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I have 2 position estimates (along with their measurement error) and a difference in time between estimates. I estimate velocity using

Velocity = (PosA - PosB)/DeltaT

I am trying to estimate the error in my velocity estimate, but I can't seem to find any ways to calculate this. I assume it has to use Sigma_PosA and Sigma_PosB. I would also assume it's relative to DeltaT and/or abs(PosA - PosB). What is the velocity measurement variance/standard deviation?

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What do you know about the error in the two positions? Do you have an explicit distribution, or just an error term? –  templatetypedef Aug 31 '11 at 21:23
    
Are the errors in the positions measurements uncorrelated? (If you don't understand that question, the answer is probably "yes".) –  Beta Aug 31 '11 at 21:30
    
templatetypedef: I am assuming a gaussian distribution with a standard deviation of Sigma_Pos –  user858146 Aug 31 '11 at 21:55
    
Beta: I'm not sure if they are. They are the same object, but that's the only relationship between the 2 measurements. –  user858146 Aug 31 '11 at 21:57

2 Answers 2

up vote 2 down vote accepted

sigmav = sqrt((sigmaA)2 + (sigmaB)2) / (DeltaT)

EDIT:

(Corrected an error above-- DeltaT should not be squared.)

It sounds as if the measurements are independent, so the errors are uncorrelated. We want the standard deviation of a linear combination of (two) variables:

$V = \frac{(B-A)}{\Delta_t} = \frac{1}{\Delta_t}B - \frac{1}{\Delta_t}A$

$\sigma_V^2= \sum_i^n a_i^2\sigma_i^2 = (\frac{1}{\Delta_t})^2\sigma_B^2 + (\frac{1}{\Delta_t})^2\sigma_A^2 = (\frac{1}{\Delta_t})^2(\sigma_A^2 + \sigma_B^2)$

$\sigma_V = \sqrt {(\frac{1}{\Delta_t})^2(\sigma_A^2 + \sigma_B^2)} = \frac{\sqrt{\sigma_A^2 + \sigma_B^2}}{\Delta_t}$

(I'm new to math.stackexchange-- gotta say I'm lovin' the MathJax.)

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3  
Can you explain where this comes from? –  templatetypedef Aug 31 '11 at 21:41
    
It looks like that would work, but I would like to see some explanation. –  Charles L. Sep 1 '11 at 2:18

You want the propagation of error, or propagation of uncertainty. Since you're subtracting them, the formula is

sqrt(sigma_posA * sigma_posA + sigma_posB * sigma_posB)

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Technically, you need to add (inside the square root) 2 * rho * sigma_posA * sigma_posB where rho is the correlation of the errors. Also, this only gives one standard deviation of the error (this is what the OP asked for explicitly). This may (and will likely) be an underestimate of the actual error, given a reasonable tolerance (say 10^-2). –  Alexandre C. Aug 31 '11 at 21:33
2  
I don't think this would work. I know that if the 2 measurements are far apart (in time/space) then the velocity estimate improves. –  Charles L. Aug 31 '11 at 21:50

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