Finding Velocity Components given Speed and Heading

[Ash_Sdr]

Homework Statement

Given the Speed of vehicle at any instant (say 30 meters/s) and Heading of a vehicle (say 270 degrees ) , how can i find the velocity components which is Vx and Vy .

Lemme tell you about the Heading ...This is the convention..

I don;t know if this pic is viewable or not... So Heading is with true north... which means that , if vehicle is facing NORTH --- heading is 0 degrees (or) 360 ... if its facing East , its 90 degrees , if its facing South , its 180 and If its facing West , its 270...

The X,Y convention is ... X is Vertical Axis and Y is Horizontal...

Homework Equations

Speed = sqrt(Vx^2 + Vy^2)

The Attempt at a Solution

Two possible Solutions which i can think of...

Since X is longitudinal and Y is Lateral ... Vx = Speed * sin(Heading)
Vy = Speed *cos(Heading)

and Second solution is to use delta heading
Vx = Speed * sin(Heading2 - Heading1)
Vy = Speed *cos(Heading2 - Heading1)

but I am not getting good results with these... can anyone tell me what I am doing wrong

 

[Ray Vickson]

Homework Statement

Given the Speed of vehicle at any instant (say 30 meters/s) and Heading of a vehicle (say 270 degrees ) , how can i find the velocity components which is Vx and Vy .

Lemme tell you about the Heading ...This is the convention..

I don;t know if this pic is viewable or not... So Heading is with true north... which means that , if vehicle is facing NORTH --- heading is 0 degrees (or) 360 ... if its facing East , its 90 degrees , if its facing South , its 180 and If its facing West , its 270...

The X,Y convention is ... X is Vertical Axis and Y is Horizontal...

Homework Equations

Speed = sqrt(Vx^2 + Vy^2)

The Attempt at a Solution

Two possible Solutions which i can think of...

Since X is longitudinal and Y is Lateral ... Vx = Speed * sin(Heading)
Vy = Speed *cos(Heading)

and Second solution is to use delta heading
Vx = Speed * sin(Heading2 - Heading1)
Vy = Speed *cos(Heading2 - Heading1)

but I am not getting good results with these... can anyone tell me what I am doing wrong

By vertical, do you mean "pointing north"? By horizontal, do you mean "pointing east"? I ask, because your North-East-South-West directions seem to be at odds with your vertical/horizontal.

 

[Ash_Sdr]

No, By vertical , its not pointing North ... nor Horizontal is pointing east... Thats the vehicle coordinate system and the heading angle is w.r.t global coordinates..

 

[Ray Vickson]

No, By vertical , its not pointing North ... nor Horizontal is pointing east... Thats the vehicle coordinate system and the heading angle is w.r.t global coordinates..

Well, your first solution gave $v_x = s \sin(\theta), v_y = s \cos(\theta)$, where $s$ is the speed and $\theta$ is the heading (measured clockwise from north). That means that $v_x$ is the velocity component in the easterly direction (actually, pointing west if $v_x < 0$) and $v_y$ is the northerly component. That is what you get if you use sines and cosines the way you did!

Your second "solution" makes no sense at all unless you specify two headings, which you have not done.

 

[Ash_Sdr]

Well, your first solution gave $v_x = s \sin(\theta), v_y = s \cos(\theta)$, where $s$ is the speed and $\theta$ is the heading (measured clockwise from north). That means that $v_x$ is the velocity component in the easterly direction (actually, pointing west if $v_x < 0$) and $v_y$ is the northerly component. That is what you get if you use sines and cosines the way you did!

Your second "solution" makes no sense at all unless you specify two headings, which you have not done.

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vx should be northerly component right ? and vy should be Easterly ?

 

[Ash_Sdr]

Well, your first solution gave $v_x = s \sin(\theta), v_y = s \cos(\theta)$, where $s$ is the speed and $\theta$ is the heading (measured clockwise from north). That means that $v_x$ is the velocity component in the easterly direction (actually, pointing west if $v_x < 0$) and $v_y$ is the northerly component. That is what you get if you use sines and cosines the way you did!

Your second "solution" makes no sense at all unless you specify two headings, which you have not done.

The second solution " heading 2 is current heading and heading 1 is previous heading

 

[Ray Vickson]

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vx should be northerly component right ? and vy should be Easterly ?

No, not the way you defined things. For speed $s$ and heading $\theta$ (measured clockwise from north) the quantity $s \sin(\theta)$ is the EASTward component of the vector; draw a picture and check this for yourself! So, when you write $v_x = s \sin(\theta)$ you are giving the eastward component, NOT the northward one. The quantity $s \cos(\theta)$ is the northward component, which you called $v_y$. So,if I look at a map with north up and east to the right, your x-axis runs from west to east and your y-axis runs from south to north. That is exactly what a standard cartesian coordinate system looks like.

Again: draw a picture; don't take my word for it.