Problem 01 - Distance Corrections It is desired to lay off certain horizontal distances for building layouts. The length of the tapes used in the cases below Problems (a) and (b) are not 100.00 ft and not 30.00m for Problem (c). Determine the field dimensions (or actual tape readings) that should be used with the incorrect tape lengths so that the correct (true) dimensions are obtained. Desired Dimension Correct Tape Length (a) 725.00 ft x 180 ft 100.06 ft (b) 625.00 ft x 715 ft 99.96 Field dimensions of the readings (L): The equations for distance correction is: CDs = L+ L * (CTL-STDL) or CDs = L (1+ L (1 (CTL-STDL STDL CTL Br CDs = L* S STDL STDL CD5=L+(L/STD₁)*(CT₁-STD₁) (OneNote Equation) or CDs= L*CTL/STDL (OneNote Equation) O Search Noteboo Problem 08- Distance Corrections A 1-acre lot (43,560ft2) is to be staked out on level ground with the dimensions 250.00 ft by 174.24 ft. The standardized length of the tape at 68°F is 100.07ft. If the tape temperature is 94°F, what field dimensions should the survey party use to lay out this lot? The equations for distance correction are: S CDs = L+ L STDL *(CTL-STD) or CDs = L 1+ = 1 (1 (1 + (CTL-STDL) STDL CTL or CDs = L * STDL CDs=L+(L/STD)*(CTL-STD₁) (OneNote Equation) or CDs= L*CT/STD (OneNote Equation) I The equations for temperature correction are: Temperature Correction: CD₁= L (1+0.0000065 (T-Ts)) CDT= L+0.0000065*(T-Ts)*L or CD₁= L*(1+0.0000065* (T-Ts)) (OneNote Equation)
Problem 01 - Distance Corrections It is desired to lay off certain horizontal distances for building layouts. The length of the tapes used in the cases below Problems (a) and (b) are not 100.00 ft and not 30.00m for Problem (c). Determine the field dimensions (or actual tape readings) that should be used with the incorrect tape lengths so that the correct (true) dimensions are obtained. Desired Dimension Correct Tape Length (a) 725.00 ft x 180 ft 100.06 ft (b) 625.00 ft x 715 ft 99.96 Field dimensions of the readings (L): The equations for distance correction is: CDs = L+ L * (CTL-STDL) or CDs = L (1+ L (1 (CTL-STDL STDL CTL Br CDs = L* S STDL STDL CD5=L+(L/STD₁)*(CT₁-STD₁) (OneNote Equation) or CDs= L*CTL/STDL (OneNote Equation) O Search Noteboo Problem 08- Distance Corrections A 1-acre lot (43,560ft2) is to be staked out on level ground with the dimensions 250.00 ft by 174.24 ft. The standardized length of the tape at 68°F is 100.07ft. If the tape temperature is 94°F, what field dimensions should the survey party use to lay out this lot? The equations for distance correction are: S CDs = L+ L STDL *(CTL-STD) or CDs = L 1+ = 1 (1 (1 + (CTL-STDL) STDL CTL or CDs = L * STDL CDs=L+(L/STD)*(CTL-STD₁) (OneNote Equation) or CDs= L*CT/STD (OneNote Equation) I The equations for temperature correction are: Temperature Correction: CD₁= L (1+0.0000065 (T-Ts)) CDT= L+0.0000065*(T-Ts)*L or CD₁= L*(1+0.0000065* (T-Ts)) (OneNote Equation)
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Transcribed Image Text:Problem 01 - Distance Corrections
It is desired to lay off certain horizontal distances for building layouts. The length of the tapes used in
the cases below Problems (a) and (b) are not 100.00 ft and not 30.00m for Problem (c). Determine the
field dimensions (or actual tape readings) that should be used with the incorrect tape lengths so that
the correct (true) dimensions are obtained.
Desired Dimension Correct Tape Length
(a)
725.00 ft x 180 ft
100.06 ft
(b)
625.00 ft x 715 ft
99.96
Field dimensions of the readings (L):
The equations for distance correction is:
CDs = L+
L * (CTL-STDL) or CDs = L (1+
L (1
(CTL-STDL
STDL
CTL
Br CDs = L*
S
STDL
STDL
CD5=L+(L/STD₁)*(CT₁-STD₁) (OneNote Equation) or CDs= L*CTL/STDL (OneNote Equation)
O
Search Noteboo
Transcribed Image Text:Problem 08- Distance Corrections
A 1-acre lot (43,560ft2) is to be staked out on level ground with the dimensions 250.00 ft by 174.24 ft.
The standardized length of the tape at 68°F is 100.07ft. If the tape temperature is 94°F, what field
dimensions should the survey party use to lay out this lot?
The equations for distance correction are:
S
CDs = L+
L
STDL
*(CTL-STD) or CDs = L 1+
= 1 (1
(1 +
(CTL-STDL)
STDL
CTL
or CDs = L *
STDL
CDs=L+(L/STD)*(CTL-STD₁) (OneNote Equation) or CDs= L*CT/STD (OneNote Equation)
I
The equations for temperature correction are:
Temperature Correction: CD₁= L (1+0.0000065 (T-Ts))
CDT= L+0.0000065*(T-Ts)*L or CD₁= L*(1+0.0000065* (T-Ts)) (OneNote Equation)
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