CE 371 Surveying EDM & Stadia
Dr. Ragab Khalil
Department of Landscape Architecture Faculty of Environmental Design
King AbdulAziz University Room LIE15
1
Lec. 19 +
Lec. 20
Overview
•
EDM•
Classification of EDM Instruments Electro-optical Instruments
Microwave Instruments
•
Fundamental Principles of EDMI Operation•
Total Station Instruments•
Accuracy of EDM Instruments•
Sources of Errors•
Stadia•
Stadia Measurements for Horizontal Sights•
Stadia Measurements for Inclined Sights2
EDM
•
An electronic distance-measuring (EDM) instrument can determine distances by measuring phase changes that occur as electromagnetic energy of known wavelength travels from the instrument and returns.•
Accurate distance measurements can be obtained rapidly and easily. Distances can be measured over bodies of water or rugged terrain that is inaccessible for taping.•
According to wavelength of transmitted electromagnetic energy, EDM instruments are classified as Electro-optical and Microwave instruments.3
Electro-optical Instruments
Reflector
Transmitter
A
B Light Beam
FIGURE 6.1: Measurement principle of an electro-optical EDM.
Composed of transmitter (main instrument) and reflector.
S
• Transmit visible and infrared waves.
• Wave lengths range from 0.7 to 1.2 micrometers.
• Range from 1 to 3 km. Maximum ranges 20 km.
• Use Amplitude Modulation (AM) wave.
Microwave instruments
A
FIGURE 6.2: Measurement principle of a microwave EDM.
B Master unit
Remote unit Microwave beam
Composed of two identical units: Master and Remote.
• These instruments transmit microwave signals.
• Wavelengths range from1.0 to 8.6 millimeters.
• Can measure much longer distances of up to 80 km.
• Use Frequency Modulation (FM) wave.
Fundamental Principles of EDM Operation
0° 90° 180° 270° 360°
Amplitude
/2
Phase
FIGURE 6.3: Phase angle of a sinusoidal wave.
= V f
Where = wavelength (distance traveled during the period of one cycle) V = velocity of propagation (V = 299792.5 0.4 km/sec in a vacuum) f = frequency in Hz (cycle/sec)
Phase Angle
0
90
180
270
Assume that = 100 m
If 1 = 80, it corresponds to a distance = (80/360) * = 22.22 m If 2 = 135, it corresponds to a distance = (135/360) * = 37.5 m If 3 = 240, it corresponds to a distance = (80/360) * = 66.67 m
Fundamental Principles of EDMI Operation
Electronic Distance Measurement
EDM
Transmitter
50°
300°
Reflector Transmitted
waves Returning
waves
FIGURE 6.4: Measurement of phase difference.
S = 1
2 (n +
360 )
Where: = wavelength
n = total number of full wavelengths
= phase difference
The instrument is not capable of measuring the term n, it can only measure the phase shift .
To solve this problem, a signal with longer wavelength is used such that the distance between the instrument and the reflector is less than .
Frequency 1
2
360°
(m)
20
200 2000 20000 F1 = 14.989625 MHz
F2 = 1.4989625 MHz F3 = 149.89625 KHz F4 = 14.989625 KHz
Measured
Phase Difference
1 = 250°
= 190°
= 91°
2 = 98°
3
4
6.944 27
527 2527
m m m m S = 6.944 + 20 + 500 + 2000 = 2526.944 m
To go around measuring n in the previous equation (which is difficult to measure with a single frequency), the EDM sends several frequencies successively and measures the phase differences and sums the fractions of the distance to find the total distance (see table below).
Example 2
•
An EDMI transmits 2 signals having wavelengths 1.000 m and 2000.0 m. When measuring a distance, the phase shifts were: 186.70o and 308.97o respectively. Compute the length of this measured distance.Solution:
• For first signal,
• measured distance = 1.000 (186.70o/360o) = 0.519 m
• For second signal,
• measured distance = 2000.0 (308.97/360)=1716.5 m
• Therefore, the distance = 0.519 + 1716 = 1716.519 m.
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Total station
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RETRO-REFLECTORS
•
Fixed and tilted•
Single and multipleFIGURE 6.7: Examples of retro-reflectors
Accuracy of EDM Instruments
•
An EDM instrument with an accuracy of ±(5 mm + 3 ppm) indicates a fixed error of 5 mm regardless of the distance measured, and a variable error of 3 parts per million (ppm) of the measured distance.•
For short distances, fixed error is more critical than variable error, while for long distances the opposite is true.14
Example
• Distance X= 5580.284 m is measured by two EDMI instruments.
Instrument A has an accuracy equals to ±(5 mm + 3 ppm), while instrument B accuracy is ±(3 mm + 5 ppm). Compute the error in distance X made by each instrument. Which instrument is better to use? Why?
•
Solution:• Error using instrument A= 5 mm + 3 (5580284 mm /1,000,000) = 21.7 mm
• Error using instrument B= 3 mm + 5 (5580284 mm /1,000,000) = 30.9 mm
• It is better to use instrument A, because it produces less error
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Sources of Errors
•
Instrumental errors:such as prism constant, electronic center constant of the instrument, frequency change of the signal. These errors can be eliminated by periodic checking of the EDM instrument against a calibrated base line.
•
Natural errors:such as variation in temperature, pressure, and humidity.
These can affect the atmospheric index of refraction, which will affect the speed of transmitted electromagnetic signal in air.
•
Human (personal):such as misreading or miswriting a measurement, or failing
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Systematic Errors
• Systematic errors can be eliminated by using certain mathematical formulas and by proper field measurements and calibration of instruments.
• Prism constant and electronic center constant can be subtracted from measured distance.
• Temperature, pressure, and humidity effect on atmospheric index of refraction can be computed from special charts or formulas and then eliminated from the measurements.
• In some total stations, temperature, pressure, and humidity are measured and directly entered into the instrument to eliminate their effects.
• By measuring distance AB from A to B and then from B to A, some systematic errors can be eliminated.
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From similarity of triangles:
d/f = I/i or d = (f/i) I= K I thus D = K I + C
Because of the signs of f and c the distance C becomes negligible, thus,
• I) rod intercept, or stadia interval
• (i) spacing between stadia hair
• (f/i) = k = 100: stadia interval factor
• C = (c + f) approximately
Stadia
Example
• It is required to determine the horizontal distance between points A and B by stadia method. A theodolite at A is used to sight a level rod at B while the telescope is horizontal. The rod readings are u =1.85 m, l =0.45 m. If the stadia interval factor K =100, what is the horizontal distance HAB? If elevation of A is 10.00 m, hiA=1.60 m, compute elevation of B.
• Solution:
• H= K I = K (u-l) = 100 (1.85 - 0.45) = 140.00 m
• ElevB = 10.00 + 1.60 - (1.85+0.45)/2 = 10.45 m.
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Stadia Measurements for Inclined Sights
•
I’=I cos (a)•
L= K I’ = K I cos (a)•
H= L cos (a)•
H= K I cos2 (a) = 100 I cos2 (a)•
H= 100 I sin2 (z)•
V= 100 I cos (a) sin (a)•
V= 50 I sin (2z)•
ElvO=ElvM + hi + V - R20
Example
•
A theodolite is used to determine horizontal distance HAB and the elevation ElevB of point B. The following stadia readings were taken: zenith angle z=82o, u =1.92 m, r=1.15 m, l =0.32 m. If K=100, ElevA=500.00 m, telescope height hiA = 1.60 m, compute horizontal distance HAB and elevation of B.
•
Solution:• Stadia interval I = u - l = 1.92 - 0.32 = 1.60 m
• HAB=K I (sin z)2 =100(1.60)(sin 82o)2 = 156.90 m
• V = 0.5 K I sin(2z) = 0.5 (100) (1.60) sin(282o) = 22.05 m
• ElevB= ElevA + hiA + V - r = 500.00 + 1.60 + 22.05 - 1.15 = 522.50 m
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Precision of Stadia Measurements
•
Stadia survey is not suitable for precise horizontal distance measurements or difference in elevations.•
Error ratio can be as low as 1/300 to 1/500 which makes it suitable for rough measurements such as in topographic surveys.•
For long distances, half the stadia interval I can be used to intercept the level rod.22
Summary
•
EDM•
Classification of EDM Instruments Electro-optical Instruments
Microwave Instruments
•
Fundamental Principles of EDMI Operation•
Total Station Instruments•
Accuracy of EDM Instruments•
Sources of Errors•
Stadia•
Stadia Measurements for Horizontal Sights•
Stadia Measurements for Inclined Sights•
Precision of Stadia Measurements23