CALCULATOR TECHNIQUES
CALCULATOR 101
Mode 1: COMP (Computation)
This is the IDLE mode of your calculator once you turn it ON.
This MODE is commonly used for GENERAL CALCULATIONS.
This is the MODE used in computing problems involving COMPLEX NUMBERS.
Mode 2 : CMPLX (Complex)
To CONFIGURE the SETTINGS on MODE 2: COMPLEX
PRESS: [ SHIFT ] – [ 2] OPTIONS (after pressing): [ 1 ] – ARG (ARGUMENT) [ 2] – CONJG (CONJUGATE) [3]–r‹θ [ 4 ] – a + bi
This is the MODE used in computing problems involving STATISTICS and REGRESSION. VARIATION [ 1 ] , LINEAR [ 2 ], QUADRATIC [ 3 ], LOGARITHMIC [ 4 ], EXPONENTIAL [ 5 ], To CONFIGURE the SETTINGS on MODE 3: STAT
Mode 3 : STAT (Statistics)
PRESS: [ SHIFT ] – [ 1] OPTIONS (after pressing): [ 1 ] – TYPE [ 2] – DATA [ 3 ] – SUM [ 4 ] – VAR (Variation) [ 5 ] – DISTR (Distribution) [ 6 ] - MinMax
Mode 5 : EQN (Equations)
This is the MODE used in computing problems involving EQUATIONS and POLYNOMIALS. OPTIONS (after pressing [ MODE ] – [ 5 ] ) : [ 1 ] – TWO – UNKNOWNS [ 2] – THREE UNKNOWNS [ 3 ] – QUADRATIC [ 4 ] – CUBIC
This is the MODE used in computing problems involving MATRIX ANALYSIS. Mat A [ 1 ], Mat B [ 2 ] & Mat C [ 3 ]
To CONFIGURE the SETTINGS on MODE 6: MATRIX PRESS:
Mode 6: MATRIX
[ SHIFT ] – [ 4] OPTIONS (after pressing): [ 1 ] – DIM (Dimension) [ 2] – DATA [ 3 ] – MATRIX A [ 4 ] - MATRIX B [ 5 ] – MATRIX C [ 6 ] – MatAns (Matrix Answer) [ 7 ] – Det (Determinant) [ 8 ] – Trn (Transpose)
Mode 7: TABLE
This is the MODE used in GENERATING A NUMBER TABLE based on ONE OR TWO FUNCTIONS.
This is the MODE used in computing problems involving VECTOR ANALYSIS. Vct A [ 1 ], Vct B [ 2 ] & Vct C [ 3 ] To CONFIGURE the SETTINGS on MODE 8: VECTOR
Mode 8: VECTOR
PRESS: [ SHIFT ] – [ 5] OPTIONS (after pressing): [ 1 ] – DIM (Dimension) [ 2] – DATA [ 3 ] – VECTOR A [ 4 ] – VECTOR B [ 5 ] – VECTOR C [ 6 ] – VctAns (Vector Answer) [ 7 ] – DOT
To CONFIGURE the SETUP of your CALCULATOR ENTER:
Configuring Your Calculator’s SETUP
[ SHIFT ] – [ MODE ] 1st OPTIONS ( after pressing) [ 1 ] – MthIO [ 3 ] – Deg [ 5 ] – Gra [ 7 ] – Sci
[ 2] - LineIO [ 4 ] – Rad [ 6 ] – Fix [ 8 ] – Norm
2nd OPTIONS (scroll down) [ 1 ] - ab/c [ 3 ] – CMPLX [ 5 ] – Disp
[ 2 ] – d/c [ 4 ] - STAT [ 6 ] - Contrast
SEMICOLON “:”
USING THE SEMICOLON “:” FUNCTION OF YOUR CALCULATOR Past CE Board Exam Problem A block of copper having a mass of 50kg is drawn out to make 500m of wire of uniform cross-section. Given that the density of copper is 819 g/cm2. 1. What is the volume of steel in cubic centimeters? 2. What is the cross-sectional area of the wire in square millimeters? 3. What is the of the wire in millimeters?
Solution: ρ = M/V ,
V = AL
Π 𝟒
A = (D2)
Input: (50000/8.91) : Ans/(500 x 100) : Press: [ = ] Press: [ = ] Press: [ = ]
𝟒𝒙𝑨𝒏𝒔 Π
Display: 5611.67 cm3 Display: 0.1122 cm2 or 11.22 mm2 Display: 0.378 cm or 3.78 mm
ENGINEERING MECHANICS
RIGHT ANGLE TECHNIQUE The “Rec” and “Pol” Functions PRESS: [ SHIFT ] – [ + ]
For “POL” Function
PRESS
[ SHIFT ] – [ - ]
For “REC” Function
The “Rec” function converts POLAR COORDINATES (r<θ) into RECTANGULAR COORDINATES(x,y). The “Pol” function converts RECTANGULAR COORDINATES(x,y). into POLAR COORDINATES (r<θ).
RIGHT TRIANGLE TECHNIQUE Example. In the figure shown, Find the remaining side and angles if A = 4 and B = 7. Solution: Let: x = 4 and y = 7 Input: [ SHIFT ] – [ + ] – [SHIFT] – [ 4 ] – [SHIFT] – [ ( ] – [ 7 ] – [)] PRESS: [ SHIFT ] – [ + ]
For “POL” Function
[ SHIFT ] – [ - ]
For “REC” Function
Display: Pol(4,7) Press [ = ] Display: r = 8.062257748 θ = 60.2551187
C β
COMPONENT OF A 2D FORCE
Example. In the figure shown, Find the x and y component of the force F if F = 4000N and θ = 32.3o Solution: Applying the RIGHT ANGLE TECHNIQUE Let: r = 4000N and θ = 32.3o Input: [ SHIFT ] – [ - ] – [SHIFT] – [ 4000 ] – [SHIFT] – [ ( ] – [ 32.3 ] – [ ) ]
PRESS: [ SHIFT ] – [ + ]
For “POL” Function
[ SHIFT ] – [ - ]
For “REC” Function
Display:
Rec(4000,32.3)
Press [ = ] Display: x = 3381.047333 y = 2137.409398
Fx Fy
MODE 2: COMPLEX Argand Diagram (Complex Plane) Similarly with Cartesian Coordinate Plane the Complex Plane has the following forms too;
r < θ is the POLAR FORM a + bi is the RECTANGULAR FORM NOTE: Θ IS ALWAYS MEASURED FROM THE ORIGIN (COUNTER CLOCKWISE) Where: r = absolute value or modulus Θ = argument or amplitude a is the real axis and b is the imaginary axis
MODE 2: COMPLEX NOTE: All angles (θ) must be referred from the x-axis.
1st Quadrant: As is 2nd Quadrant: 180 – Angle 3rd Quadrant: 180 + Angle 4th Quadrant: - Angle or 360 – angle Where: Angle is solved by either the RIGHT ANGLE TECHNIQUE or tan-1 (slope)
COMPONENTS OF A 2D FORCE
Example. In the figure shown, Find the x and y component of the force F if F = 4000N and θ = 32.3o Solution: Applying Complex Mode Let: r = 4000N and θ = 32.3o
Press: [Mode] – [ 2 ] ‘The Calculator now is in CMPLX MODE’ Input: [ 4000 ] – [ SHIFT ] – [ ( - ) ] – [ 32.3 ] Display:
4000 < 32.3
Press [ = ] Display: 3381.047333 + 2137.409398i a = 3381.047333 Fx b = 2137.409398 Fy
CONCURRENT – COPLANAR FORCE SYSTEM PAST CE BOARD: For the forces in the figure shown. a. Find the magnitude of the horizontal component of the resultant. b. Find the magnitude of the resultant. c. Find the angle that the resultant makes with the x-axis.
CONCURRENT – COPLANAR FORCE SYSTEM Solution: Enter: [ MODE ] – [ 2 ] ‘CMPLX MODE’ To get the resultant: Input: 86.6<(30) + 70.7<(180-45) + 68.6<(270) Press: [ = ] Output: 25.00535054 + 24.69244943i Press: [ SHIFT ] – [ 2 ] – [ 3 ] – [ = ] Display: Ans r<θ Output: 35.14234788<44.6392662
Resultant<θ = Summation of r<θ’s
a. Rx = 25.00535054 & Ry = 24.69244943 in Newtons b. R = 35.14234788 in Newtons c. θx = 44.6392662o
NONCONCURRENT – COPLANAR FORCE SYSTEM PAST CE BOARD: Consider the coplanar force system shown. a. Find the magnitude of the resultant. b. Compute the location of the resultant from the y – axis. c. Compute the location of the resultant from the x – axis.
NONCONCURRENT – COPLANAR FORCE SYSTEM Solution: Enter: [ MODE ] – [ 2 ] ‘CMPLX MODE’ To get the resultant: Input: 10<(90) + 6.7< (tan-1 1/2) Press: [ = ] Display: 5.9927 + 12.9963i Press: [ SHIFT ] – [ 2 ] – [ 3 ] – [ = ] Display: Ans r<θ Output: 14.31142<65.24538 Rx = 5.9927 & Ry = 12.9963 in Newtons R = 14.31142 in Newtons θx =65.24538o
NONCONCURRENT – COPLANAR FORCE SYSTEM Solution: CALCULATING THE MOMENT USING STATISTICS Enter: [ MODE ] – [ 3 ] – [ 2 ] ‘STAT MODE – LINEAR’ Enter: [ SHIFT ] – [ MODE ] – [ ] – [ 4 ] – [ 1 ] ‘FREQUENCY ON’ NOTE: X is the FORCE, Y is the MOMENT ARM and Freq is 1 for CLOCKWISE and -1 if COUNTERCLOCKWISE ROTATION. Moment @ 0 Input: X Y Freq 10 5 -1 2.996 8 -1 5.993 2 1 30 1 1 Enter: [AC] - [ SHIFT ] – [ 1 ] – [ 3 ] – [ 5 ] – [ = ] Output: 𝒙𝒚 = moment at 0 = - 31.982N-m (Ry = 12.9963 )(Ix) = 31.982 (Rx = 5.9927 ) (Iy) = 31.982
Ix = 2.461m Iy = 5.34m
x-intercept y-intercept
NONCONCURRENT – COPLANAR FORCE SYSTEM PAST CE BOARD: The force system shown consist of a Couple C and the 4 Forces. The resultant of this system is 500kNm counter-clockwise couple. a. What is the value of P? b. What is the value of Q? c. What is the value of C?
NONCONCURRENT – COPLANAR FORCE SYSTEM Solution: Enter: [ MODE ] – [ 2 ] ‘CMPLX MODE’ To get the resultant: Resultant Force = 0, since Resultant is just a Couple P<(tan-1 3/4) + 20<(270) + Q(180 + tan-1 (5/12)) + 80<0 = 0 P<(tan-1 3/4) + Q(180 + tan-1 (5/12)) = - ( 20<270 + 80<0) Simplifying: P(4/5 + 3/5i) + Q( -12/13 – 5/13i) = -80 + 20i Use [ MODE ] – [ 5 ] – [ 1 ] ‘EQN – TWO UNKNOWNS’
1 2
a 4/5 3/5
b -12/13 -5/13
c -80 20
X = P = 200kN and Y = Q = 260kN
NONCONCURRENT – COPLANAR FORCE SYSTEM Solution: Enter: [ MODE ] – [ 3 ] – [ 2 ] ‘STAT MODE – LINEAR’ Enter: [ SHIFT ] – [ MODE ] – [ ] – [ 4 ] – [ 1 ] ‘FREQUENCY ON’ Mo = 500 kN-m counterclockwise Input: X
Y
Freq
80
2
1
120
6
-1
20
3
1
240 6 -1 Enter: [ AC ] – [ SHIFT ] – [ 1 ] – [ 3 ] – [ 5 ] Output: 𝒙𝒚 = -1940 kN-m Mo = C – 1940 = -500 C = 1440 kN-m
COMPONENTS OF A 3D FORCE Example. In the figure shown, Find the x, y and z component of the force F if F = 100N and passes from point A(0,0,0) to the point B(3,4,5). Its direction cosines and the angle that it makes with the coordinate axes. Solution: Enter: [ MODE ] – [ 8 ] – [ 1 ] – [ 1 ] ‘VECTOR MODE – 3D’ Input: [ 3 4 5] Press: [ AC ] ‘VECTOR DIFFERENCE BET. B & A’ To get direction cosines or unit vector: Enter: [ SHIFT ] – [ 5 ] – [ 3 ] – [ ÷ ] – [ SHIFT ] – [ HYP ] – [ SHIFT ] – [ 5 ] – [ 3 ] – [ = ] Display: VctA ÷ Abs(VctA) Output: [ 0.42426 0.56568 0.70711 ] To get the components: Enter: [ x ] – [ 100 ] – [ = ] Display: [ 42.426 55.568 Fx = 42.426 N Fy = 55.568 N Fz = 70.711 N
70.711 ]
To get the angles it makes with the coordinate axes just take the inverse cosine of the direction cosines above.
CONCURRENT – NONCOPLANAR FORCE SYSTEM Past CE Board Exam A concurrent force system in space is composed of three forces described as follows. P1 has a magnitude of 100kN and acts through the origin and the points x = 3, y = 4 and z = 2. P2 has the magnitude of 60kN and acts through he origin and the points x = 4, y = 1 and z = - 2 P3 has the magnitude of 80kN and acts through the origin and the points x = 2, y = -3 and z = 3 Find the components of the resultant, its direction cosines and magnitude of the resultant.
CONCURRENT – NONCOPLANAR FORCE SYSTEM Solution: P1 = 100kN P2 = 60kN P3 = 80kN
(3, 4, 2) (4, 1, -2) (2, -3, 3)
Enter: [ MODE ] – [ 8 ] – [ 1 ] Input: [ 3 4 2 ] for VctA Enter: [ AC ] Enter: [ SHIFT ] – [ 5 ] – [ 1 ] – [ 2 ] – [1 ] Input: [ 4 1 -2 ] for VctB Enter: [ AC ] Enter: [ SHIFT ] – [ 5 ] – [ 1 ] – [ 3 ] – [ 1 ] Input: [ 2 -3 3 ] for VctC Enter [ AC ] To get the COMPONENTS:
Input: 100VctA ÷ Abs(VctA) + 60VctB ÷ Abs(VctB) + 80VctA ÷ Abs(VctC) Enter: [ = ] Output: [ 142.913
36.203
62.121 ]
Rx = 142. 913 kN Ry = 36.203 kN Rz = 62.121 kN This Vector is automatically stored to VctAns
Input: Abs(VctAns) Enter: [ = ] Output: R = 159.3378 kN Input: VctAns ÷ Abs(VctAns) Enter: [ = ] Output: [ 0.8924 0.2272
0.3898 ]
PARALLEL – NONCOPLANAR FORCE SYSTEM
Find the resultant of the four forces acting on the plane shown. Also find its position.
PARALLEL – NONCOPLANAR FORCE SYSTEM Solution: Enter: [ MODE ] – [ 3 ] – [ 2 ] Enter: [ SHIFT ] – [ MODE ] – [
‘ STAT MODE’ ] – [ 4 ] – [ 1 ] ‘Frequency ON’
X
Y
Freq
2
-6
70
6
-6
64
6
0
72
Resultant, R = 286 lb
0 0 80 Using Right Hand Rule the above data is obtained X is the MOMENT ARM for the Moment about the x-axis Y is the MOMENT ARM for the Moment about the y-axis Freq is the FORCE To get Mx, Enter: [SHIFT] – [ 1 ] – [ 3 ] – [ = ] Output: ∑X = 956 lb-ft = Mx To Get My, Enter: [ SHIFT ] – [ 3 ] – [ 4 ] – [ = ] Output: ∑Y = -804 lb-ft = My Then Mo = 286yi – 286xj = 956i – 804j
y = 3.343, x = 2.881
APPLICATIONS
EQUILIBRIUM OF COPLANAR FORCES CALCULATOR TECHNIQUE USING MODE 5: EQN (TWO – UNKNOWNS) NOTE: For REACTIVE FORCES all components to the RIGHT and UPWARD are POSITIVE. All components to the LEFT and DOWNWARD are NEGATIVE For APPLIED FORCES the sign convention of the REACTIVE FORCES are REVERSED. Example: Determine the Tensile Forces in each of the chords that support the 200 lb weight as shown.
EQUILIBRIUM OF COPLANAR FORCES CALCULATOR TECHNIQUE USING MODE 5: EQN (TWO – UNKNOWNS) Solution: Enter: [ MODE ] – [ 5 ] – [ 1 ] Display and Input:
‘ EQUATION MODE – 2 UNKNOWNS’
BC
BA
Applied Loads
a
b
c
Cos30
- cos60
0
X - Component
Sin30
Sin60
200
Y-Component
Enter: [ = ] Output: X = BC = 100lb
AND
Y = BA = 173.205lb
TRUSS ANALYSIS
PROBLEM: Determine the forces in all members of the Crane Truss shown. Using Method of Joints.
TRUSS ANALYSIS Solution: For Triangle ABC: AC = 13.077 ft. [ SOLVED BY COSINE LAW] By Sine Law : α = 23.4120 β = 36.5860 For Triangle CBD Θ = tan-1 12/6 = 63.4350 While, γ = 180 – ( α+β+θ ) = 56.5670
TRUSS ANALYSIS @ Joint A
Solution: Enter: [ MODE ] – [ 5 ] – [ 1 ] Let AC = X and AB = Y , where α = 23.4120 Input: Applied Loads a
b
c
-cos23.412
-1
O
XComponent
Sin23.412
0
5200
Y-Component
Enter: [ = ] Output: AC = 13,087.55 lb (Tension) AB = - 12,010.16 lb (Compression)
TRUSS ANALYSIS @ Joint B
Solution: Enter: [ MODE ] – [ 5 ] – [ 1 ] Let BC = X and BD = Y Input: Applied Loads a
b
c
-cos60
-cos30
12,010.16
X - Component
Sin60
-sin30
0
Y-Component
Enter: [ = ] Output: BC = - 6005.08 lb (Compression) BD = - 10,401.104lb (Compression)
TRUSS ANALYSIS @ Joint C
Solution: Enter: [ MODE ] – [ 5 ] – [ 1 ] Let CE = X and CD = Y , where α = 23.4120 , β = 36.5860 , Θ = 63.4350 & γ = 56.5670 Input: a -1 0
b -cos56.567 -sin56.567
c [-13,087.55cos23.412 + 6,005.08cos59.998] [13,087.55sin23.412 - 6,005.08sin59.998]
Enter: [ = ] Output: CE = 9007.19 lb (Tension) CD = 0.285 lb (Tension)
BEAMS Solution: Enter: [ MODE ] – [ 5 ] – [ 1 ]
‘EQN- 2 UNKNOWNS’
Let Ra = X and Rb = Y Input: a
b
c
8
0
397.5
Ma or Mb = 0
1
1
132.5
Fy = 0
Enter: [ = ] Output: X = Ra = 49.6875kN Y = Rb = 82.8125kN Find the magnitude of the reaction in the simply supported beam as shown in the figure.
CHECK: UPWARD FORCES = DOWNWARD FORCES 49.6875 + 82.8125 = 10 + .5(15)(3) + 25(4) 132.5 = 132.5 okay!!!
CABLES
Past CE Board For the cable shown, determine the angle β1 & β2 and the forces in each segment and the length of the cable
CABLES @ Joint C
Solution: Enter: [ MODE ] – [ 2 ]
‘Complex Mode’
BC<35 + CD<(180 – β2) + 2000< 270 = 0 Input(Right Side of the Equation to your Calculator) CD < (180 – β2) = - ( 2333.95<35 + 2000<270 ) CD < (180 – β2) = 2023 < 160.92 Thus,
CD = 2023 lb 180 - β2 = 160.92
β2 = 19.080
CABLES @ Joint D
Solution: Enter: [ MODE ] – [ 2 ]
‘Complex Mode’
CD<(- β2 or 360 – β2) + AD<(180 – β1) + 1600< 270 = 0 Input(Right Side of the Equation to your Calculator) AD < (180 – β1) = - ( 2023<(-19.08) + 1600<270 ) AD < (180 – β1) = 2961.19 < 130.2140
Thus, AD = 2961.19 lb
180 – β1 = 130.214
β1 = 49.80
CABLES Solution To get the length of the cable:
Length of the Cable = ∑L/cosβ Where β = 350 , β1 = 49.80 and β2 = 19.08
Length of the Cable = 7/cos35 + 11/cos 19.08 + 6/cos49.8 Length of the Cable = 29.48 feet
CENTROID AND MOMENT OF INERTIA Solution: Enter: [ MODE ] – [ 3 ] – [ 2 ] ‘STAT MODE – LINEAR’ Enter: [ SHIFT ] – [ MODE ] – [ ] – [ 4 ] – [ 1 ] ‘FREQUENCY ON’ To get the CENTROID Ῡ from the X-AXIS STAND BY
C.G. OF EACH AREA FROM X-AXIS
AREA
X
Y
Freq
30
60 X 200
AREA 1
60 + 250/2
250 X 40
AREA 2
Enter: [AC]
To get Ῡ: Input: [ SHIFT] – [ 1 ] – [ 3 ] – [ 4 ] – [ ÷ ] – [ SHIFT ] – [1]–[4]–[1]–[=] Display: ∑y ÷ n = Ῡ = 100.4545 mm
CENTROID AND MOMENT OF INERTIA Solution: Enter: [ MODE ] – [ 3 ] – [ 2 ] ‘STAT MODE – LINEAR’ Enter: [ SHIFT ] – [ MODE ] – [ ] – [ 4 ] – [ 1 ] ‘FREQUENCY ON’ Store Ῡ to ‘VARIABLE – A’ Enter: [ SHIFT ] – [ RCL ]
Display: Ans
A
Edit Data Enter: [ SHIFT ] – [ 1 ] – [ 2 ] Y MINUS A
C.G. OF EACH AREA FROM X-AXIS
AREA
X
Y
Freq
30 – A
30
60 X 200
AREA 1
60 + 250/2 - A
60 + 250/2
250 X 40
AREA 2
To get the Centroidal Moment of Inertia ( Igg) Input: ∑bh3 /12 + ∑x2 ‘∑x2 is [ SHIFT ] – [ 1 ] – [ 3 ] – [ 1 ]’ Output: 40(250)3 / 12 + 200(60)2 / 12 + ∑x2 Enter: [ = ] Display: Igg = 186.7287879 x 106 mm4
FRICTION A 200lb block is in contact with a plane inclined at 300 with the horizontal. A force P parallel to and acting up the plane is applied to the body. If the coefficient of static friction is 0.2. a. Find the value of P to just cause the motion to impend up the plane b. Find the value of P to just prevent the motion down the plane.
FRICTION A. Solution: Enter: [ MODE ] – [ 5 ] – [ 1 ] ‘EQN MODE – 2 UNKNOWNS’ Input: P
N
Applied Load
a
b
c
1
-0.2
200sin30
X-component
0
1
200cos30
Y-component
Enter: [ = ] Output: P = X = 134.64 Newtons N = Y = 173.2 Newtons
FRICTION A. Solution: Enter: [ MODE ] – [ 5 ] – [ 1 ] ‘EQN MODE – 2 UNKNOWNS’ Input: P
N
Applied Load
a
b
c
1
0.2
200sin30
X-component
0
1
200cos30
Y-component
Enter: [ = ] Output: P = X = 65.35 Newtons N = Y = 173.2 Newtons
END OF SLIDES Prepared By: Engr. Niño Gem M. Ngo Lee
WEIGHTED OBSERVATION The difference of elevation between B and C was taken by a survey party using different trials and paths. Compute the probable elevation of point C if it is above point B. Elevation of point B is 825m. Trial
Distance
Diff. in Elevation
1
2
0.89m
2
6
0.67m
3
4
0.78m
4
12
1.02m
Enter: [ MODE ] – [ 3 ] – [ 2 ] – [ SHIFT ] – [ MODE ] – [ 1 ] – [ ] – [ 4 ] – [ 1 ] X
Y
½
0.89
1/6
0.67
¼
0.78
1/12
1.02
‘STAT MODE – FREQUENCY ON’
WEIGHTED OBSERVATION The difference of elevation between B and C was taken by a survey party using different trials and paths. Compute the probable elevation of point C if it is above point B. Elevation of point B is 825m. X
Y
½
0.89
1/6
0.67
¼
0.78
1/12
1.02
Enter: [ AC ] Input: [ SHIFT ] – [ 1 ] – [ 3 ] – [ 5 ] – [÷] – [ SHIFT ] – [ 1 ] – [ 3 ] – [ 2 ] Display: ∑xy÷∑x Enter: [ = ]
X average = 0.83667
Thus, Elev. At C = Elev. At B + 0.83667 = 825 + 0.83667 = 825.83667m
CLOSED COMPASS TRAVERSE Given the following traverse notes taken by a survey party, find: a. Linear Error of Closure b. Relative Error c. Correct using Compass Rule d. Correct using Transit Rule
LINE
DISTANCE
BEARING
AB
410m
N 45020’ W
BC
605m
S 65010’ E
CD
600m
N 80015’E
DA
280m
S 55030’ W
SURVEYING
MISSING SIDES CASE 1: Length and Bearing of One Side Completely Missing LINE
DISTANCE
BEARING
AB
300m
S 37030’ E
BC
400m
S 43015’ W
CD
250m
N 73000’ W
DE
350m
N 12045’E
EA
?
?
MISSING SIDES CASE 2: Length of Two Adjacent Sides Missing CASE 5: Length of Two Non-Adjacent Sides Missing LINE
DISTANCE
BEARING
AB
300m
S 37030’ E
BC
400m
S 43015’ W
CD
250m
N 73000’ W
DE
X
N 12045’E
EA
Y
N 65040’E
LINE
DISTANCE
BEARING
AB
200m
N 73000’ W
BC
X
N 12045’E
CD
300m
S 37030’ E
DE
Y
N 65040’E
EA
400m
S 43015’ W
MISSING SIDES CASE 3: Length of One Side and Bearing of its Adjacent Side Missing LINE
DISTANCE
BEARING
AB
300m
S 37030’ E
BC
400m
S 43015’ W
CD
250m
N 73000’ W
DE
X
N 12045’E
EA
277.96m
βea
MISSING SIDES CASE 4: Bearing of Two Adjacent Sides Missing LINE
DISTANCE
BEARING
AB
300m
S 37030’ E
BC
400m
S 43015’ W
CD
250m
N 73000’ W
DE
350m
?
EA
277.96m
?
