Trihedral Angle
L V Nagarajan
Introduction
Solid geometry has always fascinated me and has fired up my imaginative power. I was introduced to Solid Geometry by my elder brother (L V Sundaram) when I was barely 10 years old. In his university days, he had a habit of reading aloud and explaining to himself as a part of his study process. When I am around as a young boy, he will do the explaining to me, not caring how much of it I understood. But he was rather surprised that I understood a lot of what he told me especially in geometry. As I learnt more of plane geometry, I developed a habit of extending the results of the theorems to the third dimension. Subsequent to my earlier blog on Polyhedrons, I want to now share my research on trihedral angle, as it is the basis of all solid angles
Normal Plane Angles
I am re-stating some fairly obvious facts about plane angles, since we need the same to compare with the three dimensional angle, known as Trihedral Angle. In a plane if two straight lines are drawn, they intersect each other at some point. But when they are parallel to each other, they maintain constant distance between themselves all along their infinite lengths in both directions. Hence they never intersect each other. Two non-parallel straight lines in a plane, intersecting each other, have a rotational distance between them. When one of the straight lines is rotated about the point of intersection, it will coincide with the other straight line after some rotation. This rotational distance between the straight lines is known as Angle. One full rotation is divided into 360 parts for the purpose of quantifying the amount of rotation, each part known as a degree. Consider a circle with O as the centre and R as the radius. As the radius rotates a complete round it completes a circumferential distance of 2πR. Hence a full rotation of 360 degrees is associated with 2π Radians. Any arc of the circle of length L subtends at the centre O an angle of L/R radians. When two straight lines in a plane intersect each other they actually create four angles, with opposite angles of equal magnitudes. It is easy to see that the adjacent angles add up to 180 degrees or half rotation. These four angles are actually the four internal angles of a parallelogram! We may also easily see that two intersecting straight lines in a plane divide the plane into four quadrants. If the straight lines are perpendicular (90 degrees) to each other, they divide the plane into four equal quadrants. The lines themselves become X-axis and Y-axis of the coordinate system.
Trihedral Angle
Now let us move to the third dimension. Consider two planes parallel to each other. They maintain a constant distance between them all along their infinite surface. They never intersect each other. But two non-parallel planes do intersect each other, at some place, along a straight line. They do have a rotational distance between them. This is similar to plane angle and this rotational distance is known as dihedral angle, as the angle between two planes, varying from 0 to 360 degrees (or 2π radians). Now let us consider a third plane non-parallel to either of the earlier two planes. Let this plane intersect each of the earlier two planes along two different and non-parallel straight lines. All these three planes (and their three straight lines of intersection) will meet at a common point. At this point is formed a solid angle, known as trihedral angle. In general an angle represents a rotational distance. Similarly, the trihedral angle also is a measure of the sum of rotational distances, in two directions, i.e., rotations about two of the edges of trihedral intersections. Hence some time it is called as square-angle or square degrees. Trihedral angles vary from 0 to 720 square degrees. Consider a sphere with O as the centre and R as the radius. As the radius rotates a complete round in a, say, vertical plane, we get a vertical circle. When this circle rotates a complete round about any diameter as axis, we get a complete sphere with a surface area 4πR^{2}, covered by rotation. Hence a full rotation of 720 degrees is associated with 4π Stradians. Now consider a solid angle subtended by a partial spherical segment at the centre. It is given as surface area of the segment divided by radius squared. When three planes intersect each other in space, they actually create eight trihedral angles. These eight trihedral angles are actually the eight solid angles of a parallelepiped! We can also see that three intersecting planes divide the 3-dimensional space into eight octants. If these three planes are perpendicular each other, they divide the 3-dimensional space into eight equal octants. The lines of intersections, some time called as edges, form the X-Y-Z axis of the 3-demensional co-ordinate system.
Other angles of trihedral angle
Though we are talking of three planes making a trihedral angle, we feel comfortable to talk of trihedral angle as formed by three (non-planar) straight lines meeting at a point. Hence we talk about vertex angle as the plane-angle between any two edges at the vertex of the trihedral angle. Dihedral angles are the angles between adjacent planes making up any trihedral or polyhedral angles. The solid angle (or any polyhedral angle), the dihedral angle and the vertex angle are normally denoted as, Ω, Φ and θ. It is interesting to see below, how these angles relate to each other, say, for the case of a trihedral angle.
From the theory of spherical excess, it has been derived and stated that, the solid angle at O as enclosed by vectors A, B and C as above, is given by Ω = Φa + Φb + Φc – π
Let us consider the solid angle Ω1, at O, enclosed by OA, OB and OC. The dihedral angle Φa is as shown. This is the angle between two semi circular faces of the orange-like dice ABA_{1}CA. This piece of spherical surface is sometimes known as di-angle. The spherical surface area of this dice is Φa/2π times the whole surface area of the sphere. Hence the solid angle subtended by this area at the centre O is Φa/2π times 4π, i.e., 2Φa.
Hence supplementary solid angle Ω2, enclosed by OA_{1},OB and OC is given by 2Φa – Ω1.
We may infer similar results with OB_{1} and OC_{1} and get
Ω2 = 2Φa – Ω1
Ω3 = 2Φb – Ω1
Ω4 = 2Φc – Ω1
It is easy to see that Ω1+ Ω2 + Ω3 + Ω4 = a hemisphere = 2π
i.e. 2(Φa + Φb + Φc – Ω1) = 2π
i.e. Ω = Φa + Φb + Φc – π
This proves the result we got earlier. Ω1, Ω2, Ω3 and Ω4 are the four solid angles of the parallelepiped obtained from OA,OB, OC. The other four solid angles will be equal to their opposite angles. These eight solid angles form the octants in the three dimensional space as mentioned earlier.
Trihedral Angle Ω, in terms of vertex angles α, β and γ
In a parallelogram described by sides ‘a’ and ‘b’ and angle θ, the area is given by (ab sin θ). Referring to the diagram below:
We can write tan θ/2 = b sin θ/ (b + b cos θ)
i.e., tan θ/2 = ab sin θ/ (ab + ab cos θ)
= Area of Parallelogram/ ab (1+cos θ)
i.e., tan θ/2 = (a x b)/(ab + a.b)
We have a similar relationship for a parallelepiped described by vectors and angles: a, b, c, α, β and γ, given by Oosterom and Strackee (2) as:
Tan(Ω/2) = (a . b x c) / [ abc + (a.b)c + (b.c)a + (c.a)b]
It is easy to see that (a . b x c) represents the volume of the parallelepiped a, b, c.
Volume V = (a . b x c)
= Det (a / b / c), with vectors a, b, c represented as row vectors.
= Det (a , b , c), with vectors a, b, c represented as column vectors
i.e., V = a . b x c = Det [a1,a2.a3/ b1,b2,b3/c1,c2,c3]
= Det [a1,b1,c1/a2,b2,c2/a3,b3,c3]
Hence, V*V = Det [a.a, a.b, a.c/ b.a, b.b, b.c/ c.a, c.b, c.c]
It will reduce to
V*V =
a^{2} b^{2} c^{2} (1 + 2 cos α cos β cos γ – cos^{2}α – cos^{2} β – cos^{2} γ )
The solid angle Ω does not depend on magnitudes of vectors a, b, c and hence assuming them to be unit vectors, we get the expression for solid angle Ω, as
tan (Ω/2)
= (√ (1 + 2 cos α cos β cos γ – cos^{2}α – cos^{2} β – cos^{2} γ)) / (1+cos α+cos β+cos γ)
Now we have expressed Ω in terms of vertex angles α, β and γ. We have also earlier expressed Ω in terms of dihedral angles Φa, Φb, Φc. Later in this write-up, we will also find how α, β and γ relate to Φa, Φb and Φc.
‘Tetragonometric’ ratios of trihedral angle
When we talk of tan (Ω/2) as above, we are not talking of actual trigonometric ratio of trihedral angle Ω. We are only treating the square-degree Ω, as normal angle of the same magnitude and then expressing its trigonometric ratio in the above formula. But do we have ‘trigonometric ratios’, like Sine, Cosine, Tangent, for trihedral angles also? (May be, we should call them as ‘tetragonometric ratios’)
Let us go back to our comparison with a parallelogram.
Area A of the parallelogram a, b, θ, is given by ab sinθ
So we say, Sin θ = A/ab. i.e., Area/(product of adjacent lengths)
Similarly, in case of parallelepiped, (a, b, c, α, β, γ), G-sin Ω has been defined as,
G-sin Ω = V*V/ (Product of adjacent areas)
i.e., G-sin Ω = V*V/ (ab sin γ)(bc sin α)(ca sin β);
We know V*V =
a^{2} b^{2} c^{2} (1 + 2 cos α cos β cos γ – cos^{2}α – cos^{2} β – cos^{2} γ )
Hence
G-Sin Ω =
(1+2 cosα cosβ cosγ -cos^{2}α -cos^{2}β -cos^{2}γ) /(sinα sinβ sinγ)
Volume V1 of a unit parallelepiped with angles α, β, γ and with (a=b=c=1) is given by,
V1 = √(1+2 cos α cos β cos γ – cos^{2}α -cos^{2} β -cos^{2} γ)
Hence, we can say,
G-Sin Ω = (V1*V1)/ (sin α sin β sin γ);
It is easy to see that all the eight solid angles at the eight corners of the parallelepiped will have the same G-sine value, as the same three parallelograms are enclosing them, but in a different order. This is exactly as in the case of a parallelogram where all the four angles have the same sine value.
This G-sine value appears to signify nothing, especially when you see this value not being consistent. Even as two sets of α, β, γ produces same solid angle Ω, their G-sine values will vary. For a particular value of Ω, G-sine will be maximum when α = β = γ.
Let us approach G-sine from a different perspective. We know trigonometric ratios of any angle are derived from a right angled triangle that includes this angle. Please refer to the following diagrams.
Sin θ = BC/AC; Cos θ = AB/AC; Tan θ = BC/AB
Any trihedral angle at O can be represented as a part of a semi-right-angled tetrahedron OABC as above.
Angles OCA and OCB are right angles. i.e. OC is perpendicular to plane ACB.
Angle AOC (= β) and angle COB (=α) are independent and, along with the dihedral angle Φ, completely defines the trihedral angle at O. Angle AOB (=γ) and angle Φ depend on each other. Let OC = c. Now from the definition of G-sin we get,
G-Sin O = V*V/ Product of adjacent areas
V (of parallelepiped) = (AC*BC Sin Φ) * OC
Product of adjacent parallelograms
= (OC*AC sin β)*(OC*BC sin α )*(OA*OBsin γ)
(as α =β = 90 degrees); = (AC*BC*OC*OC)*(OA*OBsin γ)
G-Sin O = AC*BC Sin^{2} Φ / (OA*OB sin γ)
= (2*Area ABC/ 2*Area ABO) Sin Φ
i.e. G-Sin O = (Area ABC / Area ABO) * Sin Φ
i.e G- Sin O = (Opposite area/ Hypotenuse area) * Sin Φ
In the special case when Φ = 90 degrees, we get,
G sin O = (Opposite area/ Hypotenuse area)
This is very similar to trigonometric ratio for plane angle. It is easy to see we require a special set of angles α, β, γ (with Φ = 90 deg) to produce a right angled tetrahedron. In all other cases, the relationship,
G- Sin O = (Opposite area/ Hypotenuse area) * Sin Φ, holds good.
For a right angles tetrahedron, it can be easily proved that, the sum of the squares of adjacent areas will be equal to the square of hypotenuse area. This is very similar to Pythagoras theorem of right-angled triangles. Using this property, we can define other ‘tetragonomertic’ ratios for trihedral angles as below:
G-Cos O = (√(Sum of the squares of the two adjacent areas)) / Hypotenuse area.
G-Tan O = Opposite area / (√(Sum of the squares of adjacent areas))
We can also write G-Sin^{2} O + G-Cos^{2} O = 1
We can also note:
G-Sin^{2} O + G-Sin^{2} A + G-Sin^{2} B = G-Sin^{2} C = 1.
Please note that this right tetrahedron is right-solid-angled at C (i.e. π/2 stradians or 90 sq. degrees).
Other interesting relationships
We started this section with the fact that any trihedral angle can be represented as a part of semi-right-angled tetrahedron as below.
We derived the relationship
G- Sin O = (Opposite area/ Hypotenuse area) * Sin Φc
Now (AC = OA Sin AOC = a sin β) and (BC = OB Sin BOC = b sin α)
Therefore Area ACB
= (1/2)AC.BC sin Φc = (1/2)(a sin β) (b sin α) Sin Φc
Similarly Area AOB = (1/2)OA.OB. Sin γ = (1/2)a.b. Sin γ
G-Sin O = (Area ACB / Area AOB)* Sin Φc
= (Sin α. Sin β. Sin^{2} Φc) / Sin γ
By extending this logic, we can write,
G-Sin O = (Sin α. Sin β. Sin^{2} Φc)/Sin γ
= (Sin β. Sin γ. Sin^{2} Φa)/Sin α
= (Sin γ. Sin α. Sin^{2} Φb)/Sin β
Thus we have established a comprehensive relationship for any trihedral angle in terms of vertex angles and dihedral angles.
Using the above along with the expression for G-Sin O,
G-Sin O = (V1*V1) / (sin α sin β sin γ); we get,
Sin Φa = V1/( sin β sin γ), Sin Φb = V1/( sin γ sin α) and
Sin Φc = V1/( sin α sin β)
Remember V1 is volume of unit parallelepiped α, β, γ, with a = b = c =1 given by:
V1 = √(1 + 2 cos α cos β cos γ – cos^{2}α – cos^{2} β – cos^{2} γ )
Once the values of Φa, Φb, Φc are calculated as above, we can find each of the eight solid angles in the eight octants formed by α, β, γ planes. Of course these are same as the eight solid angles of the parallelepiped (a,b,c, α, β, γ). We already know that
Ω1 = Φa + Φb + Φc – π = Ω
Ω2 = 2Φa – Ω1
Ω3 = 2Φb – Ω1
Ω4 = 2Φc – Ω1
These solid angles are repeated again as the diametrically opposite angles, thus forming all the eight trihedral angles of the parallelepiped. It may be observed that the above four trihedral angles add up to 2π. All eight trihedral angles add up to 4π.
As stated earlier, G-Sine of all the eight solid angles of the parallelepiped are equal.
G-Sin Ω1 = G-Sin (2Φa – Ω1)
= G-Sin (2Φb – Ω1) = G-Sin (2Φc – Ω1).
Now let us tabulate all these angles for different types of trihedral angles.
Type |
α |
β |
γ |
Ω Sq Deg |
Φa |
Φb |
Φc |
1 |
120.00 |
120.00 |
120.00 |
360.00 |
0.00 |
0.00 |
0.00 |
2 |
109.47 |
109.47 |
109.47 |
180.00 |
60.00 |
60.00 |
60.00 |
3 |
90.00 |
90.00 |
90.00 |
90.00 |
90.00 |
90.00 |
90.00 |
4 |
63.43 |
63.43 |
63.43 |
36.00 |
72.00 |
72.00 |
72.00 |
5 |
108.00 |
108.00 |
108.00 |
169.70 |
63.44 |
63.44 |
63.44 |
6 |
60.00 |
60.00 |
60.00 |
31.59 |
70.53 |
70.53 |
70.53 |
7 |
30.00 |
90.00 |
90.00 |
30.00 |
30.00 |
90.00 |
90.00 |
8 |
90.00 |
60.00 |
90.00 |
60.00 |
90.00 |
60.00 |
90.00 |
9 |
90.00 |
90.00 |
110.00 |
110.00 |
90.00 |
90.00 |
70.00 |
10 |
70.00 |
70.00 |
70.00 |
45.71 |
75.24 |
75.24 |
75.24 |
11 |
70.00 |
70.00 |
65.00 |
42.93 |
76.59 |
76.59 |
69.75 |
12 |
30.00 |
70.00 |
65.00 |
19.52 |
32.10 |
86.97 |
74.39 |
Types of trihedral angles are as listed below
Type | Type of trihedral angle |
1 |
@ Centre of a Hemi-sphere |
2 |
@ Centre of a Reg. Tetrahedron-4 |
3 |
@ Centre of Reg. Octahedron- 8, Vertex of a Cube, one octant of a sphere, Right-angled trihedral angle |
4 |
@ Centre of reg. Icosahedron – 20 |
5 |
Vertex of a reg. Dodecahedron-12 |
6 |
Vertex of a reg. Tetrahedron – 4 |
7 |
Semi Right angled trihedral angle |
8 |
Semi Right angled trihedral angle |
9 |
Semi Right angled trihedral angle |
10 |
Equilateral Trihedral angle |
11 |
Isosceles Trihedral angle |
12 |
Typical acute trihedral angle |
Conclusion
The sites given in the references had triggered my interest in this topic of trihedral angle. However most of the concepts developed in this write-up are the results of my own research. I hope this will be useful for students of geometry and for others who are doing further research on the application of solid angles.
References:
2. A. VAN OOSTEROM, J. STRACKEE: A solid angle of a plane triangle. – IEEE Trans. Biomed. Eng. 30:2 (1983); 125-126.
3. http://mathdl.maa.org/images/upload_library/22/Ford/CarlBAllendoerfer.pdf
October 17, 2011 at 2:52 pm |
Exciting generalizations in a higher dimension – I think the derived relationships can be used in engineering and physics.
May 22, 2015 at 11:09 am |
Saved as a favorite, I like your web site!
January 17, 2016 at 10:14 pm |
I like your generalization but I would like to add something
Dimension 2
tan θ/2 = sin θ/ (1 + cos θ)
tan θ/2 = √ 1- cos2 θ/ (1 + cos θ)=sqrt (det Gram(U,V))/(1 + cos θ)
Dimension 3
tan (Ω/2)
= (√ (1 + 2 cos α cos β cos γ – cos2α – cos2 β – cos2 γ)) / (1+cos α+cos β+cos γ)
tan (Ω/2)=sqrt (det Gram(U,V,W)/(1+cosα+cos β+cos γ)
Dimension 4
tan (Ω/2)
= sqrt( det Gram(U,V,W,T)) / (1+cos α+cos β+cos γ+cos Lampda + cos Mu + cos Nu)
Dimension N
tan (Ω/2)
= sqrt( det Gram(U,V,W,….TN)) / (1+cos α+cos β+cos γ+cos Lampda + cos Mu + cos Nu……….cosOmegaN)
Do you agree?
January 18, 2016 at 7:11 pm |
Thanks for your comments. However I am not very sure about your extension of the results to ‘higher’ dimension. Are U,V,W,T the vectors forming 4-hedral angle (like the solid angle subtended by a side of a cube at its center). What are these angles α, β, γ, lambda, Mu, Nu? Are they angle U-V, U-W, U-T, V-W, V-T, W-T? You may try and check your extension for the 4-hedral angle as mentioned above. It should workout to 2Pi/3 !
January 19, 2016 at 3:53 am |
Thanks you for your comments.
Yes U,V,W,T are the vectors forming 4-hedral angle (like the solid angle subtended by a side of a cube ,
α, β, γ, λ, μ, ν are angle U-V, U-W, U-T, V-W, V-T, W-T
The formula tan (Ω/2) is valid for dimension 2 and dimension 3
it is extended to dimension 4, numerator is OK (det of Gram matrix)