Percentages; Areas; Volumes; Capacities and
how to use Fractions.
Fractions
What is a Fraction?
A fraction is a part of a whole. Two and a half inches is equal to two inches plus one half of an inch. This can be represented in two ways.
Certain conversions leave five, six, seven and above numbers after the decimal point
e.g. 0.28463215
This is clumsy and should be reduced for most purposes to four figures e.g. 0.2846
As most calculations are performed on the calculator it is easy, and accurate, to use four figures. Using four figures in a hand calculation is clumsy and leads to error. Therefore, use the calculator often.
Measuring in feet and inches presents problems when tallying pipe. To ease the situation, feet and tenths of a foot are used. You will have noticed the Pipe Measuring Tape is calibrated in feet, and tenths of a foot.
Diameters are most commonly measured in feet and inches because they are usually taken on their own. In contrast, length is measured in feet and tenths of a foot for ease of addition.
When diameters are involved in calculations, for instance in volumes, the inches or vulgar fraction has to be converted to decimal.
e.g. Cylindrical Tank 6ft 4 inches diameter
To convert: There are 12 inches to 1 ft, Therefore, 4 inches = 4/12
Therefore 4 ÷ 12 = 0.3333 Diameter in decimals = 6.3333 ft
The calculation is recurring therefore four decimal places are used.
What is a Decimal Place?
When asked to calculate to four Decimal Places your inputs should have four numbers to the right of the decimal point.
e.g. What is 8.32567418 to 3, 4 and 5 decimal 3 decimal places = 8.326 4 decimal places = 8.3257 5 decimal places = 8.32567
Notice that the first three decimal places are 8.325, but the answer above is 8.326.
The technique of Rounding-Off is being used.
If the next number is five or greater, then increase your last decimal place by one.
e.g. 8.32748
To “round-off” to 4 decimal places, look at the first decimal place. Being 8 it is greater than five, therefore increase 4 to 5 = 8.3275
Examples
“Round off” to 4 decimal places
9.382416 = 9.3824 9.221134 = 9.2211 9.18796 = 9.188 9.25256 = 9.2526
Exercise: (Round off to 4 decimal places if necessary)
1) Convert 6-2/8; 3-4/16; 5-7/13; 8-2/6 to decimals.
2) Convert 42ft 7 inches to decimals.
3) Convert 10ft 6-1/2 inches to decimals.
Areas
The use of area is found in many places around the rig.
Force on a unit area Area of deck space Surface area of pits
Area is expressed as a square – a square inch, square centimetre, square foot, square mile, etc.
A square inch is the area taken up by a square, of 1-inch long sides.
There are 3 common shapes that can easily have their areas calculated.
A shape with 4 sides, each side at 90° to the other – Rectangle
Area = Length x Breadth
5
4 5 x 4 = 20 sq. ins
6
6 x 3 = 18 sq. ins 3
A shape with 3 sides, angles between each side are variable – Triangle
Area = Base x ½ vertical height
height
base
½ height
Area = base x ½ height
A shape with 4 sides, none of the angles are 90° - Trapezium Area = Sum of Parallel Sizes x ½ distance between them.
a
ht (a+b) x ½ ht
b
Cut trapezium into 3 parts
Another common shape, but not readily calculated, is the Circle.
The area is a relationship between radius, or diameter and circumference.
The Radius is the distance from the centre to the edge.
The Diameter is the distance from the edge to edge via the centre.
Radius
Diameter = 2 x Radius Radius = ½ Diameter
The Circumference is the distance round the edge of the circle. This has a fixed relationship with the diameter. The diameter of any circle will go round the circumference 3.1416 times. This value is constant and is called PI (π).
To calculate Circumference using diameter, multiply Diameter by π Circumference = π x diameter
Or Circumference = 2 π x radius
To find the formula for calculating area we can divide the circle into slices like a cake.
Circumference
= 2 π x Radius
For instance the circle has been divided into 32 equal portions - each like a triangle. Unpeeling the circle we get the shape below.
The base = circumference = π D or 2 π r
Radius 2 π r
Each triangle has an area of 1/2ht x base.
To calculate for 32 triangles –
= 32 x r x 2 π r
If using Diameter –
Find Area of Circles with the following:
a) Diameter = 12”
To aid calculation, remember π = .7854 4
Therefore, Area = .7854 x D2
π D2
One major application of 4 is the calculation of Annular Area and Volume.
The Annular Area is the area between two concentric circles. For instance hole to pipe or OD of pipe to ID of pipe.
ANNULAR AREA
The Annular area is calculated by subtracting the small circle from the larger circle
With D = diameter of large circle
d = diameter of small circle
Annular Area = π D2 π d2 4 - 4
π
Because 4 is common to both the above formula can be rewritten:
π (D2 - d2) 4
or .7854 (D2 - d2)
Example:
Find Annular Area when D = 10” and d = 5”
π
Area = 4 (102 – 52)
= .7854 (100 – 25)
= .7854 x 75 = 58.9 sq. inches
Formulas and Problems
Up to this point the formulas used show division, multiplication and brackets.
This can lead to problems unless two basic rules are practiced.
First: the use of brackets. Whenever brackets appear in a formula the calculation inside must be done prior to using the values outside.
e.g. .7854 (D2 – d2)
Calculate the bracket first
= .7854 (102 – 52)
= .7854 (100 – 25)
= .7854 (75)
The value outside can now be multiplied with that inside.
Second: Solving the equation. This means rearranging a formula to get the unknown value on one side and the known value on the other side.
Find a: a + b = c
find a: a – b = c
Move b across and change – to + a = c + b
In multiplication the technique is different. Values are moved diagonally.
a
Find a: b = c Move b diagonally across = sign
a = c x b a
Find b: b = c Move b up to c and c to b
a
a) Pressure Mud Weight x 0.52 = Depth
b) Pressure
Depth x .052 = Mud Weight
Solving an Equation with squares requires the use of Square Roots.
Example:
Area = π D2 4
Find D: Area x 4 π
to eliminate the square you must square root the other side.
D = Area x 4 π
Square roots are commonly found on calculators today.
The square root of 4 is 2 (2 x 2 = 4) The square root of 64 is 8 (8 x 8 = 64)
Volumes and Capacities
With an understanding of how to calculate areas it is a straight forward procedure to calculate the volume of a container.
Volume is the amount of space in a container.
Capacity is the amount of a substance that can be placed in that container expressed in units relating to both substance and container.
When talking about the capacity of a tank or hole we use barrels, and think of common rig substance like oil, mud or cement.
To calculate volume we multiply the surface area by the height.
Example:
A tank of 12” long x 6” wide x 8” deep
= 12 x 6 x 8
= 576 cubic inches
This means 576 cubes of 1” x 1” x 1” would fit into a tank 12” x 6” x 8”.
When calculating volume all units must be the same.
Example:
Find capacity in cubic inches of a tank 1’ 2” x 8” x 3’ 6”
1’ 2” = 14”
3’ 6 = 42”
Capacity = 14 x 8 x 42 = 4704 cubic inches
We have assumed vertical walls. If the tank had sloping walls the following volume calculations would be used.
30
View
Plan View 10
10 20 50
Side
The area of side A can be found using the formula for a trapezium.
Area = Sum of Parallel sides x half distance between them 10
Area = (50 + 30) x 2
= 400 sq. inches
Then calculate capacity as:
Area x sum of parallel sides on wall B 2
20 + 10
= 400 x 2
= 400 x 30
2
= 400 x 15
= 6000 sq. inches
Calculating volumes of Cylinders or the Annulus the formula is:
Area x height π D2
Volume = 4 x height
Make sure all units are the same
π
Annular Volume 4(D2 – d2) x height π
4 = .7854 Example:
Find a volume of cylinder in cubic feet/foot of depth if diameter is 10”
.7854 (102)
Volume = 144 x 1ft
= .5454 x 1ft
= .5454 cubic feet/foot of depth
The 144 is used to convert square inches into square feet (1 square foot = 144 square inches).
Example:
Calculate Annular Volume if D = 10” d = 6” depth = 1ft .7854 (102 – 62) x 1
Volume = 144
= .349 cubic feet/foot
The use of cubic feet is not as common as barrels. To calculate the volume in barrels, we need to convert feet to barrels.
1 barrel = 5.6146 cubic feet.
Applying this to the formula:
.7854 (D2)
Volume in barrels/ft = 144 x 5.6146 x 1
Calculating out .7854, 144 and 5.6146 we can simplify the formula to D2
Volume in bbls/ft = 1029
Or
(D2 – d2) 1029
Percentage Calculations
Calculating percentages involves simple multiplication, division and rearranging formula.
Percent is the number of parts of 100.
Example 1
Example 3
Each of the above examples tackles the problem differently.
Example 1 - What was the value of 10%
Example 2 - What was the %
Example 3 - What was the value of 100%
The above examples, although different, use the same formula.
P = R X B
P = Percentage:-the actual value equaling chosen %
R = Rate in decimals:- the part of a 100 to be found ie in 4% of 50, 4% is rate.
B = Base:- the number of which some percentage is to be found.
Example 1 (Repeat)
What is 10% of 200 logs? The question asks you to find a number that equals a %, being 10% here.
P = R x B
Rate is the parts of a 100 to be found. In this case 10 parts (10%).
Remember Rate is expressed in decimals. 10% of 100% = .1 P = .1 x B
Base is the number of which some percentage is to be found. In this case we want to find 10% of 200
P = .1 x 200
= 200 logs
10% of 200 logs = 20 logs
Example 2 (Repeat)
What % is 35 logs of 200 logs?
The question asks for an actual percentage. This being the Rate P = R x B (R is unknown)
Percentage means the actual number. In this case 35 logs.
35 = Rate x Base
Base is the whole. In this case 200.
35 = Rate x 200 Rate = 35
200
Rate = .175 whole (1.75 of 1.0) Convert decimal to % by multiplying by 100.
.175 x 100
= 17.5%
To convert % to decimal ÷ 100 To convert decimal to % x 100
Example 3 (Repeat)
If 42 logs = 75% of the total, how many logs are there?
P = R x B Percentage is number of logs = 42
Rate is parts of 100 to be found in decimals
225 is 15% of what?
P = R x B 225 = .15 x B
B = 225 = 1500
.15
To remember formula use following diagram
P
P
R R B
P = R x B R = P
B
B = P
R