Q1. An a/c receives a reply pulse from a DME 1200micro sec after transsmission of the interrogation pulse.The DME has a fixed delay of 50 micro sec. the range of the a/c from the DME station is:
a.75nm
b.63nm
c.45nm
d.93nm- ans
Radio waves travel at 300000km/sec
The delay is 50 microsec therefore time is 1150 microsec.
Distance traveled by Radio wave (To and fro) in 1150 * 10 ^ (-6) sec = 345 km = 186 nm
one way distance = 93 nm.
The time interval b/w the transmission of a given DME interrogation pulse and the reception of the appropriate response pulse at the a/c is 2 milli sec. the slant range is
2 milli sec = 2 *10^ (-3) sec
Distance covered by radio wave in 2 ms = 600 km = 324.3 nm
one way distance = 162 nm
Q4.A/c is following the ILS glidepath of 3degrees at an airfeild where the outer mrker is 4.2nm from the ILS touchdown point. the a/c approach speed is 130kt. the height of the a/c at the outer marker shud be:
a.1150ft
b.1200ft
c.1310ft---ans
d.960ft
Now ROD = Glide Angle * Ground Speed * 100/60
=> ROD = 3 * 130* 100/60 = 650 ft/min
Time to cover 4.2nm @ 130kts = 2 min
=> A/C would decend 1300 feet ~ 1310.
Q5. At 5.25nm frm threshold an a/c on an ILS Approach has a display showing it to be 4dots low on a 3 degree glidepath, using an angle of 0.15degree per dot of glideslope deviation and the 1 in 60 rule calculate the height of the a/c from the touchdown.
a.1375ft
b.1325ft
c.1450ft
d.1280ft
{as a/c is low for 4dots on ILS ...... i.e. 0.15*4=0.6
therefore, 3-0.6=2.4degree}
(now, using 1 in 60 rule, ((2.4*5.26)/60)*6080=1279.2}
Dude even I'm getting 1280ft maybe they mean the GS indication is down 4 dots => present GS is 3.6 but then the answer is some 1900 ft ..does anyone else know how to solve this.