Calculating unit vector for velocity

AI Thread Summary
The discussion revolves around calculating the unit vector for velocity and the confusion surrounding the equations used. The initial equation presented for velocity, vel = (U * Unit vector) / unit vector, is deemed nonsensical. Participants clarify that a unit vector should be derived by normalizing a vector, which involves dividing by its magnitude. The correct form for a 2D unit vector is provided, along with the formula for its magnitude. The conversation also emphasizes the importance of using the dot product to find the component of velocity in the direction of the unit vector. This is illustrated through the projection of velocity onto the unit vector, which is essential for understanding how to transfer velocity without recalculating at a specific location. There is a consensus that the goal is to determine the velocity component in the direction of the unit vector, rather than calculating the velocity itself. The discussion concludes with a suggestion to use clearer notation for vector components to enhance understanding.
andykol
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Hello,
I am trying to calculate unit vector for velocity (vel= (U*Unit vector)/unit vector). But if I consider calculation by angle change e.g. unit vector= cos(theta) at certain angle velocity becomes infinity.
Please inform me how I can take care of this problem.
Thanks in advance.
 

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vel= (U*Unit vector)/unit vector).

This equation makes no sense.

unit vector= cos(theta)

This equation makes even less sense...as cos(theta) is a scalar not a vector.

Typically you make unit vectors by normalizing them, which means to divide by the length.
 
The unit vector for velocity would be:
U=\left(\frac{V_x}{|V|},\frac{V_y}{|V|},\frac{V_z}{|V|}\right)

where the magnitude of the velocity vector is given by:

|V|=sqrt(V_x^2+V_y^2+V_z^2)

and

V_x,V_y,V_z, are the x, y, and z components of the velocity vector respectively.

However, without more information on the original problem I'm not sure if this is what you want.
 
Thanks for reply.
I am trying to multiply velocity with unit vector to transfer velocity without calculating at perticular location. Please see attached picture. Please tell me if I m using right equation and its values.

John,
I need to use following equation to calculate velocity for 2D?

Unit vector=\left(\frac{U_x}{|U|}+\frac{U_y}{|U|}\right)

|U|=sqrt(U_x^2+U_y^2)

Then I can calculate U' by (U(x,y)*Unit vector(x,y))/unit vector(x,y)
 

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andykol said:
Thanks for reply.
I am trying to multiply velocity with unit vector to transfer velocity without calculating at perticular location. Please see attached picture. Please tell me if I m using right equation and its values.

John,
I need to use following equation to calculate velocity for 2D?

Unit vector=\left(\frac{U_x}{|U|}+\frac{U_y}{|U|}\right)

|U|=sqrt(U_x^2+U_y^2)

Then I can calculate U' by (U(x,y)*Unit vector(x,y))/unit vector(x,y)

I think what you want is the dot product. The dot product will give you the component of the velocity in the direction of the unit vector.

For a unit vector u the projection of V on U is given by:

Proj_UV=U\cdot V=UxVx+UyVy+UzVz
 
means-
If we consider one direction like picture I have attached in last post.
U- Velocity(known)
U'-Velocity(unknown)
Uv-Velocity Vector

Then if I am transferring velocity value

U'=U.Uv

Where U_v=\left(\frac{U}{|U|}\right )

But this becomes U'=U. I think this is wrong as location of velocity changes.
 
andykol said:
means-
If we consider one direction like picture I have attached in last post.
U- Velocity(known)
U'-Velocity(unknown)
Uv-Velocity Vector

Then if I am transferring velocity value

U'=U.Uv

Where U_v=\left(\frac{U}{|U|}\right )

But this becomes U'=U. I think this is wrong as location of velocity changes.

To make your posts more clear, use subscripts for components of vectors. Now reread my last two posts. As far as I understand your problem, you are trying to find the component of the velocity in the direction of the unit vector. This is not the same thing as finding the velocity.
 
Thank you John. This helped alot.
 
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