Question

What is the slope correction for a length of 30.0 m along a gradient of 1 in 20? (a) 3.75 cm (b) 0.375 cm (c) 37.5 m (d) 0.0375 cm?

Answer 1

In 1 in 20 means, 20m = 1m.

So total 30m length = 1.5m.

In Gradient 1 in 20 Means.

Imagine Triangle, length-30m & Height-1.5m.

Use Pythagoras theorem.

X2 = 302+1.52.
X = 30.0375m.

Change length = 30.0375-30.

= 0.0375m.
= 3.75cm.

Alternate method :

In ABC triangle,

AC = l = 30m.
Slope = 1 : 20.

1 in20 so in 30 = (1/20) * 30 = 1.5

Therefore BC = 1.5m.

And AB extends to BB'.
And meet to points A as arc & B as a straight line.
Here BB'= correction of line,
= AC - AB.

AB = (30^2 - 1.5^2)^1/2
= 29.962m.

BB' = 30 - 29.962
= 0.0375m or 3.75cm.

Hope This helps :)

Answer 2

(a)3.75cm, is the slope correction.

Step by step explanation

Given data:

Length$=30.0m$

The length of $30.0m$ is along the gradient  $=1 in 20$

To find:

We must find the slope correction for the length given.

Formula to be used:

Slope correction$=\frac{h^{2} }{2l}$

Where:

h=height of the slope

l=length of the slope

Solution:

Using the formula :

$tan$θ$=\frac{1}{20}$

$\frac{1}{20}=\frac{h}{30}$

We can determine the height which can further be used for finding the slope correction.

$h=\frac{1}{20}$×$30$

$h=1.5m$

Now using the formula: Slope correction$=\frac{h^{2} }{2l}$

$=\frac{(1.5)^{2} }{2*30}$

$=\frac{2.25}{60}$

$=0.0375m$

Therefore the slope correction is $3.75 cm$

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