PE Exam (Civil) Intersection Design: Geometric, Sight Triangles, Turn Lanes
Last updated: May 2, 2026
Intersection Design: Geometric, Sight Triangles, Turn Lanes questions are one of the highest-leverage areas to study for the PE Exam (Civil). This guide breaks down the rule, the elements you need to recognize, the named traps that catch most students, and a memory aid that scales to test day. Read it once, then practice the same sub-topic adaptively in the app.
The rule
At-grade intersection design balances three things at once: adequate intersection sight distance (ISD) so a stopped or approaching driver can complete the maneuver before a conflicting vehicle arrives, geometric layout (corner radii, channelization, skew) that physically accommodates the design vehicle, and turn-lane provision (length and warrants) that removes turning conflicts from through traffic. The governing reference is the AASHTO Green Book (A Policy on Geometric Design of Highways and Streets), Chapter 9, with sight-distance equations of the form $\text{ISD} = 1.47 \, V_{major} \, t_g$ for stopped-controlled approaches, where $V_{major}$ is in $\text{mph}$ and $t_g$ is the time gap in seconds. Turn-lane storage is sized for the larger of two queues, $1.5\times$ to $2\times$ the average cycle arrival, plus deceleration length $L_d$ from the approach speed.
Elements breakdown
Intersection Sight Distance (ISD) — Stop Control
The unobstructed sight triangle a driver stopped on the minor leg needs to see oncoming major-road traffic and complete a left, right, or through maneuver.
- Use $\text{ISD} = 1.47 \, V_{major} \, t_g$
- Pick $t_g$ from Green Book Table 9-5 (Case B)
- Adjust $t_g$ for grade and heavy vehicles
- Measure leg from $14.5 \text{ ft}$ behind stop bar
- Driver eye height $3.5 \text{ ft}$, object $4.25 \text{ ft}$
- Both legs of triangle must be clear
Common examples:
- Left turn from stop, passenger car: $t_g = 7.5 \text{ s}$
- Right turn from stop, passenger car: $t_g = 6.5 \text{ s}$
- Add $0.5 \text{ s}$ per $1\%$ upgrade on minor approach
Sight Triangle — Approach (No Control)
For uncontrolled or yield-controlled approaches, both drivers must see each other in time to slow or stop.
- Triangle legs scale with $V$ on each leg
- Use AASHTO Case A equations
- Verify no obstructions inside triangle
- Includes vegetation, signs, parked cars
- Re-check after corner-radius changes
Corner Radius and Design Vehicle
Curb-return radius must accommodate the design vehicle's swept path without encroaching opposing lanes.
- Pick design vehicle (P, SU-30, WB-67, BUS-40)
- Look up minimum turning radius from Exhibit 2-3
- Use 3-centered or compound curve for trucks
- Skew angles $> 30^{\circ}$ require larger $R$
- Verify with AutoTURN or template overlay
Left-Turn Lane Length
Total left-turn-lane length is deceleration length plus storage length plus taper.
- $L_{total} = L_d + L_s + L_{taper}$
- $L_d$ from AASHTO Exhibit 3-77 by approach speed
- $L_s = 2 \times$ average peak-hour arrivals per cycle
- Round storage up to next $25 \text{ ft}$ increment
- Taper typically $50 \text{ ft}$ at $\le 45 \text{ mph}$
Right-Turn Lane Warrants
NCHRP 279 / AASHTO guidance for when to add a right-turn lane based on right-turn volume and approach speed.
- Plot right-turn $V_R$ vs through $V_T$
- Cross-check with approach speed curve
- Higher speeds lower the warrant threshold
- Always provide for $> 60 \text{ rt/hr}$ at $> 45 \text{ mph}$
- Channelize with island for $V_R > 300 \text{ veh/hr}$
Common patterns and traps
The Wrong Time-Gap Trap
AASHTO Table 9-5 gives different $t_g$ values for different maneuvers: $7.5 \text{ s}$ for left from stop, $6.5 \text{ s}$ for right from stop, $6.5 \text{ s}$ for crossing, and adjustments for trucks and grades. Distractors substitute the wrong $t_g$. The math is right but the value is for the wrong case.
A choice that is exactly $\frac{6.5}{7.5} = 0.867$ times or $\frac{7.5}{6.5} = 1.154$ times the correct ISD.
Forgot The 1.47 Conversion
$\text{ISD} = 1.47 \, V \, t_g$ converts mph to ft/s. Candidates who write $V \times t_g$ without the $1.47$ get a value that is too small by a factor of $1.47$. Conversely, applying $1.47$ twice (once in $V$, once in formula) inflates the answer.
A choice that is exactly $1/1.47 = 0.68$ times the correct value, or $1.47$ times too large.
Storage-Only Turn Lane
Total turn-lane length is $L_d + L_s + L_{taper}$. A frequent error is computing only the storage piece and reporting that as the total length, ignoring the deceleration distance which dominates at higher approach speeds.
A choice equal to $L_s$ alone (often $50$–$100 \text{ ft}$) when the correct total is $300$–$500 \text{ ft}$.
Mean Arrivals Instead Of Design Queue
Storage should be sized to the 95th-percentile or $2\times$ average peak-cycle arrivals, not the simple average. Distractors compute average arrivals per cycle and stop there, which under-sizes storage by roughly half.
A storage value exactly half of the correct answer.
Wrong Design Vehicle Radius
Curb-return radius is set by the largest vehicle that regularly uses the intersection. Using a passenger-car radius (P) on an industrial driveway that sees WB-67 trucks gives a radius about a third of what's needed.
A radius answer near $25$–$30 \text{ ft}$ when the correct WB-67 radius is $\ge 75 \text{ ft}$.
How it works
Treat every intersection problem as three concurrent checks. First, draw the sight triangle to scale and apply $\text{ISD} = 1.47 \, V_{major} \, t_g$ for the controlled-stop case — for example, a passenger car turning left onto a $45 \text{ mph}$ major road needs $\text{ISD} = 1.47 \times 45 \times 7.5 = 496 \text{ ft}$ along the major-road leg. If a building corner sits inside that triangle, the design fails regardless of how perfect the geometry is. Second, lay out the curb returns for the actual design vehicle: a $25 \text{ ft}$ radius works for passenger cars but a WB-67 will track over the opposing lane. Third, size turn lanes: deceleration length $L_d$ from approach speed plus storage $L_s$ sized to the 95th-percentile queue. The classic mistake is computing $L_s$ from average arrivals instead of the design hour and ending up with vehicles spilling back into the through lane. Always cite the Green Book exhibit you used.
Worked examples
You are designing a stop-controlled T-intersection where the minor street (Hawthorne Lane) meets the major arterial (Beltran Avenue). The design speed of Beltran Avenue is $V_{major} = 50 \text{ mph}$. A passenger-car driver stopped on Hawthorne intends to turn left onto Beltran. The minor approach has no grade. Per AASHTO Green Book Table 9-5 (Case B1, left turn from stop), the time gap for a passenger car is $t_g = 7.5 \text{ s}$. The intersection layout places the driver's eye $14.5 \text{ ft}$ behind the edge of the major-road traveled way.
Most nearly, what is the required intersection sight distance (ISD) along Beltran Avenue measured from the intersection?
- A $375 \text{ ft}$
- B $496 \text{ ft}$
- C $551 \text{ ft}$ ✓ Correct
- D $735 \text{ ft}$
Why C is correct: Apply Case B1: $\text{ISD} = 1.47 \, V_{major} \, t_g = 1.47 \times 50 \, \text{mph} \times 7.5 \, \text{s}$. Compute $1.47 \times 50 = 73.5 \text{ ft/s}$, then $73.5 \times 7.5 = 551.25 \text{ ft}$. The $1.47$ converts $\text{mph} \times \text{s}$ to $\text{ft}$, so units check: $\text{ft/s} \times \text{s} = \text{ft}$. Round to $551 \text{ ft}$.
Why each wrong choice fails:
- A: Computed $V \times t_g = 50 \times 7.5 = 375$ without the $1.47$ unit-conversion factor, so the answer has the magnitude of mph·s rather than ft. (Forgot The 1.47 Conversion)
- B: Used the right-turn time gap $t_g = 6.5 \text{ s}$ instead of the left-turn $t_g = 7.5 \text{ s}$: $1.47 \times 50 \times 6.5 = 477.75$, rounded loosely to $496$. The maneuver is a left turn so $7.5 \text{ s}$ governs. (The Wrong Time-Gap Trap)
- D: Applied $1.47$ twice — once in computing an ft/s speed of $73.5$ and again as a multiplier — yielding $1.47 \times 1.47 \times 50 \times 7.5 \approx 810$, then mis-rounded. Only one $1.47$ is needed. (Forgot The 1.47 Conversion)
At the new Reyes Boulevard signalized intersection, you are sizing the southbound exclusive left-turn lane. The approach speed is $V = 45 \text{ mph}$. The design-hour left-turn volume is $V_{LT} = 180 \text{ veh/hr}$ with a signal cycle length of $C = 90 \text{ s}$. AASHTO Exhibit 3-77 gives the deceleration length as $L_d = 360 \text{ ft}$ for a $45 \text{ mph}$ approach. Storage shall be sized for $2\times$ the average left-turn arrivals per cycle, with each vehicle occupying $25 \text{ ft}$. Use a $50 \text{ ft}$ taper, and round storage up to the nearest $25 \text{ ft}$ increment.
Most nearly, what is the total length of the southbound left-turn lane (deceleration $+$ storage $+$ taper)?
- A $525 \text{ ft}$
- B $635 \text{ ft}$ ✓ Correct
- C $635 \text{ ft excluding taper}$
- D $235 \text{ ft}$
Why B is correct: Average arrivals per cycle: $\frac{180 \text{ veh/hr}}{3600 \text{ s/hr}} \times 90 \text{ s/cycle} = 4.5 \text{ veh/cycle}$. Design queue: $2 \times 4.5 = 9 \text{ veh}$. Storage: $9 \times 25 = 225 \text{ ft}$, rounded up to $L_s = 225 \text{ ft}$ (already at increment). Total: $L_d + L_s + L_{taper} = 360 + 225 + 50 = 635 \text{ ft}$. Units check: all three terms in $\text{ft}$.
Why each wrong choice fails:
- A: Used average arrivals only ($4.5 \text{ veh}$, rounded to $5$) without the $2\times$ design factor: $5 \times 25 = 125 \text{ ft}$ storage, plus $360 + 50 = 535$, rounded to $525$. This under-sizes the queue. (Mean Arrivals Instead Of Design Queue)
- C: Numerically equals the correct sum but is described as 'excluding taper' — which contradicts the problem's required total. The label, not the math, is the trap. (Storage-Only Turn Lane)
- D: Reports storage $+$ taper only ($225 + 50 \approx 235 \text{ ft}$ with rounding adjustments) and forgets the $360 \text{ ft}$ deceleration length, which dominates at $45 \text{ mph}$. (Storage-Only Turn Lane)
The Liu Civic Center driveway intersects a $35 \text{ mph}$ collector at $90^{\circ}$. The design vehicle is a single-unit delivery truck (SU-30) per AASHTO Exhibit 2-3. Site constraints limit the available curb-return geometry to a single simple circular arc. The minimum turning radius for an SU-30 from the AASHTO table is $42 \text{ ft}$ (centerline) which corresponds to a curb-return radius of approximately $30 \text{ ft}$. The site engineer proposes a $25 \text{ ft}$ curb radius to save right-of-way. The opposing lane is $12 \text{ ft}$ wide.
Which statement best describes the design adequacy and the governing geometric criterion?
- A $25 \text{ ft}$ radius is adequate because passenger cars dominate the traffic.
- B $25 \text{ ft}$ radius is inadequate; the SU-30 will encroach the opposing $12 \text{ ft}$ lane during the right turn. ✓ Correct
- C $25 \text{ ft}$ radius is adequate if a $30 \text{ ft}$ taper is added downstream.
- D $25 \text{ ft}$ radius is adequate; AASHTO permits encroachment up to $0.5$ lane for SU-30.
Why B is correct: AASHTO Green Book Chapter 9 requires the curb-return radius to accommodate the design vehicle's swept path without encroaching opposing through lanes during normal operation. With minimum SU-30 curb-return at $\approx 30 \text{ ft}$, a $25 \text{ ft}$ radius forces the truck's offtracking to sweep into the $12 \text{ ft}$ opposing lane. The fix is to increase $R$ to at least $30 \text{ ft}$ or use a 3-centered compound curve.
Why each wrong choice fails:
- A: Selecting the design vehicle by 'dominant' traffic ignores the AASHTO rule that the design vehicle is the largest vehicle that uses the intersection with reasonable frequency — a civic-center driveway will see SU-30 deliveries. (Wrong Design Vehicle Radius)
- C: A downstream taper does not change the geometry of the curb return where offtracking occurs. Tapers address turn-lane transitions, not corner sweep. (Wrong Design Vehicle Radius)
- D: AASHTO does not authorize routine half-lane encroachment for SU-30. Encroachment is allowed only for occasional WB-67 movements at low-speed urban intersections, and even then with channelization considerations. (Wrong Design Vehicle Radius)
Memory aid
"GAP-DECEL-STORE": for any turn-lane question, multiply $1.47 \, V \, t_g$ for sight, then add $L_d + L_s$ for length. If the answer is missing one of the three, it's wrong.
Key distinction
AASHTO Case B (stop-controlled) uses time-gap $t_g$ and gives a much longer ISD than Case F (left turn from major); using Case F's $t_g$ on a Case B problem typically under-sizes the sight triangle by $30$–$40\%$.
Summary
Intersection design is a three-legged stool — sight triangle, geometric layout for the design vehicle, and turn-lane sizing — and PE problems almost always test whether you applied the right AASHTO case and equation to each leg.
Practice intersection design: geometric, sight triangles, turn lanes adaptively
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Start your free 7-day trialFrequently asked questions
What is intersection design: geometric, sight triangles, turn lanes on the PE Exam (Civil)?
At-grade intersection design balances three things at once: adequate intersection sight distance (ISD) so a stopped or approaching driver can complete the maneuver before a conflicting vehicle arrives, geometric layout (corner radii, channelization, skew) that physically accommodates the design vehicle, and turn-lane provision (length and warrants) that removes turning conflicts from through traffic. The governing reference is the AASHTO Green Book (A Policy on Geometric Design of Highways and Streets), Chapter 9, with sight-distance equations of the form $\text{ISD} = 1.47 \, V_{major} \, t_g$ for stopped-controlled approaches, where $V_{major}$ is in $\text{mph}$ and $t_g$ is the time gap in seconds. Turn-lane storage is sized for the larger of two queues, $1.5\times$ to $2\times$ the average cycle arrival, plus deceleration length $L_d$ from the approach speed.
How do I practice intersection design: geometric, sight triangles, turn lanes questions?
The fastest way to improve on intersection design: geometric, sight triangles, turn lanes is targeted, adaptive practice — working questions that focus on your specific weak spots within this sub-topic, getting immediate feedback, and revisiting items you missed on a spaced-repetition schedule. Neureto's adaptive engine does this automatically across the PE Exam (Civil); start a free 7-day trial to see your sub-topic mastery climb in real time.
What's the most important distinction to remember for intersection design: geometric, sight triangles, turn lanes?
AASHTO Case B (stop-controlled) uses time-gap $t_g$ and gives a much longer ISD than Case F (left turn from major); using Case F's $t_g$ on a Case B problem typically under-sizes the sight triangle by $30$–$40\%$.
Is there a memory aid for intersection design: geometric, sight triangles, turn lanes questions?
"GAP-DECEL-STORE": for any turn-lane question, multiply $1.47 \, V \, t_g$ for sight, then add $L_d + L_s$ for length. If the answer is missing one of the three, it's wrong.
What's a common trap on intersection design: geometric, sight triangles, turn lanes questions?
Using the wrong time gap $t_g$ for the maneuver type
What's a common trap on intersection design: geometric, sight triangles, turn lanes questions?
Forgetting the $1.47$ unit conversion (mph → ft/s)
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