ASP® formula sheet with worked examples
23 formulas for the calculation tasks in BCSP's ASP11 blueprint. Each one has a worked example, computed by code and checked against the source, and the source's own words.
Free, no email needed. Letter size, print-ready.
At the exam, BCSP says
BCSP, What to Expect When Sitting for a BCSP Exam (item 7)(opens ecfr.gov in a new tab)“For any question requiring a mathematical formula, the formula will be included within the exam question (unless the question is specifically testing your knowledge of a formula). You will not be provided or permitted to bring a formula sheet.”
So learn what each formula does and when to use it. This sheet is for study, not for the test room.
1. Incidence rate (TRIR)
IR = (N × 200,000) ÷ EH
N = recordable injuries and illnesses; EH = total hours worked by all employees in the year; 200,000 = 100 full-time workers × 40 h × 50 weeks.
Worked example
BLS example: 7 recordable cases and 400,000 hours worked.
(7 × 200,000) ÷ 400,000 = 3.5
Blueprint: ASP11 Domain 1, task 8: lagging indicators (incidence rates)
The formula comes from
BLS, How To Compute Your Firm's Incidence Rate for Safety Management(opens ecfr.gov in a new tab)“(Number of injuries and illnesses X 200,000) / Employee hours worked = Incidence rate (The 200,000 hours in the formula represents the equivalent of 100 employees working 40 hours per week, 50 weeks per year, and provides the standard base for the incidence rates.)”
2. DART rate
DART = (N × 200,000) ÷ EH
N = cases with days away from work, job restriction or transfer (OSHA 300 Log columns H + I); EH = total hours worked.
Worked example
BLS example: 3 such cases and 400,000 hours worked.
(3 × 200,000) ÷ 400,000 = 1.5
Blueprint: ASP11 Domain 1, task 8: lagging indicators (lost time)
The formula comes from
BLS, How To Compute Your Firm's Incidence Rate for Safety Management(opens ecfr.gov in a new tab)“The same formula can be used to compute the incidence rate for the most serious injury and illness cases, defined here as cases that result in workers taking time off from their jobs or being transferred to another job or doing lighter (restricted) duties.”
3. Noise dose
D = 100 × (C1/T1 + C2/T2 + … + Cn/Tn)
C = hours at a given sound level; T = OSHA reference duration for that level from Table G-16a (90 dBA = 8 h, 95 dBA = 4 h).
Worked example
4 h at 90 dBA and 4 h at 95 dBA.
100 × (4/8 + 4/4) = 150%
Blueprint: ASP11 Domain 1, task 5: noise hazards (TWA)
The formula comes from
29 CFR 1910.95, Appendix A(opens ecfr.gov in a new tab)“When the workshift noise exposure is composed of two or more periods of noise at different levels, the total noise dose over the work day is given by: D = 100(C1 / T1 + C2 / T2 + Cn / Tn), where Cn indicates the total time of exposure at a specific noise level, and Tn indicates the reference duration for that level as given by Table G-16a.”
4. Noise TWA from dose
TWA = 16.61 × log10(D ÷ 100) + 90
D = dose in percent. A 100% dose is a 90 dBA TWA; a 50% dose is 85 dBA, the hearing conservation action level.
Worked example
A 150% dose (the example above).
16.61 × log10(1.5) + 90 = 92.9 dBA
1910.95 Table A-1 lists 92.9 dBA for a 150% dose.
Blueprint: ASP11 Domain 1, task 5: noise hazards (TWA)
The formula comes from
29 CFR 1910.95, Appendix A(opens ecfr.gov in a new tab)“The eight-hour time-weighted average sound level (TWA), in decibels, may be computed from the dose, in percent, by means of the formula: TWA = 16.61 log10 (D/100) + 90.”
5. Reference duration for a sound level
T = 8 ÷ 2^((L − 90) ÷ 5)
L = A-weighted sound level in dBA. Each 5 dB above 90 halves the allowed time (OSHA's 5-dB exchange rate).
Worked example
L = 95 dBA.
8 ÷ 2^((95 − 90) ÷ 5) = 4.0 h
Table G-16a lists 4 hours for 95 dBA. eCFR shows this formula as an image under the table.
Blueprint: ASP11 Domain 1, task 5: noise hazards (TWA)
The formula comes from
29 CFR 1910.95, Appendix A, Table G-16a(opens ecfr.gov in a new tab)“T is the reference duration corresponding to the measured sound level, L, as given in Table G-16a or by the formula shown as a footnote to that table.”
6. 8-hour TWA exposure (air contaminants)
E = (Ca × Ta + Cb × Tb + … + Cn × Tn) ÷ 8
C = concentration during a period; T = hours at that concentration. E may not exceed the substance's 8-hour TWA in Subpart Z.
Worked example
OSHA example: 2 h at 150 ppm, 2 h at 75 ppm, 4 h at 50 ppm; limit 100 ppm.
(2 × 150 + 2 × 75 + 4 × 50) ÷ 8 = 81.25 ppm, under 100 ppm
Blueprint: ASP11 Domain 1, task 14: exposure assessments (TWA)
The formula comes from
29 CFR 1910.1000(d)(1)(i)(opens ecfr.gov in a new tab)“The cumulative exposure for an 8-hour work shift shall be computed as follows: E = (Ca Ta + Cb Tb + . . .Cn Tn) ÷ 8 Where: E is the equivalent exposure for the working shift. C is the concentration during any period of time T where the concentration remains constant. T is the duration in hours of the exposure at the concentration C.”
7. Mixture of air contaminants
Em = C1/L1 + C2/L2 + … + Cn/Ln
Em must not exceed 1
C = measured concentration of each substance; L = its exposure limit from Subpart Z.
Worked example
OSHA example: 500 ppm (limit 1,000), 45 ppm (limit 200), 40 ppm (limit 200).
0.500 + 0.225 + 0.200 = 0.925, under 1, so within limits
Blueprint: ASP11 Domain 1, task 14: exposure assessments (TLV, TWA)
The formula comes from
29 CFR 1910.1000(d)(2)(i)(opens ecfr.gov in a new tab)“In case of a mixture of air contaminants an employer shall compute the equivalent exposure as follows: Em = (C1 ÷ L1 + C2 ÷ L2) + . . .(Cn ÷ Ln) Where: Em is the equivalent exposure for the mixture. C is the concentration of a particular contaminant. L is the exposure limit for that substance specified in subpart Z of 29 CFR part 1910. The value of Em shall not exceed unity (1).”
8. NIOSH recommended weight limit (U.S. units)
RWL = 51 × HM × VM × DM × AM × FM × CM
HM = 10/H, VM = 1 − 0.0075|V − 30|
DM = 0.82 + 1.8/D, AM = 1 − 0.0032A
LI = load ÷ RWL
H, V, D in inches; A in degrees; FM and CM from the manual's frequency and coupling tables. LI above 1 means elevated risk.
Worked example
H = 20 in, V = 15 in, D = 25 in, A = 30°, one lift every 5 min for 8 h (FM 0.85), fair coupling below 30 in (CM 0.95), 30 lb load.
51 × 0.50 × 0.8875 × 0.892 × 0.904 × 0.85 × 0.95 = 14.7 lb; LI = 30 ÷ 14.7 = 2.0
Blueprint: ASP11 Domain 1, task 9: manual lift parameters (NIOSH lifting equation)
The formula comes from
NIOSH, Applications Manual for the Revised NIOSH Lifting Equation (DHHS Pub. No. 94-110)(opens ecfr.gov in a new tab)“The RWL is defined by the following equation: RWL = LC X HM X VM X DM X AM X FM X CM ... Load Constant LC 23kg 51lb; Horizontal Multiplier HM (25/H) (10/H); Vertical Multiplier VM 1-(.003|V-75|) 1-(.0075|V-30|); Distance Multiplier DM .82 + (4.5/D) .82 + (1.8/D); Asymmetric Multiplier AM 1-(.0032A) 1-(.0032A)”
9. Wet bulb globe temperature (WBGT)
Outdoors: WBGT = 0.7 tnwb + 0.2 tg + 0.1 ta
Indoors: WBGT = 0.7 tnwb + 0.3 tg
tnwb = natural wet bulb; tg = globe; ta = dry bulb (air) temperature, all in the same units.
Worked example
Outdoors, tnwb 78°F, tg 100°F, ta 88°F.
0.7 × 78 + 0.2 × 100 + 0.1 × 88 = 83.4°F
The same readings indoors (no ta term): 0.7 × 78 + 0.3 × 100 = 84.6°F.
Blueprint: ASP11 Domain 1, task 6: climate and environmental conditions
The formula comes from
NIOSH, Criteria for a Recommended Standard: Occupational Exposure to Heat and Hot Environments (Pub. 2016-106), PDF p. 145(opens ecfr.gov in a new tab)“The calculation of the WBGT for indoors is WBGT = 0.7tnwb + 0.3tg and the calculation of the WBGT for outdoors is WBGT = 0.7tnwb + 0.2tg + 0.1ta.”
10. Wind chill
WC = 35.74 + 0.6215T − 35.75V^0.16 + 0.4275T × V^0.16
T = air temperature in °F; V = wind speed in mph.
Worked example
T = 20°F, V = 15 mph.
35.74 + 12.43 − 55.14 + 13.19 = 6°F
Blueprint: ASP11 Domain 1, task 6: climate and environmental conditions (wind chill)
The formula comes from
NOAA/National Weather Service: Wind Chill calculation(opens ecfr.gov in a new tab)“WindChill = 35.74 + (0.6215 × T) − (35.75 × Wind sfc^0.16) + (0.4275 × T × Wind sfc^0.16) [T in °F, wind speed in mph]”
11. Dilution ventilation
qd = (387 × lbs) ÷ (MW × min × d)
Qd = (qd × Km × 10^6) ÷ Ca
lbs = pounds evaporated; MW = molecular weight; min = minutes; d = density correction (1 at standard air); Km = mixing factor (2 to 5); Ca = acceptable concentration in ppm (typically half the PEL).
Worked example
2 lb of toluene (MW 92.14) evaporates in 60 min, d = 1, Km = 3, Ca = 50 ppm.
qd = 774 ÷ 5,528.4 = 0.140 cfm; Qd = 0.140 × 3 × 10^6 ÷ 50 = 8,400 cfm
Blueprint: ASP11 Domain 1, task 3: flow rates (ventilation)
The formula comes from
OSHA Technical Manual, Section III, Chapter 3 (Ventilation Investigation), General Exhaust (Dilution) Ventilation Systems(opens ecfr.gov in a new tab)“q d = [(387)(lbs)]/[(MW)(min)(d)] ... Q d = [(q d)(K m)(10 6)]/C a ... K m = mixing factor to account for poor or random mixing (NOTE: K m = 2 to 5; K m = 2 is optimum) C a = acceptable airborne concentration of the material (typically half of the PEL).”
12. Velocity from a smoke tube
V = D ÷ T
D = distance the smoke travels (ft); T = time (min). Convert seconds to minutes first.
Worked example
OSHA example: smoke travels 2 ft in 2 seconds.
2 ft ÷ (2/60 min) = 60 fpm
Blueprint: ASP11 Domain 1, task 3: flow rates (ventilation)
The formula comes from
OSHA Technical Manual, Section III, Chapter 3 (Ventilation Investigation), IV.F.3.d and Appendix III:3-2(opens ecfr.gov in a new tab)“Velocity = Distance/Time, or V = D/T Squeeze off a quick burst of smoke. Time the smoke plume's travel over a two-foot distance. Calculate the velocity in feet per minute. For example, if it takes two seconds for the smoke to travel two feet, the velocity is 60 fpm. ... Slot velocity is calculated by dividing the total volume flow rate by the slot area”
13. Secondary containment capacity
Capacity ≥ the greater of 10% of total container volume or the largest container
Hazardous waste container storage areas under 40 CFR 264.175. Containers without free liquids are not counted.
Worked example
20 drums of 55 gal each.
max(0.10 × 1,100, 55) = 110 gal
Blueprint: ASP11 Domain 1, task 1: storage capacity
The formula comes from
40 CFR 264.175(b)(3)(opens ecfr.gov in a new tab)“The containment system must have sufficient capacity to contain 10% of the volume of containers or the volume of the largest container, whichever is greater. Containers that do not contain free liquids need not be considered in this determination;”
14. Excavation sloping (simple slope, 20 ft or less)
Horizontal run = depth × H:V ratio
Type A 3/4:1, Type B 1:1, Type C 1½:1
Maximum allowable slopes from Subpart P, Appendix B, for excavations 20 ft deep or less.
Worked example
A 12 ft deep trench in Type C soil.
12 ft × 1.5 = 18 ft of horizontal setback on each side
Blueprint: ASP11 Domain 1, task 4: slope angle and depth ratio for trenching and excavation
The formula comes from
29 CFR 1926 Subpart P, Appendix B (Figure B-1)(opens ecfr.gov in a new tab)“All simple slope excavation 20 feet or less in depth shall have a maximum allowable slope of 3/4 :1. ... All simple slope excavations 20 feet or less in depth shall have a maximum allowable slope of 1:1. ... All simple slope excavations 20 feet or less in depth shall have a maximum allowable slope of 1 1/2 :1.”
15. Choker hitch capacity by angle of choke
Capacity = choker rating × angle factor
Over 120° 100%, 90–120° 87%, 60–89° 74%, 30–59° 62%, 0–29° 49%
Grade 80 alloy chain choker ratings (OSHA Table 3) apply at angles of choke over 120°.
Worked example
1/2 in Grade 80 chain, single-leg choker (9,600 lb), angle of choke 75°.
9,600 lb × 0.74 = 7,104 lb
Blueprint: ASP11 Domain 1, task 2: rigging and load calculations
The formula comes from
OSHA, Guidance on Safe Sling Use: Tables and Figures, Table 3 and Fig. 2 (Angle of Choke)(opens ecfr.gov in a new tab)“TABLE 3 RATED LOAD FOR GRADE 80 ALLOY STEEL CHAIN SLINGS CHOKER HITCHES ... (2) Rated loads are for angles of choke greater than 120 degrees. ... Angle of choke, deg | Rated Capacity, % ... Over 120 | 100; 90-120 | 87; 60-89 | 74; 30-59 | 62; 0-29 | 49. NOTE: (1) Percent of sling rated capacity in a chocker hitch.”
16. Personal fall arrest distances
Free fall ≤ 6 ft
Deceleration distance ≤ 3.5 ft
Limits for personal fall arrest systems in construction, 29 CFR 1926.502(d)(16).
Worked example
The longest fall the two limits allow before the worker stops.
6 ft + 3.5 ft = 9.5 ft
Clearance below must also cover the worker's height below the D-ring, harness stretch and a margin. 1926.502 doesn't set those as numbers; use the equipment maker's figures.
Blueprint: ASP11 Domain 1, task 7: fall protection parameters (free-fall distance, clearance)
The formula comes from
29 CFR 1926.502(d)(16)(iii)-(iv)(opens ecfr.gov in a new tab)“(iii) be rigged such that an employee can neither free fall more than 6 feet (1.8 m), nor contact any lower level; (iv) bring an employee to a complete stop and limit maximum deceleration distance an employee travels to 3.5 feet (1.07 m); and,”
17. Radioactive decay
At = A0 × e^(−λt)
λ = 0.693 ÷ T½
A0 = starting activity; t = elapsed time; T½ = half-life, in the same time units as t.
Worked example
100 mCi source, half-life 8 days, after 24 days.
100 × e^(−(0.693 ÷ 8) × 24) = 12.5 mCi
24 days is three half-lives: 100 → 50 → 25 → 12.5.
Blueprint: ASP11 Domain 1, task 15: radiation exposure (half-life)
The formula comes from
OSHA, Ionizing Radiation: Introduction to Health Physics handout(opens ecfr.gov in a new tab)“A t = A o e (- λ x t) = Activity after time t ... λ= 0.693/T ½ = disintegration constant”
18. Inverse square law
I2 = I1 × (d1 ÷ d2)²
I = intensity or dose rate at distance d from a point source.
Worked example
100 mR/h at 2 ft; what at 6 ft?
100 × (2 ÷ 6)² = 11.1 mR/h
Blueprint: ASP11 Domain 1, task 15: radiation exposure (dosage)
The formula comes from
OSHA, Ionizing Radiation: Introduction to Health Physics handout(opens ecfr.gov in a new tab)“Estimating Attenuation in Air Using Inverse Square Law I 2 = I 1 [ d 1 / d 2 ] 2”
19. Shielding with half-value layers
I = I0 × e^(−μx)
μ = 0.693 ÷ HVL
I0 = unshielded intensity; x = shield thickness; HVL = thickness that halves intensity (same units as x).
Worked example
80 mR/h, lead HVL 0.5 in, 1.5 in of lead.
80 × e^(−(0.693 ÷ 0.5) × 1.5) = 10.0 mR/h
1.5 in is three HVLs: 80 → 40 → 20 → 10.
Blueprint: ASP11 Domain 1, task 15: radiation exposure (shielding)
The formula comes from
OSHA, Ionizing Radiation: Introduction to Health Physics handout(opens ecfr.gov in a new tab)“I = I o Be -µx = intensity on other side of shield ... Half-Value Layer (HVL): The shield thickness necessary to reduce intensity by half. HVL = 0.693 / µ.”
20. Mean, median and sample standard deviation
Mean = Σx ÷ n
s = √(Σ(x − mean)² ÷ (n − 1))
The median is the middle value of the sorted data (the average of the two middle values when n is even).
Worked example
Data: 2, 4, 4, 4, 5, 5, 7, 9.
mean = 40 ÷ 8 = 5; median = 4.5; s = √(32 ÷ 7) = 2.14
Blueprint: ASP11 Domain 1, task 11: descriptive statistics
The formula comes from
NIST/SEMATECH e-Handbook of Statistical Methods, 1.3.5.1 Measures of Location and 1.3.5.6 Measures of Scale(opens ecfr.gov in a new tab)“mean - the mean is the sum of the data points divided by the number of data points. ... median - the median is the value of the point which has half the data smaller than that point and half the data larger than that point. ... standard deviation - the standard deviation is the square root of the variance.”
21. Fault tree gates (independent events)
OR gate: P = P(A) + P(B) − P(A)P(B)
AND gate: P = P(A) × P(B)
P = probability of each independent basic event.
Worked example
P(A) = 0.1, P(B) = 0.2.
OR: 0.1 + 0.2 − 0.02 = 0.28; AND: 0.1 × 0.2 = 0.02
Blueprint: ASP11 Domain 1, task 12: probability of failure mode
The formula comes from
NUREG-0492, Fault Tree Handbook (U.S. NRC, 1981), Chapter VII(opens ecfr.gov in a new tab)“If A and B are independent events, then P(BIA) = P(B) and P(Q) = P(A) + P(B) - P(A) P(B); ... If A and B are independent events, then P(BIA) = P(B), P(AIB) = P(A), and P(Q) = P(A) P(B);”
22. Force from mass and deceleration
F = m × a = m × (v1 − v0) ÷ t
m = mass (kg); v = velocity (m/s); t = time (s); F in newtons. The sign shows direction; use the magnitude.
Worked example
80 kg stopped from 5 m/s in 0.25 s.
80 × (0 − 5) ÷ 0.25 = −1,600 N, so 1,600 N
Blueprint: ASP11 Domain 1, task 10: general physics (force, acceleration, velocity)
The formula comes from
NASA Glenn Research Center, Beginner's Guide to Aeronautics: Newton's Second Law of Motion(opens ecfr.gov in a new tab)“If the mass is a constant and using the definition of acceleration a as the change in velocity with time, the second law reduces to the more familiar product of a mass and an acceleration: F = m * a Since acceleration is a change in velocity with a change in time t, we can also write this equation in the third form shown on the slide: F = m * (v1 - v0) / (t1 - t0)”
23. Unit conversions
1 ft = 0.3048 m, 1 in = 2.54 cm
1 gal = 3.785412 L, 1 lb = 0.4535924 kg
°C = (°F − 32) ÷ 1.8
Exact or standard factors from NIST SP 811.
Worked example
Convert 55 gal to liters.
55 × 3.785412 = 208.2 L
95°F = (95 − 32) ÷ 1.8 = 35°C.
Blueprint: ASP11 Domain 1, task 16: unit conversions (metric/imperial)
The formula comes from
NIST Special Publication 811, Guide for the Use of the International System of Units, Appendix B.8(opens ecfr.gov in a new tab)“foot (ft) | meter (m) | 3.048 E-01; inch (in) | centimeter (cm) | 2.54 E+00; gallon (U.S.) (gal) | liter (L) | 3.785 412 E+00; pound (avoirdupois) (lb) | kilogram (kg) | 4.535 924 E-01; pound-force (lbf) | newton (N) | 4.448 222 E+00; degree Fahrenheit (temperature) (°F) | degree Celsius (°C) | t/°C = (t/°F - 32)/1.8”
How this sheet was checked
Each worked example is computed by the same code as our free calculators. A test re-runs every example and compares it with the value printed in the source, where the source gives one (BLS, 29 CFR 1910.1000, 1910.95 Table A-1, the OSHA Technical Manual), or with a hand calculation. Checked 2026-10-16. If you spot a problem, email hello@passcited.com.
Other sheets: CSP formula sheet and CHST formula sheet. Calculators: free safety exam tools.