| Calculation
of gradients using ratio
Marjorie
Henry, Contributor
 | Dianna
Blake-Bennett, marketing manager of Burger King, presents the winning trophy to
the Camperdown team at the National Schools' Debate Competition finals at the
TVJ Studios earlier this year. From second left are: Jonhoi Vaughn, Ricardo Brooks
and Michelle Danials. - Junior Dowie/Staff Photographer |
THIS
WEEK I continue my discussion of Question 1 from the recent examinations (that
is, June 2007) by focusing on the calculation of gradients using ratio. Let me
first of all remind you that the gradient is used to calculate the steepness of
a slope. It is expressed as a ratio between the highest and lowest points and
the horizontal distance between them. Your aim is to find out the rate at which
the land increases in height (the vertical) as you travel along the slope (the
horizontal). Here is the section of the question that tested that skill: (e)
Calculate the average gradient along the College Gut (square 2213) between 250
feet and 200 feet contours. Show your calculation. (Hint:
100 yds. = 300ft.) To
calculate the gradient, you must have two sets of information, namely: 1.
The difference in height (DH or VD) between the two points on the slope,
that is, between the lower and the higher points. To obtain this information,
you simply subtract the lower height from the higher one. 2.
The horizontal distance (HD) which is obtained by measuring the distance
on the map between the two points on the slope. Refer
to the map extract for this examination, that of Basseterre, St. Christopher and
identify the slope referred to in the question. The heights have already been
given so you must now subtract 200 from 250. The answer is 50 feet. This is your
DH or VD. To obtain the horizontal distance, place the straight edge of a bit
of paper between the two contours along the College Gut. Mark off where each point
falls. Place that edge along the linear scale given on the map in order to determine
the distance. Since the contours are in feet, you must use the linear scale that
is in yards. Be guided also by the hint given in the question. The distance is
500 yards. This is your HD. A
common unit of measurement must be used in your calculation, hence, 500 yards
must now be converted to feet. Since three feet make one yard, multiply 500 by
three. This equals 1,500 feet. The hint given in the question allows you to arrive
at the same answer.  | | St.
Christopher Map - Contributed | |
The
formula for gradient must also be known. This is: Vertical
Difference in Height | Horizontal
Distance |
| This
is written as | VD
or DH | | | HD |
Supply
the figures for this question. This would be 50/1,500
To determine
the ratio or gradient, the VD, that is, the top number 50 ft., is regarded as
1 unit. The 1,500ft, that is, the HD, is divided by the 50. That gives an answer
of 30. The gradient would then be written as 1:30. Please
note that a whole number is always given for the answer. If there is a remainder
after your calculation, then you must do one of the following: 1.
If the remainder is more than a half of the divisor, then add 1.
2. If
the remainder is less than half of the divisor, it is disregarded.
You
were instructed in the question to 'show your calculation.' Here are the steps
to follow:- STEP
1 | Write
out the formula, that is, | VD
or DH | | | HD |
STEP
2 Calculate
the vertical difference:-
250ft-200ft
= 50ft
Difference
in height = 50ft STEP
3 Calculate
the horizontal distance
Horizontal
distance
=
500 yds x 3 = 1,500ft STEP
4 Gradient
= 50/1500
Answer
= 1:30 Here
is an exercise for you. Refer to the same extract for Question 1. Calculate
the average gradient between the trigonometrical stations K8 (234137) and K9 (246130).
Here
are the answers for the exercise given last week on measuring of distance: 1.
The length of the main road to the nearest metre is 2km 400m 2.
The length of the secondary road to the nearest mile is one mile. Marjorie
Henry is an independent contributor. | 
