Distance from VOR Formulas

[GogglesPisano]

Does anyone have the formulas to solve these:

1) Bearing pointer moves from 5 degrees in front of wing to 5 degrees behind in 8 minutes. Speed is 360kts. How far are you from station?

2) Wind is 360 at 50. What heading wll you fly to track inbound on the 270 radial?

3) You are on the 090 radial 20 DME. What heading would you fly to go direct to the 190 radial 60 DME fix?

Thanks

 

[AC560]

1) Bearing pointer moves from 5 degrees in front of wing to 5 degrees behind in 8 minutes. Speed is 360kts. How far are you from station?

Minutes to VOR = 480 seconds (time for change) / 10 (degree of change) = 48 mins
GS is 6 NM's a min = 288 NM to station.

Do I win a prize?

 

[Grey Ghost]

who cares????

 

[GogglesPisano]

who cares????

A certain major airline.

 

[gnvav8r]

2) Wind is 360 at 50. What heading wll you fly to track inbound on the 270 radial?

Depends on your TAS. Here's how to figure Crosswinds in your head:

Use the RULE OF SIXTY, which gives the number of nm an angle of 1 deg subtends at 60 nm... Well it should really be the RULE OF FIFTY-SEVEN. You see, a circle 60 nm in radius has a circumference of 2piR = 377 nm. Each degree along that circumference therefore is 377/360 = 1.05 nm. That's a 5% error. Only if pi equals the Biblical value of 3.0 does a degree subtend 1 nm at the circumference. Use 60 when doing it in your head, but when using a whiz wheel, use 57. Figger it out... a circle with 360 nm circumference has a radius of 57 nm. (using the real value of pi). Oddly enough there are FAA test questions on which you will be marked wrong if you use pi = 3.14159

However it's not that hard to divide by 57 in your head. GS/57 = GS/60 + 5%. And 5% is half of 10%. So 120/57 = 2 + 10% or 2 + 0.2. But add in only half of the 0.2. Final result is 2.1 Precisely.... however you often don't need that subtlety.

Back to the problem. Suppose you are trucking along at 150 kts. Alter your course by 1 deg and at the end of an hour you will be 150/57 = 2.6 nm off course. Winds are reported also in kts... so a direct cross wind of 2.6 kts will be nullified by a crab angle of 1 deg. Basically your speed in kts/57 tells you how many kts of crosswind is compensated for by each deg of crab angle. Shooting your approach speed at 140 kts with a 10 kt crosswind? 140/57 = 2.5; each degree of crab will handle 2.5 kts crosswind => need 4 deg of crab.

 

[GogglesPisano]

Thanks guys.

So crab angle required = x-wind component x 60/TAS (or 57/TAS to be more accurate.)

 

[TigerDriver]

thats all well and good for gee whiz interview stuff but I for prefer to fly whatever heading it takes to keep the course centered. I'll do calculus and differential equations on the ground.