Straight Line from polar co-ordinates

[pd5009]

Hi All,
This is my first post. I am an Electronics Engineer and came by this great forum while searching something for my presently running project.
Could anyone please help me with the following:

I have two points A(magnitude1,phase1[deg]) and B(magnitude2,phase2[deg]) on the input side. Also, I have the Time input to travel from pt.A to pt.B

I have one polar plot as indicator output.


I need to show on the polar plot a line which is gradually increasing from pt.A to pt.B in a straight line on the polar plot.


You may refer the picture attached to get an idea of what I am trying to do.

Thanks!

 

[zgozvrm]

I assume what you're looking for is the magnitude and direction of the vector from point A [itex](2,\angle 10^\circ)[/tex] to point B [itex](20, \angle 60^\circ )[/tex].<br /> <br /> First, convert the coordinates of each point from polar to rectangular:<br /> <br /> Point A:<br /> $$X = 2 \cos(10), Y = 2 \sin(10) \approx (1.970, 0.347)$$<br /> <br /> Point B:<br /> $$X = 20 \cos(60), Y = 20 \sin(60) \approx (10, 17.321)$$:<br /> <br /> <br /> <br /> Next, find the change in X [itex](\Delta X)[/tex] and the change in Y [itex](\Delta Y)[/tex]:<br /> <br /> $$\Delta X = X_B - X_A \approx 10 - 1.970 \approx 8.030$$<br /> <br /> $$\Delta Y = Y_B - Y_A \approx 17.321 - 0.347 \approx 16.973$$<br /> <br /> <br /> <br /> Now, find the length of the line segment [itex]M[/tex] using the Pythagorean Theorem:<br /> <br /> $$M = \sqrt{8.030^2 + 16.973^2} \approx 18.777$$<br /> <br /> <br /> <br /> Lastly, find the angle of the line segment [itex]\alpha[/tex]:<br /> <br /> $$\alpha \approx \tan^{-1} \left( \frac{\Delta Y}{\Delta X}\right) \approx \tan^{-1} \left( \frac{16.973}{8.030} \right) \approx \tan^{-1}(2.114) \approx 64.680^\circ$$<br /> <br /> (I only rounded the numbers for display, not for intermediate calculations)<br /> <br /> <br /> <br /> <br /> So, your resulting line segment from point A to point B can be represented by the vector [itex](m, \alpha) \approx (18.777, \angle 64.680^\circ)[/tex]<br /> <br /> Hope this helps![/itex][/itex][/itex][/itex][/itex][/itex][/itex]

 

[pd5009]

^ ^ Thanks :) I implemented that in my software and it worked :)