[math-fun] Spherical trig question
David Wilson>On the unit sphere, measuring angles in radians. Place an arc with measure A on the equator. Place arcs with measures B and C at each end of A running north along meridians (perpendicular to A). Join the northern endpoints of B and C with a great circle arc. What is the area of the spherical quadrilateral thus formed? rwg>If you know the angles between those arc at those endpoints to be a, isn't it just the angular excess 2 a + 2 pi - 2 pi = 2a? If you know all four arcs, max_solid_angle(a,b,c,d):=2*acos((cos(d)+cos(c)+cos(b)+cos(a))/(4*cos(a/2)*cos(b/2)*cos(c/2)*cos(d/2))-tan(a/2)*tan(b/2)*tan(c/2)*tan(d/2)) --rwg
Ah, from 3.5 yrs ago,
From gosper@alum.mit.edu Mon Jul 16 08:47:39 2007 Date: Mon, 16 Jul 2007 10:39:48 -0400 (EDT) Message-ID: <23173122.9411184596788302.JavaMail.gbourne@brunch.mit.edu> From: Bill Gosper <gosper@alum.mit.edu> To: math-fun@mailman.xmission.com Lemma: The apical solid angle of an isosceles trapezoidal pyramid with vertex angles a,b,a,c is
2 c b c 2 b - sin (-) + (cos(a) - 1) sin(-) sin(-) - sin (-) + cos(a) + 1 2 2 2 2 (d79) 2 acos(-------------------------------------------------------------) b c (cos(a) + 1) cos(-) cos(-) 2 2 I derived this by splitting the general a,b,a,c hedral vertex into trihedrals a,b,d and d,a,c, then maximizing over d, getting b c (d38) cos(d) = cos(a) - 2 sin(-) sin(-) 2 2 Clearer might have been the difference between two isosceles trihedrals: solid_angle(d,d,b)-solid_angle(d-a,d-a,c) with d chosen to equalize the dihedrals, using dihedral(a,b,c):=acos(csc(a)*csc(b)*cos(c)-cot(a)*cot(b)), which gives sin(a) d = atan(----------------------), b c cos(a) - csc(-) sin(-) 2 2 and eventually the same solid angle formula. But the max technique generalizes from trapezoidal pyramids to arbitrary quadrilateral pyramids with apex angles a,b,c,d, whose maximal solid angle is cos(d) + cos(c) + cos(b) + cos(a) a b c d --------------------------------- - tan(-) tan(-) tan(-) tan(-) a b c d 2 2 2 2 4 cos(-) cos(-) cos(-) cos(-) 2 2 2 2 Note we get the old formula when d=0. Note also the symmetry, giving the same result for a,b,d,c. In this maximal case, the "diagonal angles", i.e. the apex angles exposed by splitting into two triangular pyramids, are a d b c b d a c 2 (sin(-) sin(-) + sin(-) sin(-)) (sin(-) sin(-) + sin(-) sin(-)) 2 2 2 2 2 2 2 2 acos(1 - -----------------------------------------------------------------) c d a b sin(-) sin(-) + sin(-) sin(-) 2 2 2 2 and same(a<->c). This is analogous to the maximal area of a quadrilateral with sides a,b,c,d (or a,b,d,c): sqrt((c + b + a - d) (d - c + b + a) (d + c - b + a) (d + c + b - a)) ---------------------------------------------------------------------, 4 with diagonals (a d + b c) (b d + a c) (b d + a c) (c d + a b) sqrt(-----------------------) and sqrt(-----------------------). c d + a b a d + b c Whose formula is this? (Superhero of Alexandria?) And what about arbitrary pentagons, etc? Are the maximal areas fixed w.r.t. permuting the sides? The formulas must be doozies. --rwg IMPORTUNATE PERMUTATION ------=_Part_509_2287320.1184596788300-- On Sun, Feb 6, 2011 at 7:13 PM, Bill Gosper <billgosper@gmail.com> wrote:
David Wilson>On the unit sphere, measuring angles in radians.
Place an arc with measure A on the equator. Place arcs with measures B and C at each end of A running north along meridians (perpendicular to A). Join the northern
endpoints of B and C with a great circle arc. What is the area of the spherical quadrilateral thus formed?
rwg>If you know the angles between those arc at those endpoints to be a, isn't it just the angular excess 2 a + 2 pi - 2 pi = 2a? If you know all four arcs,
max_solid_angle(a,b,c,d):=2*acos((cos(d)+cos(c)+cos(b)+cos(a))/(4*cos(a/2)*cos(b/2)*cos(c/2)*cos(d/2))-tan(a/2)*tan(b/2)*tan(c/2)*tan(d/2)) --rwg
On 2/7/11, Bill Gosper <billgosper@gmail.com> wrote:
Ah, from 3.5 yrs ago, ...
If there is a connection with the original thread, I'm afraid it has escaped me. However ...
This is analogous to the maximal area of a quadrilateral with sides a,b,c,d (or a,b,d,c):
sqrt((c + b + a - d) (d - c + b + a) (d + c - b + a) (d + c + b - a)) ---------------------------------------------------------------------, 4 with diagonals
(a d + b c) (b d + a c) (b d + a c) (c d + a b) sqrt(-----------------------) and sqrt(-----------------------). c d + a b a d + b c
Whose formula is this? (Superhero of Alexandria?)
See http://en.wikipedia.org/wiki/Brahmagupta's_formula http://en.wikipedia.org/wiki/Cyclic_quadrilateral
And what about arbitrary pentagons, etc? Are the maximal areas fixed w.r.t. permuting the sides? The formulas must be doozies. --rwg
It doesn't seem generally appreciated that this result leads directly to the isoperimetric property of the circle. Given the side-lengths of any plane polygon, its area may be increased by adjusting any 4 successive angles until their vertices are concyclic. Hence the area is a maximum when the polygon is concyclic; now let the number of sides go to infinity. I seem to have missed the angular analogue below (Billygosper's formula?) last time round. Does it lead to some sort of dual to the isoperimetric theorem, I wonder ... WFL
From gosper@alum.mit.edu Mon Jul 16 08:47:39 2007 Date: Mon, 16 Jul 2007 10:39:48 -0400 (EDT) Message-ID: <23173122.9411184596788302.JavaMail.gbourne@brunch.mit.edu> From: Bill Gosper <gosper@alum.mit.edu> To: math-fun@mailman.xmission.com Lemma: The apical solid angle of an isosceles trapezoidal pyramid with vertex angles a,b,a,c is
2 c b c 2 b - sin (-) + (cos(a) - 1) sin(-) sin(-) - sin (-) + cos(a) + 1 2 2 2 2 (d79) 2 acos(-------------------------------------------------------------) b c (cos(a) + 1) cos(-) cos(-) 2 2
I derived this by splitting the general a,b,a,c hedral vertex into trihedrals a,b,d and d,a,c, then maximizing over d, getting
b c (d38) cos(d) = cos(a) - 2 sin(-) sin(-) 2 2
Clearer might have been the difference between two isosceles trihedrals:
solid_angle(d,d,b)-solid_angle(d-a,d-a,c)
with d chosen to equalize the dihedrals, using
dihedral(a,b,c):=acos(csc(a)*csc(b)*cos(c)-cot(a)*cot(b)),
which gives sin(a) d = atan(----------------------), b c cos(a) - csc(-) sin(-) 2 2
and eventually the same solid angle formula. But the max technique generalizes from trapezoidal pyramids to arbitrary quadrilateral pyramids with apex angles a,b,c,d, whose maximal solid angle is
cos(d) + cos(c) + cos(b) + cos(a) a b c d --------------------------------- - tan(-) tan(-) tan(-) tan(-) a b c d 2 2 2 2 4 cos(-) cos(-) cos(-) cos(-) 2 2 2 2
Note we get the old formula when d=0. Note also the symmetry, giving the same result for a,b,d,c. In this maximal case, the "diagonal angles", i.e. the apex angles exposed by splitting into two triangular pyramids, are a d b c b d a c 2 (sin(-) sin(-) + sin(-) sin(-)) (sin(-) sin(-) + sin(-) sin(-)) 2 2 2 2 2 2 2 2 acos(1 - -----------------------------------------------------------------) c d a b sin(-) sin(-) + sin(-) sin(-) 2 2 2 2
and same(a<->c).
I have a sometimes rude habit of asking pointed questions without giving context. When I worked at ComputerVision, I implemented an algorithm for polygonal area. For each edge of the polygon, I computed the area of the quadrilateral between the edge and its shadow on the x-axis via orthogonal projection. The edges were directed counterclockwise around the polygon, allowing computation of directed normals, the areas were added or subtracted according to whether the normal pointed toward or away from the x-axis. Adding up all these areas gave the area of the polygon. Later, I generalized the algorithm to polyhedra. Here faces were projected onto the xy-plane to produce cylindrical volumes. Face vertices were oriented in a counterclockwise direction around the face, again allowing computation of directed face normals, and volumes were added or subtracted based on the whether the normal pointed toward or away from the xy-plane. Adding the volumes gave the volume of the polyhedron. Just recently, my new boss told me he had worked at a mapping company, and one of his last tasks, which he did not finish, was to compute the area of a map region (spherical polygon) with vertices given in lat-long coordinates. I was trying to apply the same technique, projecting the edges down to the equator and adding the areas of the quadrilaterals thus formed, hoping that this area might have some tractable formula in terms of the lat-longs. Fred Lunnon's analysis uncovered a simpler approach, to wit, project the edges not to the equator, but to the north pole. Instead of adding up quadrilateral areas below the edges, I could add up triangular areas above the edges. The sides of the triangle impinging on the pole are easy to obtain (complements of the vertex latitudes) as is angle between them (difference of the vertex longitudes). From there, standard spherical trigonometry should allow me to derive the remaining side (the original edge, via law of cosines), angles (via law of sines), and area (law of excess), the prize being the area in terms of edge vertex lat-longs. The sign of the area is determined by local order of the vertex latitudes. The trig, however, is too daunting for me, which is why I posted to brighter lights than myself. I was hoping that some of the computational ugliness might implode leaving a relatively tractable expression. ----- Original Message ----- From: "Fred lunnon" <fred.lunnon@gmail.com> To: "math-fun" <math-fun@mailman.xmission.com> Sent: Tuesday, February 08, 2011 9:07 PM Subject: Re: [math-fun] Spherical trig question
On 2/7/11, Bill Gosper <billgosper@gmail.com> wrote:
Ah, from 3.5 yrs ago, ...
If there is a connection with the original thread, I'm afraid it has escaped me. However ...
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On 2/9/11, David Wilson <davidwwilson@comcast.net> wrote:
... Just recently, my new boss told me he had worked at a mapping company, and one of his last tasks, which he did not finish, was to compute the area of a map region (spherical polygon) with vertices given in lat-long coordinates. I was trying to apply the same technique, projecting the edges down to the equator and adding the areas of the quadrilaterals thus formed, hoping that this area might have some tractable formula in terms of the lat-longs. ...
Heigh-ho --- conned by commerce yet again --- at this rate I'm never going to make a mathematical millionaire! WFL
There is no longer any possibility of monetary to us in this problem. It was just a problem my boss described to me that I found interesting and thought might have an elegant solution. There is no application for spherical trigonometry in our present positions. ----- Original Message ----- From: "Fred lunnon" <fred.lunnon@gmail.com> To: "math-fun" <math-fun@mailman.xmission.com> Sent: Wednesday, February 09, 2011 1:02 AM Subject: Re: [math-fun] Spherical trig question
On 2/9/11, David Wilson <davidwwilson@comcast.net> wrote:
... Just recently, my new boss told me he had worked at a mapping company, and one of his last tasks, which he did not finish, was to compute the area of a map region (spherical polygon) with vertices given in lat-long coordinates. I was trying to apply the same technique, projecting the edges down to the equator and adding the areas of the quadrilaterals thus formed, hoping that this area might have some tractable formula in terms of the lat-longs. ...
Heigh-ho --- conned by commerce yet again --- at this rate I'm never going to make a mathematical millionaire! WFL
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It's OK --- I didn't fancy an 18,000 sq ft pad with a private concert hall anyway. Not much ... WFL On 2/10/11, David Wilson <davidwwilson@comcast.net> wrote:
There is no longer any possibility of monetary to us in this problem. It was just a problem my boss described to me that I found interesting and thought might have an elegant solution. There is no application for spherical trigonometry in our present positions.
participants (3)
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Bill Gosper -
David Wilson -
Fred lunnon