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Measuring two angles of a triangle as a substitute for the third.
Posted by rfc on April 3, 2016 at 3:39 pmWe know that if you wanted to know angle A, you could make n number of observations (D+R), put them into Star*net (or just do the math), and come up with a 95% level of confidence for the observation.
But you can’t access that point, so elect to occupy both B and C, and derive A accordingly (A=180(B+C)). How many observations of B and C would you have to make to attain the same statistical reliability compared to just occupying A? 2n for each?
conrad replied 8 years, 4 months ago 12 Members · 21 Replies 
21 Replies

Starnet will answer that for you if you run some examples. Off the top of my head Id guess you need twice the usul number of observations at each of the observed points to achieve the usual confidrnce at the third point.
Strength of figure is worthy of study but doesnt directly answer the question asked. It applies more to positional uncertainty given angle uncertainties.
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Offhand, it would seem that A depends on the sum, B + C, so it looks like the variance of A would be the variance of B + the variance of C and the standard deviation of A would be SQRT(variance of B + variance of C).
For up to date insight on number of observations, look at Dr. Ghilani’s article here: http://bt.editionsbyfry.com/publication/?i=294780

(any triangle has 6 elements. So, if know any 3 you can calculate the remaining 3)
Apparently you can measure 5 elements. The distances to A are just as important. Also, the reliability of the angle at A is dependent on the 2 measured angles, in both estimated error and the anglular value. The smaller that the measured angles are, (the larger angle at A large), the greater the uncertainty.
So, an isosceles triangle, 2 measured angles, and 3 measured distances will return highest estimated accuracy at A.
The err of the unknown angle would be (roughly, not textbook, thumb rule) mean err of the measured angles times Â‰Ã¶?2, and increase as sine of the measured angles approach 0 and 1. However, the distances will effect the resulting estimated accuracy. Short distances std err 0.02′ , angles 5″ std err, near isosceles triangle, the missing angle won’t change things much. Just another degree of freedom.
And distance BC adjusted estimated err, overall network estimated error will increase or reduce the uncertainty at A as well. Barring blunder, the uncertainty at A will be marginally larger absent angle A. If the network has high noise, the uncertainty climbs, and the partial lack of redundancy at A will be proportionately larger.

It depends on the number of decimals carried in your logarithm tables or the modern equivalent. Rounding does make a difference. Been playing with numbers for about 55 years and still waiting to get the same number twice.

I carefully measured a handful of angles. And then remeasured them a good while later.
Yup, they’re different – every time.
But, if I round off to the nearest 0.25″, they’re pretty much the same. And of course that’s distance to target sensitive.
So, shoot for a standard err. And it might take that 4, 8, …, n, sets. Depending on the task at hand and gear available, and std err required.

Larry Scott, post: 365372, member: 8766 wrote: (any triangle has 6 elements. So, if know any 3 you can calculate the remaining 3)
Apparently you can measure 5 elements. The distances to A are just as important. Also, the reliability of the angle at A is dependent on the 2 measured angles, in both estimated error and the angular value. The smaller that the measured angles are, (the larger angle at A large), the greater the uncertainty.
Not so, in this exercise. I cannot measure the distance to A; No reflectorless here. I’m really only interested in the angles here anyway. Assuming the triangle is roughly equilateral (isoceles) then, the standard deviation of A would be the square root of 2 times the variance of A, which I think is exactly what you’ve stated.
If Math Teacher’s and your formula is correct then, to achieve the same precision I would need to observe angles B and C each 1.414 times the number of observations at A if I could get to it. Approximately 50% more observations at each of B and C. I guess if I had the time, I could set up a triangle all three of whose points I could occupy, then measure one angle only, then the other two only and sequentially add them to an adjustment until the numbers match, but there are only so many hours in a day.
BTW I did like Holy Cow’s suggestion to read more on Strength of Figure, There’s not much in Ghilani and Wolf on it, but it bears directly on using trilateration and triangulation together in a network (which is what I’m doing). I’ve ordered the book you suggested on Surveying India and the Himalayans by Keay…I’m certain there will be some good stuff there.

rfc, post: 365402, member: 8882 wrote: need to observe angles B and C each 1.414 times the number of observations at A
No you would need twice the observations to reduce the rms by 1/sqrt(2)
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Bill93, post: 365368, member: 87 wrote: Starnet will answer that for you if you run some examples. Off the top of my head Id guess you need twice the usul number of observations at each of the observed points to achieve the usual confidrnce at the third point.
Strength of figure is worthy of study but doesnt directly answer the question asked. It applies more to positional uncertainty given angle uncertainties.
Star*net is a fabulous planning tool. Using that tool, together with eliminating pointing and recording errors, instrument specifications and calibration… and then you will have your answer.

The strength of reliability would definitely depend upon the ratio of the length of C to B in relation to the distances from A to B and A to C.
For those important angles, my standard is 6 wraps and use the average.
Another way arrive at a suitable solution would be to sideshot point A from as many setups as possible and mean the coordinates.
o.O

Given the figure with angles A, B, C with distances a, b, c and A not accessible in all probability you will not have a precise target at A and also distances b & c cannot be measured. Therefore you have much more than a combined angular error to factor in to your precise calculation of location of A. You also have the setup error at points B & C and the error in the distance measurement from B to C, i.e. a plus the estimated error in your target point.
I estimate that to equal the precision one would have in being able to determine A, B, C by occupying A you would need to at least observe A from 3 points, B, C & D such that A is within the BCD polygon.
I had a 19 acre farm survey I did 6 years ago with a seven point traverse around the perimeter, better described as a 6 point traverse with the 7th point being an alternate traverse point between 2 & 3 tieing in TP 4. From TP 6 I closed on TP 1 within 0.006′ and 2″ of angle. I never bothered to add the reobservation from 1 back to 6 into my data collector, although it is written into my notes. That means I never saw fit to need or do an LS adjustment on the project. From 5 of the 7 TPs I recorded a D&R angles only observation to the pivot bolt on a wind mill on this farm. I also observed it from 4 external traverse points. I estimate the pivot bolt to be 0.15′ diameter and calculated a wind mill bolt coordinate such that all observations were within an 0.021′ diameter circle about those coordinates. I considered that good enough to use the wind mill as the backsight for setting property corners from several of the TPs. From the nearest TP at 203′ the max spread on windmill observations was 0.04′ or 40″ of angle. For an 804′ backsight to a 57′ corner stakeout point there was nothing measurable to quibble about.
For those obsessed with LS I can post my real world raw traverse data.
Paul in PA

adding to Bill93’s post, my favorite surveying equation: sigma over square root of n, where sigma is the std. deviation of your measurement method, be it angles or distances, measured with tape or EDM or banjos.
how well can you measure it? how good does it need to be? turn the crank until you know how much redundancy you need.
One second results with a 5 second instrument? 5/sqrt(n)=1, solve for n. Oversimplified, yes, and doesn’t account for centering errors etc. etc.
Yet probably the simplest network design “what if” available.Another illustration of this is in how EDMs work. One EDM flash gets you a measurement to a couple of tenths of a foot. So EDMs shoot the distance a few thousand times for every push of the “measure” button and use least squares internally to mean it out. [0.20′ / sqrt(400) = 0.01′ not accounting for other factors].
“strength of figure” was a big deal back in the day of expanded base nets that had one baseline in the center and were expanded outwards by many many sets of angles. Also had to do with lack of computing power at the time: Far fewer unknowns when calculating the network with condition equations (few distances, many closed figures, no skinny triangles) rather than observation equations (every observation used to calculate coordinates, statistics performed on coordinates, residuals calculated for observations). Starnet and most everything else nowadays use observation equations. It will not care how you think of the triangles because in its mind, everything is a coordinate.
OK lastly, if you can’t shoot a distance to the point you’re trying to intersect, make sure you get good redundant distances between the two points you can see it from, and if your angles are greater than 30 degrees and less than 150 it ought to work fine. Raise and lower the tripod, turn the tribrachs, etc. The beauty of observation equations is that redundancy is redundancy and it matters less what kind of measurements they are, as long as you (or the software) can calculate weighted mean coordinates from the data.

It appears this thread is about sideshots to observable unoccupiable targets adjacent to a traverse using terrestrial (optical) gear, which may improve the traverse. Yes, it will tighten up a traverse especially where the unoccupiable point is visible from many traverse stations and is stable. I assume it would tighten up a control triangulation network, even a GPS survey. But only when distance observations are considerable.
The problem is targetry and atmospherics. You would think some structure when observing a few miles away is a good target, but shooting an antenna which is 5 miles away, swaying and will be elsewhere 2 hours later doesn’t cut it. Wind, time of day (concerning distortion of the structure) complicates things. Atmospherics on a particular day can distort observations.
I did a valley survey with unlimited funds and we built plywood 8’x16′ targets that could be observed from 15 miles in daylight good conditions. The target guys were there on NGS stations to make sure the target was at right angles to the sun and right size to the observer to eliminate phase phase error. It did come out too good because the targets were 5000′ above the survey.
Obviously, a waste of time; nighttime terrestrial lamp target observations are not affected by phase shift (they are like stars), but our union said we couldn’t work at night. Now GPS is much more accurate at + 20 miles or more so it’s no longer an issue. Gone are the long distance surveys using terrestrial techniques.

half bubble, post: 365427, member: 175 wrote: adding to Bill93’s post, my favorite surveying equation: sigma over square root of n, where sigma is the std. deviation of your measurement method, be it angles or distances, measured with tape or EDM or banjos.
What’s a banjo? Put ’em end to end? Wouldn’t you have to calibrate them to get good results? Up hills do you just do slope banjoing or use one of the strings as a plumb bob to the previous point. Wouldn’t two banjos make the work go faster than one? I can’t find the syntax in the Starnet manual for Banjo Distance.:S

hello half bubble,
half bubble, post: 365427, member: 175 wrote: adding to Bill93’s post, my favorite surveying equation: sigma over square root of n,
i think this Â‰Ã¶?n thing gets overdone in surveying a bit, or applied with little thought to the types of errors it applies, and to what quantity it applies in the end.
One second results with a 5 second instrument? 5/sqrt(n)=1, solve for n. Oversimplified, yes, and doesn’t account for centering errors etc. etc.
according to my work, it’s not oversimplified, it’s just not right at all. a 5″ instrument is 5″ for a reason, and it’s not for lack of angles turned.
if i remember correctly you use a leica 1200. if it’s a 5″ gun you could set up and measure an angle until the end of time or until the bearings wore out and you could still be up to 8″ from the true angle. with this particular instrument you aren’t much closer to the true angle after your 5000th arc than you were after the first.

The assumption in the sqrt fpormula is that the errors are independent. If you measure the angle many times with the same centering error and using the same part of thr instrument circle then it doesnt apply. Of you rotate and recenter the instrument for each set then it may be valid
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This one aspect of the dial in scales of pre electronic days that is far too less appreciated than the instruments deserve.
There was the day when all the numbers after the decimal came together better by gathering your karma and experience into being one with the gun.
I traded in my microscopic eyes for computer vision since those days with the machine doing the mojo and the operator laboring away with a shovel and a target being cared for and used correctly.
Perhaps every survey gathering should have a manual instrument set up for everyone to log a reading.
:plumbbob: 
5000 angles at one setup doesn’t help. 5000 crosstied measurements in a least squares network where everything has some redundancy won’t make the individual angles “better” either. What will be better is the error ellipses of the points observed. Observation equations again. We will never know the true angle or the true std. error so we must make do with the standard deviations and whatever statistical bets we can develop from those, 95% confidence etc.
Maybe you could explain your perspective a bit?

hello half bubble,
half bubble, post: 365427, member: 175 wrote: Maybe you could explain your perspective a bit?
yes, my post was to do with the Â‰Ã¶?n rule and this comment:
One second results with a 5 second instrument? 5/sqrt(n)=1, solve for n. Oversimplified, yes, and doesn’t account for centering errors etc. etc.
it’s not oversimplified, it just isn’t right. the angle from your 5″ instrument is not 1″ quality because of biases in direction readings that are likely fixed and repeating. whether it’s injected on purpose, or something left uncorrected, it doesn’t submit to the Â‰Ã¶?n rule because the likely biases are constant and don’t cancel. that formula omits those biases. you need to employ measuring techniques to try to cancel these biases otherwise they persist in the same magnitude after 1, 5 or 5000 rounds.
We will never know the true angle or the true std. error so we must make do with the standard deviations and whatever statistical bets we can develop from those, 95% confidence etc.
the trick is to know how to model the errors to get a representative statistical picture. my point is just that the Â‰Ã¶?n rule when applied to a bunch of repeated EDM or angle observations is incomplete. hope this clarifies my earlier post.

The distance to “A” is as important and will be the dependent variable on how many sets should be turned to get the error ellipse to a real and workable number. 50′, maybe two doubles. 2000′ feet, maybe several sets of doubles. Add some elevation and pressure differences in and I may turn as many as 30 sets.
Once it’s intersected, then you can evaluate the differences in the math vs. what was calculated and maybe you have an answer.
If the angle of “A” is very small, like less than say 10Ã¥Â¡, then maybe you need another point somewhere else do work another triangle.
Triangulation can be a pain. If done correctly though, very powerful.
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