaystyle \frac{1}{2}}{\left(p\right)}^2$$

(89)

We have demonstrated that it is possible to obtain each and every quantity as a function of F₁₂ , and conversely F₂₁ , and more importantly we have demonstrated the new Cabeza-Lainez postulate for quarter of circles with a common edge [48,49].

$$F_{23}=1-2F_{21}-{\displaystyle \frac{\alpha }{\pi }}+{\displaystyle \frac{1}{2}}{\left({\displaystyle \frac{\alpha }{\pi }}\right)}^2$$

(90)

The former Cabeza-Lainez’ law for semicircles was as a function of the angle α,

$$F_{23}=1-2{\displaystyle \frac{\alpha }{\pi }}+{\left({\displaystyle \frac{\alpha }{\pi }}\right)}^2$$

(91)

However, this law was independent of any previously found factor. In terms of p,

$$F_{23}=1-2F_{21}-p+{\displaystyle \frac{p^2}{2}}$$

(92)

And the former law is:

$$F_{23}=1-2p+p^2$$

(93)

For the record, the postulate for whole circles in terms of p is,

$$F_{12}=1+2\sin {\displaystyle \frac{\pi p}{2}}\left(sin{\displaystyle \frac{\pi p}{2}}-1\right)/{cos}^2{\displaystyle \frac{\pi p}{2}}$$

(94)

Now, we can compare the three laws obtained with difficulty in Figure 18

Figure 18. Graph of comparison of the form-factor postulates, so far obtained, as a function of angle α, yellow colour is for semicircles, orange for quarter of circles and violet for whole circles. Blue colour depicts the evolution of F₁₂, the exchange between a vertical circular sector of angle α and a horizontal fixed quarter of circle.

Figure 18. Graph of comparison of the form-factor postulates, so far obtained, as a function of angle α, yellow colour is for semicircles, orange for quarter of circles and violet for whole circles. Blue colour depicts the evolution of F₁₂, the exchange between a vertical circular sector of angle α and a horizontal fixed quarter of circle.

Adding as a novelty the form factors between alternate quarter of circles we receive Figure 19

Figure 19. Depiction of the laws discovered adding the alternate quarter of circles in green colour. The maximum for this figure is 0.06935 at point x=23, which means an angle of π/4.34 or 41.4 deg.

Figure 19. Depiction of the laws discovered adding the alternate quarter of circles in green colour. The maximum for this figure is 0.06935 at point x=23, which means an angle of π/4.34 or 41.4 deg.

6.1. Check for Validity

Demonstration of validity by a simple example [50], for the quadrant, octave of a sphere or curved trihedron

In this case α= π/2, and as stated previously, we know, by virtue of Cabeza-Lainez first law, that,

$$F_{44}={\displaystyle \frac{\alpha }{4\pi }}={\displaystyle \frac{1}{8}}$$

Therefore,

$$F_{41}+F_{42}{+F}_{43}=1-F_{44}=7/8$$

Thus, being F₄₁ and F₄₂ and F₄₂ equal and symmetrical,

$$F_{41}=F_{42}=F_{43}={\displaystyle \frac{7}{24}}$$

As A₄ is π/2 and the area of each quadrant is π/4, A₄ =2 A₁

$$F_{14}={2F}_{41}={\displaystyle \frac{7}{12}}$$

Equally,

$$F_{24}={2F}_{34}={\displaystyle \frac{7}{12}}$$

$$F_{12}+F_{13}{+F}_{14}=1$$

$$F_{12}+F_{13}={\displaystyle \frac{5}{12}}$$

But, as before, F₁₂ and F₁₃ are equal and symmetrical, consequently,

$$F_{12}=F_{13}={\displaystyle \frac{5}{24}}$$

And the problem is completely solved since,

F₂₁ and F₂₃ = F₃₂ = F₃₁ = 5/24

As in abcissa 51 in Figure 17. Value 0.208.

With a similar approach, the total form factors can be found for the usual sets of precints, such as prisms or oblique cuboids. Inter-reflection within the so-generated volumes is also an issue that we have to foresee, in the case of triangular prisms we would require a set of five equations with five unknowns which is detailed at the appendix [36].

For example, when designing and solving inclined slats, the triangles that appear are not right, but usually they come as obtuse or acute angles, in these situation we cannot employ the already solved form factor for perpendicular rectangles [5], instead we need to resort to different extended formulas developed by the author [] to take into account the exchanges between inclined rectangles.

7. Examples and Applications for Tunnel Openings

We will try to show some of the relevant possibilities that the form factors, found in this article, offer for different technologies. One of the most important example is the circular opening of tunnels which presents important problem of lighting because of glare, flicker effect and excessive use of energy; necessary to counterbalance the very different levels of lighting at the threshold of the tunnel. The radiative exchange between the entrance opening, usually a semicircle or on other occasions a circular or elliptical segment and the rest of the surfaces inside the tunnel and especially the floor where the vehicles circulate, could not be determined by any means previously to obtain the results that we now possess. In Figure 20, Figure 21 and Figure 22, we present some drawings of one type of innovative tunnels that we have proposed [42] and whose performance was unavailable to assess before we arrived to the findings hereby exposed.

Having an accurate map of the radiation impinging on the diverse surfaces of the tunnel we are now in the position to control and improve the many undesirable effects that have been registered since the beginning of this technology.

The equations derived from the analysis of Figure 23 are employed to determine the distribution of radiation especially at the most critical point which is the floor level.

In Figure 24 we present some of the early results of radiation as achieved by the combined use of solar ducts and other means of lighting [43].

These early results prove that the improvements achieved from the detailed knowledge of the new form factor have just begun and more developments can be expected from such significant findings [44].

8. Conclusion

Throughout this manuscript we have demonstrated by means of accessible calculus operations how to obtain the solution to certain integral equations that provide us with more than five new postulates of radiation between circular sectors, ellipses, circular segments and the interaction between those and figures like the quarter of circle, the triangle and the rectangle. Such effort will undoubtedly contribute to the solution of numerous radiative problems that hindered the development of technology in fields as diverse as of tunneling, solar energies, lighting and aircraft design or construction among others.

Certainly, integral equations are a problematic domain of mathematics but they are mandatory to advance in the many problems of radiative heat transfer as they are frequently colligated to surface emission and demand an exact solution due to the extreme and vital magnitudes which are usually under consideration in this realm.

In the mathematical process followed to improve the level of knowledge in the ralm of radiative transfer it is possible to outline the exactitude and feasibility of the solutions, the introduction of silent variables that only become active in the last phase of integration and the new form of addressing inter-reflections for curved surfaces. Also the combinations with different geometries are a novelty that should not be disregarded.

Semicircles and other derived figures, like circular segments and sectors have been employed in a great number of artifacts and industries since early epochs, now finally are we able to determine with total accuracy their performance under a significant number of demanding situations in radiative heat transfer. However, a detailed logical approach had not been attempted in the past because of the mathematical hindrances arisen in the process. Statistical procedures were almost inapplicable in this case. The solutions provided are especially feasible for the design and retrofit of tunnels, lighting elements from windows to LED emitters and also aircrafts and components affected by radiation and systematically employed in the industry.

The quite relevant novelties that these postulates entail, lie also in the question by which they have been implemented in simulation algorithms by the author. We firmly believe that given the preparedness for application that they present, they will radically transform and sometimes ignite processes based on transfer of radiation whether natural or artificial and amongst them tunnelling, lighting, thermal machinery and radiant power instruments and apparatuses. In this fashion, fundamental issues of radiative heat transfer can be treated in an accurate and assertive manner to boost knowledge on sophisticated forms of energy as radiation undoubtedly is. Thus, we hope that consequences of great significance should by derived from the current research if time provides.