Co-ordinate systems and velocities

  • Thread starter Spanky Deluxe
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In summary, the conversation discusses the conversion between Cartesian and Cylindrical coordinates for positions and velocities. The equations for converting from Cartesian to Cylindrical coordinates for positions are given, along with the equations for converting back. However, the conversation becomes more complicated when trying to determine velocity vectors in Cylindrical coordinates. The time derivatives of the vector are used for this, which can be found in the "time derivatives of the vector" section in the provided link. The conversation also mentions the need to convert between azimuthal and radial velocities and x and y velocities, and the use of sign functions in the equations. Additionally, a sketch is included to better explain the problem at hand.
  • #1
Spanky Deluxe
19
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Can anyone help explain how to convert velocities to different co ordinate systems? I've always been shocking at converting between co ordinate systems as it is.

I know that converting from Cartesian to Cylindrical for positions it is:

r = sqrt(x^2 + y^2)
theta = atan(y/x)
z = z

and back its:

x = rcos(theta)
y = rsin(theta)
z = z

However, I'm getting confused as to what it would be for velocities.

What I've got so far is:

vx = vtheta * sin(theta) * sign(y)
vy = vtheta * cos(theta) * sign(-x)
vz = vz

Am I doing the right kind of thing or am I overcomplicating things? Any help would be greatly appreciated!
 
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  • #2
What is "vtheta'?

x is a function of two variables, r and theta and these are both functions of time, in general.

dx/dt=(dr/dt)cos(theta)-r sin(theta) d(theta)/dt
and similar for y.

dr/dt is the radial speed and d(theta)/dt is angular speed.
 
  • #3
vtheta is the rotational velocity
 
  • #4
Hi Spanky,

I think you are overcomplicating things. If you have any vector in Cartesian coordinates then to transform it to Cylindrical coordinates you use
r = sqrt(x^2 + y^2)
theta = atan(y/x)
z = z

That part is easy. What is more challenging is determining the velocity vector in Cylindrical coordinates if you have a position in Cylindrical coordinates as a function of time. For that, see the "time derivatives of the vector" section here: http://mathworld.wolfram.com/CylindricalCoordinates.html

v = (r', r theta', z')
 
  • #5
I think I'm getting muddled up with different co-ordinate systems, rotational velocities and the spin dependence of elephants when eating carrot sticks.

Looking at everything again, I basically need to be able to convert between azimuthal and radial velocities and x and y velocities for a body rotating around another body. Its all basically a mess of those diagrams I had to draw at school working out different angles and their dependence on each other, which I sucked donkey doo doo at.

I basically need to convert backwards and forwards between azimuthal and radial components of force and velocity to Cartesian co-ordinates. The origin is at (0,0).

So I guess I need something like this for anticlockwise motion:
theta = (y/x)
vx = (v_azimuthal * sin(theta) * sign(-y)) + (v_radial * cos(theta) * sign(x))
vy = (v_azimuthal * cos(theta) * sign(x)) + (v_radial * sin(theta) * sign(y))
vz = vz

Then the same should be true for forces I think. For clockwise motion the first sign functions of vx and vy would need to be sign(y) and sign(-x) respectively I think. I think those equations are right, if a bit complex. Now though, I need to work out a way to get v_azimuthal and v_radial from vx and vy, i.e. the other way around.
 
  • #6
If you have any vector in Cartesian coordinates and you want to express it in Cylindrical coordinates then you use:

r = sqrt(x²+y²)
theta = atan(y/x)
z = z

If you have any vector in Cylindrical coordinates and you want to express it in Cartesian coordinates then you use:

x = r cos(theta)
y = r sin(theta)
z = z

It makes no difference whatsoever what the units are nor whether the vector represents position, acceleration, velocity, momentum, etc.

The only trick comes if you have a position as a function of time in Cylindrical coordinates and you want to find the velocity or acceleration as a function of time in the Cylindrical coordinates without converting to and from Cartesian. Then you use the formulas I linked to above.
 
  • #7
Why are there sign functions in your expression?

To find velocities you need only differentiate withe respect to time:

[tex]v_x\equiv\frac{dx}{dt}=\frac{d}{dt}(r\cos\theta)=\frac{dr}{dt}\cos\theta-r\sin\theta\frac{d\theta}{dt}=v_r\cos\theta- v_{\theta}\sin\theta[/tex]

To get [itex]v_y[/itex] and [itex]v_z[/itex] you just do the same thing to your expressions for [itex]y[/itex] and [itex]z[/itex]


To get [itex]v_r[/itex] and [itex]v_{\theta}[/itex] in terms of Cartesians you just differentiate the expressions for [itex]r[/itex] and [itex]\theta[/itex] with respect to time:

[tex]v_r=\frac{dr}{dt}=\frac{d}{dt}\sqrt{x^2+y^2}[/tex]

[tex]v_{\theta}\equiv r\frac{d\theta}{dt}=r\frac{d}{dt} \tan^{-1}\left(\frac{y}{x}\right)=\sqrt{x^2+y^2}\frac{d}{dt} \tan^{-1}\left(\frac{y}{x}\right)[/tex]

OR you solve the 3 equations you got for v_x, v_y and v_z for v_r, v_theta and v_z the same way you would solve any system of 3 equations for 3 unknowns.
 
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  • #8
DaleSpam said:
If you have any vector in Cartesian coordinates and you want to express it in Cylindrical coordinates then you use:

r = sqrt(x²+y²)
theta = atan(y/x)
z = z

If you have any vector in Cylindrical coordinates and you want to express it in Cartesian coordinates then you use:

x = r cos(theta)
y = r sin(theta)
z = z

It makes no difference whatsoever what the units are nor whether the vector represents position, acceleration, velocity, momentum, etc.

The only trick comes if you have a position as a function of time in Cylindrical coordinates and you want to find the velocity or acceleration as a function of time in the Cylindrical coordinates without converting to and from Cartesian. Then you use the formulas I linked to above.

So radial velocity would be vr = sqrt(vx^2+vy^2). However, for an orbiting body, radial velocity is largely zero, which with that equation would imply that vx and vy are both largely zero, which is not the case.

gabbagabbahey said:
Why are there sign functions in your expression?

To find velocities you need only differentiate withe respect to time:

[tex]v_x\equiv\frac{dx}{dt}=\frac{d}{dt}(r\cos\theta)=\frac{dr}{dt}\cos\theta-r\sin\theta\frac{d\theta}{dt}=v_r\cos\theta- v_{\theta}\sin\theta[/tex]

To get [itex]v_y[/itex] and [itex]v_z[/itex] you just do the same thing to your expressions for [itex]y[/itex] and [itex]z[/itex]To get [itex]v_r[/itex] and [itex]v_{\theta}[/itex] in terms of Cartesians you just differentiate the expressions for [itex]r[/itex] and [itex]\theta[/itex] with respect to time:

[tex]v_r=\frac{dr}{dt}=\frac{d}{dt}\sqrt{x^2+y^2}[/tex]

[tex]v_{\theta}\equiv r\frac{d\theta}{dt}=r\frac{d}{dt} \tan^{-1}\left(\frac{y}{x}\right)=\sqrt{x^2+y^2}\frac{d}{dt} \tan^{-1}\left(\frac{y}{x}\right)[/tex]

OR you solve the 3 equations you got for v_x, v_y and v_z for v_r, v_theta and v_z the same way you would solve any system of 3 equations for 3 unknowns.

The signs are because x and y become negative and the signs on the velocities need to be right.

Sorry guys if I'm missing the point or I'm not explaining it well enough. Like I said, I really suck at angle conversions and co-ordinate systems etc.

To try to explain things better, I'm attaching a sketch of what I'm trying to do:

IMG_0616.jpg


I basically start with information in the form x,y,z,vx,vy,vz. I then have to calculate a force for which the equations work in azimuthal and radial values and depend on the azimuthal and radial velocities (in the same units as vx,vy,vz). So I have to first convert vx and vy into rotational velocity and radial velocity. Then the resulting force has to be converted back to the force in x,y,z.

So, vx,vy,vz,v_azimuthal,v_radial all have the same units (i.e. m/s). Theta is calculated by using y/x but since y/x changes sign depending on which quadrant it is in, I've added those sign modifiers to try to make things work.

===============================================================

Edit: Right, I think I've worked out the formulae that I need so I'll post them up on here in case anyone else finds this via search and is as similarly lacking as I am in the angles and conversions departments!

theta = atan(y/x)

v_x = abs(v_radial*cos(theta))*sign(x) + abs(v_rotational*sin(theta))*sign(Cy)
v_y = abs(v_radial*sin(theta))*sign(y) - abs(v_rotational*cos(theta))*sign(Cx)
v_radial = abs(v_x*cos(theta))*sign(x) + abs(v_y*sin(theta))*sign(y)
v_rotational = abs(v_x*sin(theta))*sign(Cy) - abs(v_y*cos(theta))*sign(Cx)

where abs() denotes the absolute value (or modulus) of a number (i.e. -1.5623 -> 1.5623 and +1.2479 -> 1.2479) and sign() gives the sign of a number (i.e. 1.2435 -> 1 and -1.7829 -> -1).

C is a sign factor depending on the direction of rotation. For anticlockwise motion, C is -1, for clockwise motion, C is 1. This is for a system where a body is rotating around another, which is at x = y = 0.

There's probably an easier way of doing this but I'm not bright enough to work it out!
 
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  • #9
Spanky Deluxe said:
So radial velocity would be vr = sqrt(vx^2+vy^2). However, for an orbiting body, radial velocity is largely zero, which with that equation would imply that vx and vy are both largely zero, which is not the case.
Ahh, I think that is your problem. What you are talking about is the first derivative of the radius, not the radial component of the velocity. Assuming that you are talking about motion in the plane (z = 0) then the radial component of the velocity vector is the speed and assuming a uniform circular orbit then theta component of the velocity vector is 90 degrees from the theta component of the position. I am not sure if this is clearly written or not, but do you understand how, in cylindrical components, the velocity will not have a radial component of zero even in a perfectly circular orbit?
 

1. What is a co-ordinate system?

A co-ordinate system is a mathematical system used to locate points in space or on a plane. It is typically made up of two or three axes, with each axis representing a different dimension.

2. What are the different types of co-ordinate systems?

There are several types of co-ordinate systems, including Cartesian, polar, cylindrical, and spherical. Each type has its own set of rules and equations for locating points in space.

3. How do I convert between different co-ordinate systems?

To convert between different co-ordinate systems, you will need to use specific formulas or equations depending on the type of co-ordinate system you are working with. It is important to understand the rules and equations for each type in order to properly convert between them.

4. What is velocity?

Velocity is a measure of the rate and direction of an object's motion. It is typically expressed in units of distance per time, such as meters per second or kilometers per hour.

5. How are co-ordinate systems and velocities related?

Co-ordinate systems are used to describe the position of an object in space, while velocity is used to describe the object's motion. By using co-ordinate systems, we can calculate an object's velocity by measuring its change in position over time.

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