BTW I just noticed a typo. Instead of :
D = SQRT{[ n(1)- n(2)]^2 + [ e(1)+ e(2) ]^2 },
make that:
D = SQRT{[ n(1)- n(2)]^2 + [ e(1)- e(2) ]^2 }
I'm sure that anyone working on this problem wasn't thrown off track by that detail, but for the record there it is.
Kent
> Now where did I say that? Nowhere. No, what I said was, the Delphi method incorporates both quantifiable (numbers) and qualifiable (experience) into the mix. Refusing to look at one, well that's a little myopic.
Kris, just think about what you're saying for a few seconds. The thrust of what you're wanting to argue is that you somehow "disagree" with numerical analysis because of some experience you had. Well, the only sort of experience you could have had that would cast the problem as I've posed it in any doubt is if you had an experience that somehow convinced you that random RTK GPS errors aren't normally distributed. Now you say that you don't dispute that. So, you're left with basically ... nothing, right?
The way that the RTK enthusiasts are running for the hills on this little problem is eyebrow raising, to say the least. Do you somehow suspect that the answer is going to be one that you don't want to know about?
You're likely to repeat within .04' but both measurements can be out 8' or more.
Interesting, and reasonable discussion overall, but I wouldn't agree with this statement.
If you initialize twice and occupy, I'd bet a month's paycheck you'll never initialize incorrectly the exact same way and have the two be within .04' of each other. Just doesn't happen.
> You're likely to repeat within .04' but both measurements can be out 8' or more.
It has nothing to do with incorrect initialization. It has to do with local multipath. And yes, I'll bet ya a month's pay that I can prove you can have to measurements within .04 of each other and still be many feet away from the true value.
Is it a bet? Shall we say, $1000?
> Maybe running from the same point being made over and over with nothing new, and with a side dose of obtuseness we don't need or deserve.
Well, why do you suppose that no one has given the answer, then, if this isn't actually something novel that they haven't thought carefully about? The poster "ashton" got close with a viable method of solution, but didn't recognize the answer in the data produced by the Monte Carlo method he employed.
This particular problem as posed assumed that the coordinates of the marker as found and of the new replacement as set had random, normally distributed errors with a standard error of +/-0.02 ft. The nature of the problem is such that if you substitute a standard error of +/-0.015 ft. or +/-0.03 ft., the most probable error of the replaced mark, the expected error, is just decreased or increased in the same ratio as the new s.e. to 0.02 ft. The fact that the most probable error isn't zero and is somewhat larger than most RTK users would expect doesn't go away.
Mmmm, modeled multipath. I succumb, and I've done it in a lab type of environment. However, I don't think that's what we're talking about here because if you're trying to use GPS in an environment where multi-path is that clean and consistent, you shouldn't be using GPS in the first place.
> It has nothing to do with incorrect initialization. It has to do with local multipath. And yes, I'll bet ya a month's pay that I can prove you can have to measurements within .04 of each other and still be many feet away from the true value.
>
> Is it a bet? Shall we say, $1000?
There is something that bothers me about the original statement of the problem:
"What is the most probable error that you will make in remarking the corner? That is, what is the most likely distance that the new marker you set with, say, network RTK that can measure coordinates with standard errors of, say, +/-0.02 ft. will be from the original position of the marker (of which no trace remains other than the previously determined coordinates)?"
In pure mathematics, all the distances will be real numbers with an infinite number of digits after the decimal, so the probability of any distance exactly equaling any value that might be proposed as the answer is 0. Or if the proposed answer is chosen to match one of the distances, there will only be one value that exactly matches.
Of course, real GPS units will have limited bits to store and report results, so the probability of an answer matching more than one distance becomes non-zero, but I still don't think that is what was intended. Perhaps the mean distance is really what Kent wants?
> It has nothing to do with incorrect initialization. It has to do with local multipath. And yes, I'll bet ya a month's pay that I can prove you can have to measurements within .04 of each other and still be many feet away from the true value.
>
> Is it a bet? Shall we say, $1000?
Maybe I'm wrong, and am willing to learn something here.
Are you saying you have taken a "fixed" shot, flipped the head over (lose initialization), regain "fix", and then got the same bad reading twice?
This has never happened to me.
> In pure mathematics, all the distances will be real numbers with an infinite number of digits after the decimal, so the probability of any distance exactly equaling any value that might be proposed as the answer is 0. Or if the proposed answer is chosen to match one of the distances, there will only be one value that exactly matches.
In this case, the most probable or expected value is the value at the peak in the distribution of the errors, just as the most probable or expected value of a normally distributed set of errors with zero mean is the value at the peak of that distribution. The outcomes cluster around that probable value.
> Get some hands on exprieence and frame the conversation as how-to-improve instead of how to slam or just win and argument, and they maybe more pople will be ready to play.
Okay, I'll mark you down as not interested in the topic from the standpoint of error analysis and we'll carry on. The idea that we can't answer the problem as posed without holding a network RTK rover is ridiculous on its face, as will be seen.
OK, I recognize "expected value" as a probability term of art with a definite definition. For nicely shaped distributions, it would coincide with the peak of the distributions. "Most probable value" is not a term of art that I recall or can find in handy texts; unless it is a term used in textbooks I haven't encountered, one could quibble about what it means.
As a practical matter, 0.003 ft., (nominally 1mm) is the least significant unit of distance for nearly all purposes of land surveying (meaning boundary surveying). A distance such as 125.31415978 ft. has no meaning beyond possibly the first three places in the decimal, so there's not much point in dealing with a continuous probability distribution anyway. A discrete distribution with 0.003 ft. steps is for all practical purposes more than sufficient.
I made the calculation using 1,000 trials from random values with a Gaussian distribution (mean = 0, sigma = 0.02 ft.) and used the results to construct a discrete distribution with 0.003 ft. steps.
How do you know you've never got the same bad reading twice?
What is you email Mr. Schrock? I have something for you.
