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Radial (not Polar!) planimeter theory…
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I am investigating the theory of a radial (not a polar) planimeter - when tracing a closed contour with the "pole" of the device outside of the contour (normally, the "pole" would be inside the contour).
This has led me to this horrible expression below, whilst trying to evaluate a line integral - and I have been unable to progress beyond it...
$$
int_0^2pi fracrldotp mathrmcosleft(tright)ldotp left(a+rldotp mathrmcosleft(tright)right)sqrtleft(a+rldotp mathrmcosleft(tright)right)^2 +left(b+rldotp mathrmsinleft(tright)right)^2 +fracrldotp mathrmsinleft(tright)ldotp left(b+rldotp mathrmsinleft(tright)right)sqrtleft(a+rldotp mathrmcosleft(tright)right)^2 +left(b+rldotp mathrmsinleft(tright)right)^2 mathrmdt
$$
Does anyone have any suggestions about how I might proceed with the solution to this? (Wolfram times out and so does the rudimentary symbolic processor which I have access to).
I therefore reverted to a numerical approach and found that the results appear to be physically plausible (also nicely confirmed by measurements using an actual instrument), but I want to find the analytical solution as this will obviously provide much more insight into the underlying process. Of course, I realise that there may not be an analytical solution...
Thanks for looking.
calculus
asked Jul 24 at 11:09
bobg1756
212
add a comment |Â
up vote
4
down vote
favorite
I am investigating the theory of a radial (not a polar) planimeter - when tracing a closed contour with the "pole" of the device outside of the contour (normally, the "pole" would be inside the contour).
This has led me to this horrible expression below, whilst trying to evaluate a line integral - and I have been unable to progress beyond it...
$$
int_0^2pi fracrldotp mathrmcosleft(tright)ldotp left(a+rldotp mathrmcosleft(tright)right)sqrtleft(a+rldotp mathrmcosleft(tright)right)^2 +left(b+rldotp mathrmsinleft(tright)right)^2 +fracrldotp mathrmsinleft(tright)ldotp left(b+rldotp mathrmsinleft(tright)right)sqrtleft(a+rldotp mathrmcosleft(tright)right)^2 +left(b+rldotp mathrmsinleft(tright)right)^2 mathrmdt
$$
Does anyone have any suggestions about how I might proceed with the solution to this? (Wolfram times out and so does the rudimentary symbolic processor which I have access to).
I therefore reverted to a numerical approach and found that the results appear to be physically plausible (also nicely confirmed by measurements using an actual instrument), but I want to find the analytical solution as this will obviously provide much more insight into the underlying process. Of course, I realise that there may not be an analytical solution...
Thanks for looking.
calculus
asked Jul 24 at 11:09
bobg1756
212
I forgot to mention that the contour being considered is a circle of radius r, with a centre at (a,b). The "pole" of the device is assumed to be at (0,0).
– bobg1756
Jul 24 at 12:42
This doesn't look like the right integral; what's the original expression you're using to find the area?
– Chappers
Jul 24 at 14:03
I'm not finding the area - a radial planimeter doesn't do that. When the pole is inside the contour it will measure the mean value of the contour (or a function of it) relative to the 0,0 point. The question I am interested in is - what does it measure when the pole is outside the shape? Of course, it may not be anything meaningful...!
– bobg1756
Jul 24 at 14:45
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I am investigating the theory of a radial (not a polar) planimeter - when tracing a closed contour with the "pole" of the device outside of the contour (normally, the "pole" would be inside the contour).
This has led me to this horrible expression below, whilst trying to evaluate a line integral - and I have been unable to progress beyond it...
$$
int_0^2pi fracrldotp mathrmcosleft(tright)ldotp left(a+rldotp mathrmcosleft(tright)right)sqrtleft(a+rldotp mathrmcosleft(tright)right)^2 +left(b+rldotp mathrmsinleft(tright)right)^2 +fracrldotp mathrmsinleft(tright)ldotp left(b+rldotp mathrmsinleft(tright)right)sqrtleft(a+rldotp mathrmcosleft(tright)right)^2 +left(b+rldotp mathrmsinleft(tright)right)^2 mathrmdt
$$
Does anyone have any suggestions about how I might proceed with the solution to this? (Wolfram times out and so does the rudimentary symbolic processor which I have access to).
I therefore reverted to a numerical approach and found that the results appear to be physically plausible (also nicely confirmed by measurements using an actual instrument), but I want to find the analytical solution as this will obviously provide much more insight into the underlying process. Of course, I realise that there may not be an analytical solution...
Thanks for looking.
calculus
asked Jul 24 at 11:09
bobg1756
212
I am investigating the theory of a radial (not a polar) planimeter - when tracing a closed contour with the "pole" of the device outside of the contour (normally, the "pole" would be inside the contour).
This has led me to this horrible expression below, whilst trying to evaluate a line integral - and I have been unable to progress beyond it...
$$
int_0^2pi fracrldotp mathrmcosleft(tright)ldotp left(a+rldotp mathrmcosleft(tright)right)sqrtleft(a+rldotp mathrmcosleft(tright)right)^2 +left(b+rldotp mathrmsinleft(tright)right)^2 +fracrldotp mathrmsinleft(tright)ldotp left(b+rldotp mathrmsinleft(tright)right)sqrtleft(a+rldotp mathrmcosleft(tright)right)^2 +left(b+rldotp mathrmsinleft(tright)right)^2 mathrmdt
$$
Does anyone have any suggestions about how I might proceed with the solution to this? (Wolfram times out and so does the rudimentary symbolic processor which I have access to).
I therefore reverted to a numerical approach and found that the results appear to be physically plausible (also nicely confirmed by measurements using an actual instrument), but I want to find the analytical solution as this will obviously provide much more insight into the underlying process. Of course, I realise that there may not be an analytical solution...
Thanks for looking.
calculus
asked Jul 24 at 11:09
bobg1756
212
asked Jul 24 at 11:09
bobg1756
212
asked Jul 24 at 11:09
bobg1756
212
asked Jul 24 at 11:09
bobg1756
212
212
I forgot to mention that the contour being considered is a circle of radius r, with a centre at (a,b). The "pole" of the device is assumed to be at (0,0).
– bobg1756
Jul 24 at 12:42
This doesn't look like the right integral; what's the original expression you're using to find the area?
– Chappers
Jul 24 at 14:03
I'm not finding the area - a radial planimeter doesn't do that. When the pole is inside the contour it will measure the mean value of the contour (or a function of it) relative to the 0,0 point. The question I am interested in is - what does it measure when the pole is outside the shape? Of course, it may not be anything meaningful...!
– bobg1756
Jul 24 at 14:45
add a comment |Â
I forgot to mention that the contour being considered is a circle of radius r, with a centre at (a,b). The "pole" of the device is assumed to be at (0,0).
– bobg1756
Jul 24 at 12:42
This doesn't look like the right integral; what's the original expression you're using to find the area?
– Chappers
Jul 24 at 14:03
I'm not finding the area - a radial planimeter doesn't do that. When the pole is inside the contour it will measure the mean value of the contour (or a function of it) relative to the 0,0 point. The question I am interested in is - what does it measure when the pole is outside the shape? Of course, it may not be anything meaningful...!
– bobg1756
Jul 24 at 14:45
I forgot to mention that the contour being considered is a circle of radius r, with a centre at (a,b). The "pole" of the device is assumed to be at (0,0).
– bobg1756
Jul 24 at 12:42
I forgot to mention that the contour being considered is a circle of radius r, with a centre at (a,b). The "pole" of the device is assumed to be at (0,0).
– bobg1756
Jul 24 at 12:42
This doesn't look like the right integral; what's the original expression you're using to find the area?
– Chappers
Jul 24 at 14:03
This doesn't look like the right integral; what's the original expression you're using to find the area?
– Chappers
Jul 24 at 14:03
I'm not finding the area - a radial planimeter doesn't do that. When the pole is inside the contour it will measure the mean value of the contour (or a function of it) relative to the 0,0 point. The question I am interested in is - what does it measure when the pole is outside the shape? Of course, it may not be anything meaningful...!
– bobg1756
Jul 24 at 14:45
I'm not finding the area - a radial planimeter doesn't do that. When the pole is inside the contour it will measure the mean value of the contour (or a function of it) relative to the 0,0 point. The question I am interested in is - what does it measure when the pole is outside the shape? Of course, it may not be anything meaningful...!
– bobg1756
Jul 24 at 14:45
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
Assuming that this is indeed the correct integral you want to evaluate, you can do the following:
Since your "pole" lies outside the contour we can work from
$$
sqrta^2 + b^2 = c > r > 0
$$
where $c$ is the distance from the origin to the point $(a,b)$ and $r$ the radius of the circles of the contour. Expanding the numerators and denominators we get
$$
int_0^2 pi frac
r^2 + a r cos t + b r sin t
sqrtr^2 + a^2 + b^2 + 2 a r cos t + 2 b r sin t
d t,
$$
where I already used that $sin^2 t + cos^2 t = 1$. Since we also can use that
$a cos t + b sin t = sqrta^2+b^2 sin(t + phi) = csin(t + phi) $ for $phi = arctan a/b$, the integral becomes simpler
$$
int_0^2 pi fracr^2 + r c sin (t+phi)sqrtr^2 + c^2 + 2 r c sin (t+phi) d t.
$$
While the integration is over a whole period, the angular shift $phi$ can be discarded, and we get
$$
int_0^2 pi fracr^2 + r c sin tsqrtr^2 + c^2 + 2 r c sin t d t
$$
An alternative way to look at this result is to rotate the frame of reference such that the point $(a,b)$ lies on the $y$-axis at a distance $c$ from the origin.
There are only two integrals to be evaluated:
$$
int_0^2 pi frac1sqrt1 + q sin t d t
$$
$$
int_0^2 pi fracsin tsqrt1 + q sin t d t
$$
with $0 < q = 2 r c/sqrtr^2+c^2<1$. These give rise to the complete elliptic integral functions $E(..)$ and $K(..)$, and will result in
$$
(c-r) left[ Eleft(frac-4 c r(c-r)^2right) - Kleft(frac4 c r(c+r)^2right) right] + (c+r) left[ Eleft(frac4 c r(c+r)^2right) - Kleft(frac-4 c r(c-r)^2right) right]
$$
answered Jul 24 at 14:25
Ronald Blaak
1,52927
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Assuming that this is indeed the correct integral you want to evaluate, you can do the following:
Since your "pole" lies outside the contour we can work from
$$
sqrta^2 + b^2 = c > r > 0
$$
where $c$ is the distance from the origin to the point $(a,b)$ and $r$ the radius of the circles of the contour. Expanding the numerators and denominators we get
$$
int_0^2 pi frac
r^2 + a r cos t + b r sin t
sqrtr^2 + a^2 + b^2 + 2 a r cos t + 2 b r sin t
d t,
$$
where I already used that $sin^2 t + cos^2 t = 1$. Since we also can use that
$a cos t + b sin t = sqrta^2+b^2 sin(t + phi) = csin(t + phi) $ for $phi = arctan a/b$, the integral becomes simpler
$$
int_0^2 pi fracr^2 + r c sin (t+phi)sqrtr^2 + c^2 + 2 r c sin (t+phi) d t.
$$
While the integration is over a whole period, the angular shift $phi$ can be discarded, and we get
$$
int_0^2 pi fracr^2 + r c sin tsqrtr^2 + c^2 + 2 r c sin t d t
$$
An alternative way to look at this result is to rotate the frame of reference such that the point $(a,b)$ lies on the $y$-axis at a distance $c$ from the origin.
There are only two integrals to be evaluated:
$$
int_0^2 pi frac1sqrt1 + q sin t d t
$$
$$
int_0^2 pi fracsin tsqrt1 + q sin t d t
$$
with $0 < q = 2 r c/sqrtr^2+c^2<1$. These give rise to the complete elliptic integral functions $E(..)$ and $K(..)$, and will result in
$$
(c-r) left[ Eleft(frac-4 c r(c-r)^2right) - Kleft(frac4 c r(c+r)^2right) right] + (c+r) left[ Eleft(frac4 c r(c+r)^2right) - Kleft(frac-4 c r(c-r)^2right) right]
$$
answered Jul 24 at 14:25
Ronald Blaak
1,52927
add a comment |Â
up vote
0
down vote
Assuming that this is indeed the correct integral you want to evaluate, you can do the following:
Since your "pole" lies outside the contour we can work from
$$
sqrta^2 + b^2 = c > r > 0
$$
where $c$ is the distance from the origin to the point $(a,b)$ and $r$ the radius of the circles of the contour. Expanding the numerators and denominators we get
$$
int_0^2 pi frac
r^2 + a r cos t + b r sin t
sqrtr^2 + a^2 + b^2 + 2 a r cos t + 2 b r sin t
d t,
$$
where I already used that $sin^2 t + cos^2 t = 1$. Since we also can use that
$a cos t + b sin t = sqrta^2+b^2 sin(t + phi) = csin(t + phi) $ for $phi = arctan a/b$, the integral becomes simpler
$$
int_0^2 pi fracr^2 + r c sin (t+phi)sqrtr^2 + c^2 + 2 r c sin (t+phi) d t.
$$
While the integration is over a whole period, the angular shift $phi$ can be discarded, and we get
$$
int_0^2 pi fracr^2 + r c sin tsqrtr^2 + c^2 + 2 r c sin t d t
$$
An alternative way to look at this result is to rotate the frame of reference such that the point $(a,b)$ lies on the $y$-axis at a distance $c$ from the origin.
There are only two integrals to be evaluated:
$$
int_0^2 pi frac1sqrt1 + q sin t d t
$$
$$
int_0^2 pi fracsin tsqrt1 + q sin t d t
$$
with $0 < q = 2 r c/sqrtr^2+c^2<1$. These give rise to the complete elliptic integral functions $E(..)$ and $K(..)$, and will result in
$$
(c-r) left[ Eleft(frac-4 c r(c-r)^2right) - Kleft(frac4 c r(c+r)^2right) right] + (c+r) left[ Eleft(frac4 c r(c+r)^2right) - Kleft(frac-4 c r(c-r)^2right) right]
$$
answered Jul 24 at 14:25
Ronald Blaak
1,52927
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Assuming that this is indeed the correct integral you want to evaluate, you can do the following:
Since your "pole" lies outside the contour we can work from
$$
sqrta^2 + b^2 = c > r > 0
$$
where $c$ is the distance from the origin to the point $(a,b)$ and $r$ the radius of the circles of the contour. Expanding the numerators and denominators we get
$$
int_0^2 pi frac
r^2 + a r cos t + b r sin t
sqrtr^2 + a^2 + b^2 + 2 a r cos t + 2 b r sin t
d t,
$$
where I already used that $sin^2 t + cos^2 t = 1$. Since we also can use that
$a cos t + b sin t = sqrta^2+b^2 sin(t + phi) = csin(t + phi) $ for $phi = arctan a/b$, the integral becomes simpler
$$
int_0^2 pi fracr^2 + r c sin (t+phi)sqrtr^2 + c^2 + 2 r c sin (t+phi) d t.
$$
While the integration is over a whole period, the angular shift $phi$ can be discarded, and we get
$$
int_0^2 pi fracr^2 + r c sin tsqrtr^2 + c^2 + 2 r c sin t d t
$$
An alternative way to look at this result is to rotate the frame of reference such that the point $(a,b)$ lies on the $y$-axis at a distance $c$ from the origin.
There are only two integrals to be evaluated:
$$
int_0^2 pi frac1sqrt1 + q sin t d t
$$
$$
int_0^2 pi fracsin tsqrt1 + q sin t d t
$$
with $0 < q = 2 r c/sqrtr^2+c^2<1$. These give rise to the complete elliptic integral functions $E(..)$ and $K(..)$, and will result in
$$
(c-r) left[ Eleft(frac-4 c r(c-r)^2right) - Kleft(frac4 c r(c+r)^2right) right] + (c+r) left[ Eleft(frac4 c r(c+r)^2right) - Kleft(frac-4 c r(c-r)^2right) right]
$$
answered Jul 24 at 14:25
Ronald Blaak
1,52927
Assuming that this is indeed the correct integral you want to evaluate, you can do the following:
Since your "pole" lies outside the contour we can work from
$$
sqrta^2 + b^2 = c > r > 0
$$
where $c$ is the distance from the origin to the point $(a,b)$ and $r$ the radius of the circles of the contour. Expanding the numerators and denominators we get
$$
int_0^2 pi frac
r^2 + a r cos t + b r sin t
sqrtr^2 + a^2 + b^2 + 2 a r cos t + 2 b r sin t
d t,
$$
where I already used that $sin^2 t + cos^2 t = 1$. Since we also can use that
$a cos t + b sin t = sqrta^2+b^2 sin(t + phi) = csin(t + phi) $ for $phi = arctan a/b$, the integral becomes simpler
$$
int_0^2 pi fracr^2 + r c sin (t+phi)sqrtr^2 + c^2 + 2 r c sin (t+phi) d t.
$$
While the integration is over a whole period, the angular shift $phi$ can be discarded, and we get
$$
int_0^2 pi fracr^2 + r c sin tsqrtr^2 + c^2 + 2 r c sin t d t
$$
An alternative way to look at this result is to rotate the frame of reference such that the point $(a,b)$ lies on the $y$-axis at a distance $c$ from the origin.
There are only two integrals to be evaluated:
$$
int_0^2 pi frac1sqrt1 + q sin t d t
$$
$$
int_0^2 pi fracsin tsqrt1 + q sin t d t
$$
with $0 < q = 2 r c/sqrtr^2+c^2<1$. These give rise to the complete elliptic integral functions $E(..)$ and $K(..)$, and will result in
$$
(c-r) left[ Eleft(frac-4 c r(c-r)^2right) - Kleft(frac4 c r(c+r)^2right) right] + (c+r) left[ Eleft(frac4 c r(c+r)^2right) - Kleft(frac-4 c r(c-r)^2right) right]
$$
answered Jul 24 at 14:25
Ronald Blaak
1,52927
answered Jul 24 at 14:25
Ronald Blaak
1,52927
answered Jul 24 at 14:25
Ronald Blaak
1,52927
answered Jul 24 at 14:25
Ronald Blaak
1,52927
1,52927
add a comment |Â
add a comment |Â
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I forgot to mention that the contour being considered is a circle of radius r, with a centre at (a,b). The "pole" of the device is assumed to be at (0,0).
– bobg1756
Jul 24 at 12:42
This doesn't look like the right integral; what's the original expression you're using to find the area?
– Chappers
Jul 24 at 14:03
I'm not finding the area - a radial planimeter doesn't do that. When the pole is inside the contour it will measure the mean value of the contour (or a function of it) relative to the 0,0 point. The question I am interested in is - what does it measure when the pole is outside the shape? Of course, it may not be anything meaningful...!
– bobg1756
Jul 24 at 14:45