I came across the following problem, while doing my last minute studying today. Can anyone explain the basics for how to determine a standard error ellipse for a total station measurement? I've taken two statistics classes in college, but that was quite a while ago.
Thanks Again! Jeff
> I came across the following problem, while doing my last minute studying today. Can anyone explain the basics for how to determine a standard error ellipse for a total station measurement? I've taken two statistics classes in college, but that was quite a while ago.
> Thanks Again! Jeff
>
>
>
Can't read the actual problem very well, but I'll assume that the s.e. of angle measurement is +/-20" and the s.e. of range measurement is +/-(0.02 + 5ppm x D) and that D = AC = 2000.07 ft. The semi-axis of the 95% confidence error ellipse perpendicular to the line from the instrument to the object will be 2.47 x sin 0-00-20 x D. If AC = 2000.07 ft., then the semi-axis is 2.47 x 2000.07 x sin 20" = 0.48 ft.
The semi-axis of the 95% confidence error ellipse from uncertainties in the distance measurement is 2.47 x [0.02 + (0.000005 x 2000.07)] = 0.07 ft.
So, the larger value, 0.48 ft. is the semi-major axis and 0.07 ft. is the semi-minor axis of the 95% confidence error ellipse. Naturally, when sketching this situation up, the semi-major axis will be perpendicular to A-C and the semi-minor will be parallel to A-C.
Kent - Error Analysis Problem
What is the 2.47 multiplier? I would have thought, off the top of my head, you would multiply the standard error times 2 (or 1.96) for the 95% confidence level? I haven't studied this in some time, and must be wrong, but that is what I would have guessed.
Error Analysis Problem - CORRECTION
> What is the 2.47 multiplier? I would have thought, off the top of my head, you would multiply the standard error times 2 (or 1.96) for the 95% confidence level?
Actually, that multiplier should have been 2.447 for the 95%-confidence error ellipse. You'll find a discussion on Page 33 of the 6th Edition of "Surveying Theory and Practice" by Davis, Foote, Anderson, and Mikhail.
On rereading the fine print of the problem, I see that the test was asking about the STANDARD ERROR ellipse, not the 95%-confidence error ellipse. For the former, delete the multiplier of 2.447 (2.47 in my post).
Error Analysis Problem - CORRECTION
Hi Kent,
Thanks for the response and the reference.
Jeff
Error Analysis Problem - CORRECTION
If you'll post a legible scan of the question, I'm sure that Tom Adams or I will do a better job of working the problem as posed. Sorry for the initial mix-up on my first iteration.
Error Analysis Problem - CORRECTION
Not sure if Jeff still needs the help but I had a copy the pdf of the past exam problem with better quality. 
Kind of interested in the answer myself. Rob
Error Analysis Problem - CORRECTION
Okay, so:
A-C = 2000.00 ft.(+/- 0.03 ft.*)
* s.e. A-C = +/-(0.02 ft. + 0.000005 x 2000.00) = +/-0.03
and CAB + 40°00'00"(+/-20" s.e.)
At a distance of 2000.00 ft. an angular uncertainty of 0°00'20" propagates into an uncertainty of +/-[2000.00 x sin (0°00'20")] = +/- 0.194 ft.
In other words, the position is estimated from the angle measurement to be 0.000 ft. from a line running N50°00'00"E from A, but with an uncertainty of +/- 0.194 ft. s.e. So, there is an approximately 68% chance that the point is actually within 0.194 ft. of that line.
The uncertainty in the angle measurements acts in a direction perpenticular to the line of sight and the distance uncertainty in a direction parallel with the line of sight. That distance uncertainty is +/- 0.03 ft. s.e. (as computed above).
To sketch the standard error ellipse, you would lay out the axis perpendicular to the line of sight (an axis oriented N40°W). The axis would have a magnitude of 0.194 ft. either way along the axis from the nominal estimate of the point positioned bearing exactly N50°00'00"E from A.
The standard error of the distance is much smaller (+/-0.03 ft.), so the major axis is that due to uncertainties in the angle measurement and the minor axis corresponds with the uncertainty in the distance measurement. The length of the semi-major axis is 0.194 ft. and that of the semi-minor axis is 0.03 ft.
To draw the standard error ellipse, you'd sketch an ellipse with a long axis oriented N50°W 2 x 0.194 ft. long and a short axis oriented N40°E, 2 x 0.03 ft. long.
The chance that the point positioned by those measurements with the stated uncertainties falls somewhere with that ellipse is 68%.
If you want to know what the 95% error ellipse looks like, you have to multiply the lengths of both axes by a factor of 2.447.
If you want to decrease the uncertainty in the position that results from the uncertainties in the angle measured to it, then repeated measurements of the angle will usually do that. The standard error of an angle taken as the mean of n repeated measurements is s/SQRT(n), where s = s.e. of one angle.
So for:
[pre]
n s.e.
1 0°00'20"
2 0°00'14"
3 0°00'12"
[/pre]
So, you'd need to take the angle as the mean of 3 repeated measurements to get the s.e. of the semi-major axis below 0.12 ft. on the standard error ellipse.
The last part of the problem isn't asked very well since it doesn't give the accuracy of the theodolite and EDM. A compass theodolite has a nominal uncertainty of a minute or two and just about any American-made transit can measure angles by repetition much, much better than that.
Error Analysis Problem - CORRECTION
:good:
Sorry about mentioning #2" multiplier confusion. I could have looked it up before I posted.
Error Analysis Problem - CORRECTION
Hi Keith,
Thanks for the explanation. The only part that I don't fully understand is how you determine the 2.447 multiplier. I always (incorrectly) thought that one should use 1.96. I looked it up in a t-table and at the 95% confidence level, 2.447 corresponds to 6 degrees of freedom. How do you come up with 6 degrees of freedom?
Thanks,
Jeff
Error Analysis Problem - CORRECTION
> Thanks for the explanation. The only part that I don't fully understand is how you determine the 2.447 multiplier. I always (incorrectly) thought that one should use 1.96. I looked it up in a t-table and at the 95% confidence level, 2.447 corresponds to 6 degrees of freedom. How do you come up with 6 degrees of freedom?
You wouldn't use a t-table for the scale factor to apply to form the 95%-confidence error ellipse from the standard error ellipse because the ellipse is a function of the squares of two normally distributed quantities, the errors attributable to distance measurement errors and the errors due to angle measurement errors. The sum of squares of normally distributed random variables follows a different distribution.
Error Analysis Problem - CORRECTION
:> Hi Keith,
>
ER........he might respond better to "Kent" :'(
Error Analysis Problem - CORRECTION
> :> Hi Keith,
> >
>
> ER........he might respond better to "Kent" :'(
Oops. Sorry, Kent.
Error Analysis Problem - CORRECTION
> > :> Hi Keith,
> > ER........he might respond better to "Kent" :'(
>
> Oops. Sorry, Kent.
Ya, right. Just remember that you CAN sell what you don't own. Counter-intuitive, but perfectly legal. Except where prohibited by law, of course.
Error Analysis Problem - CORRECTION
> Oops. Sorry, Kent.
I'm sure he's used to it. I get the two of them mixed up all the time. 😉
Error Analysis Problem - CORRECTION
😉
