1. Ventilation 1 - Program Basic Math & Problem Solving
Presented by Training Staff
Bureau of Deep Mine Safety
Basic Math & Problem Solving

2. Review of Formula Terms
a = sectional area of airway, in square feet (ft.2)
l = length of airway, in feet (ft.)
o = perimeter of airway, in feet (ft.)
s = rubbing surface, in square feet (ft2)
v = velocity of air current, in feet per minute (fpm)
q = quantity of air, in cubic feet per minute (cfm)

3. Rectangular or Square Dimension:
COMMON AREA FORMULAS
Rectangular or Square Dimension:
Area = Height X Width
Note: Please remember to convert inches into the decimal equivalent of one foot - inches divided by 12

4. Practice Problems – Area ; Rectangle
Determine the area of a mine entry that is 19 feet wide and 7 feet high:
Solution:
A = W x H
A = 19 x 7’
A = 133 sq. ft. 7’
19’

5. Practice Problems – Area ; Rectangle
Determine the area of a mine entry that is 18 feet wide and 5 feet, 6 inches high:
Solution:
A = W x H
A = 5.5’ x 18’
A = 99 sq. ft.
5’6’’
18’

6. Practice Problems Solution:
Determine the area of a mine entry that is 17 feet 3 inches wide and 6 feet 9 inches high:
Solution:
A = W x H
A = 17.25’ x 6.75’
A = sq. ft.
6’9’’
17’3’’

7. Trapezoid: COMMON AREA FORMULAS
Area = Top Width + Bottom Width X Height 2

8. Practice Problems – Area ; Trapezoid
Determine the area of a mine entry that is 6 foot high, and 18 feet wide across the top, and is 19 feet wide across the bottom.
Solution:
Area = Top Width + Bottom Width X Height 2
A = 18’ + 19’ x 6’
A = 37’ x 6’
A = 18.5’ x 6’
A = sq. ft.
18’ 6’
19’

9. Practice Problems – Area ; Trapezoid
Determine the area of a mine entry that is 5 foot high, and 20 feet wide across the top, and is 22 feet wide across the bottom.
Solution:
Area = Top Width + Bottom Width X Height 2
A = 20’ + 22’ x 5’
A = 42’ x 5’
A = 21’ x 5’
A = 105 sq. ft.
20’ 5’
22’

10. Practice Problems Solution:
Determine the area of a mine entry that is 4 foot 6 inches high, and 17 feet wide across the top, and is 20 feet wide across the bottom.
Solution:
Area = Top Width + Bottom Width X Height 2
A = 17’ + 20’ x 4.5’
A = 37’ x 4.5’
A = 18.5’ x 4.5’
A = sq. ft.
17’
4’6’’
20’

11. COMMON AREA FORMULAS - Circle
Circular:
A = ¶ x D2 4 or
A = ¶ x R2
Please use the following
For Pi………
¶ =
diameter
radius

12. Practice Problems –Area ; Circle
Determine the area of a circle that has an diameter of 20 feet 9inches.
Solution:
A = ¶ x R2
R = = 2
A = x
A = x
A = sq. ft. R

13. Area - Circle
Determine the area of a circular air shaft with a diameter of 20 feet
Solution:
A = ¶ x R2
R = 20 = 10 2
A = x 102
A = x 100
A = sq. ft.
20”

14. Practice Problems Solution: A = ¶ x r2 A = 3.1416 x 8.52
Determine the area of a circle that has an diameter of 17 feet.
Solution:
A = ¶ x r2
R = 17 = 8.5 2
A = x 8.52
A = x 72.25
A = sq. ft.
17’

15. Square or Rectangle Perimeters
o = Top Width + Bottom Width + Side 1 + Side 2
Remember, perimeter measured in linear feet

16. Practice Problem – Perimeter ; Rectangle
Determine the perimeter of an entry 7 feet high and 22 feet wide.
Solution:
o = Top Width + Bottom Width + Side 1 + Side 2
o = 22’ + 22’ + 7’ + 7’
o = 58 feet
7 ft.
22 ft.

17. Practice Problem – Perimeter ; Rectangle
Determine the perimeter of an entry 6 feet 6 inches high and 20 feet 3 inches wide.
Solution:
o = Top Width + Bottom Width + Side 1 + Side 2
o = 6.5’ + 6.5’ ’ ’
o = 53.5 feet
6ft.6in.
20ft.3in.

18. Perimeters - Circle
o = ¶ x Diameter
¶ =
Diameter

19. Solution: Perimeter - Circle
Determine the perimeter of a circular air shaft with a diameter of 17 feet, 6 inches.
Solution:
o = ¶ x Diameter
o = x 17.5 ft.
o = ft.
17’6”

20. Solution: Perimeter - Circle
Determine the perimeter of a circular air shaft with a diameter of 20 feet
Solution:
o = ¶ x Diameter
o = x 20 ft.
o = ft.
20”

21. Solution: Perimeter - Circle
Determine the perimeter of a circular air shaft with a radius of 9 feet.
Solution:
D = 2 x r
D = 2 x 9 ft.
D = 18 ft.
¶ =
o = ¶ x Diameter
o = x 18.0 ft.
o = ft. 9’

22. Formula Equations
Quantity of Air (cfm)
Q = AV
Quantity = Area X Velocity
Velocity of air (fpm)
V = _ Q_ A
Velocity = Quantity  Area
Area (when velocity and quantity a known)
A = _Q_ V
Area = Quantity  Velocity
Algebraic Circle Q A V

23. Practice Problem - Quantity
Find the quantity of air passing thru an entry 17 feet 6 inches wide and 9 feet high, with 180 fpm registered on the anemometer.
A = WH
Q = AV
Solution:
A = WH
A = 17.5’ x 9’
A = sq. ft.
Q = AV
Q = (157.5 sq.ft.)(180 fpm)
Q = 28,350 CFM

24. Practice Problem - Quantity
Find the quantity of air passing thru and entry 18 feet wide and 6 feet 6 inches high, with 110 fpm registered on the anemometer.
A = WH
Q = AV
Solution:
A = WH
A = 18’ x 6.5’
A = 117 sq. ft.
Q = AV
Q = (117 sq.ft.)(110 fpm)
Q = 12,870 CFM

25. Practice Problem - Velocity
What is the velocity in a entry 10 feet high and 22 feet wide, with a quantity of 11,380 CFM?
A = WH
V = _Q_ A
Solution:
A = WH
A = 22 ft. x 10 ft.
A = 220 sq. ft.
V = _Q_ A
V = 11,380 CFM
220 sq.ft.
V = fpm

26. Practice Problem - Area
An entry has 12,500 CFM of air with a velocity of 150 fpm. What is the area of the entry?
A = _Q_ V
Solution:
A = _Q_ V
A = 12,500 CFM
150 fpm
A = sq. ft.