Mixing
Relating Turning Angle To Arc
$begingroup$
I am trying to determine how to relate the turning angle of a vehicle to an arc. For the purposes of this vehicle, I am not taking into account Ackerman Steering, so for the intent of this question, the vehicle is essentially just one-wheeled (like a bike)
I need to know how to calculate the angle, in relation to the body of the vehicle, that the front axle needs to turn in order for the front of the car to go from point a to point b in a scenario such as
this diagram.
This is all assuming I know:
- The arc length
- The chord length
- The radius of the circle
How do I determine this steering angle?
circle kinematics
asked Jan 9 at 19:38
VectorVector
33
$endgroup$
|
show 6 more comments
$begingroup$
I am trying to determine how to relate the turning angle of a vehicle to an arc. For the purposes of this vehicle, I am not taking into account Ackerman Steering, so for the intent of this question, the vehicle is essentially just one-wheeled (like a bike)
I need to know how to calculate the angle, in relation to the body of the vehicle, that the front axle needs to turn in order for the front of the car to go from point a to point b in a scenario such as
this diagram.
This is all assuming I know:
- The arc length
- The chord length
- The radius of the circle
How do I determine this steering angle?
circle kinematics
asked Jan 9 at 19:38
VectorVector
33
$endgroup$
$begingroup$
I think the diagram isn't quite right ant the back wheel should also be on the circle, so the back wheel follows the same path as the front wheel. In which case, each axle points towsrds the centre of the circle and the single depends on the vehicle length and the circle radius.
$endgroup$
– timtfj
Jan 9 at 19:52
$begingroup$
Sorry, I wasn't clear. The rear axle is fixed--it won't turn. Only the front axle turns, so the way I drew the diagram was the way I meant it.
$endgroup$
– Vector
Jan 9 at 19:54
$begingroup$
The whole vehicle turns, though. Put the front axle at an angle to the back one, and each axle points towsrds the centre of the turning circle. That's what I mean. (Treating it as one wheel behind the other, like a bivycle)
$endgroup$
– timtfj
Jan 9 at 19:57
$begingroup$
I'm not sure what you're getting at...the way this vehicle works, the vehicle will start the way it's positioned in the diagram, except that the front axle will be angled. What I don't know is how to calculate the angle of the front axle.
$endgroup$
– Vector
Jan 9 at 20:00
$begingroup$
@timtfj In reality, unless the vehicle is on rails, every wheel follows a different path.
$endgroup$
– amd
Jan 9 at 20:03
|
show 6 more comments
$begingroup$
I am trying to determine how to relate the turning angle of a vehicle to an arc. For the purposes of this vehicle, I am not taking into account Ackerman Steering, so for the intent of this question, the vehicle is essentially just one-wheeled (like a bike)
I need to know how to calculate the angle, in relation to the body of the vehicle, that the front axle needs to turn in order for the front of the car to go from point a to point b in a scenario such as
this diagram.
This is all assuming I know:
- The arc length
- The chord length
- The radius of the circle
How do I determine this steering angle?
circle kinematics
asked Jan 9 at 19:38
VectorVector
33
$endgroup$
I am trying to determine how to relate the turning angle of a vehicle to an arc. For the purposes of this vehicle, I am not taking into account Ackerman Steering, so for the intent of this question, the vehicle is essentially just one-wheeled (like a bike)
I need to know how to calculate the angle, in relation to the body of the vehicle, that the front axle needs to turn in order for the front of the car to go from point a to point b in a scenario such as
this diagram.
This is all assuming I know:
- The arc length
- The chord length
- The radius of the circle
How do I determine this steering angle?
circle kinematics
circle kinematics
asked Jan 9 at 19:38
VectorVector
33
asked Jan 9 at 19:38
VectorVector
33
edited Jan 10 at 16:54
Vector
asked Jan 9 at 19:38
VectorVector
33
asked Jan 9 at 19:38
VectorVector
33
asked Jan 9 at 19:38
VectorVector
33
33
$begingroup$
I think the diagram isn't quite right ant the back wheel should also be on the circle, so the back wheel follows the same path as the front wheel. In which case, each axle points towsrds the centre of the circle and the single depends on the vehicle length and the circle radius.
$endgroup$
– timtfj
Jan 9 at 19:52
$begingroup$
Sorry, I wasn't clear. The rear axle is fixed--it won't turn. Only the front axle turns, so the way I drew the diagram was the way I meant it.
$endgroup$
– Vector
Jan 9 at 19:54
$begingroup$
The whole vehicle turns, though. Put the front axle at an angle to the back one, and each axle points towsrds the centre of the turning circle. That's what I mean. (Treating it as one wheel behind the other, like a bivycle)
$endgroup$
– timtfj
Jan 9 at 19:57
$begingroup$
I'm not sure what you're getting at...the way this vehicle works, the vehicle will start the way it's positioned in the diagram, except that the front axle will be angled. What I don't know is how to calculate the angle of the front axle.
$endgroup$
– Vector
Jan 9 at 20:00
$begingroup$
@timtfj In reality, unless the vehicle is on rails, every wheel follows a different path.
$endgroup$
– amd
Jan 9 at 20:03
|
show 6 more comments
$begingroup$
I think the diagram isn't quite right ant the back wheel should also be on the circle, so the back wheel follows the same path as the front wheel. In which case, each axle points towsrds the centre of the circle and the single depends on the vehicle length and the circle radius.
$endgroup$
– timtfj
Jan 9 at 19:52
$begingroup$
Sorry, I wasn't clear. The rear axle is fixed--it won't turn. Only the front axle turns, so the way I drew the diagram was the way I meant it.
$endgroup$
– Vector
Jan 9 at 19:54
$begingroup$
The whole vehicle turns, though. Put the front axle at an angle to the back one, and each axle points towsrds the centre of the turning circle. That's what I mean. (Treating it as one wheel behind the other, like a bivycle)
$endgroup$
– timtfj
Jan 9 at 19:57
$begingroup$
I'm not sure what you're getting at...the way this vehicle works, the vehicle will start the way it's positioned in the diagram, except that the front axle will be angled. What I don't know is how to calculate the angle of the front axle.
$endgroup$
– Vector
Jan 9 at 20:00
$begingroup$
@timtfj In reality, unless the vehicle is on rails, every wheel follows a different path.
$endgroup$
– amd
Jan 9 at 20:03
$begingroup$
I think the diagram isn't quite right ant the back wheel should also be on the circle, so the back wheel follows the same path as the front wheel. In which case, each axle points towsrds the centre of the circle and the single depends on the vehicle length and the circle radius.
$endgroup$
– timtfj
Jan 9 at 19:52
$begingroup$
I think the diagram isn't quite right ant the back wheel should also be on the circle, so the back wheel follows the same path as the front wheel. In which case, each axle points towsrds the centre of the circle and the single depends on the vehicle length and the circle radius.
$endgroup$
– timtfj
Jan 9 at 19:52
$begingroup$
Sorry, I wasn't clear. The rear axle is fixed--it won't turn. Only the front axle turns, so the way I drew the diagram was the way I meant it.
$endgroup$
– Vector
Jan 9 at 19:54
$begingroup$
Sorry, I wasn't clear. The rear axle is fixed--it won't turn. Only the front axle turns, so the way I drew the diagram was the way I meant it.
$endgroup$
– Vector
Jan 9 at 19:54
$begingroup$
The whole vehicle turns, though. Put the front axle at an angle to the back one, and each axle points towsrds the centre of the turning circle. That's what I mean. (Treating it as one wheel behind the other, like a bivycle)
$endgroup$
– timtfj
Jan 9 at 19:57
$begingroup$
The whole vehicle turns, though. Put the front axle at an angle to the back one, and each axle points towsrds the centre of the turning circle. That's what I mean. (Treating it as one wheel behind the other, like a bivycle)
$endgroup$
– timtfj
Jan 9 at 19:57
$begingroup$
I'm not sure what you're getting at...the way this vehicle works, the vehicle will start the way it's positioned in the diagram, except that the front axle will be angled. What I don't know is how to calculate the angle of the front axle.
$endgroup$
– Vector
Jan 9 at 20:00
$begingroup$
I'm not sure what you're getting at...the way this vehicle works, the vehicle will start the way it's positioned in the diagram, except that the front axle will be angled. What I don't know is how to calculate the angle of the front axle.
$endgroup$
– Vector
Jan 9 at 20:00
$begingroup$
@timtfj In reality, unless the vehicle is on rails, every wheel follows a different path.
$endgroup$
– amd
Jan 9 at 20:03
$begingroup$
@timtfj In reality, unless the vehicle is on rails, every wheel follows a different path.
$endgroup$
– amd
Jan 9 at 20:03
|
show 6 more comments
1 Answer
1
active
oldest
votes
$begingroup$
To answer your question you need to know two things:
1) The radius $r$ of the curve
2) The distance $l$ from a point on the curve (which I will assume is the center of the car) to the center of the front axle.
Since a circle has constant curvature, the angle will be the same no matter which two points on the circle you pick.
(Formula provided at end, you can skip the derivation if you want)
Now, there's a slight mechanical problem here: the car has minimum turning radius, and without having a lot more information (friction, weight distribution, drivetrain specs, etc.) it isn't really possible to say what that minimum is. So, for the sake of argument, let's just say that the center of the car is affixed to the circle by a rail, the car is front wheel drive, and the rear wheels have no friction.
This means that, at most, $l$ is twice $r$, and the front axle is always perpendicular to the direction the car is facing (i.e. the car is spinning in a circle without moving forward). If you treat the rear wheels as the fixed point instead of the center of the car, then you can redefine $l$ as the distance between the center of the rear axle and that of the front axle - the maximum value of $l$ remains the same.
Now you have the first value of a function which outputs the desired angle given the ratio between $r$ and $l$, namely $(r,2r)mapsto frac{pi}{2}$.
Now consider the other extreme - suppose that $r$ is infinitely greater than $l$. A circle whose radius is infinite is the same as a straight line, so the angle of the front axle would be zero. This gives a second value for your function: $(r,0r)mapsto 0$ ($l=0r$ because $r$ is infinite).
Thus, we can conclude that the angle, is directly proportional to $l$ and inversely proportional to $r$. $quadleft(thetaproptofrac{l}{r}right)$. If $x=frac{l}{r}$, then you are looking for a function $f(x)=theta$, where $theta$ is the angle you want. Furthermore, you know that $f(0)=0$ and $f(2)=frac{pi}{2}$. Since the input to your function is a ratio between two lengths and the output is an angle, we are likely dealing with an inverse trigonometric function.
(This is the part where you start drawing things, if you haven't already)
There are several ways to obtain the correct function using different trig identities, but all are equivalent to:
$$theta=sin^{-1}left(frac{x}{2}right)=sin^{-1}left(frac{l}{2r}right)$$
Note: I can provide a more formal geometric derivation, if you'd like
answered Jan 10 at 18:51
R. BurtonR. Burton
51119
$endgroup$
$begingroup$
That's exactly what I was looking for, thank you so much! I would never have derived that on my own.
$endgroup$
– Vector
Jan 11 at 16:57
add a comment |
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1 Answer
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1 Answer
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$begingroup$
To answer your question you need to know two things:
1) The radius $r$ of the curve
2) The distance $l$ from a point on the curve (which I will assume is the center of the car) to the center of the front axle.
Since a circle has constant curvature, the angle will be the same no matter which two points on the circle you pick.
(Formula provided at end, you can skip the derivation if you want)
Now, there's a slight mechanical problem here: the car has minimum turning radius, and without having a lot more information (friction, weight distribution, drivetrain specs, etc.) it isn't really possible to say what that minimum is. So, for the sake of argument, let's just say that the center of the car is affixed to the circle by a rail, the car is front wheel drive, and the rear wheels have no friction.
This means that, at most, $l$ is twice $r$, and the front axle is always perpendicular to the direction the car is facing (i.e. the car is spinning in a circle without moving forward). If you treat the rear wheels as the fixed point instead of the center of the car, then you can redefine $l$ as the distance between the center of the rear axle and that of the front axle - the maximum value of $l$ remains the same.
Now you have the first value of a function which outputs the desired angle given the ratio between $r$ and $l$, namely $(r,2r)mapsto frac{pi}{2}$.
Now consider the other extreme - suppose that $r$ is infinitely greater than $l$. A circle whose radius is infinite is the same as a straight line, so the angle of the front axle would be zero. This gives a second value for your function: $(r,0r)mapsto 0$ ($l=0r$ because $r$ is infinite).
Thus, we can conclude that the angle, is directly proportional to $l$ and inversely proportional to $r$. $quadleft(thetaproptofrac{l}{r}right)$. If $x=frac{l}{r}$, then you are looking for a function $f(x)=theta$, where $theta$ is the angle you want. Furthermore, you know that $f(0)=0$ and $f(2)=frac{pi}{2}$. Since the input to your function is a ratio between two lengths and the output is an angle, we are likely dealing with an inverse trigonometric function.
(This is the part where you start drawing things, if you haven't already)
There are several ways to obtain the correct function using different trig identities, but all are equivalent to:
$$theta=sin^{-1}left(frac{x}{2}right)=sin^{-1}left(frac{l}{2r}right)$$
Note: I can provide a more formal geometric derivation, if you'd like
answered Jan 10 at 18:51
R. BurtonR. Burton
51119
$endgroup$
$begingroup$
That's exactly what I was looking for, thank you so much! I would never have derived that on my own.
$endgroup$
– Vector
Jan 11 at 16:57
add a comment |
$begingroup$
To answer your question you need to know two things:
1) The radius $r$ of the curve
2) The distance $l$ from a point on the curve (which I will assume is the center of the car) to the center of the front axle.
Since a circle has constant curvature, the angle will be the same no matter which two points on the circle you pick.
(Formula provided at end, you can skip the derivation if you want)
Now, there's a slight mechanical problem here: the car has minimum turning radius, and without having a lot more information (friction, weight distribution, drivetrain specs, etc.) it isn't really possible to say what that minimum is. So, for the sake of argument, let's just say that the center of the car is affixed to the circle by a rail, the car is front wheel drive, and the rear wheels have no friction.
This means that, at most, $l$ is twice $r$, and the front axle is always perpendicular to the direction the car is facing (i.e. the car is spinning in a circle without moving forward). If you treat the rear wheels as the fixed point instead of the center of the car, then you can redefine $l$ as the distance between the center of the rear axle and that of the front axle - the maximum value of $l$ remains the same.
Now you have the first value of a function which outputs the desired angle given the ratio between $r$ and $l$, namely $(r,2r)mapsto frac{pi}{2}$.
Now consider the other extreme - suppose that $r$ is infinitely greater than $l$. A circle whose radius is infinite is the same as a straight line, so the angle of the front axle would be zero. This gives a second value for your function: $(r,0r)mapsto 0$ ($l=0r$ because $r$ is infinite).
Thus, we can conclude that the angle, is directly proportional to $l$ and inversely proportional to $r$. $quadleft(thetaproptofrac{l}{r}right)$. If $x=frac{l}{r}$, then you are looking for a function $f(x)=theta$, where $theta$ is the angle you want. Furthermore, you know that $f(0)=0$ and $f(2)=frac{pi}{2}$. Since the input to your function is a ratio between two lengths and the output is an angle, we are likely dealing with an inverse trigonometric function.
(This is the part where you start drawing things, if you haven't already)
There are several ways to obtain the correct function using different trig identities, but all are equivalent to:
$$theta=sin^{-1}left(frac{x}{2}right)=sin^{-1}left(frac{l}{2r}right)$$
Note: I can provide a more formal geometric derivation, if you'd like
answered Jan 10 at 18:51
R. BurtonR. Burton
51119
$endgroup$
$begingroup$
That's exactly what I was looking for, thank you so much! I would never have derived that on my own.
$endgroup$
– Vector
Jan 11 at 16:57
add a comment |
$begingroup$
To answer your question you need to know two things:
1) The radius $r$ of the curve
2) The distance $l$ from a point on the curve (which I will assume is the center of the car) to the center of the front axle.
Since a circle has constant curvature, the angle will be the same no matter which two points on the circle you pick.
(Formula provided at end, you can skip the derivation if you want)
Now, there's a slight mechanical problem here: the car has minimum turning radius, and without having a lot more information (friction, weight distribution, drivetrain specs, etc.) it isn't really possible to say what that minimum is. So, for the sake of argument, let's just say that the center of the car is affixed to the circle by a rail, the car is front wheel drive, and the rear wheels have no friction.
This means that, at most, $l$ is twice $r$, and the front axle is always perpendicular to the direction the car is facing (i.e. the car is spinning in a circle without moving forward). If you treat the rear wheels as the fixed point instead of the center of the car, then you can redefine $l$ as the distance between the center of the rear axle and that of the front axle - the maximum value of $l$ remains the same.
Now you have the first value of a function which outputs the desired angle given the ratio between $r$ and $l$, namely $(r,2r)mapsto frac{pi}{2}$.
Now consider the other extreme - suppose that $r$ is infinitely greater than $l$. A circle whose radius is infinite is the same as a straight line, so the angle of the front axle would be zero. This gives a second value for your function: $(r,0r)mapsto 0$ ($l=0r$ because $r$ is infinite).
Thus, we can conclude that the angle, is directly proportional to $l$ and inversely proportional to $r$. $quadleft(thetaproptofrac{l}{r}right)$. If $x=frac{l}{r}$, then you are looking for a function $f(x)=theta$, where $theta$ is the angle you want. Furthermore, you know that $f(0)=0$ and $f(2)=frac{pi}{2}$. Since the input to your function is a ratio between two lengths and the output is an angle, we are likely dealing with an inverse trigonometric function.
(This is the part where you start drawing things, if you haven't already)
There are several ways to obtain the correct function using different trig identities, but all are equivalent to:
$$theta=sin^{-1}left(frac{x}{2}right)=sin^{-1}left(frac{l}{2r}right)$$
Note: I can provide a more formal geometric derivation, if you'd like
answered Jan 10 at 18:51
R. BurtonR. Burton
51119
$endgroup$
To answer your question you need to know two things:
1) The radius $r$ of the curve
2) The distance $l$ from a point on the curve (which I will assume is the center of the car) to the center of the front axle.
Since a circle has constant curvature, the angle will be the same no matter which two points on the circle you pick.
(Formula provided at end, you can skip the derivation if you want)
Now, there's a slight mechanical problem here: the car has minimum turning radius, and without having a lot more information (friction, weight distribution, drivetrain specs, etc.) it isn't really possible to say what that minimum is. So, for the sake of argument, let's just say that the center of the car is affixed to the circle by a rail, the car is front wheel drive, and the rear wheels have no friction.
This means that, at most, $l$ is twice $r$, and the front axle is always perpendicular to the direction the car is facing (i.e. the car is spinning in a circle without moving forward). If you treat the rear wheels as the fixed point instead of the center of the car, then you can redefine $l$ as the distance between the center of the rear axle and that of the front axle - the maximum value of $l$ remains the same.
Now you have the first value of a function which outputs the desired angle given the ratio between $r$ and $l$, namely $(r,2r)mapsto frac{pi}{2}$.
Now consider the other extreme - suppose that $r$ is infinitely greater than $l$. A circle whose radius is infinite is the same as a straight line, so the angle of the front axle would be zero. This gives a second value for your function: $(r,0r)mapsto 0$ ($l=0r$ because $r$ is infinite).
Thus, we can conclude that the angle, is directly proportional to $l$ and inversely proportional to $r$. $quadleft(thetaproptofrac{l}{r}right)$. If $x=frac{l}{r}$, then you are looking for a function $f(x)=theta$, where $theta$ is the angle you want. Furthermore, you know that $f(0)=0$ and $f(2)=frac{pi}{2}$. Since the input to your function is a ratio between two lengths and the output is an angle, we are likely dealing with an inverse trigonometric function.
(This is the part where you start drawing things, if you haven't already)
There are several ways to obtain the correct function using different trig identities, but all are equivalent to:
$$theta=sin^{-1}left(frac{x}{2}right)=sin^{-1}left(frac{l}{2r}right)$$
Note: I can provide a more formal geometric derivation, if you'd like
answered Jan 10 at 18:51
R. BurtonR. Burton
51119
answered Jan 10 at 18:51
R. BurtonR. Burton
51119
answered Jan 10 at 18:51
R. BurtonR. Burton
51119
answered Jan 10 at 18:51
R. BurtonR. Burton
51119
51119
$begingroup$
That's exactly what I was looking for, thank you so much! I would never have derived that on my own.
$endgroup$
– Vector
Jan 11 at 16:57
add a comment |
$begingroup$
That's exactly what I was looking for, thank you so much! I would never have derived that on my own.
$endgroup$
– Vector
Jan 11 at 16:57
$begingroup$
That's exactly what I was looking for, thank you so much! I would never have derived that on my own.
$endgroup$
– Vector
Jan 11 at 16:57
$begingroup$
That's exactly what I was looking for, thank you so much! I would never have derived that on my own.
$endgroup$
– Vector
Jan 11 at 16:57
add a comment |
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$begingroup$
I think the diagram isn't quite right ant the back wheel should also be on the circle, so the back wheel follows the same path as the front wheel. In which case, each axle points towsrds the centre of the circle and the single depends on the vehicle length and the circle radius.
$endgroup$
– timtfj
Jan 9 at 19:52
$begingroup$
Sorry, I wasn't clear. The rear axle is fixed--it won't turn. Only the front axle turns, so the way I drew the diagram was the way I meant it.
$endgroup$
– Vector
Jan 9 at 19:54
$begingroup$
The whole vehicle turns, though. Put the front axle at an angle to the back one, and each axle points towsrds the centre of the turning circle. That's what I mean. (Treating it as one wheel behind the other, like a bivycle)
$endgroup$
– timtfj
Jan 9 at 19:57
$begingroup$
I'm not sure what you're getting at...the way this vehicle works, the vehicle will start the way it's positioned in the diagram, except that the front axle will be angled. What I don't know is how to calculate the angle of the front axle.
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– Vector
Jan 9 at 20:00
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@timtfj In reality, unless the vehicle is on rails, every wheel follows a different path.
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– amd
Jan 9 at 20:03