CAE-Indigo recruitment
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CASS – CAE Aircrew Selection System
---------------------------
Selection Process
All pre-selected applicants will go through the following tests:
Aviation Knowledge Test
English Language Skills Test
CASS – CAE Aircrew Selection System
Group Exercise
Personal Interview
Dates and location shall be advised when you are invited for the selection process.
The cost for this selection process is USD 375 and is non refundable. Extra costs (travel, housing, etc.) will be payable by the trainee.
https://pilot.cae.com/Programs/Indigo.aspx
---------------------------
Selection Process
All pre-selected applicants will go through the following tests:
Aviation Knowledge Test
English Language Skills Test
CASS – CAE Aircrew Selection System
Group Exercise
Personal Interview
Dates and location shall be advised when you are invited for the selection process.
The cost for this selection process is USD 375 and is non refundable. Extra costs (travel, housing, etc.) will be payable by the trainee.
https://pilot.cae.com/Programs/Indigo.aspx
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@Toucan9278 Thanks 4 the questions
is that a complete question?
Anybody know any CRM questions or resources to read
4 Flying @ FL 22000 decent to 4000 Ft. G/S 150 kts. what will be the ROD?
Anybody know any CRM questions or resources to read
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The question was as follows
TAS - 197 kts
Course - 240
Wind - 180/30
Descending from FL 200 to FL 40
Horizontal Distance travelled - 39 NMs
What is the ROD?
TAS - 197 kts
Course - 240
Wind - 180/30
Descending from FL 200 to FL 40
Horizontal Distance travelled - 39 NMs
What is the ROD?
Now find tan Ө .
tan Ө = perpendicular/base ---> tan Ө = 16000/237120
You get tan Ө = 0.06747
Therefore, Ө = 3.86 degrees
Then apply that to the formula,
Rate of descent = Groundspeed x Glide slope angle x 100/60
You will get the ROD as 1158 feet per minute.
(I hope this is correct. If not, kindly correct me but please be polite)
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well i m getting the answer as 1230 fpm
the way i calculated is that i first found out ground speed by using tas, track and wind dir and speed via flight calculator.
i got gs as 180kts
then took hori distance and divided it by speed to get time which i got as 13 mins
then i divided 16000 feet by 13 mins and i got the answer as 1230 fpm
well is this method and answer correct????
the way i calculated is that i first found out ground speed by using tas, track and wind dir and speed via flight calculator.
i got gs as 180kts
then took hori distance and divided it by speed to get time which i got as 13 mins
then i divided 16000 feet by 13 mins and i got the answer as 1230 fpm
well is this method and answer correct????
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My Solution
TAS=197Kts
Course---240
Wind---180/30
Using this data on the flight computer, GS=182Kts
Using the GS and D=39NM
Time=12.8mins
Vertical Distance flown in 12.8mins=16000ft
RoD=16000/12.8
RoD=1250fpm
@ktmb-ktmb
hey, good attempt buddy. Will always b polite
I DOn't trust that formula because the G.S. is in Kts and RoD comes out in FpM. Did u get ur GS 180 or 182 Kts?
TAS=197Kts
Course---240
Wind---180/30
Using this data on the flight computer, GS=182Kts
Using the GS and D=39NM
Time=12.8mins
Vertical Distance flown in 12.8mins=16000ft
RoD=16000/12.8
RoD=1250fpm
(I hope this is correct. If not, kindly correct me but please be polite)
hey, good attempt buddy. Will always b polite
I DOn't trust that formula because the G.S. is in Kts and RoD comes out in FpM. Did u get ur GS 180 or 182 Kts?
Rate of descent = Groundspeed x Glide slope angle x 100/60
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@aditya
thats same like me
however i took all decimals and multiplied so for me time came to say 12.9 so i rounded of to 13
and btw the exact formula wud be gs x slope angle x 6080 /3600
well this wat i have been learning and if calculated then answers mismatch by very less difference
but just wanted to confirm the original formula
thats same like me
however i took all decimals and multiplied so for me time came to say 12.9 so i rounded of to 13
and btw the exact formula wud be gs x slope angle x 6080 /3600
well this wat i have been learning and if calculated then answers mismatch by very less difference
but just wanted to confirm the original formula
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My Solution
TAS=197Kts
Course---240
Wind---180/30
Using this data on the flight computer, GS=182Kts
Using the GS and D=39NM
Time=12.8mins
Vertical Distance flown in 12.8mins=16000ft
RoD=16000/12.8
RoD=1250fpm
Quote
I hope this is correct. If not, kindly correct me but please be polite)
@ktmb-ktmb
hey, good attempt buddy. Will always b polite
I DOn't trust that formula because the G.S. is in Kts and RoD comes out in FpM. Did u get ur GS 180 or 182 Kts?
Quote:Rate of descent = Groundspeed x Glide slope angle x 100/60
TAS=197Kts
Course---240
Wind---180/30
Using this data on the flight computer, GS=182Kts
Using the GS and D=39NM
Time=12.8mins
Vertical Distance flown in 12.8mins=16000ft
RoD=16000/12.8
RoD=1250fpm
Quote
I hope this is correct. If not, kindly correct me but please be polite) @ktmb-ktmb
hey, good attempt buddy. Will always b polite
I DOn't trust that formula because the G.S. is in Kts and RoD comes out in FpM. Did u get ur GS 180 or 182 Kts?
Quote:Rate of descent = Groundspeed x Glide slope angle x 100/60
I got my groundspeed as 180 (by CX-2)
I had to mention the "polite" part coz I have seen people here getting just way out of control for just a simple comment.
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I had to mention the "polite" part coz I have seen people here getting just way out of control for just a simple comment.
@ktmb-ktmb @avalanche007
I don't trust that formula because its an approximation. Most people use it and it should give u the same answer. ktmb, my solution could be wrong.
I got GS of 182Kts from manual flight computer and verified with vector-trigonometry method. U got 180Kts from CX2.
bombayhues tells me that he got 1450fpm, he used a descend from 22000ft.
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The descent started from FL 220 and not FL 200 as mentioned earlier.
Hence the total vertical distance travelled is 18000 feet.
TAS - 197
Wind - 240/30 (Hence headwind = 30 cosine 60 = 15)
Ground Speed = 197 - 15 = 182 kts
39 NMs travelled at a speed of 182 kts will take 12.85 mins
Hence the ROD will be 18000/12.85 = 1400 ft/min
I have no clue why we get a different answer using the CX2 or the ROD formula. But the above is the way I solved it
Hence the total vertical distance travelled is 18000 feet.
TAS - 197
Wind - 240/30 (Hence headwind = 30 cosine 60 = 15)
Ground Speed = 197 - 15 = 182 kts
39 NMs travelled at a speed of 182 kts will take 12.85 mins
Hence the ROD will be 18000/12.85 = 1400 ft/min
I have no clue why we get a different answer using the CX2 or the ROD formula. But the above is the way I solved it
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@adiyta
hi, well its not about trusting
our method is correct but still someone came up with rod formula which was slightly different than mine so i wanna inquire which is actually correct:
hi, well its not about trusting
our method is correct but still someone came up with rod formula which was slightly different than mine so i wanna inquire which is actually correct:
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(I hope this is correct. If not, kindly correct me but please be polite)
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@jimmygill apologies unexpected but still accepted. u were just being urself.
@avalanche @bombayhues
only me and hues got 182 kts GS.
the formula can b used if u are comfortable with cramming .
Anyways shall we move onto discussing another question?
Any CRM question or resources.
@avalanche @bombayhues
only me and hues got 182 kts GS.
I don't trust that formula because its an approximation. Most people use it and it should give u the same answer.
Anyways shall we move onto discussing another question?
Any CRM question or resources.
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cae indigo scam
You may try all you want, you may answer all questions here. Are you aware how we all are being taken for a royal ride by cae indigo combine.
What has been the number of pilots passing each time? Only 4 out of 40 to 50. now now much have we paid? - $375 each which translates into approx Rs.17,5000/- per pilot. So they charge ALL of us appearing this amount and pass a miniscule number.
Question – how do only 4 guys get above 70% each time they hold the exams?
A simple calculation tells us that they collect Rs.8,75,000/- (17500*50) each time they hold an exam. Only 17500*4=70000 is used and rest of amount (over 8,00,000) is just free income for them – and this is for each day they conduct an exam. So if they do it twice a month – it is a neat SIXTEEN LAKH RUPEES. Now if this is not a racket than what is?
While you all can keep studying, you know that you will not be able to get in. getting in is another racket based on how much money you can pay up. The going rate is 5 to 8 lakhs. Now sixteen lakh is official income by making suckers out of us, rest is private income on the side. What a nice racket!!!
What are we all doing studying? Lets start a job fair for pilots and charge each one any amount. Considering the number in the market and all of us ready to part with our money easily, we wouldn’t ever need to work. Go air wants 10000 rupees with our application – for what?
Can DGCA or someone do anything about this recruitment scam going on? Guess no as they would all be hand in glove in it.
Out of sixteen lakh how much is cae raking it in and how much is indigo making out of it? Can anyone answer this question?
What has been the number of pilots passing each time? Only 4 out of 40 to 50. now now much have we paid? - $375 each which translates into approx Rs.17,5000/- per pilot. So they charge ALL of us appearing this amount and pass a miniscule number.
Question – how do only 4 guys get above 70% each time they hold the exams?
A simple calculation tells us that they collect Rs.8,75,000/- (17500*50) each time they hold an exam. Only 17500*4=70000 is used and rest of amount (over 8,00,000) is just free income for them – and this is for each day they conduct an exam. So if they do it twice a month – it is a neat SIXTEEN LAKH RUPEES. Now if this is not a racket than what is?
While you all can keep studying, you know that you will not be able to get in. getting in is another racket based on how much money you can pay up. The going rate is 5 to 8 lakhs. Now sixteen lakh is official income by making suckers out of us, rest is private income on the side. What a nice racket!!!
What are we all doing studying? Lets start a job fair for pilots and charge each one any amount. Considering the number in the market and all of us ready to part with our money easily, we wouldn’t ever need to work. Go air wants 10000 rupees with our application – for what?
Can DGCA or someone do anything about this recruitment scam going on? Guess no as they would all be hand in glove in it.
Out of sixteen lakh how much is cae raking it in and how much is indigo making out of it? Can anyone answer this question?
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@aditya104
@bombayhues
@avalanche007
@jimmygill
I tried to solve the problem using 3 methods.
(i) Trigonometry
(ii) 1: 60 Rule
(iii) Unitary Method
And guess what, all 3 methods give a different answer.
@bombayhues
@avalanche007
@jimmygill
I tried to solve the problem using 3 methods.
(i) Trigonometry
(ii) 1: 60 Rule
(iii) Unitary Method
And guess what, all 3 methods give a different answer.
Last edited by ktmb-ktmb; 13th Jul 2010 at 07:11.
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@ktmb
Dude, though I do not see any fault with the trig solution of yours, I cant help but feel that you can apply basic trigonometry to solve this one (why? I dont know).
As for the other two solutions. If you use 6080/3600 (instead of 100/60) in the 1:60 rule method then you will get the same answer for both the 1:60 rule method as well as the unitary method, which is 1230 ft/min.
Dude, though I do not see any fault with the trig solution of yours, I cant help but feel that you can apply basic trigonometry to solve this one (why? I dont know).
As for the other two solutions. If you use 6080/3600 (instead of 100/60) in the 1:60 rule method then you will get the same answer for both the 1:60 rule method as well as the unitary method, which is 1230 ft/min.
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@ktmb
Well..good job first of all!
but since this is a thread specifically for Cae-IndiGo process... its a must to know that the answers mentioned there would be an approximation, so you may use any method, the moment you look at the choices the right answer will be clear!
@everyone else... I would suggest go thru CRM and Human Factors too, as everything else you have anyways covered for your DGCA exams!
Happy Landings!
P.S. anybody knows when the next exam is goin to take place?
but since this is a thread specifically for Cae-IndiGo process... its a must to know that the answers mentioned there would be an approximation, so you may use any method, the moment you look at the choices the right answer will be clear!
@everyone else... I would suggest go thru CRM and Human Factors too, as everything else you have anyways covered for your DGCA exams!
Happy Landings!
P.S. anybody knows when the next exam is goin to take place?
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i) ROD / GS = tan(theta) = 16000/(39*6080),
i.e. ROD = GS*16000/(39*6080) = 1231 fpm
calculating theta was unnecessary here
ii) 1:60 rule is an approximation, actually it ought to be 1:57.3, another approximation used is
for small theta
tan(theta) = theta (radian)
one overestimates other underestimates so we have a fairly reasonable answer.
This approximation is used in calculating theta, and calculating theta is not necessary.
iii) Unitary method is by far the most accurate method you have used... and it is same as using good trigonometry without approximation hence same answer 1231 fpm..
i.e. ROD = GS*16000/(39*6080) = 1231 fpm
calculating theta was unnecessary here
ii) 1:60 rule is an approximation, actually it ought to be 1:57.3, another approximation used is
for small theta
tan(theta) = theta (radian)
one overestimates other underestimates so we have a fairly reasonable answer.
This approximation is used in calculating theta, and calculating theta is not necessary.
iii) Unitary method is by far the most accurate method you have used... and it is same as using good trigonometry without approximation hence same answer 1231 fpm..