Basic Freeway Segment Highway Capacity

Example 1

Basic Freeway Segment Highway Capacity — solve for flow rate — Basic Freeway Segment Highway Capacity

a basic freeway segment highway capacity analysis for a level-terrain freeway Given hourly volume (V) = 3,050 veh/h; peak hour factor (PHF) = 0.9100; number of lanes (N) = 4.0000; heavy vehicle adjustment factor (f_HV) = 0.8700; driver population factor (f_p) = 0.9600, determine the flow rate (v_p) in pc/h/ln.

Given

$hourly volume (V) = 3,050 veh/h$
$peak hour factor (PHF) = 0.9100$
$number of lanes (N) = 4.0000$
$heavy vehicle adjustment factor (f_HV) = 0.8700$
$driver population factor (f_p) = 0.9600$

Find

flow rate (v_p), in pc/h/ln

Start with the thinking

The governing relation printed in this handbook section is Basic Freeway Segment Highway Capacity.
Everything except v_p is given, so isolate v_p symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
The basic freeway segment highway capacity relation converts hourly volume into an equivalent passenger-car flow rate per lane.

Step-by-step solution

Step 1 — State the governing relation:
$v_p = \dfrac{V}{PHF \cdot N \cdot f_{HV} \cdot f_p}$
Step 2 — Rearrange the relation so that v_p stands alone on the left-hand side.
Step 3 — List the givens: hourly volume (V) = 3,050 veh/h, peak hour factor (PHF) = 0.9100, number of lanes (N) = 4.0000, heavy vehicle adjustment factor (f_HV) = 0.8700, driver population factor (f_p) = 0.9600.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$v_{p} = 1003\ \text{pc/h/ln}$
Step 6 — Check: returning v_p = 1,003 pc/h/ln to
$v_p = \dfrac{V}{PHF \cdot N \cdot f_{HV} \cdot f_p}$
reproduces the given quantities, and both sides carry the same units.

Answer:

v_{p} = 1003\ \text{pc/h/ln}

Why the other options are there

2,006 — kept a factor of two that cancels in the correct rearrangement.
501.6 — dropped that same factor in the other direction.
1,104 — rounded an intermediate value before the final step.

Reference: FE Handbook — Basic Freeway Segment Highway Capacity

Example 2

Basic Freeway Segment Highway Capacity — solve for hourly volume — Basic Freeway Segment Highway Capacity (2)

highway capacity analysis of a basic freeway segment during peak hour Given peak hour factor (PHF) = 0.9100; number of lanes (N) = 3.0000; heavy vehicle adjustment factor (f_HV) = 0.9400; driver population factor (f_p) = 0.9400; flow rate (v_p) = 1,650 pc/h/ln, determine the hourly volume (V) in veh/h.

Given

$peak hour factor (PHF) = 0.9100$
$number of lanes (N) = 3.0000$
$heavy vehicle adjustment factor (f_HV) = 0.9400$
$driver population factor (f_p) = 0.9400$
$flow rate (v_p) = 1,650 pc/h/\ln$

Find

hourly volume (V), in veh/h

Start with the thinking

The governing relation printed in this handbook section is Basic Freeway Segment Highway Capacity.
Everything except V is given, so isolate V symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
The basic freeway segment highway capacity relation converts hourly volume into an equivalent passenger-car flow rate per lane.

Step-by-step solution

Step 1 — State the governing relation:
$v_p = \dfrac{V}{PHF \cdot N \cdot f_{HV} \cdot f_p}$
Step 2 — Rearrange the relation so that V stands alone on the left-hand side.
Step 3 — List the givens: peak hour factor (PHF) = 0.9100, number of lanes (N) = 3.0000, heavy vehicle adjustment factor (f_HV) = 0.9400, driver population factor (f_p) = 0.9400, flow rate (v_p) = 1,650 pc/h/ln.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$V = 3980\ \text{veh/h}$
Step 6 — Check: returning V = 3,980 veh/h to
$v_p = \dfrac{V}{PHF \cdot N \cdot f_{HV} \cdot f_p}$
reproduces the given quantities, and both sides carry the same units.

Answer:

V = 3980\ \text{veh/h}

Why the other options are there

7,960 — kept a factor of two that cancels in the correct rearrangement.
1,990 — dropped that same factor in the other direction.
4,378 — rounded an intermediate value before the final step.

Reference: FE Handbook — Basic Freeway Segment Highway Capacity

Example 3

Basic Freeway Segment Highway Capacity — solve for number of lanes — Basic Freeway Segment Highway Capacity (3)

basic freeway segment capacity with heavy-vehicle traffic Given hourly volume (V) = 4,650 veh/h; peak hour factor (PHF) = 0.8600; heavy vehicle adjustment factor (f_HV) = 0.8800; driver population factor (f_p) = 0.8800; flow rate (v_p) = 1,880 pc/h/ln, determine the number of lanes (N).

Given

$hourly volume (V) = 4,650 veh/h$
$peak hour factor (PHF) = 0.8600$
$heavy vehicle adjustment factor (f_HV) = 0.8800$
$driver population factor (f_p) = 0.8800$
$flow rate (v_p) = 1,880 pc/h/\ln$

Find

number of lanes (N)

Start with the thinking

The governing relation printed in this handbook section is Basic Freeway Segment Highway Capacity.
Everything except N is given, so isolate N symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
The basic freeway segment highway capacity relation converts hourly volume into an equivalent passenger-car flow rate per lane.

Step-by-step solution

Step 1 — State the governing relation:
$v_p = \dfrac{V}{PHF \cdot N \cdot f_{HV} \cdot f_p}$
Step 2 — Rearrange the relation so that N stands alone on the left-hand side.
Step 3 — List the givens: hourly volume (V) = 4,650 veh/h, peak hour factor (PHF) = 0.8600, heavy vehicle adjustment factor (f_HV) = 0.8800, driver population factor (f_p) = 0.8800, flow rate (v_p) = 1,880 pc/h/ln.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$N = 3.7139$
Step 6 — Check: returning N = 3.7139 to
$v_p = \dfrac{V}{PHF \cdot N \cdot f_{HV} \cdot f_p}$
reproduces the given quantities, and both sides carry the same units.

Answer:

N = 3.7139

Why the other options are there

7.4278 — kept a factor of two that cancels in the correct rearrangement.
1.8570 — dropped that same factor in the other direction.
4.0853 — rounded an intermediate value before the final step.

Reference: FE Handbook — Basic Freeway Segment Highway Capacity

Example 4

Basic Freeway Segment Highway Capacity — solve for flow rate (case 2) — Basic Freeway Segment Highway Capacity (4)

a basic freeway segment highway capacity analysis for a level-terrain freeway Given hourly volume (V) = 1,500 veh/h; peak hour factor (PHF) = 0.9600; number of lanes (N) = 4.0000; heavy vehicle adjustment factor (f_HV) = 0.9000; driver population factor (f_p) = 1.0000, determine the flow rate (v_p) in pc/h/ln.

Given

$hourly volume (V) = 1,500 veh/h$
$peak hour factor (PHF) = 0.9600$
$number of lanes (N) = 4.0000$
$heavy vehicle adjustment factor (f_HV) = 0.9000$
$driver population factor (f_p) = 1.0000$

Find

flow rate (v_p), in pc/h/ln

Start with the thinking

The governing relation printed in this handbook section is Basic Freeway Segment Highway Capacity.
Everything except v_p is given, so isolate v_p symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
The basic freeway segment highway capacity relation converts hourly volume into an equivalent passenger-car flow rate per lane.

Step-by-step solution

Step 1 — State the governing relation:
$v_p = \dfrac{V}{PHF \cdot N \cdot f_{HV} \cdot f_p}$
Step 2 — Rearrange the relation so that v_p stands alone on the left-hand side.
Step 3 — List the givens: hourly volume (V) = 1,500 veh/h, peak hour factor (PHF) = 0.9600, number of lanes (N) = 4.0000, heavy vehicle adjustment factor (f_HV) = 0.9000, driver population factor (f_p) = 1.0000.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$v_{p} = 434.0\ \text{pc/h/ln}$
Step 6 — Check: returning v_p = 434.0 pc/h/ln to
$v_p = \dfrac{V}{PHF \cdot N \cdot f_{HV} \cdot f_p}$
reproduces the given quantities, and both sides carry the same units.

Answer:

v_{p} = 434.0\ \text{pc/h/ln}

Why the other options are there

868.1 — kept a factor of two that cancels in the correct rearrangement.
217.0 — dropped that same factor in the other direction.
477.4 — rounded an intermediate value before the final step.

Reference: FE Handbook — Basic Freeway Segment Highway Capacity

Example 5

Basic Freeway Segment Highway Capacity — solve for hourly volume (case 2) — Basic Freeway Segment Highway Capacity (5)

highway capacity analysis of a basic freeway segment during peak hour Given peak hour factor (PHF) = 0.9000; number of lanes (N) = 3.0000; heavy vehicle adjustment factor (f_HV) = 0.8800; driver population factor (f_p) = 0.9200; flow rate (v_p) = 900.0 pc/h/ln, determine the hourly volume (V) in veh/h.

Given

$peak hour factor (PHF) = 0.9000$
$number of lanes (N) = 3.0000$
$heavy vehicle adjustment factor (f_HV) = 0.8800$
$driver population factor (f_p) = 0.9200$
$flow rate (v_p) = 900.0 pc/h/\ln$

Find

hourly volume (V), in veh/h

Start with the thinking

The governing relation printed in this handbook section is Basic Freeway Segment Highway Capacity.
Everything except V is given, so isolate V symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
The basic freeway segment highway capacity relation converts hourly volume into an equivalent passenger-car flow rate per lane.

Step-by-step solution

Step 1 — State the governing relation:
$v_p = \dfrac{V}{PHF \cdot N \cdot f_{HV} \cdot f_p}$
Step 2 — Rearrange the relation so that V stands alone on the left-hand side.
Step 3 — List the givens: peak hour factor (PHF) = 0.9000, number of lanes (N) = 3.0000, heavy vehicle adjustment factor (f_HV) = 0.8800, driver population factor (f_p) = 0.9200, flow rate (v_p) = 900.0 pc/h/ln.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$V = 1967\ \text{veh/h}$
Step 6 — Check: returning V = 1,967 veh/h to
$v_p = \dfrac{V}{PHF \cdot N \cdot f_{HV} \cdot f_p}$
reproduces the given quantities, and both sides carry the same units.

Answer:

V = 1967\ \text{veh/h}

Why the other options are there

3,935 — kept a factor of two that cancels in the correct rearrangement.
983.7 — dropped that same factor in the other direction.
2,164 — rounded an intermediate value before the final step.

Reference: FE Handbook — Basic Freeway Segment Highway Capacity

Example 6

Basic Freeway Segment Highway Capacity — solve for number of lanes (case 2) — Basic Freeway Segment Highway Capacity (6)

basic freeway segment capacity with heavy-vehicle traffic Given hourly volume (V) = 2,250 veh/h; peak hour factor (PHF) = 0.8800; heavy vehicle adjustment factor (f_HV) = 0.9300; driver population factor (f_p) = 0.8700; flow rate (v_p) = 2,020 pc/h/ln, determine the number of lanes (N).

Given

$hourly volume (V) = 2,250 veh/h$
$peak hour factor (PHF) = 0.8800$
$heavy vehicle adjustment factor (f_HV) = 0.9300$
$driver population factor (f_p) = 0.8700$
$flow rate (v_p) = 2,020 pc/h/\ln$

Find

number of lanes (N)

Start with the thinking

The governing relation printed in this handbook section is Basic Freeway Segment Highway Capacity.
Everything except N is given, so isolate N symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
The basic freeway segment highway capacity relation converts hourly volume into an equivalent passenger-car flow rate per lane.

Step-by-step solution

Step 1 — State the governing relation:
$v_p = \dfrac{V}{PHF \cdot N \cdot f_{HV} \cdot f_p}$
Step 2 — Rearrange the relation so that N stands alone on the left-hand side.
Step 3 — List the givens: hourly volume (V) = 2,250 veh/h, peak hour factor (PHF) = 0.8800, heavy vehicle adjustment factor (f_HV) = 0.9300, driver population factor (f_p) = 0.8700, flow rate (v_p) = 2,020 pc/h/ln.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$N = 1.5644$
Step 6 — Check: returning N = 1.5644 to
$v_p = \dfrac{V}{PHF \cdot N \cdot f_{HV} \cdot f_p}$
reproduces the given quantities, and both sides carry the same units.

Answer:

N = 1.5644

Why the other options are there

3.1288 — kept a factor of two that cancels in the correct rearrangement.
0.7822 — dropped that same factor in the other direction.
1.7208 — rounded an intermediate value before the final step.

Reference: FE Handbook — Basic Freeway Segment Highway Capacity

Example 7

Basic Freeway Segment Highway Capacity — solve for flow rate (case 3) — Basic Freeway Segment Highway Capacity (7)

a basic freeway segment highway capacity analysis for a level-terrain freeway Given hourly volume (V) = 600.0 veh/h; peak hour factor (PHF) = 0.9200; number of lanes (N) = 3.0000; heavy vehicle adjustment factor (f_HV) = 0.9800; driver population factor (f_p) = 0.8700, determine the flow rate (v_p) in pc/h/ln.

Given

$hourly volume (V) = 600.0 veh/h$
$peak hour factor (PHF) = 0.9200$
$number of lanes (N) = 3.0000$
$heavy vehicle adjustment factor (f_HV) = 0.9800$
$driver population factor (f_p) = 0.8700$

Find

flow rate (v_p), in pc/h/ln

Start with the thinking

The governing relation printed in this handbook section is Basic Freeway Segment Highway Capacity.
Everything except v_p is given, so isolate v_p symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
The basic freeway segment highway capacity relation converts hourly volume into an equivalent passenger-car flow rate per lane.

Step-by-step solution

Step 1 — State the governing relation:
$v_p = \dfrac{V}{PHF \cdot N \cdot f_{HV} \cdot f_p}$
Step 2 — Rearrange the relation so that v_p stands alone on the left-hand side.
Step 3 — List the givens: hourly volume (V) = 600.0 veh/h, peak hour factor (PHF) = 0.9200, number of lanes (N) = 3.0000, heavy vehicle adjustment factor (f_HV) = 0.9800, driver population factor (f_p) = 0.8700.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$v_{p} = 255.0\ \text{pc/h/ln}$
Step 6 — Check: returning v_p = 255.0 pc/h/ln to
$v_p = \dfrac{V}{PHF \cdot N \cdot f_{HV} \cdot f_p}$
reproduces the given quantities, and both sides carry the same units.

Answer:

v_{p} = 255.0\ \text{pc/h/ln}

Why the other options are there

509.9 — kept a factor of two that cancels in the correct rearrangement.
127.5 — dropped that same factor in the other direction.
280.5 — rounded an intermediate value before the final step.

Reference: FE Handbook — Basic Freeway Segment Highway Capacity

Example 8

Basic Freeway Segment Highway Capacity — solve for hourly volume (case 3) — Basic Freeway Segment Highway Capacity (8)

highway capacity analysis of a basic freeway segment during peak hour Given peak hour factor (PHF) = 0.8800; number of lanes (N) = 3.0000; heavy vehicle adjustment factor (f_HV) = 0.9700; driver population factor (f_p) = 0.9100; flow rate (v_p) = 640.0 pc/h/ln, determine the hourly volume (V) in veh/h.

Given

$peak hour factor (PHF) = 0.8800$
$number of lanes (N) = 3.0000$
$heavy vehicle adjustment factor (f_HV) = 0.9700$
$driver population factor (f_p) = 0.9100$
$flow rate (v_p) = 640.0 pc/h/\ln$

Find

hourly volume (V), in veh/h

Start with the thinking

The governing relation printed in this handbook section is Basic Freeway Segment Highway Capacity.
Everything except V is given, so isolate V symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
The basic freeway segment highway capacity relation converts hourly volume into an equivalent passenger-car flow rate per lane.

Step-by-step solution

Step 1 — State the governing relation:
$v_p = \dfrac{V}{PHF \cdot N \cdot f_{HV} \cdot f_p}$
Step 2 — Rearrange the relation so that V stands alone on the left-hand side.
Step 3 — List the givens: peak hour factor (PHF) = 0.8800, number of lanes (N) = 3.0000, heavy vehicle adjustment factor (f_HV) = 0.9700, driver population factor (f_p) = 0.9100, flow rate (v_p) = 640.0 pc/h/ln.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$V = 1491\ \text{veh/h}$
Step 6 — Check: returning V = 1,491 veh/h to
$v_p = \dfrac{V}{PHF \cdot N \cdot f_{HV} \cdot f_p}$
reproduces the given quantities, and both sides carry the same units.

Answer:

V = 1491\ \text{veh/h}

Why the other options are there

2,983 — kept a factor of two that cancels in the correct rearrangement.
745.7 — dropped that same factor in the other direction.
1,641 — rounded an intermediate value before the final step.

Reference: FE Handbook — Basic Freeway Segment Highway Capacity

Example 9

Basic Freeway Segment Highway Capacity — solve for number of lanes (case 3) — Basic Freeway Segment Highway Capacity (9)

basic freeway segment capacity with heavy-vehicle traffic Given hourly volume (V) = 3,200 veh/h; peak hour factor (PHF) = 0.9500; heavy vehicle adjustment factor (f_HV) = 0.9400; driver population factor (f_p) = 0.9200; flow rate (v_p) = 1,630 pc/h/ln, determine the number of lanes (N).

Given

$hourly volume (V) = 3,200 veh/h$
$peak hour factor (PHF) = 0.9500$
$heavy vehicle adjustment factor (f_HV) = 0.9400$
$driver population factor (f_p) = 0.9200$
$flow rate (v_p) = 1,630 pc/h/\ln$

Find

number of lanes (N)

Start with the thinking

The governing relation printed in this handbook section is Basic Freeway Segment Highway Capacity.
Everything except N is given, so isolate N symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
The basic freeway segment highway capacity relation converts hourly volume into an equivalent passenger-car flow rate per lane.

Step-by-step solution

Step 1 — State the governing relation:
$v_p = \dfrac{V}{PHF \cdot N \cdot f_{HV} \cdot f_p}$
Step 2 — Rearrange the relation so that N stands alone on the left-hand side.
Step 3 — List the givens: hourly volume (V) = 3,200 veh/h, peak hour factor (PHF) = 0.9500, heavy vehicle adjustment factor (f_HV) = 0.9400, driver population factor (f_p) = 0.9200, flow rate (v_p) = 1,630 pc/h/ln.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$N = 2.3896$
Step 6 — Check: returning N = 2.3896 to
$v_p = \dfrac{V}{PHF \cdot N \cdot f_{HV} \cdot f_p}$
reproduces the given quantities, and both sides carry the same units.

Answer:

N = 2.3896

Why the other options are there

4.7792 — kept a factor of two that cancels in the correct rearrangement.
1.1948 — dropped that same factor in the other direction.
2.6285 — rounded an intermediate value before the final step.

Reference: FE Handbook — Basic Freeway Segment Highway Capacity

Example 10

Basic Freeway Segment Highway Capacity — solve for flow rate (case 4) — Basic Freeway Segment Highway Capacity (10)

a basic freeway segment highway capacity analysis for a level-terrain freeway Given hourly volume (V) = 3,050 veh/h; peak hour factor (PHF) = 0.9300; number of lanes (N) = 4.0000; heavy vehicle adjustment factor (f_HV) = 0.8600; driver population factor (f_p) = 0.9300, determine the flow rate (v_p) in pc/h/ln.

Given

$hourly volume (V) = 3,050 veh/h$
$peak hour factor (PHF) = 0.9300$
$number of lanes (N) = 4.0000$
$heavy vehicle adjustment factor (f_HV) = 0.8600$
$driver population factor (f_p) = 0.9300$

Find

flow rate (v_p), in pc/h/ln

Start with the thinking

The governing relation printed in this handbook section is Basic Freeway Segment Highway Capacity.
Everything except v_p is given, so isolate v_p symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
The basic freeway segment highway capacity relation converts hourly volume into an equivalent passenger-car flow rate per lane.

Step-by-step solution

Step 1 — State the governing relation:
$v_p = \dfrac{V}{PHF \cdot N \cdot f_{HV} \cdot f_p}$
Step 2 — Rearrange the relation so that v_p stands alone on the left-hand side.
Step 3 — List the givens: hourly volume (V) = 3,050 veh/h, peak hour factor (PHF) = 0.9300, number of lanes (N) = 4.0000, heavy vehicle adjustment factor (f_HV) = 0.8600, driver population factor (f_p) = 0.9300.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$v_{p} = 1025\ \text{pc/h/ln}$
Step 6 — Check: returning v_p = 1,025 pc/h/ln to
$v_p = \dfrac{V}{PHF \cdot N \cdot f_{HV} \cdot f_p}$
reproduces the given quantities, and both sides carry the same units.

Answer:

v_{p} = 1025\ \text{pc/h/ln}

Why the other options are there

2,050 — kept a factor of two that cancels in the correct rearrangement.
512.6 — dropped that same factor in the other direction.
1,128 — rounded an intermediate value before the final step.

Reference: FE Handbook — Basic Freeway Segment Highway Capacity

Basic Freeway Segment Highway Capacity

Handbook notes for this section

Core formulas for this FE topic

Worked exam-style examples

Basic Freeway Segment Highway Capacity

Handbook notes for this section

Core formulas for this FE topic

FFSCapacityBreakpoint>Breakpoint≤CapacityFFS Capacity Breakpoint > Breakpoint \le CapacityFFSCapacityBreakpoint>Breakpoint≤Capacity

752,4001,000=75−0.00001105(vp−1,000)275 2,400 1,000 = 75 - 0.00001105 (vp - 1,000)2752,4001,000=75−0.00001105(vp−1,000)2

702,4001,200=70−0.00001157(vp−1,200)270 2,400 1,200 = 70 - 0.00001157 (vp - 1,200)2702,4001,200=70−0.00001157(vp−1,200)2

652,3501,400=65−0.00001416(vp−1,400)265 2,350 1,400 = 65 - 0.00001416 (vp - 1,400)2652,3501,400=65−0.00001416(vp−1,400)2

602,3001,600=60−0.00001814(vp−1,600)260 2,300 1,600 = 60 - 0.00001814 (vp - 1,600)2602,3001,600=60−0.00001814(vp−1,600)2

552,2501,800=55−0.00002469(vp−1,800)255 2,250 1,800 = 55 - 0.00002469 (vp - 1,800)2552,2501,800=55−0.00002469(vp−1,800)2

pc/h/ln⁡=passengercarsperhourperlanepc/h/\ln = passenger cars per hour per lanepc/h/ln=passengercarsperhourperlane

A≤11A \le11A≤11

FFS=BFFS−fLW−fRLC−3.22TRD0.84FFS = BFFS - fLW - fRLC - 3.22 TRD0.84FFS=BFFS−fLW−fRLC−3.22TRD0.84

FFS=freeflowspeedofbasicfreewaysegment(mph)FFS = free flow speed of basic freeway segment (mph)FFS=freeflowspeedofbasicfreewaysegment(mph)

BFFS=basefreeflowspeedofbasicfreewaysegment(mph);defaultis75.4mphBFFS = base free flow speed of basic freeway segment (mph); default is 75.4 mphBFFS=basefreeflowspeedofbasicfreewaysegment(mph);defaultis75.4mph

fLW=adjustmentforlanewidth(mph)fLW = adjustment for lane width (mph)fLW=adjustmentforlanewidth(mph)

fRLC=adjustmentforright−sidelateralclearance(mph)fRLC = adjustment for right-side lateral clearance (mph)fRLC=adjustmentforright−sidelateralclearance(mph)

TRD=totalrampdensity(ramps/mi)TRD = total ramp density (ramps/mi)TRD=totalrampdensity(ramps/mi)

≥120.0\ge12 0.0≥120.0

≥11−121.9\ge11 - 12 1.9≥11−121.9

≥10−116.6\ge10 - 11 6.6≥10−116.6

Clearance(ft)234≥5Clearance (ft) 2 3 4 \ge_{5}Clearance(ft)234≥5​

≥60.00.00.00.0\ge_{6} 0.0 0.0 0.0 0.0≥6​0.00.00.00.0

vp = PHF # N # fNeeds technical review

vp=demandflowrateunderequivalentbaseconditions(pc/h/ln⁡)vp = demand flowrate under equivalent base conditions (pc/h/\ln )vp=demandflowrateunderequivalentbaseconditions(pc/h/ln)

V=demandvolumeunderprevailingconditions(veh/h)V = demand volume under prevailing conditions (veh/h)V=demandvolumeunderprevailingconditions(veh/h)

PHF=peak−hourfactorPHF = peak-hour factorPHF=peak−hourfactor

N=numberoflanesinanalysisdirectionN = number of lanes in analysis directionN=numberoflanesinanalysisdirection

fHV=adjustmentfactorforpresenceofheavyvehiclesintrafficstream,calculatedwithfHV = adjustment factor for presence of heavy vehicles in traffic stream, calculated withfHV=adjustmentfactorforpresenceofheavyvehiclesintrafficstream,calculatedwith

fHV =Needs technical review

fHV=heavy−vehicleadjustmentfactorfHV = heavy-vehicle adjustment factorfHV=heavy−vehicleadjustmentfactor

PT=proportionofsingleunittrucksandtractortrailersintrafficstreamPT = proportion of single unit trucks and tractor trailers in traffic streamPT=proportionofsingleunittrucksandtractortrailersintrafficstream

ET=passenger−carequivalent(PCE)ofsingleunittruckortractortrailerintrafficstreamET = passenger-car equivalent (PCE) of single unit truck or tractor trailer in traffic streamET=passenger−carequivalent(PCE)ofsingleunittruckortractortrailerintrafficstream

D=SD= SD=S

D=density(pc/mi/ln⁡)D = density(pc/mi/\ln )D=density(pc/mi/ln)

vp=demandflowrate(pc/h/ln⁡)vp = demand flow rate (pc/h/\ln )vp=demandflowrate(pc/h/ln)

S=meanspeedoftrafficstreamunderbaseconditions(mph)S = mean speed of traffic stream under base conditions (mph)S=meanspeedoftrafficstreamunderbaseconditions(mph)

Worked exam-style examples