radio aids questions

[AVIATROZ]

Q1. the approximate range of a 10KW ndb over the sea is:

ans-100nm
{but i think ans should be 3* under root of 100=3*316=948nm.
plz some one xpalin the answer}

Q2.An a/c is over flying a VOR at 30,000feet at a grd speed of 300kt.The maximum time during which no usable signal will be received{in mins and sec} is:
a.0.50
b.4.40
c.2.25- ans
d.1.40

plz expalin which method or formula is used.....i have no idea regarding this question.

Q3. using a 5 dot CDI, how many dots would show for an a/c on the edge at 100nm for VOR beacon

ans- 1.5

plz expalin which method or formula is used.....i have no idea regarding this question.

Q4.one degree at 200 nm represents a width of:
ans-3.5 nm
plz expalin which method or formula is used.....i have no idea regarding this question.

[Aviator09]

1. Range of NDB over sea is 3 sq.root of power in watt
power=10 KW=1000W
3 Sq.root 1000= 94.84 Nm (Close to 100 nm)

2. VOR beacon has a elevation angle in which the signal coverage is readily available and ICAO prescribes this to be a tangent of 50 degree.
Refer to ATPL oxford Radio Aids for the diagram which explains it more clearly.

Tan 50 =P/B
B=30000 ft=4.93 Nm
P=?
Put in the values and find P, which comes out to be 5.87 nm.
Now, double this as aircraft flies over the entire range( Diagram will explain you all) So, now this is about 11.75 nm.

Time=D/S i.e. 11.75/300= 0.03 Hr= 2.35 Mins

3. An airway has a 5nm boundary on either side, theta= 5/100 *60 (basic 1 in 60 rule) = 3 degree
On a 5 dot indicator, each dot is 2 degree,max defelction being 10degree, hence 1.5 dot

4.Same 1 in 60 rule applies here, theta=1 degree and R=200 nm S=? 1= s/200*60 = 3.33 NM

[gAMbl3]

10 kW = 10000 W

[cyrilroy21]

A1

Range of NDB over water = 3 X Square root ( Transmission power in watts )

1 KW = 1000 watts
10 KW = 10,000 watts

Range = 3 X Square root ( 10,000 )
= 3 X 100
= 300 nm

Not sure how the answer is 100nm
The closest answer would be 94 nm for 1 KW NDB ( 3 X SQ root {1000} )

---------------------------------------------------------------------------------------

A2

The time during which no usable signal will be received is the time flown over the cone of confusion
For this we have to calculate the cone of confusion at 30,000 feet

The cone of confusion for a VOR is 50 degrees to either side (left and right ) of a VOR

30,000 feet is the vertical distance . We can convert this into nm by dividing by 6080
30,000/6080 = 4.934nm = Vertical Distance from the VOR

Tan ( 50 ) = Horizontal distance / Vertical distance

horizontal distance = tan50 X Vertical distance
= tan50 X 4.934
= 5.88nm

Now this is the radius of the cone of confusion . To get the diameter we multiply by 2
5.88 X 2 = 11.76

So when the aircraft travels through the diameter of the cone of confusion no signal be received .

Now we have to calculate the time taken to fly 11.76 nm at a GS of 300 knots
Time = 11.76 / 300 = 2 minutes and 21 seconds

The closest answer is 2.25 mintues


---------------------------------------------------------------------------------------


A3

For this question i dont think you seem to have the whole question with you
The question actually is what is the Deviation shown on a 5 dot CDI indicator when the aircraft in on the edge of an airway (10nm wide )and 100nm from a VOR

If an airway is 10nm that means the aircraft is 5nm from the centreline on the airway ( since its on the edge )

Now we use the 1:60 rule to calculate the angle subtended from the VOR

Angle = ( S/R ) X 60
= ( 5 / 100 ) X 60
= 3 degree

A CDI show a max of 10 degrees
So on a 5 dot indicator , 1 dot = 10/5 = 2 degrees

So how many dots = 3 degree

The answer is 3/2 = 1.5 dots

---------------------------------------------------------------------------------------


A4

This questions is simple . Just use the 1:60 rule

Angle = ( S / R ) X 60

( Angle X R ) / 60 = S
( 1 X 200 ) / 60

Therefore S = 3.33 nm

The closest answer is 3.5nm

---------------------------------------------------------------------------------------

I am not sure about the first answer

The 2nd answer is much easier to understand if you draw them first

I hope you understood the methods explained

[PP351]

the answer should be 1 min 40 sec option d on 2nd question

[Aviator09]

cyrilroy21

Regarding the first question,I made a silly mistake of converting KW to Watt. It has to be 300 NM.

[Panama Jack]

What's an NDB? Is this an interview question for a First Officer position on a DC-3?

Do some people actually run these type of calculations whilst in the cockpit???

[AVIATROZ]

thanks everyone

thanks alot everyone for there inputs..............special thanks to cyrilroy and aviator09....

thanks friends

[AVIATROZ]

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

Q2. 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:

ans162nm

Q3. The accuracy of DME at 100 nm slant range is within:
a.3nm-- ans
b.4nm
c.2nm
d.1nm
{ i think ans shud be 1 nm as accuracy of dme is 1.25% of range.
so (1.25/100)*100=1.25}

plz shed some light.

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

{my ans does`nt match.........i did like..........i used 1 in 60 rule...
i.e. {(3degrees *4.2)/60}*6080=1276.8.................}
i think i m doing it in wrong way plz correct...thanks

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}

plz ILS hed ILS ome light on this

[DJ Flyboy]

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.

[AVIATROZ]

Q3. The accuracy of DME at 100 nm slant range is within:
a.3nm-- ans
b.4nm
c.2nm
d.1nm
{ i think ans shud be 1 nm as accuracy of dme is 1.25% of range.
so (1.25/100)*100=1.25}

plz shed some light.

[DJ Flyboy]

i think ans shud be 1 nm as accuracy of dme is 1.25% of range.

Its actually 0.25nm plus 1.25% of the distance measured. so that makes it 1.5nm plus or minus. Still dont know how 3nm is the answer. Are you sure you have the complete question?

[AVIATROZ]

LAsT set of radio aids doubts

Q1.What is the minimum level tht an a/c,at a range of113nm,must fly in order to contact the tower on R/T for a VDF bearing frm an airport sited 169ft above MSL
a.FL50
b.FL60-----ANs

Q2.An a/c travelling at330 m/sec transmit a signal at 10GHZ to a stationary receiver.if the a/c is flying directly towards the receiver and they are approximately at the same height the received frequency will be
a. 11MHz
b. 10.000011GHz

Q3.At 1000Z an a/c is overhead NDB AB enroute to NDB BC,track 075{M},hdg 082{M} at1029Z NDB AB bears 176 relative and NDB BC bears 353 relative.the hdg to steer at 1029Z to reach NDB BC is:
079{M}---ans

Q4. An a/c is mantaining track outbound from an NDB with a constant relative bearing of 184degree. to return to the NDB the relative bearing to maintain is:
356degree-------ans
{shud`nt the ans be 004degree......plz correct my concept.......i think i m making some mistake in understanding the question somewhere...}

thanks

[Shabez]

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}

plz ILS hed ILS ome light on this




it is from the threshold so you have to add 50 feet ht above threshold to get ht above touchdown 1280+50 = 1330 answer is b

[elitepilot20]

Q3. The accuracy of DME at 100 nm slant range is within:
a.3nm-- ans
b.4nm
c.2nm
d.1nm
{ i think ans shud be 1 nm as accuracy of dme is 1.25% of range.
so (1.25/100)*100=1.25}

according to gsp its 3% but in oxf. its 1.25% given according to my understanding 1.25 should be used in exams. unless there is some trick :P

[AVIATROZ]

vor question

hello aviator mates need some help with a question from radio aids.
Q1.An aircraft is on the airway boundary range 100nm from a vor marking the airway centerline.Assuming that each dot equates to 2degree how many dots deviation will be shown on deviation indicator?
a. 3 dots
b. 2.5 dots
c. 2 dots
d. 1.5dots
thanks

[keith williams]

To answer this type of question we need to know the width of the airway and then apply the 1 in 60 rule. The JAR/EASA version of this questions specifies an airway width of 10 nm.

With 10 nm airway width when the aircraft is on the boundary of the airway it is 5 nm off the centerline.

The 1 in 60 rule states that for each degree of error the cross track error will be 1/60 of the range.

This can be stated as:


Cross Track Error = 1/60 x Error angle x Range

Rearranging this gives

Error angle = Cross Track Error x 60 / Range

Inserting the data provided in the question gives the following:

Error angle = 5 nm x 60 / 100 nm = 3 degrees

Using 2 degrees per dot 3 degree error produces a 1.5 dot indication.

Using a different airway width will of course give a different answer.

[AVIATROZ]

Doppler radar numericals

Hey guy i need some help with few doppler radar questions.a detail explanation would be great
Q1.
When an aircraft is flying directly away from a transmitter, transmitting on a frequency of 15 GHz, a frequency shift of 6.0 KHz is measured. The speed of the aircraft is:

a. 120 kt
b. 222 ms-1
c. 300 kt
d. 222 kt(ans)

Q2.
An aeroplane is flying at 486 kt directly towards a beacon transmitting on a frequency of 12 GHz. The change in frequency observed will be:

a. 10 MHz
b. 6.25 Hz
c. 10 KHz (ans)
d. 6.25 Khz

Q3.
A receiver is moving directly towards a transmitter at 900kph. The transmission frequency is 6 GHz. Calculate the frequency shift in kHz caused by the relative motion between the transmitter and receiver.
Ans-5khz

Q4.
A stationary transmitter is operating on a wavelength of 3cm. A receiver moving directly away from the transmitter measures a Doppler shift of 6kHz. Calculate the speed of the receiver away from the transmitter in m/sec and knots.
Ans-180 ms-1, 350 kt

Q5.
An 8800MHz transmitter is moving directly away from a receiver at 291 kt. Calculate:

a. the speed of relative motion in m/sec
b. the frequency shift at the receiver in Khz
c. the frequency received in MHz
Ans-
a. 150 ms-1

b. 4.4 KHz

c. 8799.9956 MHz

Thanks for the help guys