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Equation for arc with decaying radius
up vote
1
down vote
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Hoping for some insight into the equation and mechanics of an arc with a decaying radius.
Say at 0 degrees / 0 rad, the radius is 1. But by 90 degrees / π/2 rad the radius is 1/2. During that sweep from 0 --> 90 degrees, the radius is decaying at a constant rate.
As I was thinking about this, I was surprised -- though I shouldn't be -- that the arc would "balloon out" a bit on its way to 1/2r from 1r. And I dig this, and would like to understand the equation that would graph this.
Attached is a sketch of what I'm attempting to graph:
Thanks.
geometry polar-coordinates arc-length
asked yesterday
ghukill
1084
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ghukill is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
up vote
1
down vote
favorite
Hoping for some insight into the equation and mechanics of an arc with a decaying radius.
Say at 0 degrees / 0 rad, the radius is 1. But by 90 degrees / π/2 rad the radius is 1/2. During that sweep from 0 --> 90 degrees, the radius is decaying at a constant rate.
As I was thinking about this, I was surprised -- though I shouldn't be -- that the arc would "balloon out" a bit on its way to 1/2r from 1r. And I dig this, and would like to understand the equation that would graph this.
Attached is a sketch of what I'm attempting to graph:
Thanks.
geometry polar-coordinates arc-length
asked yesterday
ghukill
1084
New contributor
ghukill is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
Suggest write $pi/2$ radians-- $1/2pi$ might be confused with $1/(2pi).$
– coffeemath
yesterday
Thanks - been awhile since I worked with radians, updating...
– ghukill
yesterday
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Hoping for some insight into the equation and mechanics of an arc with a decaying radius.
Say at 0 degrees / 0 rad, the radius is 1. But by 90 degrees / π/2 rad the radius is 1/2. During that sweep from 0 --> 90 degrees, the radius is decaying at a constant rate.
As I was thinking about this, I was surprised -- though I shouldn't be -- that the arc would "balloon out" a bit on its way to 1/2r from 1r. And I dig this, and would like to understand the equation that would graph this.
Attached is a sketch of what I'm attempting to graph:
Thanks.
geometry polar-coordinates arc-length
asked yesterday
ghukill
1084
New contributor
ghukill is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Hoping for some insight into the equation and mechanics of an arc with a decaying radius.
Say at 0 degrees / 0 rad, the radius is 1. But by 90 degrees / π/2 rad the radius is 1/2. During that sweep from 0 --> 90 degrees, the radius is decaying at a constant rate.
As I was thinking about this, I was surprised -- though I shouldn't be -- that the arc would "balloon out" a bit on its way to 1/2r from 1r. And I dig this, and would like to understand the equation that would graph this.
Attached is a sketch of what I'm attempting to graph:
Thanks.
geometry polar-coordinates arc-length
geometry polar-coordinates arc-length
asked yesterday
ghukill
1084
New contributor
ghukill is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
ghukill
1084
New contributor
ghukill is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
asked yesterday
ghukill
1084
New contributor
ghukill is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
ghukill
1084
asked yesterday
ghukill
1084
1084
New contributor
ghukill is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
ghukill is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
ghukill is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
Suggest write $pi/2$ radians-- $1/2pi$ might be confused with $1/(2pi).$
– coffeemath
yesterday
Thanks - been awhile since I worked with radians, updating...
– ghukill
yesterday
add a comment |
1
Suggest write $pi/2$ radians-- $1/2pi$ might be confused with $1/(2pi).$
– coffeemath
yesterday
Thanks - been awhile since I worked with radians, updating...
– ghukill
yesterday
1
1
Suggest write $pi/2$ radians-- $1/2pi$ might be confused with $1/(2pi).$
– coffeemath
yesterday
Suggest write $pi/2$ radians-- $1/2pi$ might be confused with $1/(2pi).$
– coffeemath
yesterday
Thanks - been awhile since I worked with radians, updating...
– ghukill
yesterday
Thanks - been awhile since I worked with radians, updating...
– ghukill
yesterday
add a comment |
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
What you have constructed is a spiral of Archimedes.
There are a couple of minor differences between the usual construction of the spiral of Archimedes and yours.
One difference is that in the usual spiral equations, we have the radius increasing as the angle increases, whereas you have it decreasing.
If that were the only difference between the two constructions, you would get a mirror image.
But another difference is that the usual convention is that the angle increases in a counterclockwise direction, whereas you have gone clockwise.
So that cancels the mirror-image effect.
Other things that are different in the usual convention is that the "zero" angle is conventionally pointing in the horizontal direction to the right rather than vertically, and the radius is usually zero at the zero angle.
The combined result of all these differences compared to the "standard" construction is that the "standard" construction, which would have the polar equation
$r = frac1pitheta$ (with $theta$ in radians) ends up making a spiral just like yours but rotated $90$ degrees counterclockwise.
answered yesterday
David K
51.2k340113
You have made my day, thank you. I found value in the other answer as well, but the connection to a pre-existing thing / story / example was extremely helpful. And the equation. Thanks again.
– ghukill
yesterday
add a comment |
up vote
0
down vote
You should look into polar coordinates. If you have a point at angle $theta$ from the $Ox^to$ axis, and at distance $r$ from the center at the coordinate system, it has polar coordinates $(r, theta)$.
In your case, $r$ is a function of $theta$: $r(theta) = 1 - frac{1}{pi}theta$.
Note at angle 0, $r(0) = 1$, and at angle 90 degrees, $r(frac{pi}{2}) = 1 - frac{1}{2} = frac{1}{2}$.
In polar coordinates, it is easy to transfer from polar to Cartesian coordinates $(x, y)$:
$$x = rcos(theta) = (1 - frac{1}{pi}theta)cos(theta)$$
$$y = rsin(theta) = (1 - frac{1}{pi}theta)sin(theta)$$
The above is a good parametric description of the curve. If you want an equation that relates $x$ and $y$, you need to go a step further.
Lets limit the curve to the first quadrant ($0 leq theta leq frac{pi}{2}$) to avoid dealing with multiple $y$ points on the same $x$. In that case, we can easily express $sin(theta) = sqrt{1 - cos^2(theta)}$. Then
$$y = (1 - frac{1}{pi}theta)sqrt{1 - cos^2(theta)} =
(1 - frac{1}{pi}theta)
sqrt{1 - Bigg(frac{x}{1 - frac{1}{pi}theta}Bigg)^2}$$
answered yesterday
Todor Markov
3365
New contributor
Todor Markov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Thanks, this is very helpful! I selected the other answer as it will be immediately helpful to me, but I look forward to pouring over this a bit as well.
– ghukill
yesterday
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
What you have constructed is a spiral of Archimedes.
There are a couple of minor differences between the usual construction of the spiral of Archimedes and yours.
One difference is that in the usual spiral equations, we have the radius increasing as the angle increases, whereas you have it decreasing.
If that were the only difference between the two constructions, you would get a mirror image.
But another difference is that the usual convention is that the angle increases in a counterclockwise direction, whereas you have gone clockwise.
So that cancels the mirror-image effect.
Other things that are different in the usual convention is that the "zero" angle is conventionally pointing in the horizontal direction to the right rather than vertically, and the radius is usually zero at the zero angle.
The combined result of all these differences compared to the "standard" construction is that the "standard" construction, which would have the polar equation
$r = frac1pitheta$ (with $theta$ in radians) ends up making a spiral just like yours but rotated $90$ degrees counterclockwise.
answered yesterday
David K
51.2k340113
You have made my day, thank you. I found value in the other answer as well, but the connection to a pre-existing thing / story / example was extremely helpful. And the equation. Thanks again.
– ghukill
yesterday
add a comment |
up vote
1
down vote
accepted
What you have constructed is a spiral of Archimedes.
There are a couple of minor differences between the usual construction of the spiral of Archimedes and yours.
One difference is that in the usual spiral equations, we have the radius increasing as the angle increases, whereas you have it decreasing.
If that were the only difference between the two constructions, you would get a mirror image.
But another difference is that the usual convention is that the angle increases in a counterclockwise direction, whereas you have gone clockwise.
So that cancels the mirror-image effect.
Other things that are different in the usual convention is that the "zero" angle is conventionally pointing in the horizontal direction to the right rather than vertically, and the radius is usually zero at the zero angle.
The combined result of all these differences compared to the "standard" construction is that the "standard" construction, which would have the polar equation
$r = frac1pitheta$ (with $theta$ in radians) ends up making a spiral just like yours but rotated $90$ degrees counterclockwise.
answered yesterday
David K
51.2k340113
You have made my day, thank you. I found value in the other answer as well, but the connection to a pre-existing thing / story / example was extremely helpful. And the equation. Thanks again.
– ghukill
yesterday
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
What you have constructed is a spiral of Archimedes.
There are a couple of minor differences between the usual construction of the spiral of Archimedes and yours.
One difference is that in the usual spiral equations, we have the radius increasing as the angle increases, whereas you have it decreasing.
If that were the only difference between the two constructions, you would get a mirror image.
But another difference is that the usual convention is that the angle increases in a counterclockwise direction, whereas you have gone clockwise.
So that cancels the mirror-image effect.
Other things that are different in the usual convention is that the "zero" angle is conventionally pointing in the horizontal direction to the right rather than vertically, and the radius is usually zero at the zero angle.
The combined result of all these differences compared to the "standard" construction is that the "standard" construction, which would have the polar equation
$r = frac1pitheta$ (with $theta$ in radians) ends up making a spiral just like yours but rotated $90$ degrees counterclockwise.
answered yesterday
David K
51.2k340113
What you have constructed is a spiral of Archimedes.
There are a couple of minor differences between the usual construction of the spiral of Archimedes and yours.
One difference is that in the usual spiral equations, we have the radius increasing as the angle increases, whereas you have it decreasing.
If that were the only difference between the two constructions, you would get a mirror image.
But another difference is that the usual convention is that the angle increases in a counterclockwise direction, whereas you have gone clockwise.
So that cancels the mirror-image effect.
Other things that are different in the usual convention is that the "zero" angle is conventionally pointing in the horizontal direction to the right rather than vertically, and the radius is usually zero at the zero angle.
The combined result of all these differences compared to the "standard" construction is that the "standard" construction, which would have the polar equation
$r = frac1pitheta$ (with $theta$ in radians) ends up making a spiral just like yours but rotated $90$ degrees counterclockwise.
answered yesterday
David K
51.2k340113
answered yesterday
David K
51.2k340113
answered yesterday
David K
51.2k340113
answered yesterday
David K
51.2k340113
51.2k340113
You have made my day, thank you. I found value in the other answer as well, but the connection to a pre-existing thing / story / example was extremely helpful. And the equation. Thanks again.
– ghukill
yesterday
add a comment |
You have made my day, thank you. I found value in the other answer as well, but the connection to a pre-existing thing / story / example was extremely helpful. And the equation. Thanks again.
– ghukill
yesterday
You have made my day, thank you. I found value in the other answer as well, but the connection to a pre-existing thing / story / example was extremely helpful. And the equation. Thanks again.
– ghukill
yesterday
You have made my day, thank you. I found value in the other answer as well, but the connection to a pre-existing thing / story / example was extremely helpful. And the equation. Thanks again.
– ghukill
yesterday
add a comment |
up vote
0
down vote
You should look into polar coordinates. If you have a point at angle $theta$ from the $Ox^to$ axis, and at distance $r$ from the center at the coordinate system, it has polar coordinates $(r, theta)$.
In your case, $r$ is a function of $theta$: $r(theta) = 1 - frac{1}{pi}theta$.
Note at angle 0, $r(0) = 1$, and at angle 90 degrees, $r(frac{pi}{2}) = 1 - frac{1}{2} = frac{1}{2}$.
In polar coordinates, it is easy to transfer from polar to Cartesian coordinates $(x, y)$:
$$x = rcos(theta) = (1 - frac{1}{pi}theta)cos(theta)$$
$$y = rsin(theta) = (1 - frac{1}{pi}theta)sin(theta)$$
The above is a good parametric description of the curve. If you want an equation that relates $x$ and $y$, you need to go a step further.
Lets limit the curve to the first quadrant ($0 leq theta leq frac{pi}{2}$) to avoid dealing with multiple $y$ points on the same $x$. In that case, we can easily express $sin(theta) = sqrt{1 - cos^2(theta)}$. Then
$$y = (1 - frac{1}{pi}theta)sqrt{1 - cos^2(theta)} =
(1 - frac{1}{pi}theta)
sqrt{1 - Bigg(frac{x}{1 - frac{1}{pi}theta}Bigg)^2}$$
answered yesterday
Todor Markov
3365
New contributor
Todor Markov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Thanks, this is very helpful! I selected the other answer as it will be immediately helpful to me, but I look forward to pouring over this a bit as well.
– ghukill
yesterday
add a comment |
up vote
0
down vote
You should look into polar coordinates. If you have a point at angle $theta$ from the $Ox^to$ axis, and at distance $r$ from the center at the coordinate system, it has polar coordinates $(r, theta)$.
In your case, $r$ is a function of $theta$: $r(theta) = 1 - frac{1}{pi}theta$.
Note at angle 0, $r(0) = 1$, and at angle 90 degrees, $r(frac{pi}{2}) = 1 - frac{1}{2} = frac{1}{2}$.
In polar coordinates, it is easy to transfer from polar to Cartesian coordinates $(x, y)$:
$$x = rcos(theta) = (1 - frac{1}{pi}theta)cos(theta)$$
$$y = rsin(theta) = (1 - frac{1}{pi}theta)sin(theta)$$
The above is a good parametric description of the curve. If you want an equation that relates $x$ and $y$, you need to go a step further.
Lets limit the curve to the first quadrant ($0 leq theta leq frac{pi}{2}$) to avoid dealing with multiple $y$ points on the same $x$. In that case, we can easily express $sin(theta) = sqrt{1 - cos^2(theta)}$. Then
$$y = (1 - frac{1}{pi}theta)sqrt{1 - cos^2(theta)} =
(1 - frac{1}{pi}theta)
sqrt{1 - Bigg(frac{x}{1 - frac{1}{pi}theta}Bigg)^2}$$
answered yesterday
Todor Markov
3365
New contributor
Todor Markov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Thanks, this is very helpful! I selected the other answer as it will be immediately helpful to me, but I look forward to pouring over this a bit as well.
– ghukill
yesterday
add a comment |
up vote
0
down vote
up vote
0
down vote
You should look into polar coordinates. If you have a point at angle $theta$ from the $Ox^to$ axis, and at distance $r$ from the center at the coordinate system, it has polar coordinates $(r, theta)$.
In your case, $r$ is a function of $theta$: $r(theta) = 1 - frac{1}{pi}theta$.
Note at angle 0, $r(0) = 1$, and at angle 90 degrees, $r(frac{pi}{2}) = 1 - frac{1}{2} = frac{1}{2}$.
In polar coordinates, it is easy to transfer from polar to Cartesian coordinates $(x, y)$:
$$x = rcos(theta) = (1 - frac{1}{pi}theta)cos(theta)$$
$$y = rsin(theta) = (1 - frac{1}{pi}theta)sin(theta)$$
The above is a good parametric description of the curve. If you want an equation that relates $x$ and $y$, you need to go a step further.
Lets limit the curve to the first quadrant ($0 leq theta leq frac{pi}{2}$) to avoid dealing with multiple $y$ points on the same $x$. In that case, we can easily express $sin(theta) = sqrt{1 - cos^2(theta)}$. Then
$$y = (1 - frac{1}{pi}theta)sqrt{1 - cos^2(theta)} =
(1 - frac{1}{pi}theta)
sqrt{1 - Bigg(frac{x}{1 - frac{1}{pi}theta}Bigg)^2}$$
answered yesterday
Todor Markov
3365
New contributor
Todor Markov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
You should look into polar coordinates. If you have a point at angle $theta$ from the $Ox^to$ axis, and at distance $r$ from the center at the coordinate system, it has polar coordinates $(r, theta)$.
In your case, $r$ is a function of $theta$: $r(theta) = 1 - frac{1}{pi}theta$.
Note at angle 0, $r(0) = 1$, and at angle 90 degrees, $r(frac{pi}{2}) = 1 - frac{1}{2} = frac{1}{2}$.
In polar coordinates, it is easy to transfer from polar to Cartesian coordinates $(x, y)$:
$$x = rcos(theta) = (1 - frac{1}{pi}theta)cos(theta)$$
$$y = rsin(theta) = (1 - frac{1}{pi}theta)sin(theta)$$
The above is a good parametric description of the curve. If you want an equation that relates $x$ and $y$, you need to go a step further.
Lets limit the curve to the first quadrant ($0 leq theta leq frac{pi}{2}$) to avoid dealing with multiple $y$ points on the same $x$. In that case, we can easily express $sin(theta) = sqrt{1 - cos^2(theta)}$. Then
$$y = (1 - frac{1}{pi}theta)sqrt{1 - cos^2(theta)} =
(1 - frac{1}{pi}theta)
sqrt{1 - Bigg(frac{x}{1 - frac{1}{pi}theta}Bigg)^2}$$
answered yesterday
Todor Markov
3365
New contributor
Todor Markov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered yesterday
Todor Markov
3365
New contributor
Todor Markov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered yesterday
Todor Markov
3365
answered yesterday
Todor Markov
3365
3365
New contributor
Todor Markov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Todor Markov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Todor Markov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Thanks, this is very helpful! I selected the other answer as it will be immediately helpful to me, but I look forward to pouring over this a bit as well.
– ghukill
yesterday
add a comment |
Thanks, this is very helpful! I selected the other answer as it will be immediately helpful to me, but I look forward to pouring over this a bit as well.
– ghukill
yesterday
Thanks, this is very helpful! I selected the other answer as it will be immediately helpful to me, but I look forward to pouring over this a bit as well.
– ghukill
yesterday
Thanks, this is very helpful! I selected the other answer as it will be immediately helpful to me, but I look forward to pouring over this a bit as well.
– ghukill
yesterday
add a comment |
ghukill is a new contributor. Be nice, and check out our Code of Conduct.
ghukill is a new contributor. Be nice, and check out our Code of Conduct.
ghukill is a new contributor. Be nice, and check out our Code of Conduct.
ghukill is a new contributor. Be nice, and check out our Code of Conduct.
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1
Suggest write $pi/2$ radians-- $1/2pi$ might be confused with $1/(2pi).$
– coffeemath
yesterday
Thanks - been awhile since I worked with radians, updating...
– ghukill
yesterday