AngelFuree
Well-Known Member
Thank you all for the replies!
Upon reading that section of the AIM, it seems pretty clear to me that they are stating a substitution of distance for time with regards to holding and entries, so I'll go ahead and agree w/ tgrayson with his interpretation. However, I do agree with the others in saying that flying 4nm during a teardrop procedure might take you too far away making it unnecessary to fly out so far. I completely agree with that.
The description given by the AIM may very well be poorly written, but I do understand from it that the substitution of distances applies to both entries and holding.
To take this a step further, we can use simple trigonometry to see how far each procedure would take us.
So let's say you decide to...
...fly a teardrop entry with a 1 minute outbound at 100 kts. The distance covered in 1 minute is 1.6NM. If we use simple trigonometry, we can find out how far this actually takes us away from the inbound course. Given the 30 degree teardrop entry, our 1.6NM is the hypotenuse and all we need to find is our opposite side. sin(30)*1.6 = 0.8NM. That is, the one minute outbound teardrop entry will have taken us 0.8NM.
....fly a teardrop entry with a 4NM outbound leg, we can find how far, again. sin(30)*4=2NM. That is, the 4NM outbound teardrop entry will have taken us 2NM away. Not a HUGE difference, but something to consider.
The difference comes in with an increase in speed. Say, if we hold at 200kts, by flying a one minute entry, we will have traveled 1.6NM away from the inbound course. The difference would be minimal.
Just something to think about, I guess.
Upon reading that section of the AIM, it seems pretty clear to me that they are stating a substitution of distance for time with regards to holding and entries, so I'll go ahead and agree w/ tgrayson with his interpretation. However, I do agree with the others in saying that flying 4nm during a teardrop procedure might take you too far away making it unnecessary to fly out so far. I completely agree with that.
The description given by the AIM may very well be poorly written, but I do understand from it that the substitution of distances applies to both entries and holding.
To take this a step further, we can use simple trigonometry to see how far each procedure would take us.
So let's say you decide to...
...fly a teardrop entry with a 1 minute outbound at 100 kts. The distance covered in 1 minute is 1.6NM. If we use simple trigonometry, we can find out how far this actually takes us away from the inbound course. Given the 30 degree teardrop entry, our 1.6NM is the hypotenuse and all we need to find is our opposite side. sin(30)*1.6 = 0.8NM. That is, the one minute outbound teardrop entry will have taken us 0.8NM.
....fly a teardrop entry with a 4NM outbound leg, we can find how far, again. sin(30)*4=2NM. That is, the 4NM outbound teardrop entry will have taken us 2NM away. Not a HUGE difference, but something to consider.
The difference comes in with an increase in speed. Say, if we hold at 200kts, by flying a one minute entry, we will have traveled 1.6NM away from the inbound course. The difference would be minimal.
Just something to think about, I guess.
