Tuuapache's ground school questions!
 

Hi,
Does anyone know how to calculate wind without the whistwheel?
I tried to do this question but I could not get figure out what to do with the wind.

An airplane along a track of 093 degrees, wind 020/20, cruises speed (TAS) 110 kt. Fuel-endurance 6 hours (4 hours 15 min plus reserve).

1. Find the radius of action
a.102NM
b.114NM
c.216NM
d.226NM

2. Find the time to turn to base
a. 1 hour 7 minutes
b. 2 hours 14 minutes
c. 1 hour 12 minutes
d. 2 hours 24 minutes

3. The heading to steer out is
a. 073 degrees
b. 083 degrees
c. 093 degrees
d. 103 degrees

4. The heading to steer home is
a. 273 degrees
b. 283 degrees
c. 293 degrees
d. 303 degrees

You will probably get the right answer with the formula for PSR
(point of safe return):

time to PSR= (HE)/(O+H)

where
H= groundspeed home
O= groundspeed outbound
E= endurance without reserves



You do need a wheel because the answers are close together.

If you don't have one with you use 7 knots wind component for the first question, but this doesn't give an accurate answer.

What factors affect last light?
 

Hi,
I came across this question and I am just wondering if anyone can give me an insight of what factors affect the time of last light?
Here is the question:

The time (UTC) of last light is affected by
a. latitude, date, cloud cover, reduced visibility, and terrain
b. longitude, date, cloud cover, reduced visibility, and terrain
c. latitude and date only
d. latitude, longitude, date, cloud cover, reduced visibility, and terrain

From what I read in the Navigation book, it says that the factors that affect the duration of daylight are the date and latitude, but I guess it also depends on the cloud cover, visibility, and terrain. My answer is A.
Any helps would appreciated.

Nope your books are correct its C.

But thats only for landing. Height also comes into play as well if you use the 5deg below horizon method.

What if the question asks for the duration of daylight.
Would the answer be: a. latitude, date, cloud cover, reduced visibility, and terrain or it is still C.
Thanks for the reply.

Hi

I agree with you..

Longitude is East West Movement... Which doesn't affect your EoD

Latitude is North South... closer/further away from the Sun (Earth Tilted on its axis.)
Just think.. Antarica gets 6 months of daylight and 6 months of darkness.. Opposite case at the other pole

For legal purposes - it must be lat, long and date. So, none of those answers are correct, but C is the least wrong and would be correct if the question was about endurance of daylight.

In reality, the actual availability of light is heavily dependent upon cloud cover and terrain.

I had a good example last month at EGTC - I was a late departure to EGTE, flying off 21. Taxiway A at EGTC has bu66er all lighting, so I had to ask for a follow-me vehicle from parking near the tower to the threshold and could only then see where I was going once I was right next to the runway lights (and yes, my landing/taxi light was serviceable!).

According to my GPS, whose algorithms I'm sure are correct, legal night started 1 minute after I took off! Fact is, that close to dusk and with an overcast, "real" night started about five minutes after dusk.

So, learn the right answer for the exam, but apply lots of commonsense in real flying.

G

Just want to sum up. The factors that affect duration of daylight and last light are date and latitude, is this correct?
Cheers

From my understanding, if the wind and the track is 30 degrees off, the crosswind is half, 45 degrees is 2/3, 60 degrees is 9/10, and 90 degrees is all crosswind.
Also, 10 degree is 1/6, 20 degree is 2/6, 30 degree is 3/6, 40 degree is 4/6, 50 degree is 5/6, and 60 degree is 6/6. Is this method accurate?

In this question, the wind is 73 off track then how do I solve this mathematically? The reason I want to know how to estimate crosswind by hand is because I am preparing for an airline selection process and I am sure whistwheel is not allowed in the exam room.
Thank you

Duration of daylight, yes. Time of last light, no - you need to add in longitude.

And in the real world, it's more complex than that.

G

Crosswind
 

The first you wrote, which is that if the wind is 30 less/more then the crosswind is half, and if it has 90 difference then it is the whole because it comes directly from the side. It is correct. You also said that If it is around a 60 difference, then it cannot be equal to the whole wind because it must be either less than the whole of the wind speed or more than the half, so the second calculation you mentioned is not correct as obvious.

The easiest way of calculating the crosswind is to use sin/cos stuff. For example;

Sin(the difference between the Wind direction and the aircraft's Direction) x Wind Speed = Crosswind

If the heading is 060
The wind comes from 030 with 10 kts
Sin30 x 10 = 5 kts is the crosswind

Likewise, Cosinus gives the Headwind.

Regards

What if the wind is 73 off, is there a way to do this using sin and cos?

If you really want to, you can draw the situation with the air and wind vector on scale, with a protractor. Than you can draw the ground vector and measure its angle and length, to get drift angle and groundspeed. This can also be calculated with the law of sinuses or cosinuses.

This is completely useless:} Don't do it. I cannot imagine that you would need to calculate this stuff for a selection process or something. Navigation is not an exact science.

You should be able to do this just by looking at the possible answers. Take drift. Will that be 10 or 20 degrees? Which way should be it be applied? For the range, it will be about 2 hours at 110 kts (less the headwind) but if you answer No. 2 first - you should be able to figure out No. 1. So if you are flying into a headwind, will you fly for more or less than half of your endurance? The only real question is here is by how much? Is is seven minutes (12 miles or so) or is it 14 minutes (30 miles)? The difference in the groundspeeds is about 20 knots - I'll go for 2 hours, 24 minutes.

As for headwinds and crosswinds, just remember 0.5, 0.7 and 0.9. The numbers 1 and zero or already in your brain. Use the numbers as sine or cosine, as appropriate for the intermediate angles of 30, 45 and 60 degrees. It's close enough for the real world.

So 70 degrees off track, as in this case, gives a crosswind of (0.9 x 20) 18 knots and a headwind/tailwind of (0.5 x 20) 10 kts.

I figured out number 3 and 4 by taking the TAS multiplies by wind speed and divide by 60.
Here is how I did it: 110*20/60 = 37

Then use the clock code to work out how much of that max drift actually applies. The clock code can be used by imagining a clock, taking the number of degrees the wind is off your nose and pretending these degrees are minutes on the clock face.

If the wind is 15 degrees off the nose look at the 15 minute mark on your clock face: 15 minutes is quarter of an hour so the drift you can expect is a quarter of the max drift you calculated before.

If the wind is 30 degrees off the nose, look at the 30 minute mark on your clock face: 30 minutes is half an hour so the drift you can expect is half the max drift you calculated before.

the wind is 73 degree off so it is 73 minutes on the clock and the maximum drift is (20*110)/60 = 36.6 or 37 degree.
73 minutes on the clock is about a quarter, so 37*1/4 = 9.25 or 9 degree. so the answers for number 3 is 93-9 = 84 and number 4 is 93+9+180 = 282.
But I am still having trouble with the first two questions.

Calculate the approximate time and compass heading during the climb
 

Could anyone guide me to the right direction with what to do with the wind? (no whizz wheel)
Here is the question:

An airplane departs an airport under the following condition:
Airport elevation 1,000 ft Cruise altitude 9,500 ft
Rate of climb 500 ft/min Average true airspeed 135 kts
True course 215 Average wind velocity 290/20 kts
Variation 3 W Deviation -2

Determine the approximate time and compass heading during the climb.
a. 17 minutes, 224
b. 17 minutes, 242
c. 14 minutes, 234
d. 14 minutes, 242

I can only find the time which is 17 minutes: 9500-1000 = 8500/500 = 17 minutes. As for the compass heading: C D M V T
216 -2 218 3W 215
My answer is not one of the given answers so what the heck did I do wrong?

VOR question
 

Given: HDG 085 M, OBS 255
TO/FROM TO, CDI 2 dots left
The aeroplane is on the
a. 071 radial
b. 075 radial
c. 255 radial
d. 251 radial
Could someone teach me how to solve this?
Thanks

Use this 3 step approach:
1. Pretend Heading=OBS (i.e heading = 255)
2. TO indication = You are flying towards the VOR (so on radial 255-180=75 if centered)
3. Fly LEFT = you are right of radial = lesser value.

Now to get 2 dots to a degree you need to know if its a 5 dot, 6 dot instrument
assuming 5 dot:
5 dots=10 degree
2 dots = 4 degree

So radial = 75-4 = 71..

For graphical, try setting it op on This:
Tim's Air Navigation Simulator

I just want to make sure that I have the right idea. The OBS is used to set the desired course or the radial, correct? In this question, the OBS is set to 255 radial to, therefore the radial from the station is 075 radial or it is the other way around? Does the heading 085M mean anything at all in this question? Do you know any websites I can practice this type of questions?
Thanks a lot.

For a fixed CDI indicator heading is ignored.
For an HSI heading does come into play.

For a fixed display, the 3 steps always work
1. Set heading = OBS
2. Flying towards the VOR => TO indication.
3. Need to fly left to get to radial => LEFT indication

For practise I like Tim's for the simplicity. For more graphical illustrations try luizmonteiro - Aviation Calculations Home

I have one more question to ask you. Could you please check to see if I am on the right track?
Given: Track required left 090M, HDG 102M
ADF (destination) 000R DME (destination) 21 NM
According to the "one in sixty rule" the aeroplane is
a. 12 NM right of required track
b. 4 NM right of required track
c. 12 NM left of required track
d. 4 NM left of required track

The difference between heading and track is 12 degree to the right so the drift is 12 degree left which means that the wind is coming from the right so the airplane is on the left of the track. Am I on the right track? According to the one in sixty rule, every mile off track along a sixty miles along track is equal to one degree of track error. Now what is the next step?
Cheers

Could someone kindly guide me to the right direction?
Thanks

So the correct answer is d. latitude, longitude, date, cloud cover, reduced visibility, and terrain?

What is the origin of the question?
Under whose syllabus are you being examined?

I ask because (for example) the JAA syllabus has no such term as 'time of last light' in the learning objectives. Does this term have a specific definition or is it just being used by the author to describe 'when it's too dark to see'.

BTW folks there is a big glaring RTFQ in here. (C) wouldn't even be correct for an 'end of twilight' definition because the question states UTC not LMT. So we need Lat, Long and Date for starters. Only (D) has all 3.

Problem is, discussions about multiple choice questions really don't work well on a bulletin board like Pprune. They're more for a tutor at a training provider.

There are a great many people here with a lot of experience, much of which just adds, not reduces, complexity in any given problem. Multiple choice exam questions, whilst inevitable, are often very simplistic.

Nobody on Pprune does simplistic very well!

G

Your last question is very unclear.

Maybe this helps:

The 1 in 60 rule in formula form is:

(1 degree) / 60 = (1 NM) / (60 NM)

Finding an ETA with 1 in 60
 

An aircraft obtains a pinpoint at "A", 2 NM left of flight path at 0515 UTC. At 0530 UTC, after maintaining a constant heading (and TAS) another pinpoint is obtained at "B", 3.5 NM right of the flight path and 26 NM. From "A", the alteration of heading necessary at "B" (to make good the FPT at a DR position "C", 38 NM from "B" and the ETA at that DR position and respectively.

I worked out the problem and I got 19 degree left but I could not figure out the ETA time. Does anyone know how to work this out?

Cheers

I agree with 18 left (close enough huh), but for the ETA I went with:

Time = 38 x 15 divided by 26 = 22 mins

So an extra 22 mins for the remaining 38 miles at same speed.

Added to 0530 = 0552 UTC.

Any thoughts?

Tuuapache

I echo Genghis's comments!

These are questions more suited for your tutor than a bulletin board because the spectrum of experience here really is very wide and the answers will vary in accuracy accordingly!

This thread will be locked in 24 hours!

Halfwayback

Convert GMT to Local time
 

Does anyone know how to convert GMT to local time?

GMT 0200 hrs, 30 June, local longitude 68 degrees 21 minutes, E. What is the local time? (conversion factor is +5)
What I did was I converted the arc to time. 68 degrees 21 minutes E is 4 HR 33 MIN then I got stuck. What am I supposed to do next?
Cheers

Are you serious? Just in case you are, do you know what 2+5=?

what about the longitude 68 degrees 21 minutes, E?
What does this have anything to do with the question?