Weird distance estimate? In a book about radar signal processing, they are estimating the squared distance between two objects as:
$$d^2 = \frac{(R_p-R_o)^2}{\sigma^2_R} + \frac{(\theta_p-\theta_o)^2}{\sigma^2_\theta}$$
where R is a radius, $\theta$ is an angle and $\sigma^2$ is the variance of the radius and angle respectivly. p is the predicted target and o is an observation. The book is Blackman, S. - Design and Analysis of Modern Tracking Systems
Is it just me or is the dimensional analysis for this equation a bit off?
 A: The formula is not even close.  You are right that d should have dimensions of length and the right hand side is unitless.  Worse, the formula doesn't reduce properly if the errors are zero.  Are you sure it isn't $d^2/\sigma(d)^2$ on the left?  That would fix the units.  The cosine formula gives the distance estimate you want: $d^2=R_p^2+R_o^2-2 R_p R_o \cos(\theta_p-\theta_o)$.  When you mix in sigmas you are trying to estimate the error in d based on the errors in the measurements.  So you should use the formulas you have for combining measurement errors to give the error in the measured quantity.
A: I think whuber's comment is on the spot. That formula is surely not refering to a "physical distance", but it's simply a general measure of the distance (as "dissimilarity") between two multidimensional (general) $features$. 
A classical example is statistical classification; for example when applying the nearest-neighbour rule, we must compute the distance from our observed feature (say ${\mathbf x} = [x_1 x_2]$) against some reference values (say ${\mathbf x^A} = [x_1^A x_2^A]$) . 
Here the components $x_1 x_2$ can be any measurements, often with different dimensions (eg, weight and length f some object). 
A possible way to measure the (square) distance is to compute $d^2 = (x_1 - x_1^A)^2+(x_2 - x_2^A)^2$ , but this would be dimensionally inconsistent, and hence sensitive to the scale. A more reasonable recipe is to normalize each component dividing by some "characteristic" value (eg: the standard deviation), so all components are now adimensional and approximately even distributed: 
$$ d^2 = \frac{(x_1 - x_1^A)^2}{\sigma_1^2}+\frac{(x_2 - x_2^A)^2}{\sigma_2^2}$$ 
Of course, the distance obtained by this feature normalization is adimensional, it's only meaningful when compared with other distances.
