Learn: Stopping sight distance — basic model
Stopping sight distance combines distance traveled while the driver perceives and reacts with distance traveled during braking. Road grade changes the simplified braking-distance term.
What the formula is saying
Calculate reaction distance vt. Then calculate braking distance v²/[2g(f + G)]. Add them. In this stored convention G is positive uphill and negative downhill, because climbing assists deceleration.
Read the symbols in plain language
- v
- Speed
Speed. The vehicle continues at its initial modeled speed during the reaction interval.
m/sUse m/s as the base unit shown here. Use v in m/s, reaction time t in s, gravity g in m/s², and friction f and signed grade G as decimal ratios. A 3% downgrade is G = −0.03. Both distances and the final answer are m.
- t
- Reaction time
Reaction time. The vehicle continues at its initial modeled speed during the reaction interval.
sUse s as the base unit shown here. Use v in m/s, reaction time t in s, gravity g in m/s², and friction f and signed grade G as decimal ratios. A 3% downgrade is G = −0.03. Both distances and the final answer are m.
- g
- Gravity
Gravity. An uphill positive grade assists braking; a downhill negative grade reduces this effective deceleration.
m/s²Use m/s² as the base unit shown here. Use v in m/s, reaction time t in s, gravity g in m/s², and friction f and signed grade G as decimal ratios. A 3% downgrade is G = −0.03. Both distances and the final answer are m.
- f
- Longitudinal friction
Longitudinal friction. An uphill positive grade assists braking; a downhill negative grade reduces this effective deceleration.
ratio / no unitA dimensionless ratio has no physical unit; 0.01 as a ratio is 1% when the percent option is selected.
- G
- Signed grade (+ uphill, − downhill)
Signed grade (+ uphill, − downhill). An uphill positive grade assists braking; a downhill negative grade reduces this effective deceleration.
ratio / no unitA dimensionless ratio has no physical unit; 0.01 as a ratio is 1% when the percent option is selected.
- SSD
- Result to find
Stopping sight distance — basic model. Reaction and braking occur in sequence, so both distances contribute to the total.
m
Sort out the units first
Use v in m/s, reaction time t in s, gravity g in m/s², and friction f and signed grade G as decimal ratios. A 3% downgrade is G = −0.03. Both distances and the final answer are m.
Assumptions before calculating
Assume constant initial speed during reaction, constant modeled braking deceleration, small longitudinal grade and a positive effective braking ratio f + G. The supplied reaction time and friction must suit the study conditions.
Let’s solve one together
Read the given values, follow each operation, then check what the result means.
Read the supplied values as one complete study case. Find SSD and explain the result in the stated output unit.
- v · Speed
- 22.22 m/s
- t · Reaction time
- 2.5 s
- g · Gravity
- 9.81 m/s²
- f · Longitudinal friction
- 0.35
- G · Signed grade (+ uphill, − downhill)
- 0
Calculate travel during perception and reaction
The vehicle continues at its initial modeled speed during the reaction interval.
(22.22) × (2.5) = 55.55 mApply the signed grade to braking deceleration
An uphill positive grade assists braking; a downhill negative grade reduces this effective deceleration.
(9.81) × ((0.35) + (0)) = 3.4335 m/s²Calculate the constant-deceleration braking distance
Stopping from speed v under constant deceleration requires v squared divided by twice that deceleration.
(22.22)^2 ÷ (2 × (3.4335)) ≈ 71.89870395 mAdd the two consecutive travel distances
Reaction and braking occur in sequence, so both distances contribute to the total.
(55.55) + (71.89870395) ≈ 127.4487039 m
Does this worked answer make sense?
Increasing reaction time adds v times the time increase. For equal speed and friction, downhill requires more braking distance than level ground, while uphill requires less in this model.
A second worked example — different values
A second case uses different data. Predict which way the answer will change, then calculate it without reusing the first answer.
- v · Speed
- 20 m/s
- t · Reaction time
- 2.5 s
- g · Gravity
- 9.81 m/s²
- f · Longitudinal friction
- 0.3
- G · Signed grade (+ uphill, − downhill)
- -0.03
Calculate travel during perception and reaction
The vehicle continues at its initial modeled speed during the reaction interval.
(20) × (2.5) = 50 mApply the signed grade to braking deceleration
An uphill positive grade assists braking; a downhill negative grade reduces this effective deceleration.
(9.81) × ((0.3) + (-0.03)) = 2.6487 m/s²Calculate the constant-deceleration braking distance
Stopping from speed v under constant deceleration requires v squared divided by twice that deceleration.
(20)^2 ÷ (2 × (2.6487)) ≈ 75.50874014 mAdd the two consecutive travel distances
Reaction and braking occur in sequence, so both distances contribute to the total.
(50) + (75.50874014) ≈ 125.5087401 m
Now try your own values
Change a value or its unit. The same method will show your calculation, step by step.
Results update only when you calculate. The lesson example above stays unchanged.
Your turn — check your understanding
Solve this separate case yourself. Use only the values below; the two worked examples use different data. Give the requested result in the selected unit.
- v · Speed
- 25 m/s
- t · Reaction time
- 2 s
- g · Gravity
- 9.81 m/s²
- f · Longitudinal friction
- 0.35
- G · Signed grade (+ uphill, − downhill)
- 0.02
Find: Learn: Stopping sight distance — basic model
A hint, not the answer
Calculate reaction distance vt. Then calculate braking distance v²/[2g(f + G)]. Add them. In this stored convention G is positive uphill and negative downhill, because climbing assists deceleration.
Use v in m/s, reaction time t in s, gravity g in m/s², and friction f and signed grade G as decimal ratios. A 3% downgrade is G = −0.03. Both distances and the final answer are m.
Show the full practice solution
Compare the steps with your work; revealing a solution does not mark the lesson complete.
Calculate travel during perception and reaction
The vehicle continues at its initial modeled speed during the reaction interval.
(25) × (2) = 50 mApply the signed grade to braking deceleration
An uphill positive grade assists braking; a downhill negative grade reduces this effective deceleration.
(9.81) × ((0.35) + (0.02)) = 3.6297 m/s²Calculate the constant-deceleration braking distance
Stopping from speed v under constant deceleration requires v squared divided by twice that deceleration.
(25)^2 ÷ (2 × (3.6297)) ≈ 86.09526958 mAdd the two consecutive travel distances
Reaction and braking occur in sequence, so both distances contribute to the total.
(50) + (86.09526958) ≈ 136.0952696 m
Avoid the common trap
Do not insert speed in km/h without conversion, omit reaction distance, or use a positive grade for downhill in this convention. If f + G is not positive, this constant-deceleration model cannot provide a finite stopping distance.
When this method applies — and when it does not
This simplified formula is not a safety guarantee or a substitute for the adopted road design method. Actual stopping depends on driver, tires, pavement, weather, brake response and slope variation. Geometric sight obstructions must be checked separately.
For study and understanding, not approval of a real structure, site operation or design. Apply the correct standard, National Annex and professional review to actual engineering work.
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Sources & further reading
References open in a new tab and explain the underlying principles. The teaching text and examples here are SimpleFlick’s own; the source organisations have not endorsed this calculator.
Reading focus: Stopping sight distance — basic model. Speed, friction, curvature and stopping-distance principles. Supplied friction/reaction inputs are study data, not a driving recommendation.
Lesson updated: · Both examples and the separate practice case are checked against an independent high-precision numerical implementation. This verifies arithmetic for the stated model, not engineering certification.