'Time to station' question
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'Time to station' question
Hi everyone.
Hope you all are doing great.
Someone please help me with the following question:
Q: While maintaining a constant heading, a relative bearing of 15(deg) doubles in 6 mins. The time taken to the station being used is:
A: 3mins
B: 6mins
C: 12mins
According to my working:
Time to station = 60 x number of minutes flown / change in bearing
number of minutes flown: 6
change in bearing: 15 (because the RB doubles)
60x6/15 = 24mins. How come?
Hope you all are doing great.
Someone please help me with the following question:
Q: While maintaining a constant heading, a relative bearing of 15(deg) doubles in 6 mins. The time taken to the station being used is:
A: 3mins
B: 6mins
C: 12mins
According to my working:
Time to station = 60 x number of minutes flown / change in bearing
number of minutes flown: 6
change in bearing: 15 (because the RB doubles)
60x6/15 = 24mins. How come?
Join Date: Mar 2002
Location: Euroland
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Ah the old "double bow method"!!
Wish I could draw the answer but I hope you can draw what I explain and see the reason behind it.
At "A" the relative bearing of the Station "C" is 015
The aircraft flies on a constant heading until the relative bearing has doubled (030 in this case). Call this position "B"
Draw a line from A to B to C and back to A which makes a triangle.
The angle at "A" is 15 degrees.
By simple geometry, the angle at "C" is also 15 degrees.
If two angles of a triangle are equal then the sides opposite them are also equal.
The line A-B is opposite angle C which is 15 degrees
The line B - C is opposite angle A which is also 15 degrees.
Therefore A to B = B to C
So if it takes 6 minutes to fly from A to B then (assuming constant groundspeed) it will also take 6 minutes to fly from B to C (the station).
Far easier with a diagram but I hope that helps.
Regards,
DFC
Wish I could draw the answer but I hope you can draw what I explain and see the reason behind it.
At "A" the relative bearing of the Station "C" is 015
The aircraft flies on a constant heading until the relative bearing has doubled (030 in this case). Call this position "B"
Draw a line from A to B to C and back to A which makes a triangle.
The angle at "A" is 15 degrees.
By simple geometry, the angle at "C" is also 15 degrees.
If two angles of a triangle are equal then the sides opposite them are also equal.
The line A-B is opposite angle C which is 15 degrees
The line B - C is opposite angle A which is also 15 degrees.
Therefore A to B = B to C
So if it takes 6 minutes to fly from A to B then (assuming constant groundspeed) it will also take 6 minutes to fly from B to C (the station).
Far easier with a diagram but I hope that helps.
Regards,
DFC
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@wingless aviator
Your formula is correct if you put your bearing pointer on the 90 degrees index (abeam) and note the bearing change. In your example this formula doesn't apply.
I guess DFC's method is the best answer.
Regards.
Your formula is correct if you put your bearing pointer on the 90 degrees index (abeam) and note the bearing change. In your example this formula doesn't apply.
I guess DFC's method is the best answer.
Regards.
Join Date: Jun 2006
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I made a quick sketch of the problem.
By trigonometry default when B=C then b=c also
Edit: whoops! That turned out larger than expected. I hope I'm better at flying than computers.
Edit2: I see that with regards to one of the previous explanations A and B is swapped but you get the picture.
