Calculation of Nutation and Rotation from Pitch and roll (yaw is fixed to 0)
$begingroup$
i am stuck attempting to convert two angles (pitch and roll) to represent tilt within a circle.
Take a plane $(Z text{(Vertical)}, X text{(Roll)}, Y text{(Pitch)})$
If i have $45$ degrees of roll, $Z$ is $45$ degrees from vertical. If I have $-10$ degrees of Pitch, $Z$ is $-10$ degrees from vertical.
Thats fine, but if I have BOTH $45$ degrees of roll, and $-10$ degrees of pitch, how do i find out how much i have?
The application is an inclinometer. I need to combine pitch and roll angles to show how many degrees from vertical the object is, and in which direction it is leaning. ($0$ degrees up, $90$ degrees right, $180$ degrees down, $270$ left.) I have been googling for hours and i am sure i have looked straight past the solution but my brain is cooked and i can't find the solution.
Im currently using $Z text{angle} = sqrt{text{roll^2} + text{pitch^2}}$
I think the angle i am looking for is called the angle of nutation??
trigonometry
edited Sep 28 '14 at 2:59
user179068
273
asked Sep 27 '14 at 23:55
WharbioWharbio
10112
$endgroup$
add a comment |
$begingroup$
i am stuck attempting to convert two angles (pitch and roll) to represent tilt within a circle.
Take a plane $(Z text{(Vertical)}, X text{(Roll)}, Y text{(Pitch)})$
If i have $45$ degrees of roll, $Z$ is $45$ degrees from vertical. If I have $-10$ degrees of Pitch, $Z$ is $-10$ degrees from vertical.
Thats fine, but if I have BOTH $45$ degrees of roll, and $-10$ degrees of pitch, how do i find out how much i have?
The application is an inclinometer. I need to combine pitch and roll angles to show how many degrees from vertical the object is, and in which direction it is leaning. ($0$ degrees up, $90$ degrees right, $180$ degrees down, $270$ left.) I have been googling for hours and i am sure i have looked straight past the solution but my brain is cooked and i can't find the solution.
Im currently using $Z text{angle} = sqrt{text{roll^2} + text{pitch^2}}$
I think the angle i am looking for is called the angle of nutation??
trigonometry
edited Sep 28 '14 at 2:59
user179068
273
asked Sep 27 '14 at 23:55
WharbioWharbio
10112
$endgroup$
add a comment |
$begingroup$
i am stuck attempting to convert two angles (pitch and roll) to represent tilt within a circle.
Take a plane $(Z text{(Vertical)}, X text{(Roll)}, Y text{(Pitch)})$
If i have $45$ degrees of roll, $Z$ is $45$ degrees from vertical. If I have $-10$ degrees of Pitch, $Z$ is $-10$ degrees from vertical.
Thats fine, but if I have BOTH $45$ degrees of roll, and $-10$ degrees of pitch, how do i find out how much i have?
The application is an inclinometer. I need to combine pitch and roll angles to show how many degrees from vertical the object is, and in which direction it is leaning. ($0$ degrees up, $90$ degrees right, $180$ degrees down, $270$ left.) I have been googling for hours and i am sure i have looked straight past the solution but my brain is cooked and i can't find the solution.
Im currently using $Z text{angle} = sqrt{text{roll^2} + text{pitch^2}}$
I think the angle i am looking for is called the angle of nutation??
trigonometry
edited Sep 28 '14 at 2:59
user179068
273
asked Sep 27 '14 at 23:55
WharbioWharbio
10112
$endgroup$
i am stuck attempting to convert two angles (pitch and roll) to represent tilt within a circle.
Take a plane $(Z text{(Vertical)}, X text{(Roll)}, Y text{(Pitch)})$
If i have $45$ degrees of roll, $Z$ is $45$ degrees from vertical. If I have $-10$ degrees of Pitch, $Z$ is $-10$ degrees from vertical.
Thats fine, but if I have BOTH $45$ degrees of roll, and $-10$ degrees of pitch, how do i find out how much i have?
The application is an inclinometer. I need to combine pitch and roll angles to show how many degrees from vertical the object is, and in which direction it is leaning. ($0$ degrees up, $90$ degrees right, $180$ degrees down, $270$ left.) I have been googling for hours and i am sure i have looked straight past the solution but my brain is cooked and i can't find the solution.
Im currently using $Z text{angle} = sqrt{text{roll^2} + text{pitch^2}}$
I think the angle i am looking for is called the angle of nutation??
trigonometry
trigonometry
edited Sep 28 '14 at 2:59
user179068
273
asked Sep 27 '14 at 23:55
WharbioWharbio
10112
edited Sep 28 '14 at 2:59
user179068
273
asked Sep 27 '14 at 23:55
WharbioWharbio
10112
edited Sep 28 '14 at 2:59
user179068
273
edited Sep 28 '14 at 2:59
user179068
273
edited Sep 28 '14 at 2:59
user179068
273
273
asked Sep 27 '14 at 23:55
WharbioWharbio
10112
asked Sep 27 '14 at 23:55
WharbioWharbio
10112
asked Sep 27 '14 at 23:55
WharbioWharbio
10112
10112
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Say we have pitch $ = alpha$ and roll $ = beta$. I'll assume that both are non-zero.
Start with a vertical line segment $OA$ of length $1$. "Pitch" it to $OB$ and then "roll" it to $OC$. Then we want to find angle $ gamma$ between $OC$ and the $z$-axis. Using standard trig:
begin{eqnarray*}
u &=& cos{alpha} \
v &=& ucos{beta} = cos{alpha}cos{beta} \
gamma &=& cos^{-1}v = cos^{-1}left(cos{alpha}cos{beta}right).
end{eqnarray*}
We'll measure the direction in which $OC$ is leaning as an anti-clockwise angle from the $x$-axis. That is, angle $theta$. We have,
begin{eqnarray*}
s &=& sin{alpha} \
t &=& usin{beta} = cos{alpha}sin{beta} \
theta &=& tan^{-1}left(dfrac{t}{s}right)
= tan^{-1}left(dfrac{sin{beta}}{tan{alpha}}right).
end{eqnarray*}
answered Sep 28 '14 at 8:09
Mick AMick A
8,7952825
$endgroup$
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Say we have pitch $ = alpha$ and roll $ = beta$. I'll assume that both are non-zero.
Start with a vertical line segment $OA$ of length $1$. "Pitch" it to $OB$ and then "roll" it to $OC$. Then we want to find angle $ gamma$ between $OC$ and the $z$-axis. Using standard trig:
begin{eqnarray*}
u &=& cos{alpha} \
v &=& ucos{beta} = cos{alpha}cos{beta} \
gamma &=& cos^{-1}v = cos^{-1}left(cos{alpha}cos{beta}right).
end{eqnarray*}
We'll measure the direction in which $OC$ is leaning as an anti-clockwise angle from the $x$-axis. That is, angle $theta$. We have,
begin{eqnarray*}
s &=& sin{alpha} \
t &=& usin{beta} = cos{alpha}sin{beta} \
theta &=& tan^{-1}left(dfrac{t}{s}right)
= tan^{-1}left(dfrac{sin{beta}}{tan{alpha}}right).
end{eqnarray*}
answered Sep 28 '14 at 8:09
Mick AMick A
8,7952825
$endgroup$
add a comment |
$begingroup$
Say we have pitch $ = alpha$ and roll $ = beta$. I'll assume that both are non-zero.
Start with a vertical line segment $OA$ of length $1$. "Pitch" it to $OB$ and then "roll" it to $OC$. Then we want to find angle $ gamma$ between $OC$ and the $z$-axis. Using standard trig:
begin{eqnarray*}
u &=& cos{alpha} \
v &=& ucos{beta} = cos{alpha}cos{beta} \
gamma &=& cos^{-1}v = cos^{-1}left(cos{alpha}cos{beta}right).
end{eqnarray*}
We'll measure the direction in which $OC$ is leaning as an anti-clockwise angle from the $x$-axis. That is, angle $theta$. We have,
begin{eqnarray*}
s &=& sin{alpha} \
t &=& usin{beta} = cos{alpha}sin{beta} \
theta &=& tan^{-1}left(dfrac{t}{s}right)
= tan^{-1}left(dfrac{sin{beta}}{tan{alpha}}right).
end{eqnarray*}
answered Sep 28 '14 at 8:09
Mick AMick A
8,7952825
$endgroup$
add a comment |
$begingroup$
Say we have pitch $ = alpha$ and roll $ = beta$. I'll assume that both are non-zero.
Start with a vertical line segment $OA$ of length $1$. "Pitch" it to $OB$ and then "roll" it to $OC$. Then we want to find angle $ gamma$ between $OC$ and the $z$-axis. Using standard trig:
begin{eqnarray*}
u &=& cos{alpha} \
v &=& ucos{beta} = cos{alpha}cos{beta} \
gamma &=& cos^{-1}v = cos^{-1}left(cos{alpha}cos{beta}right).
end{eqnarray*}
We'll measure the direction in which $OC$ is leaning as an anti-clockwise angle from the $x$-axis. That is, angle $theta$. We have,
begin{eqnarray*}
s &=& sin{alpha} \
t &=& usin{beta} = cos{alpha}sin{beta} \
theta &=& tan^{-1}left(dfrac{t}{s}right)
= tan^{-1}left(dfrac{sin{beta}}{tan{alpha}}right).
end{eqnarray*}
answered Sep 28 '14 at 8:09
Mick AMick A
8,7952825
$endgroup$
Say we have pitch $ = alpha$ and roll $ = beta$. I'll assume that both are non-zero.
Start with a vertical line segment $OA$ of length $1$. "Pitch" it to $OB$ and then "roll" it to $OC$. Then we want to find angle $ gamma$ between $OC$ and the $z$-axis. Using standard trig:
begin{eqnarray*}
u &=& cos{alpha} \
v &=& ucos{beta} = cos{alpha}cos{beta} \
gamma &=& cos^{-1}v = cos^{-1}left(cos{alpha}cos{beta}right).
end{eqnarray*}
We'll measure the direction in which $OC$ is leaning as an anti-clockwise angle from the $x$-axis. That is, angle $theta$. We have,
begin{eqnarray*}
s &=& sin{alpha} \
t &=& usin{beta} = cos{alpha}sin{beta} \
theta &=& tan^{-1}left(dfrac{t}{s}right)
= tan^{-1}left(dfrac{sin{beta}}{tan{alpha}}right).
end{eqnarray*}
answered Sep 28 '14 at 8:09
Mick AMick A
8,7952825
edited Sep 29 '14 at 15:25
answered Sep 28 '14 at 8:09
Mick AMick A
8,7952825
answered Sep 28 '14 at 8:09
Mick AMick A
8,7952825
answered Sep 28 '14 at 8:09
Mick AMick A
8,7952825
8,7952825
add a comment |
add a comment |
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