Polar to Rectangular conversion gives false answer

Thread Starter

tahir4awan

Joined Dec 5, 2010
5
Hi guys,
I have problem here in solving resultat polar quantity.

Example 1:
I converted two polar quantities into rectangular quantities and added them. Now I can convert that rectangular quantity again into polar quantity. Which gives same magnitude and angle when I draw that resultant vector.

Example 2:
Again I converted two polar quantities into rectangular quantities and added them but it gave false result.

What I concluded that when the angle is acute it gives correct answer. But when the angle is optuse it gives false answer.

I am confused with example no 2. Can someone explain this problem.

Thanks
 

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Thread Starter

tahir4awan

Joined Dec 5, 2010
5
Yes I have seen this website. It says the same that we can add two complex quantities to get the resultant vector. But the problem is, this law does not applicable for angle greater than 90.


You can see the attached files.
 

Papabravo

Joined Feb 24, 2006
13,491
Yes I have seen this website. It says the same that we can add two complex quantities to get the resultant vector. But the problem is, this law does not applicable for angle greater than 90.


You can see the attached files.
It might have to do with using the Sine function. That is the wrong function to be using in this process.
∠ = tan^(-1) (y/x) where x + jy is the rectangular representation
 
Last edited:

bogosort

Joined Sep 24, 2011
915
Yes I have seen this website. It says the same that we can add two complex quantities to get the resultant vector. But the problem is, this law does not applicable for angle greater than 90.


You can see the attached files.
You actually had the correct answer in the 2nd image, but you abandoned it with the note "vt is in opposite diagonal, which is false". You then took the long hyperbola as vt, which is incorrect.

To better understand what went wrong, try visualizing vector addition as directions on a map. If you were standing at the head of v1 and I told you to walk a distance equal to the magnitude of v2 in the direction of v2, would you end up at your final conclusion of vt (the long hyperbola)? No, you'd end up at 30 ∠ 60°, as in your abandoned (but correct) conclusion. That's all vector addition is: a sequence of displacements, which -- in 2D spaces -- we can think of as walking directions on a map.
 

Thread Starter

tahir4awan

Joined Dec 5, 2010
5
Yes. You are right. I have calculated it in TI89. The problem is, if I assume short diagonal is correct then it is not true for three phase vector addition.

For example V1 = 230, V2 = 230 if I choose short diagonal then the answer is,
230V with angle 60°.

But if I assume longest diagonal as a resultant vector then it gives

398.73V with angle 30° which is true for three phase.
 

bogosort

Joined Sep 24, 2011
915
Yes. You are right. I have calculated it in TI89. The problem is, if I assume short diagonal is correct then it is not true for three phase vector addition.

For example V1 = 230, V2 = 230 if I choose short diagonal then the answer is,
230V with angle 60°.

But if I assume longest diagonal as a resultant vector then it gives

398.73V with angle 30° which is true for three phase.
I'm not sure why you've introduced three-phase. In a three-phase circuit, we could take the vector sum of three source voltages. Here, in contrast, we are summing two source voltage vectors; the resultant is not in any way related to the third phase of a 3-phase circuit. It is indeed the case that 230 ∠ 0° + 230 ∠ 120° = 230 ∠ 60°.
 

MrAl

Joined Jun 17, 2014
7,477
It might have to do with using the Sine function. That is the wrong function to be using in this process.
∠ = tan^(-1) (y/x) where x + jy is the rectangular representation
Hi,

Actually that's not quite right either, although the interpretation does have something to do with it too i must say.

The correct function is:
Angle=atan2(y,x)

where y is the imaginary part and x the real part.

The reason for this is so that the signs of x and y can not cause anomalies in the result.
That is sometimes called the "two argument inverse tangent function".
That can resolve angles between 0 and 360 degrees (0 to 2pi rads) or -360 to 0 degrees (-2pi to 0 rads) or typically 0 to 180 or -180 to 0.
Angles outside that range have to be handled on a case by case basis but often are not considered.
 

Papabravo

Joined Feb 24, 2006
13,491
Hi,

Actually that's not quite right either, although the interpretation does have something to do with it too i must say.

The correct function is:
Angle=atan2(y,x)

where y is the imaginary part and x the real part.

The reason for this is so that the signs of x and y can not cause anomalies in the result.
That is sometimes called the "two argument inverse tangent function".
That can resolve angles between 0 and 360 degrees (0 to 2pi rads) or -360 to 0 degrees (-2pi to 0 rads) or typically 0 to 180 or -180 to 0.
Angles outside that range have to be handled on a case by case basis but often are not considered.
To be completely pedantic about it, the atan2 is not a well defined mathematical construct. It is an artifact of the use of programming languages. The inverse tangent function has a domain and a range that are well defined and understood. I wasn't assuming the knowledge of any particular programming language.
 
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