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Plane SurveyingElectronic Distance

Measurement and Curves

Civil Engineering Students

Year (1)

Second semester – Phase II Dr. Kamal M. Ahmed

Lecture Topics• Older techniques of distance measurement

• Electronic Distance Measurement– The idea

– History

– Types of Instruments

– Principles of EDM

– Review of wave geometry

– Methods of Measurement

– The reflectors

– Reduction of Sloped lines to horizontal

– Accuracy and Errors

Distance Measurement

• Older technologies “ quick look”– Pacing : 1:50

– Optical rangefinders: 1: 50

– Odometers: 1:200

– Tacheometry / stadia 1: 500

– Subtense bar 1: 3000

– Tapes: 1: 10,000

• Modern Technology:- Electronic Distance Measuring (EDM) devices: 2 +2 ppm

Devices and accuracy:

Tape Measurement• Accuracy and speed considerations for

civil engineers.

• Sources of Errors:• Incorrect length of the tape

• Temperature difference

• Sag

• Poor alignment

• Tape not horizontal

• Improper Plumbing

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Odometer and Subtense Bar

• The idea of an odometer.

• Subtense bar: a 2 m rod.

Distance H = cot(/2) m.

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–Subtense Bar

–Distance H = cot(/2) m.

Aiming Telescope

L

L /2

/2tan ( /2) = (L/2) / H

H = (L/2) / tan( /2)

If L = 2 m, then

H = 1 / tan (/2) = cot (/2)

H

Example:

what is the horizontal distance between A and B if the angle was 2?

D = KI + C = 100 I, for horizontal sight

Principle of Tachometry (Stadia) measurements

H = 100 I cos2(α) V = 100 I sin(α) cos(α)

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The Idea: (sec 3.8)

To measure the distance between two points (A) and (B) the EDM on point (A) sends electromagnetic waves. The waves received at (B) are reflected back or resent to (A) by a device on (B).

Electronic Distance MeasurementEDM

•Knowing the speed of electromagnetic waves in the air, the EDM computes the distance by measuring the time difference or the shift of the wave phase angle (will be explained in details later).

History of EDM• The use of electromagnetic waves to measure distances

started during WORLD War II with the development of Radar (Radio Detection and Ranging) and progressed to use other forms of energy such as visible light, Microwaves, and laser.

• LIDAR (Light Detection and Ranging) is the latest in the use of lasers to measure distances from an airborne craft. LIDARS returned the distance to an object by timing the length of time from sending of a pulse to its return.

• In 1949, Dr. Erik Bergstrand of Sweden introduced the first EDM: the Geodimeter (Geodetic Distance Measurement) that used light (0.55 micron wavelength) to measure geodetic quality distances (instrument weighed 100kg)

• Light was reflected at the end of the line by glass corner reflectors

Geodimeter

• First units circa 1959 (50 kg each for measurement unit and optics

Later model Geodimeter

• Example of a latter model Geodimeter (circa 1966)

Front and back views

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History of EDM• Distance range was about 10km during daylight and

25km at night, why?

• Greater range during daytime was achieved by using radio waves, Dr. T. L. Wadley, of South Africa introduced the Telluometer in 1957.

• Instrument used radio waves (~10GHz),

• Two similar units at the ends of the line, one sends and the other receives and resends the signal.

• Distances up to 50 km could be measured in daylight with this instrument and later models.

• EDM’s that employ microwaves applied the same technique,

Telluometer

• Example of circa 1962 model.

Back and front of instrument (9 kg with case)

1970’s version (1.7kg)

Modern versions (13.9)

• Smaller, use laser or infra red.

• Mounted on a theodolite تیودولیت or built into a digital theodolite.

• Accuracy varies between ±(1mm +1ppm) and ±(10mm+ 5ppm). A part-per-million (ppm) is one mm in one km

Corner cube reflector Modern example

(circa 2000)Triple retro-reflector

EDM Mounted on a theodolite

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Radio waves were used to measure distances up to 60 km, but GPS replaced them. Currently, electromagnetic distance measuring equipment use three different wavelength bands:

1- Microwave systems: •Range up to 150 km. •Were very successful in hydrographic surveys, GPS reduced their significance.•Wavelength 0.1 to 80 cm •Not limited to line of sight •Unaffected by visibility •Employ Active reflectors

Types of EDM (3.8.1)

2- Electro-Optical EDMs

• Light wave systems – range up to 10 km

– visible light, lasers

– Wave length 10-7 – 10-6 m

– distance reduced by visibility

– Employ Passive reflectors

• Infra red systems – range up to 3 km for short range or up to 30 km for

long range

– limited to line of sight

– Wave length 10-6 – 10-4m

– limited by rain, fog, other airborne particles.

– Employ Passive reflectors.

Principles of EDM (3.8.2)

• As mentioned before, a wave is sent and received back and the distance is estimated.

In general, the distance is estimated by1- Measuring the time of flight2- Frequency shift measurement3- Phase shift measurement.4- Phase shift and the number of full cycles measurement

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Revision of Basics of Waves Geometry

• A wave is a vibration that is propagated in space. The alternating electrical current produces an electric and magnetic field that are also alternating: the succession of cycles going "back and forth" constitutes the wave

•. The wave of the electric field and the wave of the magnetic field are propagated perpendicularly to the direction of propagation and to each other. At extremely low frequencies, the electric field and the magnetic field are specified separately.

•At higher frequencies, electric and magnetic fields are not separable, and are named "electromagnetic waves" or "electromagnetic fields".Any electromagnetic wave is characterized by two parameters: its frequency and its wavelength.

Wave length:

Frequency:

Amplitude:

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Electro Magnetic Spectrum

The symbols used in the following table mean:

KKilo (thousand: 103) m

milli (1 thousandth: 10-3)

M Mega (1 million: 106) µmicro (1 millionth : 10-6)

G Giga (1 billion: 109) nnano (1 billionth: 10-9)

TTera (thousand billion: 1012) p

pico (1 thousandth of billionth : 10-12)

Electromagnetic wave

Frequency Wavelength Applications

g-Rays>3000

THz<100 nm

Medical imaging

X-Rays>3000

THz<100 nm Radiography

Ultraviolet radiations

3000 THz400 nm to 100 nm

Sun-tanning devices

Visible spectrum385 THz to 750 THz

780 to 400 nm

Lighting

Infrared radiations

0,3 THz to 385 THz

1 mm to 780 nm

Heating

Hyperfrequencies or microwaves (MW)

0,3 GHz to 300 GHz

1 m to 1 mmGSM,

microwaves oven, radars

Radiofrequencies (RF)

0,3 MHz to 300 GHz

1 Km to 1 mm

Radio, television

Extremely low frequencies (ELF)

3 Hz to 300 kHz

>1000 km to 1 km

Electricity distribution

network50 Hz 6000 Km

Earth electromagnetic field

0 Hz (continuous)

infinite Compass

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• Velocity (V), frequency (f), and wave length (l):

V = f * l

• Electromagnetic waves slow down as they travel in a “denser” medium, such as the atmosphere or glass.

V = c / n

where c is the speed of light (299,792,458 m/sec) in vacuum, and n is the index of refraction of light pass through a medium other than a vacuum.

• Distance = Velocity * Time = ((N *l) + ) / 2

Where is a fraction of wave length = (/360) *l

N is the number of full cycles, ambiguity?

Basic relationships (3.8.2)

Since is divided by 2, so is l, we call l/2 “effective wave length”

Concept of the fraction

Phase Angle

0

90

180

270

Assume that l = 2 m If 1 = 80, it corresponds to a distance = (80/360) * l = 0.44 mIf 2 = 135, it corresponds to a distance = (135/360) * l = 0.75 mIf 3 = 240, it corresponds to a distance = (80/360) * l = 1.33 m

Wave Modulation (3.8.3)

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Methods of Measurement(3.8.4 to 3.8.7)

1- Time of Flight Measurement • Few instruments use this method, but many reflectorless

EDMs use it today because of the need of discrete high energy pulses of energy. Why??

• many individual pulses of light are emitted, each of relatively high energy when compared with phase resolving EDMs.

• When the pulse returns to the source of the emitted light. The detectors in the EDM then process the pulse, compare it with the emitted pulse, and determine the time of flight

• Since the Distance (D) = velocity (V) * Time (t), we know the speed of light, if we measure the time of flight, we can determine the TWO WAY distance.

• Accuracy of measurement depends on duration of pulse

2- Frequency shift Measurement• In many EDM systems, the modulation

frequency can be changed by small increments and this allows distances to be measured by recording the consecutive frequencies (wave lengths) that return zero phase angles.

• The frequency is set such that x/l1=N (an unknown integer)

• If we are certain that l1 and l2 represent exactly one cycle difference over the distance x then the two equations can be solved for x:

• Many EDMs work this way but notice the sensitivity to the difference in wavelength

x /l1 N

x /l2 N 1 Subtracting the two eqns

xl1 l2

l1l2

1 and x =l1l2

l1 l2

• If the frequency is slowly changed then the phase difference will be non-zero, but will return to zero again at some slightly different frequency so that x/l2=N+1. How do we use this?

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Example

• What is the distance measured using an EDM if the frequency was modulated and the following consecutive wavelengths returned zero phase angle: l1 = 0.12 and l2 = 0.1205 m

Answer

x =l1l2

l1 l2= (0.1205 * 0.12) / (0.1205 – 0.12) = 0.01205 / 0.0005 = 24.10 m

3- Phase shift Measurement• All that is measured is phase shift , different waves of

effective length: 10 100 1000 10,000 m are sent. Each fraction provides a digit(s).

• A partial wavelength is determined by the phase-shiftof the returning wave when compared to the outgoing wave. It is measured in degrees. For example, if the phase shift is 135°, then the partial wavelength is (135/360)l = 0.375l.

• Since we always need half the total two way distance, we need half that fraction, that is why we consider the effective wave length equals half the actual modulated wave length

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• Usually, the accuracy of measurement using a wave is not better than 1% of its wave length.

• The EDM sends EM waves of increasing wave lengths.

• The shortest wave provides the smallest digits to the right, or the resolution, when its fraction is measured.

• As fractions of longer waves are measured, each longer wave provides a digit to the left and a check, why?

• The EDM send the waves it is designed to send regardless of the distance, un-needed waves will simply provide zeros to the left and ignored.

Example

An EDM sent EM of frequencies 1498.5 MHz,149.85MHz, 14.985 MHz, 1.4985 MHz, 149.85 KHz, 14.985 KHz, and 1.4985 KHz and the shift of their phase angles when returned were in decimal degrees) : 298.7975, 280.7976, 169.1986, 50.39957, and 11.15991 respectively. The speed of EM in vacuum is 29,792,458 m /sec, and the refraction index in air is 1.0003.

a) What was the distance measured ?

b) Resolve the ambiguity.

Wave frequency l(m) effective l (m) Phase shift fraction

1498.5 MHz, 0.2 0.1 298.7975 0.083

149.85 MHz 2.0 1.0 280.7976 0.78

14.985 MHZ 20.0 10.0 169.1986 4.7

1.4985 MHZ 200.0 100.0 50.3996 14

149.85 KHz 2000.0 1000.0 111.5991 310

ANSWER*Speed of EM in air = 299,792,458/ 1.0003 = 299,702,547 m/sec*l= speed /frequency = C / f*Fraction of any returned wave = (Phase shift /360) * l

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• Assume that the unknown distance is (abc.def)

• The first wave (fraction) provides the digits 0.083, the e and f in our number

• The second provides the first decimal place digit 7, the d digit in our number, and check on the second (8)

• The third provides the digit for full meters (4), c digit in our number, and checks on the value (7)

• The fourth provides the tens of meters ( 1), b digit in our number, and checks on the value (4)

• The sixth provides the hundreds of meters (3), a digit in our number, and checks on the value (1)

• Distance measured is 314.783 m

• Ambiguity: Number of full cycles: 0 for l = 1000m, 3 for l = 100m, 31 for l = 10m, 314 for l = 1m,

4- Phase shift and Number of Cycles Measurement

4-1 Resolving the ambiguity by comparing the sent EM wave and a typical reference wave.

the full number of cycles might be determined through the comparison. The fraction is estimated as before. How is that done??

No need for plenty of waves sent, at least two waves. The average is used as a check and a result. Check of what???

4-2 Resolving the ambiguity by sending several waves

- The longest wave is longer than the maximum range of the instrument. For example, if the instrument can measure up to 10 KM, the longest wave used should have an effective length more than 10 km, say 20 km

- The phase shift of the longest wave determines the length to about 1%, but can also resolve the ambiguity of the next shorter wave. For example, if the phase shift of 20 km wave is 320 24’, resulted in a fraction that is 0.89l, then the length is 0.89* 20,000 = 17800 m

- If the following wave length is 5 km, then the number of full cycles “ambiguity” is 17800 / 5 = 3 waves

4- Phase shift and Number of Cycles Measurement

- Now if the fraction for the 5 km wave is 0.55, then there are: 3 waves from above + 0.55 from the phase shift, the length is refined to: 3.55 *5000 = 17,750 m.

- This way, each wave length solves the ambiguity of the following shorter wavelength and the phase shift solves for the fraction of that shorter wave and so on. Every shorter wavelength will improve the accuracy of the result

• EXAMPLE:

An EDM can measure distances up to 10 km using wavelengths of 20 km, 1 km, 200 m, 10 m, 0.5 m. Given the fractions of the waves measured in the table below, compute the total distance.

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Wavelength l Cycles Resolved wave lengths = (R)

Distance m

= R * l

20 km 0.59 0.59 11800

1 km 0.79 11.79 11790

200 m 0.93 58.93 11786

10 m 0.54 1178.54 11785.4

0.5m 0.70 23570.70 11785.350

The Distance is 11785.350 m

Reflectors• Reflectors (Prisms):

– Are made of high quality glass, excellent reflection capability.

– Accept rotation to right and left, and maybe up and down, why?? And can be mounted on a tribrach or a pole, why??

– Reflect the energy at the same path it came through. The theory is that if a ray of light hits a mirror and is reflected to hit another mirror and exit the system, the angle between the first and last rays is double the angle between the mirrors. The angle between the mirrors is 90, three of them resemble three mirrors installed on the walls at the corner of a room, That is why they are called “corner reflectors”

The angle between the rays A and C is double the angle between the two mirrors = 2 *90 = 180

Notice that the objects will look upside down, notice the box at the tail of the arrow

A

C

Aiming at a prism through the telescope of a total station in a zoo!

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Reflectors (Prisms)

Pole and bipod

Prism and sighting target

Fully rotating prism

• A reflector might include a single prism or a group of prisms

• Reflectors can be a simple reflecting paper-sticker, they are called sheet-prisms, paper prism, or reflective sheeting. Very instrumental in construction sites and deformation monitoring of structures.

• The sighting target might be lower than the prism itself, why???

V = f * l

where V is the velocity of light, f is the frequency of the light, and l is the wavelength

V = c / nwhere c is the speed of light (299,792,458 m/sec) in vacuum, and n is the index of refraction of light pass through a medium other than a vacuum. Thus a change in V will result in a change in wavelength.

EDM Accuracy• Since the light must pass through air of

different densities, temperatures, and humidity, the index of refraction is different. This will result in a systematic error in the measured length.

• Manufacturers account for this by implementing a parts-per-million (ppm) corrections to the measured distance. (review pages 86 – 88 of the book)

• Humidity has little effect on near-infrared light (0.900 to 0.930 mm.) Thus only temperature and pressure measurements are required for typical survey accuracies.

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EDM Accuracy • ±(3mm + 3ppm) is common, effect on shorter

distances is higher, why???

• Estimated error in distance =

Ei2 + er

2 + ec2 + (ppm X D)2

Where: Ei and er are the centering errors of the instrument and the reflector, ec is the constant error of the EDM, ppm is the scalar error of the EDM, D is slope distance measured in KM

Example

If the estimated errors of centering the instrument and target were ± 3mm and ± 5mm respectively, the EDM had a specified error of ± (2mm +2ppm) what is the estimated error in measuring 827.329 m?

Answer: ± mm

Adjustments in EDM Measurements

1- Prism Constant• A systematic error that should be SUBTRACTED from every distance measured.

•It results from the fact that EM waves travel slower inglass, the prism is solid glass. It takes the waves a longer time to return and that is interpreted by the EDM as a longer distance than the actual distance. the refractive index in glass is 1.517, that creates an effective distance = 1.517D

• Waves travel an actual distance inside the glass of the prism that is always = a +b + c = 2D

• Because of refraction in glass, the effective center is shifted from the plummet line ( the center) by a distance K to the back of the prism, a little longer than 0.517 D

• This shift can be as large as 7 cm.

• Although it can be adjusted for automatically if entered into a total station, it may cause serious problems, why??

• Cases of zero prism constant??

• Distance corrected for prism constant S2 = S1 - PC

Field determination of Prism constant– Setup a straight line with two endpoints and a point

about midway between the ends (a baseline)

– Then AC + K = (AB + K) + (BC + K), or

K = AC - (AB + BC)

– K generally has values of 0, -20, -30, -40 mm, and

up to 70 mm

–Example (problem 2 page 96)If the measured distances AC, AB, and BC respectively

were 2438.29, 1206.48, AND 1231.84m. Compute the prism constant and the corrected distances?

Answer: K = AC –( AB +BC) = 2438.29 –( 1206.48 + 1231.84) =

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2- Corrections for Temperature and Pressure

As mentioned before, EM speed in the air will be affected by humidity, temperature, and pressure, the effect of humidity is minimal and can be ignored.

The temperature & pressure correction ( T.P.C.) is given in ppm where:

T.P.C in ppm = 279.6 – {(80 * P)/(273 + T)}

Where

P is measured in mm bar

T is temperature in C

Distance S3 corrected for prism, Temp, and Pressure = S2 + T.P.C *S2

Adjustments in EDM Measurements

ExampleCompute the T.P.C if the temperature was

25 and the pressure was 1013 mm bar? If S2 was 1745.245 m compute the adjusted distance for temperature and pressure.

Answer:

T.P.C = 279.6 – {(80*1013)/(273.2+25)}= 7.8 ppm

Adjusted distance = 1745.245 + { (7.8/1000 000) * (11745.245) m

Adjustments in EDM Measurements3- Slope CorrectionTo compute the horizontal distance from the measured

SLOPE distance. Assuming that the EDM is built into the theodolite and that the target is at the center of the signal, the usual case today, then the horizontal distance D 1 can be computed for the measured slope distance and the vertical angle (α) as follows:

D1 = S3 * (cos α) and

D1

v

• The elevation of station B can also be computed if the elevation of A is given:

Elevation of B = Elevation of A + hi ± V – HR

Where V = S3 * (sin α)

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Adjustments in EDM Measurements4- Distance on the reference Ellipsoid (3.8.6)

• For long distances, why?• The goal of this GROUP of adjustments is to

compute the distance on the curved surface of the ellipsoid, then we will transform that onto the map.

• This adjustment accounts for the curvature of the path of the EM waves D2, the variation of the refractive index along the path D3, the transformation to the cord of the ellipsoid D4, and the transformation to the curved surface of the ellipsoid D5. All previous adjustment can be combined in the following approximated equation

S3

D1

D5

D4

De = D13 / 43R2 (equation 34 page 94 in the book)

Where R = Radius of earth (6375000 m)

Then, D5 = D1 - De

Questions: calculate the correction for referring the measured horizontal distance to the ellipsoid in mm if the distance was 1, 10, or 30 km

Adjustments in EDM Measurements

5- Adjustment for scale factor

• To project the distance onto the map or to use it in a different coordinate system.

• The earth’s surface is curved, we approximated that by an ellipsoid, or a sphere in less accurate geodetic applications, we applied adjustments to get the distance on the curved surface of the ellipsoid D5

• Because of the need for a simple Cartesian two dimensional coordinate system of X and Y, countries establish their own FLAT Cartesian system, for example the national coordinate system in Egypt is ETM (Egyptian Transverse Mercator) based on an ellipsoid called Helmert and a cylindrical projection.

Reference Ellipsoid

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Purple Belt RED Belt Blue Belt37 E ETM zones

and Origins

• The transformation of the data from the curved surface of the ellipsoid to the flat Cartesian system is called “map projection”, which results in distortions or error in scale.

• Given that correction (S.F) you simply multiply it by the distance on the ellipsoid:

Distance corrected for scale factor

D6 = D5 * (S.F)

Examples of developable surfaces used in map projections: cones and cylinders.

•Imagine a light source at the center of the ellipsoid that projects the data, or the shapes of buildings and streets, etc. onto the a film around it “developable surface” and capture an image, the picture is the map and this process is map projection.

•After the data is projected onto the flat surface, an origin is selected of zero X and Y, and all points will have X and Y values.

•Distances on the flat surface (the map or the cone or the cylinder) is not equal to the distance on

the curved surface (reality), except for when the two surface are tangent. This difference need to be accounted for if the distances will be used on a map or in calculations in a Cartesian system, the correction is called “scale Factor” or S.F, this correction will be given to you in this course.

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Summary of the Adjustments• Assume that the measured slope distance is (S1), then

• Prism constant (P.C)

S2 = S1 – P.C

• Temperature and pressure correction

T.P.C in ppm = 279.6 – {(80 * P)/(273 + T)}

S3 = S2 + T.P.C *S2

• Slope correction

Horizontal distance D1 = S3 * (cos α)

• Distance on the ellipsoid (several combined corrections)

De = D13 / 43R2

D5 = D1 - De

• Distance corrected for scale factor

• D6 = D5 * (S.F)

Example

• The distance between two points (A) and (B) was measured using an EDM. IF the distance measured was 8453.728 m, the prism constant was 3mm, the temperature was 22 C, the pressure was 1010mm bar, the vertical angle was 5 22’ 13”, radius of earth is 6375000 m, and the scale factor was 1.0003. Calculate the coordinates of B if the coordinates of A were(686336.673, 972309.525) m in the ETM coordinate system, red belt, and the azimuth of the line AB was 212 2’ 12”.

ANSWER• Measured slope distance S1 = 8453.728 m

• S2 = S1 – PC = 8453.728 – 0.003 = 8453.725m

• T.P.C = 279.6 – {(80 * P)/(273 + T)}

= 279.6 – {(80 * 1010)/(273 + 22)}

= - 5.7 ppm

S3 = S2 +T.PC. * S2 =

=8453.725 – (5.7/1000,000) * (8453.725)

= 8453.677 m

• Horizontal Distance

D1= S3 cos (α)

= 8453.677 cos (5 22’ 13”)

= 8416.571 m

• Distance on the ellipsoid D5

De = D13 / 43R2

= (8416.571)3 / (43 *(6375000) 2 )

= 0.0003 m can be ignored

D5 = 8416.571 + 0.000 = 8416.571 m

Horizontal Distance corrected for scale factor

= 1.0003 * 8416.571 = 8419.096 m

XB = XA + DAB sin (azimuth)

= 686336.673 + 8419.096 *cos (212 2’ 12”)

= 679199.731 m

YB = YA + DAB cos (azimuth) =

= 972309.525 + 8419.096 * sin (212 2’ 12”)

= 967843.516 m

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Handheld laser measuring devices

ANSWER TO SHHET PROBLEM 14

• Next slide

Wavelength l Cycles Resolved wave lengths = (R)

Distance (km)

= R * l

15 km 0.690 0.690 10.3500

1 km 0.346 10.346 10.3460

300 m 0.487 34.487 10.3461

5 m 0.18 20692.18 10.34609

0.5m 0.77 206921.77 10.3460885

The Distance is 10346.0885 m