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Joseph O'Rourke
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Not a precisely quantitative answer. Just attempting to track the specified geometry. I used $\epsilon=0.1$ radians to delimit the subset $B \subset T$.

My question is, what does $C$ look like (as a set of points)?

It "looks like" as indicated below:


  [![SphereArcQ][1]][1]
Pardon that I did not

(Assume we choose the orthogonal vector with positive x-value)

but instead showed both $\pm$.

$C$ appears to be bounded by a circular arc connected to a V whose angle is determined by $\epsilon$, in particular, the straight-line boundaries of $C$ are orthogonal to the $\epsilon$ extremes of $B$, as is the circular arc boundary., more clearly seen from $(+\infty,0,0)$:


[![Smiley][2]][2]
In any case, have a nice day! :-)

Not a precisely quantitative answer. Just attempting to track the specified geometry. I used $\epsilon=0.1$ radians to delimit the subset $B \subset T$.

My question is, what does $C$ look like (as a set of points)?

It "looks like" as indicated below:


  [![SphereArcQ][1]][1]
Pardon that I did not

(Assume we choose the orthogonal vector with positive x-value)

but instead showed both $\pm$.

$C$ appears to be bounded by a circular arc connected to a V whose angle is determined by $\epsilon$, in particular, the straight-line boundaries of $C$ are orthogonal to the $\epsilon$ extremes of $B$, as is the circular arc boundary.


[![Smiley][2]][2]
In any case, have a nice day! :-)

Not a precisely quantitative answer. Just attempting to track the specified geometry. I used $\epsilon=0.1$ radians to delimit the subset $B \subset T$.

My question is, what does $C$ look like (as a set of points)?

It "looks like" as indicated below:


  [![SphereArcQ][1]][1]
Pardon that I did not

(Assume we choose the orthogonal vector with positive x-value)

but instead showed both $\pm$.

$C$ appears to be bounded by a circular arc connected to a V whose angle is determined by $\epsilon$, in particular, the straight-line boundaries of $C$ are orthogonal to the $\epsilon$ extremes of $B$, as is the circular arc boundary, more clearly seen from $(+\infty,0,0)$:


[![Smiley][2]][2]
In any case, have a nice day! :-)
added 30 characters in body
Source Link
Joseph O'Rourke
  • 150.8k
  • 36
  • 358
  • 958

Not a precisely quantitative answer. Just attempting to track the specified geometry. I used $\epsilon=0.1$ radians to delimit the subset $B \subset T$.

My question is, what does $C$ look like (as a set of points)?

It "looks like" as indicated below:


  [![SphereArcQ][1]][1]
Pardon that I did not

(Assume we choose the orthogonal vector with positive x-value)

but instead showed both $\pm$.

$C$ appears to be bounded by a circular arc connected to a V whose angle is determined by $\epsilon$, in particular, the straight-line boundaries of $C$ are orthogonal to the $\epsilon$ extremes of $B$, as is the circular arc boundary.


[![Smiley][2]][2]
In any case, have a nice day! :-)

Not a precisely quantitative answer. Just attempting to track the specified geometry. I used $\epsilon=0.1$ radians to delimit the subset $B \subset T$.

My question is, what does $C$ look like (as a set of points)?

It "looks like" as indicated below:


  [![SphereArcQ][1]][1]
Pardon that I did not

(Assume we choose the orthogonal vector with positive x-value)

but instead showed both $\pm$.

$C$ appears to be bounded by a circular arc connected to a V whose angle is determined by $\epsilon$, in particular, the straight-line boundaries of $C$ are orthogonal to the $\epsilon$ extremes of $B$.


[![Smiley][2]][2]
In any case, have a nice day! :-)

Not a precisely quantitative answer. Just attempting to track the specified geometry. I used $\epsilon=0.1$ radians to delimit the subset $B \subset T$.

My question is, what does $C$ look like (as a set of points)?

It "looks like" as indicated below:


  [![SphereArcQ][1]][1]
Pardon that I did not

(Assume we choose the orthogonal vector with positive x-value)

but instead showed both $\pm$.

$C$ appears to be bounded by a circular arc connected to a V whose angle is determined by $\epsilon$, in particular, the straight-line boundaries of $C$ are orthogonal to the $\epsilon$ extremes of $B$, as is the circular arc boundary.


[![Smiley][2]][2]
In any case, have a nice day! :-)
added 242 characters in body
Source Link
Joseph O'Rourke
  • 150.8k
  • 36
  • 358
  • 958

Not ana precisely quantitative answer. Just Just attempting to track the specified geometry. I used $\epsilon=0.1$ radians to delimit the subset $B \subset T$.

My question is, what does $C$ look like (as a set of points)?

It looks like"looks like" as shownindicated below:


  [![SphereArcQ][1]][1]
Pardon that I did not

(Assume we choose the orthogonal vector with positive x-value)

but instead showed both $\pm$.

In any case, have$C$ appears to be bounded by a nice day!circular arc connected to a :V whose angle is determined by $\epsilon$, in particular, the straight-)line boundaries of $C$ are orthogonal to the $\epsilon$ extremes of $B$.


[![Smiley][2]][2]  
In any case, have a nice day! :-)

Not an answer. Just attempting to track the specified geometry. I used $\epsilon=0.1$ radians to delimit the subset $B \subset T$.

My question is, what does $C$ look like (as a set of points)?

It looks like as shown below:


  [![SphereArcQ][1]][1]
Pardon that I did not

(Assume we choose the orthogonal vector with positive x-value)

but instead showed both $\pm$.

In any case, have a nice day! :-)


[![Smiley][2]][2]  

Not a precisely quantitative answer. Just attempting to track the specified geometry. I used $\epsilon=0.1$ radians to delimit the subset $B \subset T$.

My question is, what does $C$ look like (as a set of points)?

It "looks like" as indicated below:


  [![SphereArcQ][1]][1]
Pardon that I did not

(Assume we choose the orthogonal vector with positive x-value)

but instead showed both $\pm$.

$C$ appears to be bounded by a circular arc connected to a V whose angle is determined by $\epsilon$, in particular, the straight-line boundaries of $C$ are orthogonal to the $\epsilon$ extremes of $B$.


[![Smiley][2]][2]
In any case, have a nice day! :-)
Source Link
Joseph O'Rourke
  • 150.8k
  • 36
  • 358
  • 958
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