After spending probably more time making suggestions to a poster who was working toward determining the standard errors of the EDM in his total station than it would take to actually do the work in the first place, I spent an hour and a half this afternoon on a demonstration using my 1995-vintage Zeiss Elta50R.
This is a total station with an EDM that has a basic measuring length of 10m.
I've previously described the test range, but here it is again. This is specifically for an instrument with a 10m measuring length, so if your instrument has a different basic measuring length, you'll want to use a different set of distances.
I laid out the range by just taping along a sidewalk, making marks at nominally:
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4 0.0m
3 2.5m
8 16.25m
7 18.75m
6 21.25m
5 23.75m
2 35.00m
1 37.50m
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The actual marks were all just a "+" in white chalk on the concrete sidewalk. No great accuracy better than a few centimeters for the layout was necessary. I used the joint in the sidewalk for alignment.
Then, I set up prisms at Stations 4, 3, 2, and 1 and proceeded to occupy stations 5, 6, 7, and 8, measuring ranges and angles, zenith and horizontal, at each to the four prisms at stations 1, 2, 3, and 4.
At each station, four ranges were measured to each prism and gave a pooled estimate of about +/-1.3mm for the apparent standard error of a single range. In other words the scatter of the repeat ranges was such that there was on average a 68% chance that a single range measurement was within 1.3mm of the mean of a much longer series of range measurements to the same prism with the same instrument from the same station
All of the stations were nominally on the same line or within a few centimeters of it. That simplified calculations.
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So, 5-1 = 13.7498m
5-4 = 23.7916
-------------
1-4 = 37.5414m calc
6-1 = 16.2552
6-4 = 21.2888
-------------
1-4 = 37.5440m calc
7-1 = 18.7572
7-4 = 18.7867
-------------
1-4 + 37.5439m calc
8-1 = 21.2673
8-4 = 16.2772
-------------
1-4 = 37.5445m calc
5-2 = 11.2482m
5-3 = 21.2807
-------------
2-3 = 32.5289m calc
6-2 = 13.7500m
6-3 = 18.7763
-------------
2-3 = 32.5263m calc
7-2 = 16.2515m
7-3 = 16.2752
-------------
2-3 = 32.5267m calc
8-2 = 18.7628m
9-3 = 13.7645
-------------
2-3 = 32.5273m calc
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So, how would one work out the cyclic errors of the EDM from the above data?
Hello Kent,
I'm going to wing-it without referring to texts because I like to try to work things out myself first.
What you seem to end up with is a series of distance for 1-4 and 2-3 that, either due to being exactly out of phase with the unit length or in phase, should produce a 'perfect' range where the out of phase nature cancels the cyclic errors, or a range that due to the 2 distances being exactly in phase should produce a range that is 'bad' by exactly twice the cyclic error amplitude at that point in the phase. (That last sentence looks waaaay too long but I didn't know what else to do with it). I would mean all your 1-4 distances and for the out of phase points I'd plot half the deviation from the mean distance against it's modulo(distance,10) position. Likewise for the 2-3 distances. Anyway I estimate visually the cyclic error to have an amplitude of about 0.7mm. My plot looks like this:
Now, why do I think I may be wrong? Well I see no reason why the average of the two out of phase 1-4 and 2-3 measurements shouldn't be exactly the same as the mean of the two in phase and opposite ranges. This doesn't seem to be the case and both groups have one range that looks out of place. I can't figure out why. Plus my plot seems to be offset from zero by some amount and I see no reason why it should be. I await the proper answer!
Perhaps to keep things simple for me, the test I've devised for cyclic error detection relies on me establishing a series of station positions to a high precision and then comparing them to the EDM ranges. I'm going to try it with our 1mm+1.5ppm instrument one day, but I think I'll have a hard time detecting anything really.
As you point out, the means of the distances 1-4 and 2-3 are the best estimates of the sums of the measured ranges. While the measured ranges contain cyclic errors, those errors cancel in the sums since they cover the full period of 10m over which the errors vary.
So, if a, b, c, and d denote the corrections required to cancel the characteristic cyclic errors of the measured ranges _1.25, _3.75, _6.75, and _8.25, where __ is some multiple of 10m, then one would expect that they recur in the test measurements as follows:
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1.25 3.75 6.25 8.25
13.7498 + 23.7916 + 2b = 37.5434
16.2552 + 21.2888 + a c = 37.5434
18.7572 + 18.7867 + 2d = 37.5434
21.2673 + 16.2772 + a c = 37.5434
11.2482 + 21.2807 + 2a = 32.5273
13.7500 + 18.7763 + b d = 32.5273
16.2515 + 16.2752 + 2c = 32.5273
18.7628 + 13.7645 + b d = 32.5273
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Those reduce to
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1.25 3.75 6.25 8.25
2b = +0.0020
a c = -0.0006
2d = -0.0005
a c = -0.0011
2a = -0.0016
b d = +0.0010
2c = +0.0006
b d = 0.0000
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> As you point out, the means of the distances 1-4 and 2-3 are the best estimates of the sums of the measured ranges. While the measured ranges contain cyclic errors, those errors cancel in the sums since they cover the full period of 10m over which the errors vary.
>
> So, if a, b, c, and d denote the corrections required to cancel the characteristic cyclic errors of the measured ranges _1.25, _3.75, _6.75, and _8.25, where __ is some multiple of 10m, then one would expect that they recur in the test measurements as follows:
>
> [pre]
> 1.25 3.75 6.25 8.25
>
> 13.7498 + 23.7916 + 2b = 37.5434
>
> 16.2552 + 21.2888 + a c = 37.5434
I think I got it right then. Reason being, you'll never see a and c in the second equation as they are a half wavelength apart and therefore should be equal and opposite in magnitude. In those cases any deviation from the mean of the distances is likely noise. That's why I drew my wave using the values derived from 2a, 2b, 2c and 2d.
> I think I got it right then. Reason being, you'll never see a and c in the second equation as they are a half wavelength apart and therefore should be equal and opposite in magnitude. In those cases any deviation from the mean of the distances is likely noise. That's why I drew my wave using the values derived from 2a, 2b, 2c and 2d.
Just...wow. You guys are way above my pay grade, that's for sure. It's everything I can do just to keep up.
I am curious to know though, why Kent was holding my feet to the fire to measure my rail to .1mm, when it looks like he's using chalk marks on the sidewalk. Whassup with that, lol?
> I am curious to know though, why Kent was holding my feet to the fire to measure my rail to .1mm, when it looks like he's using chalk marks on the sidewalk. Whassup with that, lol?
It's an entirely different method using lines of unknown length with fixed prisms at the endpoints. The purpose of the chalk marks was to guide the setups to make sure they were at distances of _1.25, _3.75, _6.25, and _8.75 or within a few centimeters of those values. The idea is that the cyclic error patterns of an instrument with a 10m basic measuring length repeat at 10m and vary as a smooth function. So the cyclic error of a range that measures 18.7763m should be essentially the same as that of a range measuring 18.7867m for most practical surveying purposes.
The "foolproof" method you were using requires a careful calibration of your rail, whereas mine does not.
Hey rfc,
Kent was just using chalk marks for positioning his instrument stations in roughly the right spot in the cyclic error wavelength. He only needs to be within a couple of cm or so. The targets he is taking ranges to are absolutely fixed. 1, 2, 3 and 4 are either on tripods or bipoles and don't move throughout the test. And despite what it looked like I don't think he attempts to reoccupy 5, 6, 7 & 8. He just presented his results in an order that looks like it. I'm sure he would have measured 1, 2, 3 & 4 from 5 all at once, likewise for 6, 7 & 8.
He did outline the same test in your thread if you still wish to do it. I can also detail a similarly basic test you could perform on a track with drilled holes, small pole and an assistant.
> Kent was just using chalk marks for positioning his instrument stations in roughly the right spot in the cyclic error wavelength.
Absolutely right. Sorry to be slow in posting the results as I work them out using both the 2a, 2b, 2c, and 2d conditions and the a + c and b + d conditions.
The punchline is that I think the instrument has software corrections in it that are already working on the problem of cyclic error corrections. So the results are not the nice first-order sinusoidal curve that I for one would otherwise expect for an EDM with a 10m basic length and for which the manufacturer only claims a standard error of +/-5mm + 2ppm.
This is the sort of thing that an accurate rail with fine resolution could answer quite well since I suspect that the cyclic error corrections are by a lookup table in the instrument that produces a discontinuous correction function as a series of straight line approximations.
The situation is that all of the measured ranges have uncertainties. Since the standard error of a single range was estimated from analysis of variance to be 1.3mm, that means that the standard error of the mean of four range measurements is 1.3mm/2 = 0.65mm. The sums of two range measurements have a standard error of:
SQRT(2) x 0.65mm = 0.92mm
The mean of four sums has a standard error of:
0.92 / SQRT(4) = 0.46mm.
[pre]
1.25 3.75 6.25 8.25
13.7498 + 23.7916 + 2b = 37.5434
16.2552 + 21.2889 + a + c = 37.5434
18.7572 + 18.7867 + 2d = 37.5434
21.2673 + 16.2772 + a + c = 37.5434
11.2482 + 21.2907 + 2a = 32.5273
13.7500 + 18.7763 + b + d = 32.5273
16.2515 + 16.2752 + 2c = 32.5273
18.7628 + 13.7645 + b + d = 32.5273
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[pre]
1.25 3.75 6.25 8.25
2b = +2.0mm +/-1.03mm (1)
a + c = -0.6mm +/-1.03mm (2)
2d = -0.5mm +/-1.03mm (3)
a + c = -1.1mm +/-1.03mm (2)
2a = -1.6mm +/-1.03mm (4)
b + d = +1.0mm +/-1.03mm (5)
2c = +0.6mm +/-1.03mm (6)
b + d = 0.0mm +/-1.03mm (5)
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Which in turn yield:
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b = +1.0mm +/-0.52mm (1)
a + c = -0.8mm +/-0.73mm (2)
d = -0.2mm +/-0.52mm (3)
a = -0.8mm +/-0.52mm (4)
b + d = +0.5mm +/-0.73mm (5)
c = +0.3mm +/-0.52mm (6)
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So:
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a + c = -0.5mm +/-0.73mm (4) & (6)
a + c = -0.8mm +/-0.73mm (2)
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which adjusted is:
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a + c = -0.65mm +/-0.51mm,
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correcting both a and c by -0.075mm gives:
[pre]
a = -0.9mm +/-0.73mm
c = +0.2mm +/-0.73mm
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And
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b + d = +0.8mm +/-0.73mm (1) & (3)
b + d = +0.5mm +/-0.73mm (5)
[/pre]
which adjusted is:
[pre]
b + d = -0.65mm +/-0.51mm
[/pre]
correcting both b and d by -0.075mm gives :
[pre]
b = +0.9mm +/-0.51mm
d = -0.3mm +/-0.51mm
[/pre]
Or if you prefer:
A=
0 2 0 0
1 0 1 0
0 0 0 2
1 0 1 0
2 0 0 0
0 1 0 1
0 0 2 0
0 1 0 1
b=
0.0020
-0.0007
-0.0005
-0.0011
-0.0016
0.0010
0.0006
0.0000
Now (AtA)-1*(Atb) gives:
a -0.0009
b 0.0009
c 0.0002
d -0.0003
I *would* prefer to do it that way, actually.