atan2 and Cardan angles - Problem with sign
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I'd like to get the three Cardan angles starting from the below matrix and by using the function atan2. The problem is that I get two conflicting results.
where $- pi/2 le beta le pi/2$.
If I consider the case $x>0$ in the above Wikipedia's webpage, I'll get (I call $R_{ij}$ the elements of the matrix):
$$ tan(gamma) = -frac{R_{31}}{R_{33}} $$
and I can use one of these two $operatorname {atan2}.$
Case 1: $x=R_{33}>0$
$$ gamma = operatorname {atan2}(-R_{31},R_{33}) = arctan(tan(gamma)) = gamma $$
if $R_{33}>0$.
Case 2: $x=-R_{33}>0$
$$ gamma = operatorname {atan2}(R_{31},-R_{33}) = arctan(tan(gamma)) = gamma $$
if $-R_{33}>0$.
How it is possible to get the same result the first time when $R_{33}>0$, the second time when $-R_{33}>0$?
Maybe I can pass from $tan$ to $operatorname {atan2}$ only when there is no sign before the ratio between the numerator and the denominator of the right term $tan(text {angle})=dfrac {text {numerator}}{text {denominator}}$?
Thank you so much in advance.
trigonometry education classical-mechanics
edited Nov 14 at 0:28
bjcolby15
1,0961916
asked Nov 13 at 17:11
Gennaro Arguzzi
321313
|
show 1 more comment
up vote
0
down vote
favorite
I'd like to get the three Cardan angles starting from the below matrix and by using the function atan2. The problem is that I get two conflicting results.
where $- pi/2 le beta le pi/2$.
If I consider the case $x>0$ in the above Wikipedia's webpage, I'll get (I call $R_{ij}$ the elements of the matrix):
$$ tan(gamma) = -frac{R_{31}}{R_{33}} $$
and I can use one of these two $operatorname {atan2}.$
Case 1: $x=R_{33}>0$
$$ gamma = operatorname {atan2}(-R_{31},R_{33}) = arctan(tan(gamma)) = gamma $$
if $R_{33}>0$.
Case 2: $x=-R_{33}>0$
$$ gamma = operatorname {atan2}(R_{31},-R_{33}) = arctan(tan(gamma)) = gamma $$
if $-R_{33}>0$.
How it is possible to get the same result the first time when $R_{33}>0$, the second time when $-R_{33}>0$?
Maybe I can pass from $tan$ to $operatorname {atan2}$ only when there is no sign before the ratio between the numerator and the denominator of the right term $tan(text {angle})=dfrac {text {numerator}}{text {denominator}}$?
Thank you so much in advance.
trigonometry education classical-mechanics
edited Nov 14 at 0:28
bjcolby15
1,0961916
asked Nov 13 at 17:11
Gennaro Arguzzi
321313
1
As $cos beta>0$, you always have that $(cos γ,sinγ)sim (R_{33},-R_{31})$ so that always $γ=arg(R_{33}-iR_{31})=atan2(-R_{31},R_{33})$.
– LutzL
Nov 13 at 17:28
Hello @LutzL, can you tell me why you wrote $(cos γ,sinγ)sim (R_{33},-R_{31})$ instead of $(cos γ,-sinγ)sim (R_{33},-R_{31})$ please?
– Gennaro Arguzzi
Nov 13 at 17:39
1
Because there is only one sign to distribute. As I said, $cosβ>0$ and $(cosγ,sinγ)=(R_{33},−R_{31})/cosβ$. This defines the polar decomposition uniquely.
– LutzL
Nov 13 at 18:21
@LutzL thank you for your extremely clear explanation.
– Gennaro Arguzzi
Nov 13 at 18:23
1
1. So what do you find confusing here exactly? Why are you expecting different results in two chains of equalities that both start with $gamma$? As it is, it is unclear what you are asking. 2. How is $gamma$ defined, and what is its domain? You know that $arctan(tangamma)=gamma$ does not hold in general, right?
– Federico Poloni
Nov 13 at 19:02
|
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'd like to get the three Cardan angles starting from the below matrix and by using the function atan2. The problem is that I get two conflicting results.
where $- pi/2 le beta le pi/2$.
If I consider the case $x>0$ in the above Wikipedia's webpage, I'll get (I call $R_{ij}$ the elements of the matrix):
$$ tan(gamma) = -frac{R_{31}}{R_{33}} $$
and I can use one of these two $operatorname {atan2}.$
Case 1: $x=R_{33}>0$
$$ gamma = operatorname {atan2}(-R_{31},R_{33}) = arctan(tan(gamma)) = gamma $$
if $R_{33}>0$.
Case 2: $x=-R_{33}>0$
$$ gamma = operatorname {atan2}(R_{31},-R_{33}) = arctan(tan(gamma)) = gamma $$
if $-R_{33}>0$.
How it is possible to get the same result the first time when $R_{33}>0$, the second time when $-R_{33}>0$?
Maybe I can pass from $tan$ to $operatorname {atan2}$ only when there is no sign before the ratio between the numerator and the denominator of the right term $tan(text {angle})=dfrac {text {numerator}}{text {denominator}}$?
Thank you so much in advance.
trigonometry education classical-mechanics
edited Nov 14 at 0:28
bjcolby15
1,0961916
asked Nov 13 at 17:11
Gennaro Arguzzi
321313
I'd like to get the three Cardan angles starting from the below matrix and by using the function atan2. The problem is that I get two conflicting results.
where $- pi/2 le beta le pi/2$.
If I consider the case $x>0$ in the above Wikipedia's webpage, I'll get (I call $R_{ij}$ the elements of the matrix):
$$ tan(gamma) = -frac{R_{31}}{R_{33}} $$
and I can use one of these two $operatorname {atan2}.$
Case 1: $x=R_{33}>0$
$$ gamma = operatorname {atan2}(-R_{31},R_{33}) = arctan(tan(gamma)) = gamma $$
if $R_{33}>0$.
Case 2: $x=-R_{33}>0$
$$ gamma = operatorname {atan2}(R_{31},-R_{33}) = arctan(tan(gamma)) = gamma $$
if $-R_{33}>0$.
How it is possible to get the same result the first time when $R_{33}>0$, the second time when $-R_{33}>0$?
Maybe I can pass from $tan$ to $operatorname {atan2}$ only when there is no sign before the ratio between the numerator and the denominator of the right term $tan(text {angle})=dfrac {text {numerator}}{text {denominator}}$?
Thank you so much in advance.
trigonometry education classical-mechanics
trigonometry education classical-mechanics
edited Nov 14 at 0:28
bjcolby15
1,0961916
asked Nov 13 at 17:11
Gennaro Arguzzi
321313
edited Nov 14 at 0:28
bjcolby15
1,0961916
asked Nov 13 at 17:11
Gennaro Arguzzi
321313
edited Nov 14 at 0:28
bjcolby15
1,0961916
edited Nov 14 at 0:28
bjcolby15
1,0961916
edited Nov 14 at 0:28
bjcolby15
1,0961916
1,0961916
asked Nov 13 at 17:11
Gennaro Arguzzi
321313
asked Nov 13 at 17:11
Gennaro Arguzzi
321313
asked Nov 13 at 17:11
Gennaro Arguzzi
321313
321313
1
As $cos beta>0$, you always have that $(cos γ,sinγ)sim (R_{33},-R_{31})$ so that always $γ=arg(R_{33}-iR_{31})=atan2(-R_{31},R_{33})$.
– LutzL
Nov 13 at 17:28
Hello @LutzL, can you tell me why you wrote $(cos γ,sinγ)sim (R_{33},-R_{31})$ instead of $(cos γ,-sinγ)sim (R_{33},-R_{31})$ please?
– Gennaro Arguzzi
Nov 13 at 17:39
1
Because there is only one sign to distribute. As I said, $cosβ>0$ and $(cosγ,sinγ)=(R_{33},−R_{31})/cosβ$. This defines the polar decomposition uniquely.
– LutzL
Nov 13 at 18:21
@LutzL thank you for your extremely clear explanation.
– Gennaro Arguzzi
Nov 13 at 18:23
1
1. So what do you find confusing here exactly? Why are you expecting different results in two chains of equalities that both start with $gamma$? As it is, it is unclear what you are asking. 2. How is $gamma$ defined, and what is its domain? You know that $arctan(tangamma)=gamma$ does not hold in general, right?
– Federico Poloni
Nov 13 at 19:02
|
show 1 more comment
1
As $cos beta>0$, you always have that $(cos γ,sinγ)sim (R_{33},-R_{31})$ so that always $γ=arg(R_{33}-iR_{31})=atan2(-R_{31},R_{33})$.
– LutzL
Nov 13 at 17:28
Hello @LutzL, can you tell me why you wrote $(cos γ,sinγ)sim (R_{33},-R_{31})$ instead of $(cos γ,-sinγ)sim (R_{33},-R_{31})$ please?
– Gennaro Arguzzi
Nov 13 at 17:39
1
Because there is only one sign to distribute. As I said, $cosβ>0$ and $(cosγ,sinγ)=(R_{33},−R_{31})/cosβ$. This defines the polar decomposition uniquely.
– LutzL
Nov 13 at 18:21
@LutzL thank you for your extremely clear explanation.
– Gennaro Arguzzi
Nov 13 at 18:23
1
1. So what do you find confusing here exactly? Why are you expecting different results in two chains of equalities that both start with $gamma$? As it is, it is unclear what you are asking. 2. How is $gamma$ defined, and what is its domain? You know that $arctan(tangamma)=gamma$ does not hold in general, right?
– Federico Poloni
Nov 13 at 19:02
1
1
As $cos beta>0$, you always have that $(cos γ,sinγ)sim (R_{33},-R_{31})$ so that always $γ=arg(R_{33}-iR_{31})=atan2(-R_{31},R_{33})$.
– LutzL
Nov 13 at 17:28
As $cos beta>0$, you always have that $(cos γ,sinγ)sim (R_{33},-R_{31})$ so that always $γ=arg(R_{33}-iR_{31})=atan2(-R_{31},R_{33})$.
– LutzL
Nov 13 at 17:28
Hello @LutzL, can you tell me why you wrote $(cos γ,sinγ)sim (R_{33},-R_{31})$ instead of $(cos γ,-sinγ)sim (R_{33},-R_{31})$ please?
– Gennaro Arguzzi
Nov 13 at 17:39
Hello @LutzL, can you tell me why you wrote $(cos γ,sinγ)sim (R_{33},-R_{31})$ instead of $(cos γ,-sinγ)sim (R_{33},-R_{31})$ please?
– Gennaro Arguzzi
Nov 13 at 17:39
1
1
Because there is only one sign to distribute. As I said, $cosβ>0$ and $(cosγ,sinγ)=(R_{33},−R_{31})/cosβ$. This defines the polar decomposition uniquely.
– LutzL
Nov 13 at 18:21
Because there is only one sign to distribute. As I said, $cosβ>0$ and $(cosγ,sinγ)=(R_{33},−R_{31})/cosβ$. This defines the polar decomposition uniquely.
– LutzL
Nov 13 at 18:21
@LutzL thank you for your extremely clear explanation.
– Gennaro Arguzzi
Nov 13 at 18:23
@LutzL thank you for your extremely clear explanation.
– Gennaro Arguzzi
Nov 13 at 18:23
1
1
1. So what do you find confusing here exactly? Why are you expecting different results in two chains of equalities that both start with $gamma$? As it is, it is unclear what you are asking. 2. How is $gamma$ defined, and what is its domain? You know that $arctan(tangamma)=gamma$ does not hold in general, right?
– Federico Poloni
Nov 13 at 19:02
1. So what do you find confusing here exactly? Why are you expecting different results in two chains of equalities that both start with $gamma$? As it is, it is unclear what you are asking. 2. How is $gamma$ defined, and what is its domain? You know that $arctan(tangamma)=gamma$ does not hold in general, right?
– Federico Poloni
Nov 13 at 19:02
|
show 1 more comment
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
The ratio $frac{sin gamma}{cosgamma}$ does not specify an angle $-pi < gamma leq pi$ uniquely, but only up to $pi$; to specify it, you need two coordinates $x,y$, not both zero, through which the variable side of the angle passes: that's the whole idea behind atan2 and you should make sure you are familiar with it before tackling these computations.
In your case, you have restricted to $-fracpi2 < gamma < fracpi2$, which means that your angle is in the right half-plane, so these coordinates have $x>0$. Focus on the first equal sign in each of those two alternative computations: only one of them is correct, the one with $cosgamma > 0$. To know which one it is, you will have to know the sign of $cos beta$ as well.
answered Nov 14 at 7:32
Federico Poloni
2,4631327
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The ratio $frac{sin gamma}{cosgamma}$ does not specify an angle $-pi < gamma leq pi$ uniquely, but only up to $pi$; to specify it, you need two coordinates $x,y$, not both zero, through which the variable side of the angle passes: that's the whole idea behind atan2 and you should make sure you are familiar with it before tackling these computations.
In your case, you have restricted to $-fracpi2 < gamma < fracpi2$, which means that your angle is in the right half-plane, so these coordinates have $x>0$. Focus on the first equal sign in each of those two alternative computations: only one of them is correct, the one with $cosgamma > 0$. To know which one it is, you will have to know the sign of $cos beta$ as well.
answered Nov 14 at 7:32
Federico Poloni
2,4631327
add a comment |
up vote
2
down vote
accepted
The ratio $frac{sin gamma}{cosgamma}$ does not specify an angle $-pi < gamma leq pi$ uniquely, but only up to $pi$; to specify it, you need two coordinates $x,y$, not both zero, through which the variable side of the angle passes: that's the whole idea behind atan2 and you should make sure you are familiar with it before tackling these computations.
In your case, you have restricted to $-fracpi2 < gamma < fracpi2$, which means that your angle is in the right half-plane, so these coordinates have $x>0$. Focus on the first equal sign in each of those two alternative computations: only one of them is correct, the one with $cosgamma > 0$. To know which one it is, you will have to know the sign of $cos beta$ as well.
answered Nov 14 at 7:32
Federico Poloni
2,4631327
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The ratio $frac{sin gamma}{cosgamma}$ does not specify an angle $-pi < gamma leq pi$ uniquely, but only up to $pi$; to specify it, you need two coordinates $x,y$, not both zero, through which the variable side of the angle passes: that's the whole idea behind atan2 and you should make sure you are familiar with it before tackling these computations.
In your case, you have restricted to $-fracpi2 < gamma < fracpi2$, which means that your angle is in the right half-plane, so these coordinates have $x>0$. Focus on the first equal sign in each of those two alternative computations: only one of them is correct, the one with $cosgamma > 0$. To know which one it is, you will have to know the sign of $cos beta$ as well.
answered Nov 14 at 7:32
Federico Poloni
2,4631327
The ratio $frac{sin gamma}{cosgamma}$ does not specify an angle $-pi < gamma leq pi$ uniquely, but only up to $pi$; to specify it, you need two coordinates $x,y$, not both zero, through which the variable side of the angle passes: that's the whole idea behind atan2 and you should make sure you are familiar with it before tackling these computations.
In your case, you have restricted to $-fracpi2 < gamma < fracpi2$, which means that your angle is in the right half-plane, so these coordinates have $x>0$. Focus on the first equal sign in each of those two alternative computations: only one of them is correct, the one with $cosgamma > 0$. To know which one it is, you will have to know the sign of $cos beta$ as well.
answered Nov 14 at 7:32
Federico Poloni
2,4631327
answered Nov 14 at 7:32
Federico Poloni
2,4631327
answered Nov 14 at 7:32
Federico Poloni
2,4631327
answered Nov 14 at 7:32
Federico Poloni
2,4631327
2,4631327
add a comment |
add a comment |
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As $cos beta>0$, you always have that $(cos γ,sinγ)sim (R_{33},-R_{31})$ so that always $γ=arg(R_{33}-iR_{31})=atan2(-R_{31},R_{33})$.
– LutzL
Nov 13 at 17:28
Hello @LutzL, can you tell me why you wrote $(cos γ,sinγ)sim (R_{33},-R_{31})$ instead of $(cos γ,-sinγ)sim (R_{33},-R_{31})$ please?
– Gennaro Arguzzi
Nov 13 at 17:39
1
Because there is only one sign to distribute. As I said, $cosβ>0$ and $(cosγ,sinγ)=(R_{33},−R_{31})/cosβ$. This defines the polar decomposition uniquely.
– LutzL
Nov 13 at 18:21
@LutzL thank you for your extremely clear explanation.
– Gennaro Arguzzi
Nov 13 at 18:23
1
1. So what do you find confusing here exactly? Why are you expecting different results in two chains of equalities that both start with $gamma$? As it is, it is unclear what you are asking. 2. How is $gamma$ defined, and what is its domain? You know that $arctan(tangamma)=gamma$ does not hold in general, right?
– Federico Poloni
Nov 13 at 19:02