You could still generate a gaussian distribution from the radial value. You would merely need to run my test mulitple times to determine the standard deviation. You don't necessarily have to have negative values. If the true standard error for setting up a tripod/prism combo is 0.5mm then you could observe a range of perhaps 0-1mm of error.
Also, I really do not care for any test that requires rotating the prism pole (or whatever is being observed). In my mind rotating the pole and only observing one component of the error at time relies too much on the second component cancelling out from the rotation. The only reason I don't care for it is because we are straining at gnats anyway talking about 0.5mm.
There is arc of complexity. Once you have removed the obvious sources of error, and start operating at the limits of the equipment, eliminating each remaining 0.5mm becomes exponentially harder that the previous one.
> You could still generate a gaussian distribution from the radial value.
No, that's the point. The radial centering error won't have a Gaussian distribution. It will have a Chi Square distribution.
You guys are nuts.
> Doesn't the centering error have 2 components, one that is proportional to the target height and one that is not? Could these be measured separately?
The centering error does increase as target height does. The target height was nominally 5 ft. in the test described above, which is the usual target height I set the prism poles to.
Testing Centering Accuracy (Example from Today)
> I have never had a rod that I would trust for anything sub hundredth, and most rods I've used if I got down to a hundredth I still didn't quite believe it.
Well, as the test and photos show, that is a pretty ordinary 15-year-old SECO prism pole and it has a centering accuracy of about 10 times better than 0.01 ft. The main criteria are (a) the sensitivity of the level vial and (b) the fact that the pole can be rotated in the prism pole tripod to check the level in more than one position.
Testing Centering Accuracy (Example from Today)
I got it, I didn't realize that you were moving the pole without moving the tripod. That makes good sense. I'll have to get one of those, you can't really do that with a bipod and be 100% sure that you didn't bump it some.
I have had better results with the poles that have more sensitive bubbles. Invariably, though, someone will come along and over-tighten the bipod or data collector clamp and distort the rod. It doesn't take much over-tightening to bend/crush the rod a mostly imperceptible amount.
I seem to remember from probability and statistics that the bell curve does not have to be centered at 0, though. It can be centered at the true value.
> I seem to remember from probability and statistics that the bell curve does not have to be centered at 0, though. It can be centered at the true value.
We're discussing centering errors, I believe. :> I think that Bill will confirm that the distribution of radial errors is not Gaussian, but follows the Chi Square distribution.
LOL
I hadn't yet Googled polar error in gaussian distribution. A quick search revealed this:
It seems that there is a mathematical model for handling polar (radial) errors in a gaussian distribution. In our scenario the angular component of the polar error is not important, because it is random.
I know that you didn't mean to imply this, but another reason I don't like to discuss centering error as perpendicular and parallel to line of sight is because some people will mistakenly infer that there is an element of systematic error to centering error.
There WOULD be an element, if you had a plummet that was out of adjustment and oriented it towards the instrument the same way each time, but it seems that the rotational orientation of the tribrach is by and large random.
> It seems that there is a mathematical model for handling polar (radial) errors in a gaussian distribution.
Except that the paper you linked gave a distribution for radial errors that wasn't Gaussian. Only the orthogonal components were.
Oops - I didn't state that very well (late at night). I was thinking about how you get the same sigma for in-line and sideways error when viewed from any direction. Some people have argued with me in the past that if you measure some value for each component, then they would expect when you measure components at 45 degrees you would get a bigger number. That doesn't happen if the components are independent.
I shouldn't have implied the radius value had the same statistics. Yes, the radial error is the square root of a Chi-squared n=2, which is also called Rayleigh distribution. You use the single-component sigma when looking up the Rayleigh distribution values, and I believe Star*Net expects that sigma value.
Well kiss my grits 🙂
Under your scenario, though, you couldn't have a Gaussian distribution for distance measurement, either, because you never record negative distances...
> I shouldn't have implied the radius value had the same statistics. Yes, the radial error is the square root of a Chi-squared n=2, which is also called Rayleigh distribution. You use the single-component sigma when looking up the Rayleigh distribution values, and I believe Star*Net expects that sigma value.
Thanks. That's exactly what I had thought to be the case.
Actually I think I see the light now. After searching for and skimming several articles, I believe it would be improper to use the radius value in a normal distribution because the formula for radius has the result already elevated to an exponent:
((X^2)+(Y^2))^.5 = Radius
So you are sort of suppressing some of the variability of the values if you do not use the Rayleigh distribution.
Am I on the right track?