Welcome Guest
HeliTorque
  
User Control Panel

Security Code: : Security Code
Type Security Code Here: :
 
Register Here
Lost Password?

Online Stats:
Visitors: 39
Members: 0
Total: 39

Membership:
New Today: 0
New Yesterday: 0
Registering: 0
Members: 6662
Latest: chrisw

Most Ever Online
Visitors: 447
Members: 10
Total: 457


HeliTorque :: View topic - VOR distance ?
Forum FAQ
Forum FAQ
Search
Search
Memberlist
Memberlist
Usergroups
Usergroups
Profile
Profile
Contact Manager
Contact Manager
Log in
Log in
Log in to check your private messages
Log in to check your private messages
HeliTorque Forum Index » Ground School

Post new topic   Reply to topic All times are GMT
VOR distance ?
View previous topic :: View next topic  
Author Message
paddywak
H Addict
H Addict


Offline
Joined: Mar 28, 2009
Posts: 537
Location: UK


uk.gif

PostPosted: Mon Dec 12, 2011 10:58 am    Post subject: VOR distance ? Reply with quote

If an airway 10nm wide is defined by 2 VORs with a bearing accuaracy of + or - 5.5 degrees. To ensure accurate track guidance within the limitations of the airway, what is the maximum distance apart of the transmitters?

Anyone else get 109nm? help
Back to top
View user's profile Send private message
flip2
High Flying 'Torquer
High Flying 'Torquer


Offline
Joined: Sep 05, 2009
Posts: 225



PostPosted: Mon Dec 12, 2011 5:52 pm    Post subject: Reply with quote

109nm is what I also get by using the 1 in 60 approximation.

If that is not the correct answer, I can only presume they want you to use trigonometry for a more accurate answer:

Tan (angle) = opposite / adjacent
Tan (5.5) = 5nm / ?
? = 5nm / Tan (5.5)
? = 52nm

52nm x 2 = 104nm... Is that closer to the "correct" answer? How far spaced are the options?
Back to top
View user's profile Send private message
paddywak
H Addict
H Addict


Offline
Joined: Mar 28, 2009
Posts: 537
Location: UK


uk.gif

PostPosted: Mon Dec 12, 2011 7:47 pm    Post subject: Reply with quote

The answer given was 105nm. But there is no explanation or suggestion of trigonometry so I used my CRP5 and came up with 109nm.
Oh well at least I'm not going crackers, "yet"


I did find another strange question, but probably a typo:


To acheive a 3 percent gradient at 80kt your rate of climb must be?

My answer 3 x 80 = 240 so 240 feet per minute

Their answer was 270 feet/ min Confused
Back to top
View user's profile Send private message
flip2
High Flying 'Torquer
High Flying 'Torquer


Offline
Joined: Sep 05, 2009
Posts: 225



PostPosted: Wed Dec 14, 2011 11:04 am    Post subject: Reply with quote

Again, I would say you are correct.

The approximation rule is gradient x speed, which as you said gives the answer as 240 fpm.

If you work it out more accurately:
First convert 80 knots into fpm:
80 x 6080 ft = 486400 feet per hour
486400 / 60 = 8107 feet per minute

A 3% gradient is a 1 : 33.3 therefore:
8107 / 33.333 = 243 fpm
Back to top
View user's profile Send private message
Display posts from previous:   
Post new topic   Reply to topic    HeliTorque Forum Index » Ground School All times are GMT

 
Page 1 of 1

 
Jump to:  
You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot vote in polls in this forum


Sponsors


Billund Air Center

Visit HeliTorque!