Homework 5 and 6
Name: Marcos López D. AVM3202 Section 1
Homework 5: Design a High Speed Exit Taxiway by
calculating:
a. Distance of the taxiway exit from the threshold (Se)
b. Required length (d2) of the high speed exit
c. Minimum separation (X) between the runway and
parallel taxiway
d. Time spent decelerating on the runway
Given design aircraft is B747-400 with
Touchdown speed = 170 miles per hour
Runway exit speed = 41 miles per hour
Taxiway exit speed = 13 miles per hour
𝑇𝑜𝑢𝑐ℎ𝑑𝑜𝑤𝑛 𝑠𝑝𝑒𝑒𝑑: 170 𝑚𝑝ℎ ∗ 1.467 𝑓𝑡/𝑠 = 249.39 𝑓𝑡/𝑠
𝑅𝑢𝑛𝑤𝑎𝑦 𝑒𝑥𝑖𝑡 𝑠𝑝𝑒𝑒𝑑: 41 𝑚𝑝ℎ ∗ 1.467𝑓𝑡/𝑠 = 60.15 𝑓𝑡/𝑠
𝑇𝑎𝑥𝑖𝑤𝑎𝑦 𝑒𝑥𝑖𝑡 𝑠𝑝𝑒𝑒𝑑: 13 𝑚𝑝ℎ ∗ 1.467 𝑓𝑡/𝑠 = 19.07 𝑓𝑡/𝑠
Assumptions
𝑎 = −5 𝑓𝑡/𝑠 2 𝑜𝑛 𝑡ℎ𝑒 𝑟𝑢𝑛𝑤𝑎𝑦
𝑎 = −3.3 𝑓𝑡/𝑠 2 𝑜𝑛 𝑡ℎ𝑒 𝑡𝑎𝑥𝑖𝑤𝑎𝑦
1500 ft from the approach threshold
a) 𝑑1 = 𝑑 = (𝑉𝑓𝑖𝑛𝑎𝑙 2 − 𝑉𝑖𝑛𝑖𝑡𝑖𝑎𝑙 2 )/2𝑎 𝑑2 = 493 𝑓𝑡
𝑑 = (60.152 − 249.392 )/2(−5)
𝑑1 = 5 858 𝑓𝑡 c) High Speed Taxiway corresponds to a
30o angle
𝑆𝑒 = 1 500 𝑓𝑡 + 5 858𝑓𝑡 𝑋 = 𝑑2 𝑆𝑖𝑛∅
𝑆𝑒 = 7 358 𝑓𝑡 𝑋 = 493 (𝑆𝑖𝑛 30)
𝑋 = 246.5 𝑓𝑡
b) 𝑑2 = 𝑑 = (𝑉𝑓𝑖𝑛𝑎𝑙 2 − 𝑉𝑖𝑛𝑖𝑡𝑖𝑎𝑙 2 )/2𝑎 𝑋 = 247 𝑓𝑡
𝑑 = (19.072 − 60.152 )/2(−3.3)
d) 𝑡 = (𝑉𝑓𝑖𝑛𝑎𝑙 − 𝑉𝑖𝑛𝑖𝑡𝑖𝑎𝑙)/𝑎
𝑡 = (60.15 − 249.39)/−5
𝑡 = 37.8 𝑠
𝑡 = 38 𝑠
Name: Marcos López D. AVM3202 Section 1
Homework 5: Design a High Speed Exit Taxiway by
calculating:
a. Distance of the taxiway exit from the threshold (Se)
b. Required length (d2) of the high speed exit
c. Minimum separation (X) between the runway and
parallel taxiway
d. Time spent decelerating on the runway
Given design aircraft is B747-400 with
Touchdown speed = 170 miles per hour
Runway exit speed = 41 miles per hour
Taxiway exit speed = 13 miles per hour
𝑇𝑜𝑢𝑐ℎ𝑑𝑜𝑤𝑛 𝑠𝑝𝑒𝑒𝑑: 170 𝑚𝑝ℎ ∗ 1.467 𝑓𝑡/𝑠 = 249.39 𝑓𝑡/𝑠
𝑅𝑢𝑛𝑤𝑎𝑦 𝑒𝑥𝑖𝑡 𝑠𝑝𝑒𝑒𝑑: 41 𝑚𝑝ℎ ∗ 1.467𝑓𝑡/𝑠 = 60.15 𝑓𝑡/𝑠
𝑇𝑎𝑥𝑖𝑤𝑎𝑦 𝑒𝑥𝑖𝑡 𝑠𝑝𝑒𝑒𝑑: 13 𝑚𝑝ℎ ∗ 1.467 𝑓𝑡/𝑠 = 19.07 𝑓𝑡/𝑠
Assumptions
𝑎 = −5 𝑓𝑡/𝑠 2 𝑜𝑛 𝑡ℎ𝑒 𝑟𝑢𝑛𝑤𝑎𝑦
𝑎 = −3.3 𝑓𝑡/𝑠 2 𝑜𝑛 𝑡ℎ𝑒 𝑡𝑎𝑥𝑖𝑤𝑎𝑦
1500 ft from the approach threshold
a) 𝑑1 = 𝑑 = (𝑉𝑓𝑖𝑛𝑎𝑙 2 − 𝑉𝑖𝑛𝑖𝑡𝑖𝑎𝑙 2 )/2𝑎 𝑑2 = 493 𝑓𝑡
𝑑 = (60.152 − 249.392 )/2(−5)
𝑑1 = 5 858 𝑓𝑡 c) High Speed Taxiway corresponds to a
30o angle
𝑆𝑒 = 1 500 𝑓𝑡 + 5 858𝑓𝑡 𝑋 = 𝑑2 𝑆𝑖𝑛∅
𝑆𝑒 = 7 358 𝑓𝑡 𝑋 = 493 (𝑆𝑖𝑛 30)
𝑋 = 246.5 𝑓𝑡
b) 𝑑2 = 𝑑 = (𝑉𝑓𝑖𝑛𝑎𝑙 2 − 𝑉𝑖𝑛𝑖𝑡𝑖𝑎𝑙 2 )/2𝑎 𝑋 = 247 𝑓𝑡
𝑑 = (19.072 − 60.152 )/2(−3.3)
d) 𝑡 = (𝑉𝑓𝑖𝑛𝑎𝑙 − 𝑉𝑖𝑛𝑖𝑡𝑖𝑎𝑙)/𝑎
𝑡 = (60.15 − 249.39)/−5
𝑡 = 37.8 𝑠
𝑡 = 38 𝑠
