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ENGINEERING SURVEYING (221 BE)
Levelling-TechnicalLevelling-Technical
Sr Dr. Tan Liat Choon
Email: [email protected]
Mobile: 016-4975551
LEVELLING
TWO PEG TEST
� Set 2 marks at 30-40 metre apart, also mark centre point in a relatively flat area
� Set level at midpoint and take readings at each end
� Determine difference in readings (difference in elevation)
Move level to one end and setup so that level is just in front of rod on point� Move level to one end and setup so that level is just in front of rod on point
� Read rod by looking backward through scope (x-hair not visible), hold pencil on rod to determine reading
� Read rod at other end in normal manner
� Difference in readings should equal of 3
� If values are not equal, there is error• Most instruments have adjustment screws• Adjust and repeat test as a check
2
Two Peg Test
x
S1
S1’
Line of Collimation
Horizontal Line
3
L / 2 L / 2
L
A
B
S2
S2’
x x
S1
S1’
Line of Collimation
Horizontal Line
Two Peg Test
L / 2 L / 2
L
A
B
= S1’ - S2’
The APPARENT height difference δ hA
The TRUE height difference hTδ
= S1 - S2
S1= S1’ + x and S2
= S2’ + x 4
x
S1
S1’
Line of Collimation
Horizontal Line
S2
S2’
x
Two Peg Test
L / 2 L / 2
L
A
B
δThe TRUE height difference hT = S1’ - S2’
= S1 - S2The APPARENT height difference δ hA
S1= S1’ + x and S2
= S2’ + x hA = (S1’ + x) - (S2’ + x )δ 5
x
S1
S1’
Line of Collimation
Horizontal Line
S2
S2’
x
Two Peg Test
L / 2 L / 2
L
A
B
δThe TRUE height difference hT = S1’ - S2’
= S1 - S2The APPARENT height difference δ hA
S1= S1’ + x and S2
= S2’ + x hA = S1’ - S2’δ = δ hT6
hA δ = δ hT
Therefore :
This is true since the instrument is the same distance from both staff positions and
the errors x are equal and cancel out
Two Peg Test
7
S3’
S3
Now move the instrument outside the “odd numbered” peg
Two Peg Test
A
B
L / 10
8
S3
S3’
S4
S4’
Two Peg Test
A
B
L / 10
= S3 - S4The APPARENT height difference δ hA
δBut the TRUE height difference hT We already know
9
S3
S3’
S4
S4’
Two Peg Test
A
B
L / 10
= S3 - S4
= S1 - S2
then the instrument is OK
If NOT then the error is e =
The APPARENT height difference δ hA
δBut the TRUE height difference hT
δ δTherefore if hA = hT
(S1 - S2) - (S3 - S4) / L mm / m10
Summary
Place two pegs about L = 30m (to 40m) apart
Set up level midway between the two pegs
Read staff on each peg, and calculate true height difference
Move level about L / 10 = 3m (or 4m) beyond one of the pegs
Read staff on each peg again, and calculate height differenceRead staff on each peg again, and calculate height difference
Collimation Error e = difference in the differences
and is expressed as a number of mm per L m
Acceptable errors
Uren and Price 1mm per 20m
Wimpey 4mm per 50m
Test should be carried out regularly say once per week or two11
L / 2 L / 2
L
S1
AB
S2
S3
S4 S3
AB
L / 10Collimation error,
e = (S1 - S2) - (S3 - S4) mm / Lm12
DATUM
Could be our own Datum - Assumed Datum
- Arbitrary Datum
- Site DatumOr
A National Datum - Ordnance Datum
Above Assumed Datum
In the UK we have a national organisation known as
The Ordnance Survey (O.S.)
The O.S. has established a ZERO Datum at Newlyn in
Cornwall.
Based on the Ordnance Datum - points of known height above
or below Zero height have been established around the U.K.
These points around the country are known as Bench Marks
Above Ordnance Datum
13
Levelling
Measured and CalculatedReducedRL A (known)ReducedLevel of BRL B
(unknown)
the Plane of CollimationHeight of (HPC)S1
Levelling Staff
S2
AB
Measured and CalculatedLevel of A ReducedLevel of A
RL A (known) Level of BRL B
DATUM
HPC = RL A + S1HPC = RL A + S1
RL B = HPC - S2
14
AB
C
Some Terminology
RL A RL BRL C
S1
Level staff on A Back Sight (BS) reading is first reading
BS
Levelling
Level staff on A Back Sight (BS) reading is first reading
15
RL A RL B
AB
C
RL C
Level staff on A Back Sight (BS) reading is first reading
S2FS
Levelling
Level staff on A Back Sight (BS) reading is first reading
Level staff on B Fore Sight (FS) reading is last reading
Move instrument to new position
16
RL A RL BRL C
AB
C
CP
S3BS Levelling
Move instrument to new position Level staff stays on B
The instrument has changed its position about point B
Point B is known as a Change Point (CP)
CP
2nd instrument position starts with BS to B
17
FSS4S3
BS
RL A RL BRL C
AB
C
Levelling
and finishes with FS to C
18
RL A RL B
AB
C
RL C
BS FS
BS FS
RL A is known
HPC
(CP)
HPCLevelling
RL A is known
HPC = RL A + BS RL B = HPC - FS
Now the RL B is known So we can repeat the process
HPC = RL B + BS RL C = HPC - FS
Generally : HPC = Known RL + Back Sight
Unknown RL = HPC - Fore Sight 19
RL A RL B
AB
C
RL C
BS FS
BS FSHPC
HPC
(CP)
Levelling
RL A is known
HPC = RL A + BS RL B = HPC - FS
RL B + BS RL C = HPC - FS HPC =
Generally : HPC = Known RL + Back Sight
Unknown RL = HPC - Fore Sight
Now the RL B is known So we can repeat the process
20
PLANE AND COLLIMATION METHOD
� This method is simple and easy
� Reduction of levels is easy
� Visualization is not necessary regarding the nature of the ground
� There is no check for intermediate sight readings
� This method is generally used where more number of readings can be taken
with less number of change points for constructional work and profile
levelling
� To check:
∑ BS - ∑ FS = Last RL – First RL
21
PLANE AND COLLIMATION METHOD
Determine the RLs of various points if the reduced level (RL) of a point on which the first reading was taken is
136.440 gives the Height of Plane and Collimation method and applies the check
Station BS IS FS HLC RL Remarks
1 0.585 137.025 136.440 BM A RL=136.440
HLC = RL + BS
= 136.440 + 0.585 = 137.025
RL = HL – BS
Check(Summation of BS)-(Summation of FS) = Last RL – First RL
2.670 – 9.275 = 129.835 – 136.440
-6.605 = -6.605
22
2
3
4
5 0.350
1.010
1.735
3.295
3.775 133.600
136.015
135.290
133.730
133.250 CPI
6
7
8
9
10 1.735
1.300
1.795
2.575
3.375
3.895 131.440
132.300
131.805
131.025
130.225
129.705 CP 2
11
12
0.635
1.605
130.805
129.835 BM B RL=129.835
Sum of BS=2.670 Sum of FS = 9.275
2.690-9.275 = -6.605 129.835-136.440= -6.605
How Levelling is Conduct
23
How Levelling is Conduct
24
Calculation checks
∑ FS - ∑ BS = 1st RL - Last RL
∑ IS + ∑ FS + ∑ (RLs except first)
= ∑ (each HPC x number of applications)
Simple check
Full check
Check Misclosure
Allowable Misclosure = 5 √N mm. ("Rule of Thumb")
When calculations are checked and
if the misclosure is allowable
Distribute the misclosure
25
RISE AND FALL METHOD
� This method is complicated and is not easy to carry out
� Reduction of levels takes more time
� Visualization is necessary regarding the nature of the ground
� Complete check is there for all readings
� This method is preferable for check levelling where number of
change points is more
� To check:
∑ BS - ∑ FS = ∑ Rise - ∑ Fall = Last RL – First RL
26
RISE AND FALL METHODDetermine the RLs of various points if the reduced level (RL) of a point on which the first reading was taken is
122.156 gives the Height of Rise and Fall method and applies the check
Station BS IS FS Rise Fall RL Remarks
1
2
1.536
0.974 2.072 0.536
122.156
121.620
BM A RL=122.156m
CP1
R/F = BS - FS= 1.536 – 2.072 = 0.536
RL = HL – IS
Check(Summation of BS)-(Summation of FS) = Last RL – First RL
12.172 – 12.725 = 121.605 – 122.156
-0.551 = -0.551
27
2
3
4
0.974
1.124
1.768
2.072
2.700
0.536
0.794
0.932
121.620
120.826
119.894
CP1
CP2
5
6
7
8
9
2.236
1.413
1.994
1.639
1.256
2.362
1.302
0.874
0.825
1.120
0.934
0.539
1.169
0.519
1.238 118.656
119.590
120.129
121.298
121.817
CP 3
CP 4
CP 5
CP 6
CP 7
10 1.468 0.212 121.605 BM B RL=121.605
Sum of BS=12.172 Sum of FS =
12.723
3.161 3.712
12.723-12.172=0.551 3.712-3.161=0.551 122.156-
121.605=0.551
How Levelling is Conduct
28
COMPARISON
� Plan and Collimation Method
• Quicker
• Good for a lot of IFSs
� Rise and Hall Method
• More accurate
• More calculation
• Intermediate RLs are known
29
ACCURACY IN LEVELLING
For normal engineering works and site surveys
Allowance misclosure = ± 5 √ N mm
Where N = Number of instrument positions
OR
Allowance misclosure = ± 12 √ K mm
Where K = length of levelling circuit in KM
If actual misclosure > allowance misclosure, levelling should be repeated
If actual misclosure < allowance misclosure, misclosure should be equally
distributed between the instrument positions
30
CORRECTION IN LEVELLING
Correction = (Misclosure / No. of Station) x n, n+1, n+2 and ………
For example:
Loop 1 = (0.117 / 3) x 1 = 0.039m
Loop 2 = (0.117 / 3) x 2 = 0.078m
Loop 3 = (0.117 / 3) x 3 = 0.117mLoop 3 = (0.117 / 3) x 3 = 0.117m
31
Example
32
Example
33
Example
34
Example
35
Example
36
Example
37
Example
38
Summary of work:
Check tripod is on stable ground or dig feet well in
Use pond bubble to set approximately vertical standing axis
LEVELLING WORK
Eliminate PARALLAX every time we sight the staff
check that the compensators are functioning every
time we sight the staff.
and
39
At every instrument set up - always start with a BS to a
point of known RL.
At every instrument set up - always finish with a FS.
Either the instrument moves or the staff moves
LEVELLING WORK
Either the instrument moves or the staff moves
NEVER BOTH
ALWAYS CLOSE levelling to a point of KNOWN RL
40
TBM
9.09 m A.A.D.
Main Gate
Burnaby Building
Approximate North
TBM
10.00 m A.A.D.
Start at a TBM outside the main entrance of Burnaby Building and obtain the
RL values of three points before closing onto another TBM near the main gate.
Point 1
Ground level at entrance to structures laboratory
Top of door level at entrance to structures laboratory
41
Site: …………………………………. Instrument: ………………………………….
Date: …………………………………. Observer: ………………………………….
Weather: …………………………………. Booker: ………………………………….
BS IS FS HPC RL Corr Corr RL Remarks
Building A 2
27/09/11 Group 1
Good Group 1
TBM 10.00m AAD
It is important to complete details at the top of booking forms or
on every page of field books.
42
TBM Level Posn. BS
Key
43
Site: …………………………………. Instrument: ………………………………….
Date: …………………………………. Observer: ………………………………….
Weather: …………………………………. Booker: ………………………………….
BS IS FS HPC RL Corr Corr RL Remarks
Building A 2
27/09/11 Group
Good Group
TBM 10.00m AAD10.0001.546 11.546
HPC = RL + BS HPC = 10.000 + 1.546 = 11.546
We now signal to the staff person to move to the next point.
As the next required point is too far away (it is also round a corner)
we will eventually need to move the instrument.
So, we must move the staff to a change point (CP), to allow us to move the
instrument to a better position later on.
44
TBM CP Level Posn. BS FS
Key
45
Site: …………………………………. Instrument: ………………………………….
Date: …………………………………. Observer: ………………………………….
Weather: …………………………………. Booker: ………………………………….
BS IS FS HPC RL Corr Corr RL Remarks
Building A 2
27/09/11 Group 1
Good Group 1
TBM 10.00m AAD1.546 10.00011.546
C.P.
New staff
position
therefore
a new row.
Each rowrepresents
1.562 9.984
represents
a staff
position.
RL = HPC - FS RL = 11.546 - 1.562 = 9.984
After we make a FS and we have calculated the new RL we are finished
with that instrument position.
Move the Instrument (about the CP) to a new position where we can see the CP
and also the next point we want the RL value of. 46
TBM CP Level Posn. BS FS ISKey
47
Site: …………………………………. Instrument: ………………………………….
Date: …………………………………. Observer: ………………………………….
Weather: …………………………………. Booker: ………………………………….
BS IS FS HPC RL Corr Corr RL Remarks
Building A 2
27/09/11 Group 1
Good Group 1
TBM 10.00m AAD1.546 10.00011.546
C.P.1.562 9.9841.418
Same staff
position as
last reading
therefore
the same row
11.402
HPC = RL + BS HPC = 9.984 + 1.418 = 11.402
48
TBM CP Level Posn. BS FS ISKey
This reading is not the first so it is not a BSIt is not the last from this position (we can see the next points) so it is not a FS
So it is known as an INTERMEDIATE SIGHT (IS)49
Site: …………………………………. Instrument: ………………………………….
Date: …………………………………. Observer: ………………………………….
Weather: …………………………………. Booker: ………………………………….
BS IS FS HPC RL Corr Corr RL Remarks
Building A 2
27/09/11 Group 1
Good Group 1
TBM 10.00m AAD1.546 10.00011.546
C.P.1.562 9.9841.418 11.402
Point 11.390 10.012
New staff
position
therefore
a new row
RL = HPC - IS RL = 11.402 - 1.390 = 10.012
50
TBM CP Level Posn. BS FS ISKey
51
Site: …………………………………. Instrument: ………………………………….
Date: …………………………………. Observer: ………………………………….
Weather: …………………………………. Booker: ………………………………….
BS IS FS HPC RL Corr Corr RL Remarks
Building A 2
27/09/11 Group 1
Good Group 1
TBM 10.00m AAD1.546 10.00011.546
C.P.1.562 9.9841.418 11.402
Point 11.390 10.012
GL Struct. Lab Door
New staff
position
therefore 1.281 10.121
therefore
a new row
RL = HPC - IS RL = 11.402 - 1.281 = 10.121
52
TBM CP Level Posn. BS FS ISKey
53
Site: …………………………………. Instrument: ………………………………….
Date: …………………………………. Observer: ………………………………….
Weather: …………………………………. Booker: ………………………………….
BS IS FS HPC RL Corr Corr RL Remarks
Building A 2
27/09/11 Group 1
Good Group 1
TBM 10.00m AAD1.546 10.00011.546
C.P.1.562 9.9841.418 11.402
Point 11.390 10.012
GL Struct. Lab Door1.281 10.121 New staff
position Top Struct. Lab Door
position
therefore
a new row
Requires an inverted staff i.e turn the staff upside down
Read and then book the staff with a sign
-2.420
The negative sign will keep all the calculations correct
RL = HPC - IS RL = 11.402 - (-2.420) = 11.402 + 2.420 = 13.822
13.822
54
TBM CP Level Posn. BS FS ISKey
The last point required is the TBM. However it is too long a sight
So we need a CP. This will be the last sighting from this position
Therefore we need a FS
55
Site: …………………………………. Instrument: ………………………………….
Date: …………………………………. Observer: ………………………………….
Weather: …………………………………. Booker: ………………………………….
Building A 2
27/09/11 Group 1
Good Group 1
BS IS FS HPC RL Corr Corr RL Remarks
TBM 10.00m AAD1.546 10.00011.546
C.P.1.562 9.9841.418 11.402
Point 11.390 10.012
GL Struct. Lab Door1.281 10.121
Top Struct. Lab Door-2.420 13.822
GL Struct. Lab Door1.281 10.121
CP New staff
position
therefore
a new
row
1.321
RL = HPC - FS RL = 11.402 - 1.321 = 10.081
10.081
Last Reading -- FS -- Move the instrument56
TBM CP Level Posn. BS FS ISKey
57
Site: …………………………………. Instrument: ………………………………….
Date: …………………………………. Observer: ………………………………….
Weather: …………………………………. Booker: ………………………………….
Building A 2
27/09/11 Group 1
Good Group 1
BS IS FS HPC RL Corr Corr RL Remarks
TBM 10.00m AAD1.546 10.00011.546
C.P.1.562 9.9841.418 11.402
Point 11.390 10.012
GL Struct. Lab Door1.281 10.121
Top Struct. Lab Door-2.420 13.822
GL Struct. Lab Door1.281 10.121
CP1.321 10.0811.011 Same staff
position as
last reading
therefore
the same row
HPC = RL + BS HPC = 10.081 + 1.011 = 11.092
11.092
58
TBM CP Level Posn. BS FS ISKey
59
Site: …………………………………. Instrument: ………………………………….
Date: …………………………………. Observer: ………………………………….
Weather: …………………………………. Booker: ………………………………….
Building A 2
27/09/11 Group 1
Good Group 1
BS IS FS HPC RL Corr Corr RL Remarks
TBM 10.00m AAD1.546 10.00011.546
C.P.1.562 9.9841.418 11.402
Point 11.390 10.012
GL Struct. Lab Door1.281 10.121
Top Struct. Lab Door-2.420 13.822
GL Struct. Lab Door1.281 10.121
CP1.321 10.0811.011 11.092
TBM 9.09m AAD
New staff
position
therefore
a new row
2.009
RL = HPC - FS RL = 11.092 - 2.009 = 9.083
9.083
60
Before we look more fully at the results we will complete the
second half of the levelling exercise
TBM CP Level Posn. BS FS ISKey
61
Top of door level at entrance to structures laboratory
Ground level at entrance to structures laboratory
Point 2
TBM CP Level Posn. BS FS ISKey
62
TBM CP Level Posn. BS FS ISKey
63
TBM CP Level Posn. BS FS ISKey
64
TBM CP Level Posn. BS FS ISKey
65
TBM CP Level Posn. BS FS ISKey
66
TBM CP Level Posn. BS FS ISKey
67
TBM CP Level Posn. BS FS ISKey
68
TBM CP Level Posn. BS FS ISKey
69
TBM CP Level Posn. BS FS ISKey
70
TBM CP Level Posn. BS FS ISKey
71
Summary of Levelling Fieldwork
� For levelling fieldwork, the following practice should be adhered to in order to
improve the accuracy of the levelling works
� Levelling should always start and finish at points of known RL so that misclosure can
be detected
� Where possible, all sight lengths should be below 60 metres� Where possible, all sight lengths should be below 60 metres
� The staff must be held vertical by suitable use of a bracket bubble
� BS lengths =/~ FS lengths
� Reading should be booked immediately after they are observed. Important readings,
particularly readings at change points, should be checked
� The rise and fall method of reduction should be used if possible, especially for control
works. This HPC is only a sample
72
BS IS FS HPC RL Corr Corr RL Remarks
Top Struct. Lab Door-2.420 13.822
TBM 10.00m AAD1.546 10.00011.546
C.P.1.562 9.9841.418 11.402
Point 11.390 10.012
GL Struct. Lab Door1.281 10.121
CP1.321 10.0811.011 11.092
Levelling Booking & Calculation
CP1.321 10.0811.011 11.092
TBM 9.09m AAD2.009 9.083
The RL value of 9.083m is our measured and calculated value.
It should be 9.09m.
This gives an actual misclosure of 9.083 - 9.09 = -0.007m
This actual misclosure may be because of calculation errors or field errors 73
BS IS FS HPC RL Corr Corr RL Remarks
TBM 10.00m AAD1.546 10.00011.546
C.P.1.562 9.9841.418 11.402
Point 11.390 10.012
GL Struct. Lab Door1.281 10.121
If it is due to calculation errors we MUST NOT continue.
Therefore the first thing we always do after reducing our field booking is:
Carry out Calculation Checks
∑ FS - ∑ BS = 1st RL - Last RLSimple Calculation Check:
Top Struct. Lab Door-2.420 13.822
GL Struct. Lab Door1.281 10.121
CP1.321 10.0811.011 11.092
TBM 9.09m AAD2.009 9.083
∑ FS - ∑ BS = 1st RL - Last RL
∑ ∑3.975 4.892
LHS = 4.892 - 3.975 = 0.917
RHS = 10.000 - 9.083 = 0.917
Therefore LHS = RHS Therefore Calculations are OK 74
BS IS FS HPC RL Corr Corr RL Remarks
TBM 10.00m AAD1.546 10.00011.546
C.P.1.562 9.9841.418 11.402
Point 11.390 10.012
GL Struct. Lab Door1.281 10.121 NOT
CHECKED
NOT
CHECKED
Full Calculation Check:
∑ IS + ∑ FS + ∑ (RLs except first)
= ∑ (each HPC x number of applications)
Top Struct. Lab Door-2.420 13.822
GL Struct. Lab Door1.281 10.121
CP1.321 10.0811.011 11.092
TBM 9.09m AAD2.009 9.083
This Simple Check does not check the calculations for RL values calculated from IS
CHECKED
NOT
CHECKED
∑ FS - ∑ BS = 1st RL - Last RL
∑ ∑3.975 4.892
75
1.418
1.546
BS IS FS HPC RL Corr Corr RL Remarks
Top Struct. Lab Door-2.420 13.822
TBM 10.00m AAD10.00011.546
C.P.1.562 9.98411.402
Point 11.390 10.012
GL Struct. Lab Door1.281 10.121
CP1.321 10.0811.011 11.092
TBM 9.09m AAD2.009 9.083
LHS = ∑ IS + ∑ FS + ∑ (RLs except first)
∑ 0.251 ∑ 4.892 ∑ 63.103
= 0.251 + 4.892 + 63.103 = 68.246
RHS = ∑ (each HPC x number of applications)
= (11.546x1+ 11.402x4 + 11.092x1) = (11.546 + 45.608 + 11.092) = 68.246
Therefore LHS = RHS
Therefore the calculations for all the RL values
are correct. 76
BS IS FS HPC RL Corr Corr RL Remarks
Top Struct. Lab Door-2.420 13.822
TBM 10.00m AAD1.546 10.00011.546
C.P.1.562 9.9841.418 11.402
Point 11.390 10.012
GL Struct. Lab Door1.281 10.121
CP1.321 10.0811.011 11.092
Now we can look at the magnitude of the misclosure
TBM 9.09m AAD2.009 9.083
We have already seen that the
Actual misclosure = 9.083 - 9.09 = -0.007mIs this acceptable ?
Rule of Thumb:
Allowable misclosure = ± 5 √N mmWhere N is the Number of Instrument Positions
which is the same as Number of BS readings
Therefore our Allowable misclosure = ± 5 √3 mm = ± 8.66 say ± 9mm
Therefore Actual < Allowable Therefore our Fieldwork is OK 77
We have carried out the calculation checks and have an acceptable misclosure
The final stage is to apply a correction procedure to distribute the actual misclosure
We assume that we made a similar error every time we set up the instrument
There are 3 backsights, so we set up the instrument 3 times
The actual misclosure was -7mm, so we need to add 7mm in order to correct it
We can add these 7mm to our Reduced Levels in any way, but it is best to assume
that the 7mm error occurred gradually as a set of small errors,
rather than all in one go.
We could divide 7 between 3 like this: 3 2 2
Or like this: 2 3 2
Let use choose the middle method. We will give 2mm to the 1st instrument position,
an extra 3mm to the 2nd position, and an extra 2mm to the 3rd position
We cannot divide our 7mm misclosure evenly between 3 positions,
but we can do our best (we do not use fractions of a millimetre)
Or like this: 2 2 3
We must not correct the initial Reduced Level
We apply the same correction to all readings up to and including each foresight78
BS IS FS HPC RL Corr Corr RL Remarks
TBM 10.00m AAD1.546 10.00011.546
C.P.1.562 9.9841.418 11.402
Point 11.390 10.012
GL Struct. Lab Door1.281 10.121
x
We cannot correct the given TBM value
10.000
2
5
5
9.986
10.017
10.126
Top Struct. Lab Door-2.420 13.822
GL Struct. Lab Door1.281 10.121
CP1.321 10.0811.011 11.092
TBM 9.09m AAD2.009 9.083
5
5
5
7
Corrections are applied with a +ve or -ve sign depending on the sign of the misclosure
10.126
13.827
10.086
9.090
We MUST end up with the correct
final reduced level.
79
T H A N K YO UT H A N K YO U
Q u e s t i o n & A n s w e rQ u e s t i o n & A n s w e rQ u e s t i o n & A n s w e rQ u e s t i o n & A n s w e r
80