A survey line was measured with a tape, believed to be 30m long and a length of 397.65m resulted. On checking, the tape was found to be 0.05m too short.
(a) What was the correct length of the line?
(b) If the line lay on a slope 1 in 20, what would be the reduced horizontal length used in the plotting of the survey?
(c) What tape reading is required to produce a horizontal distance of 15.08m between two site pegs, one being 0.66m above the other?
To solve these problems, we need to use the principles of surveying to adjust for an inaccurate tape measure and to account for the slope of the land.
**(a) What was the correct length of the line?**
The tape was 0.05m too short, meaning that for every 30 meters measured with the tape, the actual distance would be 30.05 meters. We need to find out how many 30m segments are in the measured 397.65m and then correct those segments with the 0.05m error each.
Number of tape measures in the measured length = Total measured length / Length of tape measure
Number of tape measures = 397.65m / 30m = 13.255
Now, we should correct the length by adding the error made in each segment:
Corrected length = Measured length + (Number of tape measures x Tape error)
Corrected length = 397.65m + (13.255 x 0.05m)
The 13.255 represents more than 13 whole tape lengths but less than 14. Since you can't use a partial tape measure, you would normally consider the fraction as well to account for the partial measure. However, as the tape measure is too short, we must ensure to count for every partial tape measure used.
So, we will round up the number of tape measures 13.255 to the nearest integer, which is 14.
Corrected length = 397.65m + (14 x 0.05m)
Corrected length = 397.65m + 0.7m
Corrected length = 398.35m
The correct length of the line is 398.35m.
**(b) If the line lay on a slope 1 in 20, what would be the reduced horizontal length used in the plotting of the survey?**
A slope of 1 in 20 means for every 20m horizontally, the elevation changes by 1m. We want to find the horizontal distance when the tape is laid on the slope.
We can use the Pythagorean theorem for this (since we assume the slope creates a right-angled triangle where the hypotenuse is the sloped distance measured by the tape).
Let's call the horizontal distance "d" and the vertical change in elevation "v":
d^2 + v^2 = corrected length^2
First, we need to find the change in elevation. We have already corrected the length of the measured distance considering the error in tape length. Now, we assume this is the hypotenuse of the right triangle formed by the slope.
We know that for every horizontal measure of 20m, there is a 1m elevation change. Since the distance is 398.35m, we can find the total elevation change "v" as follows:
v = (398.35m / 20m) = 19.9175m ≈ 19.92m (approximating to 2 decimal places for consistency)
Inserting this into the Pythagorean theorem gives us:
d^2 + 19.92^2 = 398.35^2
d^2 = 398.35^2 - 19.92^2
d^2 = 158686.5225 - 396.8064
d^2 = 158289.7161
d = √158289.7161
d ≈ 398.108 meters (rounded to three decimal places)
The reduced horizontal length for plotting purposes is approximately 398.108 meters.
**(c) What tape reading is required to produce a horizontal distance of 15.08m between two site pegs, one being 0.66m above the other?**
We will be using the Pythagorean theorem again to find the hypotenuse of the triangle, which will be the required tape reading:
Let's assign some variables:
x = the required tape reading
h = 15.08m (horizontal length)
v = 0.66m (vertical difference)
By the Pythagorean theorem:
x^2 = h^2 + v^2
x^2 = 15.08^2 + 0.66^2
x^2 = 227.4064 + 0.4356
x^2 = 227.842
x = √227.842
x ≈ 15.095 meters (rounded to three decimal places)
The tape reading required to produce a horizontal distance of 15.08m with a vertical separation of 0.66m is approximately 15.095 meters.