Spherical Triangles: Area and mapping to Euclidean space
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If we take a sphere of radius 1 and travel a quarter-circumference south from the north pole, turn 90 degrees, travel another quarter circumference, then return North, we form a triangle with an angle of 270$^text{o}$. According to Mathworld, the area of such a triangle would be given by $$Delta=R^{2}[(A+B+C)-pi]$$ With A, B and C the angles of the triangle.
Now if we consider lines of latitude and longitude (I understand these always intersect at $90^text{o}$), and shift the base of the triangle closer to the north pole but so that it remains on a line of latitude, this would suggest that the base of the triangle still meets the two longitudinal lines at 90$^text{o}$. This would suggest that the area stays the same, which seems very wrong.
I am also unsure of the maximum sum of angles of a spherical triangle. On the same Mathworld page it states:
The sum of the angles of a spherical triangle is between $pi$ and $3pi$ radians (180 degrees and 540 degrees; Zwillinger 1995, p. 469)
However, considering a very small triangle (s.t. the sum of interior angles tends to 180$^text{o}$, and taking this as the outside of the triangle (and the rest of the sphere as the inside) this would create a 'triangle' with angles summing 900$^text{o}$. My guess would be that this might not count as a triangle because it perhaps cannot be mapped appropriately to Euclidaen space. So then, what would be an appropriate mapping to Euclidean space? Would it result in a 'triangle' with curved sides?
geometry spherical-geometry spherical-trigonometry
asked Dec 19 '18 at 14:39
cluelessatthiscluelessatthis
402313
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add a comment |
$begingroup$
If we take a sphere of radius 1 and travel a quarter-circumference south from the north pole, turn 90 degrees, travel another quarter circumference, then return North, we form a triangle with an angle of 270$^text{o}$. According to Mathworld, the area of such a triangle would be given by $$Delta=R^{2}[(A+B+C)-pi]$$ With A, B and C the angles of the triangle.
Now if we consider lines of latitude and longitude (I understand these always intersect at $90^text{o}$), and shift the base of the triangle closer to the north pole but so that it remains on a line of latitude, this would suggest that the base of the triangle still meets the two longitudinal lines at 90$^text{o}$. This would suggest that the area stays the same, which seems very wrong.
I am also unsure of the maximum sum of angles of a spherical triangle. On the same Mathworld page it states:
The sum of the angles of a spherical triangle is between $pi$ and $3pi$ radians (180 degrees and 540 degrees; Zwillinger 1995, p. 469)
However, considering a very small triangle (s.t. the sum of interior angles tends to 180$^text{o}$, and taking this as the outside of the triangle (and the rest of the sphere as the inside) this would create a 'triangle' with angles summing 900$^text{o}$. My guess would be that this might not count as a triangle because it perhaps cannot be mapped appropriately to Euclidaen space. So then, what would be an appropriate mapping to Euclidean space? Would it result in a 'triangle' with curved sides?
geometry spherical-geometry spherical-trigonometry
asked Dec 19 '18 at 14:39
cluelessatthiscluelessatthis
402313
$endgroup$
add a comment |
$begingroup$
If we take a sphere of radius 1 and travel a quarter-circumference south from the north pole, turn 90 degrees, travel another quarter circumference, then return North, we form a triangle with an angle of 270$^text{o}$. According to Mathworld, the area of such a triangle would be given by $$Delta=R^{2}[(A+B+C)-pi]$$ With A, B and C the angles of the triangle.
Now if we consider lines of latitude and longitude (I understand these always intersect at $90^text{o}$), and shift the base of the triangle closer to the north pole but so that it remains on a line of latitude, this would suggest that the base of the triangle still meets the two longitudinal lines at 90$^text{o}$. This would suggest that the area stays the same, which seems very wrong.
I am also unsure of the maximum sum of angles of a spherical triangle. On the same Mathworld page it states:
The sum of the angles of a spherical triangle is between $pi$ and $3pi$ radians (180 degrees and 540 degrees; Zwillinger 1995, p. 469)
However, considering a very small triangle (s.t. the sum of interior angles tends to 180$^text{o}$, and taking this as the outside of the triangle (and the rest of the sphere as the inside) this would create a 'triangle' with angles summing 900$^text{o}$. My guess would be that this might not count as a triangle because it perhaps cannot be mapped appropriately to Euclidaen space. So then, what would be an appropriate mapping to Euclidean space? Would it result in a 'triangle' with curved sides?
geometry spherical-geometry spherical-trigonometry
asked Dec 19 '18 at 14:39
cluelessatthiscluelessatthis
402313
$endgroup$
If we take a sphere of radius 1 and travel a quarter-circumference south from the north pole, turn 90 degrees, travel another quarter circumference, then return North, we form a triangle with an angle of 270$^text{o}$. According to Mathworld, the area of such a triangle would be given by $$Delta=R^{2}[(A+B+C)-pi]$$ With A, B and C the angles of the triangle.
Now if we consider lines of latitude and longitude (I understand these always intersect at $90^text{o}$), and shift the base of the triangle closer to the north pole but so that it remains on a line of latitude, this would suggest that the base of the triangle still meets the two longitudinal lines at 90$^text{o}$. This would suggest that the area stays the same, which seems very wrong.
I am also unsure of the maximum sum of angles of a spherical triangle. On the same Mathworld page it states:
The sum of the angles of a spherical triangle is between $pi$ and $3pi$ radians (180 degrees and 540 degrees; Zwillinger 1995, p. 469)
However, considering a very small triangle (s.t. the sum of interior angles tends to 180$^text{o}$, and taking this as the outside of the triangle (and the rest of the sphere as the inside) this would create a 'triangle' with angles summing 900$^text{o}$. My guess would be that this might not count as a triangle because it perhaps cannot be mapped appropriately to Euclidaen space. So then, what would be an appropriate mapping to Euclidean space? Would it result in a 'triangle' with curved sides?
geometry spherical-geometry spherical-trigonometry
geometry spherical-geometry spherical-trigonometry
asked Dec 19 '18 at 14:39
cluelessatthiscluelessatthis
402313
asked Dec 19 '18 at 14:39
cluelessatthiscluelessatthis
402313
edited Dec 19 '18 at 15:28
cluelessatthis
asked Dec 19 '18 at 14:39
cluelessatthiscluelessatthis
402313
asked Dec 19 '18 at 14:39
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402313
asked Dec 19 '18 at 14:39
cluelessatthiscluelessatthis
402313
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1 Answer
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A spherical triangle must be formed by arcs of great circles, hence you cannot take a parallel as side, apart from the equator.
And of the two parts into which a sphere is divided by the sides of a spherical triangle, only the smaller (and convex) one is considered "the inner part" of the triangle. That explains the 540° limit.
answered Dec 19 '18 at 16:48
AretinoAretino
24k21443
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Right you are. It’s a common error to think that parallels of latitude can be sides of a spherical triangle.
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– Lubin
Dec 19 '18 at 18:14
add a comment |
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1 Answer
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$begingroup$
A spherical triangle must be formed by arcs of great circles, hence you cannot take a parallel as side, apart from the equator.
And of the two parts into which a sphere is divided by the sides of a spherical triangle, only the smaller (and convex) one is considered "the inner part" of the triangle. That explains the 540° limit.
answered Dec 19 '18 at 16:48
AretinoAretino
24k21443
$endgroup$
$begingroup$
Right you are. It’s a common error to think that parallels of latitude can be sides of a spherical triangle.
$endgroup$
– Lubin
Dec 19 '18 at 18:14
add a comment |
$begingroup$
A spherical triangle must be formed by arcs of great circles, hence you cannot take a parallel as side, apart from the equator.
And of the two parts into which a sphere is divided by the sides of a spherical triangle, only the smaller (and convex) one is considered "the inner part" of the triangle. That explains the 540° limit.
answered Dec 19 '18 at 16:48
AretinoAretino
24k21443
$endgroup$
$begingroup$
Right you are. It’s a common error to think that parallels of latitude can be sides of a spherical triangle.
$endgroup$
– Lubin
Dec 19 '18 at 18:14
add a comment |
$begingroup$
A spherical triangle must be formed by arcs of great circles, hence you cannot take a parallel as side, apart from the equator.
And of the two parts into which a sphere is divided by the sides of a spherical triangle, only the smaller (and convex) one is considered "the inner part" of the triangle. That explains the 540° limit.
answered Dec 19 '18 at 16:48
AretinoAretino
24k21443
$endgroup$
A spherical triangle must be formed by arcs of great circles, hence you cannot take a parallel as side, apart from the equator.
And of the two parts into which a sphere is divided by the sides of a spherical triangle, only the smaller (and convex) one is considered "the inner part" of the triangle. That explains the 540° limit.
answered Dec 19 '18 at 16:48
AretinoAretino
24k21443
answered Dec 19 '18 at 16:48
AretinoAretino
24k21443
answered Dec 19 '18 at 16:48
AretinoAretino
24k21443
answered Dec 19 '18 at 16:48
AretinoAretino
24k21443
24k21443
$begingroup$
Right you are. It’s a common error to think that parallels of latitude can be sides of a spherical triangle.
$endgroup$
– Lubin
Dec 19 '18 at 18:14
add a comment |
$begingroup$
Right you are. It’s a common error to think that parallels of latitude can be sides of a spherical triangle.
$endgroup$
– Lubin
Dec 19 '18 at 18:14
$begingroup$
Right you are. It’s a common error to think that parallels of latitude can be sides of a spherical triangle.
$endgroup$
– Lubin
Dec 19 '18 at 18:14
$begingroup$
Right you are. It’s a common error to think that parallels of latitude can be sides of a spherical triangle.
$endgroup$
– Lubin
Dec 19 '18 at 18:14
add a comment |
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