Complex Gall stereographic projection inverse [closed]
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Wikipedia gives the formula of the Gall stereographic projection as follows:
1) The projection is conventionally defined as
$$
x = fracR lambdasqrt2;
qquad
y = Rleft( 1 + fracsqrt22 right) tan fracvarphi2
cdotp
$$
2) The reverse projection is defined as
$$
lambda = fracx sqrt2R;
qquad
varphi = 2 arctan fracyRleft( 1 + fracsqrt22 right)
cdotp
$$
How does one go from 1 to 2? I'm having trouble finding the formula of the inverse.
algebra-precalculus projection
edited Aug 23 at 10:57
Jendrik Stelzner
7,57221037
asked Aug 22 at 5:18
Ravi Dhorajiya
1093
closed as off-topic by Jendrik Stelzner, amWhy, Strants, Adrian Keister, user91500 Aug 24 at 6:04
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jendrik Stelzner, amWhy, Strants, Adrian Keister, user91500
add a comment |Â
up vote
1
down vote
favorite
Wikipedia gives the formula of the Gall stereographic projection as follows:
1) The projection is conventionally defined as
$$
x = fracR lambdasqrt2;
qquad
y = Rleft( 1 + fracsqrt22 right) tan fracvarphi2
cdotp
$$
2) The reverse projection is defined as
$$
lambda = fracx sqrt2R;
qquad
varphi = 2 arctan fracyRleft( 1 + fracsqrt22 right)
cdotp
$$
How does one go from 1 to 2? I'm having trouble finding the formula of the inverse.
algebra-precalculus projection
edited Aug 23 at 10:57
Jendrik Stelzner
7,57221037
asked Aug 22 at 5:18
Ravi Dhorajiya
1093
closed as off-topic by Jendrik Stelzner, amWhy, Strants, Adrian Keister, user91500 Aug 24 at 6:04
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jendrik Stelzner, amWhy, Strants, Adrian Keister, user91500
Solve for $lambda$ in the first equation, and solve for $varphi$ in the second equation. Essentially the first set of equations is a map from $(x,y)$ to $(lambda , varphi )$ and the second set is the inverse map from $(lambda , varphi )$ to $(x,y)$.
– postmortes
Aug 22 at 6:34
Yes, @postmortes, Thanks for sharing this. I will do it.
– Ravi Dhorajiya
Aug 22 at 6:46
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Wikipedia gives the formula of the Gall stereographic projection as follows:
1) The projection is conventionally defined as
$$
x = fracR lambdasqrt2;
qquad
y = Rleft( 1 + fracsqrt22 right) tan fracvarphi2
cdotp
$$
2) The reverse projection is defined as
$$
lambda = fracx sqrt2R;
qquad
varphi = 2 arctan fracyRleft( 1 + fracsqrt22 right)
cdotp
$$
How does one go from 1 to 2? I'm having trouble finding the formula of the inverse.
algebra-precalculus projection
edited Aug 23 at 10:57
Jendrik Stelzner
7,57221037
asked Aug 22 at 5:18
Ravi Dhorajiya
1093
Wikipedia gives the formula of the Gall stereographic projection as follows:
1) The projection is conventionally defined as
$$
x = fracR lambdasqrt2;
qquad
y = Rleft( 1 + fracsqrt22 right) tan fracvarphi2
cdotp
$$
2) The reverse projection is defined as
$$
lambda = fracx sqrt2R;
qquad
varphi = 2 arctan fracyRleft( 1 + fracsqrt22 right)
cdotp
$$
How does one go from 1 to 2? I'm having trouble finding the formula of the inverse.
algebra-precalculus projection
edited Aug 23 at 10:57
Jendrik Stelzner
7,57221037
asked Aug 22 at 5:18
Ravi Dhorajiya
1093
edited Aug 23 at 10:57
Jendrik Stelzner
7,57221037
edited Aug 23 at 10:57
Jendrik Stelzner
7,57221037
edited Aug 23 at 10:57
Jendrik Stelzner
7,57221037
7,57221037
asked Aug 22 at 5:18
Ravi Dhorajiya
1093
asked Aug 22 at 5:18
Ravi Dhorajiya
1093
asked Aug 22 at 5:18
Ravi Dhorajiya
1093
1093
closed as off-topic by Jendrik Stelzner, amWhy, Strants, Adrian Keister, user91500 Aug 24 at 6:04
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jendrik Stelzner, amWhy, Strants, Adrian Keister, user91500
closed as off-topic by Jendrik Stelzner, amWhy, Strants, Adrian Keister, user91500 Aug 24 at 6:04
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jendrik Stelzner, amWhy, Strants, Adrian Keister, user91500
Solve for $lambda$ in the first equation, and solve for $varphi$ in the second equation. Essentially the first set of equations is a map from $(x,y)$ to $(lambda , varphi )$ and the second set is the inverse map from $(lambda , varphi )$ to $(x,y)$.
– postmortes
Aug 22 at 6:34
Yes, @postmortes, Thanks for sharing this. I will do it.
– Ravi Dhorajiya
Aug 22 at 6:46
add a comment |Â
Solve for $lambda$ in the first equation, and solve for $varphi$ in the second equation. Essentially the first set of equations is a map from $(x,y)$ to $(lambda , varphi )$ and the second set is the inverse map from $(lambda , varphi )$ to $(x,y)$.
– postmortes
Aug 22 at 6:34
Yes, @postmortes, Thanks for sharing this. I will do it.
– Ravi Dhorajiya
Aug 22 at 6:46
Solve for $lambda$ in the first equation, and solve for $varphi$ in the second equation. Essentially the first set of equations is a map from $(x,y)$ to $(lambda , varphi )$ and the second set is the inverse map from $(lambda , varphi )$ to $(x,y)$.
– postmortes
Aug 22 at 6:34
Solve for $lambda$ in the first equation, and solve for $varphi$ in the second equation. Essentially the first set of equations is a map from $(x,y)$ to $(lambda , varphi )$ and the second set is the inverse map from $(lambda , varphi )$ to $(x,y)$.
– postmortes
Aug 22 at 6:34
Yes, @postmortes, Thanks for sharing this. I will do it.
– Ravi Dhorajiya
Aug 22 at 6:46
Yes, @postmortes, Thanks for sharing this. I will do it.
– Ravi Dhorajiya
Aug 22 at 6:46
add a comment |Â
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Solve for $lambda$ in the first equation, and solve for $varphi$ in the second equation. Essentially the first set of equations is a map from $(x,y)$ to $(lambda , varphi )$ and the second set is the inverse map from $(lambda , varphi )$ to $(x,y)$.
– postmortes
Aug 22 at 6:34
Yes, @postmortes, Thanks for sharing this. I will do it.
– Ravi Dhorajiya
Aug 22 at 6:46