Recent content by kai sinclair

yes, the Weighted Average kind of works on small scales, low difference in ##\theta_n## mainly, and low ish difference in ##M_n##, if you add in a Bias, ##X_n##, then it corrects itself my current goal is to calculate ##X_n## just from the given initial inputs of ##\theta_n## and ##M_n##

true, but as you can see there's less Trig involved which has two advantages, easier to calculate, even for computers and because it has less Trig, and importantly no Trig powers the error rate is a lot further down in the decimals, not usually a concern in most cases but if you need high...

just a note one this one, do you know of this formula for it? $$\sum_{n=1}^N {M_n \cdot \cos(\theta_n - \theta_t)} = M_t$$

can you show me how that would work with this given example ##V_1## = 10 units at +15° from reference ##V_2## = 70 units at +90° from reference

so is what your talking about this formula? $$ \tan^{-1} \left( \frac { \sum_{n=1}^N {M_n \cdot \sin( \theta_n)} } {\sum_{n=1}^N {M_n \cdot \cos( \theta_n)} } \right)$$

I am unaware of this method, please elaborate So you think this is going to get needlessly complicated?

lets take a crack at this again this is the initial form of my Formula, and does not get the right answer only gets into the ball park $$\frac { \sum_{n=1}^N {\theta_n \cdot M_n} } { \sum_{n=1}^N {M_n} } \approx \theta_t$$ so first find a Corrective Power to this using $$\frac { \ln \left(...

I understand that the industry standard is the following formula, it's just my pattern recognition is saying there's a better way $$ \tan^{-1} \left( \frac { \sum_{n=1}^N {M_n \cdot \sin( \theta_n)} } {\sum_{n=1}^N {M_n \cdot \cos( \theta_n)} } \right)$$ so the main thing I'm trying to do at...

I would love to show you my work step by step but LaTeX is being annoying, but in short how to I calculate ##X_n## from only ##\theta_n## and ##M_n##

so I've made some progress to this endeavour and come up with this to calculate Vector Addition; a weight averaged with a bias $$\frac { \sum_{n=1}^{N} \theta_{n} \cdot M_{n}{}^{X_n}} {\sum_{n=1}^{N} M_{n}{}^{X_n}}=\theta_{t}$$ I do have a round about way to figure out the Bias, ##X_n##, but...

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...

$$ \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))...

$$ \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) =...

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...

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