Efficient Calculation of Final Magnitude and Angle in Vector Addition Method

[kai sinclair]

TL;DR Summary

making quicker ways to add vectors together

okay so I'm a Electrician I've found short method of calculating the final magnitude of a system (Lₜ), this relies mainly on Eucliud's axiom of angles within parrallel lines and is this

∑(cos(θₙ-θₜ)⋅Lₙ)=Lₜ

where Lₙ and θₙ are the initail manignitude and angles respectively, and θₜ is the final angle

now you might see the issue that being I need to know the final angle, so I was hoping that I might be able to transpose for θₜ and then simplify the forumal below, which uses my above formula and the long way round, to get the short method to calulate θₜ

[∑(cos(θₙ-θₜ)⋅Lₙ)]²=[∑(cos(θₙ)⋅Lₙ)]²+[∑(sin(θₙ)⋅Lₙ)]²

appologies if I am missusing functions, it makes sense in my head

 

[anuttarasammyak]

You seem to have suceeded in deleting $L_t$ in the equation, right ?

 

[kai sinclair]

You seem to have suceeded in deleting $L_t$ in the equation, right ?

Both sides of the equation find $L_t$, one side is Pythagoras, the other my own formula

 

[DaveE]

Sorry, I'm not following this at all. Please show your steps in a derivation. What happened to Θ_t? You said that's what you want to calculate but it's not in your last equation. A typo maybe?

This sort of calculation is normally done with a complex number representation of impedances, voltages, etc. Are you familiar with that?

Unfortunately, adding vectors in a general sense doesn't have many (any?) short cuts, unless there are some special relationships between them. It's drudgery, you just have to do it term by term.

 

[kai sinclair]

Sorry, I'm not following this at all. Please show your steps in a derivation. What happened to Θ_t? You said that's what you want to calculate but it's not in your last equation. A typo maybe?

This sort of calculation is normally done with a complex number representation of impedances, voltages, etc. Are you familiar with that?

Unfortunately, adding vectors in a general sense doesn't have many (any?) short cuts, unless there is some special relationships between them. It drudgery, you just have to do it term by term.

θₜ is on the left side of the equation in the cos

Not really no

Well I know of several ways, compass method, tip to tail, I also know you can calculate them, those calculations require entering each pair of inputs multiple times for them to work, this overwhelms my calculators character limit very quickly, which is something my formula has over the Pythagoras's theorem, but as I said it requires I know θₜ, so of we transpose and simplify then we should end up with a simpler method of calculating the final angle

 

[DaveE]

it makes sense in my head

But can you have it make sense on paper, to someone that knows the subject? Sorry, I still don't understand what your doing.

 

[berkeman]

okay so I'm a Electrician I've found short method of calculating the final magnitude of a system (Lₜ), this relies mainly on Eucliud's axiom of angles within parrallel lines and is this

I still don't understand what your doing.

Nor do I. @kai sinclair -- can you please say exactly what you are trying to calculate, and give an example of such a system? Are you calculating power factor for an industrial location and trying to figure out how big of a capacitor bank it will take to bring the PF back to acceptable levels for your inductive motor loads?

I'll also send you some tips for using LaTeX to post math equations. I don't know how you are posting your math right now, but it's mostly readable so that is good. LaTeX would be better, but you're probably okay for now if you can clear up what your application is.

Also, do you have access to Excel when you are on the job doing these calculations? If your calculator is getting overwhelmed by the entries, Excel would be a good next choice, IMO.

 

[Ibix]

I think you have several vectors $\vec v_n$, where $n=1,2,\ldots,N$, and you are trying to calculate the modulus of the resultant vector $\vec v_t=\vec v_1+\vec v_2+\ldots+\vec v_N$. You have defined the length of the $n$th initial vector to be $L_n$, and it makes an angle $\theta_n$ with the horizontal. Similarly, the total vector has length $L_t$ and makes angle $\theta_t$ with the horizontal.

Am I right so far?

If so, the right hand side of your expression is the result of writing your initial vectors in component form in Cartesian coordinates, adding them up, and taking the squared modulus of the result. $$\begin{eqnarray*}
\vec v_t&=&\vec v_1+\vec v_2+\ldots + \vec v_N\\
&=&\left(\begin{array}{c}L_1\cos\theta_1\\L_1\sin\theta_1\end{array}\right)
+\left(\begin{array}{c}L_2\cos\theta_2\\L_2\sin\theta_2\end{array}\right)
+\ldots
+\left(\begin{array}{c}L_N\cos\theta_N\\L_N\sin\theta_N\end{array}\right)\\
&=&\left(\begin{array}{c}\sum_{n=1}^N L_n\cos\theta_n\\\sum_{n=1}^N L_n\sin\theta_n\end{array}\right)\\
|\vec v_t|^2=L_t^2&=&\left(\sum_{n=1}^N L_n\cos\theta_n\right)^2+\left(\sum_{n=1}^N L_n\sin\theta_n\right)^2
\end{eqnarray*}$$That is, of course, correct. The left hand side is computing the projected length of each of your initial vectors on to the final vector and summing them up. That is also correct.

 

[kai sinclair]

Nor do I. @kai sinclair -- can you please say exactly what you are trying to calculate, and give an example of such a system? Are you calculating power factor for an industrial location and trying to figure out how big of a capacitor bank it will take to bring the PF back to acceptable levels for your inductive motor loads?

I'll also send you some tips for using LaTeX to post math equations. I don't know how you are posting your math right now, but it's mostly readable so that is good. LaTeX would be better, but you're probably okay for now if you can clear up what your application is.

Also, do you have access to Excel when you are on the job doing these calculations? If your calculator is getting overwhelmed by the entries, Excel would be a good next choice, IMO.

I know what I'm going on about in my head, but I'm smoking some weird dang mean translating into common speak is difficult, after work I'll write this up filling in with some variables and hopefully it'll make more sense

 

[berkeman]

Is this for residential or industrial settings where you're working as an electrician?

 

[alan123hk]

TL;DR Summary: making quicker ways to add vectors together

okay so I'm a Electrician I've found short method of calculating the final magnitude of a system (Lₜ), this relies mainly on Eucliud's axiom of angles within parrallel lines and is this

Since you are an electrician, I assume this question is related to electrical engineering, but I can't think of any clues.

Regardless, you want to simplify the left-hand side of equation and find the unknown $L_t$. Does this mean that all variables on the left side of the equation are known?

If these variables are known, I don't see why you would square both sides of the equation as that would complicate the calculation in my opinion, unless there is some mathematical connection between the variables on the left side of the equation that I'm not aware of however.

Could you please provide some further information?

 

[kai sinclair]

Since you are an electrician, I assume this question is related to electrical engineering, but I can't think of any clues.

Regardless, you want to simplify the left-hand side of equation and find the unknown $L_t$. Does this mean that all variables on the left side of the equation are known?

If these variables are known, I don't see why you would square both sides of the equation as that would complicate the calculation in my opinion, unless there is some mathematical connection between the variables on the left side of the equation that I'm not aware of however.

Could you please provide some further information?

Vector sums, I need to calculate if a ciduit is in phase or not most commonly this boils down to a RLC circuit, but you can also have several circuits adding to together

 

[kai sinclair]

Nor do I. @kai sinclair -- can you please say exactly what you are trying to calculate, and give an example of such a system? Are you calculating power factor for an industrial location and trying to figure out how big of a capacitor bank it will take to bring the PF back to acceptable levels for your inductive motor loads?

I'll also send you some tips for using LaTeX to post math equations. I don't know how you are posting your math right now, but it's mostly readable so that is good. LaTeX would be better, but you're probably okay for now if you can clear up what your application is.

Also, do you have access to Excel when you are on the job doing these calculations? If your calculator is getting overwhelmed by the entries, Excel would be a good next choice, IMO.

okay here's an example, you wouldn't find this anywhere but it should help explain what the hell I'm on about

inputs

L₁	L₂	L₃	θ₁	θ₂	θ₃
10	10	10	15	70	-40

Pythagoras
[∑(cos(θₙ)⋅Lₙ)]²+[∑(sin(θₙ)⋅Lₙ)]²=Lₜ²
[cos(15)⋅10+cos(70)⋅10+cos(-40)⋅10]²+[sin(15)⋅10+sin(70)⋅10+sin(-40)⋅10]²=461.02=21.47²

my magnitude formula
∑(cos(θₙ-θₜ)⋅Lₙ)=Lₜ
cos(15-15)⋅10+cos(70-15)⋅10+cos(-40-15)⋅10=21.47

inverse tan
tan⁻¹[∑(sin(θₙ)⋅Lₙ)/∑(cos(θₙ)⋅Lₙ)]=θₜ
tan⁻¹[(sin(15)⋅10+sin(70)⋅10+sin(-40)⋅10)/(cos(15)⋅10+cos(70)⋅10+cos(-40)⋅10)]=15

my angle formula
∑(θₙ⋅[Lₙ/∑{Lₙ}])=θₜ
15⋅[10/(10+10+10)]+70⋅[10/30]-40⋅[10/30]=15

while this looks like I've answered my own question for a short cut for calculating θₜ, this angle formula is incomplete and has failure states, so I'm looking to expand it's reach, but I can't think of anything to add to it to help outside of single cases, so my thinking is because Pythagoras and my magnitude formula are calculating the same thing, Pythagoras must necessarily be calculating θₜ without us knowing it so we rearrange my magnitude formula, slot in Pythagoras and simplify and bang should have a simple angle formula that works

 

[Baluncore]

The quickest vector sum is probably this.

First convert polar to rectangular.
L1; 10 V ∠15°; x=10*Cos(15°) = +9.659 V; y=10*Sin(15°) = +2.588 V
L2; 10 V ∠70°; x=10*Cos(70°) = +3.420 V; y=10*Sin(70°) = +9.397 V
L3; 10 V ∠-40°; x=10*Cos(-40°) = +7.660 V; y=10*Sin(-40°) = -6.428 V

Then sum the x terms and y terms separately.
Sum of x, Cos terms; 9.659 + 3.420 + 7.660 = 20.739 V
Sum of y, Sin terms; 2.588 + 9.397 – 6.428 = 5.557 V

Convert sum from rectangular to polar.
Magnitude; √(20.739² + 5.557²) = 21.471 V
Angle = Atan2( 5.557, 20.739) = +15°

Vector sum; L1+L2+L3 = 21.471 V ∠+15°

 

[kai sinclair]

The quickest vector sum is probably this.

First convert polar to rectangular.
L1; 10 V ∠15°; x=10*Cos(15°) = +9.659 V; y=10*Sin(15°) = +2.588 V
L2; 10 V ∠70°; x=10*Cos(70°) = +3.420 V; y=10*Sin(70°) = +9.397 V
L3; 10 V ∠-40°; x=10*Cos(-40°) = +7.660 V; y=10*Sin(-40°) = -6.428 V

Then sum the x terms and y terms separately.
Sum of x, Cos terms; 9.659 + 3.420 + 7.660 = 20.739 V
Sum of y, Sin terms; 2.588 + 9.397 – 6.428 = 5.557 V

Convert sum from rectangular to polar.
Magnitude; √(20.739² + 5.557²) = 21.471 V
Angle = Atan2( 5.557, 20.739) = +15°

Vector sum; L1+L2+L3 = 21.471 V ∠+15°

yes these two methods are the long way round I described, I have evidence that there is a faster way to calculate both of these in the two other formulas

Pythagoras
[∑(cos(θₙ)⋅Lₙ)]²+[∑(sin(θₙ)⋅Lₙ)]²=Lₜ²
[cos(15)⋅10+cos(70)⋅10+cos(-40)⋅10]²+[sin(15)⋅10+sin(70)⋅10+sin(-40)⋅10]²=461.02=21.47²

inverse tan
tan⁻¹[∑(sin(θₙ)⋅Lₙ)/∑(cos(θₙ)⋅Lₙ)]=θₜ
tan⁻¹[(sin(15)⋅10+sin(70)⋅10+sin(-40)⋅10)/(cos(15)⋅10+cos(70)⋅10+cos(-40)⋅10)]=15

 

[kai sinclair]

But can you have it make sense on paper, to someone that knows the subject? Sorry, I still don't understand what your doing.

Dave, does my example in #13 help any?

 

[kai sinclair]

for those that enjoy the code version here we go

these two are mine:

$$ \sum (L_n cos(\theta_n - \theta_t)) = L_t $$

$$ \sum (\theta_n \frac{L_n}{\sum L_n}) = \theta_t $$
this one is incomplete

these two are the long way around

$$ \tan^{-1} (\frac {\sum (L_n sin(\theta_n)) }{\sum (L_n cos(\theta_n)) } ) = \theta_t $$

$$ \left( \sum (L_n sin (\theta_n)) \right)^2 + \left( \sum (L_n cos(\theta_n)) \right)^2 = L_t^2 $$

 

[Baluncore]

these two are the long way around

What you call "the long way round" is robust, and computes reliably under all circumstances.

these two are mine:
...
...
this one is incomplete

The addition and subtraction of angles is fraught with quadrant or direction problems. Those problems are eliminated by the use of rectangular coordinates, or complex numbers.

In computation, an unreliable short-cut, is not faster than the reliable long way around.

 

[kai sinclair]

What you call "the long way round" is robust, and computes reliably under all circumstances.

The addition and subtraction of angles is fraught with quadrant or direction problems. Those problems are eliminated by the use of rectangular coordinates, or complex numbers.

In computation, an unreliable short-cut, is not faster than the reliable long way around.

you see a faliure state for the magnitude? can you give an example?

 

[Baluncore]

you see a faliure state for the magnitude? can you give an example?

You need to code both algorithms, then run a million random data sets, adding two, to say ten, vectors for each trial.
If the algorithms do not closely agree numerically, I predict the problem will be with your short-cut.

 

[kai sinclair]

You need to code both algorithms, then run a million random data sets, adding two, to say ten, vectors for each trial.
If the algorithms do not closely agree numerically, I predict the problem will be with your short-cut.

well hows a dozen or so and a proof? at least for the magnitude

 

[Baluncore]

well hows a dozen or so and a proof? at least for the magnitude

You must compute $\theta_t$ before you compute magnitude.
Yet, you show the equations in the reverse order.

It is the code for your complete algorithm that I want to see.

 

[kai sinclair]

You must compute $/theta_t$ before you compute magnitude.
Yet, you show the equations in the reverse order.

It is the code for your complete algorithm that I want to see.

that's the problem I'm trying to solve, but yes I started with magnitude and I could see the relation in the graphs, but less so with the angle

 

[Ibix]

you see a faliure state for the magnitude? can you give an example?

$L_1 = 0.672$, $\theta_1 = 353^\circ$, $L_2 = 0.614$, $\theta_2=128^\circ$.

Your method gives $\theta_t$ about $-114^\circ$ instead of $54^\circ$ and $L_t$ about $-0.485$ instead of $0.495$.

There are worse cases (that one is only about 2% off on the magnitude, ignoring sign), but that one came up fairly quickly.

 

[Ibix]

Here's the distribution of differences between your method and the full calculation when you add two random vectors 100,000 times:

And if you add ten random vectors 100,000 times:

 

[kai sinclair]

$L_1 = 0.672$, $\theta_1 = 353^\circ$, $L_2 = 0.614$, $\theta_2=128^\circ$.

Your method gives $\theta_t$ about $-114^\circ$ instead of $54^\circ$ and $L_t$ about $-0.485$ instead of $0.495$.

There are worse cases, but that one came up fairly quickly.

my final angle forumla is incomplete I know that, I'm trying to find the missing piece/s for it

but lets have a look for the magnitude

so cos(353-54.28)⋅0.672+cos(128-54.28)⋅0.614=0.495

no you haven't found a failure state for my magnitude, I suspect you used the incomplete angle forumla to get the final angle which then threw off the magnitude numbers, would I be correct in this?

 

[Ibix]

my final angle forumla is incomplete I know that, I'm trying to find the missing piece/s for it

You'll need to use $\sin$ functions at least, so it'll be at most as efficient as the usual method.

I suspect you used the incomplete angle forumla to get the final angle which then threw off the magnitude numbers, would I be correct in this?

Yes. There's nothing wrong with the first formula on its own - it's just summing the projection of each of your "n" vectors on to the total vector. The problem is that you need the direction of the final vector first, and you can't get that without summing the vectors the standard way.

 

[kai sinclair]

my angle formula only works if every instance of $L_n$ is the same, because the longer $L_n$ is the more disproportionately it's pairing $\theta_n$ matters so yes my angle forumla is not finished, it can't be used outside of very specific situations, I wish to expand it's reach

 

[kai sinclair]

You'll need to use $\sin$ functions at least, so it'll be at most as efficient as the usual method.

I do suspect that sin function will be needed of the angle formula, just don't know where and if it'll be the only thing needed to be added

 

[Baluncore]

I do not think you can expect us to correct your formula, and make it reliable. If we did, then it would be slower than the standard vector addition algorithm.

CPUs now generate both the sin() and cos() of an angle at the same time in parallel. You have replaced three sin() functions with several divisions.

Vector computation is "well trodden ground". A number of clever ways to speed up other numerical vector operations have been found, and are used. If there was a way to speed vector addition, then I think it would have been found and widely publicised during the last 100 years. By all means, keep searching, and let everyone know if you find a reliable solution.

 

[kai sinclair]

okay this is what I can get one my own

$$ \left( \sum (L_n cos(\theta_n - \theta_t)) \right)^2 = \left( \sum (L_n sin (\theta_n)) \right)^2 + \left( \sum (L_n cos(\theta_n)) \right)^2 $$

$$ \left( \sum (L_n cos(\theta_n - \theta_t)) \right)^2 / \left( \sum (L_n) \right)^2 = \left( \left( \sum (L_n sin (\theta_n)) \right)^2 + \left( \sum (L_n cos(\theta_n)) \right)^2 \right) / \left( \sum (L_n) \right)^2 $$

$$ \left( \sum (cos(\theta_n) cos(\theta_t) + sin(\theta_n) sin(\theta_t)) \right)^2 = \left( \sum (sin (\theta_n)) \right)^2 + \left( \sum (cos(\theta_n)) \right)^2 $$

$$ \left( \sum (cos(\theta_n) cos(\theta_t)) + \sum (sin(\theta_n) sin(\theta_t)) \right)^2 = \left( \sum (sin (\theta_n)) \right)^2 + \left( \sum (cos(\theta_n)) \right)^2 $$

$$ \left( cos(\theta_t) \sum (cos(\theta_n)) + sin(\theta_t) \sum (sin(\theta_n)) \right)^2 = \left( \sum (sin (\theta_n)) \right)^2 + \left( \sum (cos(\theta_n)) \right)^2 $$

$$ \left( cos(\theta_t) \sum (cos(\theta_n)) \right)^2 +2 cos(\theta_t) \sum (cos(\theta_n)) sin(\theta_t) \sum (sin(\theta_n)) + \left( sin(\theta_t) \sum (sin(\theta_n)) \right)^2 = \left( \sum (sin (\theta_n)) \right)^2 + \left( \sum (cos(\theta_n)) \right)^2 $$

 

[kai sinclair]

$$ \left( \left( cos(\theta_t) \sum (cos(\theta_n)) \right)^2 +2 cos(\theta_t) \sum (cos(\theta_n)) sin(\theta_t) \sum (sin(\theta_n)) + \left( sin(\theta_t) \sum (sin(\theta_n)) \right)^2 \right) / \left( \left( \sum (sin (\theta_n)) \right)^2 + \left( \sum (cos(\theta_n)) \right)^2 \right) = \left( \left( \sum (sin (\theta_n)) \right)^2 + \left( \sum (cos(\theta_n)) \right)^2 \right) / \left( \left( \sum (sin (\theta_n)) \right)^2 + \left( \sum (cos(\theta_n)) \right)^2 \right) $$

$$ \left( \left( cos(\theta_t) \sum (cos(\theta_n)) \right)^2 \right) / \left( \left( \sum (sin (\theta_n)) \right)^2 + \left( \sum (cos(\theta_n)) \right)^2 \right) + \left( 2 cos(\theta_t) \sum (cos(\theta_n)) sin(\theta_t) \sum (sin(\theta_n)) \right) / \left( \left( \sum (sin (\theta_n)) \right)^2 + \left( \sum (cos(\theta_n)) \right)^2 \right) + \left( \left( sin(\theta_t) \sum (sin(\theta_n)) \right)^2 \right) / \left( \left( \sum (sin (\theta_n)) \right)^2 + \left( \sum (cos(\theta_n)) \right)^2 \right) = 1 $$

$$ \left( cos(\theta_t) \right)^2 / \left( \left( \sum (sin (\theta_n)) \right)^2 + 1 \right) + \left( 2 cos(\theta_t) sin(\theta_t) \right) / \left( \left( \sum (sin (\theta_n)) \right) + \left( \sum (cos(\theta_n)) \right) \right) + \left( sin(\theta_t) \right)^2 / \left( 1 + \left( \sum (cos(\theta_n)) \right)^2 \right) = 1 $$

 

[kai sinclair]

$$ \left( \left( cos(\theta_t) \sum (cos(\theta_n)) \right)^2 + 2 cos(\theta_t) \sum (cos(\theta_n)) sin(\theta_t) \sum (sin(\theta_n)) + \left( sin(\theta_t) \sum (sin(\theta_n)) \right)^2 \right) / \left( \sum (sin (\theta_n)) \sum (cos(\theta_n)) \right) = \left( \left( \sum (sin (\theta_n)) \right)^2 + \left( \sum (cos(\theta_n)) \right)^2 \right) / \left( \sum (sin (\theta_n)) \sum (cos(\theta_n)) \right) $$

$$ \left( \left( cos(\theta_t) \sum (cos(\theta_n)) \right)^2 \right) / \left( \sum (sin (\theta_n)) \sum (cos(\theta_n)) \right) + \left( 2 cos(\theta_t) \sum (cos(\theta_n)) sin(\theta_t) \sum (sin(\theta_n)) \right) / \left( \sum (sin (\theta_n)) \sum (cos(\theta_n)) \right) + \left( \left( sin(\theta_t) \sum (sin(\theta_n)) \right)^2 \right) / \left( \sum (sin (\theta_n)) \sum (cos(\theta_n)) \right) = \left( \sum (sin (\theta_n)) \right)^2 / \left( \sum (sin (\theta_n)) \sum (cos(\theta_n)) \right) + \left( \sum (cos(\theta_n)) \right)^2 / \left( \sum (sin (\theta_n)) \sum (cos(\theta_n)) \right) $$

$$ \left( cos(\theta_t) \right)^2 \sum (cot(\theta_n)) + 2 cos(\theta_t) sin(\theta_t) + \left( sin(\theta_t) \right)^2 \sum (tan(\theta_n)) = \sum (tan (\theta_n)) + \sum (cot(\theta_n)) $$

 

[kai sinclair]

found an error in my equasion, this is fixed

$$ \frac {\left( cos(\theta_t) \sum (L_n cos(\theta_n)) \right)^2 + 2 cos(\theta_t) \sum (L_n cos(\theta_n)) sin(\theta_t) \sum (L_n sin(\theta_n)) + \left( sin(\theta_t) \sum (L_n sin(\theta_n)) \right)^2} {\sum (L_n sin (\theta_n)) \sum (L_n cos(\theta_n))} = \frac {\left( \sum (L_n sin (\theta_n)) \right)^2 + \left( \sum (L_n cos(\theta_n)) \right)^2} {\sum (L_n sin (\theta_n)) \sum (L_n cos(\theta_n))} $$

$$ \frac {\left( cos(\theta_t) \sum (L_n cos(\theta_n)) \right)^2} {\sum (L_n sin (\theta_n)) \sum (L_n cos(\theta_n))} + \frac {2 cos(\theta_t) \sum (L_n cos(\theta_n)) sin(\theta_t) \sum (L_n sin(\theta_n))} {\sum (L_n sin (\theta_n)) \sum (L_n cos(\theta_n))} + \frac {\left( sin(\theta_t) \sum (L_n sin(\theta_n)) \right)^2} {\sum (L_n sin (\theta_n)) \sum (L_n cos(\theta_n))} = \frac {\left( \sum (L_n sin (\theta_n)) \right)^2} {\sum (L_n sin (\theta_n)) \sum (L_n cos(\theta_n))} + \frac {\left( \sum (L_n cos(\theta_n)) \right)^2} {\sum (L_n sin (\theta_n)) \sum (L_n cos(\theta_n))} $$

$$ cos ^2 (\theta_t) \frac {\sum L_n cos(\theta_n)} {\sum L_n sin (\theta_n)} + 2 cos(\theta_t) sin(\theta_t) + sin ^2 (\theta_t) \frac {\sum L_n sin(\theta_n)} {\sum L_n cos(\theta_n)} = \frac {\sum L_n sin (\theta_n)} {\sum L_n cos(\theta_n)} + \frac {\sum L_n cos(\theta_n)} {\sum L_n sin (\theta_n)} $$